Lecture Notes in Complex AnalysisBased on lectures by Dr Sheng-Chi Liu

Throughout these notes, signifies end proof, N signifies end of exam-ple, and marks the end of exercise. The exposition in these notes isbased in part on [Con78]. This is also where most exercises are from, andis a good source of further exercises for the intrepid reader.

Table of Contents

Table of Contents i

Lecture 1 Review of Basic Complex Analysis 11.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Analytic functions . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Cauchy–Riemann equations . . . . . . . . . . . . . . . . . . . . . 3

Lecture 2 Möbius Transformations 52.1 Conformal mappings . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Möbius transformations . . . . . . . . . . . . . . . . . . . . . . . 6

Lecture 3 Power Series 103.1 Möbius transformations preserve lines and circles . . . . . . . . . 103.2 Power series representation of analytic functions . . . . . . . . . 11

Lecture 4 Entire Functions 134.1 Properties of analytic and entire functions . . . . . . . . . . . . . 13

Lecture 5 Cauchy’s Integral Theorem 185.1 Winding numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 185.2 Homotopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Lecture 6 Counting Zeros 216.1 Simply connected sets and Simple closed curve . . . . . . . . . . 216.2 Counting zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Lecture 7 Singularities 267.1 Classifying isolated singularities . . . . . . . . . . . . . . . . . . . 26

Lecture 8 Laurent Expansion 298.1 Laurent expansion around isolated singularity . . . . . . . . . . . 298.2 Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33Notes by Jakob Streipel. Last updated November 11, 2020.

i

ii TABLE OF CONTENTS

Lecture 9 The Argument Principle 349.1 Residue calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 349.2 The argument principle . . . . . . . . . . . . . . . . . . . . . . . 35

Lecture 10 Bounds of Analytic Functions 3810.1 Simple bounds . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3810.2 Automorphisms of the unit disk . . . . . . . . . . . . . . . . . . . 4010.3 Automorphisms of the upper half-plane . . . . . . . . . . . . . . 42

Lecture 11 Hadamard Three-Lines Theorem 4411.1 Generalising the maximum modulus principle . . . . . . . . . . . 44

Lecture 12 Phragmén–Lindelöf Principle 4612.1 Further generalising the maximum modulus principle . . . . . . . 46

Lecture 13 The Space of Analytic Functions 5013.1 The topology of C(G,C) . . . . . . . . . . . . . . . . . . . . . . . 5013.2 The space of analytic functions . . . . . . . . . . . . . . . . . . . 52

Lecture 14 Compactness in H(G) 5214.1 Compactness in the space of analytic functions . . . . . . . . . . 5214.2 Montel’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Lecture 15 The Space of Meromorphic Functions 5615.1 The topology of C(G,C∞) . . . . . . . . . . . . . . . . . . . . . . 56

Lecture 16 Compactness in M(G) 5816.1 Compactness in the space of meromorphic functions . . . . . . . 58

Lecture 17 Compactness in M(G), continued 6217.1 Compactness in the space of meromorphic functions, finalised . . 6217.2 Riemann mapping theorem . . . . . . . . . . . . . . . . . . . . . 64

Lecture 18 The Riemann Mapping Theorem 6418.1 Proving the Riemann mapping theorem . . . . . . . . . . . . . . 6418.2 Entire functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

Lecture 19 Infinite Products 6919.1 Review of infinite products . . . . . . . . . . . . . . . . . . . . . 69

Lecture 20 Weierstrass Factorisation Theorem 7420.1 Elementary factors . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Lecture 21 Rank and Genus 7921.1 Jensen’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 7921.2 The genus and rank of entire functions . . . . . . . . . . . . . . . 81

Lecture 22 Order 8422.1 The order of entire functions . . . . . . . . . . . . . . . . . . . . 8422.2 Hadamard’s factorisation theorem . . . . . . . . . . . . . . . . . 85

Lecture 23 Range of Analytic Functions 88

TABLE OF CONTENTS iii

23.1 Proof of Hadamard’s factorisation theorem . . . . . . . . . . . . 8823.2 The range of entire functions . . . . . . . . . . . . . . . . . . . . 8923.3 The range of an analytic function . . . . . . . . . . . . . . . . . . 90

Lecture 24 Bloch’s and Landau’s Constants 9224.1 Proof of Bloch’s theorem . . . . . . . . . . . . . . . . . . . . . . . 9224.2 The Little Picard theorem . . . . . . . . . . . . . . . . . . . . . . 96

Lecture 25 Toward the Little Picard Theorem 9725.1 Another lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

Lecture 26 The Little Picard Theorem 9826.1 Proof of the Little Picard theorem . . . . . . . . . . . . . . . . . 9826.2 Schottky’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 99

Lecture 27 The Great Picard Theorem 10127.1 The range of entire functions, revisited . . . . . . . . . . . . . . . 10127.2 Runge’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

Lecture 28 Mittag-Leffler’s Theorem 10528.1 Runge’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 10528.2 Mittag-Leffler’s theorem . . . . . . . . . . . . . . . . . . . . . . . 108

Lecture 29 Analytic Continuation 11029.1 Analytic continuation . . . . . . . . . . . . . . . . . . . . . . . . 11029.2 Dirichlet series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

Lecture 30 Perron’s Formula 11430.1 Uniqueness of Dirichlet series . . . . . . . . . . . . . . . . . . . . 11430.2 Perron’s formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

References 119

Index 120

iv TABLE OF CONTENTS

REVIEW OF BASIC COMPLEX ANALYSIS 1

Lecture 1 Review of Basic Complex Analysis

1.1 BasicsThroughout we will denote complex numbers by z = x+ iy, where x, y ∈ R andi =√−1. We call x = Re(z) and y = Im(y), the real and imaginary parts

of z, respectively. Moreover we call z = x − iy the complex conjugate of zand |z| =

√x2 + y2 = |z| the absolute value or modulus of z. Note, which

occasionally comes in handy, that |z|2 = zz.Also commonplace is to discuss polar representation of complex numbers.

To this end, let θ ∈ R and define

eiθ := cos(θ) + i sin(θ).

This way eiθ is a point on the unit circle in the complex plane, and so naturallywe can scale this by a real number to reach any point in the plane.

Since the complex plane C has an absolute value |·|, as per above, we alsohave an induced metric, namely for z1, z2 ∈ C we define the distance d(z1, z2) =|z1 − z2|. Under this metric (C, d) is a complete metric space.

We will take a moment to discuss some commonly useful functions on thecomplex plane. The first suspect we have almost seen above already, namely ez.If z = x+ iy, then of course

ez = ex+iy = ex · eiy = ex(cos(y) + i sin(y)),

so |ez| = ex = eRe(z).Close cousin of the exponential function are the basic trigonometric func-

tions; we would like for sin(z) and cos(z) to satisfy ez = cos(z) + i sin(z) (and,correspondingly, eiz = cos(z)− i sin(z)), and so to that end we define

cos(z) := eiz + e−iz

2

andsin(z) := eiz − e−iz

2i .

Another close cousin to the exponential function is the natural logarithm.This, for the first time in our brief exploration, is where things get a little bithairy. We of course would like for elog(z) = z, since we want log to be theinverse function of the exponential. Now, writing z = |z|eiθ, where θ = arg(z),its argument, we can try to puzzle out what log(z) should look like, by settinglog(z) = u+ iv and studying elog(z).

Since we then haveelog(z) = eu+iv = eu · eiv,

and in turn we want elog(z) = z = |z|eiθ we have, comparing sides, eu = |z|, i.e.,u = log|z|, and eiv = eiθ. This last one is where it gets hairy: this does notimply v = θ, but more loosely that v = θ + k2π for any k ∈ Z.

Thuslog(z) := log|z|+ i(θ + k2π)

Date: August 20th, 2019.

2 REVIEW OF BASIC COMPLEX ANALYSIS

for k ∈ Z, meaning this right-hand side is not unique, so log(z) is not well-defined, and hence not a function. To resolve this we need to choose a branchof log(z), by which we mean deciding in what range we let the argument live;so long as it’s any open interval of length 2π we are fine. For instance, between0 and 2π would be fine, as would −π to π.

1.2 Analytic functionsDefinition 1.2.1 (Complex differentiable). Let G ⊂ C be an open set. Afunction f : G→ C is differentiable at z ∈ G if

f ′(z) := limh→0

f(z + h)− f(z)h

exists.Note that, crucially, since h is a complex number, this limit must exist along

all paths.Definition 1.2.2 (Analytic function). Let G ⊂ C be an open set. A functionf : G → C is analytic if it is continuously differentiable in G, i.e., f ′(z) existsand is continuous for all z ∈ G.Remark 1.2.3. (i) Every (complex) differentiable function is analytic (so the

requirement of continuously above is not necessary). This, of course, isnot true in real analysis. Take, for example,

f(x) =x2 sin( 1

x ), if x 6= 0,0, if x = 0.

Then f ′(x) exists everywhere, but is not continuous at x = 0.

(ii) Every analytic function is infinitely differentiable, and has a power seriesrepresentation. This, again, is not true in the real case. Take, say,

f(x) =

0, if x ≤ 0,e−

1x2 , if x > 0.

Then f (n)(0) = 0 for all n ∈ N, meaning that the Taylor series centredat x = 0 is identically zero, but f is not identically zero near 0, so f(x)has no power series there (or, another way of putting it, the radius ofconvergence of its power series around x = 0 is zero).

Exercise 1.1. Show that f(z) = |z|2 has a derivative only at z = 0.

Proposition 1.2.4. Suppose that∞∑n=0

an(z−a)n has radius of convergence R >

0. Then f(z) :=∞∑n=0

an(z − a)n is analytic in |z − a| < R, and moreover

f (k)(z) =∞∑n=0

n(n− 1) . . . (n− k + 1)an(z − a)n−k

for |z − a| < R (meaning that the derivative can be taken termwise), and inaddition

an = f (n)(a)n! .

1.3 Cauchy–Riemann equations 3

Example 1.2.5. As expected we have

ez =∞∑n=0

zn

n!

for all z ∈ C, i.e., R =∞.Also as expected,

(ez)′ =∞∑n=1

nzn

n! =∞∑n=0

zn

n! = ez

by reindexing the first sum.Hence this is analytic. N

Example 1.2.6. Similarly, we then get

sin(z) = eiz − e−iz

2i = z − z3

3! + z5

5! − · · ·+ (−1)n z2n−1

(2n− 1)! + . . .

and

cos(z) = eiz + e−iz

2 = 1− z2

2! + z4

4! − · · ·+ (−1)n z2n

(2n)! + . . . ,

again with infinite radius of convergence. Differentiating termwise we thus, onceagain as expected, get (sin(z))′ = cos(z) and (cos(z))′ = − sin(z), making thetwo analytic. N

Example 1.2.7. Take a branch of f(z) = log(z), e.g., −π < arg(z) < π. Thenf(z) is analytic in this branch, and f ′(z) = 1

z . N

1.3 Cauchy–Riemann equationsSuppose f : G→ C is analytic, and write f as a function of the real and imagi-nary parts of z, i.e., f(z) = f(x+ iy) = f(x, y) = u(x, y) + iv(x, y), where thennaturally u, v : G→ R.

Since f is analytic, the derivative

f ′(z) = limh→0

f(z + h)− f(z)h

exists, and so in particular exists along any particular given path. We willcompute it along two paths, namely parallel to the real axis and parallel to theimaginary axis. In other words, for h = δ ∈ R and for h = iδ, δ ∈ R.

First, then,

f ′(z) = limδ→0

f(z + δ)− f(z)δ

= limδ→0

(u(x+ δ, y)− u(x, y)) + i(v(x+ δ, y)− v(x, y))δ

= ∂u

∂x(x, y) + i

∂v

∂x(x, y).

4 REVIEW OF BASIC COMPLEX ANALYSIS

Similarly, along the imaginary direction,

f ′(z) = limδ→0

(u(x, y + δ)− u(x, y)) + i(v(x, y + δ)− v(x, y))iδ

= ∂v

∂y(x, y)− i∂u

∂y(x, y).

Since those are equal, we compare real and imaginary parts and see that∂u∂x = ∂v

∂y∂u∂y = − ∂v

∂x ,

known as the Cauchy–Riemann equations, which the real and imaginaryparts of an analytic function must therefore satisfy.

Differentiating these relations again, we get

∂2u

∂x2 = ∂2v

∂x∂y

and∂2u

∂y2 = − ∂2v

∂y∂x.

Taking for granted at the moment that the imaginary parts have continuouspartials, so that the second partials in the right-hand sides are equal, we thenget

∂2u

∂x2 + ∂2u

∂y2 = 0,

i.e., letting ∆ = ∂2

∂x2 + ∂2

∂y2 , the Laplace operator , we have ∆u = 0, mean-ing that u is harmonic. Taking the opposite partials we can draw the sameconclusion about v.

Theorem 1.3.1. Let u and v be real-valued functions defines on an open setG ⊂ C. Suppose u and v have continuous partial derivatives. Then f(z) :=u(z) + iv(z) is analytic if and only if u and v satisfy the Cauchy–Riemannequations.

Remark 1.3.2. The function v above is called a harmonic conjugate of u.

Proof sketch. Use the Cauchy–Riemann equations to show that the derivativeof f(z) = u(z) + iv(z) exists. The converse direction we proved by the previousdiscussion.

Theorem 1.3.3. Let G ⊂ C be a simply connected open set. If u : G → Cis a harmonic function, then u has a harmonic conjugate, i.e., there exists av : G→ C such that f = u+ iv is analytic.

Proof sketch. Assume G is a ball, say G = B(0, R). We need to find v such thatu and v satisfy the Cauchy–Riemann equations. In other words, we first need∂v∂y = ∂u

∂x . Defining

v(x, y) =∫ y

0

∂u

∂x(x, t) dt+ ϕ(x)

MÖBIUS TRANSFORMATIONS 5

does the job, since u is given. Now using the second equation, ∂v∂x = −∂u∂y , we

can solve for ϕ(x), namely

∂v

∂x=∫ y

0

∂2u

∂x2 (x, t) dt+ ϕ′(x).

Since u is harmonic, we can rewrite this as∫ y

0−∂

2u

∂y2 (x, t) dt+ ϕ′(x) = −∂u∂y

(x, y) + ∂u

∂y(x, 0) + ϕ′(x).

On the other hand, by the second Cauchy–Riemann equation, this equals

−∂u∂y

(x, y),

whenceϕ′(x) = ∂u

∂y(x, 0),

implying

v(x, y) =∫ y

0

∂u

∂x(x, t) dt+

∫ x

0

∂u

∂y(s, 0) ds.

Of course if G isn’t a ball we might not be able to integrate along quite thispath, but similar arguments work.

Exercise 1.2. Let G be an open subset of C. Define G =z∣∣ z ∈ G. Suppose

that f : G → C is analytic. Show that f? : G → C defined by f?(z) = f(z) isalso analytic.

Lecture 2 Möbius Transformations

2.1 Conformal mappingsLet f : G→ C be some function and let z0 ∈ G. Imagine any two (differentiable)curves γ1 and γ2 going through z0. At the point z0 these curves have tangentlines, and we can measure the (anticlockwise) angle between the two, say θ.

Now imagine mapping γ1 and γ2 through f , resulting in two new curvesf(γ1) and f(γ2) going through a point f(z0). As before, we can measure theangles between the tangent lines of the two curves at this point, say α.

In the event that θ = α we say that f preserves angles at z0.

Definition 2.1.1 (Conformal mapping). A function f : G→ C is a conformalmapping if it preserves angles at each point z0 ∈ G.

Theorem 2.1.2. Let f : G → C be analytic and z0 ∈ G. Suppose f ′(z0) 6= 0.Then f preserves angles at z0.

Corollary 2.1.3. If f : G→ C is analytic and f ′(z) 6= 0 for all z ∈ G, then fis a conformal mapping.

Date: August 22nd, 2019.

6 MÖBIUS TRANSFORMATIONS

Proof of Theorem 2.1.2. Take γ : [a, b]→ G with γ(t0) = z0 and γ′(t0) 6= 0. Letσ(t) = f(γ(t)), so that σ(t0) = f(z0).

Then by the chain rule σ′(t) = f ′(γ(t)) · γ′(t), so in particular

σ′(t0) = f ′(z0) · γ′(t0).

Now by assumption and choice both terms in the right-hand side are nonzero,so σ′(t0) 6= 0 too.

Looking at the angles, we then get

arg σ′(t0) = arg f ′(z0) + arg γ′(t0),

meaning that arg σ′(t0)−arg γ′(t0) = arg f ′(z0), which is fixed and independentof γ. Therefore we get the same

arg σ′1(t0)− arg γ′1(t0) = arg f ′(z0)

for another curve γ1, and setting those equal and rearranging we see that

arg γ′1(t0)− arg γ′(t0) = arg σ′1(t0)− arg σ′(t0)

for any two curves γ and γ1, the angle between the tangent lines before applyingf is equal to the angle between the tangent lines after applying f , so f preservesangles at z0.

Example 2.1.4. Let f(z) = z2, so that f ′(z) = 2z, and in particular f ′(0) = 0:as expected, f does not preserve angles at z = 0.

To see this, imagine mapping the real line through f , ending up with a thenonnegative real line in the image space. Similarly, the imaginary line mappedthrough f results in the nonpositive real line in the image space.

The angles between the two axes before mapping through f is π/2, but theangle afterwards is π. N

Example 2.1.5. Consider f(z) = ez, for which f ′(z) = ez 6= 0 for all z ∈ C,so f is conformal. Keeping in mind that ez = ex+iy = exeiy, we see that, forexample, the vertical line x = c gets mapped to eceiy, with c fixed, so in otherwords the circle of radius ec centred on the origin

Similarly, the horizontal line y = θ gets mapped to exeiθ for a fixed angleθ, where ex takes any value on (0,∞), so this line gets mapped to the ray fromthe origin pointing outward at the angle θ.

Of course this ray lies on a radius of the above circle, so their the anglesbetween those curves is π/2, just like the angle between vertical and horizontallines. N

2.2 Möbius transformationsDefinition 2.2.1 (Möbius transformation). Let a, b, c, and d be complex num-bers with ad− bc 6= 0. Then the function

S(z) := az + b

cz + d

is called a Möbius transformation or linear fractional transformation.

2.2 Möbius transformations 7

Notice how for any complex number λ 6= 0,

S(z) = az + b

cz + d= λ

λ

az + b

cz + d= (λa)z + (λb)

(λc)z + (λd) ,

meaning that we can choose a, b, c, and d such that ad−bc = 1 (by just dividingthrough by whatever the original ad− bc 6= 0 is).

This means that we can identify the Möbius transformation S with a matrix

γ =(a bc d

)∈ SL2(C),

where by SL2(C) we mean the special linear group of 2× 2 matrices over Cwith determinant 1.

There is a small complication: this identification is not unique, since

az + b

cz + d= (−a)z + (−b)

(−c)z + (−d) ,

meaning that γ and −γ represent the same Möbius transformation. Conse-quently we should really identify S with a matrix γ in

PSL2(C) = SL2(C)/

+I,−I,

the projective special linear group of 2×2 matrices over C with determinant1 (where by I we mean the identity matrix).

Correspondingly then we define for

γ =(a bc d

)∈ SL2(C)

the actionγz = az + b

cz + d.

This is a group action on C, meaning that Iz = z for all z and (γ1γ2)z = γ1(γ2z)for all γ1, γ2 ∈ SL2(C) and z ∈ C. This second property is a fairly lengthy butstraight-forward computation.

Keeping the second property in mind, since SL2(C) is a group, γ has an in-verse element (which is precisely its matrix inverse), (γ−1γ)z = Iz = z, meaningthat the Möbius transformation S corresponding to γ has an inverse correspond-ing to γ−1, i.e.

S−1(z) = γ−1z = dz − b−cz + a

since the determinant of γ is 1.Notice moreover that since z = −dc makes the denominator zero, it is sensible

to define Möbius transformations not only on C but on the Riemann sphereC∞ = C ∪

∞. on which we have in particular

S(−dc

)=∞ and S(∞) = a

c.

One (useful) way to visualise the Riemann sphere is as the projection fromthe pole (which we label ∞) of a sphere centred on the origin of the complexplane onto said plane. An illustration of this is given in Figure 2.2.1.

8 MÖBIUS TRANSFORMATIONS

×∞

×ZZZZZZZZZZZZZZZZZ

×zzzzzzzzzzzzzzzzz

C

C∞

Figure 2.2.1: A model of the Riemann sphere, identifying a point z ∈ C with apoint Z on C∞.

In this view we see that circles on C∞ passing through ∞ correspond tostraight lines in C.

Recalling howS(z) = az + b

cz + d

we can see immediately that a Möbius transformation S has at most two fixedpoints, since S(z) = z becomes cz2 + (d − a)z − b = 0, unless S = Id, i.e.,S(z) = z for all z ∈ C∞.

A very useful consequence of this is the fact that S(z) is uniquely determinedby its values on any three given points in C∞.

To see this, suppose S and T are Möbius transformations such that S(z0) =T (z0), S(z1) = T (z1), and S(z2) = T (z2) for three distinct points z0, z1, z2 ∈C∞. Then since T is invertible, we have T−1 S(z0) = z0, T−1 S(z1) = z1,and T−1 S(z2) = z2, meaning that the Möbius transformation T−1 S (whichis a Möbius transformation by closure of matrix multiplication in SL2(C), forthe record) has more than two fixed points, meaning that T−1 S = Id, and soS = T .

As a consequence we have the following

Lemma 2.2.2. Let z2, z3, z4 ∈ C∞. Then the map

S(z) =

(z − z3)(z2 − z4)(z − z4)(z2 − z3) , if z2, z3, z4 ∈ C,z − z3

z − z4, if z2 =∞,

z2 − z4

z − z4, if z3 =∞,

z − z3

z2 − z3, if z4 =∞

is the unique Möbius transformation mapping z2 7→ 1, z3 7→ 0, and z4 7→ ∞.

Proof. This is more or less straight-forward construction. If we want z4 7→ ∞,we need to have a factor of z− z4 in the denominator but not in the numerator.Similarly, to make z3 7→ 0, we must have a z − z3 in the numerator but not the

2.2 Möbius transformations 9

denominator. Finally for z2 7→ 1, we need to make sure we have all the samefactors in the numerator and denominator when z is replaced by z2, so we needa new z2 − z4 in the numerator and z2 − z3 in the denominator.

For the infinity cases, just cancel the relevant infinite factors.

We denote this unique map S(z) as (z, z2, z3, z4), known as the cross ratioof z, z2, z3, and z4, meaning the Möbius transformation mapping z2 7→ 1,z3 7→ 0, and z4 7→ ∞

Example 2.2.3. This means that for example (z, 1, 0,∞) = z, since a Möbiustransformation agreeing with the identity map at three points must agree withit everywhere.

Similarly, (z2, z2, z3, z4) = 1 no matter what z2, z3, and z4 are. N

Exercise 2.1. Evaluate the following cross ratios:

(a) (7 + i, 1, 0,∞),

(b) (2, 1− i, 1, 1 + i),

(c) (0, 1, i,−1),

(d) (−1 + i,∞, 1 + i, 0).

Proposition 2.2.4. Let z2, z3, z4 ∈ C∞ be three distinct points. Let T be anyMöbius transformation. Then

(z1, z2, z3, z4) = (T (z1), T (z2), T (z3), T (z4))

for all z1 ∈ C∞.

Proof. Let S(z) = (z, z2, z3, z4) and set M = S T−1. Then

M(T (z2)) = S T−1 T (z2) = S(z2) = 1,

M(T (z3)) = S(z3) = 0, and M(T (z4)) = S(z4) =∞. Therefore M = S T−1 =S, and

M(z) = (z, T (z2), T (z3), T (z4)).Letting z1 = T−1(z), so that T (z1) = z, we then get

(T (z1), T (z2), T (z3), T (z4)) = (z1, z2, z3, z4)

for all z1 ∈ C∞, as desired.

Corollary 2.2.5. The unique Möbius transformation W = T (z) mapping z2,z3, and z4 to w2, w2, and w4, respectively, is given by

(w,w2, w3, w4) = (z, z2, z3, z4).

Exercise 2.2. Find the unique Möbius transformation mapping z2 = 1, z3 = 2,z4 = 7 to w2 = 1, w3 = 2, w4 = 3, respectively.

Proposition 2.2.6. Let z1, z2, z3, and z4 be four distinct points in C∞. then(z1, z2, z3, z4) is a real number if and only if the four points lie on a circle inC∞.

Exercise 2.3. Let T (z) = az + b

cz + d. Show that T (R∞) = R∞ if and only if we can

choose a, b, c, d to be real numbers.

10 POWER SERIES

Lecture 3 Power Series

3.1 Möbius transformations preserve lines and circlesRemember how lines in C are identified as circles through ∞ on the Riemannsphere. Similarly, circles in C are circles on C∞.

Let S(z, z2, z3, z4). Then S−1(R∞) =z ∈ C∞

∣∣ (z, z2, z3, z4) ∈ R∞, the

image of S−1 on R∞.The proposition then follows from the claim that a Möbius transformation

maps R∞ to a circle in C∞.

Proof. Let S(z) = az+bcz+d , and take z = x ∈ R, as well as w = S−1(x) 6= ∞, i.e.,

S(w) = x ∈ R. Then

S(w) = aw + b

cw + d= aw + b

cw + d= S(w)

since it is real.Subtracting and doing some algebra then yields

(ac− ac)|w|2 + (ad− bc)w + (ad− bc)w + (bd− bd) = 0.(3.1.1)

This is a essentially a quadratic polynomial in w, unless the leading coeffi-cient is 0, so there are two options. First, if ac ∈ R, meaning that ac− ac = 0.Then setting α = 2(ad− bc) and β = 2bd, Equation (3.1.1) becomes

Im(αw)i+ Im(β)i = 0,

or in other words Im(αw + β) = 0. This is a line in C, and so a circle in C∞(through infinity).

On the other hand if ac 6∈ R, then Equation (3.1.1) becomes

|w|2 + γw + γw − δ = 0

whereγ = ad− bc

ac− acand δ = bd− bd

ac− ac.

We then get

|w + γ| =∣∣∣∣ad− bcac− ac

∣∣∣∣ > 0,

so w lies on a circle.

Theorem 3.1.1. A Möbius transformation maps circles to circles in C∞.

Proof. Let Γ be any circle in C∞ and let S be a Möbius transformation. Letz2, z3, and z4 be three distinct points in Γ, and set wj = S(zj) for j = 2, 3, 4.Then, being three distinct points, w2, w3, and w4 determine some circle Γ′ inC∞.

We need to show that any other point z ∈ Γ then also has the property thatS(z) lies on the same circle Γ′, or in other words S(Γ) = Γ′.

Date: August 27th, 2019.

3.2 Power series representation of analytic functions 11

But we have, since Möbius transformations preserve cross ratios, that

(z, z2, z3, z4) = (S(z), S(z2), S(z3), S(z4)) = (S(z), w2, w3, w4).

By Proposition 2.2.6, we have z ∈ Γ if and only if (z, z2, z3, z4) ∈ R, if and onlyif (S(z), w2, w3, w4) ∈ R, if and only if S(z) ∈ Γ′.

3.2 Power series representation of analytic functionsWe will make use of the following calculus result.

Proposition 3.2.1. Let ϕ : [a, b]×[c, d]→ C be continuous. Define g : [c, d]→ Cby

g(t) =∫ b

a

ϕ(s, t) ds.

Then g is continuous, and moreover if ∂ϕ∂t exists and is continuous, then g ∈

C1([c, d]), and

g′(t) =∫ b

a

∂ϕ

∂t(s, t) ds.

Lemma 3.2.2. For |z| < 1 we have∫ 2π

0

eis

eis − zds = 2π.

Proof. Let ϕ(s, t) = eis

eis−tz for 0 ≤ t ≤ 1 and 0 ≤ s ≤ 2π. Then, because ofthe size of |z| and the choice of range of t, we are avoiding singularity and haveϕ ∈ C1([0, 2π]× [0, 1]). Set

g(t) =∫ 2π

0ϕ(s, t) ds,

so that our goal is to show that g(1) = 2π.To this end, first note how g(0) = 2π, being the integral from 0 to 2π of 1.We claim that g′(t) is identically 0, which would imply g(t) is constant, and

so g(1) = g(0) = 2π. Fortunately this is easy:

g′(t) =∫ 2π

0

eis

(eis − tz)2 (−1)(−z) ds =∫ 2π

0

zeis

(eis − tz)2 ds

= −zi(eis − tz)

∣∣∣∣s=2π

s=0= 0.

Proposition 3.2.3 (Simple version of Cauchy’s theorem). Let G be an openset in C and let f : G→ C be analytic. Suppose B(a, r) ⊂ G with r > 0, and letγ(t) = a+ reit for 0 ≤ t ≤ 2π. Then

f(z) = 12πi

∫γ

f(w)w − z

dw

for |z − a| < r.

12 POWER SERIES

Proof. Without loss of generality we may assume a = 0 and r = 1 (else studyg(z) = f(a+ rz)). Fix z with |z| < 1. We need to show that

f(z) = 12πi

∫γ

f(w)w − z

dw = 12π

∫ 2π

0

f(eis)eis

eis − zds,

where in the last integral we have parametrised by w = 1 · eis and dw = ieis ds.Subtracting f(z), this is equivalent to∫ 2π

0

(f(eis)eis

eis − z− f(z)

)ds = 0.

In view of our previous proposition, let

ϕ(s, t) = f(z + t(eis − z))eis

eis − z− f(z)

for 0 ≤ t ≤ 1 and 0 ≤ s ≤ 2π. Again it is the case t = 1 we are interested in.Note that since |z| < 1, |z + t(eis − z)| = |(1− t)z + teis| ≤ 1, meaning that

ϕ ∈ C1([0, 2π]× [0, 1]). Again let

g(t) =∫ 2π

0ϕ(s, t) ds,

and notice how

g(0) =∫ 2π

0

(f(z)eis

eis − z− f(z)

)ds = f(z)

∫ 2π

0

eis

eis − zds− 2πf(z) = 0

by the Lemma.We claim that g′(t) = 0 for all 0 ≤ t ≤ 1, and again it is a quick computation:

g′(t) =∫ 2π

0

∂ϕ

∂t(s, t) ds =

∫ 2π

0f ′(z + t(eis − z))(eis − z) eis

eis − z− 0 ds

= f(z + t(eis − z))it

∣∣∣∣s=2π

s=0= 0.

Hence g(t) is constant, and so g(1) = g(0) = 0, as desired.

Exercise 3.1. Evaluate∫γ

eiz

z2 dz, where γ(t) = eit, 0 ≤ t ≤ 2π.

Exercise 3.2. Evaluate∫γ

z2 + 1z(z2 + 4) dz, where γ(t) = reit, 0 ≤ t ≤ 2π, for all

possible values of 0 < r < 2 and 2 < r <∞.

Theorem 3.2.4 (Power series representation of analytic functions). Let f beanalytic in B(a,R), R > 0. Then

f(z) =∞∑n=0

an(z − a)n

for |z − a| < R, where

an = f (n)(a)n! .

ENTIRE FUNCTIONS 13

Proof. Let 0 < r < R, and set γ(t) = a + reit, 0 ≤ r ≤ 2π. For w ∈ γ and|z− a| < r, we have, for M = max

w∈γ|f(w)| (which exists since γ is a compact set)

|f(w)||z − a|n

|w − a|n+1 ≤ M |z − a|n

rn+1 = M

r

(|z − a|r

)n,

where the parenthesis at the end is less than 1 and so term goes to 0 exponen-tially. Hence

∞∑n=0

f(w)(z − a)n

(w − a)n+1

converges uniformly for w ∈ γ, |z − a| < r.But on the other hand, this sum is a geometric series in n, specifically

∞∑n=0

f(w)(z − a)n

(w − a)n+1 = f(w)w − a

11− z−a

w−a= f(w)w − z

.

By Cauchy’s formula, we then have

f(z) = 12πi

∫γ

f(w)w − z

dw = 12πi

∫γ

∞∑n=0

f(w)(z − a)n

(w − a)n+1 dw.

Since the series is uniformly continuous, we can switch the order of summationand integration, giving us

f(z) =∞∑n=0

(1

2πi

∫γ

f(w)w − a

n+1dw

)(z − a)n,

meaning that f(z) has a power series representation.

Lecture 4 Entire Functions

4.1 Properties of analytic and entire functionsLet us first establish the following corollary to the theorem at the end of lastlecture.

Corollary 4.1.1. Let G ⊂ C be open. If f : G→ C is analytic, then

(i) f is infinitely differentiable;

(ii) f (n)(a) = n!2πi

∫γ

f(w)(w − a)n+1 dw where γ(t) = a+reit for 0 ≤ t ≤ 2π, and

γ ⊂ G; and

(iii) f(z) = 12π

∫ 2π

0f(z + reit) dt, i.e., the mean value of an analytic function

on a circle is the value at the central point.Date: August 29th, 2019.

14 ENTIRE FUNCTIONS

Proof. The first two properties follow directly from the calculation at the end,noting that the integral in the last equality must be the coefficients from theTaylor expansion (since Taylor’s theorem guarantees uniqueness).

For the last property, note that when evaluating

f(z) = 12πi

∫γ

f(w)w − z

dw

over w = γ(t) = z + reit for 0 ≤ t ≤ 2π, noting that dw = ireit dt, we get

f(z) = 12πi

∫ 2π

0

f(z + reit)z + reit − z

ireit dt = 12π

∫ 2π

0f(z + reit) dt.

Corollary 4.1.2. Let f be analytic in B(a,R). Suppose |f(z)| ≤ M for everyz ∈ B(a,R). Then

|f (n)(a)| ≤ n!MRn

.

Proof. Using (ii) from the previous corollary, we have

|f (n)(a)| =∣∣∣∣ n!2πi

∫γ

f(w)(w − a)n+1 dw

∣∣∣∣ ≤ n!2π

∫γ

M

rn+1 |dw|

for 0 < r < R. Evaluating this we get

n!2π

M

rn+1 · 2πr = n!Mrn→ n!M

Rn

as we let r tend to R.

Theorem 4.1.3. Let f be analytic in B(a,R). Suppose γ is a closed rectifiable1

curve in B(a,R). Then f has a primitive and thus∫γ

f(z) dz = 0.

Proof. Since f is analytic, it has a series representation

f(z) =∞∑n=0

an(z − a)n.

Since this series converges uniformly on compact sets, we can integrate termwise,getting

F (z) =∞∑n=0

ann+ 1(z − a)n+1,

for which F ′(z) = f(z) on B(a,R). Therefore, if we say γ is parametrised fromt = 0 to t = 1, we have ∫

γ

f(z) dz = F (γ(t))∣∣∣∣t=1

t=0= 0.

Definition 4.1.4 (Entire function). A function f is an entire function if it isdefined and analytic on C.

1Meaning of bounded variation.

