Selected Solutions to Ahlfors, Exercícios de Matemática

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Selected Solutions to Complex Analysis by Lars Ahlfors

Matt Rosenzweig

Contents

Chapter 4 - Complex Integration

Cauchy’s Integral Formula

4.2.2 Exercise 1

Applying the Cauchy integral formula to f (z) = e z ,

1 = f (0) =

2 πi

|z|=

f (z)

z

dz ⇐⇒ 2 πi =

|z|=

e z

z

dz

Section 4.2.2 Exercise 2

Using partial fractions, we may express the integrand as

z 2

• 1

i

2(z + i)

i

2(z − i)

Applying the Cauchy integral formula to the constant function f (z) = 1,

2 πi

|z|=

z 2

• 1

dz =

i

2 πi

|z|=

z + i

dz −

i

2 πi

|z|=

z − i

dz = 0

4.2.3 Exercise 1

1. Applying Cauchy’s differentiation formula to f (z) = e z ,

1 = f

(n−1) (0) =

(n − 1)!

2 πi

|z|=

e z

z n

dz ⇐⇒

2 πi

(n − 1)!

|z|=

e z

z n

dz

1. We consider the following cases:

(a) If n ≥ 0 , m ≥ 0, then it is obvious from the analyticity of z

n (1 − z)

m and Cauchy’s theorem that

the integral is 0.

(b) If n ≥ 0 , m < 0, then by the Cauchy differentiation formula,

|z|=

z

n (1−z)

m dz = (−1)

m

|z|=

z

n

(z − 1) |m|

dz =

0 n < |m| − 1

(−1) m 2 πi (|m|−1)!

n! (n−|m|+1)!

|m| 2 πi

n |m|− 1

n ≥ |m|

(c) If n < 0 , m ≥ 0, then by a completely analogous argument,

|z|=

z

n (1−z)

m dz =

|z|=

(1 − z) m

z |n|

dz =

0 m < |n| − 1

(−1) |n|− 1 2 πi (|n|−1)!

m! (m−|n|+1)!

|n|− 1 2 πi

m |n|− 1

m ≥ n

(d) If n < 0 , m < 0, then sincen(|z| = 2, 0) = n(|z| = 2, 1) = 1, we have by the residue formula that

|z|=

(1 − z)

m z

n = 2πires(f ; 0) + 2πires(f ; 1) =

|z|= 1 2

(1 − z)

m z

n dz +

|z− 1 |= 1 2

(1 − z)

m z

n dz

Using Cauchy’s differentiation formula, we obtain

|z|=

(1 − z)

m z

n dz =

[

|z|= 1 2

(1 − z) −|m|

z |n|

dz +

|z− 1 |= 1 2

z −|n|

(1 − z) |m|

dz

]

2 πi

(|n| − 1)!

(|m| + |n| − 2)!

(|m| − 1)!

|m| 2 πi

(|m| − 1)!

|m|− 1 (|n| + |m| − 2)!

(|n| − 1)!

= 2πi

[(

|m| + |n| − 2

|n| − 1

|m| + |n| − 2

|n| − 1

)]

1. If ρ = 0, then it is trivial that

|z|=ρ

|z − a|

− 4 |dz| = 0, so assume otherwise. If a = 0, then

|z|=ρ

|z|

− 4 |dz| =

1

0

ρ

− 4 2 πiρdt =

2 πi

ρ 3

Now, assume that a 6 = 0. Observe that

|z − a|

4

(z − a) 2 (z − a)

2

|z|=ρ

|z − a|

− 4 |dz| =

|z|=ρ

(z − a)^2 (z − a)^2

|dz| =

0

(ρe^2 πit^ − a)^2 (ρe−^2 πit^ − a)^2

ρ

2 πie

4 πit

ie^4 πit^

dt

= −i

1

0

ρ 2 πie 4 πit

(ρe 2 πit − a) 2 (ρ − ae 2 πit ) 2

dt =

−i

ρ

|z|=ρ

z

(ρ −

a ρ z) 2 (z − a) 2

dz =

−iρ

a

2

|z|=ρ

z

(z −

ρ^2 a

2 (z − a) 2

dz

We consider two cases. First, suppose |a| > ρ. Then z(z − a)

− 2 is holomorphic on and inside {|z| = ρ}

and

ρ 2

a lies inside {|z| = ρ}. By Cauchy’s differentiation formula,

|z|=ρ

|z − a|

− 4 |dz| = 2πi

−iρ

a

2

[

(z − a)

− 2 − 2 z(z − a)

− 3

]

z= ρ^2 a

2 πρ

a

2 (

ρ^2 a − a) 2

[

ρ

2

a(

ρ^2 a − a)

]

− 2 πρ(ρ 2

• |a|

2 )

(ρ 2 − |a|

2 ) 3

2 πρ(ρ 2

• |a|

2 )

