Practice Exercise Set I * – Problems may be difficult.

(1) Describe the sets whose points satisfy the following relations

(a)∣∣∣∣z − 1z + 1

∣∣∣∣ = 2 (b) |z + 1| − |z − 1| < 2 (c) |z2 − 1| = 1 (d) argz − 1z + i

=π

3

(2) Given three distinct complex numbers α, β, γ. Show that they are collinear iff Im(αβ + βγ + γα) = 0.

(3) Prove that if |z1| = |z2| = |z3|, z1, z2, z3 distinct, then argz3 − z2

z3 − z1=

12

argz2

z1.

(4) For what complex value z will the following series converge

(a)∞Σ

n=0

(z

1 + z

)n

*(b)∞Σ

n=0

zn

1 + z2n

(5) When can equality occur in the triangle inequality? That is under what conditions on z, w will |z+w| =|z|+ |w|?

(6) What is the boundary of the set z: Re z and Im z are rational?

(7) Establish the identity∣∣∣∣

n

Σk=1

αkβk

∣∣∣∣2

=n

Σk=1

|αk|2n

Σk=1

|βk|2 − Σ1≤k<j≤n

∣∣αkβj − αjβk

∣∣2 for the case n = 2.

(This implies the Cauchy-Schwarz inequality∣∣∣∣

n

Σk=1

αkβk

∣∣∣∣2

≤√

n

Σk=1

|αk|2√

n

Σk=1

|βk|2.)

(8) Suppose 0 < a0 ≤ a1 ≤ . . . ≤ an. Prove that the polynomial P (z) = a0zn + a1z

n−1 + . . . + an has noroot in the unit disk |z| < 1. (Hint: Consider (1 − z)P (z).)

*(9) Prove that if 11z10 + 10iz9 + 10iz − 11 = 0, then |z| = 1.[ Hint: Consider z9. ]

**(10) Let P (z) = zn + c1zn−1 + . . . + cn with c1, . . ., cn real. Suppose |P (i)| < 1. Prove that there is a root

x + iy of P (z) in the set√

(x2 + y2 + 1)2 − 4y2 < 1.

Practice Exercise Set II (To receive a solution, you have to hand in some work.)

(1) (exercise #15, p.17) Show that

(a) f(z) =∞Σ

k=0kzk is continuous in |z| < 1.

(b) g(z) =∞Σ

k=1

1k2 + z

is continuous in the right half plane Re z > 0.

(2) Find limn→∞

xn, where

(a) xn = 1 + (−1)n 2n

n + 1.

(b) xn = cosnπ

4.

(3) Find the radius of convergence of the following power series:

(a)∞Σ

n=0zn!.

(b)∞Σ

n=0(n + 2n)zn.

(4) Give an example of two power series∞Σ

n=0anzn and

∞Σ

n=0bnzn with radii of convergence R1 and R2,

respectively, such that the power series∞Σ

n=0(an + bn)zn has a radius of convergence > R1 + R2.

(5) Explain why there is no power series f(z) =∞Σ

n=0cnzn such that f(z) = 1 for z =

12,13,14, . . . and

f ′(0) > 0.

(6) Does there exist a power series f(z) =∞Σ

n=0cnzn such that f(

1n

) =1n2

and f(− 1n

) =1n3

for n = 1, 2, 3,. . ..

*(7) If f(z) =∞Σ

n=0cnzn satisfies f(

1n

) =n2

n2 + 1, n = 1, 2, 3, . . ., compute the values of the derivatives

f (k)(0), k = 1, 2, 3, . . ..

Practice Exercise Set III (To receive a solution, you have to hand in some work.)

(1) Show that there are no analytic function f = u + iv with u(x, y) = x2 + y2.

(2) Suppose f is an entire function of the form f(x, y) = u(x) + iv(y). Show that f is a polynomial ofdegree at most one.

*(3) What is the range of ez if we take z to lie in the infinite strip | Im z| <π

2? What are the images for

horizontal lines and vertical segments in | Imz| <π

2under the ez mapping?

**(4) Discuss if it is possible to define log(z − 1) continuously on C \ [−1, 1]. Also discuss the possibility for

log(

z + 1z − 1

)continuously defined on C \ [−1, 1].

*(5) Let G be a region and G∗ = z: z ∈ G is the mirror image of G across the x-axis. If f : G → C isanalytic, show that f∗: G∗ → C defined by f∗(z) = f(z) is analytic.

(6) If f = u + iv is analytic on some domain, given u(x, y) below, find the possibilities of v(x, y).(a) u(x, y) = x3 − 3xy2.(b) u(x, y) = e−y cos x.(c) u(x, y) = log(x2 + y2).(d) u(x, y) =

y

(1 − x)2 + y2.

(7) Write z in polar coordinates. Then f(z) = u(z) + iv(z) = u(r, θ) + iv(r, θ). Establish the polar form ofthe Cauchy-Riemann equations:

∂u

∂r=

1r

∂v

∂θ,

∂v

∂r= −1

r

∂u

∂θ.

Practice Exercise Set IV (To receive a solution, you have to hand in some work.)

(1) Find a conformal mapping from the open unit disk D = z: |z| < 1 onto the following regions:

(a) the infinite strip 0 < Im z < 1i

*(b) the upper semidisk |z| < 1, Imz > 0

1-1 0

i

( Hint: Find a map from 1st quadrant onto the upper half semidisk. )

(c) the slit disk D \ [0, 1) 0 1

( Hint: Use (b). )

*(d) C ∪∞

\ [−1, 1]

-1 1

( Hint: Find a map from C \ (−∞, 0] to (C ∪ ∞ \ [−1, 1].)

(2) Let a < b and T (z) =z − ia

z − ib. Define L1 = z: Im z = b, L2 = z: Im z = a, L3 = z: Re z = 0.

Determine which of the regions A, B, C, D, E, F in Figure 1, are mapped by T onto the region U , V ,W , X, Y , Z in Figure 2.

0 1

Figure 1 Figure 2

DA

B

C

E

F

L

L

L 3

2

1

U W

X Z

V

Y

T

ia

ib

( Hint: Orient L3 by (∞, ia, ib). )

Practice Exercise Set V (To receive a solution, you have to hand in some work.)

(1) Suppose f(z) is analytic and |f(z) − 1| < 1 in a region Ω. Show that∫

C

f ′(z)f(z)

dz = 0 for every closed

curve C in Ω, assuming f ′ is continuous.

(2) Compute∫

|z|=1

|z − 1||dz|, where the unit circle |z| = 1 is given the counterclockwise orientation.

*(3) Define∫

C

f(z) dz =∫

C

f(z) dz. If P (z) is a polynomial and C denotes the circle |z − a| = R (counter-

clockwise), show that∫

C

P (z) dz = −2πiR2P ′(a).

(4) Find∫

C

1z2

dz, where C is a smooth curve from 1 to −1 not passing through the origin.

*(5) Show that if f is a continuous real-valued function and |f(z)| ≤ 1, then

∣∣∣∣∣∣∣

∫

|z|=1

f(z) dz

∣∣∣∣∣∣∣≤ 4.

( Hint: Consult the top half of p.45, and show that∣∣∣∣∫

f

∣∣∣∣ ≤∫ 2π

0

| sin t| dt. )

(6) Evaluate the integral∫

C

z

zdz, where C is the curve show below.

( Hint:∫

C

=∫

C1

+∫

C2

+∫

C3

+∫

C4

)

i

C4

C1

2C

C3

21-1-2 0

2i

Practice Exercise Set VI (To receive a solution, you have to hand in some work.)

*(1) The following curve C divides the plane into 4 regions. For each region, state the winding number of Caround points in that region. (Give answers by inspection, no computation needed.)

1 2 34

C

(2) Find limz→0

e

z Log(z49 + 1)(cos z25) − 1 .