4.1 Properties of analytic and entire functions 15

Proposition 4.1.5. If f is an entire function, then

f(z) =∞∑n=0

anzn

wherean = f (n)(0)

n! ,

with infinite radius of convergence.

Proof. By Theorem 3.2.4, f has a power series representation on any ballB(0, R). Let R tend to infinity, and the proposition follows.

Exercise 4.1. Let f be an entire function. Suppose there are constantsM,R > 0and an integer n ≥ 1 such that |f(z)| ≤ M |z|n for |z| > R. Show that f is apolynomial of degree ≤ n.

Note how this means that, in some sense, we can think of an entire functionas a ‘polynomial of infinite degree’. This raises a natural question: can thetheory of polynomials be generalised to entire functions?

For instance, a polynomial (over C) can be factored as a product over its zerosby the Fundamental theorem of algebra. Is the same true for entire functions?The answer, it turns out, is yes, known as the Weierstrass factorisation theorem.

Another interesting question, which is easier to answer: no non-constantpolynomial is bounded. How about entire functions? The same thing holds, itturns out:

Theorem 4.1.6 (Liouville’s theorem). If f is a bounded entire function, thenf is a constant function.

Proof. Since f is entire, meaning differentiable, it must be continuous. Henceif we can show that f ′(z) = 0 for all z ∈ C, f must be constant.

We already have the estimate we need for this. Suppose f is bounded byM , then by Corollary 4.1.2 with n = 1,

|f ′(x)| ≤ M

R→ 0

as R→∞.

As it happens, an important consequence of this is one of the most elegantproofs of the Fundamental theorem of algebra.

Theorem 4.1.7 (Fundamental theorem of algebra). If p(z) is a non-constantpolynomial, then p(a) = 0 for some a ∈ C.

Proof. Suppose p(z) 6= 0 for all z ∈ C, and let f(z) = 1/p(z). Since thedenominator is never zero, f(z) is an entire function.

Moreover, since pz) is non-constant,

lim|z|→∞

|p(z)| =∞,

16 ENTIRE FUNCTIONS

implying thatlim|z|→∞

1|p(z)| = lim

|z|→∞|f(z)| = 0.

Hence f(z) is bounded, and by Liouville’s theorem constant. But that impliesthat g(z) too is constant, which is a contradiction.

Corollary 4.1.8. Let p(z) be a polynomial of degree n. Then

p(z) = c(z − a1)k1(z − a2)k2 · · · (z − am)km

for some c ∈ C, a1, a2, . . . , am ∈ C, and k1 + k2 + · · ·+ km = n.

Theorem 4.1.9. Let G be a region2 in C. Let f : G→ C be analytic. Then thefollowing are equivalent:

(i) f(z) = 0 for all z ∈ G;

(ii) there exists some a ∈ G such that f (n)(a) = 0 for all n ≥ 0; and

(iii) the zero setZ(f) :=

z ∈ G

∣∣ f(z) = 0

of f has a limit point in G.

Proof. That (i) implies (ii) and (i) implies (iii) is trivial.Let us show that (ii) implies (iii). Let a ∈ G be a limit point of Z(f). Then

since f is continuous, f(a) = 0, since we can pass the limit inside the functionby continuity. Suppose there exists some n ∈ N such that f ′(a) = f ′′(a) = · · · =f (n−1)(a) = 0, but f (n)(a) 6= 0. Then

f(z) =∞∑k=n

ak(z − a)k = (z − a)n∞∑k=n

ak(z − a)k.

Call the series at the end g(z), which is necessarily analytic, being defined interms of its power series. Hence g(a) = an 6= 0. Since a is a limit point of Z(f),there exists zn ∈ Z(f) such that zn → a, zn 6= a, and hence

f(zn) = (zn − a)ng(zn)

where the left-hand side is 0 and the first factor of the right-hand side is nonzero,so g(zn) = 0, implying g(0) = 0, which is a contradiction.

Finally let us show that (ii) implies (i). To this end, let

A =z ∈ G

∣∣ f (n)(z) = 0 for all n ≥ 0.

By (ii) we know that at least a ∈ A, so A 6= ∅. If we can show that A = G weare done, since then f as defined locally by its power series anywhere in G iszero there.

Since G is connected, and a connected set is one in which every open andclosed subset is either empty or the entire set, we need to show that A is bothopen and closed, for it is nonempty, and so must then be G.

2Meaning an open and connected set.

4.1 Properties of analytic and entire functions 17

First, to show that A is closed, let b ∈ A, and takeak⊂ A with ak → b

as k → ∞. Then f (n)(ak) = 0 for all n ≥ 0, since ak ∈ A, and since f (n) iscontinuous we also have f (n)(b) = 0 for all n ∈ N. Hence b ∈ A too, so A = Ais closed.

Second, we need to show that A is open, which is almost trivial: take a ∈ Aand take some r > 0 such that B(a, r) ⊂ G. Then since f is analytic, we havea power representation around a,

f(z) =∞∑n=0

an(z − a)n

for all z ∈ B(a, r), where crucially

an = f (n)(a)n! = 0

meaning that f(z) = 0 for all z ∈ B(a, r), whence B(a, r) ⊂ A, so A is open.

Corollary 4.1.10. Let f and g be analytic on a region G. Then f = g if andonly if

z ∈ G

∣∣ f(z) = g(z)has a limit point in G.

Proof. Take h(z) = f(z)− g(z) in the previous theorem.

Example 4.1.11. Consider the function

f(z) = cos(1 + z

1− z

)on the region G =

z∣∣ |z| < 1

. Since the potential singularity is avoided, f is

analytic in G, with the zero set

Z(f) =πn− 2πn+ 2

∣∣∣∣ n > 0 odd.

This has a limit point 1, but it is not in G. N

Corollary 4.1.12. Let f 6= 0 be analytic in a region G. Then for each a ∈ Gwith f(a) = 0, there exists some n ∈ N and some analytic function g : G → Csuch that

f(z) = (z − a)ng(z)

with g(a) 6= 0. In other words, each zero of f has finite order.

Proof. A zero a being of order n means that f (k)(a) = 0 for all k < n, so thepower series representation of f around a starts at (z − a)n, meaning that wecan factor out (z − a)n.

Corollary 4.1.13. Let G be a region and let f : G→ C be analytic, f 6= 0 andf(a) = 0 for some a ∈ G. Then there exists some R > 0 such that B(a,R) ⊂ Gand f(z) 6= 0 for all 0 < |z − a| < R. That is to say, zeros of an analyticfunction (not identically zero) are isolated.

Proof. This is Theorem 4.1.9.

18 CAUCHY’S INTEGRAL THEOREM

Exercise 4.2. Let G be a region. Let f and g be analytic functions on G suchthat f(z)g(z) = 0 for all z ∈ G. Show that either f ≡ 0 or g ≡ 0.

Theorem 4.1.14 (Maximum modulus principle). Let G be a region and letf : G→ C be analytic. Suppose there exists some a ∈ G such that |f(a)| ≥ |f(z)|for all z ∈ G. Then f is constant.

Proof. Take B(a,R) ⊂ G. Then

f(a) = 12π

∫ 2π

0f(a+ reit) dt

for r < R. Hence

|f(a)| ≤∣∣∣∣ 12π

∫ 2π

0f(a+ reit) dt

∣∣∣∣ ≤ 12π

∫ 2π

0|f(a+ reit)| dt,

but by assumption |f(a+ reit)| ≤ |f(a)|, so this is bounded by

12π

∫ 2π

0|f(a)| dt = |f(a)|.

Hence ∣∣∣∣ 12π

∫ 2π

0f(a+ reit) dt

∣∣∣∣ = |f(a)| = 12π

∫ 2π

0|f(a)| dt,

so rearranging1

2π

∫ 2π

0|f(a)| − |f(a+ reit)| dt = 0.

Again by assumption this integrand must be nonnegative, but if the integralof a nonnegative continuous function is zero, the function must be zero, so|f(a)| = |f(a + reit)| for all 0 ≤ t ≤ 2π, implying too that |f(z)| = |f(a)| forall z ∈ B(a,R). Finally this implies that f(z) is constant on B(a,R), since f iscontinuous, meaning that f(z) is constant also in G.

Exercise 4.3. Show the last part of the above proof. In other words: Let f beanalytic on a region G. Suppose |f(z)| is a constant on G. Show that f(z) is aconstant.

Lecture 5 Cauchy’s Integral Theorem

5.1 Winding numbersLet γ1(t) = a + reit for 0 ≤ t ≤ 2π, i.e., the anticlockwise circle of radius raround a. Then

12πi

∫γ1

1z − a

dz = 12πi

∫ 2π

0

ireit

a+ reit − adt = 1

2π

∫ 2π

0dt = 1.

Date: September 3rd, 2019.

5.1 Winding numbers 19

On the other hand, for b outside the region enclosed by γ1, we get

12πi

∫γ1

1z − b

dz = 0,

since the integrand is analytic on the region enclosed by γ1.Similarly, consider another curve γ2(t) = a+ reit, but this time for 0 ≤ t ≤

4π. Then1

2πi

∫γ2

1z − a

dz = 2.

So it appears, in general, as though this kind of integral counts how manytimes the curve goes around a. Indeed this is the case, and first let us makesure it counts in a reasonable way:

Proposition 5.1.1. If γ : [0, 1]→ C is a closed rectifiable curve and a 6∈γ.

Then1

2πi

∫γ

1z − a

dz

is an integer.

Proof. We may assume that γ is a smooth curve (otherwise approximate it bya smooth curve). Define the function

g(t) =∫ t

0

γ′(s)γ(s)− a ds,

so that g(0) = 0 andg(1) =

∫γ

1z − a

dz.

The strategy is this: since e2πik = 1 for all integers k, we want to show thatsomething like eg(t) is constant. To this end, note how by the Fundamentaltheorem of calculus

g′(t) = γ′(t)γ(t)− a,

and notice howd

dt

(e−g(t)(γ(t)− a)

)= e−g(t)γ′(t)− g′(t)e−g(t)(γ(t)− a) = 0.

Hence e−g(t)(γ(t)−a) = e−g(0)(γ(0)−a) = γ(0)−a is a constant, since g(0) = 0.But γ is a closed curve, so γ(0) = γ(1), meaning that

e−g(1)(γ(1)− a) = γ(0)− a,

meaning that e−g(1) = 1, so g(1) = 2πik for some integer k.

Consequently we make the following definition.

Definition 5.1.2 (Winding number). If γ is a closed rectifiable curve in C,then for a 6∈

γ,

n(γ; a) := 12πi

∫γ

1z − a

dz ∈ Z

is called the winding number of γ around a. This simply counts how manytimes the curve γ winds around a, as suggested by the name.

20 CAUCHY’S INTEGRAL THEOREM

Example 5.1.3. If a belongs to the unbounded component of C \γ

forsome closed rectifiable curve γ, then of course n(γ; a) = 0, since 1

z−a is analyticthere. N

Theorem 5.1.4 (Cauchy’s integral formula). Let G be an open set and f : G→C be analytic. Suppose γ is a closed rectifiable curve in G such that n(γ;w) = 0for all w ∈ C \G. Then for a ∈ G \

γ,

n(γ; a)f(a) = 12πi

∫γ

f(z)z − a

dz.

Remark 5.1.5. The condition n(γ;w) = 0 for all w ∈ C \ G is to avoid caseswhere, for instance, G is an annulus and with w in the inside the inner disk andγ around it.

Often in practice we use this theorem for simply connected G, where this isnot a problem.

Corollary 5.1.6. Let G be an open set and f : G → C be analytic. Suppose γis a closed and rectifiable curve in G such that n(γ;w) = 0 for all w ∈ C \ G.Then for a ∈ G \

γ,

n(γ; a)f (k)(a) = k!2πi

∫γ

f(z)(z − a)k+1 dz.

Corollary 5.1.7 (Cauchy’s theorem). Let G be an open set and f : G→ C ananalytic function. Suppose γ is a closed and rectifiable curve in G such thatn(γ,w) = 0 for all w ∈ C \G. Then∫

γ

f(z) dz = 0.

Proof. Replace f(z) by f(z)(z − a) in Cauchy’s integral formula, making theleft-hand side 0 and the right-hand side the above integral.

It is useful to note that all of these, along with Cauchy’s residue theorem(which we might talk about in future) are equivalent.

An interesting question one might naturally ask is if there are other functionsthat satisfy ∫

γ

f(z) dz = 0

for all closed curves γ. The answer is no:

Theorem 5.1.8 (Morera’s theorem). Let G be a region and let f : G → C becontinuous such that ∫

T

f(z) = 0

for all triangular paths T in G.3 Then f is analytic in G.3By triangular path we mean simply a curve composed of three line segments meeting end

on end.

5.2 Homotopy 21

Proof. We will show that f has a primitive F , i.e., F ′(z) = f(z) in G. Thisimplies that F and f are both analytic, since one derivative implies infinitederivatives in C.

Without loss of generality we may assume that G = B(a,R), since if we canmake the theorem work on any small ball in G, then we can make it work in allof G.

For z ∈ G, defineF (z) =

∫[a,z]

f(z) dz,

where by [a, z] we mean the line segment joining a and z, with that orientation.Then for any z0 ∈ G we have

F (z) =∫

[a,z]f(z) dz =

∫[a,z0]

f(z) dz +∫

[z0,z]f(z) dz

since the interval over all three segments added together is 0 by hypothesis.Therefore

F (z)− F (z0)z − z0

=

∫[z0,z] f(z) dzz − z0

.

We expect this to be f(z0), so we study the difference∣∣∣F (z)− F (z0)z − z0

− f(z0)∣∣∣ =

∣∣∣ 1z − z0

∫[z0,z]

(f(w)− f(z0)) dw∣∣∣.

We can bound the integrand by its supremum on [z0, z], so this is bounded by1

|z − z0|

∫[z0,z]

supw∈[z0,z]

|f(w)− f(z0)| dw = supw∈[z0,z]

|f(w)− f(z0)|

since the integrand no longer depends on w. Taking the limit as z → z0, thisgoes to 0 since f is continuous, and hence F ′(z0) = f(z0) for every z0 ∈ G.

5.2 HomotopyDefinition 5.2.1 (Homotopic). We say that two curves γ1 and γ2 are homo-topic in a domain G if they can be continuously deformed from one to the otherinside of the domain.

Definition 5.2.2 (Simply connected). We say that a set G is simply con-nected if every closed curve γ in G is homotopic to a point, denoted γ ' 0.

Lecture 6 Counting Zeros

6.1 Simply connected sets and Simple closed curveProposition 6.1.1 (Independence of path). Let γ0 and γ1 be two rectifiablecurves in in an open set G, both from a to b, and suppose γ0 ' γ1. Then∫

γ0

f(z) dz =∫γ1

f(z) dz

for any analytic function f on G.Date: September 5th, 2019.

22 COUNTING ZEROS

Proof. Consider the path γ0 ∪−γ1, which is then a closed path from a to itself.Then

0 =∫γ0∪−γ1

f(z) dz =∫γ0

f(z) dz −∫γ1

f(z) dz.

Corollary 6.1.2. Let G be simply connected and let f : G → C be analytic.Then f has a primitive.

Proof. We use the method from our proof of Morera’s theorem: since G issimply connected, and f is analytic, the integral of f over any simple closedcurve is 0. So in particular the integral of f over any triangular path is 0, sowe can construct a primitive of f in the same way we did for Morera’s theoremtheorem.

Corollary 6.1.3. Let G be simply connected and f : G → C be an analyticfunction. Suppose f(z) 6= 0 for all z ∈ G. Then there exists an analyticfunction g : G→ C such that f(z) = exp(g(z)). Moreover, for fixed z0 ∈ G andf(z0) = ew0 , we can choose g such that g(z0) = w0.

Before we go on to prove this, let us briefly discuss the idea behind thisresult. If we have a nonzero thing in R, we can take its logarithm. The samething is not quite true in C, because the complex logarithm is not analytic—itrequires a branch cut, but it does make sense to take logarithms locally, justmake sure the branch cut doesn’t interfere in the local neighbourhood at hand.

Hence, since we can’t take logarithms globally, we instead express a sort ofinverse of the result, hence the exponential function.

Proof. Since f(z) 6= 0 for all z ∈ G and f ′ exists since f is analytic, we musthave that its logarithmic derivative f ′

f (z) is analytic in G.Hence by Corollary 6.1.2 f ′

f (z) has a primitive, say g1. Setting h(z) =exp(g1(z)), this must be analytic in G and moreover because of the exponentialh(z) 6= 0 for all z ∈ G. Thus f

h is analytic in G, and computing its derivativewe get

(fh

)′= f ′h− fh′

h2 = f ′eg1 − feg1g′1h2 =

f ′eg1 − feg1 f′

f

h2= 0.

Therefore fh is constant, so, say, f = c · h for some c 6= 0. But then

f(z) = c exp(g1(z)) = exp(g(z))

for some scaled version g(z) of g1(z).Note finally how, since the exponential is invariant under shifts of 2πik for

integers k, this is equal to exp(g(z) + 2πik) for all integers k, and so if we want

f(z0) = ew0 = exp(g(z) + 2πik),

we need just choose k such that w0 = g(z0) + 2πk and rename g to include thisshift.

6.2 Counting zeros 23

6.2 Counting zerosRecall how if G is a region and f : G → C is analytic with zeros a1, a2, . . . , am(repeated according to multiplicities), we can write

f(z) = (z − a1)(z − a2) · · · (z − am)g(z)

where g is analytic in G and g(z) 6= 0 for all z ∈ G. Taking the logarithmicderivative of this we get

f ′

f(z) = 1

z − a1+ 1z − a2

+ · · ·+ 1z − am

+ g′

g(z).

Notice how g′

g is analytic in G since g, by construction, is never zero. Hence ifwe integrate this expression over a closed rectifiable curve γ ' 0, the last termgoes away, and importantly, by Cauchy’s integral formula, each of the remainingterms will contribute exactly 2πi times the winding number of γ around themif they are inside the region enclosed by γ, and otherwise zero. Therefore

Theorem 6.2.1. Let G be a region and let f be an analytic function on Gwith zeros a1, a2, . . . , am, repeated according to multiplicities. Let γ be a closedrectifiable curve in G such that ai 6∈

γfor any i = 1, 2, . . . ,m, and γ ' 0 in

G. Then1

2πi

∫γ

f ′

f(z) dz =

m∑i=1

n(γ; ai).

In practice, we often care about a simpler version of this result:

Corollary 6.2.2. Let G be a region and let γ be a simple closed curve in G.Then the integral ∫

γ

f ′

f(z) dz

counts the zeros of f enclosed by γ with multiplicities (assuming no zeros of flie on γ).

Theorem 6.2.3. Let f be analytic in B(a,R), and let f(a) = α. Supposef(z)−α has a zero of order m at z = a. Then there exist ε > 0 and δ > 0 suchthat for all 0 < |ξ − α| < δ, the function f(z)− ξ has exactly m simple roots inB(a, ε).

Remark 6.2.4. This theorem implies that f(B(a, ε)) ⊃ B(α, δ). This will beimportant in just a moment.

Proof. Since the zeros of an analytic function are isolated, we can find 0 < ε < R2

such that f(z) = α has no solutions in 0 < |z − a| < 2ε, and f ′(z) 6= 0 in0 < |z − a| < 2ε. For this second part, note that it is of course possible that f ′has a zero near a, but f ′ is also analytic, and its zeros are also isolated, so inthat case shrink the ball some more.

Let γ(t) = a+ ε exp(2πit) for 0 ≤ t ≤ 1, and let σ = f γ, the image of thiscircle under f .

24 COUNTING ZEROS

Since α = f(a) 6∈σ, there exists some δ > 0 such that B(α, δ)∩

σ

= ∅.Hence B(α, δ) ⊂ C \

σ. Therefore, since both α and ξ are inside the region

enclosed by σ,

n(σ;α) = 12πi

∫σ

1z − α

dz = 12πi

∫σ

1z − ξ

dz.

But σ = fγ, so taking the change of variables z = f(w)−α, dz = (f(w)−α)′ dwin the first integral, we get

12πi

∫γ

(f(w)− α)′

f(w)− α dw,

which is the number of zeros of f(z)−α in B(a, ε), or in other words m. Doingthe same thing to the second integral we get that

12πi

∫σ

1z − ξ

dz

is the number of zeros of f(z) − ξ in B(a, ε), but we we chose ε such thatf ′(z) 6= 0 for all 0 < |z−a| < 2ε, so each of these roots of f(z)−ξ is simple.

The remark, as promised, is fairly important:

Theorem 6.2.5 (Open mapping theorem). Let G be a region. Suppose f is anon-constant analytic function on G. Then for any open set U ⊂ G, f(U) isopen.

Proof. Let U ⊂ G be open. We want to show that f(U) is open, i.e., for everyα ∈ f(U) there exists some δ > 0 such that B(α, δ) ⊂ f(U).

To this end, suppose a ∈ G such that f(a) = α. By the previous theoremthere exists ε > 0 and δ > 0 such that B(a, ε) ⊂ U and f(B(a, ε)) ⊃ B(α, δ),and f(U) ⊃ f(B(a, ε)), so f(U) is open.

As an aside, this implies that an injective analytic function has analyticinverse, since it says that preimages of open sets under the inverse are opensets. Though we know more than that about an injective analytic function:its derivative must be nonzero everywhere, so in fact it would be a conformalmapping:Exercise 6.1. Let G be a region. Suppose that f : G → C is analytic and one-to-one. Show that f ′(z) 6= 0 for all z ∈ G.

Remark 6.2.6. The converse of this exercise is not true: there exist analyticfunctions with non-vanishing derivatives that are not one-to-one. An easy ex-ample if f(z) = ez; the derivative of this is never zero, but f(z + 2πik) = f(z)for all integers k, so it is certainly not injective.

We mentioned a long time ago, in Remark 1.2.3 (i), that requiring the deriva-tive of a function to be continuous in the definition of analytic is superfluous.We are finally ready to prove this:

Theorem 6.2.7 (Goursat’s theorem). Let G be an open set and let f : G→ Cbe differentiable. Then f is analytic.

6.2 Counting zeros 25

Proof. We want to show, ultimately, that f ′ is continuous, but it turns out abetter idea, by Morera’s theorem, is to show that∫

T

f(z) dz = 0

for all triangular paths T in G, since f , being differentiable, is continuous, andby Morera’s theorem this would make it analytic.

Let T be an arbitrary triangular path in G. Then if we form new add andsubtract the line segments between the midpoints of the line segments makingup T , as per Figure 6.2.1, we have not changed the path, and we have∫

T

f(z) dz =4∑i=1

∫Ti

f(z) dz.

Now let T 1 be the triangular path among T1, T2, T3, and T4 such that∣∣∣∫T ′f(z) dz

∣∣∣ = max1≤i≤4

∣∣∣∫Ti

f(z) dz∣∣∣.

T

becomes

T1

T2

T3

T4

Figure 6.2.1: Decomposing atriangular path T into smallertriangular paths.

Then by construction∣∣∣∫T

f(z) dz∣∣∣ ≤ 4

∣∣∣∫T 1f(z) dz

∣∣∣.Repeat this process with T 1 in place of T , getting T 2, and so on. In general,then, we have ∣∣∣∫

T

f(z) dz∣∣∣ ≤ 4n

∣∣∣∫Tnf(z) dz

∣∣∣,the lengths of the paths halve every time, i.e.,

`(Tn) = 12`(T

n−1) = · · · = 12n `(T ),

and letting ∆n denote the closed triangle inside Tn, so that

∆ ⊃ ∆1 ⊃ · · · ⊃ ∆n,

we havediam(∆n) = 1

2 diam(∆n−1) = · · · = 12n diam(∆),

which goes to 0 as n→∞. Now the intersection∞⋂n=1

∆n

is nonempty, since the ∆n are closed and nested, but the diameter of the inter-section goes to 0, so the intersection contains only a single point, say z0.

Since f is differentiable at z0, we have that for every ε > 0 there exists someδ > 0 such that B(z0, δ) ⊂ G and∣∣∣f(z)− f(z0)

z − z0− f ′(z0)

∣∣∣ < ε

26 SINGULARITIES

for every 0 < |z − z0| < δ. Hence

|f(z)− f(z0)− f ′(z0)(z − z0)| < ε|z − z0|.

Now, since the functions 1 and z are analytic, by Cauchy’s theorem∫Tn

1 dz =∫Tnz dz = 0,

and thus for n sufficiently large∣∣∣∫Tnf(z) dz

∣∣∣ =∣∣∣∫Tnf(z)− f(z0)− f ′(z0)(z − z0) dz

∣∣∣ ≤ ε∫Tn|z − z0| dz.

But z is on Tn and z0 is in the interior of ∆n, so this distance is bounded bydiam(∆n). Hence the integral is bounded above by

εdiam(∆n)`(Tn) = ε

4n diam(∆)`(T ).

Hence ∣∣∣∫T

f(z) dz∣∣∣ ≤ 4n ε

4n diam(∆)`(T ) = εdiam(∆)`(T ),

but ε > 0 is arbitrary, so∣∣∣∫T

f(z) dz∣∣∣ =

∫T

f(z) dz = 0,

as desired.

Lecture 7 Singularities

7.1 Classifying isolated singularitiesDefinition 7.1.1 (Singularity). A function f has an isolated singularity atz = a if there is an R > 0 such that f is defined and analytic on B(a,R) \

a,

but not at z = a itself.Moreover the point z = a is called a removable singularity if there is an

analytic function g : B(a,R) → C such that g(z) = f(z) for all z ∈ B(a,R) \a.

Example 7.1.2. The function f(z) = sin(z)z , z 6= 0, has the limit lim

z→0f(z) = 1.

Hence we can define

g(z) =

sin(z)z , if z 6= 0

1, if z = 0,

where g is analytic. So f has a removable singularity at z = 0. N

This is an interesting point about complex analysis: if we have a functionthat is analytic apart from a single point, and we can fill that point in so thatthe resulting function is continuous, then the whole function becomes analytic.Compare that with the real case: if we have a differentiable function with ahole, filling that point in does not mean the filled in function is differentiable.For example, consider f(x) = |x|, except undefined at 0.

Date: September 10th, 2019.

7.1 Classifying isolated singularities 27

Example 7.1.3. The function f(z) = 1z has an isolated singularity at z = 0,

but it is not removable, because limz→0|f(z)| =∞. N

Example 7.1.4. The function f(z) = exp( 1z ), z 6= 0, is sort of different: this

also has an isolated singularity at z = 0, but limz→0|f(z)| does not exist, even in

the sense of infinity. N

Of course the natural question that arises is how we would determine, ingeneral, if an isolated singularity is removable or not.

Theorem 7.1.5. Suppose f has an isolated singularity at z = 0. Then z = ais a removable singularity if and only if lim

z→a(z − a)f(z) = 0.

Proof. For the forward direction, assume z = a is a removable singularity of f ,i.e., there exists some analytic g : B(a,R) → C such that f(z) = g(z) for allz ∈ B(a,R) \

a. So

limz→a

(z − a)f(z) = limz→a

(z − a)g(z) = (a− a)g(a) = 0,

where we can replace f by g in the limit since the limit never touches z = a,and in the last step we use the fact that both z − a and g(z) are continuous atz = a.

For the converse direction, define

h(z) =

(z − a)f(z), if z 6= a

0, if z = a.

Since limz→a

(z − a)f(z) = 0 = h(a), h is continuous. Then if we can show that his analytic, then we are done, since it means that h(z) = (z−a)g(z), g analytic,and so for z 6= a we have (z − a)f(z) = (z − a)g(z), and cancelling z − a wehave f(z) = g(z).

Now to prove that h is analytic, it suffices by Morera’s theorem to show that∫Th(z) dz = 0 for all triangular paths R in the domain.There are (sort of) four cases to consider here. Case 1, the point z = a lies

outside of T . In that case∫T

h(z) dz =∫T

(z − a)f(z) dz = 0

since (z − a)f(z) is analytic inside T .

×a

T

becomes

×a

P

Tε

Figure 7.1.1: Decomposinga triangular path T into asmaller triangular path Tε anda quadrilateral path P .

Case 2, z = a is a vertex of T . Then we can divide T into a quadrilateralpath P and a smaller triangular path Tε, as per Figure 7.1.1. Since h is analyticinside P , we have∫

T

h(z) dz =∫Tε

h(z) dz +∫P

h(z) dz =∫Tε

h(z) dz.

Since h is continuous at z = a and h(a) = 0, for every ε > 0 there exists someTε small enough such that |h(z)| < ε for all z ∈ Tε. Hence∣∣∣∫

Tε

h(z) dz∣∣∣ ≤ ∫

Tε

|h(z)| dz ≤∫Tε

ε = ε`(Tε) < ε,

28 SINGULARITIES

since we can choose the subdivision so that `(Tε) < 1. Hence∫Th(z) dz = 0.

Case 3, z = a lies inside of T . In this case, split T into three triangularpaths by joining the vertices up with z = a, see Figure 7.1.2. Then∫

T

h(z) dz =∫T1

h(z) dz +∫T2

h(z) dz +∫T3

h(z) dz = 0

by the previous case, since a is a vertex for each of the three triangular paths.Finally case 4, z = a might sit on T but not on a vertex. In this case, we

play the same subdivision game, only splitting into two triangular paths, andagain z = a is a vertex of each.

×aT

becomes

×a

T1

T2T3

Figure 7.1.2: Dividing a tri-angular path T three triangu-lar paths T1, T2, and T3.

Remark 7.1.6. If f(z) has a ‘true’ singularity at z = a, i.e., not removable, thenf(z) must behave badly near z = a, in the sense that |f(z)| → ∞ as z → a orthe limit doesn’t exist at all.

Definition 7.1.7 (Pole, essential singularity). Let z = a be an isolated singu-larity of a function f . We call z = a a pole of f if lim

z→a|f(z)| =∞.

If z = a is neither a pole nor a removable singularity, then we call z = a anessential singularity of f .

Example 7.1.8. The function f(z) = 1(z−a)m , m ∈ N, has a pole at z = a. N

Example 7.1.9. The example discussed previously, f(z) = exp( 1z ), has an

essential singularity at z = 0. N

Proposition 7.1.10. Let G be a region and a ∈ G. Suppose f is analytic onG \

awith a pole at z = a. Then there exists some m ∈ N and an analytic

g : G→ C such that

f(z) = g(z)(z − a)m ,

where g(a) 6= 0.

Compare this to how we can factor zeros of multiplicity m out analyticfunctions.Remark 7.1.11. We say that f as above has a pole of order m at z = a.

Proof. Let

h(z) =

1f(z) , if z 6= a,

0, if z = a.

Then h(z) is analytic on a ball B(a,R) for some R > 0 since the zeros of f areisolated, and h(a) = 0. Hence we can factor

h(z) = (z − a)mh1(z)

for some m ∈ N and h1 analytic on B(a,R) with h1(a) 6= 0. For z 6= a we cantherefore write

1f(z) = (z − a)mh1(z),

LAURENT EXPANSION 29

and since the h1 is analytic, its zeros are isolated, so there exists some ballB(a, r), r ≤ R, such that h1(z) 6= 0 on it. Thus

f(z)(z − a)m = 1h1(z) ,

for all z ∈ B(a, r) \a, where the left-hand side is undefined for z = a but

the right-hand side is. This means that z = a is a removable singularity off(z)(z − a)m, and therefore there exist some analytic g : G → C such thatf(z)(z − a)m = g(z) for all z 6= a in G, and so finally

f(z) = g(z)(z − a)m .

Note that since this resulting g is analytic, it has a power series expansion

g(z) = Am +Am−1(z − a) + · · ·+A1(z − a)m−1 + (z − a)m∞∑k=0

ak(z − a)k.

Therefore f has a sort of ‘almost’ power series representation

f(z) = Am(z − a)m + Am−1

(z − a)m−1 + · · ·+ A1

z − a+∞∑k=0

ak(z − a)k,

where the infinite series at the end is analytic, and the m terms at the front arenot. We will talk more about this representation in the near future.

Lecture 8 Laurent Expansion

8.1 Laurent expansion around isolated singularityTheorem 8.1.1 (Laurent expansion). Let f be analytic in the annulus R1 <|z − a| < R2. Then

f(z) =∞∑

n=−∞an(z − a)n

where the convergence is absolute, and uniform in compact subsets r1 ≤ |z−a| ≤r2, R1 < r1 < r2 < R2. Moreover

an = 12πi

∫γ

f(z)(z − a)n+1 dz

where γ is a circle |z − a| = r with R1 < r < R2. Finally, this series represen-tation is unique.

We call this kind of series representation of f its Laurent expansion aroundz = a.

Date: September 12th, 2019.

30 LAURENT EXPANSION

Remark 8.1.2. When we say that a series∞∑

n=−∞bn converges (absolutely) we

mean that both∞∑n=0

bn and∞∑n=1

b−n converge (absolutely) separately.

Importantly, this is different from the maybe more intuitive interpretationthat

limM→∞

M∑n=−M

bn

should converge, which would allow for cancellation between the negative andpositive indices (for instance bn = 1

n and b0 = 0 would be 0 in this sense, butnot converge at all in the first sense).

Proof. Without loss of generality we may assume a = 0, else just shift. Letγ1(t) = r1e

2πit and γ2(t) = r2e2πit, both for 0 ≤ t ≤ 1, and R1 < r1 < r2 < R2.

Define a function

g(w) =f(w)−f(z)

w−z , if w 6= z

f ′(z), if w = z.

This is analytic in R1 < |w| < R2. If we join the two concentric circles γ1 andγ2 by a line segment L and its opposite −L, and let γ = γ2 + L − L − γ1, aclosed curve.

L

γ2

−γ1×0

×x

Figure 8.1.1: Joining −γ1 andγ by a line and its opposite.Note the reverse orientationon −γ1.

Now take z to be inside γ, in other words between the two circles γ1 and γ2(and not on the connecting line L, but we can move that line to never touch z).By Cauchy’s theorem, ∫

γ

g(w) dw = 0,

meaning that ∫γ2−γ1

f(w)− f(z)w − z

dw = 0,

since the integral along L and −L cancel. Splitting this apart we see that∫γ2−γ1

f(w)w − z

dw =∫γ2−γ1

f(z)w − z

dw.(8.1.1)

The second integral is easy to compute piece by piece. On γ2, since z lies insideit, ∫

γ2

f(z)w − z

dz = f(z)∫γ2

1w − z

dw = 2πif(z),

strictly times the winding number n(γ2; z), but that is 1 in this case. Similarly,since z lies outside of γ1,∫

γ1

f(z)w − z

dw = f(z)∫γ1

1w − z

dw = 0.

So the right-hand side integral in Equation (8.1.1) above is 2πif(z).Next we wish to compute the left-hand side of same. Note that on γ2,

|w| > |z| since z lies inside the circle, so we can factor 1w−z as a geometric

series,

1w − z

= 1w

11− z

w

= 1w

(1 + z

w+ z2

w2 + . . .)

