Concise Complex Analysis, Revised EditionSolution of Exercise Problems

Yan Zeng

Version 0.2, last revised on 2019-09-01.

Abstract

This is a solution manual of selected exercise problems from Gong and Gong [4]. This version solvesthe exercise problems in Chapter 1-3, except the following: Chapter 1 problem 37-42; Chapter 2 problem47, 49; Chapter 3 problem 15 (xi). If you find any typos/errors, please email me at zypublic@hotmail.com.

Contents1 Calculus 2

2 Cauchy Integral Theorem and Cauchy Integral Formula 16

3 Theory of Series of Weierstrass 32

4 Riemann Mapping Theorem 59

5 Differential Geometry and Picard’s Theorem 60

6 A First Taste of Function Theory of Several Complex Variables 60

7 Elliptic Functions 60

8 The Riemann Zeta Function and The Prime Number Theory 60

1

1 CalculusI 1. Verify the Newton-Leibniz formula, the Green’ theorem, the Stokes’s theorem and the Gauss’s theoremby (1.9).

Proof. See, for example, Munkres [6], §38.

I 2. (1) Find the modulus and the principle values of the arguments for the following complex numbers:(i) 2i; (ii) 1− i; (iii) 3 + 4i; (iv) −5 + 12i.

Solution. |2i| = 2, arg(2i) = π2 . |1− i| =

√2, arg(1− i) = 7

4π. |3 + 4i| = 5, arg(3 + 4i) = arctan 43 ≈ 0.9273

(Matlab command: atan(4/3)). |−5+12i| = 13, arg(−5+12i) = arccos(− 5

13

)≈ 1.9656 (Matlab command:

acos(−5/13)).

(2) Express the following complext numbers in the form of x + iy where x and y are real numbers. (i)(1 + 3i)3; (ii) 10

4−3i ; (iii) 2−3i4+i ; (iv) (1 + i)n + (1− i)n, where n is a positive integer.

Solution. (1 + 3i)3 = −26− 18i. 104−3i =

85 + 6

5 i.2−3i4+i = 5

17 − 1417 i. (1 + i)n + (1− i)n = 2

n2 +1 cos n

4π.

(3) Find the absolute values (modulus) of the following complex numbers: (i) −3i(2− i)(3 + 2i)(1 + i);(ii) (4−3i)(2−i)

(1+i)(1+3i) .

Solution. | − 3i(2− i)(3 + 2i)(1 + i)| = 3√130.

∣∣∣ (4−3i)(2−i)(1+i)(1+3i)

∣∣∣ = 5·√5√

2·√10

= 52 .

I 3. Show thatπ

4= 4 arctan 1

5− arctan 1

239

by calculating (5− i)4(1 + i).

Proof. Let θ = arctan 15 , α = arctan 1

239 . Then 5 − i =√26e−iθ, 1 + i =

√2ei

π4 and (5 − i)4(1 + i) =

676√2ei(

π4 −4θ). Meanwhile (5 − i)4(1 + i) = 956 − 4i = 676

√2e−αi. So we must have π

4 − 4θ = −α + 2kπ,k ∈ Z, i.e. π

4 = 4 arctan 15−arctan 1

239+2kπ, k ∈ Z. Since 4 arctan 15−arctan 1

239+2π > 0− π2 +2π = 7

4π >π4

and 4 arctan 15 − arctan 1

239 − 2π < 4 · π2 − 2π = 0 < π

4 , we must have k = 0. Therefore π4 = 4 arctan 1

5 −arctan 1

239 .

I 4. Let z = x+ iy, where x and y are real numbers. Find the real part and the imaginary part of each ofthe following complex numbers: (1) 1

z ; (2) z2; (3) 1+z1−z ; (4) z

z2+1 .

Solution. If z = x + yi, 1z = x

x2+y2 + yx2+y2 i. z2 = x2 − y2 + 2xyi. 1+z

1−z = 1−x2−y2

(1−x)2+y2 + 2y(1−x)2+y2 i.

zz2+1 = x3+xy2+x+i(−y3−x2y+y)

(x2−y2+1)2+4x2y2 .

I 5. Solve the quadratic equationz2 + (α+ iβ)z + γ + iδ = 0,

where α, β, γ and δ are real numbers.

Solution. a = 1, b = α+ iβ, c = γ + iδ. So ∆ = b2 − 4ac = (α2 − β2 − 4γ) + i(2αβ − 4δ) and

z =−(α+ iβ)±

√∆

2.

2

I 6. Assume |z| = r > 0. Show that:

Rez = 1

2

(z +

r2

z

), Imz = 1

2i

(z − r2

z

).

Proof. Denote arg z by θ, then z + r2

z = reiθ + r2

reiθ= r(eiθ + e−iθ) = 2r cos θ = 2Rez, and z − r2

z =

reiθ − r2

reiθ= r(eiθ − e−iθ) = 2ir sin θ = 2iImz.

I 7. Show that:(1) If |a| < 1, |b| < 1, then ∣∣∣∣ a− b

1− ab

∣∣∣∣ < 1;

(2) If |a| = 1 or |b| = 1, then ∣∣∣∣ a− b

1− ab

∣∣∣∣ = 1.

Discussing the case |a| = 1 and |b| = 1.

Proof. If a = r1eiθ1 and b = r2e

iθ2 , then∣∣∣∣ a− b

1− ab

∣∣∣∣ = ∣∣∣∣ r1 − r2ei(θ2−θ1)

1− r1r2ei(θ2−θ1)

∣∣∣∣ .Denote θ2 − θ1 by θ, we can reduce the problem to comparing |r1 − r2e

iθ|2 and |1− r1r2eiθ|2. Note

|r1 − r2eiθ|2 = (r1 − r2 cos θ)2 + r22 sin2 θ = r21 − 2r1r2 cos θ + r22

and|1− r1r2e

iθ|2 = (1− r1r2 cos θ)2 + r21r22 sin2 θ = 1− 2r1r2 cos θ + r21r

22.

So |1− r1r2eiθ|2 − |r1 − r2e

iθ|2 = (r21 − 1)(r22 − 1). This observation shows∣∣∣ a−b1−ab

∣∣∣ = 1 if and only if at least

one of a and b has modulus 1;∣∣∣ a−b1−ab

∣∣∣ < 1 if |a| < 1 and |b| < 1.

I 8. Prove the Lagrange equation in complex form∣∣∣∣∣n∑

i=1

aibi

∣∣∣∣∣2

=

n∑i=1

|ai|2n∑

i=1

|bi|2 −∑

1≤i<j≤n

|aibj − aj bi|2.

Derive the Cauchy inequality ∣∣∣∣∣n∑

i=1

aibi

∣∣∣∣∣2

≤n∑

i=1

|ai|2n∑

i=1

|bi|2

from the Lagrange equation. The equality holds if and only if ak and bk are proportional, where k = 1, · · · , n.

3

Proof. We prove by induction. The equation clearly holds for n = 1. Assume it holds for all n with n ≤ N .Then∣∣∣∣∣

N+1∑i=1

aibi

∣∣∣∣∣2

=

∣∣∣∣∣N∑i=1

aibi + aN+1bN+1

∣∣∣∣∣2

=

(N∑i=1

aibi + aN+1bN+1

)(N∑i=1

aibi + aN+1bN+1

)

=

∣∣∣∣∣N∑i=1

aibi

∣∣∣∣∣2

+ |aN+1|2|bN+1|2 + aN+1bN+1

N∑i=1

aibi + aN+1bN+1

N∑i=1

aibi

=

N∑i=1

|ai|2N∑i=1

|bi|2 −∑

i≤i<j≤N

|aibj − aj bi|2 + |aN+1|2|bN+1|2 + |bN+1|2∑

1≤i<j=N+1

|ai|2

+|aN+1|2∑

1≤i<j=N+1

|bi|2

−∑

1≤i<j=N+1

[|ai|2|bN+1|2 + |aN+1|2|bi|2 − aiaN+1bibN+1 − aiaN+1bibN+1

]=

N+1∑i=1

|ai|2N+1∑i=1

|bi|2 −∑

1≤i<j≤N

|aibj − aj bi|2 −∑

1≤i<j=N+1

|aibj − aj bi|2

=

N+1∑i=1

|ai|2N+1∑i=1

|bi|2 −∑

1≤i<j≤N+1

|aibj − aj bi|2.

By method of mathematical induction, |∑n

i=1 aibi|2 =∑n

i=1 |ai|2∑n

i=1 |bi|2 −∑

1≤i<j≤n |aibj − aj bi|2 holdsfor all n ∈ N. Cauchy’s inequality follows directly from this equation.

I 9. Show that a1, a2, a3 are the vertices of a equilateral triangle if and only if

a21 + a22 + a23 = a1a2 + a2a3 + a3a1.

Proof. We first consider the special case where a1 = 0 and a2 = 1. Then for a3 = reiθ, a1, a2, a3 are verticesof an equilateral triangle if and only if r = 1 and cos θ = 1

2 . Meanwhile the equation 1 + re2iθ = reiθ isequivalent to {

1 + r cos 2θ = r cos θr sin 2θ = r sin θ,

which implies r = 1 and cos θ = 12 . So the equality is proven for the special case a1 = 0, a2 = 1.

For general case, note a21 + a22 + a23 = a1a2 + a2a3 + a3a1 if and only if (a2 − a1)2 + (a3 − a1)

2 =

(a2 − a1)(a3 − a1), which is further equivalent to 1+(

a3−a1

a2−a1

)2= a3−a1

a2−a1. This shows the general case can be

reduced to the special case of a1 = 0 and a2 = 1. Essentially, these reductions correspond to the compositionof coordinate translation and coordinate rotation.

I 10. Show that the complex numbers α, β, γ are collinear if and only if∣∣∣∣∣∣α α 1β β 1γ γ 1

∣∣∣∣∣∣ = 0.

Proof.

D :=

∣∣∣∣∣∣α α 1β β 1γ γ 1

∣∣∣∣∣∣ =∣∣∣∣∣∣α− γ α− γ 1β − γ β − γ 10 0 1

∣∣∣∣∣∣ =∣∣∣∣α− γ α− γβ − γ β − γ

∣∣∣∣ .

4

Suppose α− γ = r1eiθ1 and β − γ = r2e

iθ2 . Then∣∣∣∣α− γ α− γβ − γ β − γ

∣∣∣∣ = ∣∣∣∣r1eiθ1 r1e−iθ1

r2eiθ2 r2e

−iθ2

∣∣∣∣ = r1r2ei(θ1−θ2) − r1r2e

−i(θ1−θ2).

So D = 0 if and only if θ1− θ2 = kπ, which is equivalent to α, β, γ being colinear. Note the above reductioncorresponds to the coordination transformation that places γ at origin.

I 11. Find the corresponding points of 1− i, 4 + 3i on the Riemann sphere S2.

Solution. We apply formula (1.11) (correction: the formula for x2 should be x2 = z−z(1+|z|2)i ). If z = 1 − i,

then x1 = 23 , x2 = − 2

3 , and x3 = 23 . If z = 4 + 3i, then x1 = 4

13 , x2 = 313 , and x3 = 12

13 .

I 12. Tow points z1, z2 on the C-plane correspond to the two end points of a diagonal of the Riemannsphere S2 if and only if z1z2 = −1.

Proof. Denote by x = (x1, x2, x3) and y = (y1, y2, y3) the points on the Riemann sphere S2 correspondingto z1 and z2, respectively. Then x and y are two ends points of a diagonal of S2 if and only if x+ y = 0. Byformula (1.11), x+ y = 0 is equivalent to

z1+z11+|z1|2 + z2+z2

1+|z2|2 = 0z1−z11+|z1|2 + z2−z2

1+|z2|2 = 0|z1|2−1|z1|2+1 + |z2|2−1

|z2|2+1 = 0.

Solving the third equation gives |z1z2| = 1. So we can assume z1 = reiα and z2 = 1r e

iβ (r > 0). Then1 + |z1|2 = 1 + r2 and 1 + |z2|2 = 1 + 1

r2 = 1r2 (1 + |z1|2). So the other two equations become{

2r cosα+ 2r cosβ = 0

2r sinα+ 2r sinβ = 0,

which is equivalent to {cos α+β

2 cos α−β2 = 0

sin α+β2 cos α−β

2 = 0.

So we must have 12 (α− β) = kπ+ π

2 (k ∈ Z), i.e. α− β = 2kπ+ π. So z1z2 = ei(α−β) = ei(2kπ+π) = −1.

I 13. Show that the equation of a circle is

A|z|2 +Bz + Bz + C = 0

where A, C are real numbers and |B|2 > AC. Also show that the image of this circle on the Riemann sphereS2 is a great circle if and only if A+ C = 0.

Proof. Suppose z ∈ C is on a circle. Then there exists z0 ∈ C and R > 0, such that |z − z0| = R. Supposez = x+ iy and z0 = x0+ iy0. Then |z− z0| = R implies (x−x0)2+(y− y0)2 = R2. After simplification, thiscan be written as |z|2 + z0z + z0z + |z0|2 −R2 = 0. Let A = 1, B = z0 and C = |z0|2 −R2. Then A,C ∈ Rand |B|2 > AC.

Conversely, if A|z|2 +Bz + Bz +C = 0 is the equation satisfied by z, we have two cases to consider: (i)A 6= 0; (ii) A = 0.

In case (i), A|z|2 + Bz + Bz + C = 0 can be written as |z|2 + BAz +

BA z +

|B|2A2 = |B|2

A2 − CA > 0. Define

z0 = BA and R =

√|B|2A2 − C

A . The equation can be written as |z|2+ z0z+z0z+ |z0|2 = R2, which is equivalentto |z − z0| = R. So z is on a circle.

5

In case (ii), the equation Bz + Bz + C = 0 stands for a straight line in C. Indeed, suppose B = a + iband z = x + iy, then Bz + Bz + C = 0 becomes 2ax − 2by + C = 0. If we regard straight lines as circles,A|z|2 +Bz + B+C = 0 represents a circle in case (ii) as well.

To see the necessary and sufficient condition for the image of this circle on the Riemann sphere S2 to bea great circle, we suppose a generic point z on this circle corresponds to a point (x1, x2, x3) on the Riemannsphere S2. Then by formula (1.10) z = x1+ix2

1−x3, we can write the equation A|z|2 + Bz + Bz + C = 0 as

(B+ B)x1+ i(B−B)x2+(A−C)x3 = −(A+C). (x1, x2, x3) falls on a great circle if and only if (x1, x2, x3)lies on the intersection of S2 and a plane that contains (0, 0, 0), which is equivalent to A + C = 0 by theequation (B + B)x1 + i(B −B)x2 + (A− C)x3 = −(A+ C).

I 14. Let d(z1, z2) be the spherical distance of z1 and z2 in C, that is, the spherical distance between thecorresponding points Z1 and Z2 on the Riemann sphere S2. Prove

d(z1, z2) =2|z1 − z2|√

(1 + |z1|2)(1 + |z2|2),

andd(z1,∞) =

2√1 + |z1|2

.

Proof. Suppose Z1 = (x1, x2, x3) and Z2 = (y1, y2, y3). Then by formula (1.11),

d(z1, z2)2 = (x1 − y1)

2 + (x2 − y2)2 + (x3 − y3)

2

= 2− 2(x1y1 + x2y2 + x3y3)

= 2− 2

[z1 + z11 + |z1|2

· z2 + z21 + |z2|2

+z1 − z1

i(1 + |z1|2)· z2 − z2i(1 + |z2|2)

+|z1|2 − 1

|z1|2 + 1· |z2|

2 − 1

|z2|2 + 1

]= 4

|z1|2 + |z2|2 − z1z2 − z1z2(1 + |z1|2)(1 + |z2|2)

= 4(z1 − z2)(z1 − z2)

(1 + |z1|2)(1 + |z2|2).

So d(z1, z2) = 2|z1−z2|√(1+|z1|2)(1+|z2|2)

. Consequently,

d(z1,∞) = 2 limz2→∞

∣∣∣ z1z2 − 1∣∣∣√

(1 + |z1|2)(

1|z2|2 + 1

) =2√

1 + |z1|2.

I 15. Explain the geometric meaning of the following relations:(i)∣∣∣ z−z1z−z2

∣∣∣ ≤ 1, z1 6= z2;(ii) Re z−z1

z−z2= 0, z1 6= z2;

(iii) 0 < arg z+iz−i <

π4 ;

(iv) |z + c|+ |z − c| ≤ 2a, a > 0, |c| < a.

Solution. (i) The distance between z and z1 is no greater than the distance between z and z2.(ii) z − z1 and z − z2 are perpendicular to each other.(iii) The angle between z + i and z − i is less than π

4 .(iv) z is on an ellipse with c and −c as the foci.

I 16. Prove Heine-Borel theorem and Bolzano-Weierstrass theorem on the complex plane.

6

Proof. Regard C as R2 and apply Heine-Borel theorem and Bolzano-Weierstrass theorem for Rn (n ∈ N).

I 17. Show that the sequence {zn} converges to z0 if and only if the sequences {Rezn}, {Imzn} convergesto Rez0, Imz0 respectively.

Proof. Note max{|Rezn − Rez0|, |Imzn − Imz0|} ≤ |z − z0| ≤ |Rezn − Rez0|+ |Imzn − Imz0|.

I 18. Show that(1) If limn→∞ zn = a, limn→∞ z′n = b, then

limn→∞

1

n

n∑k=1

zkz′n−k = ab.

Proof. We suppose |a|, |b| < ∞. Then (zn)n and (z′n)n are bounded sequences. Denote by M a commonbound for both sequences. ∀ε > 0, ∃N , so that for any n > N , max{|zn − a|, |z′n − b|} < ε

M . Then for any nsatisfying n > N

(1 + M2

2

)(define z0 = z′0 = 0)∣∣∣∣∣ 1n

n∑k=1

zkz′n−k − ab

∣∣∣∣∣ <

∣∣∣∣∣ 1nN∑

k=1

zkz′n−k

∣∣∣∣∣+∣∣∣∣∣ 1n

n∑k=N+1

zkz′n−k − ab

∣∣∣∣∣<

NM2

n+

∣∣∣∣∣ 1nn∑

k=N+1

(zk − a)z′n−k

∣∣∣∣∣+∣∣∣∣∣an

n∑k=N+1

z′n−k − ab

∣∣∣∣∣< ε+ ε+ |a|

∣∣∣∣∣ 1nn−N−1∑k=0

z′k − b

∣∣∣∣∣= 2ε+ |a|

∣∣∣∣∣n−N − 1

n· 1

n−N − 1

n−N−1∑k=0

z′k − b

∣∣∣∣∣ .Taking upper limits on both sides, we get

limn→∞

∣∣∣∣∣ 1nn∑

k=1

zkz′n−k − ab

∣∣∣∣∣ ≤ 2ε+ 0 = 2ε.

Since ε is arbitrary, we conclude limn→∞∣∣ 1n

∑nk=1 zkz

′n−k − ab

∣∣ = 0, i.e.

limn→∞

1

n

n∑k=1

zkz′n−k = ab.

(2) If limn→∞ zn = A, thenlimn→∞

1

n(z1 + z2 + · · ·+ zn) = A

(using the result of (1)).

Proof. Apply the result of (1) with z′n ≡ 1.

I 19. Discuss the differentiability of the following functions.(i) f(z) = |z|; (ii) f(z) = z; (iii) f(z) = Rez.

Solution. (i) f(z) =√zz is a function of both z and z. So f is not differentiable.

(ii) f(z) = z is a function of z, so it is not differentiable.(iii) f(z) = Rez = 1

2 (z + z) is a function of both z and z. So f is not differentiable.

7

I 20. Suppose g(w) and f(z) are holomorphic functions. Show that g(f(z)) is also holomorphic.

Proof. By chain rule, ∂∂z g(f(z)) = g′(f(z))∂f(z)∂z = 0. So g(f(z)) is also holomorphic.

I 21. On a region D, show that(1) If a function f(z) is holomorphic, and f ′(z) is identically zero, then f(z) is a constant function.

Proof. Let u(x, y) = Ref(z) and v(x, y) = Imf(z). Then by C-R equations f ′(z) = ux(x, y) + ivx(x, y) =vy(x, y) − iuy(x, y). So f ′(z) ≡ 0 if and only if vx = vy = ux = uy ≡ 0. By results for functions of realvariables, we conclude u and v are constants on D. so f is a constant function on D.

(2) If f(z) is holomorphic and f ′(z) satisfies one of the following conditions: (i) Ref(z) is a constantfunction; (ii) Imf(z) is a constant function; (iii) |f(z)| is a constant function; (iv) arg f(z) is a constantfunction, then f(z) is a constant function.

Proof. (i) and (ii) are direct corollaries of C-R equations.For (iii), we assume without loss of generality that f(z) 6≡ 0 on D. We note |f(z)|2 = u2(z) + v2(z). So

by the C-R equations, 0 = ∂∂x |f(z)|

2 = 2uux + 2vvx = 2uux − 2vuy and 0 = ∂∂y |f(z)|

2 = 2uuy + 2vvy =

2uuy + 2vux, i.e.[u −vv u

] [uxuy

]= 0. Since f 6≡ 0,

[u −vv u

]is invertible on D1 := {z ∈ D : f(z) 6= 0}. So

ux = uy ≡ 0 on D1. By the C-R equations vx = vy ≡ 0 on D1. So f is a constant function on each simplyconnected component of D1. Since D2 := {z ∈ D : f(z) = 0} has no accumulation points in D (see Theorem2.13), f must be identical to the same constant throughout D.

For (iv), we assume arg f(z) ≡ θ. Then g(z) := e−iθf(z) is also holomorphic and f is a constant functionif and only if g is a constant function. So without loss of generality, we can assume arg f(z) ≡ 0. ThenImf(z) ≡ 0. By C-R equations, Ref(z) is a constant too. Combined, we can conclude f is identically equalto a constant on D.

I 22. If f(z) = u + iv is holomorphic and f ′(z) 6= 0, then the curves u(x, y) = c1 and v(x, y) = c2 areorthogonal, where c1, c2 are constants.

Proof. The tangent vector a(x, y) of the curve u(x, y) = c1 at point (x, y) is perpendicular to (ux, uy);the tangent vector b(x, y) of the curve v(x, y) = c2 at point (x, y) is perpendicular to (vx, vy). Since(vy,−vx) ⊥ (vx, vy) and (ux, uy) = (vy,−vx) by C-R equations, we must have a(x, y) ⊥ b(x, y), whichmeans the curves u(x, y) = c1 and v(x, y) = c2 are orthogonal.

I 23. Show that(1) If

f(z) = u(r, θ) + iv(r, θ), z = r(cos θ + i sin θ),

then the C-R equation becomeur =

1

rvθ, vr = −1

ruθ,

andf ′(z) =

r

z(ur + ivr).

Proof. Since{x = r cos θy = r sin θ

, we have

[∂∂r∂∂θ

]=

[cos θ sin θ

−r sin θ r cos θ

] [ ∂∂x∂∂y

]:= A(θ, r)

[ ∂∂x∂∂y

].