(|a|

2 − ρ 2 ) 3

Now, suppose |a| < ρ. Then

ρ 2

a lies outside |z| = ρ, so the function z(z −

ρ 2

a

− 2 is holomorphic on and

inside {|z| = ρ}. By Cauchy’s differentiation formula,

|z|=ρ

|z − a|

− 4 |dz| = 2πi

−iρ

a 2

[

(z −

ρ

2

a

− 2 − 2 z(z −

ρ

2

a

− 3

]

z=a

2 πρ

a

2 (a −

ρ^2 a

2

[

a

(a −

ρ^2 a

]

− 2 πρ

(|a|

2 − ρ 2 ) 2

(a +

ρ 2

a

a −

ρ^2 a

− 2 πρ(|a|

2

• ρ

2 )

(|a|

2 − ρ 2 ) 3

2 πρ(|a|

2

• ρ

2 )

(ρ 2 − |a|

2 ) 3

4.2.3 Exercise 2

Let f : C → C be a holomorphic function satisfying the following condition: there exists R > 0 and n ∈ N

such that |f (z)| < |z|

n ∀ |z| ≥ R. For every r ≥ R, we have by the Cauchy differentiation formula that for

all m > n, ∣ ∣ ∣f^

(m) (a)

∣ ≤^

m!

2 π

|z|=r

|z|

n

|z|

m+

|dz| ≤

m!

r m−n

Noting that m − n ≥ 1 and letting r → ∞, we have that f

(m) (a) = 0. Since f is entire, for every a ∈ C, we

may write

f (z) = f (a) + f

′ (a)(z − a) + · · · +

f

(n) (a)

n!

(z − a)

n

• fn+1(z)(z − a)

n+ ∀z ∈ C

where fn+1 is entire. Since fn+1(a) = f

(n+1) (a) = 0 and a ∈ C was arbitary, we have that fn+1 ≡ 0 on C.

Hence, f is a polynomial of degree at most n.

Calculus of Residues

4.5.2 Exercise 1

Set f (z) = 6z

3 and g(z) = z

7 − 2 z

5 − z + 1. Clearly, f, g are entire, |f (z)| > |g(z)| ∀ |z| = 1, and

f (z) + g(z) = z

7 − 2 z

5

• 6z

3 − z + 1. By Rouch´e’s theorem, f and f + g have the same number of zeros,

which is 3 (counted with order), in the disk {|z| < 1 }.

Section 4.5.2 Exercise 2

Set f (z) = z

4 and g(z) = − 6 z + 3. Clearly, f, g are entire, |f (z)| > |g(z)| ∀ |z| = 2. By Rouch´e’s theorem,

z

4 − 6 z +3 has 4 roots (counted with order) in the open disk {|z| < 2 }. Now set f (z) = − 6 z and g(z) = z

4 +3.

Clearly, |f (z)| > |g(z)| ∀ |z| = 1. By Rouch´e’s theorem, z

4 − 6 z + 3 = 0 has 1 root in the in the open disk

{|z| < 1 }. Observe that if z ∈ { 1 ≤ |z| < 2 } is root, then by the reverse triangle inequality,

3 = |z|

z

3 − 6

≥ |z|

∣|z|

3 − 6

So |z| ∈ (1, 2). Hence, the equation z

4 − 6 z + 3 = 0 has 3 roots (counted with order) with modulus strictly

between 1 and 2.

4.5.3 Exercise 1

1. Set f (z) =

1 z^2 +5z+

1 (z+3)(z+2)

. Then f has poles z 1 = − 2 , z 2 = −3 and by Cauchy integral formula,

res(f ; z 1 ) =

2 πi

|z+2|= 1 2

(z + 3)

− 1

(z + 2)

dz =

z + 3

|z=− 2 = 1

res(f ; z 2 ) =

2 πi

|z+3|= 1 2

(z + 2) − 1

(z + 3)

dz =

z + 2

|z=− 3 − 1

1. Set f (z) =

1 (z^2 −1)^2

1 (z−1)^2 (z+1)^2

. Then f has poles z 1 = − 1 , z 2 = −1. Applying Cauchy’s differentia-

tion formula, we obtain

res(f ; z 1 ) =

2 πi

|z+1|=

(z − 1)

− 2

(z + 1) 2

dz = −2(z − 1)

− 3 |z=− 1 =

res(f ; z 2 ) =

2 πi

|z− 1 |=

(z + 1) − 2

(z − 1) 2

dz = −2(z + 1)

− 3 |z=1 = −

1. sin(z) has zeros at kπ, k ∈ Z, hence sin(z) − 1 has poles at zk = kπ. We can write sin(z) = (z −

zk) [cos(zk) + gk(z)], where gk is holomorphic and gk(zk) = 0. By the Cauchy integral formula,

res(f ; zk) =

2 πi

|z−zk |=

[f ′ (zk) + gk(z)]

− 1

(z − zk)

dz =

f ′ (zk) + g(zk)

k

1. Set f (z) = cot(z). Since sin(z) has zeros at zk = kπ, k ∈ Z and cos(zk) 6 = 0, cot(z) has poles at

zk, k ∈ Z. We can write sin(z) = (z − zk) [cos(zk) + gk(z)], where gk is holomorphic and gk(zk) = 0.