= 1w

+ z

w2 + z2

w3 + . . . .

8.1 Laurent expansion around isolated singularity 31

Therefore∫γ2

f(w)w − z

dw =∫γ2

f(w)∞∑n=0

zn

wn+1 dw =∞∑n=0

(∫γ2

f(w)wn+1 dw

)zn

since the geometric series converges absolutely.On the other hand, on γ1, |w| < |z|, so this time

1w − z

= 1z

1wz − 1 = −1

z

11− w

z

,

which as a geometric series becomes

−(1z

+ w

z2 + w2

z3 + . . .),

meaning that∫γ1

f(w)w − z

dw = −∫γ1

f(w)−∞∑n=−1

zn

wn+1 dw = −−1∑

n=−∞

(∫γ1

f(w)wn+1 dw

)zn.

Now in order to get the kind of series we want we would like to simply combinethese two expressions, taking care of the minus sign, and be done, but currentlythe integrals are over different paths γ1 and γ2. The good news though is thatγ1 ' γ2 ' γ for any closed, rectifiable γ with winding number 1 in the annulusR1 < |z − a| < R2, and the function g(w) we started integrating is analyticthere, so we can change the paths to be equal.

Hence finally, remembering that the right-hand side of Equation (8.1.1) is2πif(z), we get

f(z) =∞∑

n=−∞

(1

2πi

∫γ

f(w)wn+1 dw

)zn,

so we have the series representation we need and we call the coefficients givenby the integral an.

It remains to show uniqueness. To this end, suppose

f(z) =∞∑

n=−∞anz

n.

Then ∫γ

f(z)zk+1 dz =

∞∑n=−∞

an

∫γ

zn−k−1 dz.

Now if n ≥ k+1, so that the power of the integrand is nonnegative, this integralis zero since the integrand is analytic. Similarly if n < k, so that the power is−2 or smaller, the integrand has a primitive, so again is analytic. The onlyoutstanding option is if n = k, so that the integrand is z−1. In that case byCauchy’s theorem the integral is 2πi, so the above sum picks up the n = k termand equals ∫

γ

f(z)zk+1 dz = 2πiak.

32 LAURENT EXPANSION

But then we see thatak = 1

2πi

∫γ

f(z)zk+1 dz,

which is the same coefficient we claimed was unique.

In summary, then: Let

f(z) =∞∑

n=−∞an(z − a)n

be the Laurent expansion of f about an isolated singularity z = a. Then

(i) if f has a removable singularity at z = a, then a−n = 0 for all n ≥ 1;

(ii) if f has a pole of order m at z = a, then a−m 6= 0 and a−n = 0 for alln > m;

(iii) if f has an essential singularity at z = a, then there are infinitely manya−n 6= 0, n ≥ 1.

Exercise 8.1. Find the Laurent expansion for

(a) 1z4 + z2 about z = 0,

(b) 1z2 − 4 about z = 2.

In much the same way as analytic functions have isolated zeros, so are polesof these creatures:Exercise 8.2. Let G be an open subset of C. Suppose f : G → C is analyticexcept for poles. Show that the poles of f can not have a limit point in G.

Since a function doesn’t have a nice limit at an essential singularity, we mightexpect it to behave strangely there, and indeed this is the case:

Theorem 8.1.3 (Casorati–Weierstrass theorem). If f has an essential singular-ity at z = a, then for any δ > 0, f(Aδ) = C where Aδ =

z ∈ C

∣∣0 < |z−a| < δ.

Proof. Suppose the theorem is false, i.e., there exists some c ∈ C and ε > 0 suchthat |f(z)− c| > ε for every z ∈ Aδ and every δ > 0. Then

limz→a

|f(z)− c||z − a|

=∞,

so f(z)−cz−a has a pole at z = a. Let m be the order of this pole, so

f(z)− cz − a

= g(z)(z − a)m

where g(a) 6= 0 and g is analytic. Hence

f(z) = g(z)(z − a)m−1 + c,

for z 6= a. Note that the constant c is analytic, so if m = 1, then f(z) has aremovable singularity at z = a, a contradiction, and if m ≥ 2, then f(z) has apole of order m− 1 at z = a, also a contradiction.

8.2 Residues 33

8.2 ResiduesDefinition 8.2.1 (Residue). Let z = a be an isolated singularity of f and

f(z) =∞∑

n=−∞an(z − a)n

be the Laurent expansion of f about z = a. The residue of z = a is thecoefficient a−1, denoted by

Res(f ; a) or Resz=a

f(z).

We know by Cauchy’s theorem that if a function is analytic in 0 < |z−a| < R,then for γ(t) = a+ reit, 0 ≤ t ≤ 2π, we have

a−1 = 12πi

∫γ

f(z) dz,

and so we should expect this to hold in some kind of generality:

Theorem 8.2.2 (Cauchy’s residue theorem). Let f be analytic in a region Gexcept for isolated singularities b1, b2, . . . , bm. Let γ be a closed rectifiable curvein G such that bk 6∈

γfor k = 1, 2, . . . ,m. Suppose γ ' 0 in G. Then

12πi

∫γ

f(z) dz =m∑k=1

n(γ; bk) Res(f ; bk).

Sketch of proof. Suppose γ encloses only one isolated singularity, b1, and let

f(z) =∞∑

n=−∞an(z − b1)n

be the Laurent expansion of f around z = b1. Then, by the same calculationsas previously,

12πi

∫γ

f(z) dz = 12πi

∫γ

a−1

z − b1dz = a−1n(γ; b1),

where a−1 = Res(f ; b1).If γ encloses more poles, then by adding and subtracting paths cutting them

off from one another, and for each resulting smaller curve around one isolatedsingularity expanding f as a Laurent series around that singularity, we get thefull result.

In practice it is often useful to have an easier way to compute residues thanthis integral:

Proposition 8.2.3. Suppose f has a pole of order m at z = a. Let g(z) =(z − a)mf(z). Then

Res(f ; a) = 1(m− 1)!g

(m−1)(a).

34 THE ARGUMENT PRINCIPLE

Proof. By the discussion at the end of last lecture, the Laurent expansion of faround z = a looks like

f(z) = b−m(z − a)m +

b−(m−1)

(z − a)m−1 + · · ·+ b−1

z − a+ b0 + b1(z− a) + b2(z− a)2 + . . . .

Our goal is to extract b−1, so

g(z) = (z−a)mf(z) = b−m+b−(m−1)(z−a)+· · ·+b−1(z−a)m−1+b0(z−a)m+. . . ,

and hence the (m− 1)st derivative is

g(m−1)(z) = (m− 1)!b−1 +m(m− 1) . . . 3 · 2b0(z − a) + . . . ,

and evaluating this at z = a, all but the constant term vanish, so g(m−1)(a) =(m− 1)!b−1.

Lecture 9 The Argument Principle

9.1 Residue calculusResidues offer a very powerful tool for evaluating real integrals, as it happens.

Example 9.1.1. We have ∫ ∞−∞

x2

1 + x4 dx = π√2.

Evaluating this as a real integral is very hard, but evaluating it as a carefullyconstructed complex integral is comparatively easy. To this end, let f(z) =z2

1+z4 . This function has poles where z4 = 1, i.e., zi = eiθi , for θ1 = π4 , θ2 = 3π

4 ,θ3 = 5π

4 , and θ4 = 7π4 .

γ

∗z1∗

z2

∗z3

∗z4

Figure 9.1.1: A positively ori-ented semicircle of radius Renclosing two poles (labelled∗) of f .

Now consider integrating this over the semicircle γ of radius R > 1 sitting onthe real axis, so that the circular part is parametrised by z = Reiθ, 0 ≤ θ ≤ π,and dz = Rieiθ dθ, see Figure 9.1.1. Then by Cauchy’s residue theorem

12πi

∫γ

f(z) dz = Resz=z1

f(z) + Resz=z2

f(z),

since those are the only poles inside γ. In particular

Resz=z1

f(z) = limz→z1

(z − z1)f(z)

= limz→z1

z2

(z − z2)(z − z3)(z − z4) = 14e−π4 i,

and by completely analogous computations

Resz=z2

f(z) = 14e

i 3π4 i.

Date: September 17th, 2019.

9.2 The argument principle 35

Hence, after some calculation,

12πi

∫γ

f(z) dz = − i

2√

2.

Now on the other hand

12πi

∫γ

f(z) dz = 12πi

∫ R

−R

x2

1 + x4 dx+ 12π

∫ π

0

R2ei2θ

1 +R4ei4θReiθ dθ.

Letting R→∞, this first integral converges to the real integral we are interestedin, and the second integral goes to 0:∣∣∣∣ 1

2π

∫ π

0

R3ei3θ

1 +R4ei4θdθ

∣∣∣∣ ≤ 12π

∫ π

0

R3

R4 − 1 dθ = 12

R3

R4 − 1 → 0

as R→∞. Hence− i

2√

2= 1

2πi

∫ ∞−∞

x2

1 + x4 dx,

which, when rearranged, yields∫ ∞−∞

x2

1 + x4 dx = π√2. N

Exercise 9.1. Show that:

(a)∫ ∞

0

1(x2 + a2)2 dz = π

4a3 for a > 0,

(b)∫ ∞−∞

eax

1 + exdx = π

sin(aπ) for 0 < a < 1.

9.2 The argument principleDefinition 9.2.1 (Meromorphic). A function f is said to be meromorphicon G is f is analytic on G except for isolated poles.

We know two things about these, from earlier: If f has a zero of order ormultiplicativity m at z = a, then f(z) = (z − a)mg(z) where g is analytic nearz = a and g(a) 6= 0. Therefore

f ′(z)f(z) = m

z − a+ g′(z)g(z) ,

and so if γ is a closed rectifiable curve around a, enclosing no other poles orzeros of f , then

12πi

∫γ

f ′(z)f(z) dz = m · n(γ; a).

Similarly, if f has a pole of order m at z = a, then f(z) = g(z)(z−a)m , where

again g(z) is analytic near z = a and g(a) 6= 0. Then

f ′(z)f(z) = −m

z − a+ g′(z)g(z) ,

36 THE ARGUMENT PRINCIPLE

so1

2πi

∫γ

f ′(z)f(z) dz = −m · n(γ; a).

We can do this in general:

Theorem 9.2.2 (Argument principle). Let f be meromorphic on F with polesp1, p2, . . . , pm and zeros z1, z2, . . . , zn, counted according to multiplicity/order.Let γ be a closed, rectifiable curve in G with γ ' 0 and pj 6∈

γand zi 6∈

γ

for all i and j. Then

12πi

∫γ

f ′(z)f(z) dz =

n∑i=1

n(γ; zi)−m∑j=1

n(γ; pj).

Proof. By the same considerations as above, we can write

f(z) = (z − z1)(z − z2) · · · (z − zn)(z − p1)(z − p2) · · · (z − pm)g(z),

where g is analytic inside γ, and g(z) 6= 0 for all z inside γ. Then

f ′(z)f(z) =

n∑i=1

1z − zi

−m∑j=1

1z − pj

+ g′(z)g(z) .

Integrating this we get the above formula, since each term has only one simplepole inside γ.

Exercise 9.2. Let f be a meromorphic function on a neighbourhood of B(a;R)with no zeros or poles on γ =

z∣∣ |z − a| = R

. Let z1, z2, . . . , zn be the zeros

of f and p1, p2, . . . , pm be the poles of f (counted according to multiplicity) inB(a;R). Show that

12πi

∫γ

zf ′

f=

n∑i=1

zi −m∑j=1

pj .

A very natural question to now ask is why we call this ‘the argument prin-ciple’. To answer this, suppose for a moment we can define log f(z). Then

(log f(z))′ = f ′(z)f(z) ,

so the integrand above has a primitive, meaning that for any closed, rectifiableγ,

12πi

∫γ

f ′(z)f(z) dz = log f(z)

∣∣∣∣z=az=a

= 0.

However, we define log z := log|z|+ i arg z, and for this to make sense we need,apart from z 6= 0, to make a branch cut, i.e., pick an open interval of length 2πfor the argument. Hence log f(z) is defined if f(z) 6= 0 and f(z) doesn’t lie onthe line removed by the branch cut. But we have zeros and poles, meaning thatf(z) = 0 for some z or f(z) = ∞ (in a manner of speaking) for some z, bothof which like on any line removed by any branch cut. So if f has any zeros orpoles inside γ, log f(z) is not defined.

9.2 The argument principle 37

But for z on γ itself there are no zeros or poles of f (by hypothesis), so forany a ∈ γ, log f(z) is defined near z = a.

In other words, if we call the little arc of γ lying near a, say, γ1, with startingpoint z = b and endpoint z = c, then∫

γ1

f ′(z)f(z) dz = log f(z)

∣∣∣∣z=cz=b

= log f(c)− log f(b),

and picking a branch this is

log|f(c)| − log|f(b)|+ i(arg f(c)− arg f(b)),

so the imaginary part of this integral measures the change of angles of f(z) asz moves from z = b to z = c.

For each a ∈ γ, f(a) 6= 0, so we can define log f(z) on B(a, ra) for somesufficiently small ra > 0. This means that

B(a, ra)

∣∣ a ∈ γ

is an opencover of γ, and γ is compact, so there exists a finite subcover. Moreover, if weparametrise γ over t ∈ [0, 1], then by Lebesgue’s number lemma, there existssome ε > 0 such that for any partition

0 = t0 < t1 < t2 < · · · < tn−1 < tn = 1

with |tj−1 − tj | < ε, we have γ([tj , tj−1]) ⊂ B(a, ra) for some a ∈ γ. Now wewant to define log f(z) on B(a, ra). So log f(z) := log|f(z)| + i argj f(z) onγ([tj , tj+1]) for j = 0, 1, . . . , k − 1. Now we choose the branches in such a waythat arg0 f(γ(t1)) = arg1 f(γ(t1)), arg1 f(γ(t2)) = arg2 f(γ(t2)), and so on, sothat at the transition from one ball to another the argument doesn’t jump.

Let γj = γ([tj , tj+1]). Then∫γj

f ′(z)f(z) dz = log|f(γ(tj+1))|−log|f(γ(tj))|+i

(argj f(γ(tj+1))− argj f(γ(tj))

).

Hence∫γ

f ′(z)f(z) dz =

k−1∑j=0

∫γj

f ′(z)f(z) dz

= log|f(1))| − log|f(γ(0))|+ i

(argk−1 f(γ(1))− arg0 f(γ(0))

)since the resulting sum is telescoping.

This means that if γ is a closed curve, then of course∫γ

f ′(z)f(z) dz = 2πi · k

for some integer k, but γ is not necessarily a closed curve. Suppose γ is a curvefrom z = a to z = b. Then

Im(∫

γ

f ′(z)f(z) dz

)measures precisely the continuous variation of the argument of f along γ. This,then, is why we call it the argument principle.

As a consequence of this discussion:

38 BOUNDS OF ANALYTIC FUNCTIONS

Theorem 9.2.3 (Rouché’s theorem). Suppose f and g are meromorphic ina neighbourhood of B(a,R) and with no zeros or poles on the boundary γ =z∣∣ |z − a| = R

. Let Z(f), Z(g), P (f), and P (g) denote the number of zeros

of f and g and the number of poles of f and g inside γ, counted according tomultiplicity or order. Suppose

|f(z) + g(z)| < |f(z)|+ |g(z)|

on γ. ThenZ(f)− P (f) = Z(g)− P (g).

Remark 9.2.4. In applications, we frequently care about the special case wherethe functions are analytic, making P (f) = P (g) = 0, and |f(z) + g(z)| < |f(z)|,then implying f and g have equally many zeros inside γ.

Proof. By hypothesis |f(z) + g(z)| < |f(z)| + |g(z)| on γ. Dividing by g(z),which we can do since by assumptions g has no zeros on γ, we get∣∣∣∣f(z)

g(z) + 1∣∣∣∣ < ∣∣∣∣f(z)

g(z)

∣∣∣∣+ 1

on γ. Let λ = f(z)g(z) for z ∈ γ. If λ ∈ R≥0, then we get λ + 1 < λ + 1, which is

impossible. Hence λ will never touch the nonnegative real axis, so we can takethat as out branch cut and define

log f(z)g(z) = log

∣∣∣∣f(z)g(z)

∣∣∣∣+ i arg f(z)g(z)

for 0 < arg f(z)g(z) < 2π. Thus (f/g)′

(f/g) has a primitive, and so

12πi

∫γ

(f/g)′

(f/g) (z) dz = 0,

but (f(z)g(z)

)′= f ′(z)g(z)− f(z)g′(z)

g(z)2 ,

so(f/g)′

(f/g) (z) = f ′(z)g(z)− f(z)g′(z)g(z)2

g(z)f(z) = f ′(z)

f(z) −g′(z)g(z) ,

so the above integral is also

0 = 12πi

∫γ

f ′(z)f(z) −

g′(z)g(z) dz = (Z(f)− P (f))− (Z(g)− P (g)).

Lecture 10 Bounds of Analytic Functions

10.1 Simple boundsThe first bound of analytic functions that is of very frequent use we have alreadymet (see page 18):

Date: September 19th, 2019.

10.1 Simple bounds 39

Theorem 10.1.1 (Maximum modulus principle). Let G be a region and letf : G → C be an analytic function. Suppose there exists some a ∈ G such that|f(a)| ≥ |f(z)| for all z ∈ G. Then f is a constant function.

An equivalent statement that we make frequent use of is

Corollary 10.1.2. Let G be a region and let f : G→ C be an analytic function.Suppose

lim supz→a

|f(z)| ≤M

for all a ∈ ∂∞G for some M > 0.4 Then

|f(z)| ≤M

for all z ∈ G.

In other words, the maximum of an analytic function on the interior of aclosed set must occur at the boundary of the set.

Theorem 10.1.3 (Schwarz lemma). Let D =z∣∣ |z| < 1

be the unit disk.

Suppose f : D → C be an analytic function such that

(i) |f(z)| ≤ 1 for all z ∈ D (i.e., f : D → D), and

(ii) f(0) = 0.

Then |f ′(0)| ≤ 1 and |f(z)| ≤ |z| for every z ∈ D (so the image can only shrinkor remain the same size, not grow). Moreover, if |f ′(0)| = 1 or if |f(z)| = |z|for some z ∈ D, then there exists a c ∈ C with |c| = 1 such that f(z) = cz forall z ∈ D.

Proof. Define g : D → C by

g(z) =f(z)z , if z 6= 0

f ′(0), if z = 0.

Then g(z) is analytic in D. On |z| = r for 0 < r < 1,

|g(z)| = |f(z)||z|

≤ 1r,

and letting r → 1 we see by the Maximum modulus principle that |g(z)| ≤ 1 forall z ∈ D.

This implies in particular that |g(0)| = |f ′(0)| ≤ 1, and | f(z)z | ≤ 1, so

|f(z)| ≤ |z|.The two special cases now follow quite readily: if f ′(0) = 1, then from the

first one |g(z)| attains its maximum in D, so by the Maximum modulus principleg(z) = c = g(0) = f ′(0) for some |c| = 1, so f(z) = cz.

For the second one, if | f(z0)z0| = 1 for some z0 ∈ D, then again |g(z)| attains

its maximum in D, so g(z) = z, and f(z) = cz, where |c| = | f(z0)z0| = 1.

4By ∂∞G we mean ∂G ∪∞

in the case where G is unbounded.

40 BOUNDS OF ANALYTIC FUNCTIONS

10.2 Automorphisms of the unit diskAn interesting consequence of this lemma is that it lets us characterise theautomorphisms of the unit disk.

Definition 10.2.1 (Automorphism). A one-to-one, bi-analytic mapping of aregion G onto G is called an automorphism of G.

In other words, f : G→ G is injective, surjective, analytic, and its inverse isalso analytic.

We will denote by Aut(G) the set of all automorphisms of G.

Note that, being analytic and one-to-one, an automorphism is also necessar-ily conformal.Remark 10.2.2. The bi-analytic condition in this definition is strictly speakingsuperfluous: a function that is one-to-one, onto, and analytic necessarily hasanalytic inverse.Exercise 10.1. Let f be an analytic function on a neighbourhood of B(a;R).Suppose that f is one-to-one on B(a;R). Let Ω := f(B(a;R)) and γ =

z∣∣ |z−

a| = R. Show that

f−1(w) = 12πi

∫γ

zf ′(z)f(z)− w dz

for all w ∈ Ω.

Let D =z ∈ G

∣∣ |z| < 1denote the unit disk in the proceeding discussion.

Let a ∈ C with |a| < 1, and define the Möbius transformation

ϕa(z) = z − a1− az .

Notice how ϕa is analytic for |z| < 1|a| , but 1

|a| > 1, so ϕa is analytic on aneighbourhood of D.

By straight-forward calculation

ϕa(ϕ−a(z)) = ϕa

( z + a

1 + az

)=

z+a1+az − a

1− a z+a1+az

= z + a− a− |a|2z1 + az − az − |a|2

= z,

and for exactly the same reason

ϕ−a(ϕa(z)) = z,

so ϕa : D → D so a one-to-one and onto analytic mapping, and its inverse ϕ−ais as well, so ϕa is an automorphism of D.

We strictly speaking already know it, being one-to-one and analytic, but inthis case it is also easy to verify that

ϕ′a(z) = 1− |a|2

(a− az)2 6= 0

for all z ∈ D, so ϕa is also conformal.On the unit circle ∂D =

z∣∣ |z| = 1

, i.e., z = eiθ for θ ∈ R,

|ϕa(eiθ)| =∣∣∣ eiθ − a1− aeiθ

∣∣∣ =∣∣∣ 1eiθ

eiθ − ae−θ − a

∣∣∣ = 1

10.2 Automorphisms of the unit disk 41

since the second fraction is a quotient of complex conjugates. In other wordsϕa(∂D) = ∂D.

So in summary, if we let |a| < 1 and define ϕa : D → D by

ϕa(z) = z − a1− az .

Then ϕa is one-to-one, onto, with its inverse being ϕ−a; ϕa(∂D) = ∂D; andϕa(a) = 0, ϕ′a(0) = 1− |a|2, and ϕ′a(a) = 1

1−|a|2 .

Exercise 10.2. (a) Let D =z∣∣ |z| < 1

. Let f : D → C be an analytic

function. Suppose that |f(z)| ≤ 1 for all z ∈ D and f(a) = α. Show that

|f ′(a)| ≤ 1− |α|2

1− |a|2 .

(b) Does there exist an analytic function f : D → D with f( 12 ) = 3

4 andf ′( 1

2 ) = 23?

Exercise 10.3. Let f : D → C be a non-constant analytic function such that|f(z)| ≤ 1 for all z ∈ D. Show that

|f(0)| − |z|1 + |f(0)||z| ≤ |f(z)| ≤ |f(0)|+ |z|

1− |f(0)||z| .

Exercise 10.4 (Borel–Carathéodory’s inequality). Let f be analytic on the closeddisk B(a;R) and let

M(r) = max|f(z)|

∣∣ |z| = r, A(r) = max

Re f(z)

∣∣ |z| = r.

(a) Show that A(r) is a monotone increasing function.

(b) Show that for 0 < r < R,

M(r) ≤ 2rR− r

A(R) + R+ r

R− r|f(0)|.

Not only are these automorphisms of D, but, up to rotations, all automor-phisms of D are one of these:

Theorem 10.2.3 (Schwarz–Pick theorem). Let f : D → D be a one-to-one,onto, analytic map. Suppose f(a) = 0. Then there exists c ∈ C such thatf(z) = cϕa(z) (i.e., a rotation of a Möbius map). Hence

Aut(D) =cϕa

∣∣ a ∈ C, |a| < 1, c ∈ C, |c| = 1.

Proof. Let g(z) = f(ϕ−a(z)) : D → D. Then since f(a) = 0 by assumption andϕa(a) = 0, meaning that ϕ−a(0) = a, we have

D D D

0 a 0,ϕ−a

g

f

42 BOUNDS OF ANALYTIC FUNCTIONS

so g(0) = f(f−a(0)) = f(a) = 0. Both of these maps are one-to-one and onto,so they have inverses, and hence g−1 exists. Moreover |g(z)| ≤ 1 for all z ∈ D(simply since g : D → D), so by Schwarz lemma |g(z)| ≤ |z| for all |z| < 1.

But the inverse map satisfies the exact same conditions, so |g−1(z)| ≤ |z| forall |z| < 1, so

|g(z)| ≤ |z| = |g−1(g(z))| ≤ |g(z)|,

meaning that |g(z)| = |z|. Hence by the second part of Schwarz lemma, g(z) =cz for some |c| = 1, so f(ϕ−a(z)) = cz. Replacing z by ϕa(z), this becomesf(z) = cϕa(z) for all |z| < 1.

10.3 Automorphisms of the upper half-planeAn analogous (and as we will see very much related) statement is about charac-terising the automorphisms of the complex upper half-plane H =

x + iy ∈

C∣∣ y > 0

.

Example 10.3.1. One example of such an automorphism is the Cayley trans-form

ψ(z) = z − iz + i

,

which is evidently a Möbius transformation. We see that ψ(R∞) = ∂D sincefor real x,

|ψ(x)| =∣∣∣x− ix+ i

∣∣∣ = 1

since, as once earlier, this is a quotient of complex conjugates. Note moreoverhow ψ(i) = 0, and since ψ is analytic, it is continuous, and a continuous functionmaps connected sets to connected sets, so the upper half-plane must map to theunit disk D, since the boundary of the upper half-plane, that is R∞, maps tothe boundary of the unit disk.

More generally, lettingϕα(z) = z − α

z − αfor Im(α) > 0, then ϕα(R∞) = ∂D, ϕα(α) = 0, and then ϕα : H→ D is a one-to-one, onto, and conformal mapping (where in particular of course ψ = ψi). N

Theorem 10.3.2. We have

Aut(H) =f(z) = az + b

cz + d

∣∣∣∣ a, b, c, d ∈ R, ad− bc > 0.

Proof. For f(z) = az+bcz+d , a, b, c, d ∈ R and ad − bc > 0, we naturally have

f(R∞) = R∞. Moreover

Im(f(i)) = ad− bcc2 + d2 > 0,

so by the same connected set argument as above f : H→ H, and so f ∈ Aut(H).Next let f ∈ Aut(H), and suppose f(i) = α with Im(α) > 0. The strategy

is to leverage what we already know about automorphisms of D, so in order to

10.3 Automorphisms of the upper half-plane 43

carry this situation over to D, consider the diagram

i 0H D

H Dα 0.

ψ

f g

ψα

In other words, let ψ(z) = z−iz+i be the Cayley transform and ψα(z) = z−α

z−a , anddefine g : D → D by g = ψα f ψ−1. Then

g(0) = ϕα f ψ−1(0) = ϕα(f(i)) = ϕα(α) = 0.

Moreover g ∈ Aut(D) since it is one-to-one, onto, and analytic, so by theSchwarz–Pick theorem g(z) = cz for some |c| = 1. Then, solving the diagramabove for f ,

f(z) = ψ−1α g ψ(z) = ψ−1

α (g(ψ(z))) = ψ−1α (cψ(z)).

Here cψα(z) = cz+cαcz−cα is a Möbius transformation, and so is ψ−1

α , so their com-position is too. Hence f is a Möbius transformation.

Now f ∈ Aut(H), so f(R∞) = R∞, and a Möbius transformation mappingthe real line to itself can be written as f(z) = az+b

cz+d with a, b, c, d ∈ R. MoreoverIm(f(i)) = Im(α) > 0 by assumption, so

Im(f(i)) = ad− bcc2 + d2 > 0,

meaning that ad− bc > 0.

As we have discussed before, Möbius transformations are invariant undermultiplying the numerator and denominator by a constant, so we can alwaysnormalise in such a way that Aut(H) is identified by(

a bc d

)∈ SL2(R) =

(a bc d

) ∣∣∣∣ a, b, c, d ∈ R, ad− bc = 1.

Making a brief detour into measure theory, this set of matrices has a groupstructure, and so Aut(H) inherits this group structure, and in fact it is a Liegroup. (Indeed a theorem by Henri Cartan guarantees that any automorphismgroup of a region is a Lie group.) Being a Lie group there exists a uniquemeasure invariant under the group action, known as the Haar measure, and thisprovides a corresponding measure on H, namely the hyperbolic measure

dµ(z) = dx dy

y2 ,

which then is invariant under group action by Aut(H).

Theorem 10.3.3 (Hadamard three-lines theorem). Let G be the vertical stripx+ iy ∈ C

∣∣ a < x < b. Suppose f : G→ C is continuous, not identical to 0,

and bounded. Suppose f is analytic in C.

44 HADAMARD THREE-LINES THEOREM

Define M : [a, b]→ R by

M(x) := sup−∞<y<∞

|f(x+ iy)|.

Then logM(x) is a convex function, i.e., for a ≤ x < u < y ≤ b,

logM(u) ≤ y − uy − x

logM(x) + u− xy − x

logM(y).

Lecture 11 Hadamard Three-Lines Theorem

11.1 Generalising the maximum modulus principleIn order to prove the Hadamard three-lines theorem stated at the end of lastlecture, we first need the following lemma:

Lemma 11.1.1. Let f and G be as in the Hadamard three-lines theorem. Sup-pose |f(z)| ≤ 1 on ∂G (not including ∞). Then |f(z)| ≤ 1 for all z ∈ G.

In other words, this is essentially the Maximum modulus principle except ona (special kind of) unbounded domain.

Proof. For any ε > 0, define the function gε(z) = 11+ε(z−a) for all z ∈ G. Then gε

is analytic in G, since the denominator is never 0 there. For any z = x+ iy ∈ G,since the magnitude of a complex number is bounded below by the magnitudeits real value,

|gε(z)| ≤1

|Re(1 + ε(z − a))| = 11 + ε(x− a) ≤ 1

since 0 ≤ x − a. This means that since |f(z)| ≤ 1 on ∂G, we also have|f(z)gε(z)| ≤ 1 on z ∈ ∂G.

The idea is that as the imaginary part of z is big, gε(z) is very small, sofor large heights we can ‘dampen’ whatever f is doing in the product. Moreprecisely, by assumption f is bounded in G, say |f(z)| ≤ B for all z ∈ G, andso

|f(z)gε(z)| ≤B

|1 + ε(z − a)| ≤B

ε|Im z|for |Im z| > 0, since the magnitude of a complex number is also bounded belowby the magnitude of its imaginary part. Hence if we take z to have imaginarypart larger than B

ε , the product is bounded by 1 in this strip, as we want, andin what remains we can use the ordinary Maximum modulus principle.

a b

Bε

−Bε

MaxMaxMaxMaxMaxMaxMaxMaxMaxMaxMaxMaxMaxMaxMaxMaxMax.mod.mod.mod.mod.mod.mod.mod.mod.mod.mod.mod.mod.mod.mod.mod.mod.mod.princ.princ.princ.princ.princ.princ.princ.princ.princ.princ.princ.princ.princ.princ.princ.princ.princ.hereherehereherehereherehereherehereherehereherehereherehereherehere

fgεsmall

fgεsmall

Figure 11.1.1: Split strip intobounded and unbounded por-tions.

All by way of saying: for z = x + iy ∈ G, with |y| ≥ Bε , we therefore have

|f(z)gε(z)| ≤ 1.In the remaining rectangle R =

x+iy

∣∣a ≤ x ≤ b, |y| ≤ Bε

, we have by the

above discussion that |f(z)gε(z)| ≤ 1 on ∂R, and so by the Maximum modulusprinciple |f(z)gε(z)| ≤ 1 also on R, and hence, combining the two parts, on G.

Thus|f(z)| ≤ 1

|gε(z)|= |1 + ε(z − a)|,

which goes to 1 as ε goes to 0. Hence finally |f(z)| ≤ 1 on G as desired.Date: September 24th, 2019.

11.1 Generalising the maximum modulus principle 45

With this we are equipped to prove the Hadamard three-lines theorem:

Proof. Note first how the statement we are to prove is equivalent to

M(u) ≤M(x)y−uy−xM(y)

u−xy−x

since the exponential function is increasing. Defining, therefore,

g(z) = M(x)y−zy−xM(y)

z−xy−x

for z ∈ C, we have an entire function. To see that this is the case, note firstthat M(x) ≥ 0 by definition, being the supremum of magnitudes. Moreover,M(x) = 0 would imply f(z) = 0 on a vertical line, and so f would have a limitpoints of zeros, and hence be zero everywhere, but we assumed f not identicallyzero. Moreover g(z) 6= 0 for z ∈ C.

Hence for z = u+ iv,

|g(z)| = M(x)y−uy−xM(y)

u−xy−x ,

which is the bound we are looking for. Now the right-hand side above is a con-tinuous function in u from [a, b] to R>0, and continuous functions send compactsets to compact sets, so the image must be a compact set in R>0, and hence theimage cannot touch 0. Therefore 1

|g(z)| is bounded in G, and so in turn f(z)g(z) is

bounded in G.So for z = x+ iv,

|f(z)||g(z)| = |f(x+ iv)|

M(x) ≤ 1,

since the bottom by definition is the supremum of the top, and similarly forz = y + iv,

|f(z)||g(z)| = |f(y + iv)|

M(y) ≤ 1.

Hence by Lemma 11.1.1 |f(z)||g(z)| ≤ 1 for x < Re z < y, and so |f(z)| ≤ |g(z)| for

all z = u+ iv with x ≤ u ≤ y. In other words

|f(u+ iv)| ≤M(x)y−uy−xM(y)

u−xy−x

and therefore

M(u) = sup−∞<v<∞

|f(u+ iv)| ≤M(x)y−uy−xM(y)

u−xy−x .

We showed, in the lemma, that if f is bounded by 1 on the boundary ofG, then it is bounded by 1 inside G. The same sort of thing is true in moregenerality:

Corollary 11.1.2. Let f and G be as in the Hadamard three-lines theorem.Then

|f(z)| ≤ supw∈∂G

|f(w)|.

46 PHRAGMÉN–LINDELÖF PRINCIPLE

Proof. LetM(x) = sup

−∞<y<∞|f(x+ iy)|.

By the Hadamard three-lines theorem, logM(x) is convex, and the maximumof a convex function occurs on the boundary, so

logM(x) ≤ max

logM(a), logM(b),

and taking exponentials M(x) ≤ maxM(a),M(b)

.

Lecture 12 Phragmén–Lindelöf Principle

12.1 Further generalising the maximum modulus princi-ple

First let us establish a variant of the Hadamard three-lines theorem:

Theorem 12.1.1 (Hadamard three-circle theorem). Let 0 < R1 < R2 < ∞.Suppose f is analytic and not identically zero on the annulus A =

z ∈ C

∣∣R1 <

|z| < R2. For R1 < r < R2, define

M(r) = max0≤θ≤2π

|f(reiθ)|.