8

It’s easy to see A−1(θ, r) = 1r

[r cos θ − sin θr sin θ cos θ

]. Writing the Cauchy-Riemann equations in matrix form, we

get [ ∂∂x∂∂y

]u =

[0 1−1 0

] [ ∂∂x∂∂y

]v.

Therefore, under the polar coordinate, the Cauchy-Riemann equations become[∂∂r∂∂θ

]u = A(θ, r)

[ ∂∂x∂∂y

]u = A(θ, r)

[0 1−1 0

] [ ∂∂x∂∂y

]v = A(θ, r)

[0 1−1 0

]A−1(θ, r)

[∂∂r∂∂θ

]v =

[0 1

r−r 0

] [∂∂r∂∂θ

]v.

Fix z = reiθ, then

f ′(z) = lim∆r→0

u(r +∆r, θ) + iv(r +∆r, θ)− [u(r, θ) + iv(r, θ)]

∆r · eiθ=∂u

∂re−iθ + i

∂v

∂ve−iθ =

r

z

[∂u

∂r+ i

∂v

∂r

].

(2) If f(z) = R(cosϕ+ i sinϕ), then the C-R equations are

∂R

∂r= −R

r

∂ϕ

∂θ,∂R

∂θ= −Rr∂ϕ

∂r.

Proof. If f(z) = R(cosϕ+ i sinϕ), then u = R cosϕ and v = R sinϕ. So

ur =∂R

∂rcosϕ−R sinϕ∂ϕ

∂r, vr =

∂R

∂rsinϕ+R cosϕ∂ϕ

∂r,

uθ =∂R

∂θcosϕ−R sinϕ∂ϕ

∂θ, vθ =

∂R

∂θsinϕ+R cosϕ∂ϕ

∂θ.

Writing the above equalities in matrix form, we have[urvr

]= A

[∂R∂r∂ϕ∂r

],

[uθvθ

]= A

[∂R∂θ∂ϕ∂θ

], where A =

[cosϕ −R sinϕsinϕ R cosϕ

].

Therefore the C-R equations become

A

[∂R∂r∂ϕ∂r

]=

[urvr

]=

[0 1

r− 1

r 0

] [uθvθ

]=

[0 1

r− 1

r 0

]A

[∂R∂θ∂ϕ∂θ

].

Note A−1 = 1R

[R cosϕ R sinϕsinϕ R cosϕ

]. So A−1

[0 1

r− 1

r 0

]A =

[0 R

r− 1

Rr 0

]. Therefore

[∂R∂r∂ϕ∂r

]=

[0 R

r− 1

Rr 0

] [∂R∂θ∂ϕ∂θ

], i.e.

{∂R∂r = R

r∂ϕ∂θ

∂R∂θ = −Rr ∂ϕ∂r

.

I 24. Suppose the function f(z) = u(x, y) + iv(x, y) (z = x+ iy) is holomorphic on a region D. Show thatthe Jacobi determinant of u and v with respect to x and y is

J =

∣∣∣∣ux uyvx vy

∣∣∣∣ = |f ′(z)|2,

and give the geometric meaning of J .

Proof. By the C-R equations, J =

∣∣∣∣ux uyvx vy

∣∣∣∣ = uxvy − vxuy = ux · ux − vx · (−vx) = u2x + v2x = |f ′(z)|2. J is

the area of f(D), see, for example, Munkres [6], §21 The volume of a parallelepiped.

9

I 25. Compute:(1) The integral ∫

γ

xdz

where z = x+ iy and γ is a directed segment from 0 to 1 + i.

Solution. We can parameterize γ with γ(t) = t+ ti (0 ≤ t ≤ 1). So∫γ

xdz =

∫ 1

0

t(dt+ idt) =1

2+

1

2i.

(2) The integral ∫γ

|z − 1||dz|

where γ(t) = eit, 0 ≤ t ≤ 2π.

Solution.∫γ

|z − 1||dz| =

∫ 2π

0

|eit − 1|dt =∫ 2π

0

√2− 2 cos tdt =

√2

∫ 2π

0

∣∣∣∣sin(π4 − t

2

)− cos

(π

4− t

2

)∣∣∣∣ dt=

√2

∫ π4

− 34π

|sin θ − cos θ| dθ =√2

∫ π4

− 34π

(cos θ − sin θ) dθ4.

(3) The integral ∫γ

1

z − adz

where γ(t) = a+Reit, 0 ≤ t ≤ 2π and a is a complex constant.

Solution. ∫γ

dz

z − a=

∫ 2π

0

iReit

Reitdt = 2πi.

I 26. (1) Find the real parts and the imaginary parts of cos(x+ iy), sin(x+ iy) where x, y are real numbers.(2) Show that

sin iz = i sinh z, cos iz = cosh z, (sin z)′ = cos z, cos(z1 + z2) = cos z1 cos z2 − sin z1 sin z2.

Solution. We first recall sinh z = ez−e−z

2 and cosh z = ez+e−z

2 . Then straightforward calculations show

cos(x+ iy) = cosh y cosx− i sinh y sinx, sin(x+ iy) = cosh y sinx+ i sinh y cosx.

By this result, we have

sin iz = ei(iz) − e−i(iz)

2i=e−z − ez

2i= sinh z, cos iz = ei(iz) + e−i(iz)

2=e−z + ez

2= cosh z,

(sin z)′ =(eiz − e−iz

2i

)′

=eiz + e−iz

2= cos z.

10

and

cos z1 cos z2 − sin z1 sin z2

=eiz1 + e−iz1

2

eiz2 + e−iz2

2− eiz1 − e−iz1

2i

eiz2 − e−iz2

2i

=ei(z1+z2) + e−i(z1+z2) + ei(z1−z2) + e−i(z1−z2)

4− ei(z1+z2) + e−i(z1+z2) − ei(z1−z2) − e−i(z1−z2)

−4

=2ei(z1+z2) + 2e−i(z1+z2)

4= cos(z1 + z2).

I 27. Evaluate

sin i, cos(2 + i), tan(1 + i), 2i, ii, (−1)2i, log(2− 3i), arccos 14(3 + i).

Solution. sin i = sinh 1.cos(2 + i) = cosh 1 cos 2− i sinh 1 sin 2.tan(1 + i) = sin 1 cos 1+i sinh 1 cosh 1 cos 2

cosh2 1 cos2 1+sinh2 1 sin2 1.

2i = e−2kπ[cos(log 2) + i sin(log 2)], k ∈ Z.ii = e−

π2 +2kπ, k ∈ Z.

(−1)2i = e−2(2k+1)π, k ∈ Z.log(2− 3i) = log

√13 + i[2kπ − arctan 3

2 ], k ∈ Z.arccos 1

4 (3 + i) = 3+i±√10ei(

π2

− 12

arctan 34)

4 .

I 28. (1) Find the values of ez for z = πi/2, −(2/3)πi.(2) Find the values of z, if z satisfies ez = i.

Solution. eπ2 i = i, e− 2π

3 i = 12 −

√32 i. log i = (2kπ + π

2 )i, k ∈ Z.

I 29. Find the real part and the imaginary part of zz, where z = x+ iy.

Solution. zz = ez log z = e(x+iy) log(x+iy) = ex2 log(x2+y2)−y(2kπ+arctan y

x ) · ei[y2 log(x2+y2)+x(2kπ+arctan y

x )], k ∈Z.

I 30. Show that the roots of zn = a are the vertices of a regular polygon.

Proof. The root of zn = a are |a| 1n ei

2kπ+arg an , k = 0, 1, · · · , n− 1.

I 31. Show that the Cauchy criterion for convergence and the Weierstrass M-test hold for complex field.

Proof. If un(z) = Refn(z) and vn(z) = Imfn(z), then by max{|un(z)− um(z)|, |vn(z)− vm(z)|} ≤ |fn(z)−fm(z)| ≤ |un(z)−um(z)|+ |vn(z)− vm(z)|, the Cauchy criterion for convergence for complex field is reducedto that for real numbers.

Note |∑m

n=k fn(z)| ≤M∑m

n=k an. So∑∞

n=1 fn(z) converges uniformly if∑∞

n=1 an converges.

I 32. Find the radius of convergence of∑∞

n=1 anzn if

(i) an = n1/n;

Solution. Since limn→∞ n1n2 = elimn→∞

ln nn2 = 1, Hadamard’s formula implies R = 1.

11

(ii) an = nln n;

Solution. Since limn→∞ nln nn = elimn→∞[ ln

n ·ln n]1, Hadamard’s formula implies R = 1.

(iii) an = n!/nn;

Solution. By Stirling’s formula n! ∼√2πn

(ne

)n, we have

limn→∞

[n!

nn

] 1n

= limn→∞

e1n ln n!

nn = limn→∞

e1n [ln(n!)−ln

√2πn(n

e )n] · lim

n→∞e

1n ln

√2πn(n

e )n−ln n = e−1.

So Hadamard’s formula implies R = e.

(iv) an = nn.

Solution. Since limn→∞ n = ∞, Hadamard’s formula implies R = 0.

I 33. Prove Theorem 1.15.

Proof. Let u(z) = Ref(z) and v(z) = Imf(z). Then f is uniformly convergent on a set A if and only if uand v are uniformly convergent on A. Since dz = dx+ idy, we can reduce the theorem to that of real-valuedfunctions of real variables.

I 34. Let f(z) be holomorphic on C and f(0) = 1. Show that:(1) If f ′(z) = f(z) for all z ∈ C, then f(z) = ez.

Proof. [f(z)e−z]′ = f ′(z)e−z − f(z)e−z = 0. So f(z) = Cez for some constant C. f(0) = 1 dictatesC = 1.

(2) If for every z ∈ C and every ω ∈ C, f(z + ω) = f(z)f(ω) and f ′(0) = 1, then f(z) = ez.

Proof. f ′(z) = limω→0f(z+ω)−f(z)

ω = limω→0 f(z)f(ω)−1

ω = f(z)f ′(0) = f(z). By part (1), we concludef(z) = ez.

I 35. Let f be a holomorphic function on C\(−∞, 0]. Show that:(1) If f(z) = e−f(z) for every z ∈ C\(−∞, 0], then f(z) = log z.

Proof. [ef(z)]′ = f ′(z)ef(z) = 1. So ef(z) = 1.

(2) If f(zω) = f(z) + f(ω) for every z ∈ C\(−∞, 0], ω ∈ (0,+∞) and f ′(1) = 1, then f(z) = log z.

Proof. It’s clear f(1) = 0. So

f ′(z) = limω→0

f(z + zω)− f(z)

zω= lim

ω→0

f(z(1 + ω))− f(z)

zω=

1

zlimω→0

f(1 + ω)

ω=

1

zf ′(1) =

1

z.

Therefore [1

zef(z)

]′= − 1

z2ef(z) +

1

zef(z)f ′(z) = 0.

This shows ef(z) = z.

I 36. Ifφ(z) =

1

2

(z +

1

z

),

then φ is univalent on the following four regions:(1) upper half plane {z ∈ C : Imz > 0}.(2) lower half plane {z ∈ C : Imz < 0}.(3) unit disc D(0, 1) except the center point {0}, D(0, 1)\{0}.(4) outside of the unit disc {z ∈ C : |z| > 1}.

12

Proof. 2[φ(z1) − φ(z2)] = (z1 − z2)(1− 1

z1z2

). So φ(z) is univalent in any region that makes z1z2 6= 1

whenever z1 6= z2. Since z1z2 = 1 implies |z1| = 1|z2| and arg z1 = − arg z2, we conclude φ is univalent in any

of the four regions in the problem.

I 43. If f(z) is a univalent holomorphic function on a region U , then the area of f(U) is∫ ∫U

|f ′(z)|2dxdy

where z = x+ iy.

Proof. This is straightforward from Problem 24.

I 44. Show that if {an} and {bn} satisfy the conditions:(1) the sequence {Sn} is bounded, where

Sn =n∑

k=1

ak,

(2) limn→∞ bn = 0.(3)

∑∞n=1 |bn − bn+1| <∞,

then the series∞∑

n=1

anbn

is convergent. Moreover, verify that this is a generalization of the Dirichlet criterion and the Abel criterionfor series in the real number field.

Proof. We first use Abel’s Lemma on partial sum: for n > m,n∑

k=m+1

akbk =

n∑k=m+1

(Sk − Sk−1)bk = Sbbn−1 +

n∑k=m+1

Sk(bk − bk+1)− Smbm+1.

Denote by K a bound of the sequence (Sn)∞n=1, then |Snbn+1| ≤ K|bn+1|, |Smbm+1| ≤ K|bm+1|, and

|∑n

k=m+1 Sk(bk − bk+1)| ≤ K∑n

k=m+1 |bk − bk+1|. Since limn→∞ bn = 0 and∑∞

n=1 |bn − bn+1| < ∞,we conclude

∑nk=m+1 akbk can be arbitrarily small if m is sufficiently large. So under conditions (1)-(3),∑∞

n=1 akbk is convergent.In the Dirichlet criterion, we require

(1) (Sn)∞n=1 is bounded;

(2) limn→∞ bn = 0;(3’) (bn)

∞n=1 is monotone.

Clearly, (2)+(3’) imply (3), so the Dirichlet criterion is a special case of the result in current problem.In the Abel criterion, we require(1”) (Sn)

∞n=1 is convergent;

(2”) (bn)∞n=1 is bounded;

(3”) (bn)∞n=1 is monotone.

Define b′n = bn − b, where b = limn→∞ bn is a finite number. Then (Sn)∞n=1 satisfies (1) and (b′n)

∞n=1

satisfies (2) and (3’). By the Dirichlet criterion,∑∞

n=1 anb′n converges. Note

∑Nn=1 anbn =

∑Nn=1 anb

′n+bSN .

By condition (1”), we conclude∑∞

n=1 anbn converges.

Remark 1. The result in this problem is the so-called Abel-Dedekind-Dirichlet Theorem.

I 45. Let∑∞

n=1 an be a complex series, and limn→∞|an|1/n = q. Prove:(1) If q < 1, then

∑∞n=1 an is absolutely convergent.

13

Proof. For any ε ∈ (0, 1− q), there exists N ∈ N, such that for any n ≥ N , |an|1n < q+ ε < 1. By comparing∑∞

n=1 |an| and∑∞

n=1(q + ε)n, we conclude∑∞

n=1 an is absolutely convergent.

(2) If q > 1, then∑∞

n=1 an is divergent.

Proof. By definition of upper limit and the fact q > 1, we can find ε > 0 and infinitely many n, such that|an|

1n > q − ε > 1. Therefore limn→∞ |an| = ∞ and

∑∞n=1 an is divergent.

I 46. Show that if an ∈ C\{0}, n = 1, 2, · · · , and

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = q,

then∑∞

n=1 an is absolutely convergent if q < 1. Discuss the convergence and divergence of the series if q > 1.

Proof. If q < 1, for any ε ∈ (0, 1− q), we can find N ∈ N, such that for any n ≥ N , |an+1||an| < q + ε < 1. So

|aN+k| ≤ |aN |(q + ε)k, ∀k ≥ 0. Since∑∞

k=1(q + ε)k is convergent, we conclude∑∞

n=1 an is also convergent.If q > 1, we have two cases to consider. In the first case, the upper limit is assumed to be a limit. Then

we can find ε > 0 such that q − ε > 1 and for n sufficiently large, |an+1| > |an|(q − ε) > |an|. This implieslimn→∞ an 6= 0, so the series is divergent. In the second case, the upper limit is assumed not to be a limit.Then we can manufacture counter examples where the series is convergent. Indeed, consider (k ≥ 0)

an =

{1

(k+1)2 , if n = 2k + 12

(k+1)2 , if n = 2k + 2.

Then∑∞

n=1 an is convergent and limn→∞

∣∣∣an+1

an

∣∣∣ = 2.

I 47. Let an ∈ C\{0} (n = 1, 2, · · · ) and

limn→∞

∣∣∣∣an+1

an

∣∣∣∣ = 1.

Show that iflimn→∞n

(∣∣∣∣an+1

an

∣∣∣∣− 1

)< −1,

then∑∞

n=1 an is absolutely convergent (Reabe criterion).

Proof. We choose ε > 0, such that limn→∞ n(|an+1

an| − 1

)< −1− 2ε. By the definition of upper limit, there

exists N1 ∈ N, such that for any n ≥ N1, n(|an+1

an| − 1

)< −1− 2ε, i.e. |an+1|

|an| < 1− 1+2εn . Define bn = 1

n1+ε .Then

bn+1

bn=

nε

(n+ 1)1+ε=

(1− 1

n+ 1

)1+ε

= 1− 1 + ε

n+ 1+O

(1

(n+ 1)2

).

So there exists N2 ∈ N, such that for n ≥ N2, bn+1

bn> 1 − 1+2ε

n+1 . Define N = max{N1, N2}, then for anyn ≥ N , an+1

an< bn+1

bn. In particular, we have∣∣∣∣aN+1

aN

∣∣∣∣ < bN+1

bN,

∣∣∣∣aN+2

aN+1

∣∣∣∣ < bN+2

bN+1, · · · ,

∣∣∣∣ aN+k

aN+k−1

∣∣∣∣ < bN+k

bN+k−1.

Multiply these inequalities, we get|aN+k| ≤

|aN |bN

bN+k.

Since∑∞

n=1 bn is convergent, we conclude∑∞

n=1 an must also converge. 1

1It seems the condition limn→∞∣∣∣an+1

an

∣∣∣ = 1 is redundant.

14

I 48. Let {an} be a series of positive numbers that converges to zero monotonically. Prove:(1) If R is the radius of convergence of the series

∑∞n=0 anz

n, then R ≥ 1.

Proof. Since (an)∞n=1 monotonically decreases to 0, for n sufficiently large, ln an < 0 and limn→∞

ln an

n ≤ 0.So limn→∞ a

1nn = elimn→∞

ln ann ≤ e0 = 1. By Hadamard’s formula, R ≥ 1.

(2) The series∑∞

n=0 anzn is convergent on ∂D(0, R)\{R}, where ∂D(0, R) is the boundary of the disc

D(0, R).

Proof. The claims seems problematic. For a counter example, assume R > 1 and an = 1Rn . If z = Reiθ with

θ 6= 0, we haveN∑

n=0

anzn =

N∑n=0

1

Rn·Rneinθ =

N∑n=0

cosnθ + i

N∑n=0

sinnθ =sin (N+1)θ

2 cos Nθ2

sin θ2

+ isin (N+1)θ

2 sin Nθ2

sin θ2

.

So∑∞

n=0 anzn is not convergent.

I 49. Show that the power series∞∑

n=0

an(z − z0)n

is uniformly convergent on its disc of convergence D if and only if it is uniformly convergent on D. Herez0 ∈ C is a fixed point and D is the closure of D.

Proof. By definition of uniform convergence, for any ε > 0, there exists N ∈ N, such that for any m ≥ N ,∑∞n=m |an||z − z0|n < ε, ∀z ∈ D. In particular, for any z∗ ∈ ∂D, by letting z → z∗, we get

∑∞n=m |an||z∗ −

z0|n ≤ ε. That is, for this given ε, we can pick up N , such that for any m ≥ N ,∑∞

n=m |an||z − z0|n ≤ ε,∀z ∈ D. This is exactly the definition of uniform convergence on D.

I 50. Suppose that the radius of convergence of the series∞∑

n=0

anzn = f(z)

is 1, and z0 ∈ ∂D(0, 1). Prove that if limn→∞ nan = 0 and limr→1 f(rz0) exists, then∑∞

n=0 anzn0 converges

to limn→∞ f(rz0).

Proof. We choose a sequence (rn)∞n=1 such that rn ∈ (0, 1) and limn→∞ rn = 1. Define βn =

∑nk=0 akz

k0 −

f(rnz0). We show limn→∞ βn = 0. Indeed, we note

|βn| =

∣∣∣∣∣n∑

k=0

akzk0 −

∞∑k=0

ak(rnz0)k

∣∣∣∣∣ =∣∣∣∣∣

n∑k=0

ak(1− rkn)zk0 −

∞∑k=n+1

ak(rnz0)k

∣∣∣∣∣≤

n∑k=0

|ak|(1− rn)(1 + rn + · · ·+ rk−1n ) +

∞∑k=n+1

k

n|ak|rkn

≤n∑

k=0

|ak|(1− rn)k +εn+1

n

1

1− rn,

where εn =∑

k≥n k|ak|. We choose rn in such a way that limn→∞ n(1 − rn) = 1, e.g. rn = 1 − 1n . Then

limn→∞εn+1

n(1−rn)= 0 since limn→∞ nan = 0. By Problem 18 (2), we have

n∑k=0

|ak|(1− rn)k = (1− rn)n ·∑n

k=0 |ak|kn

→ 0, as n→ ∞.

Combined, we conclude limn→∞ βn = 0.

15

Remark 2. The result in this problem is a special case of the so-called Landau’s Theorem, see, for example,Boos [1], §5.1 Boundary behavior of power series.

2 Cauchy Integral Theorem and Cauchy Integral FormulaI 1. Calculate the following by using the Cauchy Integral Formula.

(i)∫|z+i|=3

sin z dzz+i ;

(ii)∫|z|=2

ez

z−1dz;(iii)

∫|z|=4

cos zz2−π2 dz;

(iv)∫|z|=2

dz(z−1)n(z−3) , n = 1, 2, · · · ;

(v)∫|z|= 3

2

dz(z2+1)(z2+4) ;

(vi)∫|z|=2

dzz5−1 ;

(vii)∫|z|=R

dz(z−a)n(z−b) , where a, b are not on the circle |z| = R and n is a positive integer.

(viii)∫|z|=3

dz(z3−1)(z−2)2 .

Solution. (i) The integral is equal to 2πi · sin i = π(e−1 − e).(ii) The integral is equal to 2eπi.(iii) For sufficiently small ε > 0, the integral is equal to∫

|z−π|=ε

cos z(z + π)(z − π)

dz +

∫|z+π|=ε

cos z(z + π)(z − π)

dz = 2πi

[cosπ2π

+cos(−π)−2π

]= 0.

(iv) The integral is equal to 2πi(

1z−3

)(n−1)∣∣∣∣z=1

= 2πi · −(n−1)!2n = − (n−1)!

2n−1 πi.

(v) For sufficiently small ε > 0he integral is equal to∫|z+i|=ε

1

(z − i)(z2 + 4)· dz

(z + i)+

∫|z−i|=ε

1

(z + i)(z2 + 4)· dz

(z − i)= 2πi ·

[1

(−2i) · 3+

1

2i · 3

]= 0.

(vi) By the Fundamental Theorem of Algebra, the equation z5 − 1 = 0 has five solutions z1, z2, z3, z4, z5.For any i ∈ {1, 2, 3, 4, 5}, we have

(zi − z1) · · · (zi − zi−1)(zi − zi+1) · · · (zi − z5) = limz→zi

z5 − 1

z − zi= lim

z→zi

z5 − z5iz − zi

= 5z4i =5

zi.

