ANALYTIC FUNCTIONS Contents - LTH · Analytic functions have an extreme mathematical beauty. They show many properties of general functions in a very pure way. There-fore one often

ANALYTIC FUNCTIONS

LP 3/4 2006, JORG SCHMELING

Contents

1. Introduction 32. Preliminaries 42.1. Numbers – the algebraic approach 42.2. Complex numbers 43. Complex functions 63.1. What is differentiability? 74. Harmonic functions 115. Steady–state heat conduction in dimension 2 145.1. Assumptions 145.2. Local equations 155.3. Relations to the temperature 175.4. A note on harmonic conjugates 185.5. Complexification of the heat equation 186. Fluid flow 197. Electrostatics 208. Complex integration 218.1. Path integrals in the real plane 218.2. Complex path integrals 228.3. Complex integration 239. The Cauchy Integral formula 269.1. Path deformation 269.2. The integral formula 299.3. Some Applications 3210. The Dirichlet problem 3510.1. The Dirichlet Problem for a Circle 3510.2. The Dirichlet Problem for a Half Plane 3711. Infinite series 3911.1. Power series 4212. More on Taylor series 4512.1. Methods for obtaining Taylor series expansions 4512.2. The method of partial fractions 4713. Laurent series 5014. Properties related to Taylor series 5415. Analytic continuation 5616. Poles and essential singularities 64

1

2 LP 3/4 2006, JORG SCHMELING

16.1. Poles 6416.2. The point at infinity 6516.3. More on Partial Fractions 6516.4. Essential singularities 6717. Infinite products 7317.1. General properties of infinite products 7317.2. Meromorphic functions 7818. Residues 8118.1. Finding the residues 8418.2. Residues in integral calculus I 8618.3. Residues in integral calculus II 8718.4. Residues in integral calculus III 8818.5. Residues in integral calculus IV 9018.6. Residue calculus in summation formulas 9219. Laplace transforms 9519.1. Stability 10119.2. Principle of the argument 10520. The principle of the argument II 10721. Conformal mappings I 10921.1. One–to–one mappings 11521.2. Mobius transformations 11721.3. Conformal mapping and boundary value problems 11922. Conformal mappings II 12022.1. Proof of the Riemann mapping theorem 123

ANALYTIC FUNCTIONS 3

The course material follows closely (but not completely) several chap-ters of the book ”Complex Variables with Applications” by A. DavidWunsch, Addison–Wesley 1994

1. Introduction

The statemant sometimes made, that there exist onlyanalytic functions in nature, is to my opinion absurd.

F. Klein, Lectures on Mathematics, 1893

The idea of an analytic function ... includes the wholewealth of functions most important to science, whetherthey have their origin in number theory, in the theory ofdifferential equations or of algebraic functional equations,whether they arise in geometry or in mathematical physics;and, therefore, in the entire realm of functions, the analyticfunction justly holds undisputed supremacy.

D. Hilbert, Mathematical Problems, 1900

Why do we study functions in complex variables?

• Not every quadratic equation has a solution in real numbers– H. Cardano ∼ -400 years: started to work with imaginary

roots– Euler : i =

√−1

– rigorous theory: Hamilton, Gauß

• Analytic functions have extremely elegant and nice properties– have derivatives of all orders– can be represented as Taylor series and are determined by

their derivatives– many useful transformations (Fourier, Laplace, ...) have all

the desired properties– their series expansion can be used to calculate explicit solu-

tions to many differential equations

• Analytic functions can be widely applied– real–valued integrals are sometimes easyily solvable by com-

plexification– Heat conduction– Fluid flows– Electrostatics

Analytic functions have an extreme mathematical beauty. Theyshow many properties of general functions in a very pure way. There-fore one often can use them as ”test objects”.

Moreover, analytic functions have a variety of natural propertieswhich make them the ideal objects for applications.


2. Preliminaries

2.1. Numbers – the algebraic approach. First there were the nat-ural numbers N. They had the disadvantage that the equation

m+ x = n

is not always solvable in natural numbers. Therefore the entire numbersZ where introduced. There the equation m+ x = n is always solvablebut the slightly more complicated equation

m · x = n

is not always solvable. This lead to the rational numbers Q. Herethe equation m · x = n is solvable. But still this was not satisfactory.There are several questions which do not have an answer in rationalnumbers. The ratio of the circumference of a circle with its radius (π)is not a rational number and so is not

√2. This made the real numbers

R appear. But if we turn to a similar equation

x2 = −2

then this has no real solution. This finally lead to the notion of complexnumbers C.

2.2. Complex numbers. Complex numbers are of the form

z = x+ iy x, y ∈ R

where i is the imaginary unit with the property that it is one solutionof

x2 + 1 = 0.

x := ℜz is called the real part of z and y := ℑz is the imaginary part.In this way the complex numbers can be identified with pairs of realnumbers (x, y), i.e. C ∼ R2. But it is important to notice that thecomplex numbers have some additional structure on the pairs of realnumbers induced by i2 = −1.

Numbers of the form z = x+ i0 are reals and z = 0 + iy are purelyimaginary. We also have the polar representation

z = reiθ = r(cos θ + i sin θ)

with r = |z| =√

x2 + y2 (the positive root) and θ = arctan yx

(seefigure 1)

Here we see that the argument θ is not uniquely defined since arctanis a periodic function with period 2π. We say the principal argument

is the one with 0 ≤ θ < 2π.We can define all elementary operations. We also can solve quadratic

equations

az2 + bz + c = 0


x

y

r

θ

Figure 1

with

z1,2 =−b±

√b2 − 4ac

2a.

Because if b2 − 4ac < 0 then(

±i√

4ac− b2)2

= b2 − 4ac

and

z1,2 =−b± i

√4ac− b2

2a.

Later on we will see that any polynomial has at least one complex root(Fundamental Theorem of Algebra).

We want to note that not all elementary functions are ”nice” forcomplex numbers. The expression zz21 is more delicate. In case z1 = ethis can be expressed in the following way with z2 = x2 + iy2:

ez2 = ex2 · eiy2 = ex2(cos y2 + i sin2).

For the general case we will start with the logarithm and define it asthe inverse of the exponential

z = elog z.

If we write z = reiθ then

log z = log r + iθ


is a solution. But we have that ei(θ+2kπ) = eiθ for k ∈ Z and hence thelogarithm is multi-valued (it has several branches)

log z = log r + i(θ + 2kπ).

In this sense even positive real numbers have infinitely many values ofits logarithm. But only one value is real!

Now we can definezz21 = ez2 log z1

as a multi-valued function.

Example 1.

912 = e

12(log 9+i2kπ)

= e12

log 9+ikπ = e12

log 9 · eikπ

= ±3

since eikπ takes only the values ±1.

9π = eπ(log 9+i2kπ) = eπ log 9 · ei2kπ2

= 995, 04 · · ·ei2kπ2

has infinitely many different values.

ii = ei log i = ei[i(π2+2kπ)]

= eπ2−2kπ

has infinitely many real values.

The expression zz21 has infinitely many values unless z2 is rational!

3. Complex functions

We will consider functions f : C → C. We can write them as

f(z) = f(x+ iy) = u(x, y) + iv(x, y).

Under the identification

x+ iy →(

xy

)

this becomes

f(x+ iy) →(

u(x, y)v(x, y)

)


and the map f sends

(

xy

)

to

(

u(x, y)v(x, y)

)

. Hence it can be regarded

as a map from R2 to R2 and we can use higher-dimensional analysis.But we are interested in the special properties that come from thecomplex structure.

3.1. What is differentiability? Roughly speaking in a linear space amap f is differentiable at a point x0 if there is a unique ”approximation”of f at x0 by a linear map:

f(x0 + h) − f(x0) ∼ Ah

where A is a linear map. Now the difference of the differentiability inR2 and C comes because these spaces have different linear maps. InR2 any 2× 2–matrix gives a linear map while in C the linear maps arethe multiplication with a complex number a = a1 + ia2.

First we want to consider the linear (real) approximation of the com-plex map f interpreted as a map of R2. There we have

Df = D

(

u(x, y)v(x, y)

)

=

( ∂u∂x

∂u∂y

∂v∂x

∂v∂y

)

.

Now this should correspond to a multiplication of a complex number

az = (a1 + ia2)(x+ iy) = (a1x− a2y) + i(a2x+ a1y)

or

az →(

a1x− a2ya2x+ a1y

)

=

(

a1 −a2

a2 a1

)(

xy

)

.

Therefore we get

a1 =∂u

∂x, −a2 =

∂u

∂y, a2 =

∂v

∂x, a1 =

∂v

∂y

or∂u

∂x=∂v

∂y

∂u

∂y= −∂v

∂x. (1)

These equations are called the Cauchy–Riemann equations (C–R).

The considerations above were a bit ”heuristic”. We will make itmore precise:

Definition 1. We say a complex function is differentiable at z0 if thelimit

f ′(z0) = lim∆z→0

f(z0 + ∆z) − f(z0)

∆zexists. Here ∆z = ∆x+ i∆y.


Remark 1. The above definition is equivalent to

lim∆z→0

|f(z0 + ∆z) − f(z0) − f ′(z0)∆z||∆z| = 0.

This definition means that the above limit exists for each directionwe approach the point in the complex plane and is independent of thedirection. Let us choose the two canonical directions ”horizontal” ∆xand ”vertical” i∆y:

f ′(z0) = lim∆x→0

u(x0 + ∆x, y0) − u(x0, y0)

∆x+ i

v(x0 + ∆x, y0) − v(x0, y0)

∆x

=

(

∂u

∂x+ i

∂v

∂x

)

∣

∣

(x0,y0)

and

f ′(z0) = lim∆y→0

u(x0, y0 + ∆y) − u(x0, y0)

i∆y+ i

v(x0, y0 + ∆y) − v(x0, y0)

i∆y

=

(

−i∂u∂y

+∂v

∂y

)

∣

∣

(x0,y0)=

(

∂v

∂y+ i

(

−∂u∂y

))

∣

∣

(x0,y0)

So we get the Cauchy–Riemann equations (1) as a necessary condition.

Remark 2. If one uses higher-dimensional real analysis one knows thatthe continuity of the partial derivatives ∂u

∂x, ∂u∂y

, ∂v∂x

and ∂v∂y

ensures that

the directional derivatives are determined by the these partial deriva-tives. This allows to conclude the revers: If the partial derivatives arecontinuous then the Cauchy–Riemann equations (1) are necessary andsufficient for complex differentiability.

More precisely, the continuity of the partial derivatives implies theexistence for the directional derivatives for u and v:

For all (∆x,∆y) ∈ R2 we have

lim|(∆x,∆y)|→0

∣

∣

∣u(x0 + ∆x, y0 + ∆y) − u(x0, y0) − ∂u

∂x∆x− ∂u

∂y∆y∣

∣

∣

|(∆x,∆y)| = 0.

and

lim|(∆x,∆y)|→0

∣

∣

∣v(x0 + ∆x, y0 + ∆y) − v(x0, y0) − ∂v

∂x∆x− ∂v

∂y∆y∣

∣

∣

|(∆x,∆y)| = 0.


where |(∆x,∆y)| :=√

(∆x)2 + (∆y)2 = |∆x + i∆y|. Multiplying thesecond equality with i and adding both equations we obtain

0 = lim|(∆x,∆y)|→0

1

|(∆x,∆y)| ×

×∣

∣

∣u(x0 + ∆x, y0 + ∆y) − u(x0, y0) −

∂u

∂x∆x− ∂u

∂y∆y+

+ i

(

v(x0 + ∆x, y0 + ∆y) − v(x0, y0) −∂v

∂x∆x− ∂v

∂y∆y

)

∣

∣

∣

= lim|(∆x,∆y)|→0

1

|(∆x,∆y)|∣

∣

∣(u+ iv)(x0 + ∆x, y0 + ∆y) − (u+ iv)(x0, y0)

−(

∂u

∂x∆x+

∂u

∂y∆y + i

∂v

∂x∆x+ i

∂v

∂y∆y

)

∣

∣

∣

= lim|(∆x,∆y)|→0

1

|(∆x,∆y)|∣

∣

∣(u+ iv)(x0 + ∆x, y0 + ∆y) − (u+ iv)(x0, y0)

−(

∂u

∂x∆x− ∂v

∂x∆y + i

∂v

∂x∆x+ i

∂u

∂x∆y

)

∣

∣

∣

= lim|(∆x,∆y)|→0

1

|(∆x,∆y)|∣

∣

∣(u+ iv)(x0 + ∆x, y0 + ∆y) − (u+ iv)(x0, y0)

−(

∂u

∂x+ i

∂v

∂x

)

(∆x+ i∆y)∣

∣

∣

This is the definition of the derivative and also shows that f ′(z0) =(

∂u∂x

+ i∂v∂x

)

∣

∣

∣

(x0,y0).

Remark 3. We have∂f

∂x=∂u

∂x+ i

∂v

∂x

and∂f

∂y=∂u

∂y+ i

∂v

∂y


Using the C–R equations (1) we obtain

∂f

∂y= −∂v

∂x+ i

∂u

∂x= i

(

∂u

∂x+ i

∂v

∂x

)

= i∂f

∂x. (2)

This is another way of writing the C–R equations (1).

Remark 4. There is also another way of writing the C–R equations (1).From the formula of complete differentials we get

df =∂f

∂zdz +

∂f

∂zdz =

∂f

∂xdx+

∂f

∂ydy.

Moreover

x =1

2(z + z) y =

1

2i(z − z)

what implies

dx =1

2(dz + dz) dy =

1

2i(dz − dz).

Therefore

df =∂f

∂x

1

2(dz + dz) +

∂f

∂y

1

2i(dz − dz)

=

(

1

2

∂f

∂x+

1

2i

∂f

∂y

)

dz +

(

1

2

∂f

∂x− 1

2i

∂f

∂y

)

dz

Hence1

2

(

∂f

∂x+

1

i

∂f

∂y

)

=∂f

∂z

and1

2

(

∂f

∂x− 1

i

∂f

∂y

)

=∂f

∂z.

Together with the previous remark we get the C–R equations (1) in theform

∂f

∂z= 0 (3)

Definition 2. A function f is called analytic in an open domain Ω iff ′(z) exists for all z ∈ Ω and f ′ is continuous in Ω.

Example 2. Let

f(z) = f(x+ iy) = x2 + iy2

i.e. u(x, y) = x2 and v(x, y) = y2. The C-R equations (1) give

∂u

∂x= 2x = 2y =

∂v

∂y

∂v

∂x= 0 = −∂u

∂y.


Hence the C-R equations (1) hold only on the line x = y (or z = x+ix)and the function is differentiable there. Since the line is not an opendomain this function is not analytic.

Remark 5. As in the case of real analysis the sum, the difference, theproduct and the quotient (provided the denominator function g(z) 6= 0on Ω) of functions analytic in Ω are analytic in Ω.

Definition 3. A function that is analytic on the whole complex plane,i.e. Ω = C is called an entire function.

Example 3. Let us consider the exponential function

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

and u = ex cos y and v = ex sin y. Then

∂u

∂x= ex cos y

∂v

∂y= ex cos y

∂u

∂y= −ex sin y

∂v

∂x= ex sin y

and the C–R equations (1) are fulfilled. Hence ez is an entire function.Using that f ′(x+ iy) =

(

∂u∂x

+ i∂v∂x

)

we can evaluate the derivative:

d

dzez =

∂

∂xex cos y + i

∂

∂xex sin y = ex cos y + iex sin y = ez.

4. Harmonic functions

Assume that u and v are twice differentiable with continuous secondderivatives. Then we can differentiate the C–R equations (1).

∂2u

∂x2=

∂2v

∂x∂y

∂2u

∂y2= − ∂2v

∂y∂xand

∂2v

∂x2= − ∂2u

∂x∂y

∂2v

∂y2=

∂2u

∂y∂x

Since the second derivatives are continuous we have ∂2v∂x∂y

= ∂2v∂y∂x

and∂2u∂x∂y

= ∂2u∂y∂x

and therefore

∂2u

∂x2+∂2u

∂y2= 0

and∂2v

∂x2+∂2v

∂y2= 0.


the latter equations are called Laplace equations. A function fulfill-infg the Laplace equation in a domain Ω is called harmonic. Thereforethe real and the imaginary part of a (twice continuously differentiable)analytic function are harmonic functions.

Theorem 1. For any real harmonic function u(x, y) in a simply con-nected domain there is a harmonic function v(x, y) called the har-

monic conjugate of u such that f = u + iv is an analytic function.The harmonic conjugate is determined up to a constant.

Sketch of the proof. Let v be a differentiable function. Its completedifferential is

dv =∂v

∂xdx+

∂v

∂ydy

v is the harmonic conjugate if and only if it fulfills the C–R equa-tions (1). Therefore

dv = −∂u∂ydx+

∂u

∂xdy

From higher-dimensional analysis we know that a differential equationin a simply connected domain of the form

a(x, y)dx+ b(x, y)dy = df

has a solution if and only if

∂a

∂y=∂b

∂x. (4)

This is necessary because

df =∂f

∂xdx+

∂f

∂ydy

and∂2f

∂y∂x=

∂2f

∂x∂y.

In our case the latter equation means

∂

∂y

(

−∂u∂y

)

=∂

∂x

(

∂u

∂x

)

Hence a necessary and sufficient condition is

∂2u

∂x2+∂2u

∂y2= 0.

To see why condition (4) should hold we try to solve the equation:

adx+ bdy = df =∂f

∂xdx+

∂f

∂ydy.

This gives

a =∂f

∂x.


Hence∫

Ω

a dx+ c(y) = f(x, y).

On the other hand

b =∂

∂y

∫

Ω

a dx+ c′(y) =

∫

Ω

∂a

∂ydx+ c′(y)

Differentiating this with respect to x we get

∂a

∂y=∂b

∂x.

If the last equation is fulfilled we have

b =

∫

Ω

∂b

∂xdx+ c′(y) = b+ d(x) + c′(x)

and we can solve the ordinary differential equation

c′(x) = −d(x).

Remark 6. Any C2 solution of the Laplace equation gives rise to ananalytic function. As we will see later this implies that those solutionsare infinitely often differentiable!

Example 4. Let u(x, y) = x3 − 3xy2 + 2y. Then

∂2u

∂x2= −∂

2u

∂y2= 6x.

The C–R equations (1) are

∂u

∂x= 3x2 − 3y2 =

∂v

∂y

∂u

∂y= 2 − 6xy = −∂v

∂xIntegrating the first equation we get

v(x, y) =

∫

(3x2 − 3y2) dy = 3x2y − y3 + c(x)

Now we evaluate c(x). For this we differentiate the latter expressionwith respect to x:

6xy + c′(x) =∂v

∂x= 6xy − 2.

The harmonicity of u ensures that the terms depending on y cancel.I.e. we can solve the differential equation c′(x) = −2 up to a constant.So c(x) = −2x+ d and

v(x, y) = 3x2y − y3 − 2x+ d.


5. Steady–state heat conduction in dimension 2

5.1. Assumptions. We have a domain Ω ⊂ R2 which models a plate.At the boundary are some spots which serve as heat sources and somespots which serve as heat absorbers. For conduction the time rate of theheat energy flow can be specified by a vector Q in R2 (see figure 2. Itis turned inwards at the sources and outwards at the sinks. Otherwiseit is tangent to the boundary. We have no variation of the vector withtime (steady–state assumption).

Ω

y

x

Q(x,y)

Figure 2

We write

Q = Qxe1 +Qye2 e1 = (1, 0) e2 = (0, 1).

The direction of Q is the direction in what the heat is most rapidelytransported.

Given an infinitisimal straight line segment dl the heat flux f throughit is the thermal energy through it per unit time. Therefore

df = Qndl

where Qn is the component of Q normal to dl.To obatin the general heat flow through a non–straight boundary

one has to integrate this differential.


5.2. Local equations. Since the temperature in a conducting mate-rial is independent of time (steady–state condition) the heat flux intoa domain must be zero. In particular, by Stokes–Green equation

0 =

∫

Γ

Qn dl =

∫

Ω(Γ)

(

∂Qx

∂x+∂Qy

∂y

)

dxdy

for all (sufficiently small) closed curves Γ around (x0, y0) and whereΩ(Γ) is the domain bounded by Γ. Hence,

∂Qx

∂x+∂Qy

∂y= 0 (5)

at any point inside Ω.A heuristic explanation is to choose Γ to be the curve from figure 3.

Γ

R

T

Figure 3

We divide the domain Ω bounded by the curve Γ into small piecesand integrate separately.

We will first investigate the ”inner squares” as in figure 4


(x ,y )0 0

(ε,−ε)

(ε,ε)(−ε,ε)

(−ε,−ε)

Q

n

n

n

n

Q =Qn x

R

R

R

R

δ

δ

δ

4

3Rδ

2

1

Figure 4

First we note that the normal component is given by Qn dl = Qy dx−Qx dy. Then under some smoothness assumptions:∫

R

(

∂Qx

∂x+∂Qy

∂y

)

dxdy

=

∫ ǫ

−ǫ

∫ ǫ

−ǫ

∂Qx

∂xdxdy +

∫ ǫ

−ǫ

∫ ǫ

−ǫ

∂Qy

∂ydxdy

=

∫ ǫ

−ǫ(Qx(ǫ, y) −Qx(−ǫ, y)) dy +

∫ ǫ

−ǫ(Qy(x, ǫ) −Qy(x,−ǫ)) dx

=

∫ ǫ

−ǫQx(ǫ, y) dy +

∫ ǫ

−ǫQy(x, ǫ) dx

−∫ ǫ

−ǫQx(−ǫ, y) dy −

∫ ǫ

−ǫQy(x,−ǫ) dx

= −∫

∂R1

Qx(ǫ, y) dy +

∫

∂R4

Qy(x, ǫ) dx

−∫

∂R3

Qx(−ǫ, y) dy +

∫

∂R2

Qy(x,−ǫ) dx

=

∫

∂R

Qy dx−Qx dy =

∫

∂R

Qn dl.