Then for R1 < r1 < r < r2 < R2, we have

logM(r) ≤ log r2 − log rlog r2 − log r1

logM(r1) + log r − log r1

log r2 − log r1logM(r2),

i.e., logM(r) is a convex function of log r.

log r1log r1log r1log r1log r1log r1log r1log r1log r1log r1log r1log r1log r1log r1log r1log r1log r1 log r2

log rlog rlog rlog rlog rlog rlog rlog rlog rlog rlog rlog rlog rlog rlog rlog rlog r

becomes

r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1r1rrrrrrrrrrrrrrrrr

r2r2r2r2r2r2r2r2r2r2r2r2r2r2r2r2r2

Figure 12.1.1: Transforminga vertical strip with exp.

Proof. Noticing how the exponential function exp maps vertical lines at real partlog r to circles centred at the origin with radius r, as illustrated in Figure 12.1.1.

This result now follows immediately from the Hadamard three-lines theorem,since if we consider g = f exp, then g(log r + iθ) = f(reiθ). Hence

M(r) = max0≤θ≤2π

|f(reiθ)| = sup−∞≤θ≤∞

|g(log r + iθ)|,

the logarithm of which is convex in log r.

Exercise 12.1. Let f be analytic in an annulus R1 < |z| < R2 and not identicallyzero. Let

I2(r) = 12π

∫ 2π

0|f(reiθ)|2 dθ

Show that log I2(r) is a convex function of log r for R1 < r < R2.Moreover, if f is a non-constant analytic function on B(0;R), then I2(r) is

strictly increasing.

Date: September 26th, 2019.

12.1 Further generalising the maximum modulus principle 47

Exercise 12.2 ((A special case of) Hardy’s theorem). Let f be a non-constantanalytic function on B(0;R). Show that

I(r) = 12π

∫ 2π

0|f(reiθ)| dθ

is strictly increasing and log I(r) is a convex function of log r.

Notice how Corollary 11.1.2 of the Hadamard three-lines theorem says thatif a function f is bounded by M on vertical lines, then it is also bounded by Mbetween the vertical lines.

In general, to use the Maximum modulus principle to draw such a conclu-sion, we would require boundedness also at infinity, since the vertical strip isunbounded. The following is a powerful generalisation of this:

Theorem 12.1.2 (Phragmén–Lindelöf principle). Let G be a simply connectedregion. Let f : G → C be an analytic function, and ϕ : G → C be analytic andbounded in G and ϕ(z) 6= 0 for all z ∈ G.

Suppose ∂∞G = A ∪B such that

(i) lim supz→a

|f(z)| ≤M for all a ∈ A, and

(ii) lim supz→b

|ϕ(z)|η|f(z)| ≤ M for all b ∈ B and all η > 0 (in other words, f

isn’t necessarily bounded on B, but its growth is controlled by ϕ).

Then |f(z)| ≤M for all z in G.

Remark 12.1.3. In applications, one often takes A = ∂G and B =∞, so we

only need to control growth at infinity, and on the usual boundary we are justbounded.

Proof. Let |ϕ(z)| ≤ K for all z ∈ G. Since G is simply connected and ϕ(z) 6= 0for all z ∈ G, we can take ϕ(z) = eh(z) for some analytic function h : G → C.Now define g(z) = eηh(z) so that |g(z)| = |ϕ(z)|η.

Consider the function F : G→ C defined by

F (z) = f(z)g(z)Kη

,

which is analytic on G. We wish to use the Maximum modulus principle, and sowe should check that F (z) is bounded on ∂∞G, which by construction is A∪B.To do this, first notice how

|F (z)| = |f(z)||g(z)|Kη

≤ |f(z)|Kη

Kη= |f(z)|,

so for a ∈ A,lim supz→a

|F (z)| ≤ lim supz→a

|f(z)| ≤M.

Similarly, for b ∈ B,

lim supz→b

|F (z)| = lim supz→b

|f(z)||g(z)|Kη

≤ 1Kη

M

48 PHRAGMÉN–LINDELÖF PRINCIPLE

by choice of g. Hence by the Maximum modulus principle,

|F (z)| ≤ maxM,

M

kη

for all z ∈ G, and therefore

|f(z)| ≤ Kη

|ϕ(z)|η maxM,

M

Kη

for all z ∈ G since ϕ(z) 6= 0. By taking η → 0, this bound goes to M , and so|f(z)| ≤M for all z ∈ G.

This gives us a general tool for bounding functions on unbounded domains,more powerful than the Maximum modulus principle since we no longer requireboundedness at infinity, only some sufficiently good growth conditions.

For example, both as an example of how to use the theorem and the commonpractice of taking B =

∞as mentioned in the remark:

Corollary 12.1.4. Let G be the sector G =z∣∣ |arg z| < π

2a

for a ≥ 12 .

Suppose f is analytic in G and there exists some M > 0 such that

lim supz→w

|f(z)| ≤M

for all w ∈ ∂G. Suppose moreover there exists some p > 0 and 0 ≤ b < a suchthat

|f(z)| ≤ p exp(|z|b)

for all z ∈ G with |z| sufficiently large. Then |f(z)| ≤M for all z ∈ G.

πa

GGGGGGGGGGGGGGGGG

Figure 12.1.2: Sector G of an-gle π

a.

Remark 12.1.5. The corollary remains true if we replace G by any sector S ofangle π

a . To see this, consider

g : G S C,eiθ f

i.e., g(z) = f(eiθz). Then sup|g(z)| = sup|f(z)|, and for a given f , we canchoose θ such that g is in the original sector G of the same angle, and apply thecorollary there.

Proof. Let b < c < a, and define ϕ(z) = exp(−zc) for z ∈ G. Then ϕ(z) 6= 0since it is an exponential function, and writing z = reiθ ∈ G with |θ| < π

2a wehave

Re(zc) = Re(rceicθ) = rc cos(cθ).

Since 0 < c < a and |θ| < π2a , we have |cθ| <

π2 ·

ca <

π2 , whence cos(cθ) ≥ ρ > 0.

This means thatRe(−zc) ≤ −rcρ < 0,

which further gives us

|ϕ(z)| = exp(Re(−zc)) ≤ 1.

Hence for any η > 0, z = reiθ ∈ G, we have

|f(z)||ϕ(z)|η ≤ p exp(|z|b) exp(−rcρη) = p exp(rb − (ρη)rc).

12.1 Further generalising the maximum modulus principle 49

Since c > b by choice, the second term in the exponential is dominant, so asr →∞ the inside goes to −∞ and so the exponential as a whole goes to 0. Thus

lim sup|z|→∞

|f(z)||ϕ(z)|η = 0,

and hence by the Phragmén–Lindelöf principle,

|f(z)| ≤ maxM, 0

= M

for all z ∈ G.

Remark 12.1.6. In this corollary, for a sector of angle πa , we assume |f(z)| is

bounded by exp(|z|b), specifically with b < a. If we take b = a, the corollary isstill true, but the estimate needed is more subtle.

Looking at it, the key reason in our above proof where we use b < a is to pickb < c < a, which in the end makes rc the dominant term in exp(rb− (ρη)rc). Ifb = a (= c), by this argument we get exp((1 − ρη)ra), so for small η, this goesto infinity as r →∞.Exercise 12.3. Let G =

z∣∣ Re(z) > 0

and let f : G → C be analytic such

that f(1) = 0. Suppose that

lim supz→w

|f(z)| ≤M

for w ∈ ∂G, and suppose that for every δ > 0 with 0 < δ < 1, there is a constantP such that

|f(z)| ≤ P exp(|z|1−δ)

for all z ∈ G. Prove that

|f(z)| ≤M(

(1− x)2 + y2

(1 + x)2 + y2

)1/2

,

for all z = x+ iy ∈ G.

Corollary 12.1.7. Let G =z∣∣ |arg z| < π

2afor a ≥ 1

2 . Suppose

lim supz→w

|f(z)| ≤M

for all w ∈ ∂G. Suppose for any δ > 0 there exists a constant pδ > 0 such that

|f(z)| ≤ pδ exp(δ|z|a)

for all z ∈ G with |z| sufficiently large. Then |f(z)| ≤M for all z ∈ G.

Proof. For any ε > 0, consider F : G→ C defined by

F (z) = f(z) exp(−εza).

Choose 0 < δ < ε and pδ > 0 so that |f(z)| ≤ pδ exp(δ|z|a). Then for any x > 0,x ∈ R, we have

|F (x)| ≤ pδ exp((δ − ε)xa)→ 0

as x→∞ since δ − ε < 0. The idea here is this: the original approach requires|f(z)| ≤ p exp(|z|b) with b < a, where the sector is of angle π

a , but we currently

50 THE SPACE OF ANALYTIC FUNCTIONS

have b = a. What we have just shown is that F (z) is bounded on the positivereal axis, so we can split our sector of angle π

a into two sectors both of angleπ2a—see Figure 12.1.3—where now b = a < 2a permits us to use the previousresult.

In other words, let H+ =z∣∣0 < arg z < π

2aand H− =

z∣∣− π

2a < arg z <0, and let

M1 := sup0<x<∞

|F (x)| <∞

since by the above discussion F (x) is bounded at infinity, and define M2 =max

M1,M

.

H+

H−

Figure 12.1.3: Splitting sectorG into halves of angle π

2a .

Then for w ∈ ∂H+ or w ∈ ∂H− we have

lim supz→w

|F (z)| ≤M2

since it is bounded by M on the original boundary and by M1 on the real axis.Hence by Corollary 12.1.4 and Remark 12.1.5 about how it works for any sector,|F (z)| ≤M2 for all z ∈ H+ and z ∈ H−, and hence |F (z)| ≤M2 for all z ∈ G.

Finally, we claim thatM2 = M , i.e.,M ≥M1. To see this, supposeM < M1.Then there exists some x ∈ R such that |F (x)| ≥ |F (z)| for all z ∈ G, in otherwords F (z) attains its maximum in the interior of G. The Maximum modulusprinciple then implies F (z) is constant, which is a contradiction.

Hence |F (z)| ≤ M for all z ∈ G, and therefore |f(z)||exp(−εza)| ≤ M ,implying in the end

|f(z)| ≤M exp(εza)

which as ε→ 0 goes to M .

Lecture 13 The Space of Analytic Functions

13.1 The topology of C(G,C)For the proceeding discussion we need some results analysis.

Proposition 13.1.1. Let G ⊂ C be open. Then there exists a sequenceKn

of compact subsets of G such that G =

∞⋃n=1

Kn satisfying

(i) Kn ⊂ int(Kn+1);

(ii) if K ⊂ G is compact, then K ⊂ Kn for some n; and

(iii) every component of C∞ \Kn contains a component of C∞ \G.

Sketch of proof. Since G is open, we can define Kn to be the set of all points inG at least 1/n away from the boundary of boundary of G. That is,

Kn =z ∈ G

∣∣∣∣ |z − w| ≥ 1n

for all w ∈ C \G.

This construction satisfies all of these conditions.Date: October 1st, 2019.

13.1 The topology of C(G,C) 51

Definition 13.1.2 (Continuous functions). Let Ω be C or C∞. We define

C(G,Ω) =f : G→ Ω

∣∣ f is continuous on G,

the set of continuous from G to Ω.

Remark 13.1.3. Note that in principle this only requires G and Ω to be metricspaces—we are specifying for the purpose of doing complex analysis.

Denote the metric on Ω by d (so if Ω = C, then d(z1, z2) = |z1 − z2|. Themetric on the Riemann sphere C∞ we will talk more about later).

If G is an open set, then by Proposition 13.1.1 we can write G =∞⋃n=1

Kn for

a sequence of compact sets Kn. For f and g in C(G,Ω) we define

ρn(f, g) = supz∈Kn

d(f(z), g(z))

for all n ∈ N, and further define

ρ(f, g) =∞∑n=1

(12

)nρn(f, g)

1 + ρn(f, g) .

Then (C(G,Ω), ρ) is a metric space. (That ρ is a metric follows from d(x,y)1+d(x,y)

being a metric for any metric d, and since this is bounded above by 1, the seriesabove is bounded above by a convergent geometric series.)

Lemma 13.1.4. (i) Given ε > 0, there exists δ > 0 and a compact set K ⊂G such that for f, g ∈ C(G,Ω),

supz∈K

d(f(z), g(z)) < δ

implies ρ(f, g) < ε.

(ii) Given a δ > 0 and compact set K ⊂ G, there exists ε > 0 such that forf, g ∈ C(G,Ω), ρ(f, g) < ε implies

supz∈K

d(f(z), g(z)) < δ.

In other words, if f and g are close in some compact set K (or Kn inparticular), then they are close in the sense of ρ, and vice versa.

Remark 13.1.5. This lemma says that fn → f in (C(G,Ω), ρ) if and only iffn → f is uniformly convergent (because of the supremum) on every compactsubset K of G.

Proposition 13.1.6. The space (C(G,Ω), ρ) is a complete metric space.

Proof. A sequence of continuous functions converging uniformly must have acontinuous limit. This is a consequence of this fact.

52 COMPACTNESS IN H(G)

13.2 The space of analytic functionsIn particular we are interested not just in the continuous functions from G toC, but the continuous and differentiable ones, i.e., the analytic functions fromG to C.

That is to say, let G ⊂ C be open, and let

H(G) :=f : G→ C

∣∣ f is analytic on G.

Then H(G) ⊂ C(G,C) can be endowed with the subspace topology.5

Theorem 13.2.1. Letfn⊂ H(G). Suppose fn → f for some f ∈ C(G,C).

Then f ∈ H(G) and f (k)n → f (k) for all k ≥ 1.

In other words, H(G) is a closed subset of C(G,C).

Remark 13.2.2. This result does not hold in the real case. For example, iffn(x) = 1

nxn on x ∈ [0, 1], then fn → f = 0 uniformly in [0, 1], but f ′n(x) = xn−1

does not converge at all in the space of continuous functions (the pointwise limitexists, but is not continuous).

Lecture 14 Compactness in H(G)

14.1 Compactness in the space of analytic functionsProof of Theorem 13.2.1. For the first part, that the limit fn → f is analytic,note that differentiability is a local property, so it suffices for z ∈ G to considerB(a, r) ⊂ G. Then for any triangular path T ⊂ B(z, r),∫

T

f(z) dz = limn→∞

∫T

fn(z) dz

since fn → f uniformly on compact sets, say B(z, r) (take a slightly smaller r ifnecessary). Now all fn are analytic by assumption, so the integral is 0. Henceby Morera’s theorem f is analytic.

For the second part, again take B(z, r) ⊂ G, and let γ be the circle |w−z| =r. By Cauchy’s formula,

f (k)n (z)− f (k)(z) =

∫γ

fn(w)− f(w)(w − z)k+1 dw.

Since fn → f uniformly on B(z, r), we can choose, for any ε > 0, n large enoughso that |fn(w)− f(w)| < ε for all w ∈ B(z, r). Hence

|f (k)n (z)− f (k)(z)| ≤

∫γ

ε

rk+1 dw ≤ εr−(k+1)(2πr),

which we can hence make arbitrarily small, so f (k)n → f (k).

Immediately from this we have:5We call this space H for holomorphic, which in the case of complex functions is equivalent

to being analytic.Date: October 3rd, 2019.

14.1 Compactness in the space of analytic functions 53

Corollary 14.1.1. H(G) is a complete metric space.

Proof. It is a closed subset of a complete metric space, hence itself complete.

Corollary 14.1.2. Letfn⊂ H(G). Suppose

∞∑n=1

fn(z)

converges uniformly on compact subsets to f(z). Then

f (k)(z) =∞∑n=1

f (k)n (z).

In other words, we can take derivatives term by term.

Exercise 14.1. Let f, f1, f2, . . . be elements ofH(G). Show that fn → f inH(G)if and only if for each closed rectifiable curve γ in G, fn(z) → f(z) uniformlyfor z in

γ.

It is instructive to remark that this is not true in the real case: the specialstructure of complex analytic functions that make this work is that, in the caseof complex functions, the derivatives of a function f are controlled by f itself.

Theorem 14.1.3 (Hurwitz’s theorem). Let G be a region, and letfn⊂

H(G) be a convergent sequence, so fn → f ∈ H(G). Suppose f 6= 0 onB(a,R) ⊂ G and f(z) 6= 0 on |z − a| = R. Then there exists some N ∈ Nsuch that for n ≥ N , fn and f have the same number of zeros in B(a,R).

Proof. Since f(z) 6= 0 on |z − a| = R, which is a compact set, we must haveδ := inf

|z−a|=R|f(z)| > 0. Since fn → f uniformly on |z − a| = R, there exists

some N ∈ N such that for n ≥ N and |z− a| = R, fn(z) 6= 0, since it is close tof , which is at least δ away from 0.

So on |z − a| = R,

|fn(z)− f(z)| < δ

2 < |f(z)|,

which by Rouché’s theorem means fn and f have the same number of zeros inB(a,R).

Corollary 14.1.4. Let G be a region. Letfn⊂ H(G) and fn → f ∈ H(G).

Suppose fn(z) 6= 0 for all z ∈ G and all n ∈ N. Then either f = 0 or f(z) 6= 0for all z ∈ G.

Exercise 14.2. Letfn⊂ H(G) be a sequence of one-to-one functions. Sup-

pose fn → f in H(G). Show that either f is one-to-one or f is a constantfunction.

54 COMPACTNESS IN H(G)

14.2 Montel’s theoremIn order to do the following discussion justice we need to recall some informationfrom functional analysis.

First of all, if (M,d) is a metric space, then there are several equivalentnotions of compactness. In particular, M is compact (in the sense of every opencover having a finite subcover) if and only of M is sequentially compact (i.e.,every bounded sequence has a convergent subsequence), if and only ifM is limitpoint compact (meaning every infinite set has a limit point).

Recall how in Rn or Cn (or in fact any finite dimensional vector space), aset is compact if and only if it is closed and bounded. In general, this is notenough:

Definition 14.2.1 (Equicontinuity). A family of functions F in a metric space(M,d) is equicontinuous if for every ε > 0 there exists a δ > 0 so thatd(f(x), f(y)) < ε for all f ∈ F and d(x, y) < δ.

In other words, it is the usual definition of continuity, except that for a givenε > 0, the same δ > 0 works for every f ∈ F at once.

Theorem 14.2.2 (Arzelà–Ascoli theorem). Let (M,d) be a compact metricspace, and let C(M) =

f : M → R or C

∣∣ f is continuous. A family F ⊂

C(M) is compact if and only if F is closed, bounded, and equicontinuous.

Definition 14.2.3 (Normal). Let G ⊂ C be open and Ω = C or Ω = C∞. Aset F ⊂ C(G,Ω) is normal if every sequence in F has a subsequence whichconverges to a function f ∈ C(G,Ω).

Note how this is different from sequential compactness—the limit f of theconvergent subsequence need not be in F .Exercise 14.3. A set F ⊂ C(G,Ω) is normal if and only if its closure is compact.

Exercise 14.4. (a) Let f be an analytic function on an open neighbourhoodof B(a;R). Show that

|f(a)|2 ≤ 1πR2

∫ 2π

0

∫ R

0|f(a+ reiθ)|2r dr dθ.

(b) Let G be a region and let M > 0 be a fixed constant. Let F be the familyof all functions f in H(G) such that

∫∫G|f(z)|2 dx dy ≤M . Show that F

is normal.

Remark 14.2.4. This problem implies the following: Let f, fn ∈ H(G). If fn → fin L2(G), then fn → f in H(G).

In other words, L2 convergence for analytic functions implies uniform conver-gence on compact subsets. The same is true for L1 convergence, as it happens,though neither are true for real analytic functions:Exercise 14.5. Let f, fn ∈ H(G). Suppose fn → f in L1(G), i.e.,

∫∫G|fn(z) −

f(z)| dx dy → 0. Show that fn → f in H(G).

So in C(G,C), compact means closed, bounded, and equicontinuous. Thequestion we are looking to answer with this discussion is how we would similarlycharacterise compact sets in H(G). In other words, what does it mean to benormal in H(G)?

14.2 Montel’s theorem 55

Definition 14.2.5 (Locally bounded). A set F ⊂ H(G) is locally boundedif for each a ∈ G there exists some M > 0 and r > 0 so that for every f ∈ F ,|f(z)| ≤M for all z ∈ B(a, r).

In other words, for each a ∈ G, there exists r > 0 such that

supz∈B(a,r)f∈F

|f(z)| <∞,

or, as a final way of putting it, for every a ∈ G there exists r > 0 such that Fis uniformly bounded in B(a, r).

It follows more or less from this definition that:

Lemma 14.2.6. A set F ⊂ H(G) is locally bounded if and only if F is uni-formly bounded on every compact subset K ⊂ G.

Theorem 14.2.7 (Montel’s theorem). A family F ⊂ H(G) is normal if andonly if F is locally bounded.

Proof. For the forward direction, suppose F is not locally bounded, i.e., thereexists some compact subset K ⊂ G such that

supz∈Kf∈F

|f(z)| =∞.

So there exists somefn⊂ F such that sup

z∈K|fn(z)| ≥ n. Since F is normal,

fnhas a convergent subsequence

fnk

, say fnk → f ∈ H(G) (nominally

C(G,C), but H(G) is closed, we showed). Since K is compact and f is contin-uous,

supz∈K|f(z)| = M <∞.

Hence

nk ≤ supz∈K|fnk(z)| ≤ sup

z∈K|fnk(z)− f(z)|+ sup

z∈K|f(z)|

= supz∈K|fnk(z)− f(z)|+M

where the right-hand side approaches M as nk → ∞ since fnk → f uniformlyon K, but the left-hand side goes to infinity, which is a contradiction.

For the converse direction, let K ⊂ G be compact. We claim that when Frestricts to (C(K), ‖·‖∞), every sequence in F has a convergent subsequence inC(K).

To prove this, by the Arzelà–Ascoli theorem, we need to show that F isclosed, bounded, and equicontinuous. That it is closed in (C(K), ‖·‖∞) is clear:it is a closure. Since F is locally bounded, it is uniformly bounded on K,being compact, so F is also uniformly bounded on K. Hence F is bounded in(C(K), ‖·‖∞).

56 THE SPACE OF MEROMORPHIC FUNCTIONS

It remains to show that it is equicontinuous. Since F is locally bounded,for each a ∈ G there exists r > 0 and M > 0 such that |f(z)| ≤ M for everyz ∈ B(a, r) and every f ∈ F . For z ∈ B(a, r2 ), by Cauchy’s integral formula,

|f(a)− f(z)| = 12π

∣∣∣∫γ

f(w)w − a

− f(w)w − z

dw∣∣∣

= 12π

∣∣∣∫γ

f(w)(a− z)(w − a)(w − z) dw

∣∣∣≤ 1

2π

∫γ

M |a− z|r · r2

|dw| ≤ 1π

M

r2 2πr|a− z|,

where the constant in front of |a−z| does not depend on f . Hence F is uniformlyLipschitz, and hence equicontinuous, and so finally F is equicontinuous, finishingthe claim.

γ

r2r2r2r2r2r2r2r2r2r2r2r2r2r2r2r2r2

rrrrrrrrrrrrrrrrr×a

×z

×w

Figure 14.2.1: Schematic of wand z.

Finally, to finish the theorem, write G =∞⋃n=1

Kn as in Proposition 13.1.1.

For anyfn⊂ F ⊂ F , by the claim, when

fnrestricts to K1, it has a

convergent subsequence, satfn1

k

in H(K1).

Now considerfn1

k

restricted to K2—it has a convergent subsequence

fn2k

in H(K2), and so on, with

fnm

k

being convergent in H(Km).

Take the diagonalfnk

k

⊂fn. Then

fnk

k

converges uniformly when

restricted to any compact set K ⊂ G, so fnkk→ f in H(G), meaning that F is

normal.

This lets us answer our question:

Corollary 14.2.8. A set F ⊂ H(G) is compact if and only if F is closed andlocally bounded.

Lecture 15 The Space of Meromorphic Functions

15.1 The topology of C(G,C∞)Recall how we model the Riemann sphere C∞ = C ∪

∞as a sphere centred

on the origin of the origin of the complex plane, and we identify the north poleof the sphere as∞, and if we imagine a line connecting this north pole with anygiven point z on the complex plane, where this line intersects the sphere, sayZ, is the corresponding point. We have illustrated this in Figure 15.1.1.

If we wish to study continuous functions from G to C∞, we first need ametric on C∞. We define this metric d by

d(z1, z2) = d(Z1, Z2)

for z1, z2 ∈ C, where Z1 and Z2 are the points corresponding to z1 and z2 onthe sphere, and the right-hand side distance is measured in the usual Euclideanway in R3.

Then in all, we define the metric on C∞ by

Date: October 8th, 2019.

15.1 The topology of C(G,C∞) 57

×∞

×Z1Z1Z1Z1Z1Z1Z1Z1Z1Z1Z1Z1Z1Z1Z1Z1Z1

×z1z1z1z1z1z1z1z1z1z1z1z1z1z1z1z1z1

×z2z2z2z2z2z2z2z2z2z2z2z2z2z2z2z2z2

×Z2Z2Z2Z2Z2Z2Z2Z2Z2Z2Z2Z2Z2Z2Z2Z2Z2

C

C∞

Figure 15.1.1: A model of the Riemann sphere, identifying points z1, z2 ∈ Cwith points Z1, Z2 ∈ C∞.

(i) if z1, z2 ∈ C, then d(z1, z2) = 2|z1 − z2|(1 + |z1|2)1/2(1 + |z2|2)1/2 (this is just

d(Z1, Z2) written out); and

(ii) if z ∈ C, then d(z,∞) = 2(1 + |z|2)1/2 .

This metric has an interesting property: for z1, z2 6= 0,

d(z1, z2) = d

(1z1,

1z2

),

and similarly for z 6= 0,

d(z, 0) = d

(1z,∞).

This is a consequence of z and 1/z corresponding to symmetric points in theupper and lower hemispheres in this model.Remark 15.1.1. In order to distinguish the topology on the complex plane andthe topology on the Riemann sphere, we will (as before) use B(a, r) to meanan open ball in (C, |·|), and B∞(a, r) to mean a ball in the Riemann sphere(C∞, d).

The good news is that these topologies are essentially the same (barring thetricky point at infinity):

Proposition 15.1.2. (i) Given a ∈ C and r > 0, there exists ρ > 0 suchthat B∞(a, ρ) ⊂ B(a, r) (so open in |·| implies open in d);

(ii) Given a ∈ C and ρ > 0, there exists r > 0 such that B(a, r) ⊂ B∞(a, ρ)(so open in d implies open in |·|);

(iii) Given ρ > 0, there exists a compact set K ⊂ C such that C∞ \ K ⊂B∞(∞, ρ); and

(iv) Given a compact set K ⊂ C, there exists ρ > 0 such that B∞(∞, ρ) ⊂C∞ \K.

58 COMPACTNESS IN M(G)

Remark 15.1.3. Parts (i) and (ii) together imply that the subspace topology onC ⊂ C∞ and the usual topology (C, |·|) are the same. In other words, thingswill converge in one if and only if they converge in the other (though the speedof convergence needn’t be the same).

Parts (iii) and (iv) tell us that C∞ is, in fact, a one-point compactificationof C.

Recall how a function f : G → C is called meromorphic if it is analyticexcept for isolated poles. Now that we have a metric on the Riemann sphere,we can view this in a different light: a meromorphic function is an analyticfunction (hence continuous) which sometimes reaches infinity. In other words,f : G→ C∞ is continuous.

In this light, if we let M(G) denote the set of all meromorphic functions onG, then M(G) ⊂ C(G,C∞), where C(G,C∞) is a complete metric space in d.Remark 15.1.4. Unlike H(G), the space M(G) of meromorphic functions is notcomplete, for it is not closed.

For example, if we let fn(z) = n be constant functions, thenfn

is aCauchy sequence in C(G,∞) (they get close to ∞), and fn → f where f ≡ ∞.But this limit is not meromorphic.

It is instructive to note that this does not converge at all in C(G,C)—convergence in C(G,C∞) is not quite the same. This example then also demon-strates that in C(G,C∞), H(G) also stops being closed.

That said, M(G) is, in some sense, very close to being complete: the onlyfunction it is missing is the one that is constantly infinity:

Theorem 15.1.5. (i) Letfn⊂ M(G). Suppose fn → f in C(G,C∞).

Then either f ∈M(G) or f ≡ ∞.

(ii) Supposefn⊂ H(G) and fn → f in C(G,C∞). Then either f ∈ H(G)

or f ≡ ∞.

Consequently,

Corollary 15.1.6. M(G) = M(G) ∪∞is a complete metric space.

Corollary 15.1.7. H(G) = H(G) ∪∞is a closed in C(G,C∞) and hence

complete.

Lecture 16 Compactness in M(G)

16.1 Compactness in the space of meromorphic functionsBefore moving on, let us take a step back and prove Theorem 15.1.5 from theend of last lecture.

Proof. (i) Let a ∈ G. Since f is a function in C(G,C∞), there are essentiallytwo cases on hand: either f(a) is finite or f(a) =∞.

Let us start with the first one: assume f(a) 6=∞. Since fn → f in C(G,C∞),meaning fn(z)→ f(z) uniformly on compact subsets of G, we have in particular

Date: October 10th, 2019.

16.1 Compactness in the space of meromorphic functions 59

d(fn(z), f(z))→ 0 uniformly on B(a, r) ⊂ G. By Proposition 15.1.2, restrictingd to C becomes the usual topology on C, so |fn(z) − f(z)| → 0 uniformly onB(a, r).

Notice how the set S =f, f1, f2, . . .

⊂ C(G,C∞) is compact (since it it

sequentially compact), and hence it is equicontinuous, since the Arzelà–Ascolitheorem tells us compact is equivalent with closed, bounded, and equicontinu-ous.

Therefore given any ε > 0 there exists r1 < r such that d(fn(z), fn(a)) < εfor all z ∈ B(a, r1) and all n ∈ N. Again by Proposition 15.1.2, there then existssome ρ > 0 such that |fn(z)− fn(a)| < 1

2ρ for all z ∈ B(a, r1) and all n ∈ N.On the other hand, fn(a) → f(a) means that there exists some N ∈ N

such that for n ≥ N , |fn(a) − f(a)| < 12ρ. Putting these together, the triangle

inequality tells us that for n ≥ N and z ∈ B(a, r1),

|fn(z)| ≤ |fn(z)− fn(a)|+ |fn(a)− f(a)|+ |f(a)| ≤ 12ρ+ 1

2ρ+ |f(a)| <∞.

So the setfnn≥N is uniformly bounded on B(a, r1), and hence those fn

are analytic, not just meromorphic, on B(a, r1) (since boundedness means theycannot get close to infinity, so can’t have poles there). We already know (Corol-lary 14.1.1) that H(G) is a complete metric space, so fn being analytic meanstheir limit f must be too, so f is analytic on B(a, r1).

This leaves the case where f(a) = ∞. In other words, f(a) has a pole atz = a. We need to show that it is isolated.

To this end, define

gn(z) =

1fn(z) , if fn(z) 6= 0,∞, if fn(z) = 0.

Since d(z1, z2) = d( 1z1, 1z2

) and fn → f , we consequently have gn → 1f in

C(G,C∞).Now since 1

f(a) = 0, not infinity, the first case we considered tells us thatfn(z) and 1

f(z) are analytic on a ball B(a, r) ⊂ G for some n ≥ N sufficientlylarge. By Hurwitz’s theorem, either 1

f is identically zero or 1f(z) has the same

number of zeros as gn(z) in B(a, r), for n sufficiently large. This in turn impliesthat f ≡ ∞ or f has isolated poles (since its poles are zeros of 1

f , and it has asmany zeros as gn(z) in B(a, r), and those zeros must be isolated, being analytic).

(ii) Assumefn⊂ H(G). Then, since fn has no poles, 1

fnhas no zeros. Since

fn → f , we then have 1fn→ 1

f in C(G,C∞), which in turn means 1fn(z) →

1f(z)

uniformly on B(a, r) (in both C(G,C∞) and C(G,C)).By the first case above, all 1

fnare analytic for n large enough, so by Hurwitz’s

theorem either 1f ≡ 0 or 1

f and 1fn

have the same number of zeros in B(a, r) forn sufficiently large.

But 1fn

has no zeros, meaning that either 1f ≡ 0 or 1

f(z) 6= 0 for all z ∈ B(a, r).The first situation means f ≡ ∞, and the second situation means f has no poles,so f(z) is analytic on B(a, r).

60 COMPACTNESS IN M(G)

With this out of the way we are ready to start answering our main questionin this recent discussion: how do we classify normal families in M(G)?

It is instructive at this point to recall what we did in the case of H(G),namely Montel’s theorem. The culmination of our argument in showing that ourfamily was equicontinious was to instead show that it is uniformly Lipschitz, andto demonstrate this we showed that the derivatives f ′ are uniformly bounded.

The reason for this, of course, is that if we wish to show that |f(z)− f(a)| issmall when |z − a| is small, we can instead show that their quotient is bounded(i.e., f is Lipschitz), but their quotient is a good approximation of f ′(a), so wecan instead show that the derivative is bounded.

This is our approach also in the case of meromorphic functions, except nowthe metric we need to reconcile isn’t |·|, but d, and hence we need another wayof approximating the notion of continuity in terms of some kind of derivative.

Definition 16.1.1. Let f ∈M(G), and define µ(f) : G→ R by

µ(f)(z) = 2|f ′(z)|1 + |f(z)|2

if z is not a pole of f , and

µ(f)(a) = limz→a

2|f ′(z)|1 + |f(z)|2

if z = a is a pole of f .

The motivation is already outlined above, but concretely, this is becaused(f(z), f(w)) is well approximated by µ(f)(z)|z − w|.

Consequently, if µ(f) is bounded, then f is Lipschitz, and so if µ(f) isuniformly bounded for all f ∈ F , then F is uniformly Lipschitz.

That said, we ought to first make sure this µ(f) makes sense at all. Inparticular, is it well-defined? Does the limit in the case of z = a being a poleexist?

Suppose z = a is a pole of order m of f . It has a Laurent expansion

f(z) = Am(z − a)m + · · ·+ A1

z − a+ g(z),

where Am 6= 0 and g(z) is analytic. Then

f ′(z) = −(

mAm(z − a)m+1 + · · ·+ A1

(z − a)2

)+ g′(z),

meaning that

2|f ′(z)|1 + |f(z)|2 =

2∣∣∣ mAm

(z−a)m+1 + · · ·+ A1(z−a)2 − g′(z)

∣∣∣1 +

∣∣∣ Am(z−a)m + · · ·+ A1

z−a + g(z)∣∣∣2

= 2|z − a|m+1|mAm + · · ·+A1(z − a)m−1 − g′(z)(z − a)m+1||z − a|2m + |Am + · · ·+A1(z − a)m−1 + g(z)(z − a)m|2 .

Hence if m ≥ 2, then

limz→a

2|f ′(z)|1 + |f(z)|2 = 0

Am= 0,

16.1 Compactness in the space of meromorphic functions 61

and if m = 1, then

limz→a

2|f ′(z)|1 + |f(z)|2 = 2|A1|

|A1|2= 2|A1|

.