So for sufficiently small ε > 0, the integral is equal to5∑

i=1

∫|z−zi|=ε

dz

(z − z1)(z − z2)(z − z3)(z − z4)(z − z5)

= 2πi

5∑i=1

1

(zi − z1) · · · (zi − zi−1)(zi − zi+1) · · · (zi − z5)

= 2πi

6∑i=1

zi5

= 0,

where the last equality comes from the fact that the expansion of (z−z1) · · · (z−z5) is z5−1 and −(z1+· · ·+z5)is the coefficient of z4.

(vii) If |a|, |b| > R, the integral is equal to 0. If |a| > R > |b|, the integral is equal to 2πi(b−a)n . If

|b| > R > |a|, the integral is equal to 2πi(

1z−b

)(n−1)∣∣∣∣z=a

= 2πi · (−1)n−1(a − b)−n. If R > |a|, |b|, the

integral is equal to 2πi[

1(b−a)n + (−1)n−1 1

(a−b)n

]= 0.

16

(viii) Denote by z1, z2, z3 the three roots of the equation z3 − 1 = 0. Then the integral is equal to

2πi

[1

(z1 − z2)(z1 − z3)(z1 − 2)2+

1

(z2 − z1)(z2 − z3)(z2 − 2)2+

1

(z3 − z1)(z3 − z2)(z3 − 2)2

+

(1

z3 − 1

)(1)∣∣∣∣∣z=2

]

= 2πi

[z1

3(z1 − 2)2+

z23(z2 − 2)2

+z3

3(z3 − 2)2− 12

49

].

I 2. Prove that (zn

n!

)2

=1

2πi

∫C

znezζ

n!ζn· dζζ,

where C is a simple closed curve around the origin.

Proof.1

2πi

∫C

ezζ

ζn· dζζ

=1

2πi

∫C

∞∑k=0

zk

k!

1

ζn−k· dζζ

=

∞∑k=0

zk

k!· 1

2πi

∫C

1

ζn−k· dζζ

=zn

n!,

where the last equality is due to Cauchy’s integral formula. We need to justify the exchange of integrationand infinite series summation. Indeed, for m > n, we have∣∣∣∣∣

m∑k=0

zk

k!· 1

2πi

∫C

1

ζn−k· dζζ

− 1

2πi

∫C

∞∑k=0

zk

k!

1

ζn−k· dζζ

∣∣∣∣∣=

∣∣∣∣∣ 1

2πi

∫C

∞∑k=m+1

zk

k!

1

ζn−k· dζζ

∣∣∣∣∣≤ 1

2π

∫C

∞∑k=m+1

|zζ|k

k!

|dζ||ζ|n+1

≤ 1

2π

∫C

∞∑k=m+1

Mk

k!

|dζ||ζ|n+1

,

where M = |z|maxζ∈C |ζ|. Since∑∞

k=m+1Mk

k! is the remainder of eM , for m sufficiently large,∑∞

k=m+1Mk

k!can be smaller than any given positive number ε. So for m sufficiently large,∣∣∣∣∣

m∑k=0

zk

k!· 1

2πi

∫C

1

ζn−k· dζζ

− 1

2πi

∫C

∞∑k=0

zk

k!

1

ζn−k· dζζ

∣∣∣∣∣ ≤ ε1

2π

∫C

|dζ||ζ|n+1

.

This shows∑∞

k=0zk

k! ·1

2πi

∫C

1ζn−k · dζ

ζ converges to 12πi

∫C

∑∞k=0

zk

k!1

ζn−k · dζζ .

I 3. Let f and g be holomorphic in the unit disc |z| < 1 and continuous on |z| ≤ 1. Show that

1

2πi

∫|ζ|=1

(f(ζ)

ζ − z+

zg(ζ)

zζ − 1

)dζ =

{f(z), |z| < 1,

g(1z

), |z| > 1.

Proof. We note g(ζ)zζ−1 = g(ζ)

ζ− 1z

. If |z| < 1, we apply Cauchy’s integral formula to 12πi

∫|ζ|=1

f(ζ)ζ−z dζ and apply

Cauchy’s integral theorem to 12πi

∫|ζ|=1

g(ζ)

ζ− 1z

dζ (regarding 1z as a parameter and g(ζ)

ζ− 1z

an analytic function ofζ). If |z| > 1, we apply Cauchy’s integral theorem to 1

2πi

∫|ζ|=1

f(ζ)ζ−z dζ (regarding z as a parameter and f(ζ)

ζ−z

an analytic function of ζ) and apply Cauchy’s integral formula to 12πi

∫|ζ|=1

g(ζ)

ζ− 1z

dζ.

17

I 4. Letf(z) =

∫|ζ|=3

3ζ2 + 7ζ + 1

ζ − zdζ.

Find f ′(1 + i).

Solution. For z ∈ {z : |z| < 3}, by Cauchy’s integral formula, f(z) = 2πi(3z3 + 7z + 1). So f ′(1 + i) =2πi[6(1 + i) + 7] = −12π + 26πi.

I 5. Show that ∫ 2π

0

cos2n θdθ = 2π · 1 · 3 · 5 · · · · · (2n− 1)

2 · 4 · 6 · · · · · 2nby calculating ∫

|z|=1

(z +

1

z

)2ndz

z.

Proof. We note ∫|z|=1

(z +1

z)2n

dz

z=

∫ 2π

0

(eiθ + e−iθ)2nidθ = 22ni

∫ 2π

0

cos2n θdθ.

On the other hand,

(z +1

z)2n · 1

z=

2n∑k=0

Ck2nz

k(z−1)2n−k · z−1 =

2n∑k=0

Ck2n

z−2n+2k−1.

Therefore∫ 2π

0

cos2n θdθ = 1

22ni

∫|z|=1

(z +1

z)2n

dz

z=

1

22ni

∫|z|=1

Cn2ndz

z=

2πCn2n

22n= 2π · 1 · 3 · 5 · · · · · (2n− 1)

2 · 4 · 6 · · · · · 2n.

I 6. Letpn(z) =

1

2nn!

dn

dzn[(z2 − 1)n].

This is referred to as Legendre polynomial. Show that

pn(z) =1

2πi

∫γ

(ζ2 − 1)n

2n(ζ − z)n+1dζ,

where γ is a simple closed curve containing z.

Proof. This is straightforward from Cauchy’s integral formula.

I 7. Let f(z) be a holomorphic function on C and |f(z)| ≤Me|z|. Show that |f(0)| ≤M and

|fn(0)|n!

≤M( en

)n, n = 1, 2, · · · .

Proof. For any R > 0, we have f (n)(0) = n!2πi

∫|z|=R

f(ξ)ξn+1 dξ. So

∣∣∣∣f (n)(0)n!

∣∣∣∣ = 1

2π

∣∣∣∣∫ 2π

0

|f(Reiθ)|Rn

dθ

∣∣∣∣ ≤ 1

2π

∣∣∣∣∣∫ 2π

0

Me|Reiθ|

Rndθ

∣∣∣∣∣ =MeR

Rn.

Let R = n, we get the desired inequality.

18

Remark 3. Note h(R) = ln eR

Rn = R − n lnR takes minimal value at n. This is the reason why wechoose R = n, to get the best possible estimate. Another perspective is to assume f(z) = ez. Then∣∣∣ f(n)(0)

n!

∣∣∣ = 1n! ∼

1√2πn(n

e )n = 1√

2πn

(en

)n by Stirling’s formula. So the bound M eR

Rn is not “far” from the bestone.

I 8. (Cauchy integral formula for the outside region of a simple closed curve) Let γ be a rectifiable simpleclosed curve, D1 be the inside of γ and D2 be the outside of γ. Suppose a function f(z) is holomorphic inD2 and continuous on D2 ∪ γ. Show that

(1) If limz→∞ f(z) = A, then

1

2πi

∫γ

f(ζ)

ζ − zdζ =

{−f(z) +A, z ∈ D2,

A, z ∈ D1;

Proof. For any z ∈ D2, we can find R > 0 sufficiently large, so that z is in the region enclosed by γ and{ζ : |ζ − z| = R}. By Cauchy’s integral formula

f(z) =1

2πi

∫−γ

f(ζ)

ζ − zdζ +

1

2πi

∫|ζ|=R

f(ζ)

ζ − zdζ.

So

1

2πi

∫γ

f(ζ)

ζ − zdζ = −f(z) + 1

2πi

∫|ζ|=R

f(ζ)

ζ − zdζ

= −f(z) + 1

2πi

∫|ζ|=R

f(ζ)− f(∞)

ζ − zdζ +

1

2πi

∫|ζ|=R

f(∞)

ζ − zdζ.

We note ∣∣∣∣∣ 1

2πi

∫|ζ|=R

f(ζ)− f(∞)

ζ − zdζ

∣∣∣∣∣ ≤ supζ∈D(z,R)

|f(ζ)−A| → 0, as R→ ∞,

and Cauchy’s integral formula implies

1

2πi

∫|ζ|=R

f(∞)

ζ − zdζ = f(∞) · 1 = A.

So by letting R→ ∞ in the above equality, we have

1

2πi

∫γ

f(ζ)

ζ − zdζ = −f(z) +A.

If z ∈ D1, h(ζ) = f(ζ)ζ−z is a holomorphic function in D2, with z a parameter. So for R sufficiently large,

Cauchy’s integral theorem implies

0 =1

2πi

∫−γ

f(ζ)

ζ − zdζ +

1

2πi

∫|ζ|=R

f(ζ)

ζ − zdζ.

An argument similar to (1) can show 12πi

∫γ

f(ζ)ζ−z dζ = A.

(2) If the origin is in D1, then

1

2πi

∫γ

zf(ζ)

zζ − ζ2dζ =

{f(z), z ∈ D2,

0, z ∈ D1.

19

Proof. We note zzζ−ζ2 = 1

ζ − 1ζ−z . For R sufficiently large, Cauchy’s integral theorem implies

0 =1

2πi

∫−γ

f(ζ)

ζdζ +

1

2πi

∫|ζ|=R

f(ζ)

ζdζ.

So 12πi

∫γ

f(ζ)ζ dζ = 1

2πi

∫|ζ|=R

f(ζ)ζ dζ. Similarly,

1

2πi

∫γ

f(ζ)

ζ − zdζ =

{1

2πi

∫|ζ|=R

f(ζ)ζ−z dζ − f(z) z ∈ D2

12πi

∫|ζ|=R

f(ζ)ζ−z dζ z ∈ D1.

Therefore1

2πi

∫γ

zf(ζ)

zζ − ζ2dζ =

1

2πi

∫γ

[f(ζ)

ζ− f(ζ)

ζ − z

]dζ =

{f(z) z ∈ D2

0 z ∈ D1.

I 9. Let f(z) be holomorphic in a bounded region D and continuous on D with f(z) 6= 0. Show that if|f(z)| =M on ∂D, then f(z) =Meiα, where α is a real number.

Proof. Since f(z) 6= 0 in D, 1/f(z) is holomorphic in D. Applying Maximum Modulus Principle to 1/f(z),we conclude |f(z)| achieves its minimum on ∂D. Applying Maximum Modulus Principle to f(z), we conclude|f(z)| achieves its maximum on ∂D. Since |f(z)| ≡M on ∂D, |f(z)| must be a constant on D. By Chapter1, exercise problem 21 (iii), we conclude f(z) is a constant function.

I 10. Suppose f(z) is holomorphic and bounded in C, a, b are complex numbers. Find

limR→∞

∫|z|=R

f(z)

(z − a)(z − b)dz.

This leads to another proof of Liouville Theorem.

Proof. For R large enough, |a|, |b| < R. So we can find ε > 0 such that (assume a 6= b)∫|z|=R

f(z)

(z − a)(z − b)dz =

∫|z−a|=ε

f(z)

(z − a)(z − b)dz +

∫|z−b|=ε

f(z)

(z − a)(z − b)dz =

2πi

a− b[f(a)− f(b)].

Meanwhile, denote by M a bound of f , we have∣∣∣∣∣∫|z|=R

f(z)

(z − a)(z − b)dz

∣∣∣∣∣ ≤ M

∫|z|=R

|dz||(z − a)(z − b)|

≤ M

∫|z|=R

|dz|(R− |a|)(R− |b|)

=2πMR

(R− |a|)(R− |b|)→ 0, as R→ ∞.

Combined, we conclude 0 = 2πia−b [f(a)− f(b)], ∀a, b ∈ C. This shows f is a constant function.

I 11. If f(z) is holomorphic in the unit disc |z| < 1 and

|f(z)| ≤ 1

1− |z|(|z| < 1),

then|f (n)(0)| ≤ n!

rn(1− r)

for 0 < r < 1. Especially, if r = 1− 1/(n+ 1), then

|f (n)(0)| < e(n+ 1)!, n = 1, 2, · · · .

20

Proof. By Cauchy’s integral formula,

|f (n)(0)| ≤ n!

2π

∣∣∣∣∣∫|z|=r

f(ξ)

ξn+1dξ

∣∣∣∣∣ ≤ n!

2π

∫ 2π

0

11−r

rndθ =

n!

rn(1− r).

I 12. (Integral of Cauchy type) If the function φ(ζ) is continuous on a rectifiable curve γ, show that thefunction

Φ(z) =1

2πi

∫γ

f(ζ)dζ

ζ − z

is holomorphic in any region D which does not contain any point of γ. Moreover,

Φ(n)(z) =n!

2πi

∫γ

φ(ζ)dζ

(ζ − z)n+1, n = 1, 2, · · · .

Proof. We note

limz→z0

Φ(z)− Φ(z0)

z − z0=

1

2πilimz→z0

∫γ

φ(ζ)

(ζ − z)(ζ − z0)dζ =

1

2πi

∫γ

φ(ζ)

(ζ − z0)2dζ,

where the exchange of limit and integration can be seen as an application of Lebesgue’s Dominated Conver-gence Theorem (We can find ρ > 0 so that ∀z in a neighborhood of z0, dist(γ, z) ≥ ρ. Then 1

(ζ−z)(ζ−z0)) ≤ 1

ρ2 ).This shows Φ is holomorphic in any region D which does not contain any point of γ.

The formula for n-th derivative of Φ can be proven similarly and by induction. For details, we refer thereader to 方企勤 [3], Chapter 4, §3, Lemma 2.

Remark 4. The result still holds if φ is piecewise continuous on γ. See, for example, 方企勤 [3], Chapter7, proof of Theorem 7.

I 13. Suppose that a non-constant function f(z) is holomorphic on a bounded region D and continuous onD. Let m = infτ∈∂D |f(z)|, M = supz∈∂D |f(z)| and f(z) 6= 0. Then m < |f(z)| < M for any point z ∈ D.

Proof. Since f(z) 6= 0 in D, 1/f(z) is holomorphic in D. Applying Maximum Modulus Principle to 1/f(z),we conclude |f(z)| does not achieve its minimum in D unless it’s a constant function. Applying MaximumModulus Principle to f(z), we conclude f(z) does not achieve its maximum in D unless it’s a constantfunction. Combined, we have m < |f(z)| < M for any point z ∈ D.

I 14. If pn(z) is a polynomial with degree n and |pn(z)| ≤ M for |z| < 1, then |pn(z)| ≤ M |z|n for1 ≤ |z| <∞.

Proof. Define f(z) = pn(z)zn and g(z) = f( 1z ). Then g(z) is holomorphic and bounded on D(0, 1) \ {0}, since

limz→0 g(z) = limz→∞pn(z)zn is a finite number and g(z) is clearly continuous in a neighborhood of ∂D(0, 1).

By Theorem 2.11, Riemann’s Theorem, g(z) can be analytically continued to D(0, 1). By a corollary ofTheorem 2.18, Maximum Modulus Principle, the maximum value of |g(z)| can only be reached on ∂D(0, 1).So |g(z)| ≤ max|z|=1 |g(z)| = max|z|=1 |f(z)| = max|z|=1 |pn(z)| ≤M , ∀z ∈ D(0, 1), i.e. |f(z)| = |pn(z)|

|zn| ≤M

for 1 ≤ |z| <∞.

I 15. Suppose f(z) is a holomorphic function on the disc |z| < R, |f(z)| ≤M and f(0) = 0. Show that

|f(z)| ≤ M

R|z|, |f ′(0)| ≤ M

R,

where the equalities hold if and only if f(z) = (M/R)eiαz with α ∈ R.

21

Proof. Define g(z) = f(Rz)M (z ∈ D(0, 1)) and apply Theorem 2.19, Schwarz Lemma, to g(z), we have

|g(z)| ≤ |z| (∀z ∈ D(0, 1)) and |g′(0)| ≤ 1. So for any z ∈ D(0, R), f(z) = f(R · zR ) = Mg( z

R ) ≤ M |z|R and

|f ′(0)| = |g′(0)| · MR ≤ M

R .

I 16. Prove Corollary 2.1 by applying Theorem 2.7.

Proof. Define ρ = dist(C \ V,K). Then ρ ∈ (0,∞). By Theorem 2.7, for any z ∈ K, we have

|f (n)(z)| ≤ n!

2π

∫∂V

|f(ξ)||ξ − z|n+1

|dξ| ≤ n!

2π

∫∂V

|dξ|ρn+1

· supz∈V

|f(z)|.

Define cn = n!2π

∫∂V

|dξ|ρn+1 and take supremum on the left side of the above inequality, we have

supz∈K

|f (n)(z)| ≤ cn supz∈V

|f(z)|.

I 17. Prove that if {fn(z)} is a sequence of continuous functions on a region D that converges uniformlyto f(z), then ∫

γ

f(z)dz = limn→∞

∫γ

fn(z)dz

for any curve γ in D. (This assertion was used in the proof of Theorem 2.15.)

Proof. We have to assume γ has finite length, that is, L(γ) :=∫γ|dz| <∞. Then by the definition of uniform

convergence, for any ε > 0, there exists n0 such that ∀n ≥ n0, supz∈D |f(z) − fn(z)| ≤ εL(γ) . So for any

n ≥ n0, ∣∣∣∣∫γ

f(z)dz −∫γ

fn(z)dz

∣∣∣∣ ≤ ∫γ

|f(z)− fn(z)||dz| ≤∫γ

ε

L(γ)|dz| = ε.

Therefore limn→∞∫γfn(z)dz =

∫γf(z)dz.

I 18. Let f(z) be a holomorphic function in the disc |z| < 1 with Ref(z) > 0 and f(0) = α > 0. Show that

|f(z)| ≤ 2A|z|1− |z|

.

Proof. We note the linear fractional transformation ω(z) = z−αz+α maps {z : Rez > 0} to D(0, 1). So

h(z) = ω(f(z)) maps D(0, 1) to itself (see, for example,方企勤 [3], Chapter 3, §7) and h(0) = 0. By SchwarzLemma, h(z)| = |z| and |h′(0)| ≤ 1. This is equivalent to∣∣∣∣f(z)− α

f(z) + α

∣∣∣∣ ≤ |z|, and |f ′(0)| ≤ 2α.

I 19. Let f(z) be a holomorphic function in the disc |z| < 1 with f(0) = 0 and Ref(z) ≤ A (A > 0). Showthat

|f(z)| ≤ 2A|z|1− |z|

.

Proof. The linear fractional transformation ω(z) = zz−2A maps {z : Rez < A} to D(0, 1). So h(z) =

ω(f(z)) = f(z)f(z)−2A maps D(0, 1) to itself and h(0) = 0. By Schwarz Lemma, |h(z)| ≤ |z|. So ∀z ∈ D(0, 1)

|f(z)| ≤ |z| · |f(z)− 2A| ≤ |z||f(z)|+ 2A|z|, i.e. |f(z)| ≤ 2A|f(z)|1− |z|

.

22

I 20. Find the Taylor expansions and their radius of convergence of the following functions at z = 0.(i) ez+e−z+2 cos z

4 ;

Solution. We note

ez + e−z + 2 cos z4

=1

4[ez + e−z + eiz + e−iz] =

1

4

∞∑k=0

1

k![zk + (−z)k + (iz)k + (−iz)k] =

∞∑m=0

z4m

(4m)!.

The radius of convergence is ∞.

(ii) z2+4z4+z6

(1−z2)3 ;

Solution. By induction, it is easy to show[

1(1−ζ)3

](n)= (n+2)!

2! (1 − ζ)−(n+3). So the Taylor expansion of1

(1−ζ)3 is 1(1−ζ)3 =

∑∞n=0

(n+1)(n+2)2 ζn, with radius of convergence equal to 1. This implies

ζ + 4ζ2 + ζ3

(1− ζ)3= ζ + 7ζ4 +

∞∑n=3

(3n2 − 3n+ 1)ζn.

Replace ζ with z2, we can get the Taylor expansion at z = 0: z2 + 7z8 +∑∞

n=3(3n2 − 3n + 1)z2n, and the

radius of convergence is equal to 1 (this can be verified by calculating limn→∞(3n2 − 3n+ 1)1n = 1).

(iii) (1− z−5)−4;

Solution. By induction, it is easy to show[(1− ζ)−4

](n)= (n+3)!

3! (1−ζ)−(n+4). So for ζ ∈ D(0, 1), (1−ζ)−4 =∑∞n=0

(n+1)(n+2)(n+3)6 ζn. Replace ζ with z5, we get for z ∈ D(0, 1)

1

(1− z5)4=

∞∑n=0

(n+ 1)(n+ 2)(n+ 3)

6z5n.

Therefore(1− z−5)−4 =

z20

(1− z5)4=

∞∑n=0

(n+ 1)(n+ 2)(n+ 3)

6z5n+20.

By Hadamard’s formula, the radius of convergence is 1.

(iv) z6

(z2−1)(z+1) .

Solution. By induction, we can show[(1 + z)−2

](n)= (−1)n(n+ 1)!(1 + z)−(n+2). So ∀z ∈ D(0, 1),

1

(1 + z)2=

∞∑n=0

(−1)n(n+ 1)zn.

It is also easy to verify that

1

(z + 1)2(z − 1)= − 1

4(z + 1)+

1

4(z − 1)− 1

2(z + 1)2= −1

4

∞∑n=0

(−z)n − 1

4

∞∑n=0

zn − 1

2

∞∑n=0

(−1)n(n+ 1)zn.

Therefore

z6

(z + 1)(z2 − 1)= −z

6

4

∞∑n=0

[(−1)n + 1 + 2 · (−1)n(n+ 1)] zn = −1

4

∞∑n=0

[(−1)n(2n+ 3) + 1] zn+6.

By Hadamard’s formula, the radius of convergence is 1.

23

I 21.(1) If f(z) is holomorphic on |z| ≤ r and f(z) =

∑∞n=0 anz

n is its Taylor expansion at z = 0, then

∞∑n=0

|an|2r2n =1

2π

∫ π

−π

|f(reiθ)|2dθ.

Proof. We note

1

2π

∫ π

−π

|f(reiθ)|2dθ =1

2π

∫ π

−π

( ∞∑n=0

anrneinθ

)·

( ∞∑m=0

amrme−imθ

)dθ

=1

2π

∞∑m,n=0

∫ π

−π

anamrn+mei(n−m)θdθ

=

∞∑m,n=0

anamrn+mδnm

=

∞∑n=0

|an|2r2n,

where δmn =

{1 if m = n

0 if m 6= nand the exchange of integration and summation is justified by Theorem 1.14

(Abel Theorem) and Problem 17.

