Concise Complex Analysis

Solution of Exercise Problems

Ai Shu Xue

March 9, 2008

1

Contents

1 Calculus 3

2 Cauchy Integral Theorem and Cauchy Integral Formula 14

3 Theory of Series of Weierstrass 27

4 Riemann Mapping Theorem 52

5 Differential Geometry and Picard’s Theorem 53

6 A First Taste of Function Theory of Several Complex Variables 53

7 Elliptic Functions 53

8 The Riemann ζ-Function and The Prime Number Theory 53

2

This is a solution manual of selected exercise problems from Concise Complex Analysis, Second Edition,by Gong Sheng (World Scientific, 2007). This version solves the exercise problems in Chapter 1-3, exceptthe following: Chapter 1 problem 37-42; Chapter 2 problem 47, 49; Chapter 3 problem 15 (xi).

1 Calculus

1.

Proof. See, for example, Munkres [4], §38.

2. (1)

Proof. |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)

Proof. (1 + 3i)3 = −26− 18i. 104−3i = 8

5 + 65 i. 2−3i

4+i = 517 − 14

17 i. (1 + i)n + (1− i)n = 2n2 +1 cos n

4 π.

(3)

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

130.∣∣∣ (4−3i)(2−i)(1+i)(1+3i)

∣∣∣ = 5·√5√2·√10

= 52 .

3.

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 = 4arctan 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 = 4arctan 1

5 −arctan 1

239 .

4.

Proof. 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. z

z2+1 =x3+xy2+x+i(−y3−x2y+y)

(x2−y2+1)2+4x2y2 .

5.

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

z =−(α + iβ)±√∆

2.

6.

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.

7.

3

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 − r2eiθ|2 and |1− r1r2e

iθ|2. Note

|r1 − r2eiθ|2 = (r1 − r2 cos θ)2 + r2

2 sin2 θ = r21 − 2r1r2 cos θ + r2

2

and|1− r1r2e

iθ|2 = (1− r1r2 cos θ)2 + r21r

22 sin2 θ = 1− 2r1r2 cos θ + r2

1r22.

So |1− r1r2eiθ|2 − |r1 − r2e

iθ|2 = (r21 − 1)(r2

2 − 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.

8.

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, |∑ni=1 aibi|2 =

∑ni=1 |ai|2

∑ni=1 |bi|2 −

∑1≤i<j≤n |aibj − aj bi|2 holds

for all n ∈ N. Cauchy’s inequality follows directly from this equation.

9.

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 θ

r 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 + a2

2 + 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−a1a2−a1

)2

= a3−a1a2−a1

. This shows the general case can bereduced to the special case of a1 = 0 and a2 = 1. Essentially, these reductions correspond to the compositionof coordinate translation and coordinate rotation.

4

10.

Proof.

D :=

∣∣∣∣∣∣

α α 1β β 1γ γ 1

∣∣∣∣∣∣=

∣∣∣∣∣∣

α− γ α− γ 1β − γ β − γ 1

0 0 1

∣∣∣∣∣∣=

∣∣∣∣α− γ α− γβ − γ β − γ

∣∣∣∣ .

Suppose α− γ = r1eiθ1 and β − γ = r2e

iθ2 . Then∣∣∣∣α− γ α− γβ − γ β − γ

∣∣∣∣ =∣∣∣∣r1e

iθ1 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.

11.

Proof. 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 .

12.

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 eiβ (r > 0). Then

1 + |z1|2 = 1 + r2 and 1 + |z2|2 = 1 + 1r2 = 1

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

2r cosα + 2r cos β = 02r sin α + 2r sin β = 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.

13.

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 + B

A 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).

14.

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 + z1

1 + |z1|2 ·z2 + z2

1 + |z2|2 +z1 − z1

i(1 + |z1|2) ·z2 − z2

i(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.

15.

Proof. (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.

16.

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

17.

Proof. Note max{|Rezn − Rez0|, |Imzn − Imz0|} ≤ |z − z0| ≤ |Rezn − Rez0|+ |Imzn − Imz0|.18. (1)

6

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 n

satisfying n > N(1 + M2

2

)(define z0 = z′0 = 0)

∣∣∣∣∣1n

n∑

k=1

zkz′n−k − ab

∣∣∣∣∣ <

∣∣∣∣∣1n

N∑

k=1

zkz′n−k

∣∣∣∣∣ +

∣∣∣∣∣1n

n∑

k=N+1

zkz′n−k − ab

∣∣∣∣∣

<NM2

n+

∣∣∣∣∣1n

n∑

k=N+1

(zk − a)z′n−k

∣∣∣∣∣ +

∣∣∣∣∣a

n

n∑

k=N+1

z′n−k − ab

∣∣∣∣∣

< ε + ε + |a|∣∣∣∣∣1n

n−N−1∑

k=0

z′k − b

∣∣∣∣∣

= 2ε + |a|∣∣∣∣∣n−N − 1

n· 1n−N − 1

n−N−1∑

k=0

z′k − b

∣∣∣∣∣ .

Taking upper limits on both sides, we get

limn→∞

∣∣∣∣∣1n

n∑

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→∞

1n

n∑

k=1

zkz′n−k = ab.

(2)

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

19.

Proof. (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.

20.

Proof. By chain rule, ∂∂z g(f(z)) = g′(f(z))∂f(z)

∂z = 0. So g(f(z)) is also holomorphic.

21. (1)

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)

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

] [ux

uy

]= 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 simply

7

connected 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.

22.

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.

23. (1)

Proof. Since

{x = r cos θ

y = r sin θ, we have

[∂∂r∂∂θ

]=

[cos θ sin θ−r sin θ r cos θ

] [ ∂∂x∂∂y

]:= A(θ, r)

[ ∂∂x∂∂y

].

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)

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

ur =∂R

∂rcos φ−R sin φ

∂φ

∂r, vr =

∂R

∂rsin φ+R cosφ

∂φ

∂r, uθ =

∂R

∂θcos φ−R sin φ

∂φ

∂θ, vθ =

∂R

∂θsinφ+R cos φ

∂φ

∂θ.