Therefore we see that these line integrals on the ”inner edges” ofthe grid cancel each other since they are evaluated twice with oppositeorientation (direction). It only remains the integral along the ”outeredges” of the rectangular grid. There we consider a triangle T as infigure 5

a

b

n

T

Γab

Figure 5

If the grid is small we can assume that Q = (Qx, Qy) is constant onthe triangle. Hence∫

a

Qy dx−Qx dy +

∫

b

Qy dx−Qx dy = −∫

a

Qx dy +

∫

b

Qy dx

= |a|Qx + |b|Qy =|a|Qx + |b|Qy√

|a|2 + |b|2|(|b|,−|a|)| = Qn|(|b|,−|a|)|

=

∫

Γab

Qn dl

Here we used that the orthogonal component to the boundary vector

a + b = (|b|,−|a|) is given by Qn = |a|Qx+|b|Qy√|a|2+|b|2

. Summing over all the

small pieces we derive Stokes’–Green’s formula.

5.3. Relations to the temperature. The rate of heat conduction isrelated to the temperature differences occuring in the material and tothe distances over which these differences occur. If φ = φ(x, y) denotesthe temperature then this relation is expressed by

Qx(x, y) = −k∂φ∂x

(x, y) (6)

Qy(x, y) = −k∂φ∂y

(x, y). (7)


Here k is a constant called the thermal conductivity. These formulasmean that ”Q is minus k times the gradient of the temperature”.

Now (5) becomes

−k∂2φ

∂x2(x, y) − k

∂2φ

∂y2(x, y) = 0

or∂φ

∂2x2(x, y) +

∂2φ

∂y2(x, y) = 0. (8)

Therefore the temperature is a harmonic function.

5.4. A note on harmonic conjugates.

Theorem 2. Let f = u + iv be analytic and Cj, Kl, j, l ∈ N be realconstants. Then the family of curves (in R2) u = Cj is orthogonal tothe family v = Kl.

Proof. Fix j, l ∈ N. On the curve u = Cj the funcion u is (by definition)constant. Hence,

du =∂u

∂xdx+

∂u

∂ydy = 0.

Let (x0, y0) be a point of intersection with v = Kl. Then, writing thecurve u = Cj as y(x) we can find its slope at (x0, y0) by

dy

dx|(x0,y0) =

(

−∂u∂x∂u∂y

)

(x0,y0)

.

Similarly, the slope of the curve v = Kl is

dy

dx|(x0,y0) =

(

−∂v∂x∂v∂y

)

(x0,y0)

=

(

∂u∂y

∂u∂x

)

(x0,y0)

where we used the C–R equations in the last equality. Hence, theslopes of the two curves are negative reciprocals and the curves areorthogonal.

5.5. Complexification of the heat equation. We can find (up toa constant) the harmonic conjugate ψ to the temperature φ, i.e. thefunction

Φ(x, y) = φ(x, y) + iψ(x, y)

is analytic. Its imaginary part is called the stream function. The linesψ = const are called stream lines and the curves φ = const isothermes.These curves are orthogonal. We have at the stream lines that

dy

dx=

∂φ∂y

∂φ∂x

=Qy

Qx

We conclude


Theorem 3. The heat flux density vector is tangent to the streamlineand orthogonal to the isotherme.

The complex heat flux density is defined as

q(z) = Qx(x, y) + iQy(x, y).

Note that the associated complex vector is exactly the heat flux densityvector. We can rewrite (6)and (7) as

q = −k(

∂φ

∂x+ i

∂φ

∂x

)

.

We know thatdΦ

dz=∂φ

∂x+ i

∂ψ

∂x=∂φ

∂x− i

∂φ

∂y.

where we used the C–R equations for the last equality (the first isjust differentiating along the horizontal axis). Hence, we obtain thefollowing simple formula

q = −k(

dΦ

dz

)

.

6. Fluid flow

We assume that we have an ideal fluid, i.e. it is incompressible and nolosses to internal friction. Moreover the velocity of flow is independendon time. The fluid is 2–dimensional. The fluid velocity V is a vectorfieldwith coordinates Vx and Vy (analogously to Q). We define the complexvelocity as

v = Vx(x, y) + iVy(x, y).

Under certain physical assumptions (no ”whirlpools”) the velocity isgoverned by a potential

Vx =∂φ

∂xVy =

∂φ

∂y.

As in the heat conduction we have a conversation law (”as much fluidflows in as flows out”)

∂2φ

∂x2+∂2φ

∂y2= 0

We also have the (harmonic conjugate) stream function ψ such that

Φ = φ+ iψ

is analytic. AgainThe velocity vector is tangent to the streamlineSo the path of a droplet is a streamline and is perpendicular to the

equipotential line. We also have

v =

(

dΦ

dz

)

.


7. Electrostatics

We consider a stationary (non–moving) electrical charge which playsthe role of sources or sinks. The electric flux is described by the electricflux density vector D. We can measure it by putting a test charge ofq0 Coulombs at a point. It will experience a force due to the sinks andsources. The vector force F is given by

F = q0D

ǫwhere ǫ is a constant called permittivity. We can write

D = ǫE andF

q0= E

where E is the electrical field vector.Again we consider a 2–dimensional model (all charges, sources, sinks,

obstacles are distributed constantly on the infinite lines perpendicularto the x−y plane). From Maxwell’s equations one can obtain that Dx,Dy are created by a electrostatic potential φ.

We get

Dx = −ǫ∂φ∂x

Dy = −ǫ∂φ∂y.

or

Ex = −∂φ∂x

Ey = −∂φ∂y.

According to Maxwells first equation (which is in our terms ∂Ex∂x

+ ∂Ey∂y

=

0) the total electric flux through a volume containing no charge is zero.Hence, we have the conservation law, yielding

∂2φ

∂x2+∂2φ

∂y2= 0

and we can define a stream function ψ. The stream lines show thedirections of the electric flux.

Finally,

d = −ǫ(

dΦ

dz

)

e = −(

dΦ

dz

)

whered = Dx + iDy e = Ex + iEy.


8. Complex integration

In the real analysis the notion of integrability is ”weaker” than dif-ferentiability. If the function F (x) is differentiable and f(x) is its deriv-ative then F (x) =

∫

f(x) dx. On the other hand not every integrablefunction is differentable (while every differentiable and bounded func-tion is integrable). We will see that in the complex plane the situationis different. A complex function is analytic iff it is (complex) integrable.This seems to be surprizing since the notion of complex differentiabil-ity is stronger than real differentiability. But it carries out that theappropriate notion of complex integrability is also stronger than realintegrability.

8.1. Path integrals in the real plane. Let Γ be a curve (smooth)in R2 joining the points p and q. We can approximate this curve by apolygon (see figure 6)

x

y

si

Figure 6

The polygonsegments si, i = 1, 2, · · ·n have length ∆si. Let xi, yi bearbitrary points on si. The (real) path integral of a function f : R2 → C

is defined as

∫

Γ

f(x, y) ds = limn→∞

n∑

k=1

f(xk, yk)∆sk.


Similarly we can define

∫

Γ

f(x, y) dx = limn→∞

n∑

k=1

f(xk, yk)∆xk. (9)

and∫

Γ

f(x, y) dy = limn→∞

n∑

k=1

f(xk, yk)∆yk. (10)

where ∆xk and ∆yk are the projections of ∆sk onto the x− and y−axes, respectively.

We can compute those integrals by parametrizing the curve Γ. Wewrite Γ as (x, y(x)), x ∈ [a, b] and get

∫

Γ

f(x, y) ds =

∫ b

a

f(x, y(x))

√

1 +

(

dy

dx

)2

dx.

This does not depend on the parametrization. If x = x(t), t ∈ [0, 1] wehave

∫ b

a

f(x, y(x))

√

1 +

(

dy

dx

)2

dx

=

∫ 1

0

f(x(t), y(x(t)))

√

1 +

(

dy(t)

dx(t)

)2dx(t)

dtdt

=

∫ 1

0

f(x(t), y(x(t)))

√

(

dx(t)

dt

)2

+

(

dy(t)

dx(t)· dx(t)dt

)2

dt

=

∫ 1

0

f(x(t), y(x(t)))

√

(

dx(t)

dt

)2

+

(

dy(t)

dt

)2

dt

=

∫

Γ

f(x, y) ds.

8.2. Complex path integrals. We write

∆zk = ∆xk + i∆yk

and∫

Γ

f(z) dz = limn→∞

n∑

k=1

f(zk)∆zk.


Using (9) and (10) we can derive∫

Γ

f(z) dz =

∫

Γ

(u(x, y) + iv(x, y)) (dx+ idy)

=

∫

Γ

u(x, y) dx−∫

Γ

v(x, y) dy + i

(∫

Γ

v(x, y) dx+

∫

Γ

u(x, v)) dy

)

.

Example 5. Let Γ be a curve joining z1 and z2 parametrized by z =z(s); 0 ≤ s ≤ L. Then for n ≥ 0

∫

Γ

zn dz =

∫ L

0

z(s)ndz(s)

dsds

=

∫ L

0

1

n + 1

d[z(s)]n+1

dsds

=zn+12 − zn+1

1

n + 1.

We note that this integral does not depend on the curve Γ!

Theorem 4. Let maxΓ |f(z)| ≤M and

|Γ| =

∫

Γ

|dz| = limn→∞

n∑

k=1

|∆zk| ≤ L

. Then∣

∣

∣

∣

∫

Γ

f(z) dz

∣

∣

∣

∣

≤ ML.

Proof. We have∣

∣

∣

∣

∣

n∑

k=1

f(zk)∆zk

∣

∣

∣

∣

∣

≤n∑

k=1

|f(zk)||∆zk|

≤Mn∑

k=1

|∆zk|

≤ML.

8.3. Complex integration.

Definition 4. We say a function f(z) is complex integrable in a do-main Ω if for any simple closed curve Γ whose interior is in Ω theintegral

∫

Γ

f(z) dz = 0.

Remark 7. Let p, q be two points in Ω. Then the integral∫ q

pf(z) dz

does not depend on the curve (in Ω) joining them if f is integrable.


Theorem 5 (Cauchy–Goursat). Let Γ be a simple closed curve and fbe analytic in the domain bounded by Γ and on Γ itself. Then

∫

Γ

f(z) dz = 0.

Proof. We have∫

Γ

f(z) dz

=

∫

Γ

u(x, y) dx−∫

Γ

v(x, y) dy + i

(∫

Γ

v(x, y) dx+

∫

Γ

u(x, v) dy

)

.

We rewrite this equation with the help of Green’s formula:∫

Γ

u(x, y) dx−∫

Γ

v(x, y) dy = −∫ ∫

Ω

(

∂v

∂x+∂u

∂y

)

dxdy

and∫

Γ

u(x, y) dy +

∫

Γ

v(x, y) dx = −∫ ∫

Ω

(

−∂u∂x

+∂v

∂y

)

dxdy.

The Cauchy–Riemann equations imply that both integrals on the right–hand–side vanish.

Remark 8. Analytic functions are integrable and the integral does notdepend on the curve.

Theorem 6. Let F (z) be analytic and dFdz

= f . Then∫ z2

z1

f(z) dz = F (z2) − F (z1).

Proof. Let a parametrization z = z(t); z1 = z(0), z2 = z(1), 0 ≤ t ≤ 1be chosen.

∫ z2

z1

f(z) dz =

∫ 1

0

f(z(t))dz

dtdt

=

∫ 1

0

dF

dz

dz

dtdt

=

∫ 1

0

dF

dtdt

= F [z(1)] − F [z(1)] = F (z2) − F (z1).

We can also proof the revers:

Theorem 7 (Fundamental Theorem of Calculus). Let f be analytic ina connected domain. Then the integral

∫ z

af(w) dw defines an analytic

function satisfyingd

dz

∫ z

a

f(w) dw = f(z).


Proof. By theorem 5 we have that∫

Γ

f(z) dz = 0

for any simple closed curve Γ. Hence, F (z) =∫ z

af(w) dw is well defined

and does not depend on the curve joining a and z. Then we can proceed

F (z1 + h) − F (z1) =

∫ z1+h

a

f(w) dw−∫ z1

a

f(w) dw

=

∫ z1+h

z1

f(w) dw

Let |h| << 1 and define z(t) = z1 + th; 0 ≤ t ≤ 1. Let ǫ = ǫ(h) > 0and h sufficiently small that

|f(z1 + h) − f(z1)| = ǫ.

Then limh→0 ǫ(h) = 0 and

F (z1 + h) − F (z1)

h− f(z1) =

1

h

∫ z1+h

z1

[f(z) − f(z1)] dz.

Hence,∣

∣

∣

∣

F (z1 + h) − F (z1)

h− f(z1)

∣

∣

∣

∣

≤ 1

|h|ǫ(h)|h| → 0 (|h| → 0).


9. The Cauchy Integral formula

9.1. Path deformation.

Definition 5. A continuous deformation of a simple closed curve Γ1

into a simple closed curve Γ2 is a continuous family of diffeomorphisms(i.e. differentiable one–to–one maps whose inverses are also differen-tiable) Ft : S1 → Ft(S

1) ⊂ C; 0 ≤ t ≤ 1 such that F0(S1) = Γ1 and

F1(S1) = Γ2 (here we consider the orientations on the curves inheritet

from the orientation chosen on S1).

Theorem 8. The integrals of an analytic function f(z) along two sim-ple closed curves will be identical if one can be continuously deformedinto the other without passing any point where the function is not an-alytic.

Proof. The proof will be in two steps.First we proof that if Γ is a simple closed curve and C is a circle

inside Ω(Γ) such that f is analytic in the annulus bounded by Γ and Cthe integral along Γ is the same as the integral along C evaluated withthe same orientation (see figure 7).

Γ

C

Figure 7

To see this we introduce a cut L from Γ to C which will be giventwo opposite orientations L+ and L− (see figure 8).


Γ

C

L

L−

+

−

Figure 8

Then for

0 =

∫

Γ∪L+∪(−C)∪L−

f(z) dz

=

∫

Γ

f(z) dz +

∫

L+

f(z) dz +

∫

(−C)

f(z) dz +

∫

L−

f(z) dz

=

∫

Γ

f(z) dz +

∫

(−C)

f(z) dz

=

∫

Γ

f(z) dz −∫

C

f(z) dz

The second step: Since Γ1 can be deformed into Γ2 we can find simpleclosed curves Ci = Fti(S

1); i = 0, · · · , n; C0 = Γ1, Cn = Γ2 which are”overlapping” and ”connecting” the two curves. Moreover we can findcircles Si ⊂ Ci−1 ∩ Ci such that the points where f is not analytic lieinside these circles or ”outside” all of the curves Ci (see 9).

Then we get∫

Γ1

f(z) dz =

∫

S1

f(z) dz = · · · =

∫

Sn−1

f(z) dz =

∫

Γ2

f(z) dz


Γ

Γ

C

C

C

S

S

2

1

3

4

1

1

3

2

S2

S

Figure 9

Example 6. We want to evaluate the integral∫

|z|=1

zn dz n ∈ Z.

First we observe that by example 5∫

|z|=1

zn dz = 0 n ≥ 0.

Now let n ∈ Z. Since we integrate over |z| = 1 we can write z = eiθ

and get for n 6= −1:

∫

|z|=1

zn dz =

∫ 2π

0

einθ deiθ =

∫ 2π

0

einθ · i · eiθ dθ

= i

∫ 2π

0

ei(n+1)θ dθ =i

i(n + 1)

(

e2πi(n+1) − 1)

= 0.


For n = −1 we conclude∫

|z|=1

dz

z= i

∫ 2π

0

e0 dθ = i

∫ 2π

0

dθ = i · 2π.

More general for z0 ∈ C and r > 0∫

|z−z0|=r(z − z0)

n dz =

∫

|z−z0|=1

(z − z0)n dz =

0 n 6= −1

2πi n = −1(11)

where we used that the circle |z− z0| = r can be continuously deformedinto the circle |z − z0| = 1 without hitting a singularity (i.e. a point ofnon–analyticity).

9.2. The integral formula. We consider a simple closed curve Γ.Cauchy derived a remarkable fact. Suppose that f is analytic in thedomain Ω bounded by Γ. Then we can determine any value of f insideΩ by knowing only the values on Γ, i.e. f is completely determined bythe ”boundary values”.

Let z0 ∈ Ω and r be so small that the circle Cr0 centred at z0 and

of radius r is entirely contained in Ω. Then the deformation theoremimplies

∫

Γ

f(z)

z − z0dz =

∫

Cr0

f(z)

z − z0dz

since f(z)z−z0 is analytic in the region bounded by Γ and Cr

0 .We recall that

∫

Cr0

1

z − z0dz = 2πi.

(We proved that∫

Cr1zdz = 2πi where Cr is a circle centred at 0. The

above formula results from a linear change of variables.) Thus

2πif(z0) = f(z0)

∫

Cr0

1

z − z0dz =

∫

Cr0

f(z0)

z − z0dz.

Hence,∫

Cr0

f(z)

z − z0dz − 2πif(z0) =

∫

Cr0

f(z) − f(z0)

z − z0dz.

We will estimate this difference with the help of the ML–inequality(Theorem 4).

Let ǫ > 0 be fixed. By the continuity of f there is a radius r0 suchthat |f(z) − f(z0)| < ǫ for |z − z0| < r0. If we chose 0 < r < r0 wederive

∣

∣

∣

∣

f(z) − f(z0)

z − z0

∣

∣

∣

∣

<ǫ

r

on Cr0 . On the other hand the length of Cr

0 is 2πr. Therefore∣

∣

∣

∣

∣

∫

Cr0

f(z) − f(z0)

z − z0dz

∣

∣

∣

∣

∣

≤ ǫ

r2πr = 2πǫ.


Since ǫ was chosen arbitrary and the value of the integral does notdepend on the radius r we have that

∫

Cr0

f(z) − f(z0)

z − z0dz = 0.

So we proved:

Theorem 9 (Cauchy Integral Formula). Let f be analytic on Γ∪Ω(Γ)where Γ is a simple closed curve. If z0 ∈ Ω then

f(z0) =1

2πi

∫

Γ

f(z)

z − z0dz.

This formula has wide applications. With some manipulations wecan even show the existence of all derivatives and express them asintegrals along the simple closed curve Γ. Let us regard f(z0) as afunction of the variable z0. Assuming Leibnitz’s rule for integrals alongcurves we can proceed

d

dz0f(z0) =

d

dz0

1

2πi

∫

Γ

f(z)

z − z0dz =

1

2πi

∫

Γ

f(z)d

dz0

(

1

z − z0

)

dz

=1

2πi

∫

Γ

f(z)

(z − z0)2dz.

Continuing we get

f (2)(z0) =1

2πi

∫

Γ

f(z)d2

dz20

(

1

z − z0

)

dz =2

2πi

∫

Γ

f(z)

(z − z0)3dz

or

f (n)(z0) =1

2πi

∫

Γ

f(z)dn

dzn0

(

1

z − z0

)

dz =n!

2πi

∫

Γ

f(z)

(z − z0)n+1dz.


All this can be done in a rigorous way. Lets do it for the first derivative:

f ′(z0) = lim∆z0→0

f(z0 + ∆z0) − f(z0)

∆z0

= lim∆z0→0

1

∆z0

[

1

2πi

∫

Γ

f(z)

z − (z0 + ∆z0)dz − 1

2πi

∫

Γ

f(z)

z − z0dz

]

= lim∆z0→0

1

2πi∆z0

∫

Γ

[

1

z − (z0 + ∆z0)− 1

z − z0

]

f(z) dz

= lim∆z0→0

1

2πi∆z0

∫

Γ

∆z0[z − (z0 + ∆z0)](z − z0)

· f(z) dz

= lim∆z0→0

1

2πi

∫

Γ

f(z)

[z − (z0 + ∆z0)](z − z0)dz.

Now let b = d(Γ, z0), m = maxΓ |f(z)| and |∆z0| ≤ b/2. Then∣

∣

∣

∣

f ′(z0) −1

2πi

∫

Γ

f(z)

(z − z0)2dz

∣

∣

∣

∣

=

∣

∣

∣

∣

lim∆z0→0

1

2πi

∫

Γ

f(z)

[z − (z0 + ∆z0)](z − z0)dz − 1

2πi

∫

Γ

f(z)

(z − z0)2dz

∣

∣

∣

∣

= lim∆z0→0

1

2π

∣

∣

∣

∣

∫

Γ

f(z)

[z − (z0 + ∆z0)](z − z0)dz −

∫

Γ

f(z)

(z − z0)2dz

∣

∣

∣

∣

= lim∆z0→0

1

2π

∣

∣

∣

∣

∫

Γ

f(z)

(z − z0)2

(

∆z0z − (z0 + ∆z0)

)

dz

∣

∣

∣

∣

≤ lim∆z0→0

1

2πm

1

b21

b− (b/2)|∆z0|L = 0

where L is the length of Γ. Hence we derived

f ′(z0) =1

2πi

∫

Γ

f(z)

(z − z0)2dz.

One can use a similar procedure to derive the formula for f ′′(z0). Butthen f ′(z) itself is differentable (and hence analytic) in the same do-main. Now we can repeat the same arguments for f ′ instead of f andso forth. Hence we get the existence of the derivatives of all order:


Theorem 10 (Extended Cauchy formula). If a function f is ana-lytic in a domain bounded by a simple closed contour then it possessesderivatives of all order. These derivatives are themselves analytic and

f (n)(z0) =n!

2πi

∫

Γ

f(z)

(z − z0)n+1dz.

Corollary 1. A function f is analytic if and only if it is (complex)integrable.

Proof. We know already that the first integral F of an integrable func-tion is analytic. Hence, by the previous theorem the derivative of thefirst integral–i.e. f itself is analytic.

Since a harmonic function can be considered as the real part of ananalytic function and the derivatives can be written in terms of partialderivatives of u and v we have

Theorem 11. A harmonic function in a domain will possess all partialderivatives of all orders.

9.3. Some Applications. Let Γ = z(s) be a curve the mean value off(z) along the curve is defined as

MΓ(f) :=1

|Γ|

∫

Γ

f(z) ds.

Theorem 12 (Gauß’ mean value theorem). Let f be analytic in Ω andC be a circle with center z0 inside Ω. Then f(z0) is the mean valuealong the circle.