*(3) If the range of an entire function lies in the right half plane Re w > 0, show that the function is aconstant function.( Hint: Compose a Mobius map. )

(4) Suppose a polynomial is bounded by 1 in the unit disk. Show that all its coefficients are bounded by 1.(c.f. proof of Liouville’s Theorem.)

(5) Find∫

|z|=1

sin z

zdz,

∫

|z|=1

sin z

z2dz (counterclockwise orientation).

**(6) (Optional) Let f(z) = u(z) + iv(z) (or f(x, y) = (u(x, y), v(x, y)))be a one-to-one analytic function from the open unit disk D =z: |z| < 1 onto a domain G with finite area.

(a) Show that Jf (x, y) def=

∣∣∣∣∣∣∣

∂u

∂x

∂u

∂y∂v

∂x

∂v

∂y

∣∣∣∣∣∣∣= |f ′(z)|2.

y

x

D G

u

vf

(b) For distinct nonnegative integers m, n, show∫

D

zmzn dA = 0 (orthogonality relation), where dA =

r dr dθ = dx dy is the area differential.(c) Show that if f(z) =

∞Σ

n=0cnzn is the power series for f in D, then area of G = π

∞Σ

n=1n|cn|2.

Practice Exercise Set VII (Solutions will be distributed in class.)

(1) Show that if f(z) is analytic in |z| ≤ 1, there must be some positive integer n such that f(1n

) 6= 1n + 1

.

(2) Suppose that f is analytic in the annulus: 1 ≤ |z| ≤ 2, that |f | ≤ 1 for |z| = 1 and that |f | ≤ 4 for|z| = 2. Prove that |f(z)| ≤ |z|2 throughout the annulus.

(3) Show that f(z) =∫ 1

0

sin 2t

tdt is entire function

*(a) by applying Morera’s Theorem(b) by obtaining a power series expansion for f .

(4) Show that if f(z) is continuous on the closed unit disk z: |z| ≤ 1, analytic on the open disk z: |z| < 1and real-valued on the unit circle z: |z| = 1, then f(z) is a constant function.( Hint: The disk is conformally equivalent to the upper half plane. Reflection. )

*(5) Let D = z: |z| < 1. If f : D → D is analytic with at least two fixed points, prove that f(z) ≡ z.( Hint: May assume one of the fixed point is 0 by composing with suitable Mobius transformations. )

*(6) Let f(z) be an entire function which is real on the real axis and imaginary on the imaginary axis, showthat f(z) is an odd function, i.e. f(z) ≡ −f(−z).( Hint: Consider the coefficients of the power series of f(z) or make use the reflection property. )

**(7) If f is an entire function mapping the unit circle into the unit circle (i.e. |f(z)| = 1 for |z| = 1), thenf(z) = eiθzn.( Hint: f(z) has finitely many roots α1, . . . , αn (repeated according to multiplicities) in the unit disk.

Recall∣∣∣∣

z − αj

1 − αjz

∣∣∣∣ = 1 for |z| = 1. Use modulus theorems to show f(z) = eiθn

Πj=1

z − αj

1 − αjz. )

Practice Exercise Set VIII (To receive a solution, you have to hand in some work.)

(1) Suppose f is analytic on C \ a1, . . . , an and Γ is a simple closed curve “surrounding” a1, . . ., an asshown. For each aj, let Cj be a simple closed curve about aj inside Γ.

Γa1 a2 anC C

C

...

1 2n

Show that ∫

γ

f(z) dz =n

Σj=1

∫

Cj

f(z) dz,

where the orientation of Γ, C1, . . ., Cn are as shown.

(2) Identify the isolated singularities of the following functions and classify each as removable singularity,pole (and its order) or essential singularity:

(a)1

z4 + z2(b) cot z (c)

e1/z2

z − 1(d)

z2 − 1sin πz

.

(3) Find the Laurent series of1

z2 − 4on (a) 0 < |z − 2| < 4, (b) 2 < |z| < ∞.

(4) Find∫

|z|=r

sin1z

dz (counterclockwise orientation) for r 6= 0,1π

,12π

,13π

, . . ..

(5) Suppose f is analytic on C \0

and satisfies |f(z)| ≤√|z| + 1√

|z|. Prove f is constant.

(6) If f has a pole at 0, show that ef(z) cannot have a pole at 0.

(7) If f is analytic on R < |z| < ∞, we say ∞ is a removable singularity, pole of order k, essential singularity

of f(z) iff 0 is a removable singularity, pole of order k, essential singularity of f(1z).

(i) Prove that an entire function with a pole at ∞ is a polynomial.(ii) Prove that an analytic function on C ∪ ∞ except for isolated poles must be a rational function.

Practice Exercise Set IX (To receive a solution, you have to hand in some work.)

(1) Prove that the image of the plane under a nonconstant entire mapping f is dense in the plane.

[ Hint: If f is not a polynomial, consider f(1z). ]

(2) Suppose that f is entire and that f(z) is real if and only if z is real. Use the Argument Principle toshow that f can have at most one root.[ Hint: Let Γ be the circle |z| = R, R large, what is n(f Γ, 0)? ]

(3) Is there an analytic function f on z: |z| ≤ 1 which sends the unit circle with counterclockwise orien-tation into the unit circle with clockwise orientation?

*(4) If f is analytic on and inside a simple closed curve Γ, and f is one-to-one on Γ, then f is one-to-oneinside Γ.[ Hint: If f Γ a simple closed curve? For w 6∈ f Γ, let g(z) = f(z) − w, what is n(g Γ, 0)? ]

(5) Show that if α and β 6= 0 are real, the equation z2n + α2z2n−1 + β2 = 0 has n − 1 roots with positivereal parts if n is odd, and n roots with positive real parts if n is even.

(6) If a > e, show that the equation ez = azn has n solutions inside the unit circle.

Practice Exercise Set X (To receive a solution, you have to hand in some work.)

(1) Find∫ ∞

0

dx

1 + xn, where n ≥ 2 is a positive integer. [ Hint: Use the contour

Re

0 R

2πi/n

. ]

(2) Find∫ ∞

0

sin2 x

x2dx. [ Hint: Integrate

e2iz − 1 − 2iz

z2around a large semi-circle. ]

(3) Find∫ ∞

0

ln x

x2 + 1dx. [ Hint: Use the contour

-r r R-R

. ]

(4) Find∫ ∞

0

e−x2cos 2x dx. [ Hint: Use the contour

R

R+i

-R

-R+i

and f(z) = e−z2. ]

(You may need to know∫ ∞

0

e−x2dx =

√π

2.)

(5) Find∫ ∞

0

cos x2 dx and∫ ∞

0

sin x2 dx. [ Hint: Use the contour0 R

π/4 . ]

(6) Find∫ π/2

0

dθ

1 + sin2 θ.

(7) Suppose f is analytic on r < |z| < ∞, then we define Res (f,∞) =1

2πi

∫

|z|=R

f(z) dz (clockwise orienta-

tion).oo

0

(The clockwise orientation relative to 0 is the counterclockwise orientation relative to

∞.) Equivalently, if f(z) =∞Σ

k=−∞akzk on r < |z| < ∞, then Res (f,∞) = −a−1.

(a) If f is meromorphic on C with isolated poles at a1, . . . , an, show thatnΣ

j=1Res (f, aj)+Res (f,∞) = 0

(i.e. the sum of all residues on C ∪ ∞ is 0.)

(b) Show that Res (f,∞) = Res (− 1z2

f(1z), 0).

(c) Find∫

|z|=1

1sin(1

z)

dz.

Solution to Practice Exercise Set I

(1) (a) The locus of all points z whose distances from the two points a = 1 and b = −1 having a fixed

quotient λ = 2 is a circle with center on the line through a, b. On the real axis, z = −3, −13

satisfies

the equation. So the circle is

z:∣∣∣∣z +

53

∣∣∣∣ =43

.