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

vr

]= A

[∂R∂r∂φ∂r

],

[uθ

vθ

]= A

[∂R∂θ∂φ∂θ

], where A =

[cosφ −R sin φsin φ R cosφ

].

Therefore the C-R equations become

A

[∂R∂r∂φ∂r

]=

[ur

vr

]=

[0 1

r− 1

r 0

] [uθ

vθ

]=

[0 1

r− 1

r 0

]A

[∂R∂θ∂φ∂θ

].

8

Note A−1 = 1R

[R cos φ R sin φsin φ R cos φ

]. 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

.

24.

Proof. By the C-R equations, J =∣∣∣∣ux uy

vx vy

∣∣∣∣ = uxvy − vxuy = ux · ux − vx · (−vx) = u2x + v2

x = |f ′(z)|2. J is

the area of f(D), see, for example, Munkres [4], §21 The volume of a parallelepiped.

25. (1)

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

∫

γ

xdz =∫ 1

0

t(dt + idt) =12

+12i.

(2)

Proof.∫

γ

|z − 1||dz| =∫ 2π

0

|eit − 1|dt =∫ 2π

0

√2− 2 cos tdt =

√2

∫ 2π

0

∣∣∣∣sin(

π

4− t

2

)− cos

(π

4− t

2

)∣∣∣∣ dt

=√

2∫ π

4

− 34 π

|sin θ − cos θ| dθ =√

2∫ π

4

− 34 π

(cos θ − sin θ) dθ4.

(3)

Proof. ∫

γ

dz

z − a=

∫ 2π

0

iReit

Reitdt = 2πi.

26.

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

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.

9

and

cos z1 cos z2 − sin z1 sin z2

=eiz1 + e−iz1

2eiz2 + 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).

27.

Proof. 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 14 (3 + i) = 3+i±√10ei( π

2 −12 arctan 3

4 )

4 .

28.

Proof. eπ2 i = i, e−

2π3 i = 1

2 −√

32 i. log i = (2kπ + π

2 )i, k ∈ Z.

29.

Proof. 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.

30.

Proof. The root of zn = a are |a| 1n ei 2kπ+arg an , k = 0, 1, · · · , n− 1.

31. (1)

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.

(2)

Proof. Note |∑mn=k fn(z)| ≤ M

∑mn=k an. So

∑∞n=1 fn(z) converges uniformly if

∑∞n=1 an converges.

32. (i)

Proof. Since limn→∞ n1

n2 = elimn→∞ ln nn2 = 1, Hadamard’s formula implies R = 1.

(ii)

Proof. Since limn→∞ nln n

n = elimn→∞[ lnn ·ln n]1, Hadamard’s formula implies R = 1.

(iii)

10

Proof. 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] · limn→∞

e1n ln

√2πn(n

e )n−ln n = e−1.

So Hadamard’s formula implies R = e.

(iv)

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

33.

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.

34. (1)

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)

Proof. f ′(z) = limω→0f(z+ω)−f(z)

ω = limω→0 f(z) f(ω)−1ω = f(z)f ′(0) = f(z). By part (1), we conclude

f(z) = ez.

35. (1)

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

(2)

Proof. It’s clear f(1) = 0. So

f ′(z) = limω→0

f(z + zω)− f(z)zω

= limω→0

f(z(1 + ω))− f(z)zω

=1z

limω→0

f(1 + ω)ω

=1zf ′(1) =

1z.

Therefore [1zef(z)

]′= − 1

z2ef(z) +

1zef(z)f ′(z) = 0.

This shows ef(z) = z.

36.

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.

43.

Proof. This is straightforward from Problem 24.

44.

11

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∑N

n=1 anbn =∑N

n=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.

45. (1)

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)

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.

46.

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.

47.

12

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+1bn

> 1 − 1+2εn+1 . Define N = max{N1, N2}, then for any

n ≥ N , an+1an

< bn+1bn

. 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

48. (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→∞ a1nn = elimn→∞

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

(2)

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

θ 6= 0, we have

N∑n=0

anzn =N∑

n=0

1Rn

·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.

49.

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.

50.

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

k=0 akzk0 −

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)zk

0 −∞∑

k=n+1

ak(rnz0)k

∣∣∣∣∣

≤n∑

k=0

|ak|(1− rn)(1 + rn + · · ·+ rk−1n ) +

∞∑

k=n+1

k

n|ak|rk

n

≤n∑

k=0

|ak|(1− rn)k +εn+1

n

11− rn

,

1It seems the condition limn→∞∣∣∣ an+1

an

∣∣∣ = 1 is redundant.

13

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.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 Formula

1.

Proof. (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

[cos π

2π+

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 − 1z − zi

= limz→zi

z5 − z5i

z − zi= 5z4

i =5zi

.

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

5∑

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

zi

5

= 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.

14

(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+

z2

3(z2 − 2)2+

z3

3(z3 − 2)2− 12

49

].

2.

Proof.1

2πi

∫

C

ezζ

ζn· dζ

ζ=

12πi

∫

C

∞∑

k=0

zk

k!1

ζn−k· dζ

ζ=

∞∑

k=0

zk

k!· 12πi

∫

C

1ζn−k

· dζ

ζ=

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!· 12πi

∫

C

1ζn−k

· dζ

ζ− 1

2πi

∫

C

∞∑

k=0

zk

k!1

ζn−k· dζ

ζ

∣∣∣∣∣

=

∣∣∣∣∣1

2πi

∫

C

∞∑

k=m+1

zk

k!1

ζn−k· dζ

ζ

∣∣∣∣∣

≤ 12π

∫

C

∞∑

k=m+1

|zζ|kk!

|dζ||ζ|n+1

≤ 12π

∫

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!· 12πi

∫

C

1ζn−k

· dζ

ζ− 1

2πi

∫

C

∞∑

k=0

zk

k!1

ζn−k· dζ

ζ

∣∣∣∣∣ ≤ ε12π

∫

C

|dζ||ζ|n+1

.