Proof. On the circle C we can write z = z0 + reiφ. Then dz = reiφidφand

f(z0) =1

2πi

∫

Γ

f(z)

z − z0dz

=1

2πi

∫ 2π

0

f(z0 + reiφ)

reiφireiφ dφ

=1

2π

∫ 2π

0

f(z0 + reiφ) dφ

=1

2πr

∫ 2π

0

f(z0 + reiφ) · r dφ

=1

|C|

∫

C

f(z(s)) ds.

where we parametrized C = z(s) = z0 + reis; 0 ≤ s ≤ 2π.


A similar statement holds for the real and imaginary part. So onecan approximate the center value by a the (arithmetic) average of adiscrete set of values at the boundary (finite difference method).

Theorem 13 (Maximum Principle). Let f be continuous on Γ andanalytic in Ω(Γ). Then its maximum of |f(z)| on Γ∪Ω is obtained onΓ.

Proof. The case f ≡ const is trivial. Assume f is not constant. Thenit cannot be constant in any small neighborhood of any point in Ω (Wewill have a simple proof of this fact later). Let us assume that themaximum occurs at z0 ∈ Ω. We consider a circle C ⊂ Ω of radiusr around z0. W.l.o.g. we may assume that for at least one pointz ∈ C we have |f(z)| < |f(z0)|. By continuity there is a finite arc(α, β) ⊂ C = [0, 2π) where |f(z0 + reiφ)| ≤ |f(z0)| − b for some b > 0.hence,

|f(z0)|

=1

2π

∣

∣

∣

∣

∫ α

0

f(z0 + reiφ) dφ+

∫ β

α

f(z0 + reiφ) dφ+

∫ 2π

β

f(z0 + reiφ) dφ

∣

∣

∣

∣

≤ 1

2π(α|f(z0)| + (β − α)(|f(z0)| − b) + (2π − β)|f(z0)|)

= |f(z0)| −(β − α)b

2π< |f(z0)|

Corollary 2 (Minimum Principle). Let f be continuous on Γ and an-alytic in Ω(Γ). Moreover assume that f is nowhere zero. Then itsminimum of |f(z)| on Γ ∪ Ω is obtained on Γ.

Proof. Consider the function g = 1f.

Remark 9. Wew only needed the Cauchy integral formula (Theorem 9)to derive the maximum principle. The extended Cauchy integral for-mula (Theorem 10) has the next theorem as a consequence.

Theorem 14 (Liouville’s theorem). A bounded in modulus entire func-tion must be constant.

Proof. Let |f(z)| ≤ m on C and z0 ∈ C and C be a circle of radius raround z0. We have

f ′(z0) =1

2πi

∫

C

f(z)

(z − z0)2dz


and

|f ′(z0)| =1

2π

∣

∣

∣

∣

∫

C

f(z)

(z − z0)2dz

∣

∣

∣

∣

.

Hence,

|f ′(z0)| ≤1

2πM2πr

where M satisfies∣

∣

∣

∣

f(z)

(z − z0)2

∣

∣

∣

∣

≤M z ∈ C.

Now |f(z)| is bounded (by m) on the plane. Hence,∣

∣

∣

∣

f(z)

(z − z0)2

∣

∣

∣

∣

=

∣

∣

∣

∣

f(z)

r2

∣

∣

∣

∣

≤ m

r2

and|f ′(z0)| ≤

m

r.

Since this inequality holds in the entire plane we can derive this for all(even large) radii r. Consequently f ′(z0) = 0. Since z0 was arbitrarythe proof is complete.

As a consequence we can prove

Theorem 15 (Fundamental Theorem of Algebra). Any polynomialequation P (z) = anz

n + · · ·a1z + a0 = 0 of degree n > 0 has at leastone solution in the complex plane.

Proof. Assume that P (z) 6= 0 for all z ∈ C. Then 1P (z)

is analytic in

the entire plane, i.e. an entire function. Moreover,

1

P (z)=

1

zn1

an + an−1

z+ · · · + a0

z

and hence,

limz→∞

1

|P (z)| = 0.

Therefore 1P (z)

is a bounded (in modulus) entire function and must be

constant. That contradicts the assumption that n > 0.


10. The Dirichlet problem

An important problem in applications is to find the value of an har-monic function inside a domain bounded by a curve. F.e. the tempera-ture T (x, y) is a harmonic function in the heat convection problem. Sowe can ask what the value of the temperature is provided it is given atthe boundary. There are several different methods to solve this prob-lem. If the shape of the boundary is simple one could use seperationof variables. Another method is the conformal map method which willbe explained later. We will use the Cauchy integral formula.

10.1. The Dirichlet Problem for a Circle. Let us assume that theboundary of the domain in question is a circle C of radius R centeredat the origin in the complex plane. Let f(z) be a function analytic onthe circle and on its interior. By the Cauchy formula we have

f(z) =1

2πi

∫

C

f(w)

w − zdw. (12)

Let f(z) = U(x, y) + iV (x, y).

We consider the point z1 = R2

z. Since |z| < R we get |z1| = R2

|z| > R.

Also arg z1 = arg z (z = reiθ implies z1 = R2

re−iθ= R2

reiθ). The function

f(w)w−z1 is analytic in and on the circle C. Hence,

0 =1

2πi

∫

C

f(w)

w − z1dw =

1

2πi

∫

C

f(w)

w − R2

z

dw. (13)

Substracting (13) from (12) we obtain

f(z) =1

2πi

∫

C

f(w)

[

1

w − z− 1

w − R2

z

]

dw

=1

2πi

∫

C

f(w)

[

z − R2

z

(w − z)(

w − R2

z

)

]

dw


On the circle it is more convenient to work in polar coordinates: w =Reiφ, z = reiψ and on C we have dw = Reiφidφ. Hence,

f(r, ψ) =1

2πi

∫ 2π

0

f(R, φ)

[

reiψ − R2

re−iψ

(Reiφ − reiψ)(

Reiφ − R2

re−iψ

)

]

Reiφidφ

=1

2π

∫ 2π

0

f(R, φ)

(

reiψ − R2

reiψ)

Reiφ

(Reiφ − reiψ)(

Reiφ − R2

reiψ)

dφ

=1

2π

∫ 2π

0

f(R, φ)(R2 − r2)

R2 + r2 − 2Rr cos(φ− ψ)dφ.

In the last equality we multiplied both the nominator and denominatorby − r

Re−i(φ+ψ). Setting f(r, ψ) = U(r, ψ) + iV (r, ψ) we have

U(r, ψ) + iV (r, ψ) =1

2π

∫ 2π

0

[U(R, φ) + iV (R, φ)](R2 − r2)

R2 + r2 − 2Rr cos(φ− ψ)dφ

and for the real part

U(r, ψ) =1

2π

∫ 2π

0

U(R, φ)(R2 − r2)

R2 + r2 − 2Rr cos(φ− ψ)dφ.

This is the Poisson Integral Formula. It provides the value of an har-monic function inside a circle (We do not have to take into account theimaginary part).

Example 7. Let the temperature on the upper half unit circle be 1degree and on the lower half −1 degree. we can find the temperaturesinside by using the Poisson integral formula:

T (r, ψ) =1

2π

∫ π

0

(1 − r2) dφ

1 + r2 − 2r cos(φ− ψ)

− 1

2π

∫ 2π

π

(1 − r2) dφ

1 + r2 − 2r cos(φ− ψ)

Lets make the change of variables x = φ− ψ and use that

∫

dx

a+ b cos x=

2√a2 − b2

arctan

[√a2 − b2 tan x

2

a + b

]

+ C


we get

T (r, ψ) =1

π

[

2 arctan

(

1 + r

1 − rtan

(

π

2− ψ

2

))

− arctan

(

1 + r

1 − rtan

(

π − ψ

2

))

− arctan

(

1 + r

1 − rtan

(

−ψ2

))

]

where a = 1 + r2 and b = −2r.

10.2. The Dirichlet Problem for a Half Plane. We are looking fora harmonic function u(x, y) in the upper half plane with given boundaryvalues on the real line.

Let f = u+ iv be analytic for y ≥ 0. Consider the closed semicircleC from −R to R (see figure 10).

.

−R R

y

x

C

z

Base B

Arc A

Figure 10

Now let z be inside the semicircle and consider the point z outsidethe upper half plane. Thus

0 =1

2πi

∫

C

f(w)

w − zdw.


Therefore

f(z) =1

2πi

∫

C

f(w)

(

1

w − z− 1

w − z

)

dw

=1

2πi

∫

C

f(w)(z − z)

(w − z)(w − z)dw

Let B be the base of C and A be the arc. let w = s+ it. We have thaton B w = s. The integrand on the base B becomes

f(w)(z − z)

(w − z)(w − z)=

2iyf(s)

[s− (x− iy)][s− (x+ iy)]

=2iyf(s)

(s− x)2 + y2

and

f(z) =y

π

∫ R

−R

f(s)

(s− x)2 + y2ds+

y

π

∫

A

f(w)

(w − z)(w − z)dw

Now,∣

∣

∣

∣

f(w)

(w − z)(w − z)

∣

∣

∣

∣

≤∣

∣

∣

∣

f(w)

(|w| − |z|)(|w| − |z|)

∣

∣

∣

∣

≤ maxA |f(w)|(R− |z|)2

By the ML inequality we can estimate∣

∣

∣

∣

y

π

∫

A

f(w)

(w − z)(w − z)dw

∣

∣

∣

∣

≤ y

π· πR · maxA |f(w)|

(R− |z|)2−→ 0 (R → ∞)

provided we assume that f is bounded on the upper half plane. Hence,

f(z) =y

π

∫ ∞

−∞

f(s)

(s− x)2 + y2ds =

y

π

∫ ∞

−∞

u(s, 0) + iv(s, 0)

(s− x)2 + y2ds

and for the real part

u(x, y) =y

π

∫ ∞

−∞

u(s, 0)

(s− x)2 + y2ds

the Poisson Integral Formula.


11. Infinite series

Let us consider

u1(z) + u2(z) + · · · =

∞∑

j=1

uj(z)

where the uj’s are functions of a complex variable. We say that thisseries converges for z ∈ C if the sequence of partial sums

(Sn(z))n∈N =

(

n∑

j=1

uj(z)

)

n∈N

converges. The set of all values for which the series converges is calledthe region of convergence.

Example 8.∞∑

j=0

zj =1

1 − z|z| < 1

We have that

Sn(z) − zSn(z) = (1 + z + · · ·+ zn−1) − (z + z2 + · · ·+ zn) = 1 − zn

. So

Sn(z) =1 − zn

1 − z.

Since∣

∣

∣

∣

1

1 − z− Sn(z)

∣

∣

∣

∣

=|z|n

|1 − z|the sequence Sn(z) converges to 1

1−z provided |z| < 1.

Example 9. We can use the previous example to get∞∑

n=0

einz =∞∑

n=0

(

eiz)n

=1

1 − eiz

provided that |eiz| < 1. But |eiz| = |ei(x+iy)| = |eix||e−y| = e−y. Hencethe region of convergence is ℑz > 0.

A rather obvious statement is the following

Theorem 16. The series∑∞

j=1 uj(z) converges if and only if both

its real and its imaginary part converge, i.e. iff both∑∞

j=1Rj(z) and∑∞

j=1 Ij(z) converge, where uj(z) = R(x, y) + iI(x, y).

Therefore we can use the facts from real analysis to get

Theorem 17.∑∞

j=1 uj(z) diverges if limj→∞ uj(z) 6= 0.

Example 10.∑∞

j=1 zj diverges for |z| ≥ 1


Definition 6. A converging series∑∞

j=1 uj(z) converges absolutely if∑∞

j=1 |uj(z)| also converges. Otherwise a converging sequence is calledconditionally convergent.

Again using real analysis:

Theorem 18. The sum of an absolutely converging series is indepen-dent of the order of the terms. The product of two absolutely convergentseries

∑∞j=1 uj(z) and

∑∞j=1 vj(z) is the absolute convergent series ob-

tained by multiplying all terms whith each other:∞∑

k,j=1

uk(z)vj(z).

The following ratio test is analogous to the real case.

Theorem 19. Assume the following limit exists

r = limj→∞

∣

∣

∣

∣

uj+1(z)

uj(z)

∣

∣

∣

∣

.

Then the series∑∞

j=1 uj(z) converges absolutely if r < 1 and divergesif r > 1.

Proof. If r > 1 then limj→∞ |uj(z)| = ∞ since for all ǫ > 0 and allsufficiently large n we have |un+1| > (r−ǫ)|un(z)| and limn→∞ |un(z)| =∞. If r < 1 then |un(z)| < (r + ǫ)|un−1(z)| and the series

∑∞j=1 |uj(z)|

converges.

Example 11. Let us consider∑∞

n=0(−1)nn2n+1z2n. Then we have∣

∣

∣

∣

un+1(z)

un(z)

∣

∣

∣

∣

=

∣

∣

∣

∣

(n+ 1)2z2

n

∣

∣

∣

∣

→ 2|z2|.

Hence the series converges absolutely for |z| < 1√2.

Definition 7. A series∑∞

j=1 uj(z) is said to converge uniformly in a

region R if Sn(z) converges uniformly to S(z) in R.

Again as for real series we have

Theorem 20 (Weierstraß). Let the scalar series∑∞

j=1mj with mj ≥ 0

be convergent. The series∑∞

j=1 uj(z) converges uniformly in R if

|uj(z)| ≤ mj for all x ∈ R.

Corollary 3. Let supz∈R |u1(z)| < ∞. Then the series∑∞

j=1 uj(z)converges uniformly if

r = limj→∞

∣

∣

∣

∣

uj+1(z)

uj(z)

∣

∣

∣

∣

≤ 1 − ǫ

for all z ∈ R and some ǫ > 0.


Corollary 4. Let the series∑∞

j=1 uj(z) = S(z) converge uniformly in

R and |f(z)| ≤ c <∞ for all z ∈ R. Then

∞∑

j=1

f(z)uj(z) = f(z)S(z)

converges uniformly in R.

Proof. We have

|f(z)S(z) −N∑

n=1

f(z)un(z)| = |f(z)S(z) − f(z)SN (z)|

≤ |f(z)||S(z) − SN (z)| < cǫ

for N > N0.

Theorem 21. Let the series∑∞

j=1 uj(z) converge uniformly in R to

S(z) and let uj(z) be continuous. If Γ is a curve in R then

∫

Γ

S(z) dz =∞∑

j=1

∫

Γ

uj(z) dz.

Proof.∣

∣

∣

∣

∣

∫

Γ

S(z) dz −n∑

j=1

∫

Γ

uj(z) dz

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

n∑

j=1

∫

Γ

uj(z) dz +

∫

Γ

( ∞∑

j=n+1

uj(z)

)

dz −n∑

j=1

∫

Γ

uj(z) dz

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∫

Γ

( ∞∑

j=n+1

uj(z)

)

dz

∣

∣

∣

∣

∣

≤∫

Γ

∣

∣

∣

∣

∣

∞∑

j=n+1

uj(z)

∣

∣

∣

∣

∣

dz

≤ |Γ|∣

∣

∣

∣

∣

∞∑

j=n+1

uj(z)

∣

∣

∣

∣

∣

−→ 0.

Corollary 5. Let the series∑∞


S(z) and let all uj(z) be analytic. Then S(z) is analytic in R.


Theorem 22. Let the series∑∞


S(z) and let uj(z) be analytic in R. Then at any interior point of R

dS(z)

dz=

∞∑

j=1

duj(z)

dz.

Proof. Let z be an interior point of R. Then we can find a simpleclosed curve C around z where un and S are analytic in the domain

bounded by C. Now we have that∣

∣

∣

1(w−z)2

∣

∣

∣≤ M for some M > 0 on

a neighborhood of C (not the whole domain!). Therefore we can applyCorollary 4 and then Theorem 21 to get

S ′(z) =1

2πi

∫

C

S(w)

(w − z)2dw =

∞∑

n=0

1

2πi

∫

C

un(w)

(w − z)2dw =

∞∑

n=0

u′n(z)

11.1. Power series. A power series is a series of the form∑∞

j=1 ajzj

(or more general∑∞

j=1 aj(z − z0)j) where aj ∈ C. We are going to

apply the previous results to power series and will obtain the Taylorseries.

Theorem 23. If∑∞

j=1 ajzj converges for z = z0 then it converges

absolutely for all z with |z| < |z0|.Proof. Since ajz

j0 → 0 we have that |ajzj0| < M and

|ajzj | = |ajzj0|∣

∣

∣

∣

z

z0

∣

∣

∣

∣

j

< M

∣

∣

∣

∣

z

z0

∣

∣

∣

∣

j

.

Therefore,∑∞

j=1M∣

∣

∣

zz0

∣

∣

∣

j

converges and the proof is complete by the

Theorem of Weierstraß.

Let R = sup |z0| where the supremum is taken over all points wherethe series converges. Then the series converges absolutely for all |z| < Rand diverges for all |z| > R. R is called the radius of convergence.

Corollary 6. A power series is uniformly convergent in any disc |z| ≤R′ < R.

Theorem 24. The radius of convergence is determined by

R =1

lim supj→∞ |aj |1/j.

Proof. If the series converges for z0 (i.e. R ≥ |z0|) we must have thatfor sufficiently large j

|ajzj0| < 1 and so |z0||aj|1/j < 1.

Hence, R lim supj→∞ |aj|1/j ≤ 1.


Now assume that R < |z| ≤ 1−ǫlim supj→∞

|aj |1/j for some ǫ > 0. Then

for large j we have that |aj |1/j(

1 − ǫ2−ǫ)

< lim supj→∞ |aj|1/j . Hence

|ajzj | ≤(

1 − ǫ2

)jfor all large j. This implies the convergence of

∑∞j=1 ajz

j contradicting the assumption.

Remark 10. Since the uniform convergence of∑∞

j=1 ajzj implies the

uniform convergence of∑∞

j=1 jajzj−1 Theorem 22 follows from Theo-

rem 21. This is because1

lim supj→∞ |aj |1j

=1

lim supj→∞ |jaj |1j−1

.

Let now f(z) be analytic in Ω. Let C be a circle of radius R1 aroundz0 which is contained in the interior of Ω. Let z be also encircled by C(see figure 11).

..

R

z

C

1

z

0

Ω

Figure 11

By the Cauchy formula we have f(z) = 12πi

∫

Cf(w)w−z dw. Since |z −

z0| < R1 = |w − z0| we can expand

1

w − z=

1

(w − z0)(

1 − z−z0w−z0

) =1

w − z0

∞∑

j=0

(

z − z0w − z0

)j

.


This series is uniformly convergent for all w if we fix z. Hence,

f(z) =1

2πi

∫

C

f(w)

w − zdw =

1

2πi

∞∑

j=0

(z − z0)j

∫

C

f(w)

(w − z0)j+1dw.

This is a power series of the form∑∞

j=1 aj(z − z0)j where

aj =1

2πi

∫

C

f(w)

(w − z0)j+1dw. (14)

If we differentiate this power series we get

f (k)(z) =∞∑

j=k

j(j − 1) · · · (j − k + 1)aj(z − z0)j−k.

Setting z = z0 we obtain

f (k)(z0) = k!ak

which together with (14) implies the extended Cauchy integral formula.We also have

f(z) =

∞∑

j=0

f (j)(z0)

j!(z − z0)

j.

Clearly this representation is unique and hence, if a series is constantin a small open set contained in the circle of convergence of the powerseries it has to be constant in the whole circle of convergence.


12. More on Taylor series

We can summarize: A function f(z) is analytic in Ω if

1. f ′(z) exists in Ω,2. f(z) is complex integrable in Ω,3. f(z) has derivatives of all orders in Ω,4. f(z) has a Taylor series expansion in a neighborhood of each

point in Ω.

12.1. Methods for obtaining Taylor series expansions.

Example 12 (term–by–term differentiation). We will use the term–by–term differentiation to expand 1

z2about z = 1.

We have

1

z= 1 − (z − 1) + (z − 1)2 ∓ · · · |z − 1| < 1

since(

1z

)(n) |z=1 = (−1)nn!. Differentiating this series and multiplyingby −1 we obtain

1

z2= 1 − 2(z − 1) + 3(z − 1)2 ∓ · · ·

=∞∑

n=0

(−1)n(n+ 1)(z − 1)n |z − 1| < 1.

Example 13 (term–by–term integration). Let

Si(z) =

∫ z

0

f(w) dw

where

f(w) =sinw

ww 6= 0

and f(0) = 1.This function cannot be expressed in terms of elementary functions.

It often appears in problems of electromagnetic radiations.We have

sinw = w − w3

3!+w5

5!∓ · · ·

so

sinw

w= 1 − w2

3!+w4

5!∓ · · ·


This series converges for all values of w even for w = 0! We nowintegrate

Si(z) =

∫ z

0

dw +

∫ z

0

−w2

3!dw +

∫ z

0

w4

5!dw · · ·

= z − z3

3 · 3!+

z5

5 · 5!∓ =

∞∑

n=0

(−1)n

(2n+ 1)(2n+ 1)!z2n+1

for all z ∈ C.

Example 14 (Multivalued functions). We want to expand the principalbranch e1/2Log(z+1) of f(z) = (z + 1)1/2.

We have for this principal branch

f ′(z) = e1/2Log(z+1) 1

2(z + 1)=

(z + 1)1/2

2(z + 1)=

(z + 1)1/2−1

2.

Differentiating furtheron we get

f ′′(z) =1

2

(

1

2− 1

)

(z + 1)1/2−2,

f ′′′(z) =1

2

(

1

2− 1

)(

1

2− 2

)

(z + 1)1/2−2

and

f (n)(z) =1

2

(

1

2− 1

)(

1

2− 2

)

· · ·(

1

2− (n− 1)

)

(z + 1)1/2−n

where (z + 1)1/2−n = e1/2Log(z+1)

(z+1)n. The value at 0 is

e1/2Log1

1n= 1.

hence

(1 + z)1/2 = z +

∞∑

n=1

1

2

(

1

2− 1

)(

1

2− 2

)

· · ·(

1

2− (n− 1)

)

zn.