By Cauchy’s integral formula,

res(f ; zk) =

2 πi

|z−zk |=

cos(z) [cos(zk) + gk(z)]

− 1

(z − zk)

dz =

cos(zk)

cos(zk) + gk(zk)

1. It follows from (3) that f (z) = sin(z) − 2 has poles at zk = kπ, k ∈ Z. We remark further that

gk(z) = − cos(zk)(z − zk)

2

• hk(z), where hk(z) is holomorphic. By the Cauchy differentiation formula,

res(f ; zk) =

2 πi

|z−zk |=

[cos(zk) + gk(z)]

− 2

(z − zk) 2

dz = − 2

g ′ k (zk)

(cos(zk) + gk(zk)) 3

1. Evidently, the poles of f (z) =

1 zm(1−z)n^ are z 1 = 0, z 2 = 1. By Cauchy’s differentiation formula,

res(f ; z 1 ) =

2 πi

|z|= 1 2

(1 − z)

−n

zm^

dz =

(n + m − 2)!

(n − 1)!(m − 1)!

n + m − 2

m − 1

res(f ; z 2 ) =

n

2 πi

|z− 1 |= 1 2

z −m

(z − 1) n

dz =

n (−1) n− 1 (m + n − 2)!

(m − 1)!

n + m − 2

n − 1

4.5.3 Exercise 3

(a) Since a + sin

2 (θ) = a +

1 −cos(2θ) 2 = 2 [(2a + 1) − cos(2θ)], we have

∫ π 2

0

dθ

a + sin

2 (θ)

∫ π 2

0

dθ

(2a + 1) − cos(2θ)

∫ (^) π

0

dt

(2a + 1) − cos(t)

−π

dτ

(2a + 1) + cos(τ )

∫ (^) π

0

dτ

(2a + 1) + cos(τ )

where we make the change of variable τ = θ − π, and the last equality follows from the symmetry of

the integrand. Ahlfors p. 155 computes

∫ (^) π

0

dx α+cos(x)

√π α^2 − 1

for α > 1. Hence,

∫ π 2

0

dθ

a + sin

2 (θ)

π √ (2a + 1) 2 − 1

(b) Set

f (z) =

z 2

z 4

• 5z 2
• 6

z 2

(z 2

• 3)(z 2

z 2

(z −

3 i)(z +

3 i)(z −

2 i)(z +

2 i)

For R >> 0,

γ 1 : [−R, R] → C, γ 1 (t) = t; γ 2 : [0, π] → C, γ 2 (t) = Re

it

and let γ be the positively oriented closed curve formed by γ 1 , γ 2. By the residue formula and applying

the Cauchy integral formula to

e iz

z+ai

to compute res(f ; ai),

γ

f (z)dz = 2πires(f ;

3 i) + 2πires(f ;

2 i)

It is immediate from Cauchy’s integral formula that

2 πires(f ;

3 i) =

|z−i

√ (^3) |=

z 2 (z + i

− 1 (z 2

− 1

(z − i

dz = 2πi ·

(i

2

((i

2

• 2)(2i

3 π

2 πires(f ;

2 i) =

|z−i

√ (^2) |=

z 2 (z + i

− 1 (z 2

− 1

(z − i

dz = 2πi ·

(i

2

((i

2

• 3)(2i

2 π

Using the reverse triangle inequality, we obtain the estimate

γ 2

f (z)dz

πR 3

|R

2 − 3 | |R 2 − 2 |

→ 0 , R → ∞

Hence,

∞

0

x

2

x 4

• 5x 2
• 6

∞

−∞

x

2

x 4

• 5x 2
• 6

dx = (

2)π ⇒

∞

0

x

2 dx

x 4

• 5x 2
• 6

2)π

Harmonic Functions

4.6.2 Exercise 1

Let u : D

′ (0; ρ) → R be harmonic and bounded. I am going to cheat a bit and assume Schwarz’s theorem

for the Poisson integral formula, even though Ahlfors discusses it in a subsequent section. Let

Pu(z) =

2 π

∫ (^2) π

0

Re

re

iθ − z

re iθ

• z

u(re

iθ )dθ

denote the Poisson integral for u on some circle of fixed radius r < ρ. Since u is continuous, Pu(z) is a

harmonic function in the open disk D(0; r) and is continuous on the boundary {|z| = r}. We want to show

that u and Pu agree on the annulus, so that we can define a harmonic extension of u by setting u(0) = Pu(0).