This shows∑∞

k=0zk

k! · 12πi

∫C

1ζn−k · dζ

ζ converges to 12πi

∫C

∑∞k=0

zk

k!1

ζn−k · dζζ .

3.

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 of

ζ). If |z| > 1, we apply Cauchy’s integral theorem to 12πi

∫|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.

4.

Proof. 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.

5.

15

Proof. We note ∫

|z|=1

(z +1z)2n dz

z=

∫ 2π

0

(eiθ + e−iθ)2nidθ = 22ni

∫ 2π

0

cos2n θdθ.

On the other hand,

(z +1z)2n · 1

z=

2n∑

k=0

Ck2nzk(z−1)2n−k · z−1 =

2n∑

k=0

Ck2n

z−2n+2k−1.

Therefore∫ 2π

0

cos2n θdθ =1

22ni

∫

|z|=1

(z +1z)2n dz

z=

122ni

∫

|z|=1

Cn2ndz

z=

2πCn2n

22n= 2π · 1 · 3 · 5 · · · · · (2n− 1)

2 · 4 · 6 · · · · · 2n.

6.

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

7.

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

∫|z|=R

f(ξ)ξn+1 dξ. So

∣∣∣∣f (n)(0)

n!

∣∣∣∣ =12π

∣∣∣∣∫ 2π

0

|f(Reiθ)|Rn

dθ

∣∣∣∣ ≤12π

∣∣∣∣∣∫ 2π

0

Me|Reiθ|

Rndθ

∣∣∣∣∣ = MeR

Rn.

Let R = n, we get the desired inequality.

Remark 3. Note h(R) = ln eR

Rn = R − n ln R takes minimal value at n. This is the reason why we choose

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 best one.

8. (1)

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 − z

dζ +1

2πi

∫

|ζ|=R

f(ζ)ζ − z

dζ.

So

12πi

∫

γ

f(ζ)ζ − z

dζ = −f(z) +1

2πi

∫

|ζ|=R

f(ζ)ζ − z

dζ

= −f(z) +1

2πi

∫

|ζ|=R

f(ζ)− f(∞)ζ − z

dζ +1

2πi

∫

|ζ|=R

f(∞)ζ − z

dζ.

We note ∣∣∣∣∣1

2πi

∫

|ζ|=R

f(ζ)− f(∞)ζ − z

dζ

∣∣∣∣∣ ≤ supζ∈D(z,R)

|f(ζ)−A| → 0, as R →∞,

and Cauchy’s integral formula implies

12πi

∫

|ζ|=R

f(∞)ζ − z

dζ = f(∞) · 1 = A.

16

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

12πi

∫

γ

f(ζ)ζ − z

dζ = −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 − z

dζ +1

2πi

∫

|ζ|=R

f(ζ)ζ − z

dζ.

An argument similar to (1) can show 12πi

∫γ

f(ζ)ζ−z dζ = A.

(2)

Proof. We note zzζ−ζ2 = 1

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

0 =1

2πi

∫

−γ

f(ζ)ζ

dζ +1

2πi

∫

|ζ|=R

f(ζ)ζ

dζ.

So 12πi

∫γ

f(ζ)ζ dζ = 1

2πi

∫|ζ|=R

f(ζ)ζ dζ. Similarly,

12πi

∫

γ

f(ζ)ζ − z

dζ =

{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 − ζ2

dζ =1

2πi

∫

γ

[f(ζ)

ζ− f(ζ)

ζ − z

]dζ =

{f(z) z ∈ D2

0 z ∈ D1.

9.

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.

10.

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.

17

11.

Proof. By Cauchy’s integral formula,

|f (n)(0)| ≤ n!2π

∣∣∣∣∣∫

|z|=r

f(ξ)ξn+1

dξ

∣∣∣∣∣ ≤n!2π

∫ 2π

0

11−r

rndθ =

n!rn(1− r)

.

12.

Proof. We note

limz→z0

Φ(z)− Φ(z0)z − z0

=1

2πilim

z→z0

∫

γ

ϕ(ζ)(ζ − z)(ζ − z0)

dζ =1

2πi

∫

γ

ϕ(ζ)(ζ − z0)2

dζ,

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 Fang [3], Chapter 4, §3, Lemma 2.Remark 4. The result still holds if ϕ is piecewise continuous on γ. See, for example, Fang [3], Chapter 7,proof of Theorem 7.

13.

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.

14.

Proof. Define f(z) = pn(z)zn and g(z) = f( 1

z ). Then g(z) is holomorphic and bounded on D(0, 1) \ {0}, sincelimz→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| < ∞.

15.

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 .

16.

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)|.

18

17.

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 =

∫γ

f(z)dz.

18.

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, Fang [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.

19.

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| .

20. (1)

Proof. We note

ez + e−z + 2 cos z

4=

14[ez + e−z + eiz + e−iz] =

14

∞∑

k=0

1k!

[zk + (−z)k + (iz)k + (−iz)k] =∞∑

m=0

z4m

(4m)!.

The radius of convergence is ∞.

(2)

Proof. By induction, it is easy to show[

1(1−ζ)3

](n)

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

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

1n = 1).

(3)

19

Proof. 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)6

z5n.

Therefore

(1− z−5)−4 =z20

(1− z5)4=

∞∑n=0

(n + 1)(n + 2)(n + 3)6

z5n+20.

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

(4)

Proof. 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)

= − 14(z + 1)

+1

4(z − 1)− 1

2(z + 1)2= −1

4

∞∑n=0

(−z)n − 14

∞∑n=0

zn − 12

∞∑n=0

(−1)n(n + 1)zn.

Therefore

z6

(z + 1)(z2 − 1)= −z6

4

∞∑n=0

[(−1)n + 1 + 2 · (−1)n(n + 1)] zn = −14

∞∑n=0

[(−1)n(2n + 3) + 1] zn+6.