Since a Taylor series converges in the largest disc not containing asingularity we have that this series converges to the principal branchfor |z| < 1 (z = −1 is the branching point for f).

Example 15 (multiplication of series). We want to expand f(z) = ez

1−z .We have

ez =

∞∑

n=0

zn

n!and

1

1 − z=

∞∑

n=0

zn.

The latter series converges for |z| < 1.


Therefore

f(z) =

(

1 + z +z2

2!+z3

3!+ · · ·

)

(1 + z + z2 + z3 + · · · )

= 1 + (1 + 1)z +

(

1 + 1 +1

2!

)

z2 +

(

1 + 1 +1

2!+

1

3!

)

z3 + · · ·

=

∞∑

n=0

(

n∑

k=0

1

k!

)

zn |z| < 1.

Example 16 (quotients). Let h(z) = f(z)g(z)

and g(z0) 6= 0. Moreover let

h(z) =

∞∑

n=0

cn(z− z0)n f(z) =

∞∑

n=0

an(z− z0)n g(z) =

∞∑

n=0

bn(z− z0)n

where an, bn are known. Since h(z)g(z) = f(z) we get∞∑

n=0

cn(z − z0)n

∞∑

n=0

bn(z − z0)n =

∞∑

n=0

an(z − z0)n

and

c0b0 = a0

c0b1 + c1b0 = a1

c0b2 + c1b1 + c2b0 = a2

· · ·n∑

i=0

cibn−i = an.

This leads to a recursive procedure of calculating the coefficients.

12.2. The method of partial fractions. We consider a rationalfunction

f(z) =P (z)

Q(z)

where P and Q are polynomials. If Q(z0) 6= 0 then f(z) has a Taylorexpansion about z0. There is a better way of obtaining the coefficientsthan differentiating. We may assume that the degree of Q is higherthan that of P . Otherwise we obtain by a simple division

f(z) = P1(z) +P2(z)

Q(z)

where now the degree of P2 is less than that of Q.Under these assumptions we have for

Q(z) = C(z − a1)(z − a2) · · · (z − an)


and all aj different that

f(z) =P (z)

Q(z)=

A1

z − a1+

A2

z − a2+ · · · An

z − an

for z 6= aj and the Aj’s are constants.In case we have

Q(z) = C(z − a1)m1(z − a2)

m2 · · · (z − an)mn

and all aj different then

f(z) =A1

z − a1

+A1

(z − a1)2+ · · · A1

(z − a1)m1

+A2

z − a2+

A2

(z − a2)2+ · · · A2

(z − a2)m2

· · ·

+An

z − an+

An(z − an)2

+ · · · An(z − an)mn

for z 6= aj and the Aj’s are constants.Those functions are examples of meromorphic functions, i.e. func-

tions that are analytic everywhere in C exept in some poles. We willdiscuss those functions later.

The use of those functions are their simple Taylor series:

1

1 − z= 1 + z + z2 + · · · |z| < 1

1

1 + z= 1 − z + z2 ∓ · · · |z| < 1

1

(1 − z)2= 1 + 2z + 3z2 + · · · |z| < 1

1

(1 + z)2= 1 − 2z + 3z2 ∓ · · · |z| < 1

· · ·1

(1 − z)m=

∞∑

n=0

(m− 1 + n)!

n!(m− 1)!zn |z| < 1

Example 17. Let

f(z) =z

z2 − z − 2=

z

(z + 1)(z − 2).


we want to expand this function about z = 1, i.e. in powers of z − 1.We have to solve

z

(z + 1)(z − 2)=

a

z + 1+

b

z − 2or

z = a(z − 2) + b(z + 1) = (a+ b)z + b− 2a.

This impliesa+ b = 1 b = 2a

and a = 13

and b = 23. Therefore

f(z) =1

3

1

z + 1+

2

3

1

z − 2=

1

3

1

(z − 1) + 2+

2

3

1

(z − 1) − 1

=1

6

1

1 + z−12

− 2

3

1

1 − (z − 1)

=1

6

(

1 − z − 1

2+

(z − 1)2

4∓ · · ·

)

− 2

3

(

1 + (z − 1) + (z − 1)2 + · · ·)

where the first term converges for |z−1| < 2 and the second for |z−1| <1. So for |z − 1| < 1 we have

z

z2 − z − 2

=

(

1

6− 2

3

)

+

(

− 1

12− 2

3

)

(z − 1) +

(

1

24− 2

3

)

(z − 1)2 + · · ·

=

∞∑

n=0

(

1

6

(

−1

2

)n

− 2

3

)

(z − 1)n.


13. Laurent series

As we have seen we can expand rational functions into Taylor seriesin the domain where the denominator does not vanish. There is a wayto include such singularities.

Definition 8. A Laurent series is an expansion of the form

f(z) =

∞∑

n=−∞an(z − z0)

n.

Example 18. Consider

ew =

∞∑

n=0

wn

n!.

Set w = (z − 1)−1 we obtain

e1/(z−1) = 1 + (z − 1)−1 +(z − 1)−2

2!+

(z − 1)−3

3!· · ·

This series contains no positive powers of (z−1). The series convergesfor all z 6= 1! We also have

(z − 1)2e1/(z−1) = (z − 1)2 + (z − 1) +1

2+

(z − 1)−1

3!+

(z − 1)−2

4!· · ·

Those series expansions are important in the calculus of residues.

Theorem 25. Let f be analytic in r1 < |z − z0| < r2. Then for z inthis annulus we have

f(z) =

∞∑

n=−∞

(

1

2πi

∫

C

f(z)

(z − z0)n+1dz

)

· zn

where C is a simple closed curve ”going once around the inner circle”and lying in the annulus (see figure 12)

Proof. We deal with the case that z0 = 0. We will use the contourA = C1 ∪ s− ∪ s+ ∪ C2 as in figure 13.

We note that the point z1 is inside the simple closed curve A. Thus

f(z1) =1

2πi

∫

A

f(z)

z − z1dz.

The integrals along s+ and s− cancel so that the integral is taken alongthe two circles C1 and C2:

f(z1) =1

2πi

∫

C1

f(z)

z − z1dz +

1

2πi

∫

C2

f(z)

z − z1dz.


r

Cz0

r1

2

Figure 12

The integral along C2 can be dealt with the following trick

1

2πi

∫

|z|=l2

f(z)

z − z1dz

=1

2πi

∫

|z|=l2

f(z)

z(

1 − z1z

) dz

=1

2πi

∫

|z|=l2

f(z)

z

(

1 +z1z

+(z1z

)2

+ · · ·)

dz

=

∞∑

n=0

cnzn1

where

cn =1

2πi

∫

|z|=l2

f(z)

zn+1dz n ∈ N.

This can be done since |z1| < |z| for z ∈ C2.


r0

r1

2

l

ss−

+

2

1lA

z.

1

2

1C

C

Figure 13

For the other integral we revers the direction and get

1

2πi

∫

C1

f(z)

z − z1dz

= − 1

2πi

∫

−C1

f(z)

z − z1dz

=1

2πi

∫

|z|=l1

f(z)

z1 − zdz

=1

2πi

∫

|z|=l1

f(z)

z1

(

1 − zz1

) dz

=1

2πi

∫

|z|=l1

f(z)

z1

(

1 +z

z1+

(

z

z1

)2

+ · · ·)

dz


=z−11

2πi

∫

|z|=l1f(z) dz +

z−21

2πi

∫

|z|=l1zf(z) dz

+z−31

2πi

∫

|z|=l1z2f(z) dz + · · ·

=

−1∑

n=−∞cnz

n1

where

cn =1

2πi

∫

|z|=l1z−n−1f(z) dz =

1

2πi

∫

|z|=l1

f(z)

zn+1dz n ∈ −N.

This can be done since |z1| > |z| for z ∈ C1.

The function f(z)zn+1 is analytic in the annulus. Hence we can deform

the circles C1 and C2 into the simple closed curve C. Therefore

f(z1) =

−1∑

n=−∞cnz

n1 +

∞∑

n=0

cnzn1 =

∞∑

n=−∞cnz

n1

where

cn =1

2πi

∫

C

f(z)

zn+1dz.

This holds for any r1 < l1 < |z1| < l2 < r2 and hence for all z1 in theannulus.

Remark 11. It can be seen that the Laurent series converges uniformlyin any subannulus. Hence, we can perform term–by–term integrationand term–by–term differentiation inside the annulus.

Remark 12. The extended Cauchy formula is no longer valid! Forthis the function has to be analytic inside the circle and not only onan annulus! Therefore the coefficients cn do not have to be equal tof (n)(z0)/n!.

Definition 9. A function f is said to have an isolated singularity atz = z0 if f is analytic in some deleted disc around z0 but not analyticat z0.

Remark 13. A function with an isolated singularity can be expandedinto a Laurent series in an annulus domain around the singularity.

Example 19. Let f(z) = 1z−3

This function has an isolated singularityat z = 3. We want to expand it about z = 1, i.e. in powers of z − 1.


First we observe that the function is analytic for |z− 1| < 2 and hencehas a Taylor series expansion there. Now we proceed as follows

1

z − 3=

1

(z − 1) − 2=

1z−1

1 − 2z−1

hence for w = 2z−1

and with 11−w = 1 + w + w2 + · · · ; |w| < 1 we have

1

z − 3=

1

z − 1

(

1 +2

z − 1+

4

(z − 1)2+ · · ·

)

=1

z − 1+ 2

1

(z − 1)2+ 4

1

(z − 1)3+ · · ·

The condition |w| < 1 gives∣

∣

2z−1

∣

∣ < 1 or |z − 1| > 2. So we obtained aTaylor series expansion for |z − 1| < 2 and a Laurent series expansionfor |z − 1| > 2.

14. Properties related to Taylor series

Definition 10. f has an isolated 0 at z0 if f(z0) = 0 and f 6= 0 in adeleted disc around z0.

Theorem 26. An analytic function not identically 0 in some neigh-borhood has only isolated zeros.

Proof. Let f(z) =∑∞

n=0 an(z − z0)n. Since f(z0) = 0 we have a0 =

0. Let m be the minimal number such that am 6= 0 (this number isfinite since otherwise the Taylor series and hence the function wouldbe identically 0 in a neighborhood). Then

f(z) = (z − z0)m(

am + am+1(z − z0) + am+2(z − z0)2 + · · ·

)

= (z − z0)mφ(z)

The function φ is analytic and non-zero at z = z0. By continuity thereis a neighborhood of z0 such that

|φ(z) − φ(z0)| <∣

∣

∣

∣

φ(z0)

2

∣

∣

∣

∣

Hence,

|φ(z)| > |φ(z0)| −∣

∣

∣

∣

φ(z0)

2

∣

∣

∣

∣

> 0

in this neighborhood.

Definition 11. The number m from the previous proof is called theorder of zero at z0.


Example 20. We consider f(z) = sin(

1z

)

. This function is ana-

lytic for z 6= 0 and z = 0 is a singularity. The zero’s are z = 1nπ

;n = ±1,±2, · · · . All these zero’s are isolated but accumulate at thesingularity z = 0. Therefore we do not have a Taylor expansion aboutz = 0! Moreover f ′(z) = − 1

z2cos(

1z

)

and hence

f ′(

1

nπ

)

= −(nπ)2 cos(nπ) = −(nπ)2(−1)n 6= 0.

This means that all zero’s have order 1.

Remark 14. Let a Laurent series f(z) =∑∞

n=−∞ anzn be convergent

in r < |z| < R. Then this series is unique in this annulus. Let r < l <R and z = leiθ. We are going to multiply this series with zm = lmeimθ

and integrate along the circle:∫ 2π

0

f(leiθ)lmeimθ dθ =

∞∑

n=−∞anl

n+m

∫ 2π

0

ei(n+m)θ dθ = a−m · 2π

and a−m is uniquely defined.


15. Analytic continuation

Suppose we are given a function f that is analytic in a point z0. It canbe given by a power series

∑∞0 an(z−z0)n = f(z) that converges inside

a discD0 of radius R0. The radius is determined by the singularity lyingclosest to z0. Let z1 ∈ D0 (see figure 14)

.

.

z

z

0

’0

R

1R

R 1

1

’

Figure 14

Now we can consider the Taylor expansion of f about z1 which isgiven by

∑∞0 bn(z − z1)

n = f(z). This series gives the same valuef(z) inside the disc D′

1 around z1 and contained in D0. We can askwhether the Taylor series about z1 converges to a function g(z) in adisc D1 ⊃ D′

1 which coincides with f(z) throughout the region D0∩D1.In this case we call g(z) an analytic continuation of f(z).

In this case we can proceed as follows. We know that

bn =f (n)(z1)

n!

and if we differentiate f(z) =∑∞

0 an(z − z0)n term–by–term at the

point z1 we get

b0 =∞∑

n=0

an(z1 − z0)n,

b1 =∞∑

n=1

ann(z1 − z0)n−1,


b0 =

∞∑

n=2

ann(n− 1)(z1 − z0)n−2,

etc.If g(z) is now an analytic continuation of f(z) we can continue with

a point z2 ∈ D1\D0 and try to find an analytic continuation of g whichwe also can consider as an analytic continuation of f .

Example 21. Let f(z) = 1 − z + z2 − z3 ± · · · be given. This seriesconverges for |z| < 1 (circle C) and diverges for |z| > 1. We observethat in this region f(z) = 1

1+z. But f(z) = 1

1+zis analytic for z 6= −1.

This is an analytic continuation of the Taylor series 1−z+z2−z3±· · · .We can also proceed by expanding this function about z = 1

2.

∞∑

n=0

bn

(

z − 1

2

)

= 1 − z + z2 − z3 ± · · ·

We get by using bn =f(n)( 1

2)n!

b0 = 1 − 1

2+

1

4− 1

8± · · · =

2

3

b1 = −1 + 1 − 3

4+

4

8∓ · · · = −4

9= −

(

2

3

)2

b2 = 2 − 3 + 3 − 5

2± · · · =

8

27=

(

2

3

)3

and in general

bn = (−1)n(

2

3

)n+1

.

Hence, we get

g(z) =

∞∑

n=0

(−1)n(

2

3

)n+1(

z − 1

2

)n

.

The ratio test gives that this series converges in the circle C ′ givenby∣

∣z − 12

∣

∣ < 32. This circle contains the unit circle where the original

Taylor series converges. Hence it is an analytic continuation. Weremark that the singularity z = −1 lies on the boundary of this circle(see figure 15)

Lets make the previous reasoning rigorous.

Theorem 27. Let be two functions f , g be analytic in a connecteddomain Ω. Let moreover f(z) = g(z) in a nonempty subdomain Ω′ ⊂ Ω.Then f(z) = g(z) throughout Ω.


1 2

C C’

−1 0 1−2

Figure 15

Proof. We set h(z) = f(z) − g(z). Then h(z) ≡ 0 in Ω′. Let G ⊂ Ωbe the set of points z ∈ Ω for which f(z) = f ′(z) = f ′′(z) = · · · = 0.This set is open for since a point z ∈ G then its Taylor expansion isidentically zero in a (possibly small) disc around z and hence all thederivatives of f are zero in this disc. But since f and all its derivativesare continuous the complement Ω\G is also open. Since Ω is connectedand G 6= ∅ this is only possible when Ω \G = ∅ or G = Ω.

Definition 12. Let for i = 1, 2, 3, · · · , n nonempty connected domainsΩi and analytic (on Ωi) functions fi be given. Assume futhermore thatthere are nonempty domains Gi ⊂ Ωi ∩ Ωi+1, i = 1, · · · , n − 1 suchthat fi = fi+1 on Gi. Then we say that f1 on Ω1 and fn on Ωn areanalyticly equivalent

Remark 15. Analytic equivalence is indeed an equivalence relation.

Remark 16. For the analytic continuation it can be important on whatsubdomains Gi the functions coincide since the ”overlap” of two do-mains may consist of several connected components (see figure 16).We will illustrate this later on an example.


Ω

Ω

G

G1

1

2

1’

Figure 16

Remark 17. One and the same analytic expression may lead to nonequivalent analytic function. Consider

f(z) =

∫

|w|=1

1

w − zdw.

We have f(z) = 2πi for |z| < 1 and f(z) = 0 for |z| > 1.On the other hand different analytic expressions can give analyticly

equivalent functions. F.e.

f1(z) = 1 + z + z2 + · · · =1

1 − z|z| < 1

and for |z| > 1

f2(z) = −1

z− 1

z2− · · · = −

∞∑

n=1

(

1

z

)n

= −(

1

1 − 1z

− 1

)

=1

1 − z.

If two functions are equivalent then also are so their derivatives.Moreover if f1 on Ω1 is equivalent to f2 on Ω2 and g1 on Ω1 is equiva-lent to g2 on Ω2 then f1+g1 is equivalent to f2+g2 and f1g1 is equivalentto f2g2. Hence the polynomial (in two variables) P (f1(z), g1(z)) on Ω1

is equivalent to P (f2(z), g2(z)) on Ω2. This holds true for any algebraicrelation (which can be expressed as an relation between the coefficientsof the Taylor expansions). Furthermore, the same holds true for any


algebraic differential equation. Finally, one can go from algebraic rela-tions to analytic relations, meaning that in case a (analytic) functional(may be even differential) equation holds for some functions analyticin a domain then it also holds for analytic equivalent functions in a(possibly different) domain.

Example 22. We return to the second example in remark 17. TheTaylor series f(z) = 1 + z + z2 + z3 + · · · , |z| < 1 fulfills in the unitdisc the following functional equation:

f1(z) = 1 + zf(z).

Since it is equivalent to the Taylor series f2(z) = −1z− 1

z2− 1

z3− · · ·

on |z| > 1 it also must fulfill the same functional equation. This ischecked immediately. Moreover the functional equation leads to theanalytic function

f(z) =1

1 − zz 6= 1.

Remark 18. The collection of all functions f(z) on domains Ω whichare pairwise equivalent represent an analytic function. One can recoverthe whole family from any of its pieces by analytic continuation

Theorem 28. Let Ω be a bounded simply connected domain and G bea subdomain. If f is analytic on G and has an analytic continuationto Ω then this is a single-valued function.

Proof. We have to prove that an analytic continuation over G, G1, G2,... , Gn = G1, assigns the same values over G1, i.e. when continuatingand returning we arrive at the same value. Let Π be a closed polygoninside the domains Gi of the analytic continuations. Lets divide Πinto (small) triangles such that the radius of convergence of a Taylorexpansion of each of the continuations at any points of its definitioninside the polygon Π is larger than the diameter of the triangles (seefigure 17).

Then the value of f at a vertex of any such small triangle is givenby the taylor series expansion and hence is not altered by going oncearound the triangle. Therefore instead of going along one side of thetriangle we can go along the two other sides and arrive at the samevalue. This implies that we do not change the value at a given pointby going along any closed path along the triangles.

Example 23. Let f(z) = z1/2. This function is not analytic at z = 0.We consider the family of all analytic continuations of f(z) defined in|z−1| < 1 by the principal branch. Now points z may have different con-tinuations fi(z) as remarked before, depending on the ”way” of contin-uation. In general there can be infinitely many. In our case we can ex-tend f to a domain around z = −1 by going clockwise or counterclock-wise and we arive at fclock(−1) = e−i

π2 = −i and fcounterclock(−1) =

eiπ2 = i (see 18).


a continuationseries forDisc of convergence of a Taylor

Figure 17

1

i

−i

−1

Figure 18

This way we can assign to each point z 6= 0 two different analyticcontinuations depending on the parity of the winding number aroundz = 0 (If we go twice around z = 0 along |z| = 1 we will arrive at thesame value. By the proof of the previous theorem we can change the


path we continue along as long as we do not cross z = 0. Hence we donot change the winding number).

So the ”complete” analytic continuation of one branch gives us allbranches of a multivalued function. The same is also true for zn orlog z or other analytic multivalued functions.

Example 24. Here we want to see that not every analytic function onthe unit disc has an analytic continuation. We consider the lacunaryTaylor series

f(z) =

∞∑

n=0

z10n

with radius of convergence |z| < 1. We are going to show that thisseries has no analytic continuation. For this we consider radial limitsand show that they are infinite on a dense set of the unit circle. Hencethere is no ”way out” off the circle. The radial limit (if exists) is definedas follows. Let z = reiθ. Then

f(reiθ) =∞∑

n=0

r10nei10nθ.

The radial limit is

l(θ) := limrր1

∞∑

n=0

r10nei10nθ.

Let us consider a (rational) number θ = a0.a1a2 · · ·aN00000 · · · ; ai ∈0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Then

l(θ) = limrր1

∞∑

n=0

r10nei10nθ

= limrր1

(

N∑

n=0

r10nei10nθ +

∞∑

n=N+1

r10ne0

)

= limrր1

lN(r) + limrր1

∞∑

n=N+1

r10n

= lN(1) + limrր1

∞∑

n=N+1

r10n = ∞.

Since those rational θ are dense in [0, 2π) we proved that there is noTaylor series expansion on any subinterval of the unit circle. Hence wecannot perform an analytic continuation.

There is a general principle for the behavior of the radial limit:


Let Sn(θ) =∑n−1

j=0 ei10jθ.

Lemma 1. Let N = N(r) = maxn ≥ 0 : r10n ≥ e−1. We have∣

∣f(reiθ) − SN(r)(θ)∣

∣ < C

where the constant C is uniform on r and θ.

Proof. We write

f(reiθ) = SN(r)(θ) +N∑

n=0

(r10n − 1)ei10nθ +

∞∑

n=N+1

r10nei10nθ.

On one hand, since r10N+k ≤ e−10k for k ≥ 1, we have∣

∣

∣

∣

∣

∞∑

n=N+1

r10nei10nθ

∣

∣

∣

∣

∣

≤∞∑

n=N+1

r10n ≤∞∑

k=1

e−10k <∞.

On the other hand, since r10N ≥ e−1, we have r ≥ e−1

10N . So, usinge−x ≥ 1 − x we have∣

∣

∣

∣

∣

N∑

n=0

(1 − r10n)ei10nx

∣

∣

∣

∣

∣

≤N∑

n=0

(

1 − exp(−10n−N))

≤N∑

n=0

10−(N−n) ≤ 10

9.