(2) If r = 1 in (1), then∞∑

n=0

|an|2

n+ 1=

1

π

∫ ∫D

|f(z)|2dA,

where D is the unit disc |z| ≤ 1 and dA is the area element.

Proof. In the result of part (1), multiply both sides by r and integrate with respect to r from 0 to 1, we have∞∑

n=0

|an|2∫ 1

0

r2n+1dr =

∞∑n=0

|an|2

2n+ 2=

1

2π

∫ 1

0

∫ π

−π

|f(reiθ)|2rdrdθ = 1

2π

∫ ∫D

|f(z)|2dA.

So∑∞

n=0|an|2n+1 = 1

π

∫ ∫D|f(z)|2dA.

I 22. Verify the equations (2.32) and (2.33).

Proof. Suppose u is harmonic on D(0, R) and continuous D(0, R), then ∀z ∈ D(0, 1), v(z) := u(zR) isharmonic on D(0, 1) and continuous on D(0, 1). By formula (2.31), for any z ∈ D(0, R),

u(z) = v(z

R) =

1

2π

∫ 2π

0

v(eiτ )1−

∣∣ zR

∣∣2∣∣1− zRe

iτ∣∣2 dτ =

1

2π

∫ 2π

0

u(Reiτ )R2 − |z|2

|R− zeiτ |2dτ

=1

2π

∫ 2π

0

u(Reiτ )R2 − |z|2

|R− ze−iτ |2dτ =

1

2π

∫ 2π

0

u(ζ)R2 − |z|2

|ζ − z|2dτ.

This is formula (2.32). To verify formula (2.33), note

ReReiτ + z

Reiτ − z= Re (Re

iτ + z)(Re−iτ − z)

|Reiτ − z|2= ReR

2 − |z|2 +R(ze−iτ − zeiτ )

|Reiτ − z|2=

R2 − |z|2

|Reiτ − z|2.

24

Therefore,

u(z) =1

2π

∫ 2π

0

u(Reiτ )R2 − |z|2

|Reiτ − z|2dτ =

1

2π

∫ 2π

0

u(Reiτ )ReReiτ + z

Reiτ − zdτ

= Re[1

2π

∫ 2π

0

u(Reiτ )Reiτ + z

Reiτ − zdτ

]= Re

[1

2πi

∫|ζ|=R

u(ζ)ζ + z

ζ − z

dζ

ζ

].

This is formula (2.33).

I 23. Show that one of the zeros of the equation z4 − 6z + 3 = 0 is in the disc |z| < 1 and the other threezeros are in 1 < |z| < 2.

Proof. On ∂D(0, 1), |(−6z)−(z4−6z+3)| = |z4+3| ≤ 4 < |−6z|. By Rouché Theorem, z4−6z+3 and −6zhave the same number of zeros in D(0, 1), which is one. On ∂D(0, 2), |z4−(z4−6z+3)| = |6z−3| ≤ 15 < |z4|.So z4 − 6z + 3 and z4 have the same number of zeros in D(0, 2), which is four. Combined, we concludez4 − 6z + 3 = 0 has one root in D(0, 1) and three roots in the annulus {z : 1 < |z| < 2}.

I 24. Find the number of zeros of the equation z7 − 5z4 − z + 2 = 0 in the disc |z| < 1.

Solution. On ∂D(0, 1), |(−5z4)− (z7 − 5z4 − z + 2)| = |z7 − z + 2| ≤ 4 < | − 5z4|. So z7 − 5z4 − z + 2 and−5z4 have the same number of zeros in D(0, 1), which is four (counting multiplicity).

I 25. Show that there is exactly one zero of the equation z4 + 2z3 − 2z + 10 = 0 in each quadrant.

Proof. Let P (z) = z4 + 2z3 − 2z + 10. We note P (z) = (z2 − 1)(z + 1)2 + 11. If z ∈ R and |z| ≥ 1,P (z) ≥ 11; if z ∈ R and |z| ≤ 1, P (z) ≥ −1 · (1 + 1)2 + 11 = 7. So P (z) = 0 has no root on the realaxis. If z = iy with y ∈ R, we have P (iy) = y4 + 10 − 2iy(y2 + 1) 6= 0. So P (z) = 0 has no root on theimaginary axis. Consider the region D enclosed by the curve γ = γ1∪γ2∪γ3, where (R is a positive number)γ1 = {z : 0 ≤ Rez ≤ R, Imz = 0}, γ2 = {z : |z| = R, arg z ∈ [0, π2 ]}, and γ3 = {z : 0 ≤ Imz ≤ R,Rez = 0}.

On γ1, ∆γ1argP (z) = 0. On γ2, P (z) = z4

(1 + 2z3−2z+10

z4

). So ∆γ2

argP (z) = 4 · π2 + o(1) = 2π + o(1)

(R → ∞). On γ3, ∆γ3 argP (z) = argP (0) − argP (iR) = arg 10 − arg(R4 + 10 − 2iR(R2 + 1)) = 0 −arg(1− 2iR(R2+1)

R4+10

)= o(1) (R → ∞). Combined, we have ∆γ argP (z) =

∑3k=1 ∆γk

argP (z) = 2π + o(1)

(R→ ∞).By The Argument Principle, P (z) = 0 has only one root in the first quadrant. The root conjugate to

this root must lie in the fourth quadrant. The other two roots are conjugate to each other and are located inthe second and third quadrant. So one must lie in the second quadrant and the other must lie in the thirdquadrant.

Remark 5. The above solution can be found in 方企勤 [3], Chapter 6, §3 Example 5.

I 26. Find the number of zeros of the equation z4 − 8z + 10 = 0 in the disc |z| < 1 and in the annulus1 < |z| < 3.

Solution. On ∂D(0, 1), |(−8z + 10)− (z4 − 8z + 10)| = |z4| = 1 < |10− 8|z|| ≤ | − 8z + 10|. So z4 − 8z + 10and −8z + 10 have the same number of zeros in D(0, 1), which is zero.

On ∂D(0, 3), |z4 − (z4 − 8z+10)| = |8z− 10| < 34 < |z4|. So z4 − 8z+10 and z4 have the same numberof zeros in D(0, 3), which is four (counting multiplicity). So z4 − 8z+10 = 0 has no root in D(0, 1) and fourroots in the annulus {z : 1 < |z| < 3}.

I 27. Show that if a > e, then there are n zeros in the disc |z| < 1 for the equation ez = azn.

25

Proof. If a > e, on ∂D(0, 1), |azn − (azn − ez)| = |ez| ≤ e|z| = e < |azn|. By Rouché Theorem, ez − azn andazn have the same number of zeros in D(0, 1), which is n.

I 28. If f(z) is holomorphic in the disc |z| < 1 and continuous on |z| ≤ 1 and |f(z)| < 1, then there is aunique fixed point of f in the disc |z| < 1.

Proof. On ∂D(0, 1), |z − (z − f(z))| = |f(z)| < 1 = |z|. So z− f(z) and z have the same number of zeros inD(0, 1), which is one. So there is a unique fixed point of f in D(0, 1).

Remark 6. This is a special case of Brower’s Fixed Point Theorem.

I 29.(1) Find a holomorphic function whose real part is

ex(x cos y − y sin y),

where x+ iy = z.

Solution. Re(zez) = ex(x cos y − y sin y).

(2) Find the most general harmonic function of the form

ax3 + bx2y + cxy2 + dy3

where a, b, c, d are real numbers.

Solution. Let u(x, y) = ax3+ bx2y+ cxy2+dy3. Then ∆u = (6a+2c)x+(2b+6d)y. So u is harmonic if andonly if 3a+ c = 0 and b+3d = 0. So the most general harmonic function of the form ax3+ bx2y+ cxy2+dy3

is ax3 − 3dx2y − 3axy2 + dy3.

I 30. By applying the mean-value property of harmonic functions, show that if −1 < r < 1, then∫ π

0

ln(1− 2r cos θ + r2)dθ = 0.

Proof. Let u(r, θ) = ln(1−2r cos θ+ r2). Then using the fact ∆ = 1r

∂∂r

(r ∂∂r

)+ 1

r2∂2

∂θ2 , we can verify ∆u = 0.So by the mean-value property of harmonic functions, we have 1

2π

∫ 2π

0u(r, θ)dθ = u(0, 0) = 0. Since cos θ

is an even function of θ, it’s not hard to show∫ 2π

0ln(1 − 2r cos θ + r2)dθ = 2

∫ π

0ln(1 − 2r cos θ + r2)dθ. So∫ π

0ln(1− 2r cos θ + r2)dθ = 0.

I 31. Prove that if a harmonic function u(z) is bounded on the entire complex plane C, then u(z) is equalto a constant identically.

Proof. Let M = supz∈C |u(z)|. Then ∀z ∈ C and R > |z|, the holomorphic function defined in formula (2.34)satisfies

|f(z)| ≤ 1

2π

∫ 2π

0

|u(Reiθ)|∣∣∣∣Reiθ + z

Reiθ − z

∣∣∣∣ dθ ≤ M

2π

∫ 2π

0

R+ |z|R− |z|

dθ →M, as R→ ∞.

So f(z) is bounded on the entire complex plane, and by Liouville’s Theorem, f is a constant function.Therefore u = Re(f) is a constant function.

I 32. Find a harmonic function on |z| < 1 such that its value is 1 on an arc of |z| = 1 and zero on the restof |z| = 1.

26

Solution. If the arc is ∂D(0, 1), the desired harmonic function u ≡ 1. So without loss of generality, weassume the arc γ = {z ∈ ∂D(0, 1) : θ1 ≤ arg z ≤ θ2} with 0 < θ2 − θ1 < 2π.

For n sufficiently large, we consider γn = {z ∈ ∂D(0, 1) : θ1 − 1n ≤ arg z ≤ θ2 +

1n}. By the Partition of

Unity Theorem for R1, we can find φn ∈ C(∂D(0, 1)) such that 0 ≤ φn ≤ 1, φn ≡ 1 on γ, and φn ≡ 0 on∂D(0, 1) \ γn. Define un(z) =

∫ 2π

0P (ζ, z)φn(ζ)dθ where P (·, ·) is the Poisson kernel and ζ = Reiθ. Then un

is the unique solution of the Dirichlet problem with boundary value φn. So φn is harmonic in D(0, 1).It is easy to see that if (φn)n uniformly converges to the indicator function 1γ(z), (un)n uniformly

converges to u(z) :=∫ 2π

0P (ζ, z)1γ(ζ)dθ on D(0, ρ) (ρ ∈ (0, 1)). Since each un satisfies the local mean value

property, u must also satisfies the local mean value property in D(0, 1) and is continuous. By Theorem 2.22,u is harmonic on D(0, 1). This suggests the harmonic function we are looking for is exactly u.

What is left to prove is that for any z0 ∈ ∂D(0, 1), lim|z|=1,z→z0 u(z) = 1γ(z0). Indeed, in general,we have the following classical result: Suppose function φ is piecewise continuous on ∂D(0, 1), then thefunction u(z) :=

∫ 2π

0P (ζ, z)φ(ζ)dθ is harmonic in D(0, 1) and for any continuity point z0 of φ, we have

limz→z0,|z|<1 u(z) = φ(z0).For the proof of continuity on the boundary, we refer to 方企勤 [3], Chapter 7, Theorem 7.

I 33. Suppose U is a region, fi(z) (i = 1, 2, · · · ) are holomorphic on U and continuous on U . Show that if∞∑

n=1

fn(z)

converges uniformly on the boundary of U , then it converges uniformly on U .

Proof. ∀ε > 0, there exists N , such that for any n ≥ N , p ∈ N, we have

supζ∈∂U

∣∣∣∣∣n+p∑

k=n+1

fk(ζ)

∣∣∣∣∣ < ε.

By Maximum Modulus Principle,

supz∈U

∣∣∣∣∣n+p∑

k=n+1

fk(z)

∣∣∣∣∣ ≤ supζ∈∂U

∣∣∣∣∣n+p∑

k=n+1

fk(ζ)

∣∣∣∣∣ < ε.

So Cauchy criterion implies∑∞

n=1 fn(z) converges uniformly on U .

Remark 7. This result is the so-called Weierstrass’ Second Theorem.

I 34. Suppose f(z) is holomorphic on D(0, R) and continuous on D(0, R). Let M = max|z|=R |f(z)|. Showthat if z0 ∈ D(0, R)\{0} is a zero of f(z), then

R|f(0)| ≤ (M + |f(0)|)|z0|.

Proof. Define g(z) = f(z)−f(z0)z−z0

= f(z)z−z0

, ∀z ∈ D(0, R) \ {z0}. Then g(z) can be analytically continued toD(0, R) (Theorem 2.11, Riemann Theorem) and is therefore continuous on D(0, R). By Cauchy’s integralformula, we have

f(0)

−z0= g(0) =

1

2πi

∫|ζ|=R

g(ζ)

ζdζ =

1

2πi

∫|ζ|=R

f(ζ)

ζ(ζ − z0)dζ.

Therefore ∣∣∣∣f(0)z0

∣∣∣∣ ≤ 1

2π

∫ 2π

0

|f(Reiθ)|Reiθ − z0|

dθ ≤ M

R− |z0|,

which is equivalent to R|f(0)| ≤ (M + |f(0)|)|z0|.

27

Remark 8. The point of this problem is to give an estimate of a holomorphic function’s modulus at z = 0in terms of its zeros.

I 35. Let |z1| > 1, |z2| > 1, · · · , |zn| > 1. Show that there exists a point z0 such that∏n

k=1 |z0 − zk| > 1.

Proof. This problem seems suspicious. For example, when z0 = 0, of course∏n

k=1 |z0 − zk| =∏n

k=1 |zk| > 1.When z0 is sufficiently large, it is also clear

∏nk=1 |z0 − zk| > 1. So the point z0 can be both inside and

outside D(0, 1). I’m not sure what proof is needed for such a trivial claim.

I 36. Suppose that f(z) is holomorphic on D(0, R). Show that

M(r) = max|z|=r

|f(z)|

is an increasing function on [0, R).

Proof. By Maximum Modulus Principle, M(r) = max|z|≤r |f(z)|. So M(r) is an increasing function on[0, R).

I 37. Use the Maximum Modulus Principle to prove the Fundamental Theorem of Algebra.

Proof. Suppose there is an n-th degree polynomial p(z) = a0+a1z+ · · ·+anzn such that n ≥ 1, an 6= 0, andp(z) 6= 0 for any z ∈ C. Then q(z) = 1/p(z) is holomorphic over the whole complex plane. For any R > 0,we have by Maximum Modulus Principle

max|z|≤R

|q(z)| ≤ max|z|=R

|q(z)|.

Since on {z : |z| = R},

|p(z)| ≥ |an|Rn−|an−1|Rn−1−· · ·−|a1|R−|a0| = Rn

(|an| −

|an−1|R

− · · · − |a1|Rn−1

− |a0|Rn

)→ ∞ as R→ ∞,

we must have limR→∞ max|z|=R |q(z)| = 0. This implies supz∈C |q(z)| = 0, i.e. p(z) ≡ ∞. This is acontradiction and our assumption must be incorrect.

I 38. If f(z) is a non-constant holomorphic function on a region U and has no zeros in U , then |f(z)| cannot attain its minimal value in U .

Proof. Apply the Maximum Modulus Principle to the function g(z) = 1/f(z), which is holomorphic onU .

I 39. (Hadamard’s Three Circles Theorem) Suppose that 0 < r1 < r2 < ∞, U = {z ∈ C : r1 < |z| < r2},f(z) is holomorphic on U and continuous on U . Define M(r) = max|z|=r |f(z)|. Show that logM(r) is aconvex function of log r on [r1, r2]. In other words, the inequality

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

logM(r1) +log r − log r1log r2 − log r1

logM(r2)

holds for r ∈ [r1, r2].

Proof. Note the Laplace operator ∆ = 4∂∂

∂∂ , so log |z| is a harmonic function on C \ {0} and log |f(z)| is a

harmonic function outside the zeros of f(z). Define K = {z ∈ U : f(z) = 0} and Vε = ∪z∈KD(z, ε). Thenfor any α ∈ R, α log |z|+ log |f(z)| is harmonic in U \ Vε and continuous on U \ Vε. By Maximum ModulusPrinciple for harmonic functions, maxz∈U\Vε

(α log |z|+ log |f(z)|) = maxz∈∂(U\Vε)(α log |z|+ log |f(z)|).

28

When z approaches to zeros of f(z), log |f(z)| → −∞, so by letting ε → 0, we can further deduce thatmaxz∈U (α log |z|+ log |f(z)|) = maxz∈∂U (α log |z|+ log |f(z)|). Therefore

α log |z|+ log |f(z)| ≤ max{α log r1 + logM(r1), α log r2 + logM(r2)}, ∀z ∈ U,

which is the same as

α log r + logM(r) ≤ max{α log r1 + logM(r1), α log r2 + logM(r2)}, r ∈ [r1, r2].

Now let α be such that the two values inside the parentheses on the right are equal, that is

α =logM(r2)− logM(r1)

log r1 − log r2.

Then from the previous inequality, we get

logM(r) ≤ α log r1 + logM(r1)− α log r,

which upon substituting value for α gives

logM(r) ≤ (1− s) logM(r1) + s logM(r2),

where s = log r1−log rlog r2−log r1

.

I 40. If f(z) is holomorphic on D(0, 1) and f(0) = 0, show that∑∞

n=1 f(zn) converges absolutely on D(0, 1)

and converges uniformly on every compact subset of D(0, 1).

Proof. Fix r ∈ (0, 1). For any ρ ∈ (0, r) and any z ∈ D(0, ρ), we have zn ∈ D(0, ρ) (∀n ∈ N). By Cachy’sintegral formula, ∀n, p ∈ N, we have

n+p∑k=n+1

|f(zk)| =

n+p∑k=n+1

|f(zk)− f(0)| =n+p∑

k=n+1

1

2π

∣∣∣∣∣∫∂D(0,r)

[f(ζ)

ζ − zk− f(ζ)

ζ

]dζ

∣∣∣∣∣≤

n+p∑k=n+1

1

2π

∣∣∣∣∣∫∂D(0,r)

f(ζ)

ζ(ζ − zk)dζ · zk

∣∣∣∣∣ ≤n+p∑

k=n+1

1

2π

∫ 2π

0

|f(reiθ)|dθr − ρk

· ρk

≤ 1

2π

∫ 2π

0

|f(reiθ)|dθr − ρ

n+p∑k=n+1

ρk → 0

as n→ ∞. So∑∞

n=1 f(zn) converges absolutely and uniformly on D(0, ρ).

Remark 9. In general, there is no Mean Value Theorem for holomorphic functions. For example, f(z) =exp

(i 2z−(z1+z2)

z2−z1π)

is analytic but f(z2) − f(z1) 6= f ′(z)(z2 − z1), ∀z ∈ C (for more details, see Qazi [7]).But as the proof of Theorem 2.7 shows, the formula

f(z)− f(z0) =z − z02πi

∫∂U

f(ζ)dζ

(ζ − z)(ζ − z0)

more or less fills the void. In particular, it shows that holomorphic functions are locally Lipschitz.

I 41. Let f(z) be a holomorphic function on D(0, R) and f(0) = 0. If f(D(0, R)) ⊂ D(0,M), then(i) the inequalities

|f(z)| ≤ M

R|z|, |f ′(0)| ≤ M

R

hold for z ∈ ∂D(0, R)\{0}.(ii) the equalities in (1) hold if and only if

f(z) =M

Reiθz,

where θ is a real number.

29

Proof. This problem is the same as Problem 15.

I 42. Let f(z) be a holomorphic function on D(0, 1) and f(0) = 0. If there exists a constant A > 0 suchthat Ref(z) ≤ A for z ∈ D(0, 1), then

|f(z)| ≤ 2A|z|1− |z|

for z ∈ D(0, 1).

Proof. This problem is the same as Problem 19.

I 43. Let f(z) be a holomorphic function on D(0, 1) and f(0) = 1. If Ref(z) ≥ 0 for every z ∈ D(0, 1),using Schwarz Lemma, show that

(1) the inequality1− |z|1 + |z|

≤ Ref(z) ≤ |f(z)| ≤ 1 + |z|1− |z|

holds for z ∈ D(0, 1);

Proof. By Problem 18,∣∣∣ f(z)−1f(z)+1

∣∣∣ ≤ |z|. So |f(z)|−1 ≤ |f(z)−1| ≤ |z||f(z)+1| ≤ |z||f(z)|+ |z|. Since |z| < 1,the above inequality implies |f(z)| ≤ 1+|z|

1−|z| .From the same inequality

∣∣∣ f(z)−1f(z)+1

∣∣∣ ≤ |z|, we have

(Ref(z)− 1)2 + (Imf(z))2 ≤ |z|2[(Ref(z) + 1)2 + (Imf(z))2] ≤ |z|2(Ref(z) + 1)2 + (Imf(z))2.

So |Ref(z)− 1| ≤ |z|(Ref(z) + 1). If Ref(z) > 1, we can deduce from this inequality 1−|z|1+|z| ≤ 1 < Ref(z). If

Ref(z) ≤ 1, we can deduce from this inequality 1− Ref(z) ≤ |z|Ref(z) + |z|, i.e. 1−|z|1+|z| ≤ Ref(z).

(2) the equality in (1) holds for z 6= 0 if and only if

f(z) =1 + eiθz

1− eiθz,

where θ is an arbitrary real number.

Proof. We note f(z) = 1+eiθz1−eiθz

if and only if f(z)−1f(z)+1 = eiθ. So the claim is straightforward from Schwarz

Lemma.

I 44. Suppose f(z) is holomorphic on D(0, 1). Show that there exists a sequence {zn} in D(0, 1) thatconverges to a point z0 ∈ ∂D(0, 1) such that limn→∞ f(zn) exists.

Proof. Because D(0, 1) is pre-compact, by Theorem 1.11 (Bolzano-Weierstrass Theorem), it suffices to find asequence (zn)

∞n=1 ⊂ D(0, 1), such that limn→∞ zn = z0 ∈ ∂D(0, 1) and (f(zn))

∞n=1 is bounded. Assume this

does not hold. Then ∀z0 ∈ ∂D(0, 1) and ∀n ∈ N, there exists δ(z0) > 0 such that ∀z ∈ D(z0, δ(z0))∩D(0, 1),|f(z)| ≥ n. The family of these open sets (D(z, δ(z))z∈∂D(0,1) is an open covering of the compact set∂D(0, 1). By Theorem 1.10 (Heine-Borel Theorem), there is a finite sub-covering (D(zk, δ(zk))

pk=1 of ∂D(0, 1).

Consequently, for each n ∈ N, we can find εn > 0 such that ∀z ∈ {z : 1− εn ≤ |z| < 1}, |f(z)| ≥ n. Withoutloss of generality, we can assume (εn)

∞n=1 monotonically decreases to 0.