Hence µ(f)(z) ∈ R for all z ∈ G and is well-defined, and moreover by construc-tion µ(f) ∈ C(G,R).

Theorem 16.1.2. A set F ⊂ M(G) is normal in C(G,C∞) if and only ifµ(F) =

µ(f)

∣∣ f ∈ F is locally bounded.

Before we go on to prove this, it is instructive to compare this to our approachin the case of H(G) again. As previously discussed, we ended up showing thatF ′ is locally bounded in this case, except in the end our characterisation wasF is normal if and only if F is locally bounded. The reason for this is thatthe derivative of an analytic function is controlled by the function itself—thesalient part of the proof is still F ′ being locally bounded, it just so happensthat F ⊂ H(G) being locally bounded implies F ′ is locally bounded.Exercise 16.1. Show that if F ⊂ H(G) is normal, then F ′ :=

f ′∣∣ f ∈ F is

also normal. Is the converse true? Can you add something to the hypothesisthat F ′ is normal to insure that F is normal?

With this in mind, it should come as no great surprise that the proof, whilstin parts a bit technical, is closely related to our proof of Montel’s theorem.

Proof. For the forward direction, suppose µ(F) is not locally bounded. Thenthere exists

fn⊂ F and a compact subset K ⊂ G such that

supz∈K|µ(f)(z)| ≥ n.

Since F is normal, there exists some convergent subsequencefnk

⊂fn

such that fnk → f in C(G,C∞), and in particular the convergence is uniformon compact subsets (such as K!). Hence µ(fnk)→ µ(f) in C(G,R). Therefore

nk ≤ supz∈K|µ(fnk)(z)| ≤ sup

z∈K|µ(fnk(z)− µ(f)(z)|+ sup

z∈K|µ(f)(z)|.

The left-hand side evidently goes to infinity as nk →∞, but the right-hand sidedoes not: since µ(fnk) → µ(f), the first term in the right-hand side vanishes,and the second term does not depend on nk and so is bounded. This is acontradiction.

The converse direction is where things get technical, though the idea is fairlyapproachable. Assume µ(F) is locally bounded. We want to show that F isnormal, which is equivalent to F being compact. The Arzelà–Ascoli theoremtells us this is the case if and only if F is closed, bounded, and equicontinuous.

The first two of these are not so bad: F is definitely closed, being a closure,and moreover it is bounded; the point of working in C∞ is that it is compact—it’sa (one-point) compactification of C—in particular d(z, w) ≤ 2 for all z, w ∈ C∞(remember, the metric in C∞ is defined as the Euclidean R3 distance betweenpoints on the unit sphere).

Therefore the only tricky bit is to show that F is equicontinuous, which wewant to show by proving that it is uniformly Lipschitz, which in turn we willacquire as a consequence of µ(F) being locally bounded. The bad news is thatthis last part is nontrivial.

62 COMPACTNESS IN M(G), CONTINUED

We want to work on compact subsets of G, but for convenience we willrestrict ourselves to closed disks K = B(a, r), of which we could patch togetherseveral to get any compact subset.

Since µ(F) is locally bounded, there exists M > 0 such that µ(f)(z) ≤ Mfor all z ∈ K and all f ∈ F . Let z, z′ ∈ K and f ∈ F .

There are three cases to consider. First, suppose neither z nor z′ are polesof f . Let α > 0 be arbitrary, and choose

w0 = z, w1, w2, . . . , wn = z′

in K satisfying

(i) for w ∈ [wk−1, wk], w is not a pole of f ;

(ii)n∑k=1|wk − wk−1| ≤ 2|z − z′|;

(iii)∣∣∣ 1 + |f(wk−1)|2

(1 + |f(wk)|2)(1 + |f(wk)|2) − 1∣∣∣ < α for all 1 ≤ k ≤ n; and

(iv)∣∣∣f(wk)− f(wk−1)

wk − wk−1− f ′(wk−1)

∣∣∣ < α for all 1 ≤ k ≤ n.

That such a polygonal path can always be found requires a little bit of care.Certainly we can always find a path satisfying (i) and (ii)—start by connectingz and z′ by a straight line segment. If [z, z′] does not pass through any poles off , we are done. If it does, perturb a point on the line segment by a minusculeamount to avoid the pole, and repeat. A sketch of this process is shown inFigure 16.1.1 (though it is by no means the only approach).

×z = w0

∗∗ ×z′ = w1

∗

becomes

×z = w0

∗∗ ×z′ = w2

∗×w1

becomes

×z = w0

∗∗ ×z′ = w3

∗×w1

×w2

Figure 16.1.1: Creating apolygonal path from z to z′

avoiding poles of f , poles sig-nified by ∗.

Once a polygonal path P satisfying (i) and (ii) is found, note that conditions(iii) and (iv) are essentially convergence conditions on f(wk) and f(wk−1) beingclose in the d-sense (for (iii)) and (iv) is about the limit quotient approachingthe derivative. Since both of those do converge, we can find small enough ballsalong P that they hold—we can in fact cover P with such balls.

Now since P is compact, we can moreover select a finite subcover of thoseballs, and then specifically pick points w0, w1, . . . , wn on P such that eachline segment [wk−1, wk] lies in one of those balls. The resulting collectionw0, w1, . . . , wn

satisfies all four conditions.

Lecture 17 Compactness in M(G), continued

17.1 Compactness in the space of meromorphic functions,finalised

Proof, continued. The goal, as mentioned, is to establish that d(f(z), f(z′)) iswell approximated by µ(f)(z)|z− z′|, so that µ(f) being uniformly bounded oncompact subsets gives us that f is uniformly Lipschitz.

Date: October 15th, 2019.

17.1 Compactness in the space of meromorphic functions, finalised 63

With the polygonal as set up above, let βk = (1 + |f(wk−1)|2)1/2(1 +|f(wk)|2)1/2, and study the aforementioned distance. By the triangle inequality,

d(f(z), f(z′)) ≤n∑k=1

d(f(wk−1), f(wk)) =n∑k=1

2|f(wk−1)− f(wk)|βk

by the definition of d (see page 57) since neither the polygonal path never touchesa pole in this case. By the triangle inequality again, adding and subtractingf ′(wk)(wk − wk−1),

d(f(z), f(z′)) ≤n∑k=1

2βk

∣∣∣f(wk−1)− f(wk)wk − wk−1

− f ′(wk)∣∣∣|wk − wk−1|

+n∑k=1

2βk|f ′(wk)||wk − wk−1|.

Recall from (i),

µ(f)(z) = 2|f ′(z)|1 + |f(z)|2

if z is not a pole of f , so 2|f ′(wk)| ≤M(1 + |f(wk)|2). Hence by this and (iv),

d(f(z), f(z′)) < 2αn∑k=1

|wk − wk−1|βk

+M

n∑k=1

1 + |f(wk)|2

βk|wk − wk−1|.

Notice how by construction βk ≥ 1, and so 1βk≤ 1, and by the triangle inequality∣∣∣1 + |f(wk)|2

βk

∣∣∣ ≤ ∣∣∣1 + |f(wk)|2

βk− 1∣∣∣+ 1,

so

d(f(z), f(z′)) < 2α · 2|z − z′|+M

n∑k=1

(α|wk − wk−1|+ 1 · |wk − wk−1|)

≤ (2α+ αM +M) · 2|z − z′|,

by applying (iii) and (ii). Since α > 0 is arbitrary, let α→ 0, whence

d(f(z), f(z′)) ≤ 2M |z − z′|,

so f is Lipschitz if we avoid poles.For the second case, suppose z′ is a pole of f but z is not, and take w ∈ K

not a pole of f . Then

d(f(z), f(z′)) = d(f(z),∞) ≤ d(f(z), f(w)) + d(f(w),∞)

since f(z′) =∞. From the first case d(f(z), f(w)) < 2M |z − w|, so

d(f(z), f(z′)) < 2M |z − w|+ d(f(w),∞).

Now let w → z′, in which case the first term goes to 2m|z− z′|, and the secondterm goes to 0 since f is continuous. Hence again

d(f(z), f(z′)) ≤ 2M |z − z′|.

64 THE RIEMANN MAPPING THEOREM

Finally consider the case where both z and z′ are poles of f . Then triviallyd(f(z), f(z′)) = d(∞,∞) = 0, which is of course bounded by 2M |z − z′|.

So in any case, d(f(z), f(z′)) ≤ 2M |z−z′|, F is uniformly Lipschitz, meaningF is equicontinuous on K. This in term implies F is equicontinuous on allof G by the usual compactness argument of Proposition 13.1.1, and so F isequicontinuous on G.

The main application of this is:

17.2 Riemann mapping theoremDefinition 17.2.1 (Conformal equivalence). A region G1 is said to be confor-mally equivalent to another region G2 if there exists an analytic f : G1 → G2such that f is one-to-one and onto (i.e., f(G1) = G2).

Remark 17.2.2. By Exercise 6.1, f being one-to-one and analytic implies f ′(z) 6=0 for all z ∈ G, which implies f is conformal, hence the term conformallyequivalent.

By Exercise 10.1, f−1 is also analytic.We have not proved, but it is also true that a conformal mapping must

be analytic. Together these three remarks mean that there exist a number ofequivalent definitions of conformal equivalence.

Theorem 17.2.3 (Riemann mapping theorem). Let G be a simply connectedregion and G 6= C. Let a ∈ G. Then there exists a unique analytic functionf : G→ C satisfying

(i) f(a) = 0, f ′(a) ∈ R, and f ′(a) > 0;

(ii) f is one-to-one; and

(iii) f(G) = D :=z ∈ C

∣∣ |z| < 1.

Remark 17.2.4. Parts (ii) and (iii) mean that G is conformally equivalent toD.

Note that C is not conformally equivalent to any bounded region. If it were,i.e., we had an analytic function f : C → G, G a bounded region, then beinganalytic in C, f is entire, so by Liouville’s theorem f must be constant, havinga bounded image.

Lecture 18 The Riemann Mapping Theorem

18.1 Proving the Riemann mapping theoremWe will start by proving uniqueness, which is the easier part by a good margin:

Proof of the Riemann mapping theorem, (uniqueness). Let f and g be two an-alytic functions both satisfying the properties of the theorem. Then sincef : G→ D and g : G→ D, f g−1 : D → D.

Date: October 17th, 2019.

18.1 Proving the Riemann mapping theorem 65

Since f(a) = g(a) = 0, we have in turn that f g−1(0) = f(a) = 0. Hencef g−1 is one-to-one, onto, and analytic (so it is an automorphism of D).Then by the Schwarz–Pick theorem, there exists c ∈ C with |c| = 1 so thatf g−1(z) = cz for every z ∈ D. Hence f(z) = cg(z), so 0 < f ′(a) < cg′(a),where g′(a), implying that c ∈ R and c > 0. The only positive real number with|c| = 1 is c = 1, so f = g.

For the existence part of the proof it suffices to prove the following lemma:

Lemma 18.1.1. Let G be a region and G 6= C. Suppose every non-vanishinganalytic function on G has an analytic square root. Let a ∈ G. Then thereexists an analytic function f : G→ C such that

(i) f(a) = 0, f ′(a) ∈ R, and f ′(a) > 0;

(ii) f is one-to-one; and

(iii) f(G) = D.

That this is sufficient is motivated by an old result of ours, namely Corol-lary 6.1.3, and is the only reason we need simple connectedness in the Riemannmapping theorem: An non-vanishing analytic function f : G → C on a simplyconnected region G can be written as f(z) = exp(g(z)) where g(z) is analytic.Hence we can define an analytic square root of f(z) as exp( 1

2g(z)).All by way of saying, if G is simply connected, then the existence of an

analytic square root in the lemma is automatic.

Proof of Lemma 18.1.1. Let

F =f ∈ H(G)

∣∣ f is one-to-one, f ′(a) ∈ R, f ′(a) > 0, and f(G) ⊂ D.

In other words, F is the family of functions satisfying the the conditions (i)–(iii)we want, with the exception that f(G) ⊂ D instead of f(G) = D.

Assume for now, and we will prove momentarily, that

(a) F 6= ∅, and

(b) F = F ∪

0.

Since f(G) ⊂ D for every f ∈ F , F is locally bounded. Montel’s theoremimplies F is normal, and hence F is compact.

Consider the function H(G)→ C defined by f 7→ |f ′(a)|. This is continuous,meaning that, since F is nonempty (since F is nonempty) and compact, thereexists a maximum, i.e., there exists some f ∈ F such that |f ′(a)| ≥ |g′(a)| forall g ∈ F . So in particular, f ′(a) ≥ g′(a) for all g ∈ F , since restricted to Fthese quantities are positive.

Hence f ∈ F since, per above, F = F ∪

0.

If we can then show that, for this particular choice of f , f(G) = D, we aredone. To this end, suppose not. That is, suppose there exists some w ∈ D suchthat w 6∈ f(G). Then

f(z)− w1− wf(z)

is analytic on G (since |w| < 1 and f(z) < 1 means the denominator nevervanishes) and it never vanishes on G (since by choice f(z) 6= w for z ∈ G).

66 THE RIEMANN MAPPING THEOREM

By hypothesis this function has an analytic square root, in other words thereexists an analytic function h : G→ C such that

h(z)2 = f(z)− w1− wf(z) .(18.1.1)

Note that T (ξ) = ξ−w1−wξ is a Möbius transformation mappingD toD (specifically

what we called ϕw(ξ) when discussing automorphisms of the unit disk, seepage 40) and f(G) ⊂ D, meaning that h(G) ⊂ D.

Define a new function g : G→ C by

g(z) = |h′(a)|h′(a)

h(z)− h(a)1− h(a)h(z)

.

We want to show that g ∈ F and that g′(a) > f ′(a), which would contradictthe choice of f as having maximal derivative.

First, g is one-to-one simply because f is one-to-one, so h is one-to-one sinceit is a composition of a Möbius transformation (which is one-to-one) and f , andf in turn is a composition of a Möbius transformation and h.

That g(a) = 0 is clear—just plug in z = a. Finally, for the derivative, wecompute

g′(a) = |h′(a)|h′(a)

h′(z)(1− h(a)h(z))− (h(z)− h(a))(−h(a))(1− h(a)h(z))2

∣∣∣∣z=a

= |h′(a)|h′(a)

h′(a)(1− |h(a)|2)(1− |h(a)|2)2 = |h′(a)|

1− |h(a)|2 .

By Equation (18.1.1),

|h(a)|2 = |f(a)− w||1− wf(a)| = |−w| = |w|

since f(a) = 0, and hence differentiating 2h(a)h′(a) = f ′(a)(1− |w|2), so

|h′(a)| =∣∣∣f ′(a)(1− |w|2)

2h(a)

∣∣∣ = f ′(a)(1− |w|2)2|h(a)|

since f ′(a) > 0 and 1− |w|2 > 0 because |w| < 1, and so

|h′(a)| = f ′(a)(1− |w|2)2√|w|

.

Putting this back into g′(a), this means

g′(a) = f ′(a)(1− |w|2)2√|w|

11− |w| = f ′(a)1 + |w|

2√|w|

.

Rewriting1 + |w|2√|w|

= 12√|w|

+√|w|2 ,

18.1 Proving the Riemann mapping theorem 67

we can view this as the arithmetic mean of 1√|w|

and√|w|, and so by the

inequality of arithmetic and geometric means, we have the bound

1 + |w|2√|w|≥√

1√|w|

√|w| = 1.

In particular, equality holds only if the terms we are averaging are equal, thatis 1√

|w|=√|w|, meaning that |w| = 1. But by assumption w ∈ D, so |w| < 1,

whence |w| 6= 1, giving us strict inequality. Hence g′(a) > f ′(a) as desired,contradicting the maximality of f ′(a) in F , which means our assumption ofD \ f(G) 6= ∅ is false, so f(G) = D.

Looking back, it remains to show (a) and (b) in order to finish our proof.First, for (a), let us show F 6= ∅. Since G 6= C, there exists some b ∈ C \G,

and so the function z−b is analytic and non-vanishing in G. Hence by hypothesisthere exists an analytic square root g : G→ C such that g(z)2 = z − b.

Since z − b is one-to-one, so is g, and hence by Exercise 6.1 g′(z) 6= 0 forall z ∈ G. By the Open mapping theorem there exists some r > 0 such thatg(G) ⊃ B(g(a), r).

g(G)

×g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)g(a)

rrrrrrrrrrrrrrrrr

×−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)−g(a)

rrrrrrrrrrrrrrrrr

Figure 18.1.1: The image ofg contains a ball by the Openmapping theorem. The re-flected ball does not meet theimage.

Now, in a bit of a leap, note how it is true that g(G) ∩ B(−g(a), r) = ∅.To see this, suppose by way of contradiction that this is not the case. That is,suppose there exists some z ∈ G such that r > |g(z) + g(a)| = |−g(z) − g(a)|.This implies −g(z) ∈ B(g(a), r), and so by since g(G) ⊃ B(g(a), r), there mustexist some w ∈ G so that g(w) = −g(z). But then g(w)2 = (−g(z))2, sow − b = z − b, from which we gather that w = z. Then g(w) = −g(w), sow− b = −(w− b), so w− b = 0, whence b = w ∈ G, which is a contradiction—bwas chosen specifically not to be in G.

Now take a Möbius transformation T such that T (C\B(−g(a), r)) = D. Thiscan be done since we can pick a Möbius transformation mapping the circle on theboundary of B(−g(a), r) to the unit circle, and it maps connected componentsto connected components, so we can pick, in particular, the one that maps g(a)to zero, i.e., T (g(a)) = 0.

Let g1 = T g : G → D. Then g1 is analytic, g1(a) = 0, and since g and Tare both one-to-one, so is g1, and therefore in particular g′1(a) 6= 0.

Choose a complex number c, |c| so that cg′(a) ∈ R and cg′(a) > 0, and definea new function g2(z) = cg1(z). Then g′2(a) > 0, it is one-to-one since g1 is, andg2(a) = 0 by construction. Moreover g2(G) ⊂ D, so g2 ∈ F , so F is nonempty.

This leaves (b), namely showing F = F ∪

0. So let

fn⊂ F and

suppose fn → f in H(G) (since H(G) is complete). Since fn(a) = 0, f(a) = 0,and since f ′n(a) > 0, we know f ′n(a)→ f ′(a) ≥ 0.

Let z1 ∈ G be arbitrary and set ξ = f(z1) and ξn = fn(z1). For z2 6= z1 ∈ G,let K = B(z2, r) ⊂ G such that z1 6∈ K. Then fn(z)− ξn never vanishes on Ksince fn is one-to-one, and fn(z)− ξn → f(z)− ξ.

By Corollary 14.1.4 of Hurwitz’s theorem, either f(z)− ξ = 0 for all z ∈ K,or f(z) − ξ 6= 0 for all z ∈ K. In the first case, f(z) = ξ for all z ∈ K, sof(z) = ξ for all z ∈ G. But f(a) = 0, so ξ = 0, so f = 0.

In the second case, f(z) 6= ξ for all z ∈ K, so f(z2) 6= f(z1), meaning that fis one-to-one since z1 6= z2 are arbitrary. Hence f ′(z) 6= 0 for all z ∈ G, whichtogether with f ′(a) ≥ 0 means f ′(a) > 0. Hence f ∈ F .

68 THE RIEMANN MAPPING THEOREM

Therefore, as claimed, F = F ∪

0, finishing our proof of Lemma 18.1.1

and hence the Riemann mapping theorem.

Exercise 18.1. Let G =z∣∣ Re(z) > 0

. Let f be an analytic function on G

such that Re(f(z)) > 0 for all z in G and f(a) = a for some a in G. Show that|f ′(a)| ≤ 1.

18.2 Entire functionsA long time ago (namely on page 15) we asked several questions about how thebehaviour of entire functions compares to the behaviour of polynomials. We areabout ready to answer one of them, namely: Does there exist an entire functionwith a prescribed set of zeros?

The answer is that it depends. If the set is finite, set a1, a2, . . . , an ∈ C,then of course there exists an entire function f such that f(ai) = 0 for everyi = 1, 2, . . . , n, and f(z) 6= 0 for z 6= ai, namely the polynomial

f(z) = (z − a1)(z − a2) · · · (z − an).

A much more interesting situation, therefore is when the set is infinite, sayan∞n=1 ⊂ C. Does there exists an entire function f such that f(ai) = 0 for

i = 1, 2, . . . , and f(z) 6= 0 for z 6∈an?

We need some restrictions onan. First,

an

cannot be a boundedset, for otherwise, being a bounded sequence, it must contain some convergentsubsequence, so the set has a limit point. In other words, if

anis bounded,

f has a limit point of zeros, and so (all the way back from Theorem 4.1.9), fmust be identically zero.

Second,ancannot have a point repeated infinitely many times. If it does,

then f would have a zero of infinite order, meaning that all its derivatives at thatpoint are zero. This makes the power series around that point identically zero,so f is identically zero on an open neighbourhood of a point, so f is identicallyzero (or, alternatively, the aforementioned zero power series must have infiniteradius of convergence since f is entire). That is to say, the order of any zero ofa nonzero analytic function must be finite.

An easy way to satisfy both of these conditions is to require

limn→∞

|an| =∞.

With this assumption, as it happens, the answer is the affirmative: there doesexist some entire function f with f(ai) = 0 for every i = 1, 2, . . . , and f(z) 6= 0for z 6∈

an—this is the Weierstrass factorisation theorem.

Of course the naïve attempt is

f(z) =∞∏n=1

(z − an),

mimicking our approach in the polynomial case. The issue, and what we willspend the near discussing, is the question of convergence, and unfortunatelyf(z) thus defined does not converge.

INFINITE PRODUCTS 69

Lecture 19 Infinite Products

19.1 Review of infinite productsAs hinted at toward the end of last lecture, we will soon have to deal with thematter of convergence of infinite products. To this end we need to settle whatthat means.

Definition 19.1.1 (Infinite product). Letzn⊂ C. If

z = limk→∞

k∏n=1

zn

exists, then we denote

z =∞∏n=1

zn

as the infinite product of zn.

Remark 19.1.2. Of course there are cases in which this is trivially zero. First,if zn = 0 for some n, then z = 0.

Similarly, if zn = a for all n is constant, then if |a| < 1,∞∏n=1

zn = 0.

Suppose, in view of this,zn⊂ C \

0and

∞∏n=1

zn = z 6= 0.

Then setting

Pk =k∏

n=1zn,

we get zk = PkPk−1

. Thus zk → zz = 1 as k → ∞, so for n large enough, zn 6= 0

(being close to 1), so we can define its logarithm

log zn = log|zn|+ i arg zn,

taking the branch −π < arg zn < π.

Proposition 19.1.3. Letzn⊂ C such that Re(zn) > 0 for all n.6 Then

z =∞∏n=1

zn 6= 0

converges if and only if∞∑n=1

log zn,

taking the branch −π < arg zn < π, converges.Date: October 22nd, 2019.

6Note how this is a minor assumption: just consider the tail, in general, and it must betrue if the product converges.

70 INFINITE PRODUCTS

Proof. The meat of the proposition is the forwards direction. Write z = reiθ,taking −π < θ ≤ π. Let

Pk =k∏

n=1zn.

Since Pk → z as k → ∞, for k large enough logθ Pk = log|Pk| + iθk, where bylogθ we mean the branch cut is θ − π < θk < θ + π, i.e., the branch opposite z.

Let

Sk =k∑

n=1log zn

with the branch cut −π < arg zn < π. Then

exp(Sk) =k∏

n=1zn = Pk.

Taking logarithms, Sk = logθ Pk + 2πimk. Since Pk → z as k →∞,

Sk − Sk−1 = logθ Pk − logθ Pk−1 + 2πi(mk −mk−1)= log|zk|+ i(θk − θk−1) + 2πi(mk −mk−1).

On the one hand, Sk − Sk−1 = log zk → log 1 = 0 as k →∞ since z → 1.On the other hand, log|zk| → log 1 = 0, and θk − θk−1 → 0 since Pk → z.

Hence mk − mk−1 → 0 as k → ∞, but both mk and mk−1 are integers, somk → m for some integer m.

Thus, as k →∞, Sk → logθ z + 2πim, i.e.,∞∑n=1

log zn

converges.Note how all of this is just to make sure the angle mk converges, making

sure it doesn’t vary by multiples of 2πi as k changes.The reverse direction is trivial: let

Sk =∞∑n=1

log zn.

Assume Sk → s as k →∞. Then

exp(Sk) =∞∏n=1

zn → exp(s) 6= 0.

Remark 19.1.4. Since we need the factors to eventually be close to 1, it issometimes more convenient to consider

∞∏n=1

(1 + zn).

The proposition then says that for Re(zn) > −1,∞∏n=1

(1 + zn) 6= 0

19.1 Review of infinite products 71

if and only if∞∑n=1

log(1 + zn)

converges.Notice two things. First, this means zn → 0 as n→∞, and second, log(1 +

z) ≈ z for z near zero. Hence:Exercise 19.1. Let Re(zn) > −1. Then

∞∑n=1

log(1 + zn)

converges absolutely if and only if∞∑n=1

zn

converges absolutely.

This raises the question of how to correctly define absolute convergence forinfinite products.

The first naïve attempt is to say that∞∏n=1

zn

converges absolutely if∞∏n=1|zn| <∞.

However this is not a terribly useful definition, because the latter product con-verging does not imply the former does. For instance, let zn = (−1)n. Then

∞∏n=1|(−1)n| = 1,

but∞∏n=1

(−1)n

does not converge—it oscillates between −1 and 1.A better choice, inspired by the above proposition, is:

Definition 19.1.5 (Absolute convergence). Letzn⊂ C with Re(zn) > 0

for all n. We say∞∏n=1

zn

converges absolutely if∞∑n=1

log zn

converges absolutely.

72 INFINITE PRODUCTS

With this definition, fortunately, absolute convergence does imply ordinaryconvergence.

Combining this with the observations from Exercise 19.1, we get

Corollary 19.1.6. Letzn⊂ C with Re(zn) > 0 for all n. Then

∞∏n=1

zn

converges absolutely if and only if∞∑n=1

(1− zn)

converges absolutely.

Since our method for translating information from infinite series to infiniteproducts is the exponential function, we naturally ask the following question:Let X be a metric space, and let fn : X → C, n = 1, 2, . . . . Suppose fn → funiformly on X. Do we have

exp(fn)→ exp(f)

uniformly on X? The answer, in general, is no—as it happens, though fn andf are ‘close’, the exponential can amplify these small differences a lot.

Counterexample 19.1.7. Let fn(x) = xn + 1

x for x ∈ (0, 1). As n → ∞,fn(x)→ 1

x uniformly on (0, 1).However exp(fn) does not converge uniformly to exp(f):

exp(fn(x))− exp(f(x)) = exp( 1x

)(exp(xn

)− 1).

Take x = 1n . Then

exp(fn(x))− exp(f(x)) = exp(n)(

exp( 1n2

)− 1)≈ exp(n) 1

n2 →∞

as n→∞. (To see this, write exp( 1n2 ) as its power series.) Hence this does not

converge uniformly (though it does converge pointwise). N

In looking for a sufficient condition to alleviate this issue, consider exp(fn)→exp(f) pointwise. This means, roughly,∣∣∣exp(fn

exp(f) − 1∣∣∣ < ε,

and so multiplying by exp(f),

|exp(fn)− exp(f)| < ε exp(f).

Hence if exp(f) is uniformly bounded, then exp(fn) → exp(f) does convergeuniformly. Formally:

19.1 Review of infinite products 73

Lemma 19.1.8. Let X be a metric space, and let f, fn : X → C, n = 1, 2, . . . .Suppose fn → f uniformly on X. Suppose there exists some a ∈ R such thatRe(f(x)) < a for all x ∈ X (so |exp(f(x))| = exp(Re(f(x))) is uniformlybounded). Then exp(fn)→ exp(f) uniformly on X.

Proof. Since ez → 1 as z → 0, for any ε > 0 there exists δ > 0 such that if|z| < δ, then |ez − 1| < εe−a.

Since fn → f uniformly on X, there exists N ∈ N such that for n > N ,|fn(x)−f(x)| < δ for all x ∈ X. Hence for n > N , |exp(fn(x)−f(x))−1| < εe−a,so ∣∣∣exp(fn(x))

exp(f(x)) − 1∣∣∣ < εe−a.

Multiplying both sides by exp(f(x)), this becomes

|exp(fn(x))− exp(f(x))| < εe−a|exp(f(x))| ≤ εe−aea = ε

for all x ∈ X since |exp(f(x))| = exp(Re(f(x))) ≤ ea for all x ∈ X.

Lemma 19.1.9. Let X be a compact metric space. Let gn : X → C, n =1, 2, . . . , be continuous. Suppose

∞∑n=1

gn(x)

converges absolutely and uniformly on X. Then

f(x) :=∞∏n=1

(1 + gn(x))

converges absolutely and uniformly on X.Moreover, there exists an N ∈ N such that f(x) = 0 if and only if fn(x) = −1

for some n with 1 ≤ n ≤ N .

Proof. The series∞∑n=1

gn(x)

converging uniformly means there exists some N ∈ N such that for n > N ,|gn(x)| < 1

2 for all x ∈ X. Hence Re(1+gn(x)) > 0 for all n > N , x ∈ X. Thus,for n > N , log(1 + gn(x)) is defined.

Looking at the power series of log(1 + z), we see that |log(1 + z)| ≤ 32 |z| for

all |z| ≤ 12 , so moreover |log(1 + gn(x))| ≤ 3

2 |gn(x)| for all n > N and x ∈ X.Hence

∞∑n=N+1

log(1 + gn(x)),

being bounded by32

∞∑n=N+1

gn(x),

74 WEIERSTRASS FACTORISATION THEOREM

converges absolutely and uniformly on X. Therefore

h(x) :=∞∑

n=N+1log(1 + gn(x)),

being the limit of a uniformly convergent sequence of continuous functions, iscontinuous. Since X is compact, h(x) is bounded on X, so Re(h(x)) < a forsome a ∈ R for all x ∈ X. By Lemma 19.1.8, taking exponentials,

exp(h(x)) =∞∏

n=N+1(1 + gn(x))

converges uniformly on X.Thus

f(x) = (1 + g1(x))(1 + g2(x)) . . . (1 + gn(x)) exp(h(x))

converges absolutely and uniformly on X since a finite number of factors doesn’taffect convergence.

Moreover, exp(h(x)) 6= 0 for all x ∈ X because of the exponential, hence iff(x) = 0, then fn(x) = −1 for some 1 ≤ n ≤ N since the zero must come fromthe finite part.

Translating this back to analytic functions on C, this becomes

Theorem 19.1.10. Let G ⊂ C be a region. Letfn⊂ H(G) such that no fn

is identically zero. Suppose∞∑n=1

(fn(z)− 1)

converges absolutely and uniformly on compact subsets of G. Then

f(z) =∞∏n=1

fn(z)

converges and f(z) ∈ H(G).Moreover, if z = a is a zero of f , then z = a is a zero of a finite number of

fn(z), and the multiplicity of the zero of f is the sum of the multiplicities of thezeros of fn.

Lecture 20 Weierstrass Factorisation Theorem

20.1 Elementary factorsDefinition 20.1.1 (Elementary factors). An elementary factor is one of thefollowing functions En(z), n = 0, 1, 2, . . . ,

E0(z) = 1− z,

En(z) = (1− z) exp(z + z2

2 + · · ·+ zn

n

), for n ≥ 1.

Date: October 24th, 2019.

20.1 Elementary factors 75

Remark 20.1.2. These functions En(z) have simple zeros at z = 1, so En( za ) hasa simple zero at z = a, and no other zeros.

Our goal is to construct an entire function with a prescribed set of zerosan, so naturally we will want to consider something like

∞∏i=1

Eni

( zai

).

By Corollary 19.1.6 from last lecture, such a quantity converges if and only if

∞∑i=1

(1− Eni

( zai

))

converges. Hence we want to estimate:

Lemma 20.1.3. For |z| ≤ 1, |1− En(z)| ≤ |z|n+1 for all n ≥ 0.

Proof. For n = 0, 1− E0(z) = z, so the result holds.Let n ≥ 1. Since En(z) is entire, it has a power series expansion at z = 0,

say

En(z) = 1 +∞∑n=1

anzn,

and so

E′n(z) =∞∑n=1

kakzk−1.

On the other hand, from the definition of En(z), the product rule gives us

E′n(z) = −zn exp(z + z2

2 + · · ·+ zn

n

),

and since

exp(z+ z2

2 + · · ·+ zn

n

)= 1 +

(z+ z2

2 + · · ·+ zn

n

)+(z+ z2

2 + · · ·+ zn

n

)2+ . . . ,

the power series expansion of E′n(z) starts at zn (i.e., a1 = a2 = · · · = an = 0)and all the coefficients ak ≤ 0. Therefore |ak| = −ak, and so

0 = En(1) = 1 +∞∑k=1

ak,

implying that∞∑k=1|ak| = 1.

Hence for |z| ≤ 1, combining the facts that a1 = a2 = · · · = an = 0 and that

76 WEIERSTRASS FACTORISATION THEOREM

the sum of the magnitude of the coefficients is 1, we get

|1− En(z)| =∣∣∣ ∞∑k=1

akzk∣∣∣ ≤ ∞∑

k=1|ak||z|k =

∞∑k=n+1

|ak||z|k

= |z|n+1∞∑

k=n+1|ak||z|k−(n+1)

≤ |z|n+1∞∑

k=n+1|ak| = |z|n+1.

Recalling the discussion on page 68 and having the above in mind, we get

Theorem 20.1.4. Let |an| ⊂ C such that limn→∞

|an| = ∞, and an 6= 0 for alln ≥ 1. Let

pn⊂ N ∪

0such that∞∑n=1

( r

|an|

)pn+1<∞(20.1.1)

for all r > 0. Then

f(z) =∞∏n=1

Epn

( zan

)converges in H(C), i.e., f is entire. Moreover f has zeros exactly at an, n =1, 2, . . . , and if z = a occurs in

anexactly m times, then f has a zero at

z = a of multiplicity m.

Remark 20.1.5. Note that the condition in (20.1.1) always holds for pn ≥ n− 1.This is not hard to see. For any fixed r > 0, there exists an N ∈ N such thatfor n > N , r

|an| <12 (since |an| → ∞), and therefore the tail∑

n>N

( r

|an|

)pn+1<∑n>N

(12

)n<∞,

converges, and the finite part does not affect the convergence.Remark 20.1.6. Since |an| → ∞ as n→∞, no point in

ancan be repeated

infinitely many times.