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

21. (1)

Proof. We note

12π

∫ π

−π

|f(reiθ)|2dθ =12π

∫ π

−π

( ∞∑n=0

anrneinθ

)·( ∞∑

m=0

amrme−imθ

)dθ

=12π

∞∑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)

20

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|22n + 2

=12π

∫ 1

0

∫ π

−π

|f(reiθ)|2rdrdθ =12π

∫ ∫

D

|f(z)|2dA.

So∑∞

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

π

∫ ∫D|f(z)|2dA.

22.

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) =

12π

∫ 2π

0

v(eiτ )1−

∣∣ zR

∣∣2∣∣1− z

Reiτ∣∣2 dτ =

12π

∫ 2π

0

u(Reiτ )R2 − |z|2|R− zeiτ |2 dτ

=12π

∫ 2π

0

u(Reiτ )R2 − |z|2

|R− ze−iτ |2 dτ =12π

∫ 2π

0

u(ζ)R2 − |z|2|ζ − z|2 dτ.

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

ReReiτ + z

Reiτ − z= Re

(Reiτ + z)(Re−iτ − z)|Reiτ − z|2 = Re

R2 − |z|2 + R(ze−iτ − zeiτ )|Reiτ − z|2 =

R2 − |z|2|Reiτ − z|2 .

Therefore,

u(z) =12π

∫ 2π

0

u(Reiτ )R2 − |z|2|Reiτ − z|2 dτ =

12π

∫ 2π

0

u(Reiτ )ReReiτ + z

Reiτ − zdτ = Re

[12π

∫ 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).

23.

Proof. On ∂D(0, 1), |(−6z)−(z4−6z+3)| = |z4 +3| ≤ 4 < |−6z|. By Rouche 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}.24.

Proof. 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).

25.

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, π

2 ]}, and γ3 = {z : 0 ≤ Imz ≤ R, Rez = 0}.On γ1, ∆γ1 arg P (z) = 0. On γ2, P (z) = z4

(1 + 2z3−2z+10

z4

). So ∆γ2 arg P (z) = 4 · π

2 + o(1) = 2π + o(1)

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

(1− 2iR(R2+1)

R4+10

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

∑3k=1 ∆γk

arg P (z) = 2π + o(1)(R →∞).

21

By The Argument Principle, P (z) = 0 has only one root in the first quadrant. The root conjugate tothis 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 Fang [3], Chapter 6, §3 Example 5.

26.

Proof. On ∂D(0, 1), |(−8z + 10)− (z4− 8z + 10)| = |z4| = 1 < |10− 8|z|| ≤ |− 8z + 10|. So z4− 8z + 10 and−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}.27.

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

28.

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.

29. (1)

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

(2)

Proof. 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.

30.

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.

31.

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

|f(z)| ≤ 12π

∫ 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.

32.

22

Proof. If the arc is ∂D(0, 1), the desired harmonic function u ≡ 1. So without loss of generality, we assumethe 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 + 1

n}. By the Partition ofUnity 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 Fang [3] Chapter 7, Theorem 7.

33.

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.

34.

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)ζ

dζ =1

2πi

∫

|ζ|=R

f(ζ)ζ(ζ − z0)

dζ.

Therefore ∣∣∣∣f(0)z0

∣∣∣∣ ≤12π

∫ 2π

0

|f(Reiθ)|Reiθ − z0|dθ ≤ M

R− |z0| ,

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

35.

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.

23

36.

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

37.

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.

38.

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

39.

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)|).When z approaches to zeros of f(z), log |f(z)| → −∞, so by letting ε → 0, we can further deduce that

maxz∈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 + log M(r1), α log r2 + log M(r2)}, ∀z ∈ U,

which is the same as

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

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

α =log M(r2)− log M(r1)

log r1 − log r2.

Then from the previous inequality, we get

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

which upon substituting value for α gives

log M(r) ≤ (1− s) log M(r1) + s log M(r2),

where s = log r1−log rlog r2−log r1

.

40.

24

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

12π

∣∣∣∣∣∫

∂D(0,r)

[f(ζ)

ζ − zk− f(ζ)

ζ

]dζ

∣∣∣∣∣

≤n+p∑

k=n+1

12π

∣∣∣∣∣∫

∂D(0,r)

f(ζ)ζ(ζ − zk)

dζ · zk

∣∣∣∣∣ ≤n+p∑

k=n+1

12π

∫ 2π

0

|f(reiθ)|dθ

r − ρk· ρk

≤ 12π

∫ 2π

0

|f(reiθ)|dθ

r − 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 [5]).But as the proof of Theorem 2.7 shows, the formula

f(z)− f(z0) =z − z0

2πi

∫

∂U

f(ζ)dζ

(ζ − z)(ζ − z0)

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

41.

Proof. This problem is the same as Problem 15.

42.

Proof. This problem is the same as Problem 19.

43. (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)

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.

44.

25

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 thisdoes 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))p

k=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.

45.

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

j=1(z − zj)kj · h(z), where h(z) is a holomorphicfunction 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 holomorphicon 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)|∏n

j=1

∣∣∣ z−zj

1−zjz

∣∣∣kj≤ max|z|=1−ε

1∏n

j=1

∣∣∣ z−zj

1−zjz

∣∣∣kj

.

Since each z−zj

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

∣∣∣ z−zj

1−zjz

∣∣∣kj

.

46.

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.

48.

Proof. The Mobius 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)

.

50.

Proof. This problem is the same as Problem 10.

26

3 Theory of Series of Weierstrass

Throughout 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.1.

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

2.

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∫

∂Uf(z)dz = 2πi

∑nk=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

∑nk=1 Res(f, zk) + Res(f,∞) = 0.

3. (i)

Proof. 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

· 1ζ3(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)

Proof. We note

z2

(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 +1z· 11 + 1

z

− 2 · 11 + z

2

=∞∑

k=0

(−1)k

zk+1− 1−

∞∑

k=1

(−1)k

2k−1zk.

(iii)

Proof. We note

log(

z − a

z − b

)= log

(1− a

z

1− bz

)= log

(1− a

z

)− log

(1− b

z

)

= −∞∑

n=1

1n

(a

z

)n

+∞∑

n=1

1n

(b

z

)n

=∞∑

n=1

1n

(bn − an)z−n.