Since f(z) uniformly approaches to ∞ as z → ∂D(0, 1), by Theorem 2.13, f has finitely many zeros inD(0, 1). So f can be written as f(z) =

∏mi=1(z − zi)

ki · h(z), where h is a holomorphic function on ∂D(0, 1)with no zeros in D(0, 1). So h satisfies the Minimum Modulus Principle: ∀ρ > 0, |h(z)| ≥ min|ζ|=ρ |h(ζ)| forany z ∈ D(0, ρ). Therefore, for any z ∈ D(0, 1− εn),

|h(z)| ≥ min|ζ|=1−εn

|h(ζ)| = min|ζ|=1−εn

|f(ζ)||∏m

i=1(ζ − zi)ki |≥ n

max|ζ|=1−εn

∏mi=1 |ζ − zi|ki

≥ n

2∑m

i=1 ki.

Fix z ∈ D(0, 1− εn) and let n→ ∞, we get h(z) = ∞. Contradiction.

30

I 45. Let f(z) be a holomorphic function on D(0, 1) and f(D(0, 1)) ⊂ D(0, 1). If z1, z2, · · · , zn are all ofthe different zeros of f(z) in D(0, 1) and their multiplicities are k1, k2, · · · , kn respectively, then

|f(z)| ≤n∏

j=1

∣∣∣∣ z − zj1− zjz

∣∣∣∣kj

for all z ∈ D(0, 1). Especially,

|f(0)| ≤n∏

j=1

|zj |kj .

Proof. By assumption, we can write f(z) as f(z) =∏n

j=1(z − zj)kj · h(z), where h(z) is a holomorphic

function in D(0, 1) with no zeros in D(0, 1). Define G(z) = h(z) ·∏n

j=1(1− zjz)kj . Then G is holomorphic

on D(0, 1). Apply Maximum Modulus Principle to G(z) on {z : |z| ≤ 1− ε} (ε > 0), we get

|G(z)| ≤ max|z|=1−ε

|h(z)|n∏

j=1

|1− zjz|kj = max|z|=1−ε

|f(z)|∏nj=1

∣∣∣ z−zj1−zjz

∣∣∣kj≤ max

|z|=1−ε

1∏nj=1

∣∣∣ z−zj1−zjz

∣∣∣kj.

Since each z−zj1−zjz

maps ∂D(0, 1) to ∂D(0, 1), by letting ε→ 0, we get G(z) ≤ 1, i.e. |f(z)| ≤∏n

j=1

∣∣∣ z−zj1−zjz

∣∣∣kj

.

I 46. If f(z) is holomorphic on D(0, 1) and f(D(0, 1)) ⊂ D(0, 1), then

M |f ′(0)| ≤M2 − |f(0)|2.

Proof. Let a = f(0) and define φa(ζ) = −ζ+a1−aζ . Then h(z) = φa(f(z)) maps D(0, 1) to D(0, 1) with

h(0) = φa(a) = 0. So Schwarz Lemma implies |h′(0)| ≤ 1. We note h′(z) = φa(f(z)) · f ′(z). Since

φ′a(ζ) =

−(1− aζ)− (−ζ + a)(−a)(1− aζ)2

=−1 + |a|2

(1− aζ)2,

we have h′(0) = φ′a(a)f

′(0) = f ′(0)|a|2−1 . So we have

|f ′(0)|1− |a|2

≤ 1, i.e. |f ′(0)| ≤ 1− |a|2.

I 48. Find the group of holomorphic automorphisms of the upper half plane C+ = {z ∈ C : Imz > 0},Aut(C+).

Proof. The Möbius transformation w(z) = z−iz+i maps C+ to D = D(0, 1). So f ∈ Aut(C+) if and only if

w ◦ f ◦ w−1 ∈ Aut(D). By Theorem 2.20, ∃a ∈ D and τ ∈ R, such that w ◦ f ◦ w−1(ζ) = φa ◦ ρτ (ζ). Plaincalculation shows

f(z) = w−1 ◦ φa ◦ ρτ ◦ w(z) = (1 + a)(z + i)− (1 + a)eiτ (z − i)

(1− a)(z + i)− (a− 1)eiτ (z − i).

I 50. If f(z) is a bounded entire function and z1, z2 are arbitrary points in D(0, r), then∫|z|=r

f(z)

(z − z1)(z − z2)dz = 0,

and this implies the Liouville Theorem.

Proof. This problem is the same as Problem 10.

31

3 Theory of Series of WeierstrassThroughout this chapter, all the integration paths will take the following convention on orientation: all thearcs take counterclockwise as their orientation and all the segments take the natural orientation of R1 astheir orientation.

A very readable presentation of the power of analytic functions is Complex Proofs of Real Theorems [5].I 1. Let a1, a2, · · · be a sequence of distinct points and limn→∞ |an| = ∞. If

ψn(z) =

∞∑j=1

cn,j(z − an)j

, n = 1, 2, · · ·

are holomorphic on C except at points an (n = 1, 2, · · · ), then there exists a holomorphic function f(z) onC\{a1, a2, · · · } such that the principle parts of the Laurent expansion of f(z) at an (n = 1, 2, · · · ) are ψn(z)(n = 1, 2, · · · ) respectively.

Proof. This is just Mittag-Leffler Theorem (Theorem 3.8).

I 2. Prove Theorem 3.11 and Theorem 3.12.

Proof. To prove Theorem 3.1, choose varepsilon > 0 sufficiently small so that the circles {z : |z − zk| = ε}(1 ≤ k ≤ n) don’t intersect with each other or with ∂U . Then by Cauchy’s integration theorem,∫

∂U

f(z)dz −n∑

k=1

∫|z−zk|=ε

f(z)dz = 0,

which is∫∂U

f(z)dz = 2πi∑n

k=1 Res(f, zk).To prove Theorem 3.12, we choose R > 0 sufficiently large so that U ⊂ D(0, R). Then∫

∂U

f(z)dz = 2πi

n∑k=1

Res(f, zk) and∫∂D(0,R)

f(z)dz −∫∂U

f(z)dz = 0.

Since∫∂D(0,R)

f(z)dz = −2πiRes(f,∞), we conclude∑n

k=1 Res(f, zk) + Res(f,∞) = 0.

I 3. Find the Laurent series of the following function son the indicated regions.(i) 1

z3(z+i) , 0 < |z + i| < 1;

Solution. Suppose 1z3(z+i) has Laurent series

∑∞n=−∞ cn(z+ i)n. Then by Theorem 3.2 and Cauchy integral

theorem, for ε ∈ (0, 1),

cn =1

2πi

∫|ζ+i|=ε

1

(ζ + i)n+1· 1

ζ3(ζ + i)dζ =

1

(n+ 1)!

dn+1

dζn+1(ζ−3)

∣∣∣∣ζ=−i

=1

(n+ 1)!(−1)n+1 (n+ 3)!

2!(−i)−(n+4) = (−1)n+1 (n+ 2)(n+ 3)

2in.

(ii) z2

(z+1)(z+2) , 1 < |z| < 2;

Solution. We notez2

(z + 1)(z + 2)=

(z + 1)(z + 2)− 3z − 2

(z + 1)(z + 2)= 1− 3(z + 1)− 1

(z + 1)(z + 2)= 1 +

1

z + 1− 4

z + 2

= 1 +1

z· 1

1 + 1z

− 2 · 1

1 + z2

=

∞∑k=0

(−1)k

zk+1− 1−

∞∑k=1

(−1)k

2k−1zk.

32

(iii) log(

z−az−b

), max(|a|, |b|) < |z| <∞;

Solution. We note

log(z − a

z − b

)= log

(1− a

z

1− bz

)= log

(1− a

z

)− log

(1− b

z

)

= −∞∑

n=1

1

n

(az

)n+

∞∑n=1

1

n

(b

z

)n

=

∞∑n=1

1

n(bn − an)z−n.

(iv) z2e 1z , 0 < |z| <∞;

Solution. We note

z2e1z = z2

∞∑k=0

z−k

k!= z2 + z +

1

2+

∞∑k=1

z−k

(k + 2)!.

(v) sin z1+z , 0 < |z + 1| <∞.

Solution. We note sin z =∑∞

k=0(−1)k z2k+1

(2k+1)! . So sin zz+1 =

∑∞k=0

(−1)k

(2k+1)!z2k+1

(z+1)2k+1 . Suppose sin zz+1 has

Laurent series∑∞

n=−∞ cn(z + 1)n. Then by Theorem 3.2, for ε > 0,

cn =1

2πi

∫|ζ+1|=ε

1

(ζ + 1)n+1

∞∑k=0

(−1)k

(2k + 1)!

ζ2k+1

(ζ + 1)2k+1dζ

=

∞∑k=0

(−1)k

(2k + 1)!· 1

2πi

∫|ζ+1|=ε

ζ2k+1

(ζ + 1)2k+n+2dζ

=∑

k≥−n−12

(−1)k

(2k + 1)!

1

(2k + n+ 1)!

d2k+n+1

dζ2k+n+1

(ζ2k+1

)∣∣ζ=−1

.

Since

d2k+n+1

dζ2k+n+1

(ζ2k+1

)∣∣ζ=−1

=

0 n ≥ 1,

(2k + 1)! n = 0,(2k+1)!(−n)! (−1)−n n ≤ −1,

we have

cn =

0 n ≥ 1,∑

k≥0(−1)k

(2k+1)! n = 0,(−1)−n

(−n)!

∑k≥−n−1

2

(−1)k

(2k+n+1)! n ≤ −1

=

0 n ≥ 1,

sin 1 n = 0,1

(−n)! cos 1 n ≤ −1, n is odd,1

(−n)! sin 1 n ≤ −1, n is even.

Thereforesin z

z + 1=

∞∑k=0

[sin 1

(2k)!(z + 1)−2k +

cos 1(2k + 1)!

(z + 1)−(2k+1)

].

33

I 4. Find and classify the singularities of the following functions. Find the order if the singularity is a pole.(i) sin z

z ;

Solution. z = 0 is a removable singularity.

(ii) 1z2−1 cos πz

z+1 ;

Solution. z = 1 is a pole of order 1. The Laurent series of the function in a neighborhood of z = −1 can beobtained by

1

z2 − 1cos πz

z + 1=

1

(z − 1)(z + 1)

∞∑n=0

(−1)n

(2n)!

(πz

z + 1

)2n

.

By discussion (3) of page 91, z = −1 is an essential singularity.

(iii) z(e 1z − 1);

Solution. Since e 1z =

∑∞n=0 z

−n, we have z(e 1z −1) =

∑∞n=1 z

−n+1. Therefore z = 0 is an essential singularity.

(iv) sin 11−z ;

Solution. z = 1 is an essential singularity.

(v) exp{ 11−z}

ez−1 ;

Solution. z = 1 is an essential singularity. z = 0 is a pole of order 1.

(vi) tan z.

Solution. z = kπ + π2 (k ∈ Z) are poles of order 1.

I 5. Prove:(1) If a is an essential singularity of f(z) and f(z) 6= 0, then a is also an essential singularity of 1/f(z).

Proof. The limit limz→a f(z) exists (finite or ∞) if and only if limz→a1

f(z) exists. By discussion (3) on page93, a is an essential singularity of f(z) if and only if a is an essential singularity of 1

f(z) .

(2) If a is an essential singularity of f(z) and P (ζ) is a non-constant polynomial, then a is also an essentialsingularity of P (f(z)).

Proof. Clearly, a is an isolated singularity of P (f(z)). We note by Fundamental Theorem of Algebra, P (ζ)maps C to C. Consequently, P (ζ) maps a dense subset of C to a dense subset of C. By WeierstrassTheorem (Theorem 3.3), f(z) maps a neighborhood of a to a dense subset of C. Therefore, P (f(z)) maps aneighborhood of a to a dense subset of C. So it is impossible for a to become a removable singularity or apole of P (f(ζ)). Hence a must be also an essential singularity of P (f(z)).

I 6. Prove:(i)∏∞

n=0(1 + z2n

) = 11−z .

Proof. We note any positive integer can be uniquely represented in the form∑∞

k=0 ak · 2k, where eachak ∈ {0, 1} and only finitely many ak’s are non-zero. In the expansion of

∏∞n=0(1+ z

2n), each representationz∑∞

k=0 ak·2k appears once and only once. Therefore, after a rearrangement of the terms, we must have∞∏

n=0

(1 + z2n

) = 1 + z + z2 + z3 + z4 + · · · = 1

1− z.

34

(ii) sinhπz = πz∏∞

n=1

(1 + z2

n2

);

Proof. sinhπz = 12 [e

πz − e−πz] has zeros an = ni (n ∈ Z), where a0 = 0 is a zero of order 1. Since for any

R > 0, we have∑∞

n=−∞,n=0

(R

|an|

)2=∑∞

n=−∞,n=0R2

n2 <∞, by Weierstrass Factorization Theorem, we canfind an entire function h(z) such that

sinhπz = zeh(z)n=∞∏

n=−∞,n=0

{(1− z

ni

)e

zni

}= zeh(z)

∞∏n=1

(1 +

z2

n2

).

Since limz→0sinh πz

z = π, we conclude eh(0) = π. Meanwhile we have

π cothπz =(sinhπz)′sinhπz =

1

sinhπz

[zeh(z)

∞∏n=1

(1 +

z2

n2

)]′

=1

z+ h′(z) +

∞∑n=1

2zn2

1 + z2

n2

=1

z+ h′(z) +

∞∑n=1

2z

n2 + z2.

Let γn be the rectangular path [n+ (n+ 12 )i,−n+ (n+ 1

2 )i,−n− (n+ 12 )i, n− (n+ 1

2 )i, n+ (n+ 12 )i]. Then

for any given a, when n is large enough, we have∫γn

cothπzz2 − a2

dz = I + II + III + IV,

where

I =

∫ −n

n

e2π[x+(n+12)i]+1

e2π[x+(n+12)i]−1[

x+ (n+ 12 )i]2 − a2

dx =

∫ −n

n

−e2πx+1−e2πx−1[

x+ (n+ 12 )i]2 − a2

dx,

II =

∫ −(n+ 12 )

n+ 12

e2π(−n+yi)+1e2π(−n+yi)−1

(−n+ yi)2 − a2idy,

III =

∫ n

−n

e2π[x−(n+12)i]+1

e2π[x−(n+12)i]−1

[x− (n+ 12 )i]

2 − a2dx =

∫ n

−n

−e2πx+1−e2πx−1

[x− (n+ 12 )i]

2 − a2dx

and

IV =

∫ n+ 12

−(n+ 12 )

e2π(n+yi)+1e2π(n+yi)−1

(n+ iy)2 − a2idy.

We note ex−1ex+1 ∈ (−1, 1) for any x ∈ R. So we have

|I| ≤∫ n

−n

1

x2 + (n+ 12 )

2 − a2dx =

2√(n+ 1

2 )2 − a2

arctan n√(n+ 1

2 )2 − a2

→ 0,

as n → ∞. Similarly we can conclude limn→∞ III = 0. We also note for x ∈ R large enough, ex+1ex−1 < 2. So

for n large enough,

|II| ≤∫ n+ 1

2

−(n+ 12 )

e1+e−2πn

1−e−2πn

n2 + y2 − a2dy ≤ 4√

n2 − a2arctan

n+ 12√

n2 − a2→ 0,

as n→ ∞. Similarly we can conclude limn→∞ IV = 0. Combined, we have shown

limn→∞

∫γn

cothπzz2 − a2

dz = 0.

35

Define f(z) = coth πzz2−a2 . By Residue Theorem, for a 6∈ iZ, we have∫

γn

f(z)dz = 2πiRes(f ; a) + 2πiRes(f ;−a) + 2πi

n∑k=−n

Res(f ; an).

It’s easy to see Res(f ; a) = Res(f ;−a) = coth πa2a . Since

limz→an

(z − an)f(z) = limz→an

e2πz + 1

z2 − a2z − ane2πz − 1

=2

2π(a2n − a2)lim

z→an

2π(z − an)

e2πz − e2πan= − 1

π(n2 + a2),

we must have Res(f ; an) = − 1π(n2+a2) . Therefore

0 =1

2πilimn→∞

∫γn

f(z)dz = limn→∞

(cothπa

a−

n∑k=−n

1

π(a2 + k2)

)=

cothπaa

−∞∑

k=−∞

1

π(a2 + k2)

This implies for a 6∈ iZ,

π cothπa =1

a+

∞∑k=1

2a

a2 + n2.

Plugging this back into the formula π cothπz = 1z + h′(z) +

∑∞n=1

2zn2+z2 , we conclude h′(z) = 0 for z 6∈ iZ.

By Theorem 2.13, h′(z) ≡ 0 for any z ∈ C. So

sinhπz = zeh(0)∞∏

n=1

(1 +

z2

n2

)= πz

∞∏n=1

(1 +

z2

n2

).

Remark 10. The above solution is a variant of the proof for factorization formula of sinπz. See Conway[2]VII, §6 for more details.

(iii) cosπz =∏∞

n=0

(1−

(z

n+ 12

)2);

Proof. From problem (v), we have π2

∏∞n=1

(1− 1

4n2

)= 1 and

cosπz = sinπ(1

2− z

)= π

(1

2− z

) ∞∏n=1

(1−

(12 − z

)2n2

)

= π

(1

2− z

) ∞∏n=1

{(1 +

12 − z

n

)(1−

12 − z

n

)}

= π

(1

2− z

) ∞∏n=1

{(1 +

1

2n

)(1− z

n+ 12

)(1− 1

2n

)(1 +

z

n− 12

)}

= limN→∞

π

2

(1− z

12

) N∏n=1

(1− 1

4n2

)·

N∏n=1

(1− z

n+ 12

)·

N∏n=1

(1 +

z

n− 12

)

= limN→∞

{π

2

N∏n=1

(1− 1

4n2

)}·(1− z

N + 12

)·

N∏n=1

[1− z2

(n− 12 )

2

]

=

∞∏n=0

[1−

(z

n+ 12

)2].

36

(iv) ez − 1 = zez2

∏∞n=1

(1 + z

4π2n2

);

Proof. The function eπz − 1 has zeros an = 2ni (n ∈ Z), where a0 = 0 is a zero of order 1. Since for anyR > 0, we have

∑∞n=−∞,n=0

(R

|an|

)2=∑∞

n=−∞,n=0R2

4n2 <∞, by Weierstrass Factorization Theorem, we canfind an entire function h(z) such that

eπz − 1 = zeh(z)∞∏

n=−∞,n=0

{(1− z

2ni

)e

z2ni

}= zeh(z)

∞∏n=1

(1 +

z2

4n2

).

To determine h(z), we note

πeπz

eπz − 1=

(eπz − 1)′

eπz − 1=

1

z+ h′(z) +

∞∑n=1

2z4n2

1 + z2

4n2

=1

z+ h′(z) +

∞∑n=1

2z

4n2 + z2.

Let γn be the rectangular path [2n+(2n+1)i,−2n+(2n+1)i,−2n− (2n+1)i, 2n− (2n+1)i, 2n+(2n+1)i].Define f(z) = πeπz

(z2−a2)(eπz−1) . Then for any given a, when n is large enough, we have∫γn

f(z)dz = I + II + III + IV,

whereI =

∫ −2n

2n

f(x+ (2n+ 1)i)dx =

∫ −2n

2n

πeeπx

[(x+ (2n+ 1)i)2 − a2](eπx + 1)dx,

II =

∫ −(2n+1)

2n+1

f(−2n+ yi)idy =

∫ −(2n+1)

2n+1

πeπ(−2n+yi)

[(−2n+ yi)2 − a2](eπ(−2n+yi) − 1)idy,

III =

∫ 2n

−2n

f(x− (2n+ 1)i)dx =

∫ 2n

−2n

πeeπx

[(x− (2n+ 1)i)2 − a2](eπx + 1)dx,

andIV =

∫ 2n+1

−(2n+1)

f(2n+ yi)idy =

∫ 2n+1

−(2n+1)

πeπ(2n+yi)

[(2n+ yi)2 − a2](eπ(2n+yi) − 1)idy.

It is easy to see

|I| ≤∫ 2n

−2n

πdx

x2 + (2n+ 1)2 − a2≤ 2π√

(2n+ 1)2 − a2arctan 2n√

(2n+ 1)2 − a2→ 0

as n→ ∞. Similarly, we can show limn→∞ III = 0. Also, we note

|II| ≤∫ 2n+1

−(2n+1)

πdy

(4n2 + y2 − a2)(e2nπ − 1)≤ 2π

(e2nπ − 1)√4n2 − a2

arctan 2n+ 1√4n2 − a2

→ 0,

as n→ ∞. For n sufficiently large, e2nπ

e2nπ−1 ≤ 2, so

|IV | ≤∫ 2n+1

−(2n+1)

πe2πndy

(4n2 + y2 − a2)(e2πn − 1)≤ 4π√

4n2 − a2arctan 2n+ 1√

4n2 − a2→ 0,

as n→ ∞. Combined, we have shownlimn→∞

∫γn

f(z)dz = 0.

By Residue Theorem, for a 6∈ 2Zi, we have∫γn

f(z)dz = 2πiRes(f ; a) + 2πiRes(f ;−a) + 2πi

n∑k=−n

Res(f ; an).

37

It’s easy to see Res(f ; a) = πeπa

2a(eπa−1) and Res(f ;−a) = π2a(eπa−1) . Since

limz→an

(z − an)f(z) = limz→an

πeπz

(z2 − a2)· z − aneπz − 1

= − 1

4n2 + a2,

Res(f ; an) = − 14n2+a2 . Therefore

0 =1

2πilimn→∞

∫γn

f(z)dz = limn→∞

(π

2a

eπa + 1

eπa − 1−

n∑k=−n

1

4n2 + a2

).

So π2a

eπa+1eπa−1 = 1

a2 +∑∞

n=12

4n2+a2 for a 6∈ 2Zi. Therefore for z 6∈ 2Zi, we have

h′(z) =πeπz

eπz − 1− 1

z−

∞∑n=1

2z

4n2 + z2=

πeπz

eπz − 1− π

2

eπz + 1

eπz − 1=π

2.

So h(z) = πz2 + h(0). It’s easy to see limz→0

eπz−1z = π. So eh(0) = π. Combined, we conclude eh(z) = πe

πz2 .

So

eπz − 1 = πzeπz2

∞∏n=1

(1 +

z2

4n2

).

Equivalently, we have

ez − 1 = zez2

∞∏n=1

(1 +

z2

4π2n2

).

(v) sinπz = πz∏

n=0

(1− z

n

)e

zn , hence sinπz =

∏∞n=1

(1− z2

n2

).

Proof. Using the formula sinπz = −i sinh(iz) and the factorization formula for sinhπz, we have

sinπz = −iπ(iz)∞∏

n=1

(1− z2

n2

)= πz

∞∏n=1

(1− z2

n2

).

I 7. (Blaschke Product) Suppose a sequence of complex numbers {ak} satisfies 0 < |ak| < 1 (k = 1, 2, · · · )and

∑∞k=1(1− |ak|) <∞. Show that the infinite product

f(z) =

∞∏k=1

ak − z

1− akz· |an|an

converges uniformly on the disc {z : |z| ≤ r} (0 < r < 1), and f(z) is holomorphic in |z| < 1 with |f(z)| ≤ 1.Also show that the zeros of f(z) are ak (k = 1, 2, · · · ) and there is no other zeros.