Proof. Supposepn⊂ N∪

0satisfies (20.1.1). By Lemma 20.1.3, for |z| ≤ r

and n ≥ N large enough such that |an| ≥ r,∣∣∣1− Epn( zan)∣∣∣ ≤ ∣∣∣ z

an

∣∣∣pn+1≤∣∣∣ ran

∣∣∣pn+1,

then ∑n≥N

∣∣∣1− Epn( zan)∣∣∣ ≤ ∑

n≥N

∣∣∣ ran

∣∣∣pn+1<∞

by assumption.By Theorem 19.1.10, this means

f(z) =∞∏n=1

Epn

( zan

)

20.1 Elementary factors 77

converges uniformly on B(0, r), with r arbitrary, so f(z) converges uniformlyon any compact subset of C, whence f ∈ H(C). Moreover the same Theo-rem 19.1.10 means f(z) has zeros exactly at z = an, n = 1, 2, . . . .

Theorem 20.1.7 (Weierstrass factorisation theorem). Let f be an entire func-tion and let

anbe the nonzero zeros of f (repeated according to multiplicity).

Suppose f(z) has a zero of order m at z = 0. Then there exists an entirefunction g(z) and a sequence

pn⊂ N ∪

0such that

f(z) = zm∞∏n=1

Epn

( zan

)exp(g(z)).

Remark 20.1.8. Per the above remark we can take pn = n (or pn = n−1). Thenthe theorem says

f(z) = zm∞∏n=1

En

( zan

)exp(g(z)).

Proof. By Theorem 20.1.4, the function

h(z) = zm∞∏n=1

Epn

( zan

)is entire and has exactly the same zeros as f , including multiplicities. Thismeans that f(z)

h(z) has only removable singularities, specifically at z = 0 (ifm ≥ 1),z = a1, a2, . . . . Hence there exists an entire function k(z) such that k(z) = f(z)

h(z)for all z 6= 0, a1, a2, . . . , and k(z) 6= 0 for all z ∈ C. Since C is simply connected,we can write k(z) = exp(g(z)) for some g(z) ∈ H(C) (this is Corollary 6.1.3).Consequently

f(z) = k(z)h(z) = exp(g(z))zm∞∏n=1

Epn

( zan

).

Originally this argument only tells us these are equal for z 6= 0, a1, a2, . . . , buttwo analytic functions being equal on a dense set must be equal everywhere(this is Corollary 4.1.10, for the record).

Exercise 20.1. Show that

sin(πz) = πz

∞∏n=1

(1− z2

n2

)for every z ∈ C.

Notice how we specifically group the negative and positive zeros, writing∞∏n=1

(1− z2

n2

)instead of ∏

n∈Z

(1− z

n

).

There is good reason for this—think about the convergence of these two.

78 WEIERSTRASS FACTORISATION THEOREM

Note that this analysis is for functions analytic on the entire complex planeC. A natural question to ask is whether we can have the same factorisation forf ∈ H(G), where G 6= C is a region.

The answer to this question is yes, but the proof is more technical. Themain reason for this is that our key estimate, namely Lemma 20.1.3, no longerholds in this case—it relies on | zan | ≤ 1 for n large enough, which we can nolonger guarantee since we no longer know that |an| → ∞ as n→∞.

All we know in this case is that for f ∈ H(G) with zerosan, the set |an|

does not have a limit point in G.The correct elementary factor to consider now is not En( za ), for the reason

just discussed, but instead En(a−bz−b ) for b ∈ C \G, which then has a simple zeroat z = a, no other zeros in G, and is analytic in G—all properties we want.

Theorem 20.1.9. Let G 6= C be a region and letan⊂ C be a sequence

of distinct points with no limit point in G. Letmn

⊂ N. Then there ex-

ists a function f ∈ H(G) whose zeros are exactly z = an, n = 1, 2, . . . , withmultiplicities mn, respectively.

Sketch of proof. Letzn

be the sequence of an repeated with multiplicitiesmn. Take a sequence

wn⊂ C\G. Then En( zn−wnz−wn ) is analytic in G and has

a simple zero at z = zn and no other zeros. The goal then is to show that

f(z) =∞∏n=1

En

(zn − wnz − wn

)∈ H(G).

The technical part of showing this depends on the choice of wn, namely choosingwn such that lim

n→∞|zn − wn| = 0. We can do this since

znhas a limit point

in C \G, so on ∂G. The idea, then, is to use this limit in order to bound

1− En(zn − wnz − wn

).

Exercise 20.2. Let G be a region. Let f, g : G→ C be analytic functions. Showthat there exist analytic functions f1, g1, and h1 onG such that f(z) = h(z)f1(z)and g(z) = h(z)g1(z) for all z ∈ G; and f1 and g1 have no common zeros.

Exercise 20.3. (a) Let 0 < |a| < 1 and |z| ≤ r < 1. Show that∣∣∣ a+ |a|z(1− az)a

∣∣∣ ≤ 1 + r

1− r .

(b) Letan⊂ C with 0 < |an| < 1 and

∞∑n=1

(1− |an|) <∞. Show that

B(z) =∞∏n=1

|an|an

( an − z1− anz

)converges in H(B(0; 1)) and that |B(z)| ≤ 1. What are the zeros of B(z)?

(c) Find a sequenceanin B(0; 1) such that

∞∑n=1

(1− |an|) < ∞ and every

number eiθ is a limit point ofan.

RANK AND GENUS 79

Remark 20.1.10. B(z) is called a Blaschke product.

Corollary 20.1.11. Let f ∈ M(G). Then there exists function g, h ∈ H(G)such that f(z) = g(z)

h(z) . In other words, viewed as algebraic structures, M(G) isthe quotient field of the integral domain H(G) (this is an integral domain sinceit has no zero divisors, see Exercise 4.2).

Proof. Letanbe the poles of f with orders mn. Then by Theorem 20.1.9

there exists h ∈ H(G) such that h has zeros exactly at z = an with multiplicitymn. Then f(z)h(z) has only removable singularities. Hence there exists afunction g ∈ H(G) such that f(z)h(z) = g(z) for every z 6∈

an, meaning

that f(z) = g(z)h(z) .

Lecture 21 Rank and Genus

21.1 Jensen’s formulaSuppose f(z) is an entire function. By Weierstrass factorisation theorem we canwrite

f(z) = zm exp(g(z))∞∏n=1

Epn

( zan

).

The choice of pn depends on how fast |an| → ∞, because we need

∞∑n=1

( r

|an|

)pn+1<∞.

So if |an| → ∞ very fast, we can pick pn quite small. That is to say, we wantto control the growth rate of zeros of b, and we will show that the growth rateof zeros of f in B(0, r), as r →∞, is related to the growth rate of

M(r) = sup0≤θ≤2π

|f(reiθ)|.

The setup for this is not too complicated: let f be an analytic function in aneighbourhood of B(0, r), and suppose f does not vanish on B(0, r). Then wecan define the logarithm of f(z), and it is analytic on B(0, r) (that is to say,there exists an analytic function g(z) such that f(z) = eg(z)).

The mean value property of analytic functions, this means

log f(0) = 12π

∫ 2π

0log f(reiθ) dθ.

Taking real parts, this becomes

log|f(0)| = 12π

∫ 2π

0log|f(reiθ)| dθ,

if f is non-vanishing. The crucial insight here is that the left-hand side is fixed,and the integrand in the right-hand side is bounded by M(r).

Date: October 29th, 2019.

80 RANK AND GENUS

If f does have zeros, one’s first thought is to divide them away and study anew, non-vanishing function. This is essentially the right idea, except dividingby z − an directly does not give us any control of the size of those factors. Toremedy this, we instead divide by carefully chosen Möbius transformations, thegrowth of which we have very good control over. This gives rise to:

Theorem 21.1.1 (Jensen’s formula). Let f(z) be an analytic function on aneighbourhood of B(0, r). Let a1, a2, . . . , an be the zeros of f in B(0, r), repeatedaccording to multiplicity.7 Suppose f(0) 6= 0 and f(z) 6= 0 for all |z| = r. Then

log|f(0)| = −n∑k=1

log( r

|an|

)+ 1

2π

∫ 2π

0log|f(reiθ)| dθ.

Proof. Recall how for |α| < 1, the Möbius transformation

ϕα(z) = z − α1− αz

maps the unit disk D to D, and the unit circle ∂D to ∂D, and moreover ϕα(α) =0. We want to use this on B(0, r), so consider the diagram

D D

B(0, r).

ϕα

Ψa

The natural way to move from B(0, r) to D is to divide by r, so we define

Ψa(z) := ϕα

(zr

)=

zr − α

1− α zr

= z − αrr − αz

= r(z − a)r2 − az

,

where we have relabelled αr = a, so that a ∈ B(0, r). Then Ψa(z) : B(0, r)→ D,and it maps ∂B(0, r) to ∂D. This is useful, because it means the magnitude ofΨa(z) for |z| = r is exactly 1.

Now define

F (z) = f(z)n∏k=1

r2 − akzr(z − ak) .

This has removable singularities at z = ak, so just let F (z) represent the analyticfunction on B(0, r) we get by removing those singularities. Moreover F (z) 6= 0in B(0, r), by construction, and therefore

log|F (0)| = 12π

∫ 2π

0log|F (reiθ)| dθ.

7There are only finitely many zeros on B(0, r), since otherwise they we would have aninfinite bounded sequence of zeros, so they would have a limit point, forcing f to be identicallyzero.

21.2 The genus and rank of entire functions 81

Replacing F (z) per above, this yields

log∣∣∣f(0)

n∏k=1−( rak

)∣∣∣ = 12π

∫ 2π

0log∣∣∣f(reiθ)

n∏k=1

Ψak(reiθ)∣∣∣ dθ.

By the discussion above, |Ψak(reiθ)| = 1; the product on the left-hand sidebecomes a sum when taken out of the logarithm, so

log|f(0)|+n∑k=1

log∣∣∣ rak

∣∣∣ = 12π

∫ 2π

0log|f(reiθ)| dθ.

Moving the sum to the other side completes the proof.

Exercise 21.1. In the hypothesis of Jensen’s formula, do not suppose that f(0) 6=0. Show that if f has a zero at z = 0 of multiplicity m, then

log∣∣∣f (m)(0)

m!

∣∣∣+m log r = −n∑k=1

log( r

|ak|

)+ 1

2π

∫ 2π

0log|f(reiθ)| dθ.

Exercise 21.2. Let f be an entire function with f(0) 6= 0, and let

M(r) = sup0≤θ≤2π

|f(reiθ)|.

Let n(r) denote the number of zeros of f in B(0, r), counted according to mul-tiplicities. Suppose f(0) = 1 (else normalise, since f(0) 6= 0). Then

n(r) log 2 ≤ logM(2r).

21.2 The genus and rank of entire functionsSuppose f is entire. Again by Weierstrass factorisation theorem, we can write

f(z) = zm exp(g(z))∞∏n=1

Epn

( zan

),

where g(z) is an entire function.Let us consider, in some sense, the simplest form this can take (without

becoming trivial). The first nontrivial entire functions are polynomials, so weare interested in the case where g(z) is a polynomial.

Similarly, the simplest form pn can take is constant—so pn = p for all n.This is equivalent to

∞∑n=1

1|an|p+1 <∞,

since in the original condition of Weierstrass factorisation theorem we have∞∑n=1

( r

|an|

)pn+1<∞,

but with pn = p a constant, rp+1 is a constant, so we can take that out of thesum.

82 RANK AND GENUS

Definition 21.2.1 (Rank). Let f ∈ H(C) with nonzero zerosa1, a2, . . .

,

repeated according to multiplicities, and arranged such that 0 < |a1| ≤ |a2| <. . . . We say that f is of finite rank if there exists p ∈ N ∪

0such that

∞∑n=1

1|an|p+1 <∞.

If p is the smallest nonnegative integer such that this condition holds, then wesay that f is of rank p.

Remark 21.2.2. (i) If f has only finitely many zeros, then the sum abovealways converges, and in particular converges for p = 0, so we say that fhas rank 0.

(ii) We say that f is of infinite rank if it is not of finite rank. In other words,it is not of rank p for any p.

Remark 21.2.3. Suppose f is of finite rank p. Then

f(z) = zm exp(g(z))∞∏n=1

Ep

( zan

).

Here m is the order of vanishing of f at z = 0, so m is unique. Similarly, the Epterms are now uniquely determined, since we have chosen p as small as possible.Finally, exp(g(z)) is also unique, meaning that the entire expression is uniqueexcept g(z) may be replaced by g(z) + 2πik for k ∈ Z.

Finally, we call

P (z) =∞∏n=1

Ep

( zan

)the standard form of f .

Definition 21.2.4 (Genus). An entire function f has finite genus if f hasfinite rank and g(z) is a polynomial, where f(z) = zm exp(g(z))P (z) and P (z)is the standard form of f .

Let p be the rank of f and q be the degree of g(z). Then µ := maxp, q

is

called the genus of f .

Looking at this definition, this tells us that the genus of an entire functioncontrols the factorisation. As it turns out, the genus also controls the growth ofthe function:

Theorem 21.2.5. Let f be an entire function of finite genus µ. Then for anyε > 0,

f(z)ε,f exp(|z|µ+1+ε).

In particular, we have for some M > 0

log|Eµ(z)| ≤M |z|µ+1

as well aslog|Eµ(z)| ≤M |z|µ

for all z ∈ C.

21.2 The genus and rank of entire functions 83

Remark 21.2.6. By the notation f g we mean that there exists some c > 0such that |f(z)| ≤ c|g(z)| for all z ∈ C, known as Vinogradov notation.

Note that, in the case of smooth f and g, this means we only need to consider|z| large, since on bounded sets f and g are both bounded, so we can alwaysfind a c sufficiently large.Remark 21.2.7. Which one of the bounds on log|Eµ(z)| from the theorem onewants to use depends on |z|—if |z| < 1 we would prefer a bigger power, sincethat results a smaller bound, and conversely for |z| ≥ 1 we would want a smallerpower.

Proof. Since f(z) is of finite genus µ,

f(z) = zm exp(g(z))∞∏n=1

Eµ

( zan

),

where g(z) is a polynomial of degree at most µ. Taking logarithms, for z 6∈an∪

0,

log|f(z)| = m log|z|+ Re g(z) +∞∑n=1

log∣∣∣Eµ( z

an

)∣∣∣.We proceed to bound these three parts one at a time.

First, log|z| ≤M1|z|ε for any ε > 0, so definitely log|z| ≤M1|z|µ+1 for someM1 > 0 and |z| large.

Second, |Re g(z)| ≤ |g(z)| ≤ M2|z|µ since g(z) is a polynomial of degree atmost µ, so in addition |Re g(z)| ≤M2|z|µ+1 for some M2 > 0 and |z| large.

Third, and most important, we study log|Eµ(z)| as set out in the statementof the theorem. First, let |z| < 1

2 . Here since |z| is small, we have the powerseries expansion

log(1− z) = −(z + z2

2 + · · ·+ zµ

µ+ . . .

),

and so sinceEµ(z) = (1− z) exp

(z + z2

2 + · · ·+ zµ

µ

)we get

log|Eµ(z)| = log|1− z|+ Re(z + z2

2 + · · ·+ zµ

µ

)= −Re

( zµ+1

µ+ 1 + zµ+2

µ+ 2 + . . .)

= |z|µ+1(1 + |z|+ |z|2 + . . . )

where we have bounded the real part by the modulus, and noted that thedenominators in the middle step are all bigger than 1. But now |z| < 1

2 , sothis is bounded by a geometric sum, namely

log|Eµ(z)| ≤ |z|µ+1(

1 + 12 + 1

22 + . . .)

= 2|z|µ+1.

84 ORDER

Since |z| < 12 < 1, making the power smaller enlarges the bound, so this also

says log|Eµ(z)| ≤ 2|z|µ.On the other hand, for |z| > 2, we have

log|Eµ(z)| ≤ log|1− z|+ |z|+ |z|2

2 + · · ·+ |z|µ

µ≤ C|z|µ

for some constant C > 0 since, |z| > 2 > 1 is large, the last term dominates.This time, |z| being large, means enlarging the power enlarges the bound, so wealso have log|Eµ(z)| ≤ C|z|µ+1.

Finally, consider 12 ≤ |z| ≤ 2. Here, notice how log|Eµ(z)| is a continuous

function, except at z = 1, and there

limz→1|Eµ(z)| = −∞,

meaning that log|Eµ(z)| is bounded above, say log|Eµ(z)| ≤ D for some D > 0.Hence since |z| lives in a bounded range, we have

log|Eµ(z)| ≤ Da|z|a

for any power a, so in particular for the powers µ and µ+ 1.Now

∞∑n=1

log∣∣∣Eµ( z

an

)∣∣∣ ≤M3

∞∑n=1

( z

|an|

)µ+1= M3|z|µ+1

∞∑n=1

1|an|µ+1 ,

where the final sum converges since f is of genus µ. Hence

∞∑n=1

log∣∣∣Eµ( z

an

)∣∣∣ ≤M4|z|µ+1

for some M4 > 0 and |z| large.Putting the three pieces together, this means

log|f(z)| ≤ K|z|µ+1

for some K > 0 and |z| large, so

f(z) exp(|z|µ+1+ε),

where the ε > 0 comes from absorbing K into the exponential.

Lecture 22 Order

22.1 The order of entire functionsWe showed last time that if f is an entire function of finite genus µ, thenf(z) exp(|z|µ+1+ε) for any ε > 0.

Date: October 31st, 2019.

22.2 Hadamard’s factorisation theorem 85

Definition 22.1.1 (Order). An entire function f is of finite order if thereexists a ≥ 0 such that f(z) exp(|z|a).

If f is not of finite order, then we say f is of infinite order .If f is of finite order, then λ := inf

a∣∣f(z) exp(|z|a)

is called the order

of f .

That is to say, if f is of order λ, then for any ε > 0 we have f(z) exp(|z|λ+ε).

An equivalent way of saying this is that f being of order λ means thatlog|f(z)| ≤ Cε|z|λ+ε for |z| large. This is slightly delicate: when taking loga-rithms of f(z) exp(|z|λ+ε) we cannot keep the sign, for f(z) could be verylarge in the negative direction, making |f(z)| large, reversing the inequality.Exercise 22.1. Look back at Exercise 21.2 and use it to prove that if f is offinite order λ, then n(r)ε r

λ+ε.

Exercise 22.2. Let f1 and f2 be entire functions of finite order λ1 and λ2 re-spectively. Let f = f1 + f2.

(a) Show that f has finite order λ ≤ maxλ1, λ2

.

(b) Show that λ = maxλ1, λ2

if λ1 6= λ2.

(c) Give an example where λ < maxλ1, λ2

with f 6= 0.

Exercise 22.3. Let f1 and f2 be entire functions of finite order λ1 and λ2 re-spectively. Let f = f1f2. Show that f has finite order λ ≤ max

λ1, λ2

.

By last lecture’s discussion we then also have that finite genus implies finiteorder:

Corollary 22.1.2. If f is entire of finite genus µ, then f is of finite orderλ ≤ µ+ 1.

This means that the genus of an entire function, which by definition controlsthe factorisation, in fact also controls the growth of the function. The converseis also true.

22.2 Hadamard’s factorisation theoremProposition 22.2.1. Let f be an entire function of order λ which does notvanish in C. Then f(z) = exp(g(z)) where g(z) is a polynomial of degree deg g ≤λ. Thus f is of integral order deg g.

Proof. By the Weierstrass factorisation theorem, f(z) = exp(g(z)) with g(z)being entire (since f has no zeros, the other two parts of the Weierstrass fac-torisation disappear). Hence let

g(z) =∞∑n=0

anzn

be the power series expansion of g at z = 0. In order to show that g(z) is apolynomial of degree at most λ we therefore need to show that an = 0 for alln > λ.

Since f is of order λ, for any ε > 0 there exists some Cε > 0 such thatRe g(z) = log|f(z)| ≤ Cε|z|λ+ε for |z| large.

86 ORDER

Remark 22.2.2. It is instructive, at this point, to note that this is not quite aseasy as invoking Exercise 4.1, for the order only gives us a bound on log|f(z)| =Re g(z) ≤ Cε|z|λ+ε, not |g(z)|.

Writing C 3 an = αn + iβn, where αn, βn ∈ R, and setting z = Re2πiθ, wehave from the power series

Re g(z) =∞∑n=0

αnRn cos(2πnθ)−

∞∑n=1

βnRn sin(2πnθ).

Note that the second sum starts at n = 1 since sin(0) = 0.Now we can use the orthogonality of sine and cosine to extract αn and βn.

In particular, multiplying by cos(2πnθ) and integrating from 0 to 1, we extract

α0 =∫ 1

0Re g(Re2πiθ) dθ

andαnR

n = 2∫ 1

0Re g(Re2πiθ) cos(2πnθ) dθ

for n ≥ 1, and multiplying by sin(2πnθ) and integrating from 0 to 1 produces

βnRn = 2

∫ 1

0Re g(Re2πiθ) sin(2πnθ) dθ

for n ≥ 1. Since we want to show that an = 0 for n > λ, we need to show thatαn = βn = 0 for n > λ.

Since |cos(2πnθ)| ≤ 1, we have

|αnRn| ≤ 2∫ 1

0|Re g(Re2πiθ)| dθ.

As per the remark, we couldn’t use the bound from the order directly, since theinside of the absolute value might be large in the negative direction. The trickhere is to add and subtract Re g(Re2πiθ) in the integrand, resulting in

|αnRn| ≤ 2∫ 1

0|Re g(Re2πiθ)|+ Re g(Re2πiθ) dθ − 2

∫ 1

0Re g(Re2πiθ) dθ.

The second integral is nothing but α0 = a0, and the first integrand is 0 if thereal part is negative, and else just twice times itself, so

|αnRn| ≤ 2∫ 1

0max

0, 2CεRλ+ε dθ − 2a0 ≤ 4CεRλ+ε − 2a0

since Re g(Re2πiθ) ≤ CεRλ+ε because f is of order λ.Hence

|αn| ≤ 4CεRλ−n+ε − 2a0

Rn,

which goes to 0 as R→∞ if n > λ. Hence αn = 0 for n > λ.In precisely the same way, we see that βn = 0 for n > λ, whence an =

αn+βn = 0 for n > λ. Hence g(z) is a polynomial of degree at most λ, meaningthat f(z) = exp(g(z)) is of integral order deg g.

22.2 Hadamard’s factorisation theorem 87

Theorem 22.2.3 (Hadamard’s factorisation theorem). Let f be an entire func-tion of finite order λ. Then f has finite genus µ ≤ λ.

In other words, the growth of an entire function also controls its factorisation.

Proof. Let p be the integer such that p ≤ λ < p+ 1, i.e., the integer part of λ.We claim that f has rank at most p.

To show this, letan

be the nonzero zeros of f , repeated according tomultiplicity, and arranged such that 0 < |a1| ≤ |a2| ≤ . . . . To show that f hasrank at most p we need to show that

∞∑n=1

1|an|p+1 <∞.

First, without loss of generality we may assume f(0) = 1, because if f has azero of order m at z = 0, then f(z)

zm does not vanish at z = 0, and, for z 6= 0,

log∣∣∣f(z)zm

∣∣∣ = log|f(z)| −m log|z| ≤ Cε|z|λ+ε,

since f is of order λ and log|z| |z|ε. Note that this estimate also holds for zclose to an, because then log|f(z)| is large negative, so the left-hand side is evensmaller. Hence f(z)

zm has the same order as f(z), and since it does not affect therank since it touches only the first part of the Weierstrass factorisation.

Let n(r) be the number of zeros of f in B(0, r) counted according to multi-plicity, and let

M(r) = max|z|=r|f(z)|.

Then, by Exercise 21.2,

n(r) ≤ logM(2r)log 2 .

Since f has order λ, for any ε > 0, logM(2r) rλ+ε, and consequently n(r)rλ+ε.

Now having arranged 0 < |a1| ≤ |a2| ≤ · · · ≤ |ak| ≤ . . . , we have for anyδ > 0 that n(|ak|+ δ) ≥ k. On the order hand, n(|ak|+ δ) ≤ C(|ak|+ δ)λ+ε forsome constant C > 0. Letting δ → 0, this implies k ≤ C|ak|λ+ε.

Therefore1

|ak|p+1 ≤ k− p+1λ+ε ,

and since p ≤ λ < p + 1 we can choose ε small enough so that this power p+1λ+ε

is bigger than 1. So for some ε′ > 0,

∞∑n=1

1|an|p+1 ≤

∞∑n=1

n−(1+ε′)

which is convergent. Hence f has finite rank at most p ≤ λ.

88 RANGE OF ANALYTIC FUNCTIONS

Lecture 23 Range of Analytic Functions

23.1 Proof of Hadamard’s factorisation theoremWe start by finishing the proof started last time:

Proof of Hadamard’s factorisation theorem, continued. By the Weierstrass fac-torisation theorem, we therefore have

f(z) = zm exp(g(z))∞∏n=1

Ep

( zan

),

where for ease of discussion we will call the product at the end P (z). We nextwish to show that g(z) is a polynomial of degree at most λ.

Consider the function f(z)zmP (z) . It is entire (or has only removable singular-

ities, so remove them) and never vanishes in C. Then, studying the orders ofthese terms,

log∣∣∣ f(z)zmP (z)

∣∣∣ = log|f(z)| −m log|z| − log|P (z)|

≤ Cε(|z|λ+ε + |z|ε + |z|p

)≤ Kε|z|λ+ε

for |z| large, since p ≤ λ and the terms in P (z) are of the form

Ep(z) = (1− z) exp(z + z2

2 + · · ·+ zp

p

),

at least for p ≥ 1 (for p = 0 it is just 1− z).In other words f(z)

zmP (z) is of finite order at most λ, and never vanishes, whichby Proposition 22.2.1 means

f(z)zmP (z) = exp(g(z))

where g(z) is a polynomial of degree at most λ, meaning that f(z) is of finitegenus at most λ, finishing the proof.

As a small historic interlude, this theorem gives in particular a factorisationof the Riemann zeta function in terms of its zeros, a product representationfirst conjectured to exist by Riemann. It was in proving the existence of thisproduct representation of the zeta function that Hadamard happened to provethe existence for all entire functions of finite order.

Aside from genus and order, there is a third related, and sometimes pow-erful, notion. To see how this is occassionally more useful, see in particularExercise 23.2 (b).

Date: November 5th, 2019.

23.2 The range of entire functions 89

Definition 23.1.1 (Exponent of convergence). Letan

be a sequence ofnon-zero complex numbers. Let

ρ = infr

∣∣∣∣∣∑n

1|an|r

<∞

,

called the exponent of convergence ofan.

Exercise 23.1. (a) Let f be an entire function of rank p. Show that the expo-nent of convergence ρ of the non-zero zeros of f satisfies: p ≤ ρ ≤ p+ 1.

(b) Let f be an entire function of order λ and letan

be the non-zerozeros of f repeated according to multiplicity. Let ρ be the exponent ofconvergence of

an. Show that ρ ≤ λ.

(c) Let P (z) =∞∏n=1

Ep

( zan

)be a canonical product of rank p. Let ρ be the

exponent of convergence ofan. Show that the order of P (z) is ρ.

Exercise 23.2. (a) Let f and g be entire functions of finite order λ. Supposethat f(an) = g(an) for

ansuch that

∑n

|an|−(λ+1) = ∞. Show that

f = g.

(b) Replace the condition in (a) by∑n

|an|−(λ+ε) =∞ for some ε > 0. Show

that f = g.

(c) Find all entire functions f of finite order such that f(logn) = n for n =1, 2, . . . .

23.2 The range of entire functionsConsider a polynomial p(z), say of degree n, and let c ∈ C. Then the poly-nomial p(z) − c is also of degree n, and hence has precisely n roots (countingmultiplicity), according to the Fundamental theorem of algebra. This meansthat the range of a polynomial is all of C, and more precisely every point c ∈ Cis attained exactly n times.

We wish to investigate the same question, that of the range, for entire func-tions in general.

Theorem 23.2.1 (Special case of Picard’s theorem). Let f be a non-constantentire function of finite order. Then f(z) assumes each complex number withonly one possible exception.

Proof. Suppose there exist α, β ∈ C be distinct points such that f(z) 6= α andf(z) 6= β for all z ∈ C. Then f(z)− α is entire and never vanishes in C, so byProposition 22.2.1 we can write f(z)−α = exp(g(z)) where g(z) is a polynomial.That is to say, f(z) = exp(g(z)) + α.

Similarly, since f(z) 6= β for all z ∈ C we have exp(g(z)) + α 6= β for allz ∈ C, meaning that exp(g(z)) 6= β − α.

Hence g(z) 6= log(β−α) (note how β−α 6= 0). But g(z) is a polynomial, soits range is all of C, making this a contradiction.

90 RANGE OF ANALYTIC FUNCTIONS

This means that an entire function can miss only one point in C.The polynomial example brings up another natural question: if f is an entire

function, how many times can f(z) assume a ∈ C?As discussed, in the case of a degree n polynomial, the answer is precisely n

times. We answer the question for entire functions analogously:

Theorem 23.2.2. Let f be an entire function of finite order λ, where λ is notan integer. Then f has infinitely many zeros.

Note that the order not being integral implies f is non-constant, since(nonzero) constants are of order 0.

Proof. Suppose f has only finitely many zeros, saya1, a2, . . . , an

, repeated

according to multiplicity. Then we can write

f(z) = (z − a1)(z − a2) · · · (z − an) exp(g(z)),

with g entire. Consider

log∣∣∣ f(z)(z − a1) · · · (z − an)

∣∣∣ = log|f(z)| − log|z− a1| − · · · − log|z− an| ≤ Cε|z|λ+ε

for |z| large. Hence f(z)(z−a1)···(z−an) has order at most λ, so g(z) is a polynomial

of degree λ by Proposition 22.2.1. Hence the order of f is the degree of g, which,being a polynomial, is an integer. This is a contradiction.

It is worth noting that the key insight in the above proof is that scaling bypolynomials don’t affect the order of an entire function.

Corollary 23.2.3. Let f be an entire function of order λ 6∈ Z. Then f assumeseach complex number an infinite number of times (hence there are no exceptionalpoints).

Proof. Consider g(z) = f(z) − α, α ∈ C. The order of g is the same as theorder of f , i.e., λ 6∈ Z. Hence by Theorem 23.2.2, g(z) has infinitely many zeros,meaning that f(z) assumes α infinitely many times.

That is to say, order λ 6∈ Z guarantees the presence of the infinite productpart of the Weierstrass factorisation theorem.

The question of the range of analytic functions gets much more delicate ifwe move from all of C (so entire functions) to regions G 6= C.

23.3 The range of an analytic function

Let D =z∣∣ |z| < 1

, and suppose f : D → C is analytic, with f(0) = 0 and

f ′(0) = 1. (By the Riemann mapping theorem we know any region G 6= C isconformal to D, so if suffices to study D.)

We want to investigate how ‘big’ f(D) can be,

Lemma 23.3.1. Let f : D → C be analytic, f(0) = 0 and f ′(0) = 1. Suppose|f(z)| ≤M for every z ∈ D. Then M ≥ 1 and f(D) ⊃ B(0, 1

6M ).

23.3 The range of an analytic function 91

Proof. First we show that M ≥ 1. Consider the power series expansion of f(z)at z = 0,

f(z) = z + a2z2 + a3z

3 + . . . ,

since a0 = f(0) = 0 and a1 = f ′(0) = 1 by hypothesis. But 0 < r < 1, Cauchy’sintegral formula gives us |an| ≤ M

rn ≤M for all n by letting r → 1.Hence in particular a1 = f ′(0) = 1, so 1 ≤ M

r , or r ≤ M , for 0 < r < 1, soletting r → 1 we see that M ≥ 1.

Next let us show that f(D) ⊃ B(0, 16M ). For any w ∈ B(0, 1

6M ), considerthe function g(z) = f(z)−w. We wish to show that g(z) has a zero in D, sincethat corresponds to w ∈ f(D).

The idea is to apply Rouché’s theorem on the circle |z| = 14M . For |z| = 1

4M ,

|f(z)| ≥ |z| −∞∑n=2|an||z|n ≥

14m −

∞∑n=2

M( 1

4M

)nsince |an| ≤M for all n. The sum at the end is geometric, and works out to be

116M−4 , so

|f(z)| ≥ 14M − 1

16M − 4 ≥1

6M .

Hence on |z| = 14M , since w ∈ B(0, 1

6M ),

|f(z)− g(z)| = |w| < 16M ≤ |f(z)|.

Hence by Rouché’s theorem f(z) and g(z) have the same number of zeros inB(0, 1

4M ), and since f(0) = 0, we must consequently have g(z) = 0 for somez ∈ B(0, 1

4M ) ⊂ D, whence w ∈ f(D) as claimed.

Remark 23.3.2. It is useful to note that the appearance of the constant 6 seemsa bit out of the blue, and it sort of is.

As is often the case in situations like this, the way one originally discoversthe theorem is to consider some constant C in place of 6, prove the theorem insome generality, and then, once done, optimise the choice of C.

The appearance of 6 only looks vaguely magical because the exposition hereskips the hard work in finding it.

Heuristically, we can tell this lemma cannot be optimal: when M is large,i.e., we have a large bound on f(z), so f takes on many values between 0 andM , then B(0, 1

6M ) is tiny. This hints at there being a better result, and indeedwe will work on achieving it in the near future.

For reference we also work out what this theorem becomes on arbitrary ballscentred at z = 0:

Lemma 23.3.3. Suppose g is analytic on B(0, R), g(0) = 0, and |g′(0)| = µ > 0and |g(z)| ≤M for all z ∈ B(0, R). Then g(B(0, R)) ⊃ B(0, R

2µ2

6M ).

Proof. Consider the function f(z) = g(Rz)Rg′(0) : D → C. We verify that f(0) = 0,

f ′(0) = g′(Rz)RRg′(0)

∣∣∣∣z=0

= g′(0)RRg′(0) = 1,

92 BLOCH’S AND LANDAU’S CONSTANTS

and |f(z)| ≤ MRµ . Hence Lemma 23.3.1 we get

f(D) ⊃ B(

0, 16( MRµ )

)= B

(0, Rµ6M

),

sog(B(0, R)) ⊃ B

(0, R

2µ2

6M

).

We need one more lemma in order to prove the result we are really after.

Lemma 23.3.4. Let f be analytic on B(a, r) such that |f ′(z)− f ′(a)| < |f ′(a)|for all z ∈ B(a, r), z 6= a. Then f is one-to-one.

Proof. This is essentially the fundamental theorem of calculus. For z1, z2 ∈B(a, r), z1 6= z2, write

|f(z2)− f(z1)| −∣∣∣∫

[z1,z2]f ′(z) dz

∣∣∣,where by [z1, z2] we mean the line segment joining z1 and z2 (since the ballB(a, r) is convex, this line segment is in the ball). Adding and subtracting f ′(z)inside the integral, we get

|f(z2)− f(z1)| =∣∣∣∫

[z1,z2]f ′(z)− f ′(a) + f ′(a) dz

∣∣∣≥∣∣∣∫

[z1,z2]f ′(a) dz

∣∣∣− ∣∣∣∫[z1,z2]

f ′(z)− f ′(a) dz∣∣∣

> |f ′(a)||z2 − z1| −∣∣∣∫

[z1,z2]f ′(a) dz

∣∣∣= |f ′(a)||z2 − z1| − |f ′(a)||z2 − z1| = 0.