27

(iv)

Proof. We note

z2e1z = z2

∞∑

k=0

z−k

k!= z2 + z +

12

+∞∑

k=1

z−k

(k + 2)!.

(v)

Proof. 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)!· 12πi

∫

|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.

Therefore

sinz

z + 1=

∞∑

k=0

[sin 1(2k)!

(z + 1)−2k +cos 1

(2k + 1)!(z + 1)−(2k+1)

].

4. (i)

Proof. z = 0 is a removable singularity.

(ii)

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

1z2 − 1

cosπz

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.

28

(iii)

Proof. Since e1z =

∑∞n=0 z−n, we have z(e

1z − 1) =

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

(iv)

Proof. z = 1 is an essential singularity.

(v)

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

(vi)

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

5. (1)

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)

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)).

6. (i)

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+ z2n

), 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 + · · · = 11− z

.

(ii)

Proof. sinhπz = 12 [eπz − 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 6=0

(R|an|

)2

=∑∞

n=−∞,n 6=0R2

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

sinhπz = zeh(z)n=∞∏

n=−∞,n 6=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=

1sinh πz

[zeh(z)

∞∏n=1

(1 +

z2

n2

)]′

=1z

+ h′(z) +∞∑

n=1

2zn2

1 + z2

n2

=1z

+ h′(z) +∞∑

n=1

2z

n2 + z2.

29

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

cothπz

z2 − a2dz = 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 − a2dx =

∫ −n

n

−e2πx+1−e2πx−1[

x + (n + 12 )i

]2 − a2dx,

II =∫ −(n+ 1

2 )

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 − a2

dx =∫ n

−n

−e2πx+1−e2πx−1

[x− (n + 12 )i]2 − a2

dx

and

IV =∫ n+ 1

2

−(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

1x2 + (n + 1

2 )2 − a2dx =

2√(n + 1

2 )2 − a2arctan

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

cothπz

z2 − a2dz = 0.

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 + 1z2 − a2

z − an

e2πz − 1=

22π(a2

n − a2)lim

z→an

2π(z − an)e2πz − e2πan

= − 1π(n2 + a2)

,

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

0 =1

2πilim

n→∞

∫

γn

f(z)dz = limn→∞

(cothπa

a−

n∑

k=−n

1π(a2 + k2)

)=

cothπa

a−

∞∑

k=−∞

1π(a2 + k2)

This implies for a 6∈ iZ,

π coth πa =1a

+∞∑

k=1

2a

a2 + n2.

30

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

sinh πz = 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)

Proof. From problem (v), we have π2

∏∞n=1

(1− 1

4n2

)= 1 and

cosπz = sin π

(12− z

)= π

(12− z

) ∞∏n=1

(1−

(12 − z

)2

n2

)

= π

(12− z

) ∞∏n=1

{(1 +

12 − z

n

)(1−

12 − z

n

)}

= π

(12− z

) ∞∏n=1

{(1 +

12n

)(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]

.

(iv)

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

R > 0, we have∑∞

n=−∞,n 6=0

(R|an|

)2

=∑∞

n=−∞,n 6=0R2

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

eπz − 1 = zeh(z)∞∏

n=−∞,n 6=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=

1z

+ h′(z) +∞∑

n=1

2z4n2

1 + z2

4n2

=1z

+ 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,

31

where

I =∫ −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,

and

IV =∫ 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 − a2arctan

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 shown

limn→∞

∫

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).

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 − an

eπz − 1= − 1

4n2 + a2,

Res(f ; an) = − 14n2+a2 . Therefore

0 =1

2πilim

n→∞

∫

γn

f(z)dz = limn→∞

(π

2a

eπa + 1eπa − 1

−n∑

k=−n

14n2 + 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

2eπz + 1eπz − 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

).

32

Equivalently, we have

ez − 1 = zez2

∞∏n=1

(1 +

z2

4π2n2

).

(v)

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

sin πz = −iπ(iz)∞∏

n=1

(1− z2

n2

)= πz

∞∏n=1

(1− z2

n2

).

7. For a clear presentation of convergence of infinite products, we refer to Conway [2], Chapter VII, §5. Inparticular, 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) isa complete metric space. Furthermore, we have the following theorem (proof omitted).

Theorem 11. 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.

Proof. (Proof of Blaschke Product) Fix r ∈ (0, 1). Since∑∞

k=1(1 − |ak|) < ∞, limk→∞ |ak| = 1. So thereexists k0 ∈ N, such that for any k ≥ 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|

akis absolutely and uniformly convergent on {z : |z| ≤ r}. By

Weierstrass Theorem (Theorem 3.1), f(z) =∏∞

k=1ak−z1−akz · |ak|

akrepresents a non-zero holomorphic function on

{z : |z| < 1}. Since the mapping z 7→ ak−z1−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).

8.

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

Fε(z) = f(z) ·∏t

j=1Rε(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)

R2ε−aiz

maps D(0, Rε) onto itself and takes the boundary tothe 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)| = 12π

∫ 2π

0

log |Fε(Rεeiφ)|Re

Rεeiφ + z

Rεeiφ − zdφ =

12π

∫ 2π

0

log |f(Rεeiφ)|Re

Rεeiφ + z

Rεeiφ − zdφ.

33

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

log |f(z)| = 12π

∫ 2π

0

log |f(Rεeiφ)|Re

Rεeiφ + z

Rεeiφ − zdφ +

t∑

j=1

log∣∣∣∣

R2ε − bjz

Rε(z − bj)

∣∣∣∣−s∑

i=1

log∣∣∣∣

R2ε − aiz

Rε(z − ai)

∣∣∣∣ .

Letting ε → 0 yields the desired formula.

9.

Proof. 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 + 2z−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

=(

11− ζ

)′=

( ∞∑n=0

ζn

)′

=∞∑

n=0

(n + 1)ζn,

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

f(z) =1

z + 1+

2z − 2

+3

(z − 2)2+ 1

=1z

11 + 1

z

− 11− z

2

+34· 1(1− z

2 )2+ 1

=1z

∞∑n=0

(−1

z

)n

−∞∑

n=0

(z

2

)n

+34

∞∑n=0

(n + 1)(z

2

)n

+ 1

= −∞∑

n=1

(−1)n

zn+

34

+∞∑

n=1

3n− 12n+2

zn.