Proof. Fix r ∈ (0, 1). Since∑∞

k=1(1− |ak|) <∞, limk→∞ |ak| = 1. So there exists k0 ∈ N, such that for anyk ≥ k0, 1+r

2 < |ak| < 1. So for any k ≥ k0,∣∣∣∣ ak − z

1− akz· |ak|ak

− 1

∣∣∣∣ ≤ |ak|(1− |ak|) + |z||ak|(1− |ak|)|1− akz||ak|

≤ (1 + r)(1− |ak|)1+r2 (1− r)

=2

1− r(1− |ak|).

Since∑∞

k=1(1 − |ak|) < ∞, we conclude∑∞

k=1

∣∣∣ ak−z1−akz

· |ak|ak

− 1∣∣∣ is absolutely and uniformly convergent on

{z : |z| ≤ r}. Therefore∏∞

k=1ak−z1−akz

· |ak|ak

is absolutely and uniformly convergent on {z : |z| ≤ r}. ByWeierstrass Theorem (Theorem 3.1), f(z) =

∏∞k=1

ak−z1−akz

· |ak|ak

represents a non-zero holomorphic function on{z : |z| < 1}. Since the mapping z 7→ ak−z

1−akz(0 < |ak| < 1) maps D(0, 1) to D(0, 1), we conclude |f(z)| ≤ 1.

By the theorem quoted at the beginning of the solution, it’s clear that (ak)∞k=1 are the only zeros of f(z).

38

Remark 11. For a clear presentation of convergence of infinite products, we refer to Conway [2], ChapterVII, §5. In particular, we quote the following theorem (Conway [2], Chapter VII, §5, Theorem 5.9)

Let G be an open subset of the complex plane. Denote by H(G) the collection of analytic functions onG, equipped with the topology determined by uniform convergence on compact subsets of G. Then H(G) is acomplete metric space. Furthermore, we have the following theorem (proof omitted).

Theorem 3.1. Let G be a region in C and let (fn)∞n=1 be a sequence in H(G) such that no fn is identicallyzero. If

∑[fn(z)−1] converges absolutely and uniformly on compact subsets of G, then

∏∞n=1 fn(z) converges

in H(G) to an analytic function f(z). If a is a zero of f then a is a zero of only a finite number of thefunctions fn, and the multiplicity of the zero of f at a is the sum of the multiplicities of the zeros of thefunctions fn at a.

I 8. (Poisson-Jensen Formula) Suppose f(z) is a meromorphic function on the disc {z : |z| ≤ R} (0 < R <∞), and is not equal to zero identically, the points a1, a2, · · · , as and b1, b2, · · · , bt are zeros and poles off(z) on |z| < R respectively. Suppose that if a is a zero with multiplicity n, then a appears n times in a1,a2, · · · , as. Same assumption for the poles. Then we have

log |f(z)| = 1

2π

∫ 2π

0

log |f(Reiφ)|ReReiφ + 2

Reiφ − zdφ+

t∑j=1

log∣∣∣∣ R2 − bjz

R(z − bj)

∣∣∣∣− s∑i=1

log∣∣∣∣ R2 − aiz

R(z − ai)

∣∣∣∣for any points in the disc |z| < 1 which is different from ai (i = 1, 2, · · · , s) and bj (j = 1, 2, · · · , t).

This formula is the starting point of Nevanlinna Value Distribution theory.

Proof. Let Rε = R− ε, ε ∈ (0, R). Define

Fε(z) = f(z) ·

∏tj=1

Rε(z−bj)

R2ε−bjz∏s

i=1Rε(z−ai)R2

ε−aiz

, ∀z ∈ D(0, Rε).

It’s easy to verify that each of R2ε(z−bj)

R2ε−bjz

and R2ε(z−ai)R2

ε−aizmaps D(0, Rε) onto itself and takes the boundary to

the boundary. Therefore Fε(z) is analytic on D(0, Rε), has no zeros in D(0, Rε), and |Fε(z)| = |f(z)| for|z| = Rε. Since log |Fε(z)| is a harmonic function, by formula (2.33) on page 71,

log |Fε(z)| =1

2π

∫ 2π

0

log |Fε(Rεeiφ)|ReRεe

iφ + z

Rεeiφ − zdφ =

1

2π

∫ 2π

0

log |f(Rεeiφ)|ReRεe

iφ + z

Rεeiφ − zdφ.

Plugging the formula of Fε into the above equality, we get

log |f(z)| = 1

2π

∫ 2π

0

log |f(Rεeiφ)|ReRεe

iφ + z

Rεeiφ − zdφ+

t∑j=1

log∣∣∣∣ R2

ε − bjz

Rε(z − bj)

∣∣∣∣− s∑i=1

log∣∣∣∣ R2

ε − aiz

Rε(z − ai)

∣∣∣∣ .Letting ε→ 0 yields the desired formula.

I 9. Suppose a meromorphic function f(z) only has two poles on the extended complex plane C∗. The pointz = −1 is a pole of multiplicity 1 and with principle part 1/(z + 1); the point z = 2 is a pole of multiplicity2 and with principle part 2/(z − 2) + 3/(z − 2)2, and f(0) = 7/4. Find the Laurent expansion of f(z) in1 < |z| < 2.

Solution. By Theorem 3.4 (correction: “holomorphic function f(z)” in the theorem’s statement should be“meromorphic function f(z)”), f(z) has the form of 1

z+1 +2

z−2 +3

(z−2)2 + c, where c is a constant. f(0) = 74

implies c = 1. Since for any ζ ∈ D(0, 1),

1

(1− ζ)2=

(1

1− ζ

)′

=

( ∞∑n=0

ζn

)′

=

∞∑n=0

(n+ 1)ζn,

39

the Laurent expansion of f(z) in 1 < |z| < 2 is

f(z) =1

z + 1+

2

z − 2+

3

(z − 2)2+ 1

=1

z

1

1 + 1z

− 1

1− z2

+3

4· 1

(1− z2 )

2+ 1

=1

z

∞∑n=0

(−1

z

)n

−∞∑

n=0

(z2

)n+

3

4

∞∑n=0

(n+ 1)(z2

)n+ 1

= −∞∑

n=1

(−1)n

zn+

3

4+

∞∑n=1

3n− 1

2n+2zn.

I 10. Suppose a meromorphic function f(z) has poles at z = 1, 2, 3, · · · with multiplicity 2, and the principleparts of Laurent expansions in the neighborhood of z = n are ψn(z) = n/(z − n)2 (n = 1, 2, · · · ). Find ageneral form of f(z).

Solution. We follow the construction outlined in the proof of Mittag-Leffler Theorem (Theorem 3.9). Foreach n ∈ N, when |z| < n

2 , ψn(z) has Taylor expansion

ψn(z) =n

(z − n)2=

1

n· 1(

1− zn

)2 =1

n

∞∑k=0

(k + 1)( zn

)k.

Let λn be a positive integer to be determined later. We estimate the tail error obtained by retaining onlythe first λn terms of the Taylor expansion of ψn(z):∣∣∣∣∣ψn(z)−

1

n

λn∑k=0

(k + 1)( zn

)k∣∣∣∣∣ ≤ 1

n

∞∑k=λn+1

(k + 1)

2k=

1

n

∞∑k=λn

(k + 2)

2k+1=

1

n

( ∞∑k=λn

k

2k+1+

1

2λn−1

)

≤ 1

n · 2λn−1+

1

n

∞∑k=λn

∫ k+1

k

x

2xdx =

1

n · 2λn−1+

1

n

∫ ∞

λn

x

log 12

d

(1

2

)x

=1

n · 2λn−1+

1

n · 2λn

[λn

log 2 +1

(log 2)2

].

If we let λn = n, then εn := 1n·2λn−1 + 1

n·2λn

[λn

log 2 + 1(log 2)2

]= 1

n·2n−1 + 1n·2n

[n

log 2 + 1(log 2)2

]satisfies∑∞

n=1 εn <∞. By the proof of Mittag-Leffler Theorem, we can write f(z) as

f(z) = U(z) +

∞∑n=1

[n

(z − n)2− 1

n

n∑k=0

(k + 1)( zn

)k],

where U(z) is an entire function.

I 11.(i) Expand f(z) = 1/(ez − 1) to a partial fraction.

Solution. Define f(z) = 1(z2−a2)(eπz−1) , where a ∈ C \ 2iZ. Let γn be the rectangular path [2n + (2n +

1)i,−2n+ (2n+ 1)i,−2n− (2n+ 1)i, 2n− (2n+ 1)i, 2n+ (2n+ 1)i]. Then for any given a, when n is largeenough, we have ∫

γn

f(z)dz = I + II + III + IV,

40

whereI =

∫ −2n

2n

f(x+ (2n+ 1)i)dx =

∫ 2n

−2n

dx

[(x+ (2n+ 1)i)2 − a2](eπx + 1),

II =

∫ −(2n+1)

2n+1

f(2n+ yi)idy =

∫ −(2n+1)

2n+1

idy

[(−2n+ yi)2 − a2](eπ(−2n+yi − 1),

III =

∫ 2n

−2n

f(x− (2n+ 1)i)dx =

∫ 2n

−2n

−dx[(x− (2n+ 1)i)2 − a2](eπx + 1)

,

andIV =

∫ 2n+1

−(2n+1)

f(2n+ iy)idy =

∫ 2n+1

−(2n+1)

idy

[(2n+ iy)2 − a2](eπ(2n+iy) − 1).

It’s easy to see

|I| ≤∫ 2n

−2n

dx

x2 + (2n+ 1)2 − a2≤ 2√

(2n+ 1)2 − a2arctan 2n√

(2n+ 1)2 − a2→ 0

as n→ ∞. Similarly, we can show limn→∞ III = 0. Also, we note

|II| ≤∫ 2n+1

−(2n+1)

dy

(4n2 + y2 − a2)(1− e−2nπ)=

2

(1− e−2nπ)√4n2 − a2

arctan 2n+ 1√4n2 − a2

→ 0

as n→ ∞, and

|IV | ≤∫ 2n+1

−(2n+1)

dy

(4n2 + y2 − a2)(e2πn − 1)=

2

(e2πn − 1)√4n2 − a2

arctan 2n+ 1

4n2 − a2→ 0

as n→ ∞. Combined, we have shown limn→∞∫γnf(z)dz = 0. By Residue Theorem, for a 6∈ 2iZ, we have

∫γn

f(z)dz = 2πiRes(f ; a) + 2πiRes(f ;−a) + 2πi

n∑k=−n

Res(f ; 2ni).

It’s easy to see Res(f ; a) = 12a(eπa−1) and Res(f ;−a) = 1

2a(1−e−πa) . Since

limz→2ni

(z − 2ni)f(z) = limz→2ni

π(z − 2ni)

π(z2 − a2)(eπz − 1)= − 1

π(4n2 + a2),

we have Res(f ; 2ni) = − 1π(4n2+a2) . Therefore

0 =1

2πilimn→∞

∫γn

f(z)dz = limn→∞

(1

2a· e

πa + 1

eπa − 1− 1

π

n∑k=−n

1

4k2 + a2

)=

1

2a· e

πa + 1

eπa − 1− 1

π

∞∑k=−∞

1

4k2 + a2.

Replacing πa with z, we get after simplification

1

ez − 1=

1

z− 1

2+

∞∑k=1

2z

4k2π2 + z2, ∀z ∈ C \ 2Zπi.

This is the partial fraction of 1/(ez − 1).

(ii) Show that1

sin2 πz=

1

π2

∞∑n=−∞

1

(z − n)2;

41

Proof. Define f(z) = 1(z−a) sin2(πz) , where a ∈ C\Z. Let γn be the rectangular path [

(n+ 1

2

)+ni,−

(n+ 1

2

)+

ni,−(n+ 1

2

)− ni,

(n+ 1

2

)− ni,

(n+ 1

2

)+ ni]. Then when n is large enough, we have∫

γn

f(z)dz = I + II + III + IV,

where

I =

∫ −(n+ 12 )

n+ 12

dx

(x+ ni− a) sin2[π(x+ ni)]=

∫ (n+ 12 )

−(n+ 12 )

4dx

(x+ ni− a)(e−2nπ+2ixπ + e2nπ−2ixπ − 2),

II =

∫ −n

n

idy

[−(n+ 12 ) + yi− a] sin2[−π(n+ 1

2 ) + πyi]=

∫ n

−n

−4idy

[−(n+ 12 )− a+ yi](e2πy + e−2πy + 2)

,

III =

∫ (n+ 12 )

−(n+ 12 )

dx

(x− ni− a) sin2[π(x− ni)]=

∫ (n+ 12 )

−(n+ 12 )

−4dx

(x− a− ni)(e2nπ+2iπx + e−2nπ−2iπx − 2),

and

IV =

∫ n

−n

idy

[(n+ 12 ) + yi− a] sin2[π(n+ 1

2 ) + πyi]=

∫ n

−n

4idy

[(n+ 12 )− a+ yi](e−2yπ + e2yπ + 2)

.

It’s easy to see

|I| ≤∫ (n+ 1

2 )

−(n+ 12 )

4dx√(x− a)2 + n2(e2nπ − e−2nπ − 2)

≤ 4(2n+ 1)

n(e2nπ − e−2nπ − 2)→ 0

as n→ ∞,

|II| ≤∫ n

−n

4dy√(n+ 1

2 + a)2 + y2(e2πy + e−2πy + 2)=

∫ n

0

8dy√(n+ 1

2 + a)2 + y2(e2πy + e−2πy + 2)

≤∫ n

0

8dy

(n+ 12 + a)e2πy

=8

(n+ 12 + a)

1− e−2πn

2π→ 0

as n→ ∞,

|III| ≤∫ (n+ 1

2 )

−(n+ 12 )

4dx√(x− a)2 + n2(e2nπ − e−2nπ − 2)

≤ 4(2n+ 1)

n(e2nπ − e−2nπ − 2)→ 0

as n→ ∞, and lastly

|IV | ≤∫ n

−n

4dy√(n+ 1

2 − a)2 + y2(e2yπ + e−2yπ + 2)=

∫ n

0

8dy√(n+ 1

2 − a)2 + y2(e2yπ + e−2yπ + 2)

≤∫ n

0

8

(n+ 12 − a)e2yπ

=8

(n+ 12 − a)

1− e−2πn

2π→ 0

as n→ ∞. Combined, we have shown limn→∞∫γnf(z)dz = 0. By Residue Theorem, we have∫

γn

f(z)dz = 2πiRes(f ; a) + 2πi

n∑k=−n

Res(f ; k).

It’s easy to see Res(f ; a) = 1sin2(πa) . Define hk(z) = (z−k)2

sin2(πz) . Since limz→k hk(z) =1π2 , hk is holomorphic in

a neighborhood of k and for ε > 0 small enough, and we have

Res(f ; k) =1

2πi

∫|z−k|=ε

dz

(z − a) sin2(πz)=

1

2πi

∫|z−k|=ε

hk(z)

(z − a)· dz

(z − k)2=

d

dz

[hk(z)

z − a

]∣∣∣∣z=k

.

42

We note

d

dz

[hk(z)

z − a

]∣∣∣∣z=k

= limz→k

h′k(z)(z − a)− h(z)

(z − a)2= lim

z→k

h′k(z)

z − a− 1

(k − a)2π2,

and

limz→k

h′k(z)

z − a= lim

z→k

2(z − k) sin(πz)− 2π(z − k)2 cos(πz)sin3(πz)

= limz→k

2 sin(πz) + 2π(z − k) cos(πz)− 4π(z − k) cos(πz) + 2π2(z − k)2 sin(πz)3 sin2(πz) · π cos(πz)

= limz→k

2 sin(πz)− 2π(z − k) cos(πz)3π sin2(πz)

= limz→k

2π cos(πz)− 2π cos(πz) + 2π2(z − k) sin(πz)6π2 sin(πz) cos(πz)

= 0.

So for a ∈ C \ Z, ∫γn

f(z)dz =2πi

sin2(πa)−

n∑k=−n

2πi

(k − a)2π2.

Let n→ ∞, we getπ2

sin2(πz)=

∞∑n=−∞

1

(z − n)2.

Remark 12. Another choice is to let f(z) = cot(πz)(z+a)2 , see Conway [2], page 122, Exercise 6. The trick used

in this problem and problem 6 (ii) will be generalized in problem 12.

(iii) Show that if α 6= 0 and (β/α) 6= ±1,±2, · · · , then

π

αcot πβ

α=

∞∑n=0

{1

nα+ β− 1

nα+ (α− β)

}.

From the above equation show that

1

1 · 2+

1

4 · 5+

1

7 · 8+ · · ·+ 1

(3n− 2)(3n− 1)+ · · · = π

3√3.

Proof. By the solution of problem 6 (ii), we have

coth z = 1

z+

∞∑n=1

2z

z2 + π2n2.

Since coth z = i cot(iz), we have

cot z = i coth(iz) = 1

z−

∞∑n=1

2z

π2n2 − z2=

1

z−

∞∑n=1

(1

πn− z− 1

πn+ z

).

43

Therefore

π

αcot πβ

α

=1

β−

∞∑n=1

(1

αn− β− 1

αn+ β

)= lim

N→∞

[1

β−

N∑n=1

(1

αn− β− 1

αn+ β

)]

= limN→∞

[1

β−(

1

α− β+

1

2α− β+ · · ·+ 1

Nα− β

)+

(1

α+ β+

1

2α+ β+ · · ·+ 1

Nα+ β

)]= lim

N→∞

[(1

β− 1

α− β

)+

(1

α+ β− 1

2α− β

)+ · · ·+

(1

(N − 1)α+ β− 1

Nα− β

)+

1

Nα+ β

]=

∞∑n=0

[1

nα+ β− 1

nα+ (α− β)

]=

∞∑n=0

α− 2β

(nα+ β)[nα+ (α− β)].

By letting α = 3 and β = 1, we get

π

3√3=π

3cot π

3=

∞∑n=1

1

(3n− 2)(3n− 1).

Remark 13. For a direct proof without using problem 6 (ii), we can let f(z) = cot(πz)z2−a2 and apply Residue

Theorem to it. For details, see Conway [2], page 122, Exercise 8. The line of reasoning used in this approachwill be generalized in problem 12 below.

I 12. Suppose a meromorphic function f(z) only has finitely many poles α1, α2, · · · , αm and αk is not aninteger (1 ≤ k ≤ m). the infinity z = ∞ is a zero of f(z) with multiplicity p ≥ 2. Show that

(1) limn→∞∑n

k=−n(−1)kf(k) = −π∑m

k=1 Res(f(z) cotπz, αk);

Proof. Since z = ∞ is a zero of f(z) with multiplicity p, f(z) can be written as h(z)zp where h is holomorphic

in a neighborhood of ∞. Consequently, h is bounded in a neighborhood of ∞. In the below, we shall workwith a sequence of neighborhoods of ∞ that shrinks to ∞. So without loss of generality, we can assume his bounded and denote its bounded by M .

Let γn be the rectangular path [(n + 12 ) + ni,−(n + 1

2 ) + ni,−(n + 12 ) − ni, (n + 1

2 ) − ni, (n + 12 ) + ni].

Then ∫γn

f(z) cot(πz)dz = I + II + III + IV,

where

I =

∫ −(n+ 12 )

(n+ 12 )

h(x+ ni)

(x+ ni)p

eiπ(x+ni)+e−iπ(x+ni)

2eiπ(x+ni)−e−iπ(x+ni)

2i

dx =

∫ −(n+ 12 )

(n+ 12 )

h(x+ ni)

(x+ ni)peiπx−nπ + e−iπx+nπ

eiπx−nπ − e−iπx+nπidx,

II =

∫ −n

n

h(−(n+ 12 ) + yi)

[−(n+ 12 ) + yi]p

eiπ[−(n+12)+yi]+e−iπ[−(n+1

2)+yi]

2

eiπ[−(n+12)+yi]−e−iπ[−(n+1

2)+yi]

2i

idy

=

∫ −n

n

h(−(n+ 12 ) + yi)

[−(n+ 12 ) + yi]p

(−1)ne−yπ(−i) + (−1)neyπi

(−1)ne−yπ(−i)− (−1)neyπi(−1)dy,

III =

∫ (n+ 12 )

−(n+ 12 )

h(x− ni)

(x− ni)peiπx+nπ + e−iπx−nπ

eiπx+nπ − e−iπx−nπidx,

44

and

IV =

∫ n

−n

h((n+ 12 ) + yi)

[(n+ 12 ) + yi]p

(−1)ne−yπi+ (−1)neyπ(−i)(−1)ne−yπi− (−1)neyπ(−i)

(−1)dy.

We note

|I| ≤∫ (n+ 1

2 )

−(n+ 12 )

M

(x2 + n2)p2

e−nπ + enπ

enπ − e−nπdx =

e2nπ + 1

e2nπ − 1

2M

narctan

n+ 12

n→ 0

as n→ ∞,|II| ≤

∫ n

−n

M[(n+ 1

2 )2 + y2

] p2

∣∣∣∣eyπ − e−yπ

eyπ + e−yπ

∣∣∣∣ dy ≤ 2M

n+ 12

arctan n

n+ 12

→ 0

as n→ ∞,

|III| ≤∫ (n+ 1

2 )

−(n+ 12 )

M

(x2 + n2)p2

enπ + e−nπ

enπ − e−nπdx =

enπ + e−nπ

enπ − e−nπ

2M

narctan

n+ 12

n→ 0

as n→ ∞, and

|IV | ≤∫ n

−n

M[(n+ 1

2 )2 + y2

] p2

∣∣∣∣eyπ − e−yπ

eyπ + e−yπ

∣∣∣∣ dy ≤ 2M

n+ 12

arctan n

n+ 12

→ 0

as n → ∞. Combined, we can conclude limn→∞∫γnf(z) cot(πz)dz = 0. Meanwhile, for n sufficiently large,

by Residue Theorem, we have∫γn

f(z) cot(πz)dz = 2πi

m∑k=1

Res(f(z) cot(πz), αk) + 2πi

n∑k=−n

Res(f(z) cot(πz), k).

We note

Res(f(z) cot(πz), k) =1

2πi

∫|z−k|=ε

f(z)cos(πz)sin(πz) dz

=1

2πi

∫|z−k|=ε

(−1)kf(z)

z − k· cos(πz)(z − k)

sin[π(z − k)]dz

=f(k)

π.

So ∫γn

f(z) cot(πz)dz = 2πi

m∑k=1

Res(f(z) cot(πz), αk) + 2πi

n∑k=−n

f(k)

π.

Letting n→ ∞ gives us

limn→∞

n∑k=−n

f(k) = −πn∑

k=1

Res(f(z) cot(πz), αk).