Alternatively, for z = x + iy, (x − 1)2 + y2 = |z − 1|2 = 4|z + 1|2 = 4[(x + 1)2 + y2].

Simplifying we get (x +53)2 + y2 =

169

.

(b) The locus of all points z whose distances from the two points a = −1 and b = 1 having a fixeddifference λ = 2 (≤ distance between a, b) is a branch of a hyperbola having a, b as foci. If λ =distance between a, b (as is the case here), the branch degenerate to a ray (or an infinite slit). Theset is the whole complex plane minus all real numbers greater than or equal to 1.Alternatively, for z = x+iy,

√(x + 1)2 + y2−

√(x − 1)2 + y2 = |z+1|−|z−1| < 2 ⇐⇒ (x+1)2+

y2 < [2 +√

(x − 1)2 + y2]2 = 4 + 4√

(x − 1)2 + y2 + (x − 1)2 + y2 ⇐⇒ x − 1 <√

(x − 1)2 + y2.If x < 1, then x − 1 < 0 ≤

√(x − 1)2 + y2. If x ≥ 1, then 0 ≤ (x − 1)2 < (x − 1)2 + y2 implies

y 6= 0. So the set isx + iy: x < 1 or (x ≥ 1 and y 6= 0)

.

(c) |z2−1| = |z−1||z +1|. The locus of all points z whose distances from two points a = 1 and b = −1

having a fixed product λ = 1 is a lemniscate with foci at a and b. (The case λ >

√|a − b|

2results

in two curves, each about a focus (and as λ → 0, the two curves shrink toward the foci); the case

λ =

√|a − b|

2results in a figure-eight curve with double point at

a + b

2.) The set is a lemniscate

with foci at 1 and −1 having a double point at 0.

21/21/2-2

- 45

o45

o

Alternatively, for z = r cis θ, (r2 cos 2θ−1)2+(r2 sin 2θ)2 = |z2−1|2 = 1.Simplifying we get r = 0 or r2 = 2 cos 2θ. The range of θ possible are

−π

4≤ θ ≤ π

4or

3π

4≤ θ ≤ 5π

4.

α−β

α−β

α β

z

z

z - z 31

12z

3z2 - z 3

O

(d) To interpret argz3 − z2

z3 − z1

(= arg

z2 − z3

z1 − z3

), write z2 − z3 = R cis α,

z1 − z3 = r cis β. Then argz2 − z3

z1 − z3= arg

[R

rcis(α − β)

]= α − β is

the angle 6 z1z3z2 (measured from the ray −−→z3z1 counterclockwise tothe ray −−→z3z2).

-i

1

The set is the open arc of the circle containing all points z suchthat 6 − iz1 = 60.

(2) The line through β and γ is Im( z − γ

β − γ

)= 0. So α, β, γ collinear ⇐⇒ Im

(α − γ

β − γ

)= 0 ⇐⇒

Im((α − γ)(β − γ)

|β − γ|2)

=Im(αβ − γβ − αγ + |γ|2)

|β − γ|2 = 0 ⇐⇒ Im(αβ − γβ − αγ) = 0.

For any complex z = x + iy, Im(−z) = −y = Im z. So Im(αβ − γβ − αγ) = Im(αβ) + Im(−γβ − αγ) =Im(αβ + βγ + γα) = 0 is the condition.

(3)

O

z

zz

1

2

3

(c.f. exercise 1(d)) This is just the complex way of expressing the geometry theorem

that 6 z1z3z2 =126 z10z2. (You should check also the case z1, z2. z3 are oriented

clockwise.)

(4) (a) Let w =z

1 + z, then we know

∞Σ

n=0wn converges iff |w| =

∣∣∣∣z − 0

z − (−1)

∣∣∣∣ < 1 ( ⇐⇒ z is closer to 0 then

−1 ⇐⇒ Re z > −12).

(b) Case 1: (|z| = r < 1).∣∣∣∣

zn

1 + z2n

∣∣∣∣ ≤ |z|n

1 − |z|2n=

rn

1 − r2n(because 1 − |z2n| ≤ |1 + z2n|). Apply

ratio test to∞Σ

n=1

rn

1 − r2n, we have lim

n→∞

rn+1

1 − r2n+1

/rn

1 − r2n= r < 1. So Σ

rn

1 − r2nconverges

⇒ Σ

∣∣∣∣zn

1 + z2n

∣∣∣∣ converges ⇒ Σzn

1 + z2nconverges.

Case 2: (|z| = 1).∣∣∣∣

zn

1 + z2n

∣∣∣∣ ≥|z|n

1 + |z|2n=

12

(because |1 + z2n| ≤ 1 + |z2n|). Sozn

1 + z2ncannot

converge to 0 as n → ∞. Hence Σzn

1 + z2ndiverges.

Case 3: (|z| > 1). For w =1z, |w| < 1, so by case 1, Σ

zn

1 + z2n= Σ

(1w

)n

1 + (1w

)2n= Σ

wn

1 + w2n

converges.

(5) Equality holds iff either z = 0. w = 0 orz

wis real.

(6) For any complex w, every neighborhood B(w, r) of w contains a point z0 with Re z0 and Im z0 rationaland also a point z1 not both Re z1 and Im z1 rational. So any complex w is in the boundary of the set.Therefore. the boundary of the set is all complex numbers.

(7) L.H.S. = |α1β1+α2β2|2 = (α1β1+α2β2)(α1β1+α2β2) = |α1|2|β1|2+α1β2α2β1+α2β1α1β2+ |α2|2|β2|2.R.H.S. = (|α1|2+ |α2|2)(|β1|2+ |β2|2)−|α1β2−α2β1|2 = |α1|2|β1|2+ |α1|2|β2|2+ |α2|2|β1|2+ |α2|2|β2|2−(α1β2 −α2β1)(α1β2 −α2β1) = |α1|2|β1|2 + |α1|2|β2|2 + |α2|2|β1|2 + |α2|2|β2|2− (|α1|2|β2|2−α1β2α2β1 −α2β1α1β2 + |α2|2|β2|2).

Alternative solution for n ∈ N, n ≥ 2 by Chow Chak-On.

|αkβj − αjβk|2 = (αkβj − αjβk)(αkβj − αjβk) = |αk|2|βj|2 − αkβkαjβj − αjβjαkβk + |αj|2|βk|2.nΣ

k=1

nΣ

j=1|αkβj − αjβk|2 =

nΣ

k=1

nΣ

j=1(|αk|2|βj |2 − αkβkαjβj − αjβjαkβk + |αj|2|βk|2).

Sincen

Σk=1

n

Σj=1

|αkβj − αjβk|2 = Σ1≤k=j≤n

|αkβj − αjβk|2 + Σ1≤k<j≤n

|αkβj − αjβk|2 + Σ1≤j<k≤n

|αkβj − αjβk|2

= 2 Σi≤k<j≤n

|αkβj − αjβk|2

andn

Σk=1

n

Σj=1

αkβkαjβj =( n

Σk=1

αkβk

)( n

Σj=1

αjβj

)=

∣∣∣∣n

Σk=1

αkβk

∣∣∣∣2

.

Therefore, 2 Σ1≤k<j≤n

|αkβj −αjβk|2 = 2n

Σk=1

|αk|2n

Σk=1

|βk|2−2∣∣∣∣

n

Σk=1

αkβk

∣∣∣∣2

. Cancelling the factor of 2 on

both sides and rearranging terms we get the desired result.

(8) Suppose z is a root of P (z) and |z| < 1, then

0 = |(1 − z)P (z)| = | − a0zn+1 + (a0 − a1)zn + (a1 − a2)zn−1 + · · ·+ (an−1 − an)z + an|

= |an − [a0zn+1 + (a1 − a0)zn + · · ·+ (an − an−1)z]|

(*)

≥ an − |a0zn+1 + (a1 − a0)zn + · · ·+ (an − an−1)z|

(**)

≥ an − [a0|z|n+1 + (a1 − a0)|z|n + · · ·+ (an − an−1)|z|](***)

> an − [a0 + (a1 − a0) + · · ·+ (an − an−1)] = 0,

a contradiction (where (*) |α − β| ≥ |α| − |β|, (**) |α + β| ≤ |α|+ |β|, and (***) |z| < 1).