Hence f(z2) 6= f(z1), so f is one-to-one.

With these lemmata we are equipped to prove the main theorem we areafter, the proof of which is very technical and little lengthy:

Theorem 23.3.5 (Bloch’s theorem). Let f be an analytic function on a neigh-bourhood of D =

z∣∣ |z| ≤ 1

such that f(0) = 0 and f ′(0) = 1. Then there

exists a disk S ⊂ D such that f is one-to-one on S and f(S) contains a disk ofradius 1

72 .

Remark 23.3.6. Note how this constant 172 , quite remarkably, is uniform in f .

Lecture 24 Bloch’s and Landau’s Constants

24.1 Proof of Bloch’s theoremProof. The broad strategy is this: we want a new function with derivative aslarge as possible at some point a, so that we can apply Lemma 23.3.4, and

Date: November 7th, 2019.

24.1 Proof of Bloch’s theorem 93

moreover we want its derivative and the upper bound of the function to be ofsimilar size, so that in Lemma 23.3.3 we get a constant.

To this end, letK(r) = max

|z|=r|f ′(z)|

and let h(r) = (1 − r)K(r). Then h : [0, 1] → R is continuous, h(0) = 1 andh(1) = 0, by construction. Let r0 = sup

r∣∣ h(r) = 1

.

Notice three things: first, since h is continuous, h(r0) = 1. Second, 0 ≤r0 < 1 since h(1) = 0 6= 1, and third, for any r > r0, h(r) < 1 because of thesupremum.

Now take a ∈ D with |a| = r0 such that K(r0) = |f ′(a)|. This must bepossible since the supremum on |z| = r0, which is compact, must be attained.Then

1 = h(r0) = (1− r0)K(r0) = (1− r0)|f ′(a)|,

or in other words|f ′(a)| = 1

1− r0.

Our goal is to apply Lemma 23.3.4 near z = a, so consider a ball B(a, ρ0) whereρ0 := 1

2 (1− r0). Notice how this by construction means

|f ′(a)| = 12ρ0

.(24.1.1)

Then for |z − a| < ρ0, we have

|z| < |a|+ 12(1− r0) = r0 + 1

2(1− r0) = 12(1 + r0).

Notice how r0 <12 (1 + r0) since r0 < 1, and how therefore h( 1

2 (1 + r0)) < 1.Hence for |z − a| < ρ0,

|f ′(z)| ≤ K(1

2(1 + r0))

=h( 1

2 (1 + r0))1− 1

2 (1 + r0),

where the inequality in the first step is the Maximum modulus principle, andthe second step is the definition of K(r). Now since h( 1

2 (1 + r0)) < 1, this isbounded by

|f ′(z)| < 112 (1− r0)

= 1ρ0

for all |z − a| < ρ0.To apply Lemma 23.3.4 we want |f ′(z)− f ′(a)| < |f ′(a)|, so let us compute

the former: for |z − a| < ρ0,

|f ′(z)− f ′(a)| ≤ |f ′(z)|+ |f ′(a)| < 1ρ0

+ 12ρ0

= 32ρ0

.

Unfortunately this is clearly not less than |f ′(a)| = 12ρ0

—the disk is too large—so we need to shrink ρ0.

Toward this, consider

Ψ(z) = 2ρ0

3 (f ′(ρ0z + a)− f ′(a)),

94 BLOCH’S AND LANDAU’S CONSTANTS

so that Ψ: D → C. Now Ψ(0) = 0 and

|Ψ(z)| ≤ 2ρ0

33

2ρ0= 1

by the above calculations, so we can apply Schwarz lemma, which tells us that|Ψ(z)| ≤ |z| for all |z| ≤ 1. Consequently

|f ′(ρ0z + a)− f ′(a)| ≤ 32ρ0|z|,

and making the change of variables w = ρ0z+ a ∈ B(a, ρ0), so z = w−aρ0

, we get

|f ′(w)− f ′(a)| ≤ 3|w − a|2ρ2

0

for w ∈ B(a, ρ0). We want this to be less than |f ′(a)| = 12ρ0

, so

3|w − a|2ρ2

0<

12ρ0

means|w − a| < 1

3ρ0.

Therefore we take z ∈ S = B(a, 13ρ0) and there we get

|f ′(z)− f ′(a)| < 32ρ0

13ρ0 = 1

2ρ0= |f ′(a)|,

so we can apply Lemma 23.3.4, guaranteeing that f is one-to-one on S.It remains to show that f(S) contains a disk of radius 1

72 . To accomplishthis we want to apply Lemma 23.3.3, meaning that we need a function g withg(0) = 0, and we need information about its derivative at 0 and a bound for iton a ball.

Let g : B(0, 13ρ0)→ C be defined by g(z) = f(z+ a)− f(a), so that g(0) = 0

and |g′(0)| = |f ′(a)| = 12ρ0

.To get a bound on |g(z)| we need use the Fundamental theorem of calculus,

since this way we can leverage our knowledge of the derivative. Consider theline segment γ = [a, z + a] ⊂ S = B(a, 1

3ρ0) ⊂ B(a, ρ0), for which

|g(z)| =∣∣∣∫γ

g′(w) dw∣∣∣ ≤ ∫

γ

|g′(w)| dw.

By definition g′(w) = f ′(w+a), and we know that |f ′(z)| < 1ρ0

for all |z−a| < ρ0,and hence also |f ′(w + a)| < 1

ρ0, so that

|g(z)| ≤ 1ρ0|z| < 1

ρ0

13ρ0 = 1

3 .

Therefore by Lemma 23.3.3, g(B(0, 13ρ0)) ⊃ B(0, σ) with

σ =( 1

3ρ0)2( 12ρ0

)2

6 · 13

=19 ·

14

2 = 172 .

24.1 Proof of Bloch’s theorem 95

Shifting back to f , this means

f(S) ⊃ B(f(a), 1

72

),

finishing the proof.

We can readily translate this to other disks centred at z = 0:

Corollary 24.1.1. Let f be an analytic function on a neighbourhood of B(0, R).Then f(B(0, R)) contains a disk of radius 1

72R|f′(0)|.

Proof. If f ′(0) = 0, then the result is trivially true, so assume f ′(0) 6= 0.Consider the function

g(z) = f(Rz)− f(0)Rf ′(0) ,

where Rz serves to move us from B(0, R) to D, subtracting f(0) is to makeg(0) = 0, and dividing by Rf ′(0) makes g′(0) = 1.

Hence we can apply Bloch’s theorem on g, so that g(D) contains a disk ofradius 1

72 , and so f(B(0, R)) contains a disk of radius 172R|f

′(0)|.

The constant 172 in Bloch’s theorem is not best possible.

Definition 24.1.2 (Bloch’s constant). Let F be the set of all functions f thatare analytic on a neighbourhood of D =

z∣∣ |z| < 1

, with f(0) = 0 and

f ′(0) = 1.For each f ∈ F , let β(f) denote the supremum of all r such that there exists

a disk S ⊂ D where f is one-to-one on S and f(S) contains a disk of radius r.Bloch’s constant is the number B defined by

B = inff∈F

β(f)

.

Remark 24.1.3. Notice howBloch’s theorem implies that B ≥ 172 . On the other

hand, considering the function f(z) = z, we see that β(f) = 1, so B ≤ 1.The state of the art is

0.4332 ≈√

34 + 3× 10−4 ≤ B ≤

√√3− 12 ·

Γ( 13 )Γ( 11

12 )Γ( 1

4 )≈ 0.4719,

the lower bound due to Chen and Gauthier in [CG96] and then marginallyimproved by Xiong in [Xio98], and the upper bound is due to Ahlfors andGrunsky in [AG37].

It is further conjectured in the Ahlfors and Grunsky paper is the true valueof B.

We can ask a related, but slightly relaxed question:

Definition 24.1.4 (Landau’s constant). For each f ∈ F , define λ(f) to bethe supremum of all r such that f(D) contains a disk of radius r. Landau’sconstant L is defined by

L = inff∈F

λ(f)

.

96 BLOCH’S AND LANDAU’S CONSTANTS

Remark 24.1.5. Naturally λ(f) ≥ β(f) since λ drops the requirement of f beingone-to-one on the disk in question. Hence since λ(f) ≥ β(f) for all f ∈ F wemust have B ≥ L, and as before, taking f(z) = z, L ≥ 1.

The best known bounds for L are

0.5 < L ≤Γ( 1

3 )Γ( 56 )

Γ( 16 )

≈ 0.5433

Note that in particular this tells us that B < L, so the equality is ruled out.Notice how, being defined by an infimum, it is possible, in principle, no

function ever attains L exactly. As it happens, this is not actually the case:

Proposition 24.1.6. Let f be analytic on a neighbourhood of D with f(0) = 0and f ′(0) = 1. Then f(D) contains a disk of radius L.

Proof. We will show something slightly stronger, namely that f(D) contains adisk of radius λ = λ(f), and since λ(f) ≥ L, this implies the proposition.

By the definition of λ(f), in terms of a supremum, we have that for eachn ∈ N there exists some αn ∈ f(D) such that f(D) ⊃ B(αn, λ− 1

n ).Notice now how

αn⊂ f(D) ⊂ f(D). Crucially, D is compact, and f is

continuous, whence maps compact sets to compact sets, so f(D) is compact.Therefore

αnhas a limit point.

For instance, take a subsequenceαnk

such that αnk → α ∈ f(D). Then

we claim that f(D) ⊃ B(α, λ).This is an exercise in the triangle inequality: for w ∈ B(0, λ), choose M

large enough so that |w − a| < λ− 1M .

λλλλλλλλλλλλλλλλλ

λ− 1Mλ− 1Mλ− 1Mλ− 1Mλ− 1Mλ− 1Mλ− 1Mλ− 1Mλ− 1Mλ− 1M

λ− 1M

λ− 1M

λ− 1M

λ− 1M

λ− 1M

λ− 1Mλ− 1M

×α

×w×αnk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

λ− 1nk

Figure 24.1.1: The setup ofα, w, and αnk .

Since αnk → α, there must exist some N ∈ N large enough so that N > 2Mand for nk > N , |αnk − α| < 1

2M . Then

|w − αnk | ≤ |w − α|+ |α− αnk | < λ− 1M

+ 12M = λ− 1

2M < λ− 1nk

since 2M < N < nk. Hence w ∈ B(αnk , λ− 1nk

) ⊂ f(D), so B(α, λ) ⊂ f(D).

Again we can translate this to arbitrary disks centred on z = 0 (by exactlythe same method Corollary 24.1.1):

Corollary 24.1.7. Let f be analytic on a neighbourhood of B(0, R). Thenf(B(0, R)) contains a disk of radius |f ′(0)|RL.

Proof. As in Corollary 24.1.1, f ′(0) = 0 is trivial, so assume f ′(0) 6= 0 andconsider

g(z) = f(Rz)− f(0)Rf ′(0) ,

applying the Proposition 24.1.6.

24.2 The Little Picard theoremWe showed, as a consequence of Hadamard’s factorisation theorem, a Specialcase of Picard’s theorem, saying that a non-constant entire function of finiteorder can miss at most one point in C. Our next goal is to show that theassumption of finite order is not necessary.

To accomplish this we need the following small calculation:

TOWARD THE LITTLE PICARD THEOREM 97

Lemma 24.2.1. Let G be a simply connected region. Let f : G→ C be analytic.Suppose f does not assume the values 0 or 1. Then there exists an analyticfunction g : G→ C such that

f(z) = − exp(πi cosh(2g(z)))

for all z ∈ G.

It is worth recalling here that

cosh(z) := ez + e−z

2 .

Proof. The proof essentially boils down to solving the expression for f(z) abovefor g(z), then working backwards.

Since f does not vanish on G, which is simply connected, we can take log-arithms. By way of saying, there exists some analytic h : G → C such thatf(z) = exp(h(z)).

Let F (z) = 12πih(z), and notice how for any n ∈ Z, we have F (z) 6= n for

all z ∈ G. If not, h(z) = 2πin, implying that f(z) = exp(h(z)) = 1, which is acontradiction.

In particular, F (z) does not assume 0 or 1, so both F (z) and F (z)− 1 haveanalytic square roots, so we can define

H(z) =√F (z)−

√F (z)− 1,

analytic on G. Moreover H(z) 6= 0 for all z ∈ G, since otherwise we would have0 =

√F (z)−

√F (z)− 1, which rearranged and squared gives us 0 = 1.

So there exists, finally, an analytic g : G → C such that H(z) = exp(g(z)),and therefore, retracing our steps,

cosh(2g(z)) + 1 = e2g(z) + e−2g(z)

2 + 1 = (eg(z) + e−g(z))2

2

=(H(z) + 1

H(z) )2

2 = 2F (z) = 1πih(z).

Hence

f(z) = exp(h(z)) = exp(πi(cosh(2g(z)) + 1)) = − exp(πi cosh(2g(z)))

by pulling out the factor of exp(πi) = −1.

Lecture 25 Toward the Little Picard Theorem

25.1 Another lemmaAs a consequence of Lemma 24.2.1 we have the following related result:

Lemma 25.1.1. Let G be a simply connected region. Let f : G→ C be analytic.Suppose f does not assume the values 0 or 1. (In other words, G, f , and g asin Lemma 24.2.1.)

Then g(G) contains no disk of radius 1.Date: November 12th, 2019.

98 THE LITTLE PICARD THEOREM

Proof. First, let us show that g(z) does not assume any values of

A =± log(

√n+√n− 1) + 1

2 imπ∣∣∣ Z 3 n ≥ 1, m = 0,±1,±2, . . .

.

This is a simple computation. If g(z) = ± log(√n +√n− 1) + 1

2 imπ forsome z ∈ G, then

2 cosh(2g(z)) = e2g(z) + e−2g(z)

= eimπ(√n+√n− 1)±2 + e−imπ(

√n+√n− 1)∓2

= (−1)m((√n+√n− 1)2 + (

√n−√n− 1)2)

= (−1)m(2 · (2n− 1)).

Hence f(z) = − exp(πi(−1)m(2n− 1)) = 1, which contradicts Lemma 24.2.1.Now the points in A form the vertices of a grid of rectangles in the plane

with widthlog(√n+ 1 +

√n)− log(

√n+√n− 1) < 1

since it is a decreasing function in n and at n = 1 it is log(√

2 + 1) ≈ 0.88 < 1.The height of the rectangles are 1

2π, so the diameter of the rectangles isbounded by √

12 +(1

2π)2≈ 1.86 < 2,

so any disk of radius 1 or bigger (hence diameter 2 or bigger) must contain atleast one of the points in A, and hence f(G) cannot contain any such disk.

Lecture 26 The Little Picard Theorem

26.1 Proof of the Little Picard theoremWith these two lemmata in hand we are equipped to prove

Theorem 26.1.1 (Little Picard theorem). Let f be an entire function. Supposef misses two values in C. Then f is a constant function.

Proof. Suppose f(z) 6= a and f(z) 6= b for all z ∈ C, with a 6= b—that is, fmisses a and b. Then we can normalise to f(z) = f(z)−a

b−a which misses 0 and 1.Hence by Lemma 24.2.1 we can write

f(z) = − exp(π cosh(2g(z)))

for some entire function g, and by Lemma 25.1.1 we know that g(C) containsno disk of radius 1.

By way of contradiction, suppose f is not constant, meaning that g is notconstant. A non-constant entire function must have some nonzero derivative(else its power series, which converges everywhere, is zero), so there exists somez0 ∈ C such that g′(z0) 6= 0. For convenience we may assume z0 = 0, sog′(0) 6= 0 (else shift to g(z) = g(z + z0)).

Date: November 14th, 2019.

26.2 Schottky’s theorem 99

Now by Corollary 24.1.1 of Bloch’s theorem, g(B(0, R)) contains a disk ofradius 1

72R|g′(0)|. But since g′(0) 6= 0, this goes to infinity as R → ∞, so we

can pick R large enough so that g(C) ⊃ g(B(0, R)) contains a disk of radius 1,contradicting Lemma 25.1.1.

Hence f is constant, finishing the proof.

Exercise 26.1. Let f be a meromorphic function on C such that such thatC∞ \ f(C) has at least three points. Show that f is constant.

26.2 Schottky’s theoremAs a consequence of Hadamard’s factorisation theorem we showed a Special caseof Picard’s theorem, saying that a non-constant entire function of finite orderassumes each complex number with only one possible exception. Little Picardtheorem above tells us that the assumption of finite order is not necessary.

From Hadamard’s factorisation theorem we also inferred Corollary 23.2.3,saying that an entire function of finite, non-integral order assumes each complexnumber an infinite number of times. We wish to generalise this result too.

The strategy boils down to that of Lemma 24.2.1, but with slightly morecare taken, to derive precise bounds. In particular, the way we constructed thefunction g coming from f : G → C, analytic and missing 0 and 1, was to takeh(z) to be a branch of log f(z) (i.e., eh(z) = f(z)), set F (z) = 1

2πih(z), andtake H(z) =

√F (z)−

√F (z)− 1. Then we finally took g(z) to be a branch of

logH(z), i.e., eg(z) = H(z).The way we are going to make this more precise is to specify the branches

at each of these stages. In particular, we will take 0 ≤ Im h(0) ≤ 2π and0 ≤ Im g(0) ≤ 2π.

Theorem 26.2.1 (Schottky’s theorem). Let f be analytic on some simply con-nected region containing B(0, 1), and suppose f(z) misses 0 and 1.

Then for each α and β, 0 < α < ∞ and 0 ≤ β < 1, there exists a constantC(α, β) such that if |f(0)| < α, then |f(z)| ≤ C(α, β) for |z| ≤ β.

Remark 26.2.2. Notice how this constant C(α, β) is independent of f—this givesa uniform bound for all functions f so long as |f(0)| < α.

Proof. As outlined in the discussion above, the strategy is to perform the cal-culations in Lemma 24.2.1 to obtain a bound for g depending only on α and β,and then leverage this to bound f likewise.

As will soon become apparent, the calculations are easier if 2 ≤ α <∞, butof course if |f(0)| < γ for some γ ≤ 2, then it is also bounded by α, so it sufficesto study such α.

We will consider two cases: when f(0) has a lower bound, namely 12 ≤

|f(0)| ≤ α, and when 0 < |f(0)| < 12 .

Start by supposing 12 ≤ |f(0)| ≤ α. Let h, F , H, and g be defined as in

Lemma 24.2.1. Then by definition

|F (0)| = 12π |h(0)| = 1

2π |log f(0)| = 12π |log|f(0)|+ i Im log f(0)|

= 12π |log|f(0)|+ i Im h(0)| ≤ 1

2π (|log|f(0)||+ 2π)

100 THE LITTLE PICARD THEOREM

by choice of the branch 0 ≤ Im h(0) ≤ 2π. But 12 ≤ |f(0)| ≤ α, so − log 2 ≤

log|f(0)| ≤ logα, and hence |log|f(0)|| ≤ 2 ≤ α. Therefore

|F (0)| ≤ 12π (logα+ 2π) = C0(α).

Next, we have by the triangle inequality that

|√F (0)±

√F (0)− 1| ≤ |

√F (0)|+ |

√F (0)− 1|

Notice in general, writing√z = exp( 1

2 log z), we have

|√z| = exp

(12 Re log z

)= exp

(12 log|z|

)=√|z|,

so we can bring the modulus inside, whence

|√F (0)±

√F (0)− 1| ≤ |F (0)| 12 + |F (0)− 1| 12 ≤ C0(α) 1

2 + (C0(α) + 1) 12 .

Now let C1(α) = C0(α) 12 +(C0(α)+1) 1

2 so that |√F (0)±

√F (0)− 1| ≤ C1(α).

There are two cases to consider here: if |H(0)| ≥ 1, then

|g(0)| = |log|H(0)|+ i Im g(0)| ≤ |log|H(0)||+ 2π,

since we chose 0 ≤ Im g(0) ≤ 2π. Hence in particular

|g(0)| ≤ logC1(α) + 2π

since |H(0)| ≥ 1. On the other hand, if |H(0)| < 1, then

|g(0)| ≤ − log|H(0)| = 2π = log∣∣∣ 1H(0)

∣∣∣+ 2π

= log|√F (0) +

√F (0)− 1|+ 2π ≤ logC1(α) + 2π,

so we get the same bound. Thus let C2(α) = logC1(α) + 2π, so that |g(0)| ≤C2(α).

In other words, we now have control of g(0).By shifting the function in Corollary 24.1.1 of Bloch’s theorem, for |a| < 1

we see g(B(a, 1− |a|)) contains a disk of radius 172 |g

′(a)|(1− |a|).

11111111111111111 ×0×a

1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|1− |a|

Figure 26.2.1: The setup of ain the unit disk.

On the other hand, by Lemma 25.1.1, g(B(0, 1)) contains no disk of radius1, so we must have

172 |g

′(a)|(1− |a|) < 1,

meaning that we can control the derivative:

|g′(a)| < 721− |a|

for any |a| < 1.The idea now is that, since we have control of the function g at a point

(namely 0) and we have control of its derivative, we can the function at any

THE GREAT PICARD THEOREM 101

point by means of the Fundamental theorem of calculus. In particular, for|a| < 1 we have

|g(a)| ≤ |g(0)|+ |g(a)− g(0)| ≤ C2(α) +∣∣∣∫

[0,a]g′(z) dz

∣∣∣≤ C2(α) + |a| max

z∈[0,a]|g′(z)| ≤ C2(α) + |a| 72

1− |a| .

So if we letC3(α, β) = C2(α) + 72β

1− βfor 0 ≤ β < 1, we have for all |z| ≤ β that

|g(z)| ≤ C2(α) + 72|z|1− |z| ≤ C3(α, β)

since the expression 72|z|1−|z| is increasing in |z| < 1. Notice how this bound is

uniform—it doesn’t depend on g.Consequently, we get

|f(z)| = |exp(πi cosh(2g(z)))| ≤ C(α, β)

for a corresponding constant C(α, β) independent of f .This leaves the case when 0 < |f(0)| < 1

2 . In this case, consider 1 − f(z),which misses 0 and 1 since f(z) does. Then 1

2 ≤ |1−f(0)| ≤ 32 , so taking α = 3

2and applying the previous case, we get

|1− f(z)| ≤ C(α, β)

for all |z| ≤ β, and |1− f(z)| ≥ |f(z)| − 1, meaning that

|f(z)| ≤ 1 + C(α, β)

for all |z| ≤ β.

As is now our custom, we can easily generalise results on the unit ball toresults on larger balls:

Corollary 26.2.3. Let f be analytic on a simply connected region containingB(0, R). Suppose f(z) misses 0 and 1, and |f(0)| ≤ α. Let C(α, β) be theconstant in Schottky’s theorem. Then |f(z)| ≤ C(α, β) for all |z| ≤ βR.

Proof. Consider f(Rz) for |z| and apply Schottky’s theorem.

Lecture 27 The Great Picard Theorem

27.1 The range of entire functions, revisitedThe Great Picard theorem will follow almost immediately from the followingslightly technical theorem:

Date: November 19th, 2019.

102 THE GREAT PICARD THEOREM

Theorem 27.1.1 (Montel–Carathéodory theorem). Let G be a region. Let

F =f : G→ C analytic

∣∣ f(z) omits 0 and 1.

Then F is normal in C(G,C∞).

Proof. Fix z0 ∈ G and let G =f ∈ F

∣∣ |f(z)| ≤ 1and H =

f ∈ F

∣∣ |f(z)| >1. Then F = G∪H and if we can show that G and H are normal in C(G,C∞),

then so is F .We will show in particular that G is normal in H(G), and H is normal in

C(G,C∞), one at a time.First, to show that G is normal in H(G), recall how by Montel’s theorem it

suffices to show that G is locally bounded.To this end, let a ∈ G and let γ be a path from z0 to a. Take disks

D0, D1, . . . , Dn in G, centred at z0, z1, . . . , zn = a ∈γ, so that zk−1, zk ∈

Dk−1 ∪Dk for all 1 ≤ k ≤ n. This is possible by taking zk−1 and zk sufficientlyclose to one another, and since γ is compact, we can find a finite subset of themthat does the job.

Now by applying a shifted version of Schottky’s theorem on D0, we see thatthere exists some constant C0 such that |f(z)| ≤ C0 for all z ∈ D0 and all f ∈ G.

In particular, since by construction z1 ∈ D0, this gives us |f(z1)| ≤ C0 for allf ∈ G. This lets us apply Schottky’s theorem again to D1, whence |f(z)| ≤ C1for all f ∈ G and all z ∈ D1 for some constant C1, and in particular |f(z2)| ≤ C1for all f ∈ G.

Repeat this argument on D2, D3, and so on, until we get for Dn that Gis uniformly bounded on Dn = B(a, r) for some radius r. Hence G is locallybounded, since a ∈ G is arbitrary, and therefore G is normal.

This leaves showing thatH is normal in C(G,C∞). If f ∈ H is analytic, then1f is analytic on G (since f omits zero, being bounded below by 1). Moreover,

1f(z) 6= 0 since f is analytic, hence having no poles, and 1

f(z) 6= 1 since f(z) 6= 1.Finally, | 1

f(z) | < 1 since |f(z) > 1|.All by way of saying: H :=

1f

∣∣ f ∈ H ⊂ G. Since G is normal, H must benormal too.

In other words, iffn⊂ H, then there exists a subsequence

fnk

such

that 1fnk→ h ∈ H(G).

By Corollary 14.1.4 of Hurwitz’s theorem, this means either h = 0 identicallyor h(z) 6= 0 for all z ∈ G. In the first case we get fnk → ∞ in C(G,C∞), andin the second case we see that 1

h is analytic on G, and fnk → 1h in H(G), so

in particular in C(G,C∞). In either case we have a subsequence offn⊂ H

converging in C(G,C∞), so H is normal.

With this in hand we are equipped to prove

Theorem 27.1.2 (Great Picard theorem). Let G be a region and a ∈ G. Letf be analytic on G \

aand suppose g has an essential singularity at z = a.

Then in each neighbourhood of z = a, f(z) assumes each complex number, withone possible exception, infinitely many times.

Remark 27.1.3. This improves the Casorati–Weierstrass theorem, which saysthat the image of each neighbourhood of an essential singularity is dense in C.

27.1 The range of entire functions, revisited 103

Proof. Without loss of generality, assume f has an essential singularity at z = 0(else shift it). Suppose there exists some R > 0 such that f(z) omits two valueson 0 < |z| < R. Again we can assume those two values are 0 and 1, i.e., f(z) 6= 0and f(z) 6= 1 for all 0 < |z| < R, else normalise as in the proof of the LittlePicard theorem.

Let G = B(0, R) \

0and define fn : G→ C by fn(z) = f( zn ). Then fn is

analytic on G since f is and fn(z) 6= 0 for all z ∈ G since f(z) 6= 0 on G.By Montel–Carathéodory theorem,

fnis normal in C(G,C∞), meaning

that there exists a subsequencefnk

such that fnk → ϕ in C(G,C∞). In

particular, fnk → ϕ uniformly on any compact subset of G, so in particular on|z| = 1

2R.Note how, since

fnk

⊂ H(G), we have again by Corollary 14.1.4 of Hur-

witz’s theorem that either ϕ is analytic on G or ϕ =∞ identically.In the former case, let

M = max|z|= 1

2R|ϕ(z)|,

which exists since ϕ is analytic and |z| = 12R is compact. Then for |z| = 1

2R,∣∣∣f( znk

)∣∣∣ = |fnk(z)| ≤ |fnk(z)− ϕ(z)|+ |ϕ(z)|.

The first term in the right-hand side goes to 0 uniformly, and the second termis bounded uniformly by M , so for nk sufficiently large we have |f( z

nk)| ≤ 2M .

Hence |f(z)| ≤ 2M for |z| = R2nk for nk large, so f(z) is uniformly bounded

on B(0, r) \

0

for some 0 < r < R. This means f(z) has a removablesingularity at z = 0, which is a contradiction.

Similarly, in the latter case, assume ϕ =∞. In this case, fnk →∞ uniformlyon |z| = 1

2R, which means f( znk

)→∞ uniformly on |z| = 12R. In other words,

limz→∞

|f(z)| =∞,

meaning that f has a pole at z = 0, which is again a contradiction.Hence f cannot omit two values in C, meaning it can omit at most one value.For the second part of the theorem we need to show that, apart from this

possible exceptional point, all points are attained infinitely many times.Suppose, therefore, that two values are assumed by f only finitely many

times. This means that there are some finite set of preimages of those twopoints in G, which in turn means we must be able to find some sufficiently smallpunctured disk B(0, r)\

0, 0 < r < R, not containing any of those preimages.

Hence on this smaller punctured disk, f omits those two values, which by theprevious discussion is impossible.

An immediate corollary of this is

Corollary 27.1.4. If f has an isolated singularity at z = 0 and if there are twovalues that are not assumed by f(z) infinitely many times, then z = a is eithera pole or a removable singularity (i.e., z = a cannot be an essential singularity).

An almost as immediate corollary, and the result we promised a bit ago, isthis:

104 THE GREAT PICARD THEOREM

Corollary 27.1.5. If f is an entire function that is not a polynomial, then f(z)assumes every complex number, with one possible exception, infinitely manytimes.

Exercise 27.1. Suppose f is a one-to-one entire function. Show that f(z) = az+bfor some a, b ∈ C with a 6= 0.

Proof. Since f is entire and not a polynomial, it has a power series expansionabout z = 0 that never ends. Consequently, the Laurent expansion of g(z) =f( 1

z ) extends infinitely in the negative direction, so g has an essential singularityat z = 0.

Therefore by the Great Picard theorem g(z) assumes each complex number,with one possible exception, infinitely many times. But f and g have the sameimage, so f(z) does too.

27.2 Runge’s theoremRecall how in real analysis, Weierstrass theorem says that every continuousfunction f on a compact set in R can be approximated uniformly by polynomials.(One way to prove this is by way of harmonic analysis: show that they canbe approximated by Fourier series, i.e., linear combinations of trigonometricfunctions, and then approximate those trigonometric functions by their Taylorpolynomials).

A natural question to ask, then, is this: let K ⊂ C be compact and let G bea neighbourhood of K. Suppose f ∈ H(G). Can f be approximated uniformlyby polynomials on K?

The answer, in general, is no. Consider the following two examples:

Example 27.2.1. Let G = B(0, R). For any f ∈ H(G), we have a power seriesexpansion

f(z) =∞∑n=0

anzn,

which converges uniformly on any compact subset K ⊂ G. Taking the partialsums

Pk(z) =k∑

n=0anzn,

we then have Pk → f uniformly on K as k →∞. Hence in this particular case,the answer to the above question is yes. N

On the other hand,

Example 27.2.2. Let G = B(0, R) \

0, and let f(z) = 1

z ∈ H(G) (it isholomorphic since G specifically excludes the pole). Let K =

z∣∣ |z| = 1

2R⊂

G, which is compact.For any polynomial P (z), ∫

K

P (z) dz = 0

MITTAG-LEFFLER’S THEOREM 105

since K encloses no pole of P (z) (it has no poles). On the other hand,∫K

f(z) dz =∫K

1zdz = 2πi

since K encloses the pole at z = 0 of f .Now if Pn → f uniformly on K, we must have∫

K

Pn(z) dz →∫K

f(z) dz

but the left-hand side is constantly 0, and the right-hand side if 2πi 6= 0, so fcannot be approximated uniformly by polynomials on K. N

What these examples show is that, in fact, whether f can be approximateduniformly by polynomials or not is a topological property: it depends on whetherC \K is connected or not.

What Runge’s theorem says is this, which we will work toward proving:

(i) If f ∈ H(G), then f can be approximated uniformly by rational functionson K; and

(ii) If C \K is connected, then f ∈ H(G) can be approximated uniformly bypolynomials on K.

Lecture 28 Mittag-Leffler’s Theorem

28.1 Runge’s theoremWe will prove Runge’s theorem, outlined above, in three steps.

G

KKKKKKKKKKKKKKKKK

Figure 28.1.1: A square gridcovering K.

Proposition 28.1.1. Let G be a region and let K ⊂ G be compact. Then thereexists finitely many segments γ1, γ2, . . . , γn in G \K (depending only on G andK) such that for any f ∈ H(G),

f(z) =n∑k=1

12πi

∫γk

f(z)w − z

dw

for all z ∈ K.

Proof. Let δ := c · d(K,G \K) > 0 with 0 < c < 12 . Consider a grid of squares

with sides parallel to the axes and of side lengths δ.Let R =

R1, R2, . . . , Rm

denote the squares intersection K—there are

only finitely many such squares since K is compact.Let γ1, γ2, . . . , γn denote the sides of squares in R that do not belong to

any adjacent squares in R. This has two implications: first, by the choice of δ,γk⊂ G, since the distance from any point on γk to K is less than δ; and

second, γk ∩K = ∅, since otherwise γk belongs to two squares intersecting K.

G

KKKKKKKKKKKKKKKKK

Figure 28.1.2: Constructingthe γk from the square grid.

Finally, assign each square in R positive orientation, so that integration overadjacent squares cancels on the common side.

Date: November 21st, 2019.

106 MITTAG-LEFFLER’S THEOREM

Now for each z ∈ K that is not on the boundary of Rj , 1 ≤ j ≤ m, thereexists some j0 such that z ∈ Rj0 . By Cauchy’s integral formula this means

12πi

∫∂Rj

f(w)w − z

dw =f(z), if j = j0,

0, if j 6= j0.

If Ri and Fj are adjacent, then as discussed the integral over their common sideis cancelled, so

f(z) =m∑k=1

12πi

∫∂Rk

f(w)w − z

dw =n∑k=1

12πi

∫γk

f(w)w − z

dw

for z ∈ K \ (m⋃k=1

∂Rk). Bot the above expression is holomorphic on (a neigh-

bourhood) of K since all poles are in γk, which are outside K, and the equationholds on a dense subset of K, so it holds everywhere in K.

Our goal is to approximate f ∈ H(G) by a rational function on K. Theexpression we end up at above,

f(z) =n∑k=1

12πi

∫γk

f(w)w − z

dw,

is close to a rational function: it is the integral of a rational function in z. Theidea is to write this as a Riemann sum:

Lemma 28.1.2. Let G be a region and K ⊂ G compact. Let γ be a line segmentin G \ K. Then for any ε > 0, there exists a rational function R(z) with allpoles on γ such that ∣∣∣∫

γ

f(w)w − z

dw −R(z)∣∣∣ < ε

for all z ∈ K.

Proof. We parametrise γ : [0, 1]→ C. Then∫γ

f(w)w − z

dw =∫ 1

0

f(γ(t))γ(t)− z γ

′(t) dt.

Define F : K × [0, 1]→ C by

F (z, t) = f(γ(t))γ(t)− z γ

′(t).