(2) limn→∞∑n

k=−n(−1)kf(k) = −π∑m

k=1 Res(

f(z)sin πz , αk

);

Proof. As argued in the solution of (1), we can assume f(z) = h(z)zp where h is holomorphic in a neighborhood

of ∞ and has bound M in that neighborhood. Let γn denote the same rectangular path as in our solutionof (1). Then ∫

γn

f(z)

sin(πz)dz = I + II + III + IV,

45

where

I =

∫ −(n+ 12 )

n+ 12

h(x+ ni)

(x+ ni)p2idx

eiπx−nπ − e−iπx+nπ,

II =

∫ −n

n

h(−(n+ 12 ) + yi)

[−(n+ 12 ) + yi]p

2i

e−yπ(−1)n(−i)− eyπ(−1)niidy,

III =

∫ n+ 12

−(n+ 12 )

h(x− ni)

(x− ni)p2idx

eiπx+nπ − e−iπx−nπ,

andIV =

∫ n

−n

h((n+ 12 ) + yi)[

(n+ 12 ) + yi

]p 2i

e−πy(−1)ni− eπy(−1)n(−i)idy.

We note

|I| ≤ 2M

enπ − e−nπ

∫ n+ 12

−(n+ 12 )

dx

(x2 + n2)p2

≤ 4M

enπ − e−nπ

∫ n+ 12

0

dx

x2 + n2=

2M

(enπ − e−nπ)narctan

n+ 12

n→ 0

as n→ ∞, and

|II| ≤∫ n

−n

1[(n+ 1

2 )2 + y2

] p2

2Mdy

eyπ + e−yπ≤∫ n

−n

Mdy

(n+ 12 )

2 + y2=

2M

n+ 12

arctan n

n+ 12

→ 0

as n → ∞. By similar argument, we can prove limn→∞ III = limn→∞ IV = 0. Combined, we can concludethat limn→∞

∫γn

f(z)sin(πz)dz = 0. Meanwhile, for n sufficiently large, by Residue Theorem, we have

∫γn

f(z)

sin(πz)dz = 2πi

m∑k=1

Res(

f(z)

sin(πz) , αk

)+ 2πi

n∑k=−n

Res(

f(z)

sin(πz) , k).

It’s easy to see

Res(

f(z)

sin(πz) , k)

=1

2πi

∫|z−k|=ε

(−1)k

z − k· f(z)(z − k)

sin[π(z − k)]dz =

(−1)kf(k)

π.

Therefore, ∫γn

f(z)

sin(πz)dz = 2πi

m∑k=1

Res(

f(z)

sin(πz) , αk

)+ 2πi

n∑k=−n

(−1)kf(k)

π.

Let n→ ∞, we get limn→∞∑n

k=−n(−1)kf(k) = −π∑m

k=1 Res(

f(z)sin(πz) , αk

).

(3) Find the sums of the following series by using (1) and (2).(i)∑∞

n=−∞1

(a+n)2 , where a is not an integer.

Solution. Let f(z) = 1(z+a)2 . Then by result of (1),

∞∑n=−∞

1

(n+ a)2= −πRes

(cot(πz)(z + a)2

,−a)

= −π d

dz[cot(πz)]|z=−a =

π2

sin2(πa).

This result is verified by problem 11 (ii).

(ii)∑∞

n=0(−1)n

n2+a2 , where a is a non-zero real number.

46

Solution. Let f(z) = 1z2+a2 . Then by result of (2),

∞∑n=−∞

(−1)n

n2 + a2= −π

[Res

(1

(z2 + a2) sin(πz) , ai)+ Res

(1

(z2 + a2) sin(πz) ,−ai)]

= − π

ai · sin(πai) =πcsch(πa)

a.

This result can be verified by Mathematica.

I 13. Find the residues of the following functions at their isolated singularities (including the infinity pointif it is an isolated singularity).

(i) 1z2−z4 ;

Solution. f(z) = 1z2−z4 has poles 0, 1, and −1. ∞ is a removable singularity.

Res(f ; 0) = 1

2πi

∫|z|=ρ

dz

z2(1− z2)=

d

dz

(1

1− z2

)|z=0 =

−(−2z)

(1− z2)2|z=0 = 0,

Res(f ; 1) = 1

2πi

∫|z−1|=ρ

dz

(1− z)(1 + z)z2= − 1

(1 + z)z2|z=1 = −1

2,

andRes(f ;−1) =

1

2πi

∫|z+1|=ρ

dz

(1 + z)(1− z)z2=

1

2.

(ii) z2+z+2z(z2+1)2 ;

Solution. We note

f(z) :=z2 + z + 2

z(z2 + 1)2=

2

z+

−1 + i

4(z − i)2+

−4− i

4(z − i)+

−1− i

4(z + i)2+

−4 + i

4(z + i).

Therefore, the function f(z) has poles 0, i and −i. ∞ is a removable singularity of f(z). Furthermore, wehave

Res(f ; 0) = 2, Res(f ; i) = −4− i

4= −1− i

4, Res(f ;−i) = −4 + i

4= −1 +

i

4.

(iii) zn−1

zn+an , where a 6= 0 and n is a positive integer;

Solution. Suppose a1, · · · , an are the roots of the equation zn + an = 0. Then each ai is a pole of order 1for the function f(z) = zn−1

zn+an . ∞ is a removable singularity of f(z). Then

Res(f ; ai) =an−1i∏n

j=1,j =i(ai − aj)= lim

z→ai

an−1i (z − ai)∏nj=1(z − aj)

= limz→ai

an−1i (z − ai)

zn + an=

1

n.

(iv) 1sin z ;

Solution. f(z) = 1sin z has πZ as poles of order 1. ∞ is not an isolated singularity of f(z). And

Res(f, nπ) = 1

2πi

∫|z−nπ|=ε

dz

sin z =1

2πi

∫|z−nπ|=ε

1

z − nπ

(−1)n(z − nπ)

sin(z − nπ)dz = (−1)n.

47

(v) z3 cos 1z−2 ;

Solution. First, we note as cos(

1z−2

)= 1

2

(e

iz−2 + e−

iz−2

)=∑∞

n=0(−1)n

(z−2)2n . So f(z) := z3 cos(

1z−2

)=∑∞

n=0(−1)nz3

(z−2)2n . This shows z = 2 is an essential singularity of f(z), and

Res(f ; 2) =∞∑

n=0

(−1)n

2πi

∫|z−2|=ε

z3

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

3!· 6|z=2 = −11.

Also, ∞ is an isolated singularity of f(z), and we have

Res(f ;∞) = − 1

2πi

∫|z|=R

z3 cos(

1

z − 2

)dz = − 1

2πi

∫|z−2|=R

z3 cos(

1

z − 2

)dz

= − 1

2πi

∫|ζ|= 1

R

(1

ζ+ 2

)3

cos ζ(−dζζ2

)=

1

2πi

∫|ζ=ε|

(ζ + 2)3

ζ5cos ζdζ

=1

4!

d4

dζ4[(ζ + 2)3 cos ζ

]|ζ=0 = −8

3.

(vi) ez

z(z+1) .

Solution. Let f(z) = ez

z(z+1) = ez(

1z − 1

z+1

). Then Res(f ; 0) = 1, Res(f ;−1) = −e−1, and Res(f ;∞) =

1− e−1.

I 14. Suppose that f(z) and g(z) are holomorphic at z = a, f(a) 6= 0 and z = a is a zero of g(z) withmultiplicity 2. Find Res(f(z)/g(z), a).

Solution. We can write g(z) as (z − a)2h(z) where h is holomorphic and h(a) 6= 0. Then h(a) = g(a)(z−a)2 and

h′(a) =g′(a)(z − a)− 2g(a)

(z − a)3.

Therefore

Res(f(z)

g(z), a

)= Res

(1

(z − a)2· f(z)h(z)

, a

)=

d

dz

[f(z)

h(z)

]∣∣∣∣z=a

=f ′(a)h(a)− f(a)h′(a)

h2(a)

=f ′(a) g(a)

(z−a)2 − f(a) g′(a)(z−a)−2g(a)

(z−a)3

g2(a)(z−a)4

=g(a)(z − a)[f ′(a)(z − a) + 2f(a)]− f(a)g′(a)(z − a)2

g2(a).

I 15. Evaluate the following integrals:

Remark 14. We make an observation of some simple rules that facilitate the evaluation of integrals viaResidue Theorem.

Rule 1 (rule for integrand function). The poles of the integrand function should be easy to find, such thatthe integrand function can be easily represented as f(z)

(z−a)n , where f is holomorphic.The holomorphic function f(z) can be multi-valued function like logarithm function and power function.

When it’s difficult to make f holomorphic in the desired region, check if it is the real or imaginary part of aholomorphic function.

48

Rule 2 (rule for integration path). The choice of integration path is highly dependent on the propertiesof the integrand function, where symmetry and multi-valued functions (log and power functions) are oftenhelpful.

Oftentimes, the upper and lower limits of integration either form a full circle (i.e. [0, 2π]) so thatsubstitution for trigonometric functions can be easily done or have ∞ as one of the end points.

(i)∫∞0

x2dx(x2+1)2 ;

Solution. Let γ1 = {z : −R ≤ Rez ≤ R, Imz = 0}, γ2 = {z : |z| = R, 0 ≤ arg z ≤ π}, and f(z) = z2

(z2+1)2 .Then ∫

γ1

f(z)dz +

∫γ2

f(z)dz = 2πi · Res(f, i) = 2πi · limz→i

d

dz

[z2

(z + i)2

]=π

2,

and ∣∣∣∣∫γ2

f(z)dz

∣∣∣∣ = ∣∣∣∣∫ π

0

R2e2θi

(R2e2θi + 1)2Reiθ · idθ

∣∣∣∣ ≤ ∫ π

0

R3

(R2 − 1)2dθ → 0

as R→ ∞. So by letting R→ ∞, we have∫∞0

x2dx(x2+1)2 = 1

2

∫∞−∞

x2dx(x2+1)2 = π

4 .

(ii)∫ π

2

0dx

a+sin2 x , a > 0;

Solution. We note sin2 x = 1−cos 2x2 , so∫ π

2

0

dx

a+ sin2 x=

1

2

∫ π

0

dx

a+ sin2 x=

1

2

∫ 2π

0

dθ

(2a+ 1)− cos θ .

Let b = 2a+ 1, then∫ π2

0

dx

a+ sin2 x=

1

2

∫|z|=1

dziz

b− 12 (z + z−1)

=1

2i

∫|z|=1

dz

− 12z

2 + bz − 12

.

The equation − 12z

2 + bz − 12 = 0 has two roots: z1 = b −

√b2 − 1 and z2 = b +

√b2 − 1. Since b > 1, we

have z1 ∈ D(0, 1) and z2 6∈ D(0, 1). Therefore,∫ π2

0

dx

a+ sin2 x= i

∫|z|=1

dz

(z − z1)(z − z2)= i · 2πi · 1

z1 − z2=

π√b2 − 1

=π

2√a(a+ 1)

.

Remark 15. If R(x, y) is a rational function of two variables x and y, for z = eiθ, we have

R(sin θ, cos θ) = R

(1

2

(z +

1

z

),1

2i

(z − 1

z

)), dθ =

dz

iz.

Therefore ∫ 2π

0

R(sin θ, cos θ)dθ =∫|z|=1

R

(1

2(z +

1

z),

1

2i(z − 1

z)

)dz

iz.

Remark 16. When the integrand function is a rational function of trigonometric functions, in view ofthe previous remark, it is desirable to have [0, 2π] as the integration interval. For this reason, we first usesymmetry to expand the integration interval from [0, π2 ] to [0, π], and then use double angle formula to expand[0, π] to [0, 2π]. This solution is motivated by Rule 2.

(iii)∫∞0

x sin xx2+1 dx;

49

Solution. Let γ1 = {z : −R ≤ Rez ≤ R, Imz = 0}, γ2 = {z : |z| = R, 0 ≤ arg z ≤ π}, and f(z) = zeiz

z2+1 . Then∫γ1

f(z)dz +

∫γ2

f(z)dz = 2πiRes(f, i) =∫|z−i|=ε

zeizdz

(z − i)(z + i)= 2πi · ie

−1

2i=π

ei,

and ∣∣∣∣∫γ2

f(z)dz

∣∣∣∣ =∣∣∣∣∣∫ π

0

ReiReiθ

R2e2θi + 1Reiθ · idθ

∣∣∣∣∣ ≤∫ π

0

R2e−R sin θ

R2 − 1dθ → 0

as R→ ∞ by Lebesgue’s Dominated Convergence Theorem. Therefore by letting R→ ∞, we have∫ ∞

−∞

xeix

x2 + 1dx =

∫ ∞

−∞

x cosxx2 + 1

dx+ i

∫ ∞

−∞

x sinxx2 + 1

dx =π

ei,

which implies∫∞0

x sin xx2+1 dx = π

2e .

(iv)∫∞0

log x(x2+1)2 dx;

Solution. Let r,R be two positive numbers such that r < 1 < R. Let γ1 = {z : r ≤ |z| ≤ R, arg z = 0},γ2 = {z : r ≤ |z| ≤ R, arg z = π}, γR = {z : |z| = R, 0 ≤ arg z ≤ π}, and γr = {z : |z| = r, 0 ≤ arg z ≤ π}.Define f(z) = log z

(1+z2)2 where log z is defined on C \ [0,∞) and takes the principle branch of Logz witharg z ∈ (0, 2π) (see page 23).

By Residue Theorem,∫γ1+γR+γ2−γr

f(z)dz = 2πiRes(f, i) = 2πi limz→i

d

dz

[log z

(z + i)2

]= −π

2+π2

4i.

We note∣∣∣∣∫γR

f(z)dz

∣∣∣∣ = ∣∣∣∣∫ π

0

log(Reiθ)(R2e2iθ + 1)2

Reiθ · idθ∣∣∣∣ ≤ ∫ π

0

R√

(logR)2 + θ2

(R2 − 1)2dθ ≤ πR(logR+ π)

(R2 − 1)2→ 0

as R→ ∞,∣∣∣∣∫γr

f(z)dz

∣∣∣∣ = ∣∣∣∣∫ π

0

log(reiθ)(r2e2iθ + 1)2

reiθ · idθ∣∣∣∣ ≤ ∫ π

0

r√

(log r)2 + θ2

(1− r2)2dθ ≤ πr(− log r + π)

(1− r2)2→ 0

as r → 0, and ∫γ2

f(z)dz =

∫ −r

−R

logx(1 + x2)2

dx =

∫ R

r

log(−x)(1 + x2)2

dx =

∫ R

r

logx+ πi

(1 + x2)2dx.

Therefore by letting r → 0 and R→ ∞, we have∫ ∞

0

2 logx+ πi

(1 + x2)2dx = −π

2+π2

4i,

i.e.∫∞0

log x(1+x2)2 dx = −π

4 .

(v)∫∞0

x1−α

1+x2 dx, 0 < α < 2;

Solution. Let γ1, γ2, γr and γR be defined as in (iv). Define f(z) = z1−α

1+z2 , where z1−α = e(1−α) log z is definedon C \ [0,∞) with log z taking the principle branch of Logz. By Residue Theorem,∫

γ1+γR+γ2−γr

f(z)dz = 2πiRes(f, i) = 2πi · e(1−α) log z

z + i|z=i = πe(1−α)π

2 i.

50

We note ∣∣∣∣∫γR

f(z)dz

∣∣∣∣ = ∣∣∣∣∫ π

0

R1−αeiθ(1−α)

1 +R2e2iθ·Reiθ · idθ

∣∣∣∣ ≤ R2−α

R2 − 1π → 0

as R→ ∞, ∣∣∣∣∫γr

f(z)dz

∣∣∣∣ = ∣∣∣∣∫ π

0

r1−αeiθ(1−α)

1 + r2e2iθ· reiθ · idθ

∣∣∣∣ ≤ r2−α

1− r2π → 0

as r → 0, and ∫γ2

f(z)dz =

∫ −r

−R

x1−α

1 + x2dx =

∫ R

r

e(1−α) log(−x)

1 + x2dx =

∫ R

r

x1−αei(1−α)π

1 + x2dx.

Therefore by letting r → 0 and R→ ∞, we have∫ ∞

0

x1−α(1 + ei(1−α)π)

1 + x2dx =

∫ ∞

0

x1−α

1 + x2dx · 2 cos (1− α)π

2ei(1−α)π

2 = πei(1−α)π2 ,

i.e.∫∞0

x1−α

1+x2 dx = π2 csc

(απ2

).

(vi)∫∞0

dx1+xn , where n is an integer and n ≥ 2;

Solution. This problem is a special of problem (x). See the solution there.

(vii)∫ π

0dθ

a+cos θ , where a is a constant and a > 1;

Solution. We use the method outlined in the remark of Problem (ii).∫ π

0

dθ

a+ cos θ =1

2

∫ π

−π

dθ

a+ cos θ =1

2

∫|z|=1

dziz

a+ z+z−1

2

=1

2i

∫|z|=1

dz12z

2 + az + 12

.

The equation 12z

2 + az + 12 = 0 has two roots: z1 = −a +

√a2 − 1 and z2 = −a −

√a2 − 1. Clearly,

|z1| = 1a+

√a2−1

< 1 and |z2| > 1. So∫ π

0

dθ

a+ cos θ =1

i

∫|z|=1

dz

(z − z1)(z − z2)= 2π · 1

z1 − z2=

π√a2 − 1

.

Remark 17. The expansion of integration interval from [0, π] to [−π, π] is motivated by Rule 2.

(viii)∫∞0

( sin xx

)2dx;

Solution. Using the result on Dirichlet integral:∫∞0

sin xx dx = π

2 , we have∫ ∞

0

(sinxx

)2

dx =

∫ ∞

0

sin2 xd

(− 1

x

)=

∫ ∞

0

2 sinx cosxdxx

=

∫ ∞

0

sin(2x)2x

d(2x) =π

2.

(ix)∫∞0

xp

x2+2x cos λ+1dx, where −1 < p < 1 and −π < λ < π;

Solution. It is easy to see that when λ = p = 0, the integral is equal to 1. So without loss of generality, weonly consider the cases where λ and p are not simultaneously equal to 0.

Choose r,R so that 0 < r < R, and r is sufficiently small and R is sufficiently large. Define γ1 = {z :r ≤ |z| ≤ R, arg z = 0}, γ2 = {z : r ≤ |z| ≤ R, arg z = 2π}, γr = {z : |z| = r, 0 < arg z < 2π}, and γR = {z :

|z| = R, 0 < arg z < 2π}. Let ρ = cosλ, then ρ ∈ (−1, 1]. Finally, define f(z) = zp

z2+2z cos λ+1 = ep log z

z2+2ρz+1 .

51

For R sufficiently large and r sufficiently small, the two roots of z2 + 2rz + 1 = 0 are contained in{z : r < |z| < R}. These two roots are, respectively, z1 = −ρ + i

√1− r2 = − cosλ + i| sinλ| and z2 =

−ρ− i√1− r2 = − cosλ− i| sinλ|. So |z1| = |z2| = 1, and arg z1+ arg z2 = 2π. Denote arg z1 by θ. If λ 6= 0,

we have by Residue Theorem∫γ1+γR−γ2−γr

f(z)dz = 2πi[Res(f, z1) + Res(f, z2)] = 2πi

[ep log z1

z1 − z2+ep log z2

z2 − z1

]=

π

| sinλ| [epθi − ep(2π−θ)i] = − 2πi

sinλ sin(λp)epπi.

If λ = 0, we have by Residue Theorem∫γ1+γR−γ2−γr

f(z)dz = 2πiRes(f,−1) = −2pπi · epπi.

Meanwhile, we have the estimates∣∣∣∣∫γR

f(z)dz

∣∣∣∣ ≤ Rp

R2 − 2R− 1·R · 2π → 0

as R→ ∞, ∣∣∣∣∫γr

f(z)dz

∣∣∣∣ ≤ rp

1− 2ρr − r2· r · 2π → 0

as r → 0, and ∫γ2

f(z)dz = −∫ R

0

(xe2πi)pdx

x2 + 2ρx+ 1= −

∫ R

0

xpdx

x2 + 2ρx+ 1e2pπi.

Since 1− e2pπi = 2 sin2(pπ)− 2 sin(pπ) cos(pπ)i = −2i sin(pπ)epπi, by letting R→ ∞ and r → 0, we have∫ ∞

0

xpdx

x2 + 2x cosλ+ 1=

{π sin(λp)

sin λ sin(pπ) if λ 6= 0

pπ csc (pπ) if λ = 0 and p 6= 0.

If we take the convention that αsin α = 1 for α = 0, then the above three formulas can be unified into a

single one: π sin(λp)sinλ sin(pπ) .

Remark 18. We choose the above integration path in order to take advantage of the multi-valued functionxp (Rule 1). The no symmetry in x2 + 2x cosλ+ 1 is handled by using the full circle instead of half circle.

(x)∫∞0

11+xp dx, where p > 1;

Solution. Choose two positive numbers r and R such that 0 < r < 1 < R. Let γ1 = {z : r ≤ |z| ≤ R, arg z =0}, γ2 = {z : r ≤ |z| ≤ R, arg z = 2π}, γR = {z : |z| = R, 0 < arg z < 2π} and γr = {z : |z| = r, 0 < arg z <2π}. Define f(z) = z1/p

p(z+1)z where z1/p = elog zp is defined on C \ [0,∞). Note by substituting y

1p for x, we

get ∫ ∞

0

dx

1 + xp=

∫ ∞

0

y1p dy

p(y + 1)y.

By Residue Theorem,∫γ1+γR−γ2−γr

f(z)dz = 2πRes(f,−1) = 2πi · (−1)1p

−p= −2πi

pe

log eπi

p = −2iαeαi,

where α = πp . We have the estimates∣∣∣∣∫

γR

f(z)dz

∣∣∣∣ =∣∣∣∣∣∫ 2π

0

e1p log(Reiθ)

p(Reiθ + 1)ReiθReiθ · idθ

∣∣∣∣∣ ≤ 2πR1p

p(R− 1)→ 0

52

as R→ ∞, ∣∣∣∣∫γr

f(z)dz

∣∣∣∣ =∣∣∣∣∣∫ 2π

0

e1p log(reiθ)

p(reiθ + 1)reiθreiθ · idθ

∣∣∣∣∣ ≤ 2πr1p

p(1− r)→ 0

as r → 0, and ∫γ2

f(z)dz = −∫ R

r

(xe2πi)1p dx

p(x+ 1)x= −

∫ R

r

x1p dx

p(x+ 1)x· e2αi.

Therefore by letting r → 0 and R→ ∞, we have∫ ∞

0

dx

1 + xp=

∫ ∞

0

x1pdx

p(x+ 1)x=

−2iαeαi

1− e2αi=

−2iαeαi

−2 sinαe(π2 +α)i

=α

sinα =π

pcsc(π

p

).

Remark 19. The roots of 1 + xp = 0 are not very expressible. So following Rule 1, we make the change ofvariable xp = y. Then we choose the above integration path to take advantage of the multi-valued functiony

1p . Since there is no symmetry in the denominator (y + 1)y, we used a full circle instead of a half circle.

The change of variable xp = y is equivalent to integration on an arc of angle 2π/p instead of 2π.

(xi)∫ 1

0x1−p(1−x)p

1+x2 dx, where −1 < p < 2;

Solution. (Oops! looks like my solution has a bug. Catch it if you can.) By the change of variable x = 1y+1 ,

we have ∫ 1

0

x1−p(1− x)p

1 + x2dx =

∫ ∞

0

ypdy

(y + 1)3 + (y + 1).