(9) Let z = x + iy, then |z|9 = |z9| =∣∣∣∣11− 10iz

11z + 10i

∣∣∣∣ =

√121 + 220y + 100y2 + 100x2

121x2 + 121y2 + 220y + 100.

If |z| < 1, then x2 + y2 < 1 and 121 + 220y + 100y2 + 100x2 > 121x2 + 121y2 + 220y + 100, forcing|z|9 > 1, a contradiction.If |z| > 1, then x2 + y2 > 1 and 121 + 220y + 100y2 + 100x2 < 121x2 + 121y2 + 220y + 100, forcing|z|9 < 1, a contradiction.

(10) (We first observe that√

(x2 + y2 + 1) − 4y2 =√

x2 + y2 + 1 + 2y√

x2 + y2 + 1 − 2y = |i− (x+ iy)||i−(x − iy)|.) Suppose the roots of P (z) are r1, r2, . . ., rn. Because the coefficients are real, complexroots occur in conjugate pairs if any. Since 1 > |P (i)| = |i − r1||i − r2| . . . |i − rn|. For a real root r,|i − r| =

√1 + r2 ≥ 1. So P (z) must have complex roots.

Now |P (i)| =(

Πreal roots

|i − r|)(

Πcomplex roots in pairs

|i − r||i − r|). So there must be a pair of complex roots

r = x+iy and r = x−iy such that |i−r||i−r| < 1. By the observation above,√

(x2 + y2 + 1)2 − 4y2 < 1as desired.

Solution to Practice Exercise Set II

(1) (a) For fixed z0 with |z0| <, there is a disk B(0, R) containing z0 (R < 1). It suffices to show f iscontinuous on B(0, R).

0z

For z ∈ B(0, R), |z| < R < 1,∣∣kzk

∣∣ ≤ kRk = Mk. Now∞Σ

k=0Mk =

∞Σ

k=0kRk

converges by the root test since limk→∞

k√kRk = R < 1. So by Weierstrass M-test,

∞Σ

k=0kzk converges uniformly on B(0, R) to a continuous function f(z). Since z0 is

arbitrary, f(z) is continuous on |z| < 1.

(b) For x = Re z > 0,∣∣∣∣

1k2 + z

∣∣∣∣ =1√

(k2 + x)2 + y2≤ 1

k2= Mk. Since

∞Σ

k=1Mk =

∞Σ

k=1

1k2

converges,

∞Σ

k=1

1k2 + z

converges uniformly to a continuous function on Re z > 0.

(2) (a) If n is odd, xn = 1 − 2n

n + 1→ −1 as n → ∞. If n is even, xn = 1 +

2n

n + 1→ 3 as n → ∞.

So limn→∞

xn = 3.

(b) limn→∞

xn = lim

√2

2, 0,−

√2

2,−1,−

√2

2, 0,

√2

2, 1, 0, . . .

= lim

1, 1, 1, . . .

= 1.

(3) (a) ak =

1 if k = n!0 if k 6= n! , lim

k→∞k√

|ak| = lim1, 1, 0, 0,0, 1, . . .

= 1, R =

11

= 1.

(b) R =1

limn→∞

n√

n + 2n=

1

limn→∞

n√

2n n

√n

2n+ 1

=12.

(4)∞Σ

n=0zn has radius of convergence R1 = 1,

∞Σ

n=0−zn has radius of convergence R2 = 1.

However∞Σ

n=0(1 − 1)zn =

∞Σ

n=00 has radius of convergence ∞ > R1 + R2.

(5) The center of the power series is at 0. Since limn→∞

1n

= 0, f(z) ≡ 1 + 0z + 0z2 + . . .. Then f ′(z) ≡ 0. So

f ′(0) 6> 0.

(6) The center of the power series is at 0. Observe that f(z) = 0 + 0z + z2 + 0z3 + . . . for z =1n

, n = 1, 2,

3, . . .. Since limn→∞

1n

= 0, f(z) ≡ z2. Similarly, f(− 1n

) =1n3

forces f(z) ≡ 0 + 0z + 0z2 − z3 + 0z4 + . . ..

However z2 6≡ −z3, so no such f(z).

(7) Set z =1n

, then f(z) =

1z2

1z2

+ 1=

11 + z2

for z =1n

, n = 1, 2, 3, . . .. Now limn→∞

1n

= 0. So f(z) =

∞Σ

n=0cnzn must be the power series of

11 + z2

about the center 0.

Since∞Σ

n=0wn =

11 − w

for |w| < 1,1

1 + z2=

11 − (−z2)

=∞Σ

n=0(−z2)n =

∞Σ

n=0(−1)nz2n, then f (k)(0) =

k!ck =

0 if k is odd(−1)k/2k! if k is even .

Solution to Practice Exercise Set III

(1) ∂v

∂x= −∂u

∂y= −2y ⇒ v = −2xy + C1(y)

∂v

∂y=

∂u

∂x= 2x ⇒ v = 2xy + C2(x)

⇒ not possible.

(2) Observe that∂u

∂xis a function of x and

∂v

∂yis a fucntion of y. Then

∂u

∂x=

∂v

∂ymust be a constant. Then

u(x) = ax + b, v(y) = ay + c, f(z) = az + (b + ic).

(3) If | Im z| <π

2, then z = x + iy (−π

2< y <

π

2) and ez = ex(cos y + i sin y). So Re ez = ez cos y > 0.

Conversely, if Re w > 0, then w = r cis θ with −π

2< θ <

π

2and w = ez, where z = ln r + iθ is the

infinite strip. Therefore, the range of ez for | Im z| <π

2is the right half plane Re w > 0.

w = e zz wπi/2

−πi/2

A horizontal line in | Im z| <π

2has equation Im z = c with

−π

2< c <

π

2. Its image is the set ez = ex+ic = ex(cos c +

i sin c) where −∞ < x < ∞, i.e. the semicircle with the originas center and radius ec.

(4)

31-1-3

It’s not possible to define log(z−1) continuously on C\[−1, 1] because arg(z−1)on the circle, say |z| = 3, cannot be made continuous. (If arg(3− 1) is defined,then going around |z| = 3 once and returning to z = 3 would force redefinitionof arg(3 − 1).)

Yes, it is possible to define log(

z + 1z − 1

)continuously on C \ [−1, 1] as

follows: define log(

z + 1z − 1

)= ln

∣∣∣∣z + 1z − 1

∣∣∣∣ + i Arg(

z + 1z − 1

). The only possible

place where this can be discontinuous is whenz + 1z − 1

is negative or 0 (where

Arg w is discontinuous). Howeverz + 1z − 1

≤ 0 ⇐⇒ −1 ≤ z ≤ 1. This segment

is removed!

(5) f = u + iv is analytic on G ⇐⇒ ∂u

∂x(x0, y0) =

∂v

∂y(x0, y0) and

∂u

∂y(x0, y0) = −∂v

∂x(x0, y0) for all (x0, y0)

in G and fx, fy continuous on G.

For (a0, b0) in G∗, f∗(a0, b0)f(a0,−b0) = u(a0,−b0) − iv(a0,−b0) = U (a0, b0) + iV (a0, b0). [That isthe real part of f∗ is U (a0, b0) = u(a0,−b0) and the imaginary part of f∗ is V (a0, b0) = −v(a0,−b0).]Clearly, f∗

x , f∗y are continuous on G∗ because fx, fy are continuous on G.