10.

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

2 , ψn(z) has Taylor expansion

ψn(z) =n

(z − n)2=

1n· 1(

1− zn

)2 =1n

∞∑

k=0

(k + 1)( z

n

)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)− 1n

λn∑

k=0

(k + 1)( z

n

)k∣∣∣∣∣ ≤ 1

n

∞∑

k=λn+1

(k + 1)2k

=1n

∞∑

k=λn

(k + 2)2k+1

=1n

( ∞∑

k=λn

k

2k+1+

12λn−1

)

≤ 1n · 2λn−1

+1n

∞∑

k=λn

∫ k+1

k

x

2xdx =

1n · 2λn−1

+1n

∫ ∞

λn

x

log 12

d

(12

)x

=1

n · 2λn−1+

1n · 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)( z

n

)k]

,

where U(z) is an entire function.

34

11. (i)

Proof. 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 large enough, wehave ∫

γn

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

where

I =∫ −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),

and

IV =∫ 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

arctan2n + 1√4n2 − a2

→ 0

as n →∞, and

|IV | ≤∫ 2n+1

−(2n+1)

dy

(4n2 + y2 − a2)(e2πn − 1)=

2(e2πn − 1)

√4n2 − a2

arctan2n + 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)

= − 1π(4n2 + a2)

,

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

0 =1

2πilim

n→∞

∫

γn

f(z)dz = limn→∞

(12a· eπa + 1eπa − 1

− 1π

n∑

k=−n

14k2 + a2

)=

12a· eπa + 1eπa − 1

− 1π

∞∑

k=−∞

14k2 + a2

.

Replacing πa with z, we get after simplification

1ez − 1

=1z− 1

2+

∞∑

k=1

2z

4k2π2 + z2, ∀z ∈ C \ 2Zπi.

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

35

(ii)

Proof. Define f(z) = 1(z−a) sin2(πz)

, where a ∈ C\Z. Let γn be the rectangular path [(n + 1

2

)+ni,− (

n + 12

)+

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+ 1

2 )

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+ 1

2 )

−(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 + 1

2 − a)e2yπ=

8(n + 1

2 − 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).

36

It’s easy to see Res(f ; a) = 1sin2(πa)

. Define hk(z) = (z−k)2

sin2(πz). Since limz→k hk(z) = 1

π2 , hk is holomorphic ina 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)=

12πi

∫

|z−k|=ε

hk(z)(z − a)

· dz

(z − k)2=

d

dz

[hk(z)z − a

]∣∣∣∣z=k

.

We noted

dz

[hk(z)z − a

]∣∣∣∣z=k

= limz→k

h′k(z)(z − a)− h(z)(z − a)2

= limz→k

h′k(z)z − a

− 1(k − a)2π2

,

and

limz→k

h′k(z)z − a

= limz→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)

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

coth z =1z

+∞∑

n=1

2z

z2 + π2n2.

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

cot z = i coth(iz) =1z−

∞∑n=1

2z

π2n2 − z2=

1z−

∞∑n=1

(1

πn− z− 1

πn + z

).

Therefore

π

αcot

πβ

α=

1β−

∞∑n=1

(1

αn− β− 1

αn + β

)= lim

N→∞

[1β−

N∑n=1

(1

αn− β− 1

αn + β

)]

= limN→∞

[1β−

(1

α− β+

12α− β

+ · · ·+ 1Nα− β

)+

(1

α + β+

12α + β

+ · · ·+ 1Nα + β

)]

= limN→∞

[(1β− 1

α− β

)+

(1

α + β− 1

2α− β

)+ · · ·+

(1

(N − 1)α + β− 1

Nα− β

)+

1Nα + β

]

=∞∑

n=0

[1

nα + β− 1

nα + (α− β)

]=

∞∑n=0

α− 2β

(nα + β)[nα + (α− β)].

37

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.

12. (1)

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+ 1

2 )

(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+ 1

2 )

(n+ 12 )

h(x + ni)(x + ni)p

eiπx−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+ 1

2 )

−(n+ 12 )

h(x− ni)(x− ni)p

eiπx+nπ + e−iπx−nπ

eiπx+nπ − e−iπx−nπidx,

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 + 1e2nπ − 1

2M

narctan

n + 12

n→ 0

as n →∞,

|II| ≤∫ n

−n

M[(n + 1

2 )2 + y2] p

2

∣∣∣∣eyπ − e−yπ

eyπ + e−yπ

∣∣∣∣ dy ≤ 2M

n + 12

arctann

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] p

2

∣∣∣∣eyπ − e−yπ

eyπ + e−yπ

∣∣∣∣ dy ≤ 2M

n + 12

arctann

n + 12

→ 0

38

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) = −π

n∑

k=1

Res(f(z) cot(πz), αk).

(2)

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,

where

I =∫ −(n+ 1

2 )

n+ 12

h(x + ni)(x + ni)p

2idx

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+ 1

2

−(n+ 12 )

h(x− ni)(x− ni)p

2idx

eiπx+nπ − e−iπx−nπ,

and

IV =∫ n

−n

h((n + 12 ) + yi)[

(n + 12 ) + yi

]p2i

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] p

2

2Mdy

eyπ + e−yπ≤

∫ n

−n

Mdy

(n + 12 )2 + y2

=2M

n + 12

arctann

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

).

39

It’s easy to see

Res(

f(z)sin(πz)

, k

)=

12πi

∫

|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) (i)

Proof. 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)

Proof. 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.

13. (i)

Proof. 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) =

12πi

∫

|z+1|=ρ

dz

(1 + z)(1− z)z2=

12.

(ii)

Proof. We note

f(z) :=z2 + z + 2z(z2 + 1)2

=2z

+−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.