Define

g(z) = u(z) − Pu(z)

and for  > 0 define

g(z) = g(z) +  log

|z|

r

∀ 0 < |z| ≤ r

Then g is harmonic in D ′ (0; r) and continuous on the boundary. Furthermore, since u is bounded by

hypothesis and Pu is bounded by construction on D(0; r), we have that g is bounded on D(0; r). g(z) is

harmonic in D ′ (0; r) and continuous on the boundary since both its terms are. Since log

r − 1 |z|

→ −∞, z →

0, we have that

lim sup z→ 0

g(z) < 0

Hence, there exists δ > 0 such that 0 < |z| ≤ δ ⇒ g(z) ≤ 0. Since g is harmonic on the closed annulus

{δ ≤ |z| ≤ r}, we can apply the maximum principle. Hence, g assumes its maximum in {|z| = δ} ∪ {|z| = r}.

But, g(z) ≤ 0 ∀ |z| = δ, by our choice of δ, and since u, Pu agree on {|z| = r}, we have that g(z) = 0 ∀ |z| = r.

Hence,

g(z) ≤ 0 ∀ 0 < |z| ≤ r

Letting  → 0, we conclude that g(z) ≤ 0 ∀ 0 < |z| ≤ r, which shows that u ≤ Pu on the annulus. Applying

the same argument to h = Pu − u, we conclude that u = Pu on 0 < |z| ≤ r. Setting u(0) = Pu(0) defines a

harmonic extension of u on the closed disk.

4.6.2 Exercise 2

If f : Ω = {r 1 < |z| < r 2 } → C is identically zero, then there is nothing to prove. Assume otherwise. Since

the annulus is bounded, f has finitely many zeroes in the region. Hence, for λ ∈ R, the function

g(z) = λ log |z| + log |f (z)|

is harmonic in Ω \ {a 1 , · · · , an}, where a 1 , · · · , an are the zeroes of f. Applying the maximum principle to

g(z), we see that |g(z)| takes its maximum in ∂Ω. Hence,

λ log |z| + log |f (z)| = g(z) ≤ max {λ log(r 1 ) + log(M (r 1 )), λ log(r 2 ) + log(M (r 2 ))} ∀z ∈ Ω \ {a 1 , · · · , an}

Thus, if |z| = r, then we have the inequality

λ log(r) + log (M (r)) ≤ max {λ log(r 1 ) + log(M (r 1 )), λ log(r 2 ) + log(M (r 2 ))}

We now find λ ∈ R such that the two inputs in the maximum function are equal.

λ log(r 1 ) + log(M (r 1 )) = λ log(r 2 ) + log(M (r 2 )) ⇒ λ log

r 1

r 2

= log

M (r 2 )

M (r 1 )

Hence, λ = log

M (r 2 ) M (r 1 )

log

r 1 r 2

. Exponentiating both sides of the obtained inequality,

M (r) ≤ exp

log(M (r 2 )) + log

M (r 2 )

M (r 1 )

log

r 2 r

log

r 1 r 2

 (^) = exp

[

log(M (r 2 ) + log

M (r 1 )

M (r 2 )

α

]

= M (r 1 )

α M (r 2 )

1 −α

where α = log

r 2 r

log

r 2 r 1

. I claim that equality holds if and only if f (z) = az

λ , where a ∈ C, λ ∈ R.

It is obvious that equality holds if f (z) is of this form. Suppose quality holds. Then by Weierstrass’s extreme

value theorem, for some |z 0 | = r, we have

|f (z 0 )| = M (r) =

r 1

r

)λ

M (r 1 ) ⇒

z

λ 0 f^ (z^0 )

= r

λ 1 M^ (r^1 )

But since the bound on the RHS holds for all r 1 < |z| < r 2 , the Maximum Modulus Principle tells us that

z λ f (z) = a ∈ C ∀r 1 < |z| < r 2. Hence, f (z) = az −λ

. But λ is an arbitrary real parameter, from which the

claim follows.

4.6.4 Exercise 1

We seek a conformal mapping of the upper-half plane H

onto the unit disk D. lemma The map φ given by

φ(z) = i

1 + z

1 − z

is a conformal map of D onto H

and is a bijective continuous map of ∂D onto R ∪ {∞}, where 1 7 → ∞. Its

inverse is given by

φ

− 1 (w) =

w − i

w + i

Proof. The statements about conformality and continuity follow from a general theorem about the group of

linear fractional transformations of the Riemann sphere (Ahlfors p. 76), so we just need to verify the images.

For z ∈ D,

Im(φ(z)) = Im

i

1 + z

1 − z

1 − z

1 − z

1 − |z|

2

| 1 − z|

2

since |z| < 1. Furthermore, observe that Im(φ(z)) = 0 ⇐⇒ z ∈ ∂D. In particular, φ(1) = ∞. For w ∈ H

,

∣φ−^1 (w)

2

w − i

w + i

w + i

w − i

|w|

2 − 2Im(w) + 1

|w|

2

• 2Im(w) + 1

by hypothesis that Im(z) > 0. Furthermore, observe that

φ

− 1 (w)

= 1 ⇐⇒ Im(w) = 0.