Notice that

N(r) ∼ log10

1

1 − r.


16. Poles and essential singularities

In this section we want to investigate what happens at the pointswhere a function is not analytic.

Definition 13. Let f be analytic in a deleted disc around z0 with Lau-rent expansion

∑∞n=−∞ anz

n. Then f has a pole of order n at z0 ifa−m = 0 for all m > n and a−n 6= 0. We say that f has an essential

singularity at z0 if an 6= 0 for infinitely many negative n.The part of the Laurentseries containing only negative powers of (z−

z0) is called the principal part.

Remark 19. If we substract the principal part from an analytic in adeleted disc function we end up with a function g(z) which has a Taylorseries expansion in this deleted disc. The singularity at z0 can only beof the nature that g(z0) 6= a0. In this case we can alter the value of gat z0 to g(z0) = a0 and get a function which is analytic in the entiredisc. In this case we say that the singularity for g at z0 is removable.

16.1. Poles. If f has a pole of order n at 0 then

f(z) =a−nzn

+a−n+1

zn−1+ · · ·+ a0 + a1z + · · ·

= z−n (a−n + a−n+1z + · · ·a0zn + · · · )

in a disc of radius R where a−n 6= 0. Therefore we can find a smallerdisc such that

|a−n+1z + a−n+2z2 + · · ·a0z

n + · · · | < 1

2|a−n|.

Hence,

|a−n + a−n+1z + a−n+2z2 + · · ·a0z

n + · · · | > |a−n| −1

2|a−n| =

1

2|a−n|

and

|f(z)| > |a−n|2

1

|z|n .

Hence

limz→0

f(z) = ∞or

limz→0

f(z)zn = a−n 6= 0.

Clearly

f(z) − a−nz−n − a−n+1z

−n+1 · · · − a−1z−1 =

∞∑

n=0

anzn

is bounded in a disc around 0.


Since a−n 6= 0 we have that a−n + a−n+1z + · · ·a0zn + · · · 6= 0 for

|z| < R for some R > 0. We consider the function 1f(z)

in 0 < |z| < R:

1

f(z)= zn

1

a−n + a−n+1z + · · ·a0zn + · · ·

= zn1

a−n

1

1 + a−n+1

a−nz + · · · a0

a−nzn + · · ·

= zn(

b0 + b1z + b2z2 + · · ·

)

= b0zn + b1z

n+1 + b2zn+2 + · · ·

where b0 = 1a−n

6= 0. This can be done since 1a−n+a−n+1z+···a0zn+··· is

analytic. So 1f(z)

has a zero of order n for z = 0. Clearly the converse

is also true: if f has a zero of order n for z = 0 then 1f(z)

has a pole of

order n at z = 0.

16.2. The point at infinity. For the point at infinity we set w = 1z

and consider the expansion with respect to w:

f(z) =

∞∑

n=−∞anz

n =

∞∑

n=−∞a−nw

n.

This function is for |z| > R regular if there are no positive powers ofz. It has a pole of order n at infinity if

f(z) = anzn + an−1z

n−1 + · · ·+ a0 + a−11

z+ · · ·

16.3. More on Partial Fractions. Let us consider rational functionsf(z) = P (z)

Q(z)= p0+p1z+···pnzn

q0+q1z+···qmzm . They are characterized in the way that

they are univalent and up to a finite number of poles analytic in C. Arational function is analytic except the points where Q(z) = 0 or atz = infty. These are only finitely many points. Let z0 be a root oforder k of Q. Then

Q(z) = (z − z0)k (ak + ak+1(z − z0) + · · · ) ak 6= 0


Then f has a pole of order at most k at z0

P (z)

Q(z)=

b0 + b1(z − z0) + · · ·(z − z0)k (ak + ak+1(z − z0) + · · · )

=b0

(z − z0)k (ak + ak+1(z − z0) + · · · )+

+b1

(z − z0)k−1 (ak + ak+1(z − z0) + · · · ) + · · ·

where the pole is of order k if b0 6= 0.For z = ∞ we have for z = 1

w

P (z)

Q(z)=p0 + p1z + · · · pnznq0 + q1z + · · · qmzm

=zn (pn + pn−1z

−1 + · · · p0z−n)

zm (qm + qm−1z−1 + · · · q0z−m)

= wm−n pn + pn−1w + · · · p0wn

qm + qm−1w + · · · q0wm

with pn, qm 6= 0. This means z = ∞ is a pole (of order n−m) if m < nor otherwise it is regular at infinity.

Let z0, z1, · · · , zl be the poles of P (z)Q(z)

, i.e. the zeros of Q and assume

that degP < degQ, i.e. f is regular at infinity. Let H0(z), · · · , Hl(z)be the correponding principal parts of f . Then we have

Hi(z) = a(i)−ki

1

(z − zi)ki+ a

(i)−ki+1

1

(z − zi)ki−1+ · · ·a(i)

−1

1

(z − zi)

where ki is the order of the pole at z = zi and a(i)n are the coefficients

of the Laurent series about z = zi.Now we consider

g(z) =P (z)

Q(z)−H0(z) −H1(z) − · · · −Hl(z).

This function is now analytic in the entire plane and also regular atinfinity, in particular it is a bounded entire function. By Liouville’sTheorem 14 it must be constant. Hence,

P (z)

Q(z)=P (z)

Q(z)−H0(z) −H1(z) − · · · −Hl(z) + C.

This is the partial fraction expansion of P (z)Q(z)

.


16.4. Essential singularities. Weierstraß proved the following the-orem which shows the completely different behavior of an essentialsingularity.

Theorem 29. If f has an essential singularity at z = 0 then in a discaround the values of f come close to any pregiven value. I.e. for anyw ∈ C, arbitrary ǫ > 0 and arbitrary ρ > 0 there is a z with 0 < |z| < ρsuch that |f(z) − w| < ǫ.

Proof. Assume the contrary, i.e. there is a complex number w ∈ C anǫ > 0 and a ρ > 0 such that for all 0 < |z| < ρ we have |f(z) − w| > ǫor

∣

∣

∣

∣

1

f(z) − w

∣

∣

∣

∣

≤ 1

ǫ.

Then 1f(z)−w is bounded and analytic in 0 < |z| < ρ and can be ex-

panded into a Laurent series

1

f(z) − w=

∞∑

n=−∞cnz

n

where

cn =1

2πi

∫

|z|=r

dz

(f(z) − w)wn+1

where 0 < r < ρ. Hence,

|cn| ≤1

2π· 1

ǫ

1

rn+1· 2πr =

1

ǫr−n.

For negative n the right–hand–side is arbitrary small if we choose r > 0small. Hence, cn = 0 for n ∈ −N and

1

f(z) − w= c0 + c1z + c2z

2 + · · ·

If c0 6= 0 then f(z) − w is analytic since 1f(z)−w 6= 0 (possibly after

removing the singularity) and so is f(z) in |z| < ρ. If c0 = c1 =· · · ck−1 = 0 and ck 6= 0 then f(z) − w and hence f(z) has a pole of oforder k at z = 0. Both these cases contradict the assumption.

Corollary 7. If f is analytic for 0 < |z| < R univalent and boundedthen f is analytic in |z| < R after possibly removing the singularity atzero, i.e. changing the value at 0 to a0.

Example 25. Let f(z) = cos zsin z

. Then this function has simple poles atz = nπ and no other singularities. If we expand both cos and sin into


its Taylor expansion about nπ we have

f(z) =cos(nπ) − sin(nπ) · (z − nπ) − cos(nπ) · (z − nπ)2 + · · ·sin(nπ) + cos(nπ) · (z − nπ) − sin(nπ) · (z − nπ)2 − · · ·

=(−1)n − (−1)n · (z − nπ)2 ± · · ·

(−1)n · (z − nπ) − (−1)n · (z − nπ)3 ± · · ·

=1

z − nπ· 1

1 − (z − nπ)2 ± · · · +−(z − nπ) + (z − nπ)3

1 − (z − nπ)2 ± · · ·

=1

z − nπ+ terms which are analytic.

This implies that Hn(z) = 1z−nπ is the principal part at z = nπ. Un-

fortunately the sum over all principal parts is not summable but byreordering we can use

H−n(z) +Hn(z) =2z

z2 − n2π2n 6= 0

Then

H0(z) +∞∑

n=1

(H−n(z) +Hn(z)) =1

z+

∞∑

n=1

2z

z2 − n2π2= H(z).

This series converges uniformly in any disc not containing points ofthe form z = nπ. Also we have

Hn+1(z + π) =1

z + π − (n + 1)π= Hn(z)


and

H(z + π) −H(z) = H0(z + π) +

∞∑

n=1

(H−n(z + π) +Hn(z + π))−

−H0(z) +∞∑

n=1

(H−n(z) +Hn(z))

= H−1(z) +∞∑

n=1

(Hn−1(z) +H−n−1(z))−

−H0(z) +

∞∑

n=1

(Hn(z) +H−n(z))

= limN→∞

[

H−1(z) +

N∑

n=1

(Hn−1(z) +H−n−1(z))−

−H0(z) +

N∑

n=1

(Hn(z) +H−n(z))]

= limN→∞

(HN(z) +H−N−1(z)) = 0.

Hence, H(z+π) = H(z). Therefore f(z)−H(z) is an analytic functionwhich is periodic with period π. Hence it is an entire function. It is alsobounded (by periodicity we need this to check only for 0 ≤ ℜ(z) ≤ πand there only for |y| > 1 (where z = x+ iy) by continuity):

|f(z)| =|eiz + e−iz||eiz − e−iz| ≤

e|y| + e−|y|

e|y| + e−|y|


which is bounded for |y| > 1. Also with 0 ≤ x ≤ π and |y| > 1:

∣

∣

∣

∣

∣

∞∑

n=1

2z

z2 − n2π2

∣

∣

∣

∣

∣

≤∞∑

n=1

2|y|+ 2π

|z − nπ||z + nπ|

=

∞∑

n=1

2|y|+ 2π

|x+ iy − nπ||x+ iy + nπ|

≤∞∑

n=1

2|y|+ 2π

|iy − (n− 1)π||iy + nπ|

≤∞∑

n=1

2|y| + 2π

y2 + (n− 1)2π2

=

∞∑

n=0

2|y| + 2π

y2 + n2π2

≤∞∑

0

2|y|y2 + n2π2

+

∞∑

n=0

2π

1 + n2π2

≤∞∑

0

2

|y| + n2π2

|y|+

∞∑

n=0

2π

1 + n2π2

<∞

(|y| > 1). Hence f(z) − H(z) ≡ c. But since both f and H are oddfunctions:

f(z) −H(z) = − (f(−z) −H(−z))

we have c = 0. Hence,

cot z =1

z+

∞∑

n=1

2z

z2 − n2π2.


Let Γ be a curve that does not cross any of the singularities. Then∫

Γ

f(z) dz =

∫

Γ

d log sin z

dzdz

=

∫

Γ

d log sin z

=

∫

Γ

(

1

z+

∞∑

n=1

2z

z2 − n2π2

)

dz

=

∫

Γ

d log z +

∞∑

n=1

∫

Γ

d log(z2 − n2π2)

or∫

Γ

d logsin z

z=

∞∑

n=1

∫

Γ

d log(z2 − n2π2).

Now if the curve goes from 0 to w (we can do this by continuity) wederive that

logsinw

w= log

sin z

z

∣

∣

∣

w

0=

∫ w

0

d logsin z

z

=

∞∑

n=1

∫ w

0

d log(z2 − n2π2)

=∞∑

n=1

log(z2 − n2π2)∣

∣

∣

w

0

=∞∑

n=1

[

log(w2 − n2π2) − log(−n2π2)]

=

∞∑

n=1

log

(

1 − w2

n2π2

)

.

If we exponentiate this equality we would like to write the right–hand–side as an infinite product:

sin z = z∞∏

n=1

(

1 − z2

n2π2

)

.


This is a product representation of the sin–function. We will investigatethe properties of infinite products later.


17. Infinite products

In this section we deal with a special form of representations of anentire function. Let f(z) be an entire function which has a pole oforder n at infinity. By section 16.2 its Laurentseries has an 6= 0 andan+j = 0 for all j > 0. Moreover its Laurentseries is actually a taylorseries. Hence f(z) =

∑nk=0 akz

k is a polynomial and has also n zerosz1, · · · , zn. Then

f(z) = c(z − z1) · · · (z − zn).

We can try to find a similar product representation of other entire func-tions. In general those functions have infinitely many positive termsin their Taylor expansion and hence an essential singularity at infinity.On the other hand the functions eg(z) where g is an entire function donot have zeros. Therefore one can only expect that an entire functionis determined by their zeros only up to a factor eg(z). On the otherhand there are entire functions with infinitely many zeros, like sin z.However, these zeros must be isolated. We will show that this is theonly condition we need for the following.

Before that we will need some properties of infinite products.

17.1. General properties of infinite products. We are going toconsider expressions of the form

∞∏

n=1

pn(z).

Definition 14. We say that this infinite product converges at the pointz iff the finite products

Pn =

m∏

n=1

pn(z) −→ P (z) 6= 0

converge or for at most a finite number of terms pnk(z) = 0, k =1, · · · , K. Otherwise the infinite product is called divergent. Thereforethe infinite product vanishes only if a factor is zero.

Example 26. We consider

∞∏

n=1

(1 + an).

Let an = 1n. Then Pn = 2

132

43· · · n+1

n= n + 1 → ∞ and the product is

divergent.Let an = −1

n+1. Then Pn = 1

223

34· · · n

n+1= 1

n+1→ 0 and the product is

divergent.


Theorem 30. Let ak ≥ 0. Then∞∏

n=1

(1 + an) and

∞∑

n=1

an

converge or diverge simultaneously.

Proof. Since 1 + x ≤ ex for x ≥ 0 we getm∑

n=1

an ≤m∏

n=1

(1 + an) ≤ exp

(

m∑

n=1

an

)

.

where the first inequality follows from that this sum is only one specialterm in the product. Both the partial sums and the partial productsare monotone increasing with the number of terms. This completes theproof.

Example 27. Let an = 1ks

. Then the product converges if and only ifs > 1.

Definition 15. We say tat the infinite product is absolutely conver-

gent if the product∞∏

n=1

(1 + |an|).

is convergent.

Lemma 2. If a product is absolutely convergent then it is convergent.

Proof. We first note that by Theorem 30∑∞

n=1 |an| converges and hencedoes

∑∞n=1 an. Now∣

∣

∣

∣

∣

m+1∏

n=1

(1 + an) −m∏

n=1

(1 + an)

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

m∏

n=1

(1 + an)

∣

∣

∣

∣

∣

|am+1|

≤ |am+1|m∏

n=1

(1 + |an|) =

m+1∏

n=1

(1 + |an|) −m∏

n=1

(1 + |an|).

Hence limn→∞ Pn = P exist. We need to check that P 6= 0. But the

series∑∞

n=1|an|

|1+an| converges since an → 0 and hence we may assume

that an 6= −1. Therefore∞∏

n=1

(

1 +|an|

|1 + an|

)

converges and hence the limit

limn→∞

1

Pn= lim

m→∞

m∏

n=1

(

1 − an1 + an

)

= limm→∞

m∏

n=1

(1 + an)−1

exists. So P 6= 0.


Definition 16. We call an infinite product∏∞

n=1 pn(z) uniformly

convergent in a domain Ω if the functions Pn(z) converge uniformlyin Ω and the limit function has no other zeros than those of a finitenumber of pn(z), i.e. there is an N ∈ N such that for all n > N andall z ∈ Ω we have pn(z) 6= 0 .

If the functions pn(z) are analytic the same holds true for the uni-formly convergent product, since

0 =

∫

Γ

Pn(z) dz −→∫

Γ

P (z) dz

for any closed curve Γ (since for uniformly convergent functions theintegral of the limit is the limit of the integrals). In this case we canrearrange the terms in the product and also perform differentiation andintegration term by term.

Theorem 31. If∑∞

n=1 supΩ |an(z)| = M < ∞ the product∏∞

n=1(1 +an(z)) is uniformly convergent in Ω provided an(z) 6= −1 in Ω for alln > N .

Proof. As before we derive

|Pm+1(z) − Pm(z)| ≤ |am+1|m∏

n=1

(1 + |an(z)|)

≤ exp

(

m∑

n=1

|an(z)|)

|am+1(z)|.

Hence, the limit limn→∞ Pn(z) exists uniformly in Ω and it cannot bezero since the product is absolutely convergent for all z ∈ Ω unless oneof the factors is zero.

Example 28. Let z ∈ C. Then

∞∏

n=1

(

1 − z2

π2n2

)

converges uniformly in any domain Ω = |z| ≤ R to an analyticfunction whose only zeros are z = ±πn.

Theorem 32 (Weierstraß). Let (zk)k be a sequence of complex numbersnowhere accumulating in C. Then there is an entire function which hasthe zk’s as its only zeros.

We remark that the zk’s do not have to be different. However onlya finite number can be equal. This number is the multiplicity of thezero.


Proof. Let R > 0. Then for |z| < R we have

log(

1 − z

R

)

= − z

R− z2

2R2− · · · = −

∞∑

n=1

zn

nRn.

Hence, for any l ∈ N(

1 − z

R

)

ezR

+ z2

2R2 +···+ zl

lRl = e−P

∞

n=1zn

nRn ezR

+ z2

2R2 +···+ zl

lRl

= exp

(

−∞∑

n=l+1

zn

nRn

)

= G(z, R, l).

The left –hand–side is an entire function which vanishes only for z =R and equals for |z| < R the right–hand–side. For |z| < R/2 andsufficiently large l the right–hand–side is close to 1: We have

∣

∣

∣

∣

∣

∞∑

n=l+1

zn

nRn

∣

∣

∣

∣

∣

<∞∑

n=l+1

1

2n=

1

2l.

Hence,

G(z, R, l) = 1 + η

with |η| < e2−l − 1. Let zk 6= 0 for k > k0 and z1 = z2 = · · · = zk0 = 0.

We set

G(z) := zk0∞∏

n=1

G(z, zn, ln)

= zk0∞∏

n=1

(

1 − z

zn

)

ezzn

+ z2

2z2n+···+ zln

lnzlnn

where we chose ln so large that for any R > 0∞∑

n=1

1

ln + 1

∣

∣

∣

∣

R

zn

∣

∣

∣

∣

ln+1

<∞.

This is possible since limn→∞ zn = ∞. Under this choice of ln we havethat

∣

∣

∣

∣

∣

∞∑

n=ln+1

zn

nRn

∣

∣

∣

∣

∣

<1

ln + 1

∣

∣

∣

∣

R

zn

∣

∣

∣

∣

ln+1

and for |z| ≤ R < |zn|/2

G(z, zn, ln) = 1 + ηn with |ηn| < e1

ln+1 | Rzn |ln+1

− 1.

and∑

n |ηn| converges. Hence, the infinite product converges uniformlyin any domain |z| < R for any positive R. Therefore G(z) is an entirefunction with the prescribed zeros.


Remark 20. If∑∞

n=11

|zn| is convergent then we can set ln = 0. If we

look for a function with zeros ±nπ, n ∈ N then we can choose ln = 1since

∑

1n2 is convergent. We then get

G(z) = z

∞∏

n=1

(

1 − z

nπ

)

eznπ

∞∏

n=1

(

1 +z

nπ

)

e−znπ

= z∞∏

n=1

(

1 − z2

n2π2

)

= sin z.

Next we show that there are not so many different entire functionswith pregiven zeros.

Theorem 33. Any entire function without zeros has the form

f(z) = eh(z)

where h(z) is an entire function.

Proof. We notice that under these conditions the function

f ′(z)

f(z)=

∞∑

n=0

anzn

is an entire function and its Taylor expansion converges in the entireplane. Also the function

g(z) =∞∑

n=0

ann+ 1

zn+1

is entire since it has the same radius of convergence and g′(z) = f ′(z)f(z)

,

i.e. f ′(z) − g′(z)f(z) = 0. This implies that(

f(z)e−g(z))′

= f ′(z)e−g(z) − f(z)g′(z)e−g(z) = 0

and for some constant c

f(z)e−g(z) = c

orf(z) = elog c+g(z).

Now f(z)G(z)

is an entire function without zeros provided f(z) has also

the sequence (zn)n as its only zeros. Hence,

Corollary 8. Let f be a function with zero’s (zn)n. Then

f(z) = eh(z)G(z).

where h(z) is an entire function.


17.2. Meromorphic functions. Meromorphic functions are general-izations of entire functions.

Definition 17. A function is called meromorphic if it is analytic in theentire complex plane except on some (may be infinite) isolated poles.

Theorem 34. Any meromorphic function is the quotient of two entirefunctions without common zeros.

Proof. Let zn be the poles of order kn for the meromorphic functionf . By the theorem of Weierstraß we can find an entire function h(z)which has zeros at zn of multiplicity kn. Moreover, one can writeh(z) = (z − zn)

knhn(z) where hn(zn) 6= 0 is entire. On the other handf(z) = (z − zn)

−knfn(z), where fn(zn) 6= 0 is analytic near zn. Hence,f(z)h(z) = fn(z)hn(z) is analytic near zn and not vanishing. The onlypossible singularities of the function g(z) = f(z)h(z) can occur only atthe places where f has singularities. But at zn they are removable. Sog is entire and g(zn) 6= 0.

Remark 21. We have a ”correspondence” polynomials ⇐⇒ entirefunctions and rational functions ⇐⇒ meromorphic functions.

We can ask whether we can prescribe the poles for a meromorphicfunction.

Theorem 35 (Mittag–Leffler). Let (zn)n be a sequence not accumulat-ing in C. For each n there is given a rational function

Hn(z) =a

(n)1

z − zn+

a(n)2

(z − zn)2+ · · ·

a(n)ln

(z − zn)ln.

Then there is a meromorphic function with the only poles at zn and theprincipal parts of the poles are Hn(z).

Remark 22. If we have two meromorphic functions with the sameprescribed principal parts of their poles (the poles are at the same pointstoo) then their difference f(z)− g(z) has removable singularities at zn.Therefore any two such functions differ only by an entire function.