40

(iii)

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

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

Res(f ; ai) =an−1

i∏nj=1,j 6=i(ai − aj)

= limz→ai

an−1i (z − ai)∏nj=1(z − aj)

= limz→ai

an−1i (z − ai)zn + an

=1n

.

(iv)

Proof. 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=

12πi

∫

|z−nπ|=ε

1z − nπ

(−1)n(z − nπ)sin(z − nπ)

dz = (−1)n.

(v)

Proof. 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 ;∞) = − 12πi

∫

|z|=R

z3 cos(

1z − 2

)dz = − 1

2πi

∫

|z−2|=R

z3 cos(

1z − 2

)dz

= − 12πi

∫

|ζ|= 1R

(1ζ

+ 2)3

cos ζ

(−dζ

ζ2

)=

12πi

∫

|ζ=ε|

(ζ + 2)3

ζ5cos ζdζ

=14!

d4

dζ4

[(ζ + 2)3 cos ζ

] |ζ=0 = −83.

(vi)

Proof. 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.

14.

Proof. 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.

41

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).

15. We make an observation of some simple rules that facilitate the evaluation of integrals via ResidueTheorem.

Rule 1 (rule for integrand function). The poles of the integrand function should be easy to find, suchthat the 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.

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)

Proof. 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)

Proof. We note sin2 x = 1−cos 2x2 , so

∫ π2

0

dx

a + sin2 x=

12

∫ π

0

dx

a + sin2 x=

12

∫ 2π

0

dθ

(2a + 1)− cos θ.

Let b = 2a + 1, then

∫ π2

0

dx

a + sin2 x=

12

∫

|z|=1

dziz

b− 12 (z + z−1)

=12i

∫

|z|=1

dz

− 12z2 + bz − 1

2

.

The equation − 12z2 + bz − 1

2 = 0 has two roots: z1 = b − √b2 − 1 and z2 = b +√

b2 − 1. Since b > 1, wehave 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).

42

Remark 14. 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

(12

(z +

1z

),

12i

(z − 1

z

)), dθ =

dz

iz.

Therefore ∫ 2π

0

R(sin θ, cos θ)dθ =∫

|z|=1

R

(12(z +

1z),

12i

(z − 1z))

dz

iz.

Remark 15. When the integrand function is a rational function of trigonometric functions, in view of theprevious 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, π

2 ] to [0, π], and then use double angle formula toexpand [0, π] to [0, 2π]. This solution is motivated by Rule 2.

(iii)

Proof. 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 cos x

x2 + 1dx + i

∫ ∞

−∞

x sin x

x2 + 1dx =

π

ei,

which implies∫∞0

x sin xx2+1 dx = π

2e .

(iv)

Proof. 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√

(log R)2 + θ2

(R2 − 1)2dθ ≤ πR(log R + 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

log x

(1 + x2)2dx =

∫ R

r

log(−x)(1 + x2)2

dx =∫ R

r

log x + πi

(1 + x2)2dx.

43

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

0

2 log x + πi

(1 + x2)2dx = −π

2+

π2

4i,

i.e.∫∞0

log x(1+x2)2 dx = −π

4 .

(v)

Proof. 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.

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 + x2

dx =∫ ∞

0

x1−α

1 + x2dx · 2 cos

(1− α)π2

ei(1−α) π2 = πei(1−α) π

2 ,

i.e.∫∞0

x1−α

1+x2 dx = π2 csc

(απ2

).

(vi)

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

(vii)

Proof. We use the method outlined in the remark of Problem (ii).

∫ π

0

dθ

a + cos θ=

12

∫ π

−π

dθ

a + cos θ=

12

∫

|z|=1

dziz

a + z+z−1

2

=12i

∫

|z|=1

dz12z2 + az + 1

2

.

The equation 12z2 + az + 1

2 = 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=

1i

∫

|z|=1

dz

(z − z1)(z − z2)= 2π · 1

z1 − z2=

π√a2 − 1

.

Remark 16. The expansion of integration interval from [0, π] to [−π, π] is motivated by Rule 2.

(viii)

44

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

sin xx dx = π

2 , we have

∫ ∞

0

(sin x

x

)2

dx =∫ ∞

0

sin2 xd

(− 1

x

)=

∫ ∞

0

2 sin x cosxdx

x=

∫ ∞

0

sin(2x)2x

d(2x) =π

2.

(ix)

Proof. It is easy to see that when λ = p = 0, the integral is equal to 1. So without loss of generality, we onlyconsider 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 = cos λ, then ρ ∈ (−1, 1]. Finally, define f(z) = zp

z2+2z cos λ+1 = ep log z

z2+2ρz+1 .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 = − cos λ + i| sin λ| and z2 =−ρ− i

√1− r2 = − cosλ− i| sin λ|. 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

sinλsin(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)sin λ sin(pπ) .

Remark 17. 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)

45

Proof. 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 z

p is defined on C \ [0,∞). Note by substituting y1p 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

∣∣∣∣∣ ≤2πR

1p

p(R− 1)→ 0

as R →∞, ∣∣∣∣∫

γr

f(z)dz

∣∣∣∣ =

∣∣∣∣∣∫ 2π

0

e1p log(reiθ)

p(reiθ + 1)reiθreiθ · idθ

∣∣∣∣∣ ≤2πr

1p

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

x1p dx

p(x + 1)x=−2iαeαi

1− e2αi=

−2iαeαi

−2 sin αe( π2 +α)i

=α

sinα=

π

pcsc

(π

p

).

Remark 18. 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)

Proof. (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)+

1i · 2i

+1

−i · i= 0.

46

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.”

(xii)

Proof. Choose r, R such that 0 < r < R and r is sufficiently small and R is sufficiently large. 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} 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 2

2+

34πi

).

We note ∣∣∣∣∫

γR

f(z)dz

∣∣∣∣ =∣∣∣∣∫ 2π

0

log(Reiθ)R2e2iθ + 2Reiθ + 2

Reiθ · idθ

∣∣∣∣ ≤log R + 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

log x + πi

x2 − 2x + 2dx.