U˜ = U ◦ φ : ∂D → C is a piecewise continuous function since U is bounded and we therefore can ignore

the fact that φ(1) = ∞. By Poisson’s formula, the function

P ˜

U (z) =

2 π

∫ (^2) π

0

Re

e

iθ

• z

e iθ − z

U˜ (eiθ^ )dθ

is a harmonic function in the open disk D. By Lemma 1, the function

PU (z) = P (^) ˜ U ◦ φ

− 1 (z) =

2 π

2 π

0

Re

e

iθ

• φ

− 1 (z)

e iθ − φ − 1 (z)

U (e

iθ )dθ

is harmonic in H

. Fix w 0 ∈ D and let x 0 + iy 0 = z 0 = φ

− 1 (w 0 ). Let Pw 0 (θ) denote the Poisson kernel. We

apply the change of variable t = ϕ

− 1 (e

iθ ) to obtain

2 π

Pw 0 (θ)

dθ

dt

2 π

z 0 −i z 0 +i

2

z 0 −i z 0 +i

t−i t+i

2

t−i t+i

t−i t+i

2 π

|z 0 + i|

2 − |z 0 − i|

2

∣ ∣ ∣(z 0 − i) −

t−i t+i (z 0 + i)

2

((t + i) − (t − i))

2

− 2 |t + i|

2

π

y 0

|(z 0 − i)(t + i) − (t − i)(z 0 + i)|

2

π

y 0

2 |z 0 − t|

2

π

y 0

(x 0 − t) 2

• y 2 0

4.6.4 Exercise 6

Let f : C → C be an entire holomorphic function, and suppose that z − 1 Re(f (z)) → 0 , z → ∞. By Schwarz’s

formula (Ahlfors (66) p. 168), we may write

f (z) =

2 πi

|ζ|=R

ζ + z

ζ − z

Re(f (ζ))

dζ

ζ

∀ |z| < R

Let  > 0 be given and R 0 > 0 such that ∀R ≥ R 0 ,

Re(f (z)) z

∣ < . Let^ R^ be sufficiently large that^ R >^

R 2

> R 0.

By Schwarz’s formula, ∀

R 2

≤ |z| < R,

|f (z)| ≤

R

2 π

∫ (^2) π

0

Re

iθ

• z

Re iθ − z

dθ ≤

R

2 π

∫ (^2) π

0

R + |z|

R − |z|

dθ = R ·

R + R

R −

R 2

= 4R

Fix z ∈ C and let

R 2

> max {R 0 , |z|}. By Cauchy’s differentiation formula,

|f

′ (z)| =

2 π

|w|= R 2

f (w)

(w − z) 2 dw

2 π

∫ (^2) π

0

R 2

∣f ( R 2

e iθ )

∣ R

2

e iθ − z

2 dθ

2 π

R

· 4 R

∫ (^2) π

0

∣ R

2

− |z|

2

dθ = 8

R

2

(R − 2 |z|) 2

Letting R → ∞, we conclude that |f ′ (z)| ≤ 8 . Since z ∈ C was arbitrary, we conclude that |f ′ (z)| 8  ∀z ∈ C.

Since  > 0 was arbitrary, we conclude that f ′ (z) = 0, which shows that f is constant.

4.6.5 Exercise 1

Let f : C → C be an entire holomorphic function satisfying f (R) ⊂ R and f (i · R) ⊂ i · R. Since f (R) ⊂ R,

f (z) − f (z) vanishes on the real axis. By the limit-point uniqueness theorem that

f (z) = f (z) ∀z ∈ C

Since f (iR) ⊂ iR, f (z) + f (−z) vanishes on the imaginary axis. By the limit-point uniqueness theorem that

f (z) = −f (−z) ∀z ∈ C

Combining these two results, we have

f (z) = −f (−z) = −f (−z) = −f (−z) ∀z ∈ C

4.6.5 Exercise 3

Let f : D → C be holomorphic and satisfy |f (z)| = 1 ∀ |z| = 1. Let φ : C ∪ {∞} → C ∪ {∞} be the linear

fractional transformation

φ(z) =

z − i

z + i

Consider the function g = φ

− 1 ◦ f ◦ φ : H

→ C. By the maximum modulus principle, |f (z)| ≤ 1 ∀ |z| ≤ 1.

Hence, g : H

→ H

. Since |f (z)| = 1 ∀ |z| = 1, φ

− 1 (f (z)) ∈ R ∀ |z| = 1. Hence,

f (R) ⊂ R. By the Schwarz

Reflection Principle, g extends to an entire function g : C → C satisfying g(z) = g(z). Define

f˜ = φ ◦ g ◦ φ−^1 : C → C

Then f˜ is meromorphic in C since φ has a pole at z = −i and φ − 1 has a pole at z = 1. In particular,

f˜ has finitely many poles. We proved in Problem Set 1 (Ahlfors Section 4.3.2 Exercise 4) that a function

meromorphic in the extended complex plane is a rational function, so we need to verify that f˜ doesn’t have

an essential singularity at ∞. But in a neighborhood of 0,

f˜

z

= φ ◦ g

i

1 z

1 − 1 z

= φ ◦ g

i

z + 1

z − 1

which is evidently a meromorphic function. Alternatively, we note that ∀ |z| ≥ 1,

f (z)

∣ ≥ 1 since g maps

H

− onto H

−

. So the image of

f in a suitable neighborhood of ∞ is not dense in C. The Casorati-Weierstrass

theorem then tells us that

f cannot have an essential singularity at ∞.