Finally, we check Cauchy-Riemann equations for f∗:∂U

∂x(a0, b0) =

∂u

∂x(a0,−b0) =

∂v

∂y(a0,−b0) =

∂V

∂y(a0, b0),

∂U

∂y(a0, b0) = −

(∂u

∂y(a0,−b0)

)=

∂v

∂x(a0,−b0) =

∂V

∂x(a0, b0). Therefore, f∗ is analytic

on G∗.

(6) (a) ∂v

∂x= −∂u

∂y= 6xy

∂v

∂y=

∂u

∂x= 3x2 − 3y2

⇒ v(x, y) = 3x2y + C1(y)v(x, y) = 3x2y − y3 + C2(x)

⇒ v(x, y) = 3x2y − y3 + C.

(b) ∂v

∂x= −∂u

∂y= e−y cos x

∂v

∂y=

∂u

∂x= −e−y sin x

⇒ v(x, y) = e−y sin x + C1(y)v(x, y) = e−y sin x + C2(x)

⇒ v(x, y) = e−y sin x + C.

(c) ∂v

∂x= −∂u

∂y=

−2y

x2 + y2

∂v

∂y=

∂u

∂x=

2x

x2 + y2

⇒v(x, y) = 2 arctan(

y

x) + C1(y)

v(x, y) = 2 arctan(y

x) + C2(x)

⇒ v(x, y) = 2 arctan(

y

x) + C.

(d) ∂v

∂x= −∂u

∂y=

y2 − (1 − x)2

((1 − x)2 + y2)2∂v

∂y=

∂u

∂x=

2(1 − x)((1 − x)2 + y2)2

⇒v(x, y) =

x − 1(1 − x)2 + y2

+ C1(y)

v(x, y) =x − 1

(1 − x)2 + y2+ C2(x)

⇒ v(x, y) =x − 1

(1 − x)2 + y2+ C.

(7) x = r cos θ, y = r sin θ,∂u

∂x=

∂v

∂y,

∂u

∂y= −∂v

∂x.

∂u

∂r=

∂u

∂x

∂x

∂r+

∂u

∂y

∂y

∂r=

∂u

∂xcos θ +

∂u

∂ysin θ.

1r

∂v

∂θ=

1r

(∂v

∂x

∂x

∂θ+

∂v

∂y

∂y

∂θ

)=

1r

(−∂u

∂y(−r sin θ) +

∂u

∂x(r cos θ)

)=

∂u

∂xcos θ +

∂u

∂ysin θ =

∂u

∂r.

∂v

∂r=

∂v

∂x

∂x

∂r+

∂v

∂y

∂y

∂r=

∂v

∂xcos θ +

∂v

∂ysin θ.

−1r

∂u

∂θ= −1

r

(∂u

∂x

∂x

∂θ+

∂u

∂y

∂y

∂θ

)= −1

r

(∂v

∂y(−r sin θ) − ∂v

∂x(r cos θ)

)=

∂v

∂xcos θ +

∂v

∂ysin θ =

∂v

∂r.

Solution to Practice Exercise Set IV

(1) (a) z 7→ i(1 − z

1 + z

)z 7→ 1

πLog z

T (z) =1π

Log(i(1 − z

1 + z

))

(b) z 7→√

i(1 − z

1 + z

)z 7→ z − 1

z + 1

T (z) =(√

i(1 − z

1 + z

)− i

) / (−

√i(1− z

1 + z

)+ i

)

(c) T (z)=

(√i( 1−z

1+z )−i

)/(−√

i( 1−z1+z )+i

)z 7→ z2

T (z) =

[(√i(1 − z

1 + z

)− i

) / (−

√i(1 − z

1 + z

)+ i

)]2

(d) z 7→(1 − z

1 + z

)2

z 7→ z − 1 z 7→ 1z

z 7→ 2z + 1

0 -1 -1 0 -1 1

T (z) = 2((1 − z

1 + z

)2

− 1)−1

+ 1 = −(z2 + 1

2z

)= −

12

(z +

1z

).

(2) Observe that T (∞) = 1, T (ia) = 0, T (ib) = ∞. Orient L3 by (∞, ia, ib). The right side of L3 isD ∪E ∪F . The right side of T (L3) (with respect to (1, 0,∞)) is U ∪V ∪W . The part of L3 from ∞ toia is mapped onto the segment from 1 to 0. So we have the correspondence F ↔ V , E ↔ U , D ↔ W ,C ↔ Y , B ↔ X, A ↔ Z.

Solution to Practice Exercise Set V

(1)

1

Observe that the range of f(z) is in the disk |w − 1| < 1, which is contained inthe slit plane C \ (−∞, 0) (Complex plane minus the negative real axis and 0).Now Log w is analytic on the slit plane. So Log f(z) is defined and analytic. Then∫

C

f ′(z)f(z)

dz =∫

C

(Log f(z))′ dz = Log f(z(b))−Log f(z(a)) = 0 because C is closed

(i.e. z(a) = z(b)).

(2) The unit circle (counterclockwise direction) is given by z(t) = eit = cos t + i sin t (0 ≤ t ≤ 2π),|dz| = |ieit| dt = dt, so

∫

|z|=1

|z − 1||dz| =∫ 2π

0

√(cos t − 1)2 + sin2 t dt =

∫ 2π

0

√2 − 2 cos t dt = 2

∫ 2π

0

∣∣∣∣sint

2

∣∣∣∣ dt = 8.

(3) Observe that∫

C

(f(z) + g(z)) dz =∫

C

f(z) dz +∫

C

g(z) dz and∫

C

αf(z) dz = α

∫

C

f(z) dz. So it suf-

fices to consider the special case P (z) = zn (the general case is obtained by using the linearity propertiesabove). C is given by z(t) = a + Reit, 0 ≤ t ≤ 2π, dz = iReit dt.

∫

C

P (z) dz =∫

C

zn dz =∫ 2π

0

(a + Re−it)niReit dt

=∫ 2π

0

(an + nan−1Re−it + · · ·+ Rne−int)iReit dt

=∫ 2π

0

nan−1iR2 dt (∗)

= −2πinan−1R2

= −2πiR2P ′(a).

(*) where we use the fact∫ 2π

0

eint dt =

∫ 2π

0

(cos nt + i sin nt) dt = 0 if n 6= 0∫ 2π

0

1 dt = 2π if n = 0.

(4) Since1z2

has the antiderivative −1z

in C \ 0,∫

C

1z2

dz = −1z

∣∣∣z=−1

z=1= 2.

(5) Suppose∫

|z|=1

f(z) dz =∫ 2π

0

f(eit)ieit dt = Reiθ. Then

∣∣∣∫

|z|=1

f(z) dz∣∣∣ = R =

∫ 2π

0

f(eit)iei(t−θ) dt = Re(∫ 2π

0

f(eit)(− sin(t − θ) + i cos(t − θ)) dt)

= −∫ 2π

0

f(eit) sin(t − θ) dt

≤∫ 2π

0

∣∣f(eit) sin(t − θ)∣∣ dt

≤∫ 2π

0

| sin(t − θ)| dt =∫ 2π

0

| sin t| dt = 4.

(6)

i

C4

C1

2C

C3

21-1-2 0

2iC1: z(t) = t, −2 ≤ t ≤ −1C2: z(t) = eit, t from π down to 0C3: z(t) = t, 1 ≤ t ≤ 2C4: z(t) = 2eit, 0 ≤ t ≤ π

∫

C1

z

zdz =

∫ −1

−2

1 dt = 1,∫

C2

z

zdz =

∫ 0

π

eit

e−itieit dt = i

∫ 0

π

e3it dt =e3it

3

∣∣∣0

π=

23.

∫

C3

z

zdz =

∫ 2

1

1 dt = 1,∫

C4

z

zdz =

∫ π

0

2eit

2e−it2ieit dt = 2i

∫ π

0

e3it dt =23e3it

∣∣∣π

0= −4

3.