Proof. First we assume that z = 0 is not a pole. then we can for|z| < |zn| expand the corresponding principal part into a Taylor series

Hn(z) = a(n)0 + a

(n)1 z + a

(n)2 z2 + · · ·

that converges uniformly for |z| < |zn|/2. Now we modify the principalparts in order to make them summable. Let ln be so large that

∣

∣

∣Hn(z) − a

(n)0 − a

(n)1 z − a

(n)2 z2 − · · · − a

(n)lnzln∣

∣

∣<

1

2n

for |z| < |zn|/2. The function

Hn(z) − a(n)0 − a

(n)1 z − a

(n)2 z2 − · · · − a

(n)lnzln


is a rational function with exactly one pole at zn and principle partHn. Then

g(z) =

∞∑

n=1

(

Hn(z) − a(n)0 − a

(n)1 z − a

(n)2 z2 − · · · − a

(n)lnzln)

is a meromorphic function with the prescribed principal parts. It con-verges uniformly in any disc of radius R > 0 after removing the poleszn. This is because for any R > 0 there is anm ∈ N such that |zn| > 2Rfor all n ≥ m. Hence,

∞∑

n=m

(

Hn(z) − a(n)0 − a

(n)1 z − a

(n)2 z2 − · · · − a

(n)lnzln)

converges uniformly for |z| < R.If z = 0 is a pole we have to modify the function g to

g0(z) = H0(z) + g(z)

.

Example 29. Let the poles zn = n be at the integers and

Hn(z) =1

z − zn=

−1

zn

1

1 − zzn

= − 1

zn− z

z2n

− · · ·

We can choose the approximations − 1zn

and get∣

∣

∣

∣

Hn(z) −(

− 1

zn

)∣

∣

∣

∣

=

∣

∣

∣

∣

1

z − zn+

1

zn

∣

∣

∣

∣

=

∣

∣

∣

∣

z

(z − zn)zn

∣

∣

∣

∣

≤ C

k2.

We then have

g0(z) =1

z+

∞∑

n=1

(

1

z − zn+

1

zn

)

=1

z+ lim

N→∞

N∑

n=1

(

1

z − n+

1

n+

1

z + n− 1

n

)

=1

z+

∞∑

n=1

(

1

z − n+

1

z + n

)

=1

z+

∞∑

n=1

2z

z2 − n2

The general form is

f(z) = g(z) +1

z+

∞∑

n=1

(

1

z − n+

1

z + n

)

= g(z) +1

z+

∞∑

n=1

2z

z2 − n2


where g(z) is an entire function.If we consider the function

f(z) = π cot πz

it has a principal part at the integer m of the form

1

z −m.

So in order to get a nice series expansion we have to find the functiong. One can check that g ≡ 0 (compare with example 25). Hence,

π cotπz =1

z+

∞∑

n=1

(

1

z − n+

1

z + n

)

=1

z+

∞∑

n=1

2z

z2 − n2.


18. Residues

Laurent series are a way to deal with isolated singularities. We willsee that the calculation of residues is connected to Laurent series andhow it can be applied in many questions.

Definition 18. Let f(z) be analytic in a domain bounded by a simpleclosed curve Γ and punctured at z0. Then the residue of f at z0 isdefined as

Res(f, z0) :=1

2πi

∫

Γ

f(z) dz

where the orientation on Γ is taken positive.

Remark 23. The residue does not depend on the particular choice ofΓ.

Under the above assumption we can expand f about z0 into a Laurentseries

∑∞n=−∞ an(z− z0)

n which is convergent in a deleted disc of someradius R. If the curve Γ is contained in this disc we can performintegration term–by–term

Res(f, z0) =1

2πi

∫

Γ

∞∑

n=−∞an(z − z0)

n dz

=

∞∑

n=−∞

(

an ·1

2πi

∫

Γ

(z − z0)n dz

)

Using that

1

2πi

∫

Γ

(z − z0)n dz =

0 n 6= −1

1 n = −1

we conclude thatRes(f, z0) = a−1.

Example 30. We want to calculate∫

|z|=2

z sin1

zdz

This integral equals the residue at z = 0 which is the only singularity.From

sin z = z − z3

3!+z5

5!− z7

7!+ · · ·

we obtain

z sin1

z= 1 −

(

1z

)2

3!+

(

1z

)4

5!−(

1z

)6

7!+ · · ·

Hence,

Res(z sin1

z, 0) = 0.


So we have an example of a function with an isolated singularity butnevertheless the residue is 0.

Theorem 36 (Residue theorem). Let Γ be a simple closed curve bound-ing a domain on which the function f is analytic except for isolatedsingularities z1, z2, · · · , zn. Then

1

2πi

∫

Γ

f(z) dz =n∑

k=1

Res(f, zk).

Proof. We only need to consider the following figures 19 and 20:

.

. .

.

C

C

C

C

z

z

2

Γ

1z

3

1

2

z

4

4

3

Figure 19

There we see

1

2πi

∫

Γ

f(z) dz = −n∑

k=1

1

2πi

∫

−Ckf(z) dz

=

n∑

k=1

1

2πi

∫

Ck

f(z) dz

=n∑

k=1

Res(f, zk)


.

. .

.

C

C

C

z

z

2

Γ

1z

3

1

2

z

4

4

−

−

−

−3

C

Figure 20

Example 31.∫

|z|=6

1

z(z − 1)dz = Res

(

1

z(z − 1), 0

)

+Res

(

1

z(z − 1), 1

)

= −1 + 1

= 0.

For the residue at z = 1 we proceed as follows. First we use the partialfraction expansion

1

z(z − 1)= −1

z+

1

z − 1and the expand

−1

z=

−1

1 + (z − 1)= −(1 − (z − 1) + (z − 1)2 ∓ · · · ) |z − 1| < 1

to get the Laurentseries about z = 1

1

z(z − 1)= (z − 1)−1 − 1 + (z − 1) − (z − 1)2 ± · · · 0 < |z − 1| < 1.

Similarly, we obtain

1

z(z − 1)= −z−1 − 1 − z − z2 − · · · 0 < |z| < 1.


18.1. Finding the residues. Let f have a pole of order N at z0. Then

φ(z) := (z − z0)Nf(z) = a−N + a−N+1(z − z0) + · · ·+ a0(z − z0)

N + · · ·If the pole is of order 1 then N = 1 and

Res(f, z0) = a−1 = limz→z0

φ(z) = limz→z0

(z − z0)f(z).

Suppose now that f has a pole of order 2. Then

dφ

dz= a−1 + 2a0(z − z0) + · · ·

so that


dφ(z)

dz= lim

z→z0

d

dz

[

(z − z0)2f(z)

]

.

If f has a pole of order N we derive successively


1

(N − 1)!

dN−1φ(z)

dzN−1

=1

(N − 1)!limz→z0

dN−1

dzN−1

[

(z − z0)Nf(z)

]

.

Remark 24. If we don’t know the order of the pole we can proceedsimilar and guess a number M . If 0 < M < N then

1

(M − 1)!limz→z0

dM−1

dzM−1

[

(z − z0)Mf(z)

]

= limz→z0

dM−1

dzM−1

[

a−N(z − z0)−N+M + a−N+1(z − z0)

−N+M+1 + · · ·(M − 1)!

]

= limz→z0

a−N(M −N) · · · (M −N +M − 2)

(M − 1)!(z − z0)

M−N−M+1 + · · ·

= ∞.

On the other hand if M > N then


1

(M − 1)!limz→z0

dM−1

dzM−1

[

(z − z0)Mf(z)

]

= limz→z0

dM−1

dzM−1

(

a−N (z − z0)M−N + · · ·+ a−1(z − z0)

M−1 + · · ·)

(M − 1)!

= limz→z0

(

a−N(M −N) · · · (N −N) · · · (2M −N − 2)(z − z0)−N+1+

+ · · ·+ a−1(M − 1)! + a0M !(z − z0) + · · ·) 1

(M − 1)!

= a−1

We will consider the special case that f(z) = g(z)h(z)

where h has a zero

of order 1 at z0 and g(z0) 6= 0, i.e. f has a pole of order 1 at z0. Thenby the rule of L’Hopital

Res(f, z0) = limz→z0

(z − z0)g(z)

h(z)= lim

z→z0

(z − z0)g′(z) + g(z)

h′(z)=

g(z0)

h′(z0)(15)

Example 32. We consider

f(z) =ez

z2(z2 + 1)=

ez

z2(z + i)(z − i).

We use different methods:

Res(f, i) = limz→i

(z − i)ez

(z + i)(z − i)z2=

ei

−2i,

with g(z) = ez

z2and h(z) = z2 + 1 we get h′(z) = 2z and

Res(f,−i) =ez

z2

2z=e−i

2i

and finally

Res(f, 0) = limz→0

d

dz

z2ez

(z2 + 1)z2= lim

z→0

ez(z2 + 1) − 2zez

(z2 + 1)2= 1.


18.2. Residues in integral calculus I. Integrals involving trigono-metric functions are usually hard to compute. Here residual calculusis helpful.

Let us consider the integral type

∫ 2π

0

R(sinφ, cosφ) dφ.

Here we substitute

z = eiφ dz = ieiφdφ

so

sinφ =z − z−1

2icosφ =

z + z−1

2dφ =

dz

iz

and we have to integrate along the unit circle in positive direction.

Example 33.

I =

∫ 2π

0

dφ

k + sinφk > 1

This is the kind of integral that occurs in the Poisson formula. We get

I =

∫

|z|=1

dziz

k + z−z−1

2i

=

∫

|z|=1

2 dz

z2 + 2ikz − 1.

We get two poles in the plane:

z1,2 = i(

−k ±√k2 − 1

)

.

We have (by Vieta’s rule) z1z2 = −1 and |z1| = 1|z2| so that one is in

the circle the other outside. Since k > 1 then z1 = i(

−k +√k2 − 1

)

isinside the circle. Hence,

I = 2πiRes(f, z1) = 2πi2

2z + 2ik

∣

∣

∣

z=z1=

2π√k2 − 1

where we used (15). If we put k = ab

we get

∫ 2π

0

dφab

+ sinφ=

2π√

a2

b2− 1

or∫ 2π

0

dφ

a+ b sinφ=

2π√a2 − b2

.


−R R x

y

CA

. .ei3 π

/4e

i π/4

Figure 21

18.3. Residues in integral calculus II. We also can evaluate im-proper integrals.

Example 34.

I =

∫ ∞

−∞

x2

x4 + 1dx

Consider the closed contour C as in figure 21Then the complexified function z2

z4+1has 4 singularities

eiπ/4, ei3π/4, e−iπ/4, e−i3π/4

whereof only the first two are in the upper half plane. So∫

C

z2

z4 + 1dz = 2πiRes

(

z2

z4 + 1, ei3π/4

)

+ 2πiRes

(

z2

z4 + 1, eiπ/4

)

Therefore∫ R

−R

x2

x4 + 1dx+

∫

A

z2

z4 + 1dz

= 2πiRes

(

z2

z4 + 1, ei3π/4

)

+ 2πiRes

(

z2

z4 + 1, eiπ/4

)

Further we can derive∣

∣

∣

∣

∫

A

z2

z4 + 1dz

∣

∣

∣

∣

≤ πRmaxR

∣

∣

∣

∣

z2

z4 + 1

∣

∣

∣

∣

≤ πRR2

R4 − 1−→ 0


if R→ ∞.Hence,

I = 2πiRes

(

z2

z4 + 1, ei3π/4

)

+ 2πiRes

(

z2

z4 + 1, eiπ/4

)

.

Again by (15) (with g(z) = z2, h(z) = z4 + 1 and h′(z) = 4z3) we caneasily calculate the residues:

I = 2πi

(

1

4e−i3π/4 +

1

4e−iπ/4

)

=π√2.

We also could have used a semi-circle in the lower half plane.

We can derive the following general principle.

Theorem 37. Let P (x) and Q(x) be two polynomials and let the degreeof P exceed the one of Q by at least 2. Moreover let Q have no realroots and all other roots are simple. Then

∫ ∞

−∞

P (x)

Q(x)dx = 2πi

∑

singularities in the upper half plane

P (zj)

Q′(zj).

Remark 25. In the above formula we assumed that the zeros of Q aresimple. In the case of multiple zeros the formula has to be modifiedaccordingly.

18.4. Residues in integral calculus III. Often one encounters in-tegrals of the form

∫∞−∞ f(x) cos px dx. This happens often in Fourier

transforms. Unfortunately we cannot directly use the previous method.The following example shows a quite general method.

Example 35.

I =

∫ ∞

−∞

cos 3x

(x− 1)2 + 1dx

We have∫ R

−R

cos 3x

(x− 1)2 + 1dx+

∫

A

cos 3z

(z − 1)2 + 1dz

= 2πi∑

singularities in the upper half plane

Res

(

cos 3z

(z − 1)2 + 1, zj

)

In difference to the previous case the integral along the arc A does nottend to zero since the cos–function is unbounded along the imaginaryaxis. We have to evaluate a different integral.

We first see that

J =

∫

A

ei3z

(z − 1)2 + 1dz → 0.


For this we put z = Reiφ and dz = Reiφidφ. We get

J =

∫ π

0

ei3Reiφ

(z − 1)2 + 1Rieiφ dφ.

Since∣

∣

∣ei3Re

iφ∣

∣

∣=∣

∣ei3R cosφ∣

∣

∣

∣e−3R sinφ∣

∣ = e−3R sinφ.

Therefore

|J | ≤ R

∫ π

0

∣

∣

∣

∣

1

(z − 1)2 + 1

∣

∣

∣

∣

∣

∣

∣ei3Re

iφ∣

∣

∣dφ

= R

∫ π

0

∣

∣

∣

∣

1

(z − 1)2 + 1

∣

∣

∣

∣

e−i3R sinφ dφ

Since∣

∣

∣

∣

1

(z − 1)2 + 1

∣

∣

∣

∣

<1

(R− 1)2.

By the symmetry of sinφ and the fact that sin φ ≥ 2φπ

on 0 ≤ φ ≤ π2

wehave that

|J | ≤ 1

R− 1

∫ π

0

e−i3R sinφ dφ

=2

R− 1

∫ π2

0

e−i3R sinφ dφ

≤ 2

R− 1

∫ π2

0

e−i3R2φπ dφ

≤ π

3R2 − R

(

1 − e−3R)

→ 0.

This implies that∫ ∞

−∞

ei3x

(x− 1)2 + 1dx = 2πi

∑

singularities in the u. h. p.

Res

(

ei3z

(z − 1)2 + 1, zj

)

Since, ei3z = cos 3x + i sin 3x we can compare the real and imaginaryparts:∫ ∞

−∞

cos 3x

(x− 1)2 + 1dx = ℜ

[

2πi∑

sing. in the u. h. p.

Res

(

ei3z

(z − 1)2 + 1, zj

)

]

and∫ ∞

−∞

sin 3x

(x− 1)2 + 1dx = ℑ

[

2πi∑

sing. in the u. h. p.

Res

(

ei3z

(z − 1)2 + 1, zj

)

]


We have only one pole at z1 = 1+ i in the upper half plane. Using (15)we get

∫ ∞

−∞

cos 3x

(x− 1)2 + 1dx = ℜ

[

2πi limz→1+i

e3iz

2(z − 1)

]

ℜ[

πe3i(1+i)]

= ℜ[

πe3i−3]

= ℜ[

πe−3(cos 3 + i sin 3)]

= πe−3 cos 3

and∫ ∞

−∞

sin 3x

(x− 1)2 + 1dx = ℑ

[

2πi limz→1+i

e3iz

2(z − 1)

]

= ℑ[

πe−3(cos 3 + i sin 3)]

= πe−3 sin 3

18.5. Residues in integral calculus IV. In this section we allow theintegrand to have poles at the real line.

Theorem 38. Let f(z) have a simple pole at z0. Let C = Cα(r) be ansub-arc of angle α of the circle of radius r centered at z0 (see figure 22).Then

limr→0

∫

Cα(r)

f(z) dz = 2πi[ α

2πRes(f, z0)

]

.

αθ

θ

C

r

z

r

0

1

2

α (r)

Figure 22


Proof. Since f has a simple pole we get

f(z) =a−1

(z − z0)+

∞∑

n=0

an(z − z0)n =

a−1

(z − z0)+ g(z)

Hence,∫

Cα(r)

f(z) dz =

∫

Cα(r)

a−1

(z − z0)dz +

∫

Cα(r)

g(z) dz.

Since g is continuous its absolute value is bounded in a neighborhoodof z0 by some M . Therefore

∣

∣

∣

∣

∫

Cα(r)

g(z) dz

∣

∣

∣

∣

≤Mrα→ 0 (r → 0).

Changing to polar coordinates z = z0 + reiφ, dz = ireiφdφ we conclude∫

Cα(r)

a−1

(z − z0)dz =

∫ θ2

θ1

a−1ireiφ

reiφdφ = a−1iα = 2πi

α

2πa−1.

Since a−1 is the residue at z0 this concludes the proof.

Example 36.

I =

∫ ∞

−∞

cos 3x

x− 1dx

As in the previous example we will evaluate

J =

∫

C

e3iz

z − 1dz

but we have to change the contour since we have a pole on the real line.We will choose C as in figure 23

We get∫ 1−ǫ

−R

e3ix

x− 1dx+

∫

Cǫ

e3iz

z − 1dz +

∫ R

1+ǫ

e3ix

x− 1dx+

∫

A

e3iz

z − 1dz = 0

We can conclude as in the previous section that∫

A

e3iz

z − 1dz → 0

By using the previous theorem with α = −π and Res(

e3iz

z−1, 1)

= e3i we

have that∫

Cǫ

e3iz

z − 1dz = −iπe3i.

Therefore

I = limR→∞

limǫ→0

[∫ 1−ǫ

−R

e3ix

x− 1dx+

∫ R

1+ǫ

e3ix

x− 1dx

]

= iπ(cos 3 + i sin 3)

and∫ ∞

−∞

cos 3x

x− 1dx = ℜ(iπ(cos 3 + i sin 3)) = −π sin 3


−R R x

y

CA

.1 1+ εε1−

Figure 23

and∫ ∞

−∞

sin 3x

x− 1dx = ℑ(iπ(cos 3 + i sin 3)) = π cos 3.

18.6. Residue calculus in summation formulas.

Theorem 39. Let P and Q be polynomials where the degree of Q ex-ceeds the degree of P by at least two. Assume that Q(n) 6= 0 for alln ∈ N. Then

∞∑

−∞

P (n)

Q(n)= −π

∑

zeros of Q

Res

(

cot πzP (n)

Q(n), zj

)

Proof. Let CN be the square with corners (N+1/2)(1−i), (N+1/2)(1+i), −(N + 1/2)(1 − i) and −(N + 1/2)(1 + i). Let N be so large that

all poles of P (n)Q(n)

are contained in CN . The enclosed poles of cotπz are

0,±1,±2, · · · ,±N . Hence,∫

CN

π cot πzP (z)

Q(z)dz = 2πi

∑

zeros of Q

Res

(

π cot πzP (z)

Q(z), zj

)

+ 2πi

N∑

n=−NRes

(

π cot πzP (z)

Q(z), n

)


Formula (15) gives with g(z) = π cosπz · P (z)Q(z)

, h(z) = sin πz and

h′(z) = π cosπz

Res

(

π cot πzP (z)

Q(z), n

)

= Res

(

πcosπz

sin πz

P (z)

Q(z), n

)

=P (n)

Q(n). (16)

Let k = degQ− deg P . Then one can estimate on CN∣

∣

∣

∣

π cot πzP (z)

Q(z)

∣

∣

∣

∣

≤ π

∣

∣

∣

∣

eiπz + e−iπz

eiπz − e−iπz

∣

∣

∣

∣

∣

∣

∣

∣

P (z)

Q(z)

∣

∣

∣

∣

≤ π|eiπz| + |e−iπz|

|eiπxe−πy − e−iπxeπy|

∣

∣

∣

∣

P (z)

Q(z)

∣

∣

∣

∣

≤ π coth(π

2

) 1(

N + 12

)k

and∣

∣

∣

∣

∫

CN

π cot πzP (z)

Q(z)dz

∣

∣

∣

∣

≤ πcoth(π/2)

(N + 1/2)k4(2N + 1) → 0. (17)

Hence

limN→∞

∫

CN

π cot πzP (z)

Q(z)dz = 0

and

0 =∑

zeros of Q

Res

(

π cotπzP (z)

Q(z), zj

)

+N∑

n=−NRes

(

π cotπzP (z)

Q(z), n

)

Now equation (16) gives the theorem.

Example 37.

S =

∞∑

n=0

1

n2 + 1

We have that S0 =∑∞

n=−∞1

n2+1satisfies the assumptions of the theo-

rem and (S0 + 1)/2 = S. The theorem gives

S0 = −π(

Res

(

cot πz

z2 + 1, i

)

+Res

(

cot πz

z2 + 1,−i))

= −π[

cot πi

2i+

cot(−πi)−2i

]

= −πi

cot πi = π coth π.


Hence,∞∑

n=0

1

n2 + 1=π coth π + 1

2.


19. Laplace transforms

Laplace transforms play an important role in the solution of differen-tial equations. They can be analyzed with the help of residue calculus.

Definition 19. Let f(t) be a (real or complex valued) function in thereal variable t. Let s = σ+ iω be a complex variable. Then the Laplacetransform of f is defined as

F (s) = Lf(t) =

∫ ∞

0

f(t)e−st dt.

The function f is said to be the inverse Laplace transform of F and iswritten as L−1F (s).

We have that for b = α+ iβ

Le−bt =

∫ ∞

0

e−bte−st dt =e−(s+b)t

−(s+ b)

∣

∣

∣

∞

0=e−(σ+α)te−i(ω+β)t

−(s + b)

∣

∣

∣

∞

0=

1

s+ b

if ℜ(s) = σ > −α = −ℜ(b).We want to list some properties of the Laplace transformation.

1. L and L−1 are linear, i.e.

L(af(t) + bg(t)) = aL(f(t)) + bL(g(t))

and

L−1(aF (s) + bG(s)) = aL−1(F (s)) + bL−1(G(s)).