The equation (y + 1)3 + (y + 1) = 0 has three roots: z0 = −1, z1 = −1 + i, and z2 = −1 − i. Definef(z) = zp

(z+1)3+(z+1) . Then

2∑i=0

Res(f, zi) =1

(z + 1)(z − z1)

∣∣∣∣z=z2

+1

(z + 1)(z − z2)

∣∣∣∣z=z1

+1

(z − z1)(z − z2)

∣∣∣∣z=z0

=1

−i · (−2i)+

1

i · 2i+

1

−i · i= 0.

Let R and r be two positive numbers with R > r. Define γ1 = {z : r < |z| ≤ R, arg z = 0}, γ2 = {z : r <|z| ≤ R, arg z = 2π}, γR = {z : |z| = R, 0 < arg z < 2π}, and γr = {z : |z| = r, 0 < arg z < 2π}. Then byResidue Theorem ∫

γ1+γR−γ2−γr

f(z)dz = 2πi

2∑i=0

Res(f, zi) = 0.

We note ∣∣∣∣∫ΓR

f(z)dz

∣∣∣∣ = ∣∣∣∣∫ 2π

0

(Reiθ)p

(Reiθ + 1)3 + (Reiθ + 1)Reiθ · idθ

∣∣∣∣ ≤ 2πRp+1

R3 − 3R2 − 4R− 2→ 0

as R→ ∞, and ∣∣∣∣∫ 2π

0

(reiθ)p

(reiθ + 1)3 + (reiθ + 1)reiθ · idθ

∣∣∣∣ ≤ 2πrp+1

2− 4r − 3r2 − r3→ 0

as r → 0. So by letting r → 0 and R→ ∞, we have∫ ∞

0

yp

(y + 1)3 + (y + 1)dy + 0−

∫ ∞

0

yp

(y + 1)3 + (y + 1)dy · e2pπi − 0 = 0.

Therefore..., “Houston, we got a problem here.”

53

(xii)∫∞0

log xx2+2x+2dx;

Solution. Choose r, R such that 0 < r < R and r is sufficiently small and R is sufficiently large. Letγ1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ2 = {z : r ≤ |z| ≤ R, arg z = π}, γR = {z : |z| = R, 0 < arg z < π} andγr = {z : |z| = r, 0 < arg z < π}. Define f(z) = log z

z2+2z+2 . By Residue Theorem,∫γ1+γR+γ2−γr

f(z)dz = 2πi · Res(f,−1 + i) = π

(log 22

+3

4πi

).

We note ∣∣∣∣∫γR

f(z)dz

∣∣∣∣ = ∣∣∣∣∫ 2π

0

log(Reiθ)R2e2iθ + 2Reiθ + 2

Reiθ · idθ∣∣∣∣ ≤ logR+ 2π

R2 − 2R− 22πR→ 0

as R→ ∞, ∣∣∣∣∫γr

f(z)dz

∣∣∣∣ = ∣∣∣∣∫ 2π

0

log(reiθ)r2e2iθ + 2reiθ + 2

reiθ · idθ∣∣∣∣ ≤ log r + 2π

2− 2r − r22πr → 0

as r → 0, and ∫γ2

f(z)dz =

∫ R

r

log(−x)x2 − 2x+ 2

dx =

∫ R

r

logx+ πi

x2 − 2x+ 2dx.

Therefore by letting r → 0 and R→ ∞, we have∫ ∞

0

[logx

x2 − 2x+ 2+

logxx2 + 2x+ 2

]dx+ i

∫ ∞

0

πdx

x2 − 2x+ 2=π

2log 2 + 3

4π2i,

i.e.∫∞0

[log x

x2−2x+2 + log xx2+2x+2

]dx = π

2 log 2.

Meanwhile, we note∫∞0

[log x

x2−2x+2 − log xx2+2x+2

]dx =

∫∞0

logx 4x(x2+2)2−4x2 dx =

∫∞0

log yy2+4dy. To calculate

the value of∫∞0

log yy2+4dy, we apply again Residue Theorem. Let’s define g(z) = log z

z2+4 and then we have∫γ1+γR+γ2−γr

g(z)dz = 2πi · Res(g, 2i) = π

2

[log 2 + π

2i].

We note ∣∣∣∣∫γR

g(z)dz

∣∣∣∣ = ∣∣∣∣∫ 2π

0

log(Reiθ)R2e2iθ + 4

Reiθ · idθ∣∣∣∣ ≤ logR+ 2π

R2 − 42πR→ 0

as R→ ∞, ∣∣∣∣∫γr

g(z)dz

∣∣∣∣ = ∣∣∣∣∫ 2π

0

log(reiθ)r2e2iθ + 4

reiθ · idθ∣∣∣∣ ≤ log r + 2π

4− r22πr → 0

as r → 0, and ∫γ2

g(z)dz =

∫ R

r

log(−x)x2 + 4

dx =

∫ R

r

logx+ πi

x2 + 4dx.

So by letting r → 0 and R→ ∞, we have

2

∫ ∞

0

logxx2 + 4

dx+ πi

∫ ∞

0

dx

x2 + 4=π

2log 2 + π2

4i.

Hence∫∞0

log xx2+4dx = π

4 log 2. Solving the equation{∫∞0

log xx2−2x+2dx+

∫∞0

log xx2+2x+2dx = π

2 log 2∫∞0

log xx2−2x+2dx−

∫∞0

log xx2+2x+2dx = π

4 log 2,

we get ∫ ∞

0

logxx2 + 2x+ 2

dx =π

8log 2.

54

Remark 20. Note how we handled the combined difficulty caused by no symmetry in the denominator ofthe integrand function and log function: because x2 + 2x + 2 is not symmetric, we want to use full circle;but this will cause the integrals produced by different integration paths to cancel with each other. Thereforewe are forced to choose half circle and use two equations to solve for the desired integral.

(xiii)∫∞0

√x log xx2+1 dx

Solution. We choose r and R such that 0 < r < R, and r is sufficiently small and R is sufficiently large. Letγ1 = {z : r ≤ |z| ≤ R, arg z = 0}, γ2 = {z : r ≤ |z| ≤ R, arg z = π}, γR = {z : |z| = R, 0 < arg z < π}, andγr = {z : |z| = r, 0 < arg z < π}. Define f(z) =

√z log zz2+1 . Then∫

γ1+γR+γ2−γr

f(z)dz = 2πiRes(f, i) = π2

2ie

π4 i =

π2

2√2(−1 + i).

We note ∣∣∣∣∫γR

f(z)dz

∣∣∣∣ =∣∣∣∣∣∫ 2π

0

(Reiθ)12 log(Reiθ)

R2e2iθ + 1Reiθ · idθ

∣∣∣∣∣ ≤ 2πR√R(logR+ 2π)

R2 − 1→ 0

as R→ ∞, ∣∣∣∣∫γr

f(z)dz

∣∣∣∣ =∣∣∣∣∣∫ 2π

0

(reiθ)12 log(reiθ)

r2e2iθ + 1reiθ · idθ

∣∣∣∣∣ ≤ 2πr√r(log r + 2π)

1− r2→ 0

as r → 0, and∫γ2

f(z)dz =

∫ R

r

√−x log(−x)x2 + 1

dx =

∫ R

r

√xe

π2 i(logx+ πi)

x2 + 1dx = e

π2 i

∫ R

r

√x logxx2 + 1

dx+ eπ2 iπi

∫ R

r

√x

x2 + 1dx.

By letting r → 0 and R→ ∞, we have

(1 + i)

∫ ∞

0

√x logxx2 + 1

dx− π

∫ ∞

0

√x

x2 + 1dx =

π2

2√2(−1 + i).

Therefore,∫∞0

√x log xx2+1 dx = π2

2√2.

(xiv)∫∞0

log(

ex+1ex−1

)dx;

Solution. We note the power series expansion ln(1− z) = −∑∞

n=1zn

n and ln(1+ z) =∑∞

n=1(−1)n+1 zn

n holdin |z| < 1 and the convergence is uniform on |z| ≤ 1− ε, for any ε ∈ (0, 1). Therefore, for any δ > 0, we have∫ ∞

δ

log(ex + 1

ex − 1

)dx =

∫ ∞

δ

[ ∞∑n=1

(−1)n+1 e−nx

n+

∞∑n=1

e−nx

n

]dx

= 2

∫ ∞

δ

[ ∞∑k=1

e−(2k−1)x

2k − 1

]dx = 2

∞∑k=1

e−(2k−1)δ

(2k − 1)2.

Therefore ∫ ∞

0

log(ex + 1

ex − 1

)dx = lim

δ→02

∞∑k=1

e−(2k−1)δ

(2k − 1)2= 2

∞∑k=1

1

(2k − 1)2,

where the last equality can be seen as an example of Lebesgue’s Dominated Convergence Theorem appliedto counting measure. By Problem 11 (ii) with z = 1

2 , we know∑∞

k=11

(2k−1)2 = π2

8 . So the integral is equalto π2

4 .

(xv)∫∞0

xex+1dx;

55

Solution. Since we have argued rather rigorously in (xiv), we will argue non-rigorously in this problem.∫ ∞

0

x

ex + 1dx =

∫ ∞

0

e−xx

1 + e−xdx =

∫ ∞

0

e−xx

∞∑n=0

(−1)ne−nxdx

=

∞∑n=0

(−1)n∫ ∞

0

e−(n+1)xxdx =

∞∑n=0

(−1)n

(n+ 1)2.

Using the function ππ2 sin πz and imitating what we did in Problem 12, we can easily get

∑∞n=1

(−1)n−1

n2 = π2

12 .Therefore

∫∞0

xex+1dx = π2

12 .

(xvi)∫ π

2

0log(sin θ)dθ;

Solution. We provide two solutions, which are essentially the same one.Solution 1. In problem (xvii), we shall prove (let a = 1)∫ π

0

x sinx2− 2 cosxdx = π log 2.

Note ∫ π

0

x sinx2− 2 cosxdx =

∫ π

0

x · 2 sin x2 cos x

2dx

2 · 2 sin2 x2

= 2

∫ π2

0

θ cos θdθsin θ = −2

∫ π2

0

log(sin θ)dθ.

So∫ π

2

0log(sin θ)dθ = −π

2 log.

Solution 2. We provide a heuristic proof which discloses the essence of the calculation. Note how Rule1-3 lead us to this solution.∫ π

2

0

log(sin θ)dθ =

∫ π2

0

log(cosα)dα =1

2

∫ π2

−π2

log(cosα)dα

=1

4

∫ π2

−π2

log(1 + cos 2α

2

)dα

=1

8

∫ π

−π

log(1 + cosβ)dβ − π

4log 2.

(Note how Rule 2 leads us to extended the integration interval from [0, π2 ] to [−π, π].) Motivated by thesimilar difficulty explained in the remark of Problem (xvii), we try Rule 1 and note∫ π

−π

log(1 + eiβ)dβ =

∫|z|=1

log(1 + z)

izdz = 2π log 1 = 0.

For the above application of Cauchy’s integral formula to be rigorous, we need to take a branch of logarithmfunction that is defined on C \ (−∞, 0] and take a small bypass around 0 for the integration contour, andthen take limit.

Using the above result, we have

0 = Re∫ π

−π

log(1 + eiβ)dβ =1

2

∫ π

−π

log[(1 + cosβ)2 + sin2 β

]dβ =

1

2

∫ π

−π

[log 2 + log(1 + cosβ)] dβ.

Therefore∫ π

−πlog(1 + cosβ)dβ = −2π log 2 and∫ π

2

0

log(sin θ)dθ = 1

8

∫ π

−π

log(1 + cosβ)dβ − π

4log 2 = −π

2log 2.

56

(xvii)∫ π

0x sin x

1−2a cos x+a2 dx, where a > 0;

Solution. We first assume a > 1. Then by integration-by-parts formula we have∫ π

0

x sinx1− 2a cosx+ a2

dx =1

2

∫ π

−π

x sinxdx1− 2a cosx+ a2

=1

2

∫ π

−π

xd(1− 2a cosx+ a2)

(1− 2a cosx+ a2) · 2a

=1

4a

[2π log(1 + a)2 −

∫ π

−π

log(1− 2a cosx+ a2)dx

].

Since a > 1, a − ζ ∈ {z : Rez > 0} for any ζ ∈ ∂D(0, 1). Therefore we can take a branch of the logarithmfunction such that log(a−z) is a holomorphic function on D(0, 1) and is continuous on D(0, 1). For example,we can take the branch log z such that it is defined on C \ (−∞, 0] with log(eiπ) = π and log(e−iπ) = −π.By Cauchy’s integral formula,∫ π

−π

log(a− eiθ)dθ =

∫|z|=1

log(a− z)

izdz = 2π log a.

Therefore,∫ π

−πlog(1 − 2a cosx + a2)dx = 2Re

∫ π

−πlog(a − eiθ)dθ = 4π log a. Plug this equality into the

calculation of the original integral, we get πa log

(1+aa

).

If 0 < a < 1, we have∫ π

0

x sinx1− 2a cosx+ a2

dx =1

a2

∫ π

0

x sinx1− 2

a cosx+ 1a2

dx =1

a2· π1

a

log(1 + 1

a1a

)=π

alog(a+ 1).

Assuming the integral as a function of a is continuous at 1, we can conclude for a = 1, the integral isequal to π log 2. The result agrees with that of Mathematica. To prove the continuity rigorously, we splitthe integral into

∫ π2

0x sin xdx

1−2a cos x+a2 and∫ π

π2

x sin xdx1−2a cos x+a2 . For the second integral, we have (x ∈ [π2 , π])

x sinx1− 2a cosx+ a2

=x sinx|a− eix|

≤ x sinx|a− ei

π2 |

≤ x sinx.

So by Lebesgue’s dominated convergence theorem, the second integral is a continuous function of a fora ∈ (0,∞). For the first integral, we note (1− 2a cosx+ a2) takes its minimum at cosx. So

x sinx1− 2a cosx+ a2

≤ x

sinx.

Again by Lebesgue’s dominated convergence theorem, the first integral is a continuous function of a fora ∈ (0,∞). Combined, we conclude the integral∫ π

0

x sinx1− 2a cosx+ a2

dx

as a function of a ∈ (0,∞) is a continuous function.

Remark 21. We did not take the transform cosx = z+z−1

2 because log(1− 2a cosx+ a2) would then becomelog[w(a) − w(z)] − log(2a), where w(z) = z+z−1

2 maps D(0, 1) to U := C \ [−1, 1] and we cannot find aholomorphic branch of logarithm function on U .

The application of integration-by-parts formula and the extension of integration interval from [0, π] to[−π, π] are motivated by Rule 2. But here a new trick emerged: if the integrand function or part of itcould come from the real or imaginary part of a holomorphic function, apply Cauchy’s integral formula (orintegration theorem) to that holomorphic function and try to relate the result to our original integral (Rule1).

57

(xviii)∫∞0

log(1+x2)1+x2 dx

Solution. The function log(z + i) is well-defined on C+ := {z : Imz ≥ 0}. We take the branch of log(z + i)that evaluates to log 2+ π

2 i at i. Define γR = {z : |z| = R, 0 ≤ arg z ≤ π}. Then for R > 1, Residue Theoremimplies ∫ R

−R

log(z + i)

1 + z2dz +

∫γR

log(z + i)

1 + z2dz = 2πi · i+ i

i+ i= π log 2 + π2

2i.

Note ∣∣∣∣∫γR

log(z + i)

1 + z2dz

∣∣∣∣ = ∣∣∣∣∫ π

0

log(Reiθ + 1)

1 +R2e2iθReiθ · idθ

∣∣∣∣ ≤ πR[log(R+ 1) + 2π]

R2 − 1→ 0

as R→ ∞. So∫∞−∞

log(x+i)1+x2 dx = π log 2 + π2

2 i. Hence

π log 2 = Re∫ ∞

−∞

log(x+ i)

1 + x2dx =

∫ ∞

−∞

log√x2 + 1

1 + x2dx =

∫ ∞

0

log(1 + x2)

1 + x2dx.

Remark 22. The main difficulty of this problem is that the poles of 11+x2 coincide with the branch points

of log(x2 + 1), so that we cannot apply Residue Theorem directly. An attempt to avoid this difficultymight be writing log(x2+1)

1+x2 as the sum of log(x+i)1+x2 and log(x−i)

1+x2 and integrate them separately along differentpaths which do not contain their respective branch points. But log(x + i) and log(x − i) cannot be definedsimultaneously in a region of a Riemann surface which contains both integration paths. But the observationthat Re[log(x± i)] = log

√x2 + 1 gives us a simple solution (Rule 1).

I 16. Explain whether the function defined by the series

−1

z− 1− z − z2 − · · ·

in 0 < |z| < 1 can be analytically continued to the function defined by the series

1

z2+

1

z3+

1

z4+ · · ·

in |z| > 1.

Solution. The power series∑∞

n=0 zn is divergent on every point of ∂D(0, 1): if z = 1,

∑∞n=0 z

n = 1+1+1+· · · ;if z 6= 1 and z = eiθ,

N∑n=1

zn =

N∑n=0

cos(nθ) + i

N∑n=0

sin(nθ) =sin (N+1)θ

2 cos Nθ2

sin θ2

+ isin (N+1)θ

2 sin Nθ2

sin θ2

.

Similarly, the series∑∞

n=0 z−n−2 is divergent on every point of ∂D(0, 1). So the functions represented by

these two power series cannot be analytic continuation of each other.

I 17. Show that the function defined by the series

1 + αz + α2z2 + · · ·+ αnzn + · · ·

and the function defined by1

1− z− (1− α)z

(1− z)2+

(1− α)2z2

(1− z)3− · · ·

are analytic continuations of each other.

58

Proof. It’s clear we need the assumption that α 6= 0. The series∑∞

n=0(αz)n is convergent in U1 = {z : |z| <

1|α|}. The series 1

1−z

∑∞n=0(−1)n [(1−α)z]n

(1−z)n is convergent in U2 = {z : |1− α||z| < |1− z|}. On U3 = U1 ∩ U2,both of the series represent the analytic function 1

1−αz . So the functions represented by these two series areanalytic continuation of each other.

I 18. Show that the function f1(z) defined by the series

z − 1

2z2 +

1

3z3 − · · ·

in |z| < 1, and the function f2(z) defined by the series

ln 2− 1− z

2− (1− z)2

2 · 22− (1− z)3

3 · 23− · · ·

in |z − 1| < 2 are analytic continuations of each other.

Proof. On the line segment {z : 0 < Rez < 1, Imz = 0}, Taylor series of ln(1 + z) is z − 12z

2 + 13z

3 − · · · .On the same line segment, 0 < 1−z

2 < 12 . So Taylor series of ln(1− x) (|x| ≤ 1, x 6= 1) gives

ln 2− 1− z

2− (1− z)2

2 · 22− (1− z)3

3 · 23− · · · = ln 2 + ln

(1− 1− z

2

)= ln(1 + z).

Since the two holomorphic functions represented by these two series agree on a line segment, they must agreeon the intersection of their respective domains (Theorem 2.13). Therefore they are analytic continuations ofeach other.

I 19. Prove that the power series∑∞

n=0 z2n cannot be analytically continued to the outside of its circle of

convergence.

Proof. Clearly the series is convergent in D(0, 1) and is divergent for z = 1. So its radius of convergentR = 1. Let f(z) =

∑∞n=0 z

2n . Then f(z) is analytic in D(0, 1) and z = 1 is a singularity of f(z). We notef(z) = z +

∑∞n=1 z

2n = z +∑∞

n=1 z2·2n−1

= z +∑∞

n=1(z2)2

n−1

= z + f(z2). Therefore, we have

f(z) = z + f(z2) = z + z2 + f(z4) = z + z2 + z4 + f(z8) = · · · .

So the roots of equations z2 = 1, z4 = 1, z8 = 1,· · · , z2n = 1, · · · , etc. are all singularities of f . Theseroots form a dense subset of ∂D(0, 1), so f(z) can not be analytically continued to the outside of its circleof convergence D(0, 1).

4 Riemann Mapping TheoremI 3. Show that if the entire function f(z) =

∑∞n=0 cnz

n assumes real values on the real axis, then cn(n =0, 1, · · · ) are real numbers.

Proof. Let g(z) =∑∞

n=0 cnzn. Then g(z) is holomorphic and has the same radius of convergence as f(z).

∀z∗ ∈ R ∩ convergence domain of g(z), we have

g(z∗) =

∞∑n=0

cnzn∗ =

∞∑n=0

cn (¯z∗)n= f(z∗) = f(z∗) = f(z∗),

where the next to last equality is due to the fact that z∗ ∈ R and the last equality is due to the fact thatf(R) ⊂ R. By Theorem 2.13, f ≡ g in the intersection of their respective domains of convergence. Socn = cn, i.e. cn(n = 0, 1, · · · ) are real numbers.

59

I 11. In the Riemann Mapping Theorem, suppose that z0 is a real number and U is symmetric about thereal axis. Show that f satisfies the symmetric property f(z) = f(z) by using the uniqueness of the theorem.

Proof. Let g(z) = f(z) (first take conjugate of z, then take conjugate of f(z)). Then

∂g(z)

∂z=∂f(z)

∂z= conjugate

(∂f(z)

∂z

)= 0.

So g is a holomorphic function. Clearly g is univalent. Since U is symmetric w.r.t. real axis, g(U) = f(U) =f(U) = conjugate(D(0, 1)) = D(0, 1). Furthermore, g(z0) = f(z0) = f(z0) = 0, and

g′(z0) = limz→z0

f(z)− f(z0)

z − z0

= limz→z0

conjugate(f(z)− f(z0)

z − z0

)= conjugate(f ′(z0))= conjugate(f ′(z0))= f ′(z0) > 0.

By the uniqueness of Riemann Mapping Theorem, g ≡ f .

5 Differential Geometry and Picard’s Theorem

6 A First Taste of Function Theory of Several Complex VariablesThere are no exercise problems for this chapter.

7 Elliptic FunctionsThere are no exercise problems for this chapter.

8 The Riemann Zeta Function and The Prime Number TheoryThere are no exercise problems for this chapter.

References[1] Johann Boos. Classical and Modern Methods in Summability, Oxford University Press, 2001. 16

[2] John B. Conway. Functions of One Complex Variable, 2nd edition, Springer, 1978. 36, 39, 43, 44

[3] 方企勤：《复变函数教程》，北京：北京大学出版社，1996.12。21, 22, 25, 27

[4] Sheng Gong and Yuhong Gong. Concise Complex Analysis (revised edition), World Scientific, Singapore,2007. 1

[5] Peter D. Lax and Lawrence Zalcman. Complex Proofs of Real Theorems, American Mathematical Society(December 21, 2011). 32

[6] James Munkres. Analysis on Manifolds, Westview Press, 1997. 2, 9

[7] M. A. Qazi. “The mean value theorem and analytic functions of a complex variable”. J. Math. Anal.Appl. 324(2006) 30-38. 29

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