∫

C

z

zdz = 1 +

23

+ 1 − 43

=43.

Solution to Practice Exercise Set VI

(1)

a a a

a

n(C, a) = −1 n(C, a) = 1 n(C, a) = −1 n(C, a) = 0

(2) Log(w + 1) = w − w2

2+

w3

3− w4

4+ . . . for w near 0. cos w − 1 = −w2

2!+

w4

4!− w6

6!+ . . . for all w.

For z near 0,z Log(z49 + 1)

cos z25 − 1=

z(z49 − z98

2+ . . .)

−z50

2+

z100

24− . . .

=1 − z49

2+ . . .

−12

+z50

24+ . . .

.

Therefore, limz→0

exp(z Log(z49 + 1)

cos z25 − 1) = e−2.

(3)f

1

T

Let f(z) be an entire function, whose range lies in the

right half plane. Let T (z) =z − 1z + 1

, then T f(z) is

entire and has range in the unit disk, i.e. |T f(z)| ≤ 1for all z. By Liouville’s theorem, T f(z) ≡ constant.Therefore f ≡ T−1 T f ≡ constant.

(4) Let f(z) = a0 + a1z + . . . + anzn be a polynomial such that |z| ≤ 1 ⇒ |f(z)| ≤ 1. By corollary (1),

|ak| =∣∣k!f (k)(0)

∣∣ =

∣∣∣∣∣∣∣1

2πi

∫

|z|=1

f(z)zk+1

dz

∣∣∣∣∣∣∣≤ 1

2π

11︸︷︷︸M

2π︸︷︷︸L

= 1.

(5) Let f(z) sin z and g(z) =

sin z

zif z 6= 0

1 if z = 0, then by Cauchy’s integral formula,

∫

|z|=1

sin z

zdz =

∫

|z|=1

f(z)z − 0

dz = 2πif(0) = 0 and∫

|z|=1

sin z

z2dz =

∫

|z|=1

g(z)z

dz = 2πig(0) = 2πi.

(6) (a) Jf (x, y) =∂u

∂x

∂v

∂y− ∂u

∂y

∂v

∂x=

(∂u

∂x

)2

+(

∂v

∂x

)2

=∣∣f ′(z)

∣∣2 because f ′(z) =∂u

∂x+ i

∂v

∂xand by the

Cauchy-Riemann equations.

(b)∫∫

D

zmzn dAz=reiθ

=∫ 2π

0

∫ 1

0

rmeimθrne−inθr dr dθ =∫ 2π

0

ei(m−n)θ dθ

∫ 1

0

rm+n+1 dr

=ei(m−n)θ

i(m − n)

∣∣∣∣2π

θ=0

1m + n + 2

= 0.

(c) area of G =∫∫

D

∣∣Jf (x, y)∣∣ dx dy

(a)=

∫∫

D

∣∣f ′(z)∣∣2 dA =

∫∫

D

∣∣∣∣∞Σ

n=1ncnzn−1

∣∣∣∣2

dA

=∫∫

D

(∞Σ

m=1mcmzm−1

)(∞Σ

n=1ncnzn−1

)dA

=∫∫

D

(∞Σ

m=n=1n2|cn|2|z|2n−2 + Σ

m 6=nmncmcnzm−1zn−1

)dA

(b)=

∞Σ

n=1n2|cn|2

∫ 2π

0

∫ 1

0

r2n−2r dr dθ =∞Σ

n=1n2|cn|22π

12n

= π∞Σ

n=1n|cn|2.

Solution to Practice Exercise Set VII

(1) Suppose f(1n

) =1

n + 1for all positive integer n. Since

1n

has limit point 0, which is inz: |z| ≤ 1

,

so the identity theorem implies f(z) =1

1z

+ 1=

z

1 + z. However, this is not analytic at z = −1, a

contradiction.

(2) Define g(z) =f(z)z2

for 1 ≤ |z| ≤ 2. Then g is continuous on 1 ≤ |z| ≤ 2 (which is closed and bounded)and analytic on 1 < |z| < 2. So the maximum modulus theorem implies that for 1 ≤ |z| ≤ 2,

|f(z)||z|2 = |g(z)| ≤ max

|w|=1,2|g(w)| = max

|w|=1,2

|f(w)||w|2 = 1.

(3) (a) Let Γ be the boundary of a rectangle. For a fixed t, sin zt is entire, so∫

Γ

sin zt = 0. Then∫

Γ

f(z) dz =∫

Γ

∫ 1

0

sin zt

tdt dz =

∫ 1

0

∫

Γ

sin zt

tdz dt =

∫ 1

0

1t

(∫

Γ

sin zt dz

)dt = 0.

(Details: Since limw→0

sin w

w= 1,

sin w

wis bounded for |w| ≤ |z|. Suppose

∣∣∣∣sin w

w

∣∣∣∣ ≤ M for |w| ≤ |z|.

Then∣∣∣∣sin zt

t

∣∣∣∣ ≤ M |z| for 0 < t ≤ 1 and∫

Γ

∫ 1

0

∣∣∣∣sin zt

t

∣∣∣∣ dt dz < ∞. These imply f(z) is continuous

and interchange of integration is possible.)

(b) sin zt =∞Σ

n=0

(−1)n(zt)2n+1

(2n + 1)!. For a fixed z0, 0 < t ≤ 1, tz0 lies in the closed disk B(0, |z0|). So

∞Σ

n=0

(−1)nz2n+10 t2n

(2n + 1)!converges uniformly to

sin z0t

tas a function of t (power series converges uni-

formly in closed subdisks of domain of convergence). Then

f(z0) =∫ 1

0

sin z0t

tdt =

∫ 1

0

∞Σ

n=0

(−1)nz2n+10 t2n

(2n + 1)!dt =

∞Σ

n=0

(−1)nz2n+10

(2n + 1)!

∫ 1

0

t2n dt =∞Σ

n=0

(−1)nz2n+10

(2n + 1)!(2n + 1).

(4) Sincez: |z| ≤ 1

is closed and bounded, let M = max

|z|≤1|f(z)|. Also let T (z) be a Mobius transformation

mapping UHP =z: Im z ≥ 0

onto

z: |z| ≤ 1

and R onto

z: |z| = 1

(e.g. T (z) =

z − i

z + i). Then

f T (z) is continuous on UHP, analytic on UHF =z: Im z > 0

and real-valued on R. By the Schwarz

reflection principle, f T (z) can be extended to an entire function.So |f T (z)| ≤ M for all z, Liouville’s theorem implies f T (z) is constant. Therefore, f(z) is a constant.

(5) Let w1, w2 be two distinct fixed points of f in D. Let T (z) =z + w1

1 + w1z, then T is one-to-one map from

D onto D. So there is z0 such that T (z0) = w2. The function T−1 f T (z) is an analytic function fromD onto D and T−1 f T (0) = 0 and T−1 f T (z0) = z0. The equality case of Schwarz lemma impliesT−1 f T (z) ≡ eiθz. Using T−1 f T (z0) = z0, we get eiθ = 1. So T−1 f T (z) = z. Therefore,f T (z) = T (z) for all z ∈ D. Therefore f(w) ≡ w for all w ∈ D.

(6) Since f(z) is entire, f(z) =∞Σ

n=0anzn for all z, where an =

f (n)(0)n!

. Since f(z) is real on the real

axis, Schwarz reflection principle and the identity theorem imply f(z) = f(z) for all z. So∞Σ

n=0anzn =

∞Σ

n=0anzn =

∞Σ

n=0anzn. By the uniqueness of power series, an = an for all n.

Define g(z) = if(iz). Since f(z) is imaginary on the imaginary axis, g(z) is real on the real axis.

Now g(z) =∞Σ

n=0in+1anzn. By the above argument, we have in+1an = in+1an. If n is even, then

in+1an = −in+1an. This implies an = an = −an for even n. So an = 0 for all even n. Thereforef(z) =

∞Σ

k=0a2k+1z

2k+1 is odd.