Note that since γ ∈ G \K, γ(t) 6= z for z ∈ K, so F is continuous. Now sinceK × [0, 1] is compact, F is moreover uniformly continuous on K × [0, 1]. So forany ε > 0 there exists some δ > 0 such that if |s− t| < δ, then

|F (z, s)− F (z, t)| < ε.

Now approximate the integral we are interested in—the integral of F (z, t)—by Riemann sums.

28.1 Runge’s theorem 107

Choose a partition

0 = t0 < t1 < t2 < · · · < tn = 1

with ti−1 − ti < δ for all i = 0, 1, . . . , n− 1. Define

R(z) =n−1∑i=0

F (z, ti)(ti+1 − ti) =n−1∑i=1

f(γ(ti)γ(ti)− z

γ′(ti)(ti+1 − ti),

the left Riemann sum of the integral at hand. This is a finite sum of rationalfunctions in z, so it is rational with all poles on γ.

Moreover∣∣∣∫γ

f(w)w − z

dw −R(z)∣∣∣ =

∣∣∣n−1∑n=0

∫ ti+1

ti

( f(γ(t))γ(t)− z γ

′(t)− f(γ(ti))γ(ti)− z

γ′(ti))dt∣∣∣,

where the length in the Riemann sum has been absorbed by the integral. Theintegrand above is really just F (z, t)−F (z, ti), and |t− ti| < δ, so it is boundedby ε, whence ∣∣∣∫

γ

f(w)w − z

dw −R(z)∣∣∣ ≤ n−1∑

i=0

∫ ti+1

ti

ε dt = ε.

Hence this finishes the first part of Runge’s theorem: every holomorphicfunction f can be approximated by rational functions on K with poles outsideK.

If C \K we want to push the poles of ∞, since we can think of a polynomialas a rational function with poles at ∞.

Lemma 28.1.3. Let G be a region and K ⊂ G compact such that C \ K isconnected. Let a ∈ G ∈ K. Then 1

z−a can be approximated uniformly bypolynomials on K. (That is, for any ε > 0 there exists a polynomial P (z) suchthat | 1

z−a − P (z)| < ε for all z ∈ K.)

Proof. Since K is compact, and hence bounded, there exists some M > 0 suchthat |z| < M for every z ∈ K.

For any b ∈ C with |b| > M and z ∈ K, we have, since | zb | < 1,

1z − b

= −1b

11− z

b

= −1b

∞∑n=0

(zb

)nconverges uniformly on K.

Taking partial sums we then see that 1z−b can be approximated uniformly

by polynomials.The strategy now is to approximate 1

z−a , where a ∈ G \ K is arbitrary,uniformly on K by polynomials in 1

z−b .Since C \ K is connected, there exists a path γ in C \ K from a to b. Let

δ = 12d(K, γ). MMMMMMMMMMMMMMMMM

KKKKKKKKKKKKKKKKK×z

×a

γ ×b

Figure 28.1.3: Connecting aand b avoiding K.

Choose a partition a = z0, z1, z2, . . . , zn = b ∈γsuch that |zi− zi+1| < δ

for all i = 0, 1, . . . , n− 1. Then for z ∈ K we have

1z − a

= 1z − z0

= 1z − z1

11− z0−z1

z−z1

.

108 MITTAG-LEFFLER’S THEOREM

By construction and choice of δ,∣∣∣z0 − z1

z − z1

∣∣∣ ≤ δ

d(z,K) ≤12 ,

so that1

z − a= 1z − z1

∞∑n=0

(z0 − z1

z − z1

)nconverges uniformly on K. Taking partial sums, this means we can approximate

1z−a uniformly in K by polynomials in 1

z−z1. Repeating this argument with z1

in place of z0 = a, we can approximate 1z−z1

uniformly on K by polynomials in1

z−z2, and so on.

In the end, we can approximate 1z−a uniformly on K by polynomials in 1

z−b ,and as discussed 1

z−b can be approximated uniformly on K by polynomials, sowe are done.

This proposition and the two lemmata together finishes the proof of

Theorem 28.1.4 (Runge’s theorem). Let G be a region and K ⊂ G be compact.Let f ∈ H(G).

(i) For any ε > 0, there exists a rational function R(z) with all poles in G\Ksuch that |f(z)−R(z)| < ε for all z ∈ K.

(ii) Suppose C \K is connected. Then for any ε > 0 there exists a polynomialP (z) such that |f(z)− P (z)| < ε for all z ∈ K.

28.2 Mittag-Leffler’s theoremWeierstrass factorisation theorem, along with Theorem 20.1.9, tell us that thereexist holomorphic functions with a prescribed set of zeros.

We want to answer a related question: does there exist a meromorphic func-tion with a prescribed set of poles or, more specifically, does there exist a mero-morphic function with prescribed singular part?8

The answer is the affirmative:

Theorem 28.2.1 (Mittag-Leffler’s theorem). Let G be a region and letak⊂

G be a sequence of distinct points such thatakhas no limit points. For each

k ∈ N, let

Sk(z) =mk∑j=1

Ajk(z − ak)j

where mk ∈ N and Ajk ∈ C. Then there exists f ∈ M(G) whose poles areexactly

akand the singular part of f at z = ak is Sk(z).

Proof. Note that, ideally, we simply want to sum Sk(z) over k, however thissum might be divergent, so we need to be a little bit more delicate.

8Meaning the negative powers in the Laurent expansion.

28.2 Mittag-Leffler’s theorem 109

Write

G =∞⋃n=1

Kn,

where all Kn are compact and Kn ⊂ int(Kn+1), so that the Kn are growing.Note how since

akhas no limit points and Kn is compact,

ak∩Kn is a

finite set.Define I1 =

k∣∣ ak ∈ K1

, I2 =

k∣∣ ak ∈ K2 \ K1

, and so on, with

In =k∣∣ ak ∈ Kn \Kn−1

in general.

With this, definefn(z) =

∑k∈In

Sk(z).

By convention we take fn = 0 if In = ∅. Then fn(z) is a rational function withpoles exactly at

ak∣∣ k ∈ In ⊂ Kn \Kn−1.

Since fn has no poles in Kn−1, fn is analytic on a neighbourhood of Kn−1,so by Runge’s theorem there exists a rational function Rn(z) with poles in C\Gsuch that

|fn(z)−Rn(z)| < 12n

for all z ∈ Kn−1. (Note how originally Runge’s theorem only says the polesare in C \ K, but since C \ K is larger than C \ G, we can push the polesfurther to specifically lie in C \G by the same argument as used in the proof ofLemma 28.1.3.)

Now define

f(z) = f1(z) +∞∑n=2

(fn(z)−Rn(z))

for z ∈ G. We claim that f is analytic on G \ak.

To see this, let K ⊂ (G \ak

) be compact. Then K ⊂ KN for somesufficiently large N , whereby for n ≥ N we have

|fn(z)−Rn(z)| < 12n

for all z ∈ K. Therefore∞∑n=N

(fn(z)−Rn(z))

is uniformly convergent on K, and hence analytic on K.Since K∩

ak

= ∅, f1(z), f2(z), . . . , fN−1(z) are analytic on K (K doesn’ttouch any of their poles). Hence f is analytic on K, and K ⊂ (G \

ak

) isarbitrary, so f is analytic on G \

ak.

Finally, we claim that each ak is a pole of f with singular part exactly Sk(z).To show this, fix ak. Then there exists some R > 0 such that ak 6∈ B(ak, R) forj 6= k. Write, for z ∈ B(ak, R),

f(z) = Sk(z) + g(z),

with g analytic, since f has no other poles in B(ak, R). Then this is a Laurentexpansion about z = ak, which by uniqueness means that Sk(z) is the singularpart of f at z = ak, finishing the proof.

110 ANALYTIC CONTINUATION

Exercise 28.1. Letan⊂ C be a sequence of distinct points such that |an| →

∞ as n→∞. Letbn⊂ C and

kn⊂ Z. Suppose that

∞∑n=1

( r

an

)kn bnan

converges absolutely for all r > 0.Show that

∞∑n=1

( zan

)kn bnz − an

converges in M(C) to a function f with poles at each point z = an.

Exercise 28.2. Find a meromorphic function f with poles of order 2 at√n,

(n = 1, 2, . . . ), such that the residue at each pole is 2 and

limz→√n(z −

√n)2f(z) = 1

for all n.

Lecture 29 Analytic Continuation

29.1 Analytic continuationDefinition 29.1.1 (Analytic continuation). Let G1 ⊂ G be regions. Letf : G1 → C be analytic. A function g : G→ C is an analytic continuation off to G if g is analytic and g

∣∣G1

= f .

G

G

Figure 29.1.1: Two regionsG and G meeting on the realaxis.

Recall from Exercise 1.2 how if G is a region and G =z∣∣ z ∈ G, then if

f : G→ C is analytic, so is f? : G→ C defined by f?(z) = f(z).Suppose, as a special case, that G and G touch (meaning they meet, at least,

somewhere on the real axis). Then f?(z) = f(z) = f(z) on the real line. Buton the real line z = z, so f(z) = f(z), meaning that f(z) is real on the real line.

This idea gives rise to the following:

Theorem 29.1.2 (Schwarz reflection principle). Let G be a region such thatG = G.9 Let G+ =

z ∈ G

∣∣ Im(z) > 0, G0 =

z ∈ G

∣∣ Im(z) = 0, and

G− =z ∈ G

∣∣ Im(z) < 0.

Let f : G+ ∪ G0 → C be a continuous function which is analytic on G+.Suppose f(z) is real on G0.

Then there exists an analytic function g : G → C such that g(z) = f(z) forall z ∈ G+ ∪G0.

G

G+G+G+G+G+G+G+G+G+G+G+G+G+G+G+G+G+

G−G−G−G−G−G−G−G−G−G−G−G−G−G−G−G−G−

G0G0G0G0G0G0G0G0G0G0G0G0G0G0G0G0G0

Figure 29.1.2: A schematic ofG+, G−, and G0. Note howG does not need to be simplyconnected.

Proof. The discussion above gives us an outline of the proof, in that the naturalg is the correct one:

g(z) =f(z), if z ∈ G+ ∪G0

f(z), if z ∈ G−.

Date: December 3rd, 2019.9That is, G is symmetric under reflection across the real axis.

29.1 Analytic continuation 111

Since f is analytic on G+ and f(z) is analytic on G+ = G− (by Exercise 1.2),g is continuous on those, and by hypothesis g is also continuous on G0. Henceg is continuous on all of G, and analytic on G+ and G−.

We want to show that g is analytic on G, so it remains to consider G0.By Morera’s theorem, it suffices to show that∫

T

g(z) dz = 0

for any triangular path T ⊂ G. Now since g is analytic on G+ and G−, we knowalready that if T ⊂ G+ or T ⊂ G− then the above integral is indeed zero, so weonly need to consider the case when the triangle crosses the real axis.

Now importantly, analyticity is a local property, so to show that g is analyticon G, it suffices to show that it is analytic on any ball contained in G. Thereason we make this distinction is that, were we not to do that, it is possibleG has holes in it, and we would have to consider triangular paths around theseholes. So instead consider only triangular paths T such that the region theyenclose is in G.

T

becomes

T−

Tε

T+

× ×× ×

βb

αa

Figure 29.1.3: Partitioninga triangular path crossing thereal axis.

Given any such triangle T , we partition it by cutting the triangle horizontallyclose to the real axis, resulting in a piece in G+ (call it T+), a piece in G−(say T−), and a piece (of small height) in both (call it Tε), as illustrated inFigure 29.1.3. (Note how, strictly speaking there is another case to consider,namely where one or more of the vertices are on the real axis—these can bedone similarly as to what follows.)

Then∫T

g(z) dz =∫T+

g(z) dz +∫T−

g(z) dz +∫Tε

g(z) dz =∫Tε

g(z) dz.

Since g is continuous on G it is uniformly continuous on compact subsets ofG, and so in particular it is uniformly continuous on the closure of the regionenclosed by T , call it R. So given any ε > 0 there exists some δ > 0 such thatfor any z1, z2 ∈ R, if |z1 − z2| < δ we have |g(z1)− g(z2)| < ε.

Now let us specify where we cut the triangle T to get Tε. In particular, letaα, a, β, b ∈

Tbe the corners of Tε such that |α− a| < δ

2 and |β − b| < δ2 .

Parametrising the two horizontal lines, call them γ1 and γ2, we have for0 ≤ t ≤ 1

|(tβ + (1− t)α)− (tb+ (1− t)a)| ≤ t|β − b|+ (1− t)|α− a| < δ

2 .

Then∣∣∣∫γ1

g(z) dz +∫γ2

g(z) dz∣∣∣ =

=∣∣∣∫ 1

0g(tβ + (1− t)α)(β − α) dt−

∫ 1

0g(tb+ (1− t)a)(b− a) dt

∣∣∣,where we subtract the second integral because we parametrised γ2 in the oppo-site direction we wish to integrate it. We bound this by∣∣∣∫γ1

g(z) dz +∫γ2

g(z) dz∣∣∣ ≤ |β − α|∫ 1

0|g(tβ + (1− t)α)− g(tb+ (1− t)a)| dt+∫ 1

0|g(tb+ (1− t)a)((β − α)− (b− a))| dt.

112 ANALYTIC CONTINUATION

The integrand of the first integral is bounded by ε by our uniform continuity,and we can get rid of the constant at the end of the second integral since

|(β − α)− (b− a)| = |(β − b)− (a− α)| ≤ δ

2 + δ

2 = δ.

Finally letM = max

z∈T|g(z)|

so that we can bound away the integrand in the second integral, finally yielding∣∣∣∫γ1

g(z) dz +∫γ2

g(z) dz∣∣∣ ≤ |β − α| · ε+ δ ·M.

This goes to 0 as ε→ 0, taking δ → 0 at the same time.This takes care of the integral over γ1 and γ2, leaving the other two parts of

Tε. This is easier:∣∣∣∫[α,a]

g(z) dz +∫

[b,β]g(z) dz

∣∣∣ ≤ |α− a| ·M + |b− β| ·M ≤ δM,

which also goes to 0 as δ → 0.Hence the integral over any triangular path is 0, so by Morera’s theorem g

is analytic.

29.2 Dirichlet seriesDefinition 29.2.1 (Dirichlet series). A series of the form

D(s) =∞∑n=1

anns

with s ∈ C andan⊂ C is called a Dirichlet series.

Example 29.2.2. The prototypical example of a Dirichlet series is the Rie-mann zeta function

ζ(s) =∞∑n=1

1ns.

N

An important, and natural, question to ask about these series is this: forwhat s ∈ C do they converge, and converge in what sense? For instance, ζ(s)converges absolutely for Re(s) > 1.

Proposition 29.2.3. Suppose

D(s) =∞∑n=1

converges absolutely at s0 = σ0 + it0. Then D(s) converges absolutely anduniformly on Re(s) ≥ σ0.

29.2 Dirichlet series 113

Proof. This is a straight-forward computation. For Re(s) ≥ σ0 we have∞∑n=1

∣∣∣anns

∣∣∣ =∞∑n=1

|an|nRe(s) ≤

∞∑n=1

|an|nσ0

<∞.

Since the convergence of this last sum does not depend on s, the convergence isuniform.

Remark 29.2.4. (i) Because of this proposition, the region of absolute con-vergence of a Dirichlet series is always a half-plane.

(ii) There exists a unique σac ∈ [−∞,∞] such that D(s) is absolutely con-vergent for Re(s) > σac and is not absolutely convergent for Re(s) < σac.Such a σac is called the abscissa of absolute convergence of D(s).This means the series converges absolutely to the right of the vertical lineRe(s) = σac, and does not converge absolutely to the left of the line. Whathappens on the line is a mystery, and the behaviour of a given Dirichletseries can vary for points on this line.

If we only know about convergence, not absolute convergence, at a point, weneed much more delicate calculations:

Proposition 29.2.5. Suppose

D(s) =∞∑n=1

anns

converges (not necessarily absolutely) at s0 = σ0 + it0. Then D(s) convergesuniformly for |arg(s− s0)| ≤ 1

2π − δ where 0 < δ < 12π. Hence D(s) converges

uniformly on any compact subset K ⊂s∣∣ Re(s) > σ0

.

δ

δ

×s0

Figure 29.2.1: The sector inthe setup.

Proof. Without loss of generality we may assume s0 = 0 (else shift). Then

D(0) =∞∑n=1

an

converges. Let

rn =∞∑

k=n+1ak

be the tail of this sum. Since D(0) converges, this means rn → 0 as n→∞.For N > M , consider the partial sum

N∑n=M

anns

=N∑

n=M

rn−1 − rn.

By partial summation (really just the discrete version of integration by parts,see e.g., [Apo10, Chapter 3]) this is equal to

N∑n=M

rn

( 1(n+ 1)s −

1ns

)+ rM−1

Ms− rn

(N + 1)s .(29.2.1)

114 PERRON’S FORMULA

Note how∣∣∣ 1(n+ 1)s −

1ns

∣∣∣ =∣∣∣s∫ n+1

n

1us+1 du

∣∣∣≤ |s|

∫ n+1

n

1uσ+1 du = |s|

σ

( 1nσ− 1

(n+ 1)σ),

where Re(s) = σ.Since rn → 0, for any ε > 0 there exists some N0 ∈ N such that for n > N0,

|rn| < ε, and so for s = σ + it with σ > 0 and N > M > N0, we have, lookingback at Equation (29.2.1),

∣∣∣ N∑n=M

anns

∣∣∣ ≤ |s|εσ

N∑n=M

∣∣∣ 1nσ− 1

(n+ 1)σ∣∣∣+ ε

Mσ+ ε

(N + 1)σ .

In the sum on the right-hand side we can replace the absolute values withparentheses, since the 1

nσ ≥1

(n+1)σ are real, and so we get a telescoping sum,meaning that

∣∣∣ N∑n=M

anns

∣∣∣ ≤ ε|s|σ

( 1Mσ− 1

(N + 1)σ)

+ 2ε ≤ 2ε |s|σ

+ 2ε

by bounding the fractions by 1.

δ

δ

×s0s0s0s0s0s0s0s0s0s0s0s0s0s0s0s0s0

×s

σ

t|s||s||s||s||s||s||s||s||s||s||s||s||s||s||s||s||s|

Figure 29.2.2: The trigonom-etry of |s|

σ.

It remains to study |s|σ . With a bit of trigonometry (as laid out in Fig-ure 29.2.2), if |arg(s)| ≤ 1

2π − δ, thentσ ≤ tan( 1

2π − δ), so

s

σ=√σ2 + t2

σ=√

1 +( tσ

)2≤√

1 + tan(1

2π − δ).

This bounds |s|σ uniformly in the sector, so letting ε→ 0, this means

∣∣∣ N∑n=M

anns

∣∣∣goes to 0 uniformly in N > M , giving us the convergence we are after.

That this implies uniform convergence on compact subsets is simply a con-sequence of any compact subset being contained in some sector of this form.

Lecture 30 Perron’s Formula

30.1 Uniqueness of Dirichlet seriesRemark 30.1.1. As a consequence of Proposition 29.2.5, we infer:

(i) The region of convergence of a Dirichlet series is a half-plane.

(ii) There exists a unique σc ∈ [−∞,∞] such that D(s) converges for Re(s) >σc and diverges for Re(s) < σc. Such a σc is called the abscissa ofconvergence of D(s).

Date: December 5th, 2019.

30.1 Uniqueness of Dirichlet series 115

Notice how, since absolute convergence implies convergence, σc ≤ σac. How-ever they need not be equal.

Example 30.1.2. Consider the alternating harmonic series

D(s) =∞∑n=1

(−1)n

n2 .

This has abscissa of absolute convergence σac = 1, but abscissa of convergenceσc = 0 (from Dirichlet’s test: an = (−1)n has bounded average, and 1

ns isdecreasing to 0 in magnitude for Re(s) > 0). N

In this example σac and σc differ by one. This is the worst case scenario:

Proposition 30.1.3. Let

D(s) =∞∑n=1

anns

with abscissa of absolute convergence σac and abscissa of convergence σc. Then0 ≤ σac − σc ≤ 1.

Proof. The left-hand side inequality is trivial since σc ≤ σac. Hence it remainsto show σac ≤ 1 + σc, i.e., show that

∞∑n=1

ann1+σc+ε

converges absolutely for all ε > 0.Since σc is the abscissa of convergence,

D(σc + ε

2

)=∞∑n=1

annσc+

ε2

converges, meaning that the individual terms must go to zero. In other words,

limn→∞

annσc+

ε2

= 0.

Hence there exists some N ∈ N such that for n > N ,∣∣∣ annσc+

ε2

∣∣∣ < 1.

Consequently we consider the tail∞∑n=N

∣∣∣ ann1+σc+ε

∣∣∣ =∞∑n=N

∣∣∣ annσc+

ε2

∣∣∣ · 1n1+ ε

2≤∞∑n=N

1n1+ ε

2<∞,

so we have the absolute convergence we were looking for.

Theorem 30.1.4 (Uniqueness theorem for Dirichlet series). Let

D(s) =∞∑n=1

anns

and F9s) =∞∑n=1

bnns

with abscissae of absolute convergence less than σ. Suppose there existsk⊂

s∣∣ Re(s) > σ

such that

116 PERRON’S FORMULA

(i) Re(sk) = σk →∞ as k →∞ and

(ii) D(sk) = F (sk) for all k.

Then an = bn for all n ∈ N.

Proof. Define a new Dirichlet series

G(s) = D(s)− F (s) =∞∑n=1

an − bnns

=∞∑n=1

cnns,

i.e., let cn = an − bn. We wish to show that cn = 0 for all n ∈ N .Notice how, by assumption, G(sk) = 0 for all k. Now suppose cn 6= 0 for

some n. Then there must be a smallest index for which cn is nonzero, so let Nbe the smallest integer such that cN 6= 0, meaning that cn = 0 for all n < N .Then

G(s) =∞∑n=N

cnns

= cNNs

+∞∑

n=N+1

cnns.

Since G(sk) = 0, we must have

cNNsk

+∞∑

n=N+1

cnnsk

= 0,

which, if we rearrange and take absolute values, becomes

|cN | ≤ Nσk

∞∑n=N+1

|cn|nσk≤ Nσk

(N + 1)σk−σ−ε∞∑

n=N+1

abscnnσ+ε .

The factor in front of the summation goes to 0 as σk →∞, and the sum is finitesince we are in the region of absolute convergence. Hence all of this goes to 0,implying that cN = 0, which is a contradiction.

30.2 Perron’s formulaLemma 30.2.1. Let c > 0 and y > 0. Then we have

12πi

∫(c)ywdw

w=

1, if y > 10, if 0 < y < 1.

Remark 30.2.2. By integrating over (c) we mean to integrate along the verticalline at Re(s) = c, i.e., in this example,

12πi

∫(c)ywdw

w= 1

2πi

∫ c+i∞

c−i∞ywdw

w.

γR

×c

×c− iR

×c+ iR

(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)

∗0

Figure 30.2.1: The choice ofcontour when y > 1.

Proof. Let us consider y > 1 first. The integrand yw

w is analytic except for asimple pole at w = 0 with residue 1, so integrating over the semicircle describedin Figure 30.2.1 by Cauchy’s integral formula

1 = 12πi

∫γR

ywdw

w+ 1

2πi

∫ c+iR

c−iRywdw

w.

30.2 Perron’s formula 117

Sending R → ∞ the second integral becomes the integral we want, so ideallythe first integral goes to 0 as R→∞.

So for w ∈γRwe parametrise as w = c + Reiθ, with π

2 ≤ θ ≤ 3π2 . Here

also dw = Rieiθ dθ, and we estimate |w| = |c + Reiθ| ≥ R − c (since c is fixedand R is large). Then

∣∣∣ 12πi

∫γR

ywdw

w

∣∣∣ ≤ 12π

∫ 3π/2

π/2ycyR cos(θ) Rdθ

R− c.

Since π2 ≤ θ ≤ 3π

2 , cos(θ) < 0, and so R cos(θ) is negative meaning that, sincey > 1, yR cos(θ) → 0 as R→∞. Consequently the entire integral goes to 0 sinceRR−c → 1 as R→∞.

Next we consider the case 0 < y < 1. The estimate is similar, only this timewe want to integrate over a semicircle extending to the right (see Figure 30.2.2),since for 0 < y < 1 we want positive powers to make the integrand small.

γR

×c

×c− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iRc− iR

×c+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iRc+ iR

(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)(c)

∗0

Figure 30.2.2: The choice ofcontour when 0 < y < 1.

Again there is a simple pole of yw

w at w = 0, but this lies outside of ourcontour, so by Cauchy’s integral formula

0 = 12πi

∫γR

ywdw

w+ 1

2πi

∫ c+iR

c−iRywdw

w.

As before, the second integral goes to the integral we want when R → ∞, andwe show that the first integral goes to 0.

Indeed, with the same parametrisation, only with −π2 ≤ θ ≤π2 , we get

∣∣∣ 12πi

∫γR

ywdw

w

∣∣∣ ≤ 12π

∫ −π/2π/2

ycyR cos(θ) Rdθ

R− c.

Since cos(θ) > 0 in this range, the exponent is positive, but 0 < y < 1 soyR cos(θ) → 0 as R→∞. Hence the integral goes to 0 as R→∞, again.

In order to prove Perron’s formula we need the following technical lemma,the proof of which is not difficult, but is a bit lengthy:

Lemma 30.2.3. Let

D(s) =∞∑n=1

anns

with abscissa of convergence σc. For s = σ + it, σc < σ < σc + 1, we have

D(s) |t|1−(σ−σc)+ε

for any ε > 0.

The idea of the proof is similar to that of Proposition 29.2.5.

Theorem 30.2.4 (Perron’s formula). Let

D(s) =∞∑n=1

anns

118 PERRON’S FORMULA

with abscissa of convergence σc. Let X > 0, X 6∈ N, and c > 0. For anys = σ + it with σ + c > σc, we have∑

n<X

anns

= 12πi

∫(c)D(s+ w)Xw dw

w.

In particular, ∑n<X

an = 12πi

∫(c)D(w)Xw dw

w

for c > σc.

This is a remarkable formula: it translates a discrete average into an analyticformula. Hence by understanding analytic properties of the Dirichlet series, wecan glean understanding about the sequence

an.

Sketch of proof. The main idea of the proof is to writeD(s+w) in the right-handside in terms of its series representation, then switch the order or summationand integration, like so:

12πi

∫(c)D(s+ w)Xw dw

w= 1

2πi

∫(c)

( ∞∑n=1

anns+w

)Xw dw

w

=∞∑n=1

anns

12πi

∫(c)

(Xn

)w dww.

There are two steps of note here: first, we can write D(s + w) in terms of itsseries representation since Re(s + w) = σ + c > σc, so we are in the region ofconvergence. Second, being able to switch the order of summation and integra-tion is not trivial. In this case we can do this because we have control of thesize of the tails courtesy of Lemma 30.2.3.

With those details out of the way, we are done: the resulting integral at theend can be evaluated using Lemma 30.2.1:

12πi

∫(c)

(Xn

)w dww

=

1, if X > n

0, if X < n.

Hence this truncates the series to

12πi

∫(c)D(s+ w)xw dw

w=∑n<X

anns

as desired.

Remark 30.2.5. Note that both Lemma 30.2.1 and Perron’s formula can beworked out in the case where y = 1 (or correspondingly X ∈ N). In this casethe integral in Lemma 30.2.1 evaluates to 1

2 (from plain calculation, no trickyintegration required), and consequently the resulting sum in Perron’s formulawould add only half of the final term if X ∈ N.

All by way of saying: requiring X 6∈ N simply makes the formula slightlycleaner looking; it isn’t a technical limitation.

REFERENCES 119

References[AG37] Lars V. Ahlfors and Helmut Grunsky. Über die blochsche konstante.

Mathematische Zeitschrift, 42(1):671–673, 1937. 95

[Apo10] Tom M. Apostol. Introduction to Analytic Number Theory. Under-graduate Texts in Mathematics. Springer Verlag, New York, 2010. 113

[CG96] Huaihui Chen and Paul M. Gauthier. On bloch’s constant. Journald’Analyse Mathématique, 69(1):275–291, 1996. 95

[Con78] John B. Conway. Functions of One Complex Variable, volume 11 ofGraduate Texts in Mathematics. Springer Verlag, New York, 2nd edi-tion, 1978. i

[Xio98] Chengji Xiong. Lower bound of bloch’s constant. Nanjing Daxue Xue-bao Shuxue Bannian Kan, 15(2):174–179, 1998. 95

120 INDEX

IndexAabscissa of absolute convergence 113abscissa of convergence . . . . . . . .114absolute convergence . . . . . . . . . . . 71absolute value . . . . . . . . . . . . . . . . . . . 1analytic continuation . . . . . . . . . .110analytic function . . . . . . . . . . . . . . . . 2argument principle . . . . . . . . . .36–37Arzelà–Ascoli theorem . . . . . . . . . 54automorphism . . . . . . . . . . . . . . . . . 40

BBlaschke product . . . . . . . . . . . . . . .79Bloch’s constant . . . . . . . . . . . . . . . 95Bloch’s theorem . . . . . . . . . . . . . . . . 92Borel–Carathéodory’s inequality 41

CCasorati–Weierstrass theorem . . 32Cauchy’s integral formula . . . . . . 20Cauchy’s residue theorem . . . . . . 33Cauchy’s theorem . . . . . . . . . . 11, 20Cauchy–Riemann equations . . . . . 4Cayley transform . . . . . . . . . . . . . . 42compactification . . . . . . . . . . . . . . . 58complex conjugate . . . . . . . . . . . . . . 1complex differentiability . . . . . . . . . 2conformal equivalence . . . . . . . . . . 64conformal mapping . . . . . . . . . . . . . . 5continuous functions . . . . . . . . . . . 51cross ratio . . . . . . . . . . . . . . . . . . . . . . .9

DDirichlet series . . . . . . . . . . . . . . . . 112

uniqueness theorem . . 115–116Dirichlet’s test . . . . . . . . . . . . . . . . 115

Eentire function . . . . . . . . . . . . . . . . . 14equicontinuity . . . . . . . . . . . . . . . . . .54exponent of convergence . . . . . . . .89

FFundamental theorem of algebra 15

Ggenus . . . . . . . . . . . . . . . . . . . . . . . . . . 82

finite . . . . . . . . . . . . . . . . . . . . . . 82Goursat’s theorem . . . . . . . . . . . . . 24

Great Picard theorem . . . . . . . . . 102

HHaar measure . . . . . . . . . . . . . . . . . . 43Hadamard three-circle theorem .46Hadamard three-lines theorem 43–

44Hadamard’s factorisation theorem

87Hardy’s theorem . . . . . . . . . . . . 46–47harmonic

conjugate . . . . . . . . . . . . . . . . . . . 4function . . . . . . . . . . . . . . . . . . . . 4

holomorphic functions . . . . . . . . . .52homotopic . . . . . . . . . . . . . . . . . . . . . 21Hurwitz’s theorem . . . . . . . . . . . . . 53hyperbolic measure . . . . . . . . . . . . 43

Iimaginary part . . . . . . . . . . . . . . . . . . 1infinite product . . . . . . . . . . . . . . . . 69

JJensen’s formula . . . . . . . . . . . . . . . 80

LLandau’s constant . . . . . . . . . . . . . .95Laplace operator . . . . . . . . . . . . . . . . 4Laurent expansion . . . . . . . . . . . . . 29Lebesgue’s number lemma . . . . . 37Lie group . . . . . . . . . . . . . . . . . . . . . . 43linear fractional transformation see

Möbius transformationLiouville’s theorem . . . . . . . . . . . . . 15Little Picard theorem . . . . . . . . . . 98locally bounded . . . . . . . . . . . . . . . . 55logarithmic derivative . . . . . . . . . . 22

Mmaximum modulus principle . . . 18,

38–39mean value property . . . . . . . . . . . 13meromorphic function . . . . . . 35, 58metric space . . . . . . . . . . . . . . . . . . . 54Mittag-Leffler’s theorem . . . . . . 108modulus . . . . . . . . see absolute valueMontel’s theorem . . . . . . . . . . . . . . 55Montel–Carathéodory theorem 102Morera’s theorem . . . . . . . . . . . 20–21Möbius transformation . . . . . . . . . . 6

INDEX 121

Nnormal . . . . . . . . . . . . . . . . . . . . . . . . . 54

Oopen mapping theorem . . . . . . . . . 24order . . . . . . . . . . . . . . . . . . . . . . . . . . .85

finite . . . . . . . . . . . . . . . . . . . . . . 85infinite . . . . . . . . . . . . . . . . . . . . 85

PPerron’s formula . . . . . . . . . 117–118Phragmén–Lindelöf principle . . . 47Picard’s theorem

great . . . . . . . .see Great Picardtheorem

little see Little Picard theoremspecial case . . . . . . . . . . . . . . . . 89

polar form . . . . . . . . . . . . . . . . . . . . . . 1pole . . . . . . . . . . . . . . . . . . . . . . . . . . . .28

order of . . . . . . . . . . . . . . . . . . . 28power series . . . . . . . . . . . . . . . . . . . . 12preserve angle . . . . . . . . . . . . . . . . . . . 5

Rrank . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

finite . . . . . . . . . . . . . . . . . . . . . . 82infinite . . . . . . . . . . . . . . . . . . . . 82

real part . . . . . . . . . . . . . . . . . . . . . . . . 1rectifiable . . . . . . . . . . . . . . . . . . . . . . 14region . . . . . . . . . . . . . . . . . . . . . . . . . .16residue . . . . . . . . . . . . . . . . . . . . . . . . . 33Riemann mapping theorem . . . . .64Riemann sphere . . . . . . . . . . . . . 7, 56Riemann zeta function . . . . 88, 112Rouché’s theorem . . . . . . . . . . . . . . 37Runge’s theorem . . . . . . . . . 105, 108

SSchottky’s theorem . . . . . . . . . . . . .99Schwarz lemma . . . . . . . . . . . . . . . . 39Schwarz reflection principle . . . 110Schwarz–Pick theorem . . . . . . . . . 41simply connected . . . . . . . . . . . . . . .21singular part . . . . . . . . . . . . . . . . . . 108singularity

essential . . . . . . . . . . . . . . . . . . . 28isolated . . . . . . . . . . . . . . . . . . . .26removable . . . . . . . . . . . . . . . . . 26

special linear group . . . . . . . . . . . . . 7projective . . . . . . . . . . . . . . . . . . .7

standard form . . . . . . . . . . . . . . . . . .82

Ttriangular path . . . . . . . . . . . . . . . . .20

Uupper half-plane . . . . . . . . . . . . . . . 42

VVinogradov notation . . . . . . . . . . . 83

WWeierstrass factorisation theorem

77Weierstrass theorem . . . . . . . . . . 104winding number . . . . . . . . . . . . . . . .19

Zzero set . . . . . . . . . . . . . . . . . . . . . . . . 16