Chapter 5 - Series and Product

Developments

Power Series Expansions

5.1.1 Exercise 2

We know that in the region Ω = {z : Re(z) > 1 } , ζ(z) exists since

∣ ∣ ∣ ∣

n z

n Re(z)

∣nIm(z)i

n Re(z)

∣elog(n)Im(z)i

n Re(z)

and therefore

∞ n=

nz

is a convergent harmonic series; absolute convergence implies convergence by com-

pleteness. Define ζN (z) =

N n=

1 nz^

. Clearly, ζN is the sum of holomorphic functions on the region Ω. I claim

that (ζN )N ∈N converge uniformly to ζ on any compact subset K ⊂ Ω. Since K is compact and z 7 → Re(z)

is continuous, by Weierstrass’s Extreme Value Theorem ∃z 0 ∈ K such that Re(z 0 ) = infz∈K Re(z). In

particular, Re(z 0 ) > 1 since z 0 ∈ Ω. Hence,

n z

n Re(z)

n Re(z 0 )

. So by the Triangle Inequality,

∀z ∈ Ω,

N ∑

n=

n z

N ∑

n=

n z

N ∑

n=

n Re(z 0 )

∞ ∑

n=

n Re(z 0 )

By Weierstrass’s M-test, we attain that ζn → ζ uniformly on K. Therefore by Weierstrass’s theorem, ζ is

holomorphic in Ω and

ζ

′ (z) = lim N →∞

ζ

′ N (z) =^ lim N →∞

N ∑

n=

− log(n)e

− log(n)z = lim N →∞

N ∑

n=

− log(n)

n z

∞ ∑

n=

− log(n)

n z

Section 5.1.1 Exercise 3

Lemma 2. Set an = (−1) n+

. If

n=

an nz^

converges for some z 0. Then

n=

an nz^

converges uniformly on

∀z ∈ C with Re(z) ≥ Re(z 0 ).

Proof. If

∞ n=

an nz 0 conveges, there exists an M > 0 which bounds the partial sums. Let m ≤ N ∈ N. Using

summation by parts, we may write

N ∑

n=m

an

n z

N ∑

n=m

an

n z 0

n z−z 0

N

z−z 0

m− 1 ∑

n=

an

n z 0

N − 1 ∑

n=m

n ∑

k=

ak

k z 0

(n + 1) z−z 0

n z−z 0

Hence, ∣ ∣ ∣ ∣ ∣

N ∑

n=m

an

n z

≤ M

|N

z−z 0 |

+ M

|n z−z 0 |

+ M

N − 1 ∑

n=m

(n + 1) z−z 0

n z−z 0

Observe that

(n + 1) z−z 0

n z−z 0

∣e

− log(n+1)(z−z 0 ) − e

− log(n)(z−z 0 )

z − z 0

∫ (^) log(n+1)

log(n)

e

−t(z−z 0 ) dt

Hence,

∞ ∑

n=

αPn(α)z

n −

∞ ∑

n=

Pn(α)z

n+

∞ ∑

n=

nPn(α)z

n− 1 −

∞ ∑

n=

2 αnPn(α)z

n

∞ ∑

n=

nPn(α)z

n+

Invoking elementary limit properties and using the fact that a function is zero if and only if all its Taylor

coefficients are zero, we may equate terms to obtain the recurrence

αPn+1(α) − Pn(α) = (n + 2)Pn+2(α) − 2 α(n + 1)Pn+1(α) + nPn(α)

⇒ Pn+2(α) =

n + 2

[(2n + 3)αPn+1(α) − (n + 1)Pn(α)]

So,

P 2 (α) =

3 α

2 − 1

P 3 (α) =

5 α

(3α

2 − 1) − 2 α

5 α

3 − 3 α

P 4 (α) =

7 α

(5α

3 − 3 α) − 3

(3α

2 − 1)

(35α

4 − 30 α

2

5.1.2 Exercise 3

Observe that

sin(z)

z

z

∞ ∑

n=

n

(2n + 1)!

z

2 n+

∞ ∑

n=

n

(2n + 1)!

z

2 n

So,

sin(z) z

= 0 in some open disk about z = 0. Hence, the function z 7 → log

sin(z) z

is holomorphic in an open

disk about z = 0, where we take the principal branch of the logarithm. Substituting,

log

sin(z)

z

= log

sin(z)

z

∞ ∑

m=

∞ n=

(−1) n

(2n+1)! z

2 n

)m

m

∞ ∑

m=

1 3! z

2 −

1 5! z

4

1 7! z

6 − [z

8 ]

)m

m

Set P (z) =

1 3!

z

2 −

1 5!