Alternatively, f(z) = f(z) for all z and g(w) = g(w) for all w⇒ f(z) = f(z) for all z and if(iw) = if(iw) = −if(iw) for all w⇒ f(z) = f(z) and f(z) = −f(−z) for all z (set z = iw)⇒ f(z) = f(z) = −f(−z) for all z⇒ f(u) = −f(−u) for all u (set u = z)

(7) f(z) has finitely many roots α1, . . . , αn (repeated according to multiplicities) in the open unit disk andno roots on the unit circle since |f(z)| = 1 for |z| = 1 (otherwise the roots have a limit point in the

closed disk forcing f(z) ≡ 0). By the theorem following the orientation principle,∣∣∣∣

z − αj

1 − αjz

∣∣∣∣ = 1 for

|z| = 1. The function g(z) =f(z)

n

Πj=1

z − αj

1 − αjz

is analytic on D. (This is clear for z 6= α1, . . . , αn because

n

Πj=1

z − αj

1 − αjz6= 0. If αk is a root of multiplicity m, then lim

z→αk

f(z)(z − αk)m

=f (m)(αk)

m!6= 0 (by l’Hopital’s

rule) and m of the αk’s equal αk so that limz→αk

g(z) exists and is nonzero, which is used to define g(αk).

It follows that g is analytic at αk.) By the maximum modulus theorem, for |z| ≤ 1,

|g(z)| ≤ max|w|=1

|g(w)| = max|w|=1

|f(w)|∣∣∣∣n

Πj=1

w − αj

1 − αjz

∣∣∣∣= max

|w|=1|f(w)| = 1.

Since g has no root in D =z: |z| < 1

, the minimum modulus theorem implies, for |z| ≤ 1, |g(z)| ≥

min|w|=1

|g(w)| = 1. Therefore |g(z)| = 1 for |z| ≤ 1. Then g(z) = eiθ because |g(z)| = 1. Hence

f(z) = eiθn

Πj=1

z − αj

1− αjzby the identity theorem. Since f is entire, αj = 0 (otherwise f is not defined at

1αj

). Therefore f(z) ≡ eiθzn.

Solution to Practice Exercise Set VIII

(1)

......

a1 a2 an

Introducing n+1 cross-cuts as shown below and applying Cauchy’s theoremto the upper and lower simple closed curves, we get (after cancelling theintegrals over the cross-cuts)

0 =∫

Γ

f(z) dz −n

Σj=1

∫

Cj

f(z) dz.

The result follows.

(2) (a) Since1

z4 + z2=

1z2(z + i)(z − i)

, there are a pole of order 2 at 0, a pole of order 1 at i and −i,

respectively.(b) Since cot z =

cos z

sin zand lim

z→nπ(z − nπ)

cos z

sin z= 1, lim

z→nπ(z − nπ)2

cos z

sin z= 0, there is a pole of order 1

at nπ, where n is any integer.

(c) The isolated singularities are at 0 and 1. There is a pole of order 1 at 1. Sincee1/z

z − 1doesn’t have

a limit as z → 0, so 0 is an essential singularity.

(d) The isolated singularities are at all integers. Since limz→1

z2 − 1sinπz

= − 2π

, limz→−1

z2 − 1sin πz

=2π

,

limz→nπ,n6=±1

(z − nπ)z2 − 1sinπz

= ±n2π2 − 1π

6= 0, there are removable singularities at 1 and −1, respec-

tively, and pole of order 1 at all other integers.

(3) (a)1

z2 − 4=

14(z − 2)

1

(1 +z − 2

4)

=1

4(z − 2)(1 − z − 2

4+ (

z − 24

)2 − . . .) =∞Σ

k=−1

(z − 2)k

4k+2, where

∣∣∣∣z − 2

4

∣∣∣∣ < 1.

(b)1

z2 − 4=

1

z2(1 − 4z2

)=

1z2

(1 +4z2

+16z4

+ . . .) =∞Σ

k=1

4k−1

z2k, where

∣∣∣∣4z2

∣∣∣∣ < 1.

(4) Suppose1

(n + 1)π< r <

1nπ

. Since sin1z

=1z− 1

3!z3+

15!z5

− . . . for1

(n + 1)π< |z| <

1nπ

,∫

|z|=r

sin1z

dz = 2πa−1 = 2πi. (This is the same if1π

< r < ∞.)

(5) Let f(z) =∞Σ

k=−∞akzk for C \

0

=z: 0 < |z| < ∞

. For k = 0,

|ak| ≤12π

∫

|z|=r

|f(z)||z|k+1

|dz| ≤

√r +

1√r

rk=

r + 1rk+ 1

2→ 0

as r → ∞ if k > 0as r → 0 if k < 0

Therefore, f ≡ a0.

(6) If f has a pole of order k at 0, then f(z) =a−k

zk+

a−k+1

zk−1+ . . . =

a−k

zk(1 +

a−k+1

a−k+ . . .

︸ ︷︷ ︸=g(z)

) =a−kg(z)

zkfor

0 < |z| < ε, where g is analytic on |z| < ε, g(0) = 1 and a−k 6= 0. Let z = k√

ta−k with t a real variable,then lim

t→0+ef(z) = e+∞ = +∞ and lim

t→0−ef(z) = e−∞ = 0. So 0 is an essential singularity of ef(z).

(7) (i) f(z) entire ⇒ f(z) =∞Σ

k=0akzk. f(

1z) has a pole of order N ⇒ f(

1z) =

0Σ

k=N

ak

zk. So f(z) =

NΣ

k=0akzk

is a polynomial.(ii) Since C ∪

∞

is closed and bounded, and the poles are isolated, there are only finitely many

poles. Let the finite poles be p1, . . . , pk and the respectiveorders be N1, . . . , Nk. Let P (z) =(z − p1)N1 . . . (z − pk)Nkf(z), then P is entire. Since ∞ is either a removable singularity or a poleof f(z), the same is true for P . If ∞ is a removable singularity, then P is analytic at ∞ and P (∞)is a complex number. This implies P is bounded on C, so by Liouville’s theorem, P (z) ≡ P (∞)

and f(z) =P (∞)

(z − p1)N1 . . . (z − pk)Nkis a rational function. If ∞ is a pole, then by part (i), P (z)

is a polynomial, then f(z) =P (z)

(z − p1)N1 . . . (z − pk)Nkis a rational function.

Solution to Practice Exercise Set IX

(1) Let g(z) = f(1z), then 0 is an isolated singularity of g. If it is removable, then f has a removable

singularity at ∞ and f will be bounded, forcing it to be a constant. If 0 is a pole of g, then g(z) =∞Σ

k=−Nakzk near 0 and f(z) =

N

Σj=−∞

a−jzj . Since f is analytic at 0, a−j = 0 for j < 0, then f is

a polynomial. By the fundamental theorem of algebra, the image of the plane under f is the wholeplane (hence the image is dense). If 0 is an essential singularity of g, then Casorati-Weierstrass theoremimplies the image of the plane under g (or f) is dense in the plane.

(2) Suppose f has more than one roots. Let R be so large that there are more than one roots inside Γ, thecircle |z| = R. Since f(R), f(−R) are the only real valued on f Γ, the curve

f(Reiθ): 0 < θ < π

andf(Reiθ): π < θ < 2π

lie entirely on the upper or lower half plane. Then n(f Γ, 0) = 0 or 1.

Therefore, f can have at most one root by the Argument Principle, a contraction.(Note f 6≡ 0, so we may assume f(R) 6= 0 and f(−R) 6= 0 by the identity theorem.)

(3) No, otherwise the Argument Principle implies f has poles inside the unit circle.