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

0

[log x

x2 − 2x + 2+

log x

x2 + 2x + 2

]dx + i

∫ ∞

0

πdx

x2 − 2x + 2=

π

2log 2 +

34π2i,

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

log x 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

∣∣∣∣ ≤log R + 2π

R2 − 42πR → 0

47

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

log x + πi

x2 + 4dx.

So by letting r → 0 and R →∞, we have

2∫ ∞

0

log x

x2 + 4dx + 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− ∫∞

0log x

x2+2x+2dx = π4 log 2,

we get ∫ ∞

0

log x

x2 + 2x + 2dx =

π

8log 2.

Remark 19. Note how we handled the combined difficulty caused by no symmetry in the denominator of theintegrand function and log function: because x2 + 2x + 2 is not symmetric, we want to use full circle; butthis will cause the integrals produced by different integration paths to cancel with each other. Therefore weare forced to choose half circle and use two equations to solve for the desired integral.

(xiii)

Proof. 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(log R + 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(log x + πi)x2 + 1

dx = eπ2 i

∫ R

r

√x log x

x2 + 1dx + e

π2 iπi

∫ R

r

√x

x2 + 1dx.

By letting r → 0 and R →∞, we have

(1 + i)∫ ∞

0

√x log x

x2 + 1dx− π

∫ ∞

0

√x

x2 + 1dx =

π2

2√

2(−1 + i).

Therefore,∫∞0

√x log xx2+1 dx = π2

2√

2.

48

(xiv)

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

zn

n and ln(1 + z) =∑∞

n=1(−1)n+1 zn

n hold in|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 + 1ex − 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 + 1ex − 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 equal

to π2

4 .

(xv)

Proof. 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)

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

∫ π

0

x sin x

2− 2 cos xdx = π log 2.

Note ∫ π

0

x sin x

2− 2 cos xdx =

∫ π

0

x · 2 sin x2 cos x

2dx

2 · 2 sin2 x2

= 2∫ π

2

0

θ cos θdθ

sin θ= −2

∫ π2

0

log(sin θ)dθ.

So∫ π

20

log(sin θ)dθ = −π2 log.

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

∫ π2

0

log(sin θ)dθ =∫ π

2

0

log(cos α)dα =12

∫ π2

−π2

log(cos α)dα

=14

∫ π2

−π2

log(

1 + cos 2α

2

)dα

=18

∫ π

−π

log(1 + cos β)dβ − π

4log 2.

(Note how Rule 2 leads us to extended the integration interval from [0, π2 ] to [−π, π].) Motivated by the

similar 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)iz

dz = 2π log 1 = 0.

49

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 =12

∫ π

−π

log[(1 + cos β)2 + sin2 β

]dβ =

12

∫ π

−π

[log 2 + log(1 + cos β)] dβ.

Therefore∫ π

−πlog(1 + cos β)dβ = −2π log 2 and

∫ π2

0

log(sin θ)dθ =18

∫ π

−π

log(1 + cos β)dβ − π

4log 2 = −π

2log 2.

(xvii)

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

0

x sin x

1− 2a cosx + a2dx =

12

∫ π

−π

x sin xdx

1− 2a cosx + a2

=12

∫ π

−π

xd(1− 2a cosx + a2)(1− 2a cosx + a2) · 2a

=14a

[2π log(1 + a)2 −

∫ π

−π

log(1− 2a cos x + 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)iz

dz = 2π log a.

Therefore,∫ π

−πlog(1 − 2a cos x + 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+a

a

).

If 0 < a < 1, we have∫ π

0

x sin x

1− 2a cos x + a2dx =

1a2

∫ π

0

x sin x

1− 2a cosx + 1

a2

dx =1a2· π

1a

log(

1 + 1a

1a

)=

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 ∈ [π

2 , π])

x sin x

1− 2a cos x + a2=

x sin x

|a− eix| ≤x sin x

|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 cos x + a2) takes its minimum at cos x. So

x sin x

1− 2a cosx + a2≤ x

sin x.

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 sinx

1− 2a cos x + a2dx

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

50

Remark 20. We did not take the transform cos x = z+z−1

2 because log(1− 2a cos x + 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).

(xviii)

Proof. The function log(z + i) is well-defined on C+ := {z : Imz ≥ 0}. We take the branch of log(z + i) thatevaluates 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 + z2

dz +∫

γR

log(z + i)1 + z2

dz = 2πi · i + i

i + i= π log 2 +

π2

2i.

Note ∣∣∣∣∫

γR

log(z + i)1 + z2

dz

∣∣∣∣ =∣∣∣∣∫ π

0

log(Reiθ + 1)1 + R2e2iθ

Reiθ · 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 + x2

dx =∫ ∞

−∞

log√

x2 + 11 + x2

dx =∫ ∞

0

log(1 + x2)1 + x2

dx.

Remark 21. 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).

16.

Proof. The power series∑∞

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

n=0 zn = 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 bythese two power series cannot be analytic continuation of each other.

17.

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.

18.

51

Proof. On the line segment {z : 0 < Rez < 1, Imz = 0}, Taylor series of ln(1 + z) is z − 12z2 + 1

3z3 − · · · .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.

19.

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 z2n

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

∑∞n=1 z2n

= z +∑∞

n=1 z2·2n−1= z +

∑∞n=1(z

2)2n−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 Theorem

3.

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.

11.

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 .

52

5 Differential Geometry and Picard’s Theorem

6 A First Taste of Function Theory of Several Complex Variables

There are no exercise problems for this chapter.

7 Elliptic Functions

There are no exercise problems for this chapter.

8 The Riemann ζ-Function and The Prime Number Theory

There are no exercise problems for this chapter.

References

[1] Johann Boos. Classical and modern methods in summability, Oxford University Press, 2001.

[2] John B. Conway. Functions of one complex variable, 2nd Edition, Springer, 1978.

[3] Fang Qi-Qin. A course on complex analysis (in Chinese), Peking University Press, 1996.

[4] James Munkres. Analysis on manifolds, Westview Press, 1997.

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

53