2. If f is piecewise continuous for t ≥ 0 and there are numbersk, p, T such that

|f(t)| < kept for all t ≥ T (18)

then its Laplace transform exists and is analytic in ℜ(s) > p3. If f(t) and g(t) have the same Laplace transform then for t ≥ 0

they coincide almost everywhere. Nothing can be said aboutnegative t.

4. We consider Ldfdt

=∫∞0

dfdte−st dt. Then

∫ ∞

0

df

dte−st dt = e−stf(t)|∞0 +

∫ ∞

0

sf(t)e−st dt.

If f satisfies (18) then e−stf(t) → 0 for t → ∞ and ℜ(s) > p.hence,

Ldf

dt= 0 − f(0) + sL(f(t)) = sF (s) − f(0).

We can use this formula even in the case when f has a jumpsingularity at zero if we substitute f(0) by f(0+).


5. In general we can derive by induction

Ld2f

dt2=

∫ ∞

0

d2f

dt2e−st dt

= e−stdf

dt

∣

∣

∣

∞

0+ s

∫ ∞

0

df

dte−st dt

= −f ′(0) + s(sF (s) − f(0))

= s2F (s) − sf(0) − f ′(0)

and

Ldnf

dtn= snF (s) − sn−1f(0) − sn−2f ′(0) − · · · − f (n−1)(0)

provided all the derivatives are of order (18).6.

L

∫ t

0

f(x) dx =1

sLf(t) =

F (s)

s.

This can be seen as follows. Let g(t) =∫ t

0f(x) dx. Then f(t) =

g′(t) and g(0) = 0. By the previous formula we have

L(f(t)) = L(g′(t)) = sG(s) −G(0) = sG(s) = sL

(∫ t

0

f(x) dx

)

.

Example 38. We will use the Laplace transform to solve differentialequations. Let

df

dt+ 2f(t) = e−3t t ≥ 0

and f(0) = 4.The Laplace transform gives

sF (s) − 4 + 2F (s) =1

s+ 3

and

F (s) =1

(s+ 3)(s+ 2)+

4

s+ 2.

one can show that

L−1F (s) = 5e−2t − e3t

satisfies the differential equation.


The main difficulty in this approach is to determine the inverse ofthe Laplace transform. We will consider the case that F (z) is analyticin ℜ(z) ≥ a and there are positive numbers m, k,R0 such that for|z| > R0 and ℜ(z) ≥ a we have

|F (z)| ≤ m

|z|k .

We will integrate around the curve shown in figure 24.

.

a a+b

C

zs

b

1

C

b−(s−a)

Figure 24

F (s) =1

2πi

∫

C

F (z)

z − sdz =

1

2πi

[∫ a−ib

a+ib

F (z)

z − sdz +

∫

C1

F (z)

z − sdz

]

We have on C1 that |z| ≥ b or 1/|z| ≤ 1/b and |F (z)| ≤ m/bk. Theminimal distance between s and z on C1 is b−|s−a|. Hence, |z− s| ≥b− (|s| + a). This yields

∣

∣

∣

∣

∫

C1

F (z)

z − sdz

∣

∣

∣

∣

≤ πb · m

bk(b− |s| − a)→ 0

as b→ ∞. Hence,

F (s) =1

2πi

∫ a+i∞

a−i∞

F (z)

s− zdz.

Using z = a+ iy and dz = idy we get


Theorem 40. Under the above assumptions on F we have for ℜ(s) > a

F (s) =1

2π

∫ +∞

−∞

F (a+ iy)

s− (a + iy)dy.

Let f satisfy (18). Let a > p. We are going to show that

f(t) =1

2πi

∫ a+i∞

a−i∞F (s)est ds

where we integrate along a vertical line. Lets compute the Laplacetransform of f given in this form. We put s = σ + iω and set σ = a.Thus ds = idω and

f(t) =1

2π

∫ +∞

−∞F (a+ iω)e(a+iω)t dω.

Then

Lf(t) =1

2π

∫ +∞

0

e−st∫ +∞

−∞F (a+ iω)e(a+iω)t dωdt

=1

2π

∫ +∞

0

∫ +∞

−∞e−(s−a)tF (a+ iω)eiωt dωdt

=1

2π

∫ +∞

−∞F (a+ iω)

[∫ +∞

0

e−ste(a+iω)t dt

]

dω

=1

2π

∫ +∞

−∞F (a+ iω)

[

L(

e−(s−a−iω)t)]

dω

=1

2π

∫ +∞

−∞F (a+ iω)

[

1

s− a− iω

]

dω

=1

2π

∫ +∞

−∞

F (a+ iω)

s− (a+ iω)dω = F (s).

We can perform the change of order of integration provided that ℜ(s) >

a and∫ +∞−∞ |F (a + iω)| dω < ∞. Now we use the previous theorem to

derive

Theorem 41 (Laplace inversion formula). Under the above assump-tions

f(t) = L−1F (s) =

1

2πi

∫ a+i∞

a−i∞F (s)est ds


Example 39. Let us compute L−1 1(s+1)2

. We have

f(t) =1

2πi

∫ a+i∞

a−i∞

est

(s+ 1)2ds

with a pole at −1. So we take a = 0. Let us consider the curve C infigure 25.

R

−R

ω

σ.−1

AC

Figure 25

We get

1

2πi

∫

C

est

(s+ 1)2ds =

1

2πi

∫ iR

−iR

est

(s+ 1)2ds+

1

2πi

∫

A

est

(s+ 1)2ds

Then, since the pole is of order 2,

1

2πi

∫

C

est

(s+ 1)2ds =

2πi

2πiRes

(

est

(s+ 1)2,−1

)

= lims→−1

d

dsest = te−t

and

limR→∞

1

2πi

∫

A

est

(s+ 1)2ds = 0 t ≥ 0.

Hence,

te−t =1

2πi

∫ i∞

−i∞

est

(s+ 1)2ds = L

−1 1

(s + 1)2.


This example can be generalized

Theorem 42. Let F be analytic in the s–plane except for a finite num-ber of poles that lie to the left of some vertical line ℜ(s) = a. Supposethere exist m, k,R0 such that for all s with ℜ(s) ≤ a, |s| > R0 we have|F (s)| ≤ m/|s|k. Then for t > 0

L−1F (s) =

∑

poles of F

Res(F (s)est, sj).

Remark 26. We have to have all singularities to the left since theinverse Laplace formula is an integral with respect to a line such thatthe function is analytic to the right! Moreover we need that the integralalong A tends to zero which is only for large negative real parts.

Remark 27. The preceding theorem can be used to compute the inverseLaplace transform for rational functions P/Q where the degree of Q ishigher than that of P .

In engineering one often encounters functions

F (s) = (P (s)/Q(s))e−sτ = G(s)e−sτ

where τ > 0. Here we cannot use the formula directly since e−sτ isunbounded in any left half–plane. We are first computing the inverse

Laplace transform L−1(

P (s)Q(s)

)

= L−1(G(s)) = g(s). Let u(t) = 0 for

t < 0 and u(t) = 1 for t ≥ 0. We then have

L [g(t− τ)u(t− τ)] =

∫ ∞

0

e−stg(t− τ)u(t− τ) dt

=

∫ ∞

−τe−s(r+τ)g(r)u(r) dr

=

∫ ∞

0

e−s(r+τ)g(r)u(r) dr

= e−sτ∫ ∞

0

e−srg(r)u(r) dr

= e−sτL [g(t)u(t)] = e−sτLg(t) = e−sτG(s).

Conversely,

L−1(F (s)) = L

−1[

e−sτG(s)]

= g(t− τ)u(t− τ), τ ≥ 0.

Sometimes one can deal with infinitely many poles if the residue aresummable.


19.1. Stability.

Example 40. Consider a mass point of mass m attached to a springwithout friction and elasticity constant k. Let there be an external forceof the form F0 cosω0t, t ≥ 0 acting in the same line as the spring force.Let y be the displacement of the mass point to the point the spring exertsno force (see figure 26).

.y(t)

k m F cos( t)0

ω0

Figure 26

Then Newton’s second law gives

md2y

dt2+ ky = F0 cos(ω0t) = F0

(

eiω0t + e−iω0t

2

)

.

with dy/dt|t=0 = 0 and y(0) = 0. The Laplace transform is

ms2Y (s) + kY (s) =F0

2

(

1

s− iω0+

1

s+ iω0

)

= F0s

s2 + ω20

or

Y (s) =

(

1

s2 + km

)

(

F0s

m(s2 + ω20)

)

.

If ω0 6=√

k/m then there are four simple poles at s = ±iω0 and s =

±i√

k/m. If ω0 =√

k/m then there is a pole of order two at s = ±iω0.


Using the residue calculus theorem we get that in case ω0 6=√

k/m

Res

(

Y (s)est, i

√

k

m

)

+Res

(

Y (s)est,−i√

k

m

)

=

(

F0s

m(s2 + ω20)

)

est1

2s

∣

∣

∣

s=i√

km

+

(

F0s

m(s2 + ω20)

)

est1

2s

∣

∣

∣

s=−i√

km

=F0

m(ω20 − k

m)

ei√

km + e−i

√km

2

and

Res(

Y (s)est, iω0

)

+Res(

Y (s)est,−iω0

)

=

(

F0s

m(s2 + km

)

)

est1

2s

∣

∣

∣

s=iω0

+

(

F0s

m(s2 + km

)

)

est1

2s

∣

∣

∣

s=−iω0

= − F0

m(ω20 − k

m)

eiω0 + e−iω0

2

So

y(t) =F0

(

ω20 − k

m

)

m

[

− cos(ω0t) + cos

(

√

k

m

)]

and in case ω0 =√

k/m we have

Res(

Y (s)est, iω0

)

= lims→iω0

d

ds

(

(s− iω0)2Y (s)est

)

=d

ds

(

F0

msest

1

(s + iω0)2

)

∣

∣

∣

s=iω0

=F0

m

(

est

(s + iω0)2+ (

stest

(s+ iω0)2− (

2sest

(s+ iω0)3

)

∣

∣

∣

s=iω0

=F0

m

(

eiω0t

−4ω20

+ (iω0te

iω0t

−4ω20

− (2iω0e

iω0t

8iω30

)

=F0

2mω0

teiω0t

2i


and similar

Res(

Y (s)est,−iω0

)

= − F0

2mω0

te−iω0t

2i

Hence,

y(t) =F0

2mω0

t sin(ω0t).

While the first case produces two cos–waves and is bounded, the secondcase has unbounded solutions.

Let us consider the more general case

Y (s) =P (s)

Q(s)deg P < degQ.

Then

L−1P (s)

Q(s)= f(t) =

∑

zeros of Q

Res

(

P (s)

Q(s)est, zj

)

.

If Q has a root of order N at s = a+ ib the residue has one of the forms

tN−1eat cos(bt), · · · , eat cos(bt)

or

tN−1eat sin(bt), · · · , eat sin(bt)

corresponding to the partial fraction expression. Note that always theterm of order N − 1 occurs. Then we have three possibilities

1. The root is in the right half of the s–plane. Then a > 0 and eachterm is unbounded in t.

2. The root is in the left half of the s–plane. Then a < 0 and eachterm tends to zero as t→ +∞.

3. a = 0 then the functions are bounded if and only if N = 1.

Often one encounters systems which produce an output correspond-ing to some input. One of the questions is whether a bounded inputx(t) produces a bounded output y(t). We will use Laplace transformsto solve this question. Let the Laplace transforms be related as follows

Y (s) = G(s)X(s).

The function G is called the transfer function of the system. As anexample consider a system given by

andny

dtn+ · · ·+ a1

dy

dt+ a0y = x(t)

with initial conditions to be zero at t = 0. The Laplace transform is

ansnY (s) + · · ·+ a1sY (s) + a0Y (s) = X(s).

Then

G(s) =1

ansn · · ·+ a1s+ a0.


Some more general systems have a transfer function of the form

G(s) =A(s)

B(s)

where A,B are polynomials.

Definition 20. A system which produces for any bounded input a boun-ded output is called stable.

If x(t) is bounded then X(s) has no poles to the right of the imagi-nary axis and only simple poles at the imaginary axis. If now G(s) hasall its poles to the left of the imaginary axis Y (s) cannot have polesto the right of the imaginary axis and the poles on the imaginary axisare those of X(s). Hence the system is stable. If G has poles to theright of the imaginary axis so has Y . If G has a pole (simple or not) atthe imaginary axis we consider an input x with Laplace transform Xhaving a simple pole at the same place. This produces a pole of higherorder for Y . We have proven

Theorem 43. A system is stable if and only if all the poles of itstransfer function are to the left of the imaginary axis.

Remark 28. Note that the conditions are different to the previoustheorem.

Example 41.

d3y

dt3− a

d2y

dt2+ b2

dy

dt+ ab2y(t) = x(t)

with all initial conditions equal to zero. The Laplace transform is

(s3 − as2 + b2s− ab2)Y (s) = X(s)

and

G(s) =1

(s− a)(s2 + b2).

For a > 0 G has a pole to the right and the system is unstable. Leta < 0. Then there are no poles to the right. But G has poles at ±iband the system is unstable.


19.2. Principle of the argument. In the previous section we haveseen that the location of poles (or equivalently zeros for rational func-tions) play a fundamental role. Unfortunately it is not always simpleto find them. We will present a method which can be used.

Consider a function which is analytic inside and on a simple closedcurve C with may be a finite number of poles inside the contour. Ifwe go around this contour we end up with the same value we startedwith. However if we write

f(z) = |f(z)|ei arg f(z)

the value of the argument does not need to be the same. It can changein accordance with how often we encircle the origin (note that thereis a sign involved according to the direction we move!). Let us write∆C arg f(z) for the increase in the argument as we go around once inthe positive direction.

Example 42.

∆|z|=1 arg z = 2π

while

∆|z|=1 arg1

z2= −4π

This is easily checked by using polar coordinates: z = eiφ. Then 1z2

=

e−2iφ and we go around twice in negative direction.

We note that ∆C arg f(z) is always a multiple of 2π since the valueof f is unchanged.

Now consider

1

2πi

∫

C

f ′(z)

f(z)dz =

1

2πi

∫

C

d

dzlog f(z) dz

=1

2πi

∫

C

d(log f(z))

=1

2πi[ increase in log f(z) in going around C ]

=1

2πi[ increase in (log |f(z)| + i arg f(z)) in going around C ]

=1

2π∆C arg f(z).

if we fix some branch of the logarithm and C does not pass through anypole or zero of f . We can calculate this value differently using residue.

The quotient f ′(z)f(z)

is analytic except at the poles of f ′ (which cannot


occur at other places than those of f since f ′ is analytic where f is)and the zeros of f . Assume that f has a zero of order n at z0. Thenwe can write

f(z) = (z − z0)nφ(z) φ(z0) = an.

By differentiating we get

f ′(z)

f(z)=

n

z − z0+φ′(z)

φ(z)

becausef ′(z) = n(z − z0)

n−1φ(z) + (z − z0)nφ′(z).

The first expression is a simple pole with residue n The second term

has no singularity at z0. This means that the residue of f ′(z)f(z)

at z0equals the multiplicity of the zero.

Suppose that f has a pole of order p at z1 inside C. Similarly we

can obtain that the residue of f ′(z)f(z)

at z1 equals (−1)p:

f(z) =φ(z)

(z − z1)pφ(z1) 6= 0

Hence,

f ′(z) =−p

(z − z1)p+1φ(z) +

φ′(z)

(z − z1)p

andf ′(z)

f(z)=

−pz − z1

+φ′(z)

φ(z).

Now the residue theorem tells that

1

2πi

∫

C

f ′(z)

f(z)dz =

∑

poles inside C

Res

(

f ′(z)

f(z), zi

)

= NP

where N is the sum of the multiplicities of the zeros of f inside C andP is the sum of the orders of the poles of f inside C. This togetherwith earlier arguments proves

Theorem 44 (Principle of the argument). Let f be analytic on andinside a simple closed curve C except at a finite number of poles. Alsoassume that f has no zeros at C. Then

1

2π∆C arg f(z) = N − P

where N and P are the number of zeros or poles, respectively, lying in-side the circle and counted with their multiplicity or order, respectively.


20. The principle of the argument II

Example 43. How many zeros has the function f(z) = ez − 2z insidethe circle |z| = 3?

We will map the circle |z| = 3 into the w = f(z)–plane by meansof the map f(z). This is somewhat tedious but might be done with asimple computer program. We get the following picture 27

14

|z|=3

z−plane

w=f(z)−plane

f(z)=e −2z

3

z

a

b

c

d

f(a)

f(b)

f(c)

f(d)

Figure 27

We see that f moves this circle twice around zero in positive direc-tion. Hence,

∆|z|=3 arg f(z) = 4π.

The principle of the argument gives N − P = 2. Since there are nopoles inside the circle it has to have two zeros there.

Theorem 45 (Rouche). Let f, g be analytic on and inside a simpleclosed curve C. Suppose |f(z)| > |g(z)| on C. Then f and f + g havethe same number of zeros inside C.

Proof. Since there are no poles inside C we have by the principle of theargument that

∆C arg f(z) = 2πNf

and

∆C arg(f(z) + g(z)) = 2πNf+g.


From

f + g = f

[

1 +g

f

]

it follows

2πNf+g = ∆C arg f(z) + ∆C arg

(

1 +g(z)

f(z)

)

.

(Let h = fg then arg h = arg f + arg g.) Now, by the assumption onC we have

ℜ(

1 +g(z)

f(z)

)

= 1 + ℜ g(z)f(z)

> 1 −∣

∣

∣

∣

g(z)

f(z)

∣

∣

∣

∣

> 0.

Therefore the image of C under 1+ g(z)f(z)

cannot circle around the origin

since otherwise the image has to intersect the negative real axis. Hence,

∆C arg

(

1 +g(z)

f(z)

)

= 0

concluding the theorem.

We can now give another proof of the fundamental theorem of alge-bra:

Theorem 46 (Fundamental Theorem of Algebra II). Any polynomial

P (z) = anzn + · · · + a1z + a0 n ≥ 1, an 6= 0

has exactly n complex roots counted with multiplicity.

Proof. Let us considerf(z) = anz

n

andg(z) = an−1z

n−1 + · · ·a1z + a0.

Then on |z| = r > max(1,Pn−1k=0 |ak ||an| ) we have

|g(z)||f(z)| ≤

∑n−1k=0 |ak|rk|an|rn

≤∑n−1

k=0 |ak||an|r

< 1.

Hence, by the principle of the argument we have that f(z)+g(z) = P (z)has the same number of zeros inside z : |z| < r as the functionanz

n = f(z). But anzn has a root of mutiplicity n at zero. Therefore

P (z) has exactly n roots with absolute value less than r. Since r canbe chosen arbitrarily large the theorem follows.


21. Conformal mappings I

In this section we will consider an analytic function as a map f : D ⊂C → C. In most of the literature conformal mapping means a one–to–one mapping. We will not do so here but restrict to the geometricproperty first and then introduce one–to–one mappings.

Definition 21. A mapping w = f(z) that preserves the size and ori-entation of the angle of intersection of any two intersecting curves ina given point z0 is said to be conformal at z0. A mapping that is con-formal at any point (and one-to-one onto) in a domain D is calledconformal in D.

Theorem 47. Let f(z) be analytic in a domain D. Then it is confor-mal at any point z ∈ D where f ′(z) 6= 0.

Proof. Let

z(t) = x(t) + iy(t) w(t) = f(z(t)) = u(x(t), y(t)) + iv(x(t), y(t))

be a smooth curve C and its image C ′ under f . Then the tangent of Cat z0 is given by dz

dt|t=t0 and the tangent of of C ′ at w0 = f(z0) is given

by dwdt|t=t0 . By the chain rule we get

dw

dt

∣

∣

∣

t=t0=dw

dz

dz

dt

∣

∣

∣

t=t0= f ′(z0)

dz

dt

∣

∣

∣

t=t0.

Equating the arguments we derive

argdw

dt

∣

∣

∣

t=t0= arg f ′(z0) + arg

dz

dt

∣

∣

∣

t=t0.

Therefore any (tangent) vector at z0 will be increased by the valuearg f ′(z0) independent of the original angle. Hence, for two curvesz1(t) and z2(t) the angle between w1(t) and w2(t) at t0 is given by

dw1

dt

∣

∣

∣

t=t0− dw2

dt

∣

∣

∣

t=t0= arg f ′(z0) + arg

dz1dt

∣

∣

∣

t=t0

− arg f ′(z0) − argdz2dt

∣

∣

∣

t=t0

= argdz1dt

∣

∣

∣

t=t0− arg

dz2dt

∣

∣

∣

t=t0

and the angle is preserved. If f ′(z0) = 0 the argument is not definedand we cannot guarantee the conformality.

Remark 29. One can show that if f ′(z0) = 0 the map cannot be con-formal at this point. Such points are called critical. To see this we havethat f(z) = f(z0) + (z − z0)

nφ(z) where n > 1 and φ is analytic. But


then the argument arg dzdt

∣

∣

∣

t=t0is changed to n · arg dz

dt

∣

∣

∣

t=t0and the angle

between two curves will be increased by a factor n.

Let ∆z be a small line segment it is mapped under f into a (almostlinear) curve connecting f(z0) and f(z0+∆z). The ratio of their lengthtends to

lim∆z→0

∣

∣

∣

∣

f(z0 + ∆z) − f(z0)

∆z

∣

∣

∣

∣

= |f ′(z0)|

independent of the direction of the segment. Hence, a conformal map-ping scales a sufficiently small ”geometric figure” by the factor |f ′(z0)|and then rotates it around w0 = f(z0) by an angle of size arg f ′(z0).This does not necessarily hold on large scales since the derivative of fmay deviate substantially.

We are now going to show that either of the two properties imply an-alyticity. Before that we recall some facts from elementary differentialgeometry.

Let z(t) = x(t) + iy(t) be a curve. The length of its tangent vectoris given by

∣

∣

∣

∣

dz(t)

dt

∣

∣

∣

t=t0

∣

∣

∣

∣

= |(x′(t0), y′(t0))| =√

(x′(t))2 + (y′(t))2.