(4) Since f is one-to-one on Γ, f Γ has no self-intersections, so it is a simple closed curve. For w 6∈ f Γ,let g(z) = f(z)−w, then n(g Γ, 0) = n(f Γ, w) = 1 or 0 depends whether w is inside or outside f Γ.If z0 is inside Γ, then f(z0) is not on f Γ by the open mapping theorem. Now z0 is a root ofg(z) = f(z)−f(z0), hence n(g Γ, 0) = 1. Then g has no other root. This implies f is one-to-one insideΓ.

(5)

γ2

γ1

R

-iR

iR

f γ2

f o γ2

o

Consider the contour on the left with R large. On γ1, z(t) = Reit,−π

2≤ t ≤ π

2, f(z) =

z2n + α2z2n−1 + β2 ≈ z2n, so ∆γ1 arg f(z) = 2nπ. On γ2, z(t) = it, wher t decreasesfrom R to −R.

If n is odd, f(it) = −t2n + β2 + iα2t2n−1, f(iR) ≈ −∞ + i0+, f(−iR) ≈ −∞ + i0−.(Note f(i|β| 1

n ) = iα2|β|2n−1

n , f(0) = β2, f(−i|β| 1n ) = −iα2|β|

2n−1n .)

A sketch shows ∆γ2 arg f(z) = −2π. So f has(2n − 2)π

2π= n − 1 roots inside the

contour for R large, so f has n − 1 roots with positive real parts.

If n is even, f(it) = t2n + β2

︸ ︷︷ ︸>0

−iα2t2n−1, f(iR) ≈ +∞− i0+, f(−iR) ≈ +∞− i0−.

A sketch shows ∆γ2 arg f(z) = 0. So f has2nπ

2π= n roots inside the contour for R

large, so f has n roots with positive real parts.

(6) Let f(z) = ez − azn and g(z) = azn, then for |z| = 1, |f(z) + g(z)| = |ez| = eRe z ≤ e1 < a = |azn| =|g(z)| < |f(z)| + |g(z)|. So by Rouche’s theorem f(z) has n roots inside the unit circle (because g hasn roots inside the unit circle).

Solution to Practice Exercise Set X

(1)

Re

0 R

2πi/n

γ1

− γ2

By residue theorem,

(∫ R

0

+∫

γ1

+∫

γ2

) dz

1 + zn

(*)= 2πi Resz=eπi/n

( 11 + zn

)= 2πi

1n(eπi/n)n−1

.

On γ1, z = Reiθ, 0 ≤ θ ≤ 2π

n,∣∣∣∣∫

γ1

dz

1 + zn

∣∣∣∣ ≤ ML =1

Rn − 12πR

n→ 0 as R → +∞.

On γ2, z = xe2πi/n, 0 ≤ θ ≤ 2π

n,∫

−γ2

dz

1 + zn= −

∫ R

0

e2πi/n

1 + xndx = −e2πi/n

∫ R

0

dx

1 + xn.

Letting R → +∞, (*) ⇒ (1 − e2πi/n)∫ ∞

0

dx

1 + xn= 2πi

1n(eπi/n)n−1

=2πi

−neπi/n.

Therefore,∫ ∞

0

dx

1 + xn=

2πi

n(eπi/n − e−πi/n)=

π

n sinπ

n

.

(2)

R-R

CR e2iz = 1 + 2iz +(2iz)

2!+ · · ·, e2iz − 1 − 2iz

z2= −2− 4i

3z + · · ·. (At 0,

e2iz − 1 − 2iz

z2

has a removable singularity, where we assign the value −2 to make it analytic.)Consider the contour shown.

By Cauchy’s theorem,

(∫ R

−R

+∫

CR

)e2iz − 1 − 2iz

z2dz

(*)= 0.

∫ 0

−R

e2iz − 1 − 2iz

z2dz

x=−z=∫ R

0

e−2ix − 1 + 2ix

x2dx.

On CR,∣∣∣∣e2iz − 1

z2

∣∣∣∣ ≤ e−2 Im z + 1R2

≤ 2R2

. So∣∣∣∣∫

CR

e2iz − 1z2

dz

∣∣∣∣ ≤ 2R2

πR → 0 as R → +∞. Then

limR→+∞

∫

CR

e2iz − 1 − 2iz

z2dz = lim

R→+∞

(−2i

∫

CR

dz

z

)= 2π.

Taking R → +∞, (*) implies∫ ∞

0

e2ix + e−2ix − 2x2

dx + 2π = 0. Since e2ix + e−2ix − 2 = −4 sin2 x, we

get∫ ∞

0

sin2 x

x2dx =

π

2.

(3)

-r r R-R

CR

Cr

iRecall log z = ln |z|+ i arg z, 0 ≤ arg z ≤ π and so log i = i

π

2.

By residue theorem,(∫ −r

−R

−∫

Cr

+∫ R

r

+∫

CR

) log z

z2 + 1dz

(*)= 2πiResz=i

log z

z2 + 1= i

π2

2.

∣∣∣∣∫

Cr

log z

z2 + 1dz

∣∣∣∣ ≤| ln r| + π

1 − r2︸ ︷︷ ︸

M

πr︸︷︷︸L

→ 0 as r → 0+,∣∣∣∣

log z

z2 + 1dz

∣∣∣∣ ≤| lnR| + π

R2 − 1πR → 0 as R → +∞.

∫ −r

−R

log z

z2 + 1dz =

∫ R

r

lnx + iπ

x2 + 1dx. Letting r → 0+, R → +∞, (*) ⇒ 2

∫ ∞

0

lnx

x2 + 1dx+iπ

∫ ∞

0

dx

x2 + 1︸ ︷︷ ︸π/2

=

iπ

2. Therefore,

∫ ∞

0

ln x

x2 + 1dx = 0.

(4)

R

R+i

-R

-R+i

−γ3

γ1

γ2Consider f(z) = e−z2

. By Cauchy’s theorem,(∫ R

−R

+∫

γ1

+∫

γ2

−∫

γ3

)e−z2

dz(*)= 0.

∣∣∣∣∣∣

∫

γ1,γ3

e−z2dz

∣∣∣∣∣∣≤ e−R2+1

︸ ︷︷ ︸M

→ 0 as R → +∞.∫

γ2

e−z2dz =

∫ −R

R

e−(x+i)2 dx = −∫ R

−R

e−x2−2ix+1 dx.

Letting R → +∞, (*) implies∫ ∞

−∞e−x2

dx− e

∫ ∞

−∞e−x2

(cos 2x− i sin 2x) dx = 0. Since∫ ∞

−∞e−x2

dx =

√π, taking the real part, we get

∫ ∞

0

e−x2cos 2x dx =

√π

2e.

(5)

0 Rπ/4

γ1

−γ2

Consider f(z) = eiz2. By the Cauchy’s theorem,

(∫ R

0

+∫

γ1

−∫

γ2

)eiz2

dz(*)= 0. On

γ1, z = Reiθ, 0 ≤ θ ≤ π

4,∣∣eiz2 ∣∣ = e−R2 sin 2θ ≤ e

−4R2θπ by Jordan’s inequality.

So∣∣∣∣∫

γ1

eiz2dz

∣∣∣∣ ≤∫ π

4

0

e−4R2θ

π R dθ =π

4R(1 − e−R2

) → 0 as R → +∞.

On γ2, z = xeiπ4 , 0 ≤ x ≤ R,

∫

γ2

eiz2dz =

∫ R

0

e−x2(ei π

4 dx). Letting R → +∞, (*) ⇒∫ ∞

0

eix2dx −

ei π4

∫ ∞

0

e−x2dx = 0. Since

∫ ∞

0

e−x2dx =

√π

2, taking real and imaginary parts, we get

∫ ∞

0

cos x2 dx =√

π

2cos

π

4=

√2π

4,∫ ∞

0

sin x2 dx =√

2π

4.

(6)∫ π/2

0

dθ

1 + sin2 θ=

14

∫ 2π

0

dθ

1 + sin2 θ