z

4

. Then

log

sin(z)

z

[

z

6

P (z) + [z

8 ]

P (z)

2

• [z

8 ]

P (z)

3

• [z

8 ]

]

[

z

2

z

4

z

6

z

4

(3!)^2

2 z

6

z

6

3(3!)^3

• [z

8 ]

]

z

2 −

z

4 −

z

6

• [z

8 ]

Partial Fractions and Factorization

5.2.1 Exercise 1

From Ahlfors p. 189, we obtain for |z| < 1,

zπ cot(πz) = z

z

∞ ∑

n=

z

z 2 − n 2

= 1 − 2 z

2

∞ ∑

n=

n 2

z^2 n^2

= 1 − 2 z

2

∞ ∑

n=

n 2

∞ ∑

k=

z 2

n 2

)k

where we expand

z 2

n^2 using the geometric series. Since both series are absolutely convergent, we may inter-

change the order of summation to obtain

zπ cot(πz) = 1 − 2 z

2

∞ ∑

k=

∞ ∑

n=

n 2(k+1)

z

2 k = 1 − 2

∞ ∑

k=

ζ(2k)z

2 k

We now compute the Taylor series for πz cot(πz).

πz cot(πz) = πz

cos(πz)

sin(πz)

= πiz

e iπz

• e −iπz

e iπz − e −iπz

= πiz

e i 2 πz

• 1

e i 2 πz − 1

2 πiz

e 2 πiz − 1

πiz(e 2 πiz − 1)

e 2 πiz − 1

= πiz +

2 πiz

e 2 πiz − 1

Let |z| <

1 2 π

. Then

zπ cot(πz) = πiz +

2 πiz ∑∞

k=

(2πiz)k k!

= πiz +

k=

(2πiz)k (k+1)!

) (^) = πiz +

∞ ∑

n=

∞ ∑

k=

(2πiz) k

(k + 1)!

)n

= πiz +

∞ ∑

k=

Bk

k!

(2πiz)

k

where we may use the geometric expansion since

∞ k=

(2πiz) k

(k+1)!

∞ k= | 2 πz|

k < 1 (|z| <

1 2 π ), and the

change in the order of summation is permitted since the series are absolutely convergent. According to

Ahlfors, the numbers Bk are called Bernoulli numbers, the values of which one can look up. Since the two

series representations for πz cot(πz) are equal, the coefficients must agree. Hence,

ζ(2) =

(2πi)

2 B 2

π

2

ζ(4) =

(2πi) 4 B 4

16 π 4

π 4

ζ(6) =

(2πi)

6 B 6

32 π

6

π

6

π

6

5.2.1 Exercise 2

We first observe that ∞ ∑

n=−∞

z^3 − n^3

converges absolutely, being comparable to

n=

1 n^3

. For z 6 = 0, we may write (after some laborious compu-

tation, which can be found at the end of the solutions)

z 3 − n 3

(z − n)(z − ne i 2 π (^3) )(z − nei^

4 π (^3) )

(z − n)(e i 2 π (^3) z − n)(ei^

4 π (^3) z − n)

A

z − n

B

ze i 4 π (^3) − n

C

ze i 2 π (^3) − n

where

C =

e

2 π 3 i

3 z 2

B =

e

4 π 3 i

3 z 2

A =

3 z 2

Ahlfors p. 189 shows that limm→∞

∑m

−m

1 z−n

= π cot(πz), 0 < |z| < 1. Hence, for 0 < |z| < 1,

lim m→∞

m ∑

−m

z 3 − n 3

3 z 2

lim m→∞

m ∑

−m

z − n

e

2 π 3 i

3 z 2

lim m→∞

m ∑

−m

ze i 2 π (^3) − n

e

4 π 3 i

3 z 2

lim m→∞

m ∑

−m

ze i 4 π (^3) − n

π cot(πz)

3 z 2

πe

2 π 3 i cot(πe

2 π 3 i z)

3 z 2

πe

4 π 3 i cot(πe

4 π 3 i )

3 z 2

5.2.3 Exercise 4

Let f (z) be an entire function of genus h. Let {an 6 = 0} n∈N denotes the (at most countable) set of nonzero

zeroes of f and hc denote the genus of the canonical product. We may write

f (z) = z

m e

g(z)

∞ ∏

n=

z

an

e

z an

• 1 2 (^

z an )

2 +···+ 1 hc (^

z an )

hc

where g(z) is a polynomial and h = max (deg(g(z)), hc). Hence,

f (z

2 ) = z

2 m e

g(z 2 )

∞ ∏

n=

z

2

an

e

z^2 an +^

1 2

( z^2 an

) 2 +···+ 1 hc

( z^2 an

)hc

= z

2 m e

g(z 2 )

∞ ∏

n=

z √ an

z √ an

e

√z an

• 1 2

( √z an

) 2 +···+ 1 2 hc+

( √z an

) 2 hc+

e

z − √ an

• 1 2

( z − √ an

) 2 +···+ 1 2 hc+

( z − √ an

) 2 hc+

where we’ve chosen some branch of the square root. If we define b 1 =

a 1 , b 2 = −

a 1 , · · ·. Then

f˜ (z) = f (z^2 ) = z^2 meg(z

2 )