Its arclength σ(t) is given by

σ(t) =

∫ t+t0

t0

√

(x′(t))2 + (y′(t))2 dt

ordσ =

√

(x′(t))2 + (y′(t))2 dt.

Assume that the curve is parametrised by its arclength. Then σ(t) = tand

dt =√

(x′(t))2 + (y′(t))2 dt

or√

(x′(t))2 + (y′(t))2 = 1.

Let α be the angle of the tangent vector

dz(t)

dt

∣

∣

∣

t=t0= (x′(t0), y

′(t0))

with the real axis. Then

cosα =x′(t0)

√

(x′(t0))2 + (y′(t0))2=

dxdtdσdt

=dx

dσ= x′(t0)

and

sinα =y′(t0)

√

(x′(t0))2 + (y′(t0))2=

dydtdσdt

=dy

dσ= y′(t0)


Theorem 48. Let f(x, y) = u(x, y) + iv(x, y) be a mapping that hascontinuous derivatives and is one-to-one on a domain Ω and the dif-ferential has full rank. Moreover let f preserve the angle between anytwo curves. Then f is analytic.

Proof. We are going to prove that f is differentiable at any point z0 ∈Ω.

Let us consider a pair of straight lines zi(t) throug z0 having angleαi, i = 1, 2, with the real axis, i.e. (in arclength presentation)

xi(t) = x0 + cosαi · t yi(t) = y0 + sinαi · t.

Now we remark that the differential has full rank, i.e. the determinant

∂u

∂x· ∂v∂y

− ∂u

∂y· ∂v∂x

6= 0.

Hence,

(

dσidt

)2

=

(

du(xi(t), yi(t))

dt

)2

+

(

dv(xi(t), yi(t))

dt

)2

=

(

∂u

∂xx′i +

∂u

∂yy′i

)2

+

(

∂v

∂xx′i +

∂v

∂yy′i

)2

6= 0


wher σi is the arclength of the curve f(zi(t)). Therefore we can proceedwith βi the angle between f(zi(t)) and the real axis at t = t0

sin(β2 − β1) = sin β1 cosβ2 − sin β2 cosβ1

=du(x1(t), y1(t)

dσ1· dv(x2(t), y2(t))

dσ2− du(x2(t), y2(t))

dσ2· dv(x1(t), y1(t))

dσ1

=

(

du(x1, y1)

dt· dv(x2, y2)

dt− du(x2, y2)

dt· dv(x1, y1)

dt

)

dt

dσ1

dt

dσ1

=[

(

∂u

∂x· x′1 +

∂u

∂y· y′1)(

∂v

∂x· x′2 +

∂v

∂y· y′2)

−

−(

∂u

∂x· x′2 +

∂u

∂y· y′2)(

∂v

∂x· x′1 +

∂v

∂y· y′1)

] dt

dσ1

dt

dσ1

=[

(

∂u

∂xcosα1 +

∂u

∂ysinα1

)(

∂v

∂xcosα2 +

∂v

∂ysinα2

)

−

−(

∂u

∂xcosα2 +

∂u

∂ysinα2

)(

∂v

∂xcosα1 +

∂v

∂ysinα1

)

] dt

dσ1

dt

dσ1

=

(

∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

)

(cosα1 sinα2 − sinα1 cosα2)dt

dσ1

dt

dσ1

=

(

∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

)

sin(α2 − α1)dt

dσ1

dt

dσ1.

Since we have that sin(β2 − β1) = sin(α2 − α1) we derive that

∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x=dσ1

dt

dσ2

dt.

Hence the arcelement is independend of the curve

(

dσ

dt

)2

=∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x.


Now(

dσ

dt

)2

=

(

du

dt

)2

+

(

dv

dt

)2

=

(

∂u

∂x· x′ + ∂u

∂y· y′)2

+

(

∂v

∂x· x′ + ∂v

∂y· y′)2

= x′2

[

(

∂u

∂x

)2

+

(

∂y

∂x

)2]

+ 2x′y′[

∂u

∂x

∂u

∂y+∂v

∂x

∂v

∂y

]

+

+ y′2

[

(

∂u

∂y

)2

+

(

∂y

∂y

)2]

=∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

For a curve with x′ = 1, y′ = 0 we have

(

∂u

∂x

)2

+

(

∂y

∂x

)

=∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

and for a curve with x′ = 0, y′ = 1 we have

(

∂u

∂y

)2

+

(

∂y

∂y

)2

=∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

Adding these inequalities we get

(

∂u

∂x− ∂v

∂y

)2

+

(

∂u

∂y+∂v

∂x

)2

= 0

what implies the Cauchy–Riemann equations.

Theorem 49. Let the map f = u+ iv be an infinitisimal isometry, i.e.the arcelement of any mapped curve is a multiple of the original curve,independend of the curve:

(

dσ

dt

)2

= M(x, y) 6= 0.

Then f is differentiable at z = x+iy and f ′(z) 6= 0 or f is differentiable

with f′ 6= 0.


Proof. Let t be the arclength of the curve z(t), i.e. x′2 + y′2 = 1. Thenthe condition implies

(

dσ

dt

)2

= x′2

[

(

∂u

∂x

)2

+

(

∂y

∂x

)2]

+ 2x′y′[

∂u

∂x

∂u

∂y+∂v

∂x

∂v

∂y

]

+

+ y′2

[

(

∂u

∂y

)2

+

(

∂y

∂y

)2]

= M

Choosing a curve with x′ = 1 and y′ = 0 we get(

∂u

∂x

)2

+

(

∂y

∂x

)2

= M

Choosing a curve with x′ = 0 and y′ = 1 we get(

∂u

∂y

)2

+

(

∂y

∂y

)2

= M

This implies that∂u

∂x

∂u

∂y+∂v

∂x

∂v

∂y= 0.

Neither of the pairs ∂u∂x

, ∂v∂x

and ∂u∂y

, ∂v∂y

can be simutanuously zero. Let

us assume ∂u∂y

6= 0 and ∂v∂y

6= 0. Then we can write

a :=∂u∂x∂v∂y

= −∂v∂x∂u∂y

according to the third equation. This implies

0 6= M =

(

∂u

∂x

)2

+

(

∂v

∂x

)2

=

(

∂v∂x∂u∂y

)2(

∂u

∂y

)2

+

(

∂u∂x∂v∂y

)2(

∂v

∂y

)2

= a2

[

(

∂u

∂x

)2

+

(

∂v

∂x

)2]

.

This implies a2 = 1.


If ∂u∂y

6= 0 and ∂v∂y

= 0 then ∂u∂x

= 0 and hence, ∂v∂x

6= 0. We can write

a :=∂u∂x∂v∂x

= −∂v∂y

∂u∂y

= 0.

This gives

0 6= M =

(

∂v

∂x

)2

=

(

∂u

∂y

)2

=

(

∂v∂y

∂u∂y

)2(

∂u

∂y

)2

= a2

(

∂u

∂y

)2

= M.

Again a2 = 1 contradicting its definition. The other cases are similar.therefore we only have to deal with the first case. Let a = 1. Then itsdefinition implies the Cauchy–Riemann equations

∂u

∂x=∂v

∂y

∂u

∂y= −∂v

∂x.

Hence, f is differentiable.If a = −1. Then this implies the Anti–Cauchy–Riemann equations

∂u

∂x= −∂v

∂y

∂u

∂y=∂v

∂x.

This implies that f = u− iv is differentiable.

21.1. One–to–one mappings. If a mapping f : R → R′ is one–to–one from a region R into a region R′, i.e. each image point in R′ hasexactly one preimage in R, we can define its inverse g : R′ → R byg(w) = z iff f(z) = w. In real analysis it is shown that for the map(x, y) → (u, v) with f(z) = u(x, y) + i(x, y) a sufficient condition forbeing one–to–one in a neighborhood of (x0, y0) is that the Jacobian

det

(

∂α

∂β

)

α=u,v;β=x,y

is not equal to zero at (x0, y0). This means that[

∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x

]

x0,y0

6= 0.


Using the Cauchy–Riemann equations we conclude[

(

∂u

∂x

)2

+

(

∂v

∂x

)2]

x0,y0

6= 0.

Since

f ′(z0) =

(

∂u

∂x+ i

∂v

∂x

)

x0,y0

The previous condition transforms into |f ′(z0)|2 6= 0.

Theorem 50. Let f(z) be analytic at z0 and f ′(z0) 6= 0 then there isa neighborhood of z0 where this mapping is one–to–one.

Remark 30. This theorem is only local. In particular the function ez

has everywhere in C a non–vanishing derivative. Nevertheless it is notone–to–one on the complex plane.


21.2. Mobius transformations. Mobius transformations or linearfractional transformations are defined by

w =az + b

cz + da, b, c, d ∈ C.

It assigns a finite value w to any point except z = −d/c. We assumethat ad 6= bc. Otherwise w = a/c for all z ∈ C. In general

dw

dz=a(cz + d) − c(az + b)

(cz + d)2=

ad− bc

(cz + d)2

which is non–zero for ad 6= cb. Solving the equation for z we obtainthe equation for the inverse transformation

z =−dw + b

cw − a

which is defined for all w 6= a/c. If c = 0 the ”point” ∞ is fixed.Otherwise the point −d/c is mapped into ∞ and ∞ is mapped intoa/c. So we can include the ”point at infinity” into the definition of ourmap. This setup is called the extended complex plane. Here it makessense to consider straight lines as circles of infinite radius. Then thefollowing theorem holds.

Theorem 51. The Mobius transformations always transform circlesinto circles.

Proof. If c = 0 the Mobius transformation is just a linear transforma-tion, i.e. a translation and a rotation (by angle arg(a/d)) with a scalingfactor |b/d|. If c 6= 0 we can rewrite the transformation as

w =a

c+bc− ad

c

1

cz + d

This transformation can be decomposed into

w1 = cz

w2 = w1 + d = cz + d

w3 =1

w2

=1

cz + d

w4 =bc− ad

cw3 =

bc− ad

c(cz + d)

w =a

c+ w4 =

a

c+

bc− ad

c(cz + d)

The first transformation is a rotation and a rescaling. This clearlypreserves circles and straight lines. The second id just a translation,also preserving straight lines and circles. The third is an inversionaround the unit circle. we will show that it also preserves circles andstraight lines. The general equation for generalized circles is

A(x2 + y2) +Bx+ Cy +D = 0 B2 + C2 ≥ 4AD.


If A = 0 then this is the equation of a straight line. We now substitutex2 + y2 = zz; x = z+z

2; y = z−z

2i. Then

Azz +B

2(z + z) +

C

2i(z − z) +D = 0.

If we replace z by 1/w we get

A

(

1

ww

)

+B

2

(

1

w+

1

w

)

+C

2i

(

1

w− 1

w

)

+D = 0.

Multiplying by ww we derive a new equation for a circle or a straightline:

Dww +B

2(w + w) − C

2i(w − w) +D = 0.

which is a straight line if D = 0. In this case the original circle passesthrough the origin.

Remark 31. Mobius transformations form a group with respect tocomposition, i.e. the composition of two Mobius transformations isa Mobius transformation, the identity is a Mobius transformation andthe inverse transformation is a Mobius transformation.

Sometimes a Mobius transformation is wanted that transports givenpoints into given images. One can just insert the values into the equa-tion

w =az + b

cz + dand obtains that there are 3 free parameters out of a, b, c, d and hencethe general condition is on 3 points and their images. The procedureof solving these 3 linear equations can be made simpler with the helpof

Theorem 52. The Mobius transformations preserve the cross-ratios,i.e.

(w1 − w2)(w3 − w4)

(w1 − w4)(w3 − w2)=

(z1 − z2)(z3 − z4)

(z1 − z4)(z3 − z2)

Proof. We get

wi − wj =azi + b

czi + d− azj + b

czj + d=

(ad− bc)(zi − zj)

(czi + d)(czj + d)

Inserting this into the cross–ratios gives the desired result.

Example 44. Find the Mobius transformation that maps z1 = 1, z2 =i, z3 = 0 into w1 = 0, w2 = −1, w3 = −i. We use the preservation ofcross–ratios:

(0 − (−1))(−i− w)

(0 − w)(−i− (−1))=

(1 − i)(0 − z)

(1 − z)(0 − i)

yielding

w =i(z − 1)

z + 1.


Example 45. Find a conformal transformation that maps the quadrant0 < arg z < π/2 into the unit disc |z| < 1.

A Mobius transformation can map a straight line into a circle but nota line with a corner of π/2. So it cannot be a Mobius transformation.But the map u = z2 maps the quadrant into the upper half plane!

Now we can look for a Mobius transformation that transforms theupper half plane into the unit disc. Let

w =(a

c

) z + b/a

z + d/c.

Such a transformation must map the real axis into the unit circle. Letx = ℜz → ±∞ then |a|/|c| = 1, i.e. a/c = eiφ. Again on the real axiswe have

1 =

∣

∣

∣

∣

x+ b/a

x+ d/c

∣

∣

∣

∣

or∣

∣

∣

∣

x+b

a

∣

∣

∣

∣

=

∣

∣

∣

∣

x+d

c

∣

∣

∣

∣

.

Hence,

b

a=d

cor

b

a=d

c

(|x + z| = |x + w| for all real x implies that z1 = w1 and z22 = w2

2).The first choice maps the whole quadrant to a single point and has tobe ignored. the second possibility gives

w = eiφz − p

z − p

and it maps the upper half plane into the interior of the fit circle pro-vided ℑp > 0. Combining these two results we get

w = eiφz2 − p

z2 − p

transferring the quadrant into the unit disc. A similar method can beused to transform domains like 0 < arg z < α.

21.3. Conformal mapping and boundary value problems.

Theorem 53. Let the analytic function w = f(z) map the domainD to the domain D′. Suppose ψ(u, v) is harmonic in D′. Then bychanging variables

u(x, y) + iv(x, y) = f(z) = w

we have that φ(x, y) = ψ(u(x, y), v(x, y)) is harmonic in D.

Remark 32. A conformal map preserves solutions of the Laplace equa-tion.


Proof. We consider only the case that D′ is simply connected. Thenthere is an analytic in D′ function

Ψ(w) = ψ(u, v) + iη(u, v).

Since f is analytic we have that Ψ(f(z)) = Φ(z) is analytic in D. Now

Ψ(w) = ψ(f(z)) = Φ(z) = φ(x, y) + iθ(x, y).

By comparing the two last equalities with the form of f = u + iv wederive

φ(x, y) = ψ(u(x, y), v(x, y)) θ(x, y) = η(u(x, y), v(x, y))

This theorem allows us to transform a Laplace problem from a com-plicated domain into a simpler one if we can find a conformal mapping.If the simpler domain is a disc or the upper half plane we can use Pois-son’s formula to find the solutions. But it is not always easy to findthe conformal mapping. A very strong result of Riemann ensures theexistence of conformal mappings.

Theorem 54 (Riemann mapping theorem). Any domain whose bound-ary is connected and consists of at least 2 points can be mapped con-formally into the interior of the unit disc.

22. Conformal mappings II

Lemma 3 (Schwarz). Let f be analytic in |z| < 1 and |f(z)| ≤ 1 inthe unit disc. Furthermore assume that f(0) = 0. Then |f(z)| ≤ |z|.Moreover, if there is a z 6= 0 such that |f(z)| = |z| then f(z) = cz and|c| = 1.

Proof. Since f(0) = 0 it follows that f(z) = zg(z) with g analytic in|z| < 1. The maximum principle for |z0| < r < 1 gives

|g(z0)| ≤ max|z|=r

|g(z)| = max|z|=r

|f(z)||z| ≤ 1

r.

The first part follows by letting r → 1.The maximum principle imlies that in case |g(z)| = 1 for some point

inside the circle the function g has to be constant (and of absolutevalue equal to 1).

Remark 33. Since f ′(0) = g(0) we get that under the assumtions ofth previous lemma that |f ′(0)| ≤ 1 and in case |f ′(0)| = 1 the functionf has to be linear.

Theorem 55. Any conformal map that maps the interior of the unitdisc into itself is a Mobius transformation.


Proof. Let f be such a map. Let |a| < 1. The Mobius transformation

w = g(z) = eiβz − a

az − 1

maps the unit disc into itself and sends the point a into 0. Let a = f(0)and consider h = g f . The map h meets the assumptions of theprevious lemma. Hence, |h(z)| ≤ |z| inside the unit disc. The sameproperty has the inverse function (it maps the unit disc into the unitdisc and fixes z = 0). Hence, |z| ≤ |h(z)| and h(z) = cz with |c| = 1.Therefore, f(z) = g−1(cz) – a Mobus transformation.

Example 46. The exponential function w = f(z) = ez = exeiy. Thisfunction has period 2πi. It maps the strip b < ℑz < b + 2π into thecomplex plane with the removed ray argw = b. For b = −π we havethe following picture.

−i

i π

π

z=a+it

w=e e a it

w=ez

Figure 28

The logarithmic function w = f(z) = log z = ln |z| + i arg z, −π <arg z < π is the inverse to the exponential function and maps the com-plex plane without the negative ray into the horizontal strip of width2π.


Example 47. Two cylinders are configurated as in figure 29 and havetemperature 0 respectively 100 degrees. Find the temperature in thedomain between the cylinders.

0o

0 .5

100o

1

0 100

0 1

0 0

8

w=f(z)

Figure 29

We will apply a transformation that changes the domain as in fig-ure 29. The transformation

w =1 − z

z

sends 0 → ∞, 12→ 1 and 1 → 0 and transform s the domain between

the circles into a vertical strip. We transformed the problem into asimpler one:

Find a harmonic function φ(u, v) in the domain 0 < u < 1 (w =u+ iv) and has the boundary conditions φ(0, v) = 0 and φ(1, v) = 100.From symmetry we see that the harmonic function should not dependon v and then we guess that

φ(u, v) = 100u

is a solution which is harmonic. We now can use the method to findits harmonic conjugate or we simply guess that

φ(u, v) = ℜ(100w)


and hence, Φ(w) = 100w. To get the solution of our original problemwe have to transform the solution back. We have

w = u+ iv =1

z− 1 =

1

x+ iy− 1 =

x

x2 + y2− 1 − iy

x2 + y2

which implies

u =x

x2 + y2− 1 v =

−yx2 + y2

This gives

φorig(x, y) = 100

(

x

x2 + y2− 1

)

ψorig(x, y) = 100y

x2 + y2

and

Φorig(z) = 1001 − z

zThe isothermal lines are defined by

T0 = 100

(

x

x2 + y2− 1

)

0 ≤ T0 ≤ 100

and can be rewritten as(

x−12

(

1 + T0

100

)

)2

+ y2 =

(

12

1 + T0

100

)2

which are circles with centers at

y0 = 0 x0 =

(

12

1 + T0

100

)2

.

22.1. Proof of the Riemann mapping theorem.

Proof. Let a 6= b two points at the boundary of the domain D. We

consider a branch of the function f1(z) =√

z−az−b at some point inside

D and consider its analytic continuation. Since D is simply connectedand the branch points of f1 are a, b 6∈ D we get a unique continuationand the function f1 is univalent (also the values f1 and −f1 are notsimultanously assigned in D since otherwise we should go once aroundthe branch point within D). Hence, if f1(z0) = c then there is an ǫ > 0such that all points |w − c| < ǫ are values of f1 and all the values|w+ c| < ǫ are not assigned in D. Hence, the function 1

f1+cis analytic,

bounded and one–to–one in D.Let M 6= ∅ (some linear transformation of 1

f1+cis in this set) be the

set of functions which are analytic, one–to–one in D, |f(z)| < 1 andf(z0) = 0. Let

supM

|f ′(z0)| = M.


Since |f(z)| < 1 the number 0 < M is finite (it is actually not greater

than M(r)r

where r > 0 is the radius of the largest circle around z0contained in D and M(r) is the maximum of f on this circle).

Now we consider a sequence of functions fn(z) ∈ M with |f ′(z0)| →M and a sequence of points zk → z0. Let fn1k

be a subsequencethat converges at z1. Out of this subsequence we choose anothersubsequencefn2k

that converges at z2 and so on. Then the diagonalsequence fnkk = gk converges at all points zk. By the uniqueness ofthe Taylor series expansion we get that the functions gj converge inD to an analytic function g and this convergence is uniform on eachdisc inside D. Also |g′(z0)| = M and g is not constant, |g(z)| < 1 andg(z0) = 0. We need to prove that g is one–to–one in D.

Let z′ ∈ D. Then we can choose a small circle of radius R aroundthat point that lies in D and on which |g′(z)| ≥ η (otherwise thefunction would be constant). Since the functions gn are one–to–onewe have that g′n(z) 6= 0 in D and 1

g′nis analytic in D. The maximum

principle and uniform convergence tell us that∣

∣

∣

∣

1

g′n(z′)

∣

∣

∣

∣

≤ max|z−z′|=R

∣

∣

∣

∣

1

g′n(z)

∣

∣

∣

∣

=2

η<∞.

Hence, |g′(z)| 6= 0.Assume that g(z′1) = g(z′2). Then there are two discs K1, K2 around

z′1, z′2 respectively which are disjoint. If n is sufficiently large the value

g(z′1) is assumed by gn(z) for some z ∈ K1. The same holds true forthe disc K2. This contradicts the assumption that gn is one–to–one.Therefore g is one–to–one and belongs to M.

Since |g(z)| < 1 the map g maps the domain D into the unit disc.We have to show that this is onto. Let d inside the unit disc such thatg(z) 6= d for all |z| < 1. Then we can consider

w = h(z) =

√

g(z) − d

1 − dg(z)

This function also transforms the unit disc into itself (analytically andone–to–one) and sends d → 0 (note that the branching points d and1/d are not assumed by g). We can define the Mobius transformationof h by

f(z) =h(z) − h(z0)

1 − h(z0)h(z)

This function is in M. If we differentiate this function at z0 we get

f ′(z0) =1 + |d|2√−d

g′(z0)

and because |d| < 1 we have

|f ′(z0)| > |g′(z0)| = M.


This is a contradiction to the definition of M .

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ANALYTIC FUNCTIONS Contents - LTH · Analytic functions have an extreme mathematical beauty. They show many properties of general functions in a very pure way. There-fore one often