∞ ∏

n=

z

bn

e

z bn +^

1 2 (^

z bn )

2 +···+ 1 2 hc+1 (^

z bn )

2 hc+

the breaking up of the product being justified since the individual products converge absolutely by virtue of

|bn|

2 hc+1+

|an|

hc+

I claim that the genus of

f is bounded from below by h. If h = 0, then there is nothing to prove; assume

otherwise. If h = deg(g(z)) > 0, then

h ≥ deg(g(z

2 )) > h; so assume that h = hc. We will show that the

genus

hc of the canonical product associated to (bn) is bounded from below by 2hc. Suppose

hc < 2 hc. Since

an → ∞ and therefore bn → ∞ by continuity, we have that for all n sufficiently large |bn| > 1. So it suffices

to consider the case

hc = 2hc − 1. Then

∞ ∑

n=

|bn|

˜h c+^

∞ ∑

n=

|bn|

2 hc−1+

∞ ∑

n=

|an|

hc

But this shows that the genus of the canonical product associated to (an) is at most hc −1, which is obviously

a contradiction. Taking f to be a polynomial shows that this bound is sharp.

I claim that the genus of f˜ is bounded from above by 2h + 1. Indeed, 2h + 1 ≥ 2 deg(g(z)) = deg(g(z 2 )), and

we showed above that ˜hc ≤ 2 hc + 1 ≤ 2 h + 1. This bound is also sharp since we can take

f (z) =

∞ ∏

n=

z

n 2

⇒ f (z

2 ) =

n 6 =

z

n

e

z n

f (z) is clearly an entire function of genus 0, and the genus of the canonical product associated to (n)n∈Z is

1, from which we conclude the genus of f (z 2 ) is 1.

5.2.4 Exercise 2

Using Legendre’s duplication formula for the gamma function (Ahlfors p. 200),

πΓ

1 − 2 · 1 (^6) Γ

π 2

2 (^3) Γ

Applying the formula Γ(z)Γ(1 − z) =

π sin(πz) (Ahlfors p. 199), we obtain

π 2

2 (^3) Γ

sin

π ·

1 3

π

− 1 3

π

2

5.2.4 Exercise 3

It is clear from the definition of the Gamma function that for each k ∈ Z ≤ 0 ,

f (z) =

z k

Γ(z) k 6 = 0

zΓ(z) k = 0

extends to a holomorphic function in an open neighborhood of k. We abuse notation and denote the extension

also by

z k

Γ(z) and zΓ(z). lemma For any k ∈ Z

> 0 ,

Γ(z) =

Γ(z + k) ∏k

j= (z + j − 1)

∀z /∈ Z

Proof. Recall that Γ(z) has the property that the Γ(z + 1) = zΓ(z). We proceed by induction. The base case

is trivial, so assume that Γ(z) =

Γ(z+k) ∏k j=1(z+j−1)^

for some k ∈ N. Then

Γ(z + (k + 1)) ∏k+

j= (z + j − 1)

Γ((z + k) + 1) ∏k+

j= (z + j − 1)

(z + k)Γ(z + k) ∏k+

j= (z + j − 1)

Γ(z + k) ∏k

j= (z + j − 1)!

= Γ(z)

Corollary 4. For any k ∈ Z

≤ 0 ,

lim z→k

(z − k)Γ(z) =

k

|k|!

Proof. Fix k ∈ Z

≤ 0

. Immediate from the preceding lemma is that

lim z→−|k|

(z + |k|)Γ(z) = lim z→−|k|

(z + |k|)

Γ(z + |k| + 1) ∏ k+ j= (z + |j| − 1)

(−1) (−2) · · · (− |k|)

k

|k|!

Let k ∈ Z

≤ 0

. Then

res (Γ; k) =

2 πi

|z−k|= 1 2

Γ(z)dz =

2 πi

|z−k|= 1 2

z k

)Γ(z)

z k

dz =

2 πi

|z−k|= 1 2

(z − k) Γ(z)

z − k

dz

Since the function

z k

Γ(z) extends to a holomorphic function in a neighborhood of k, by Cauchy’s

integral formula,

1

2 πi

|z−k|= 1 2

(z − k) Γ(z)

z − k

dz = (z − k) Γ(z)|z=k =

k

|k|!

where use the preceding lemma to obtain the last equality. Thus,

res (Γ; k) =

k

|k|!

∀k ∈ Z

≤ 0

5.2.5 Exercise 2

Lemma 5. (^) ∫ ∞

0

log

1 − e−^2 πx

dx =

π

Proof. Let 1 >> δ > 0. Consider the function

log(1−z) z , which has the power series representation

log(1 − z)

z

z

∞ ∑

n=

n

z

n

∞ ∑

n=

n

z

n− 1 ∀ |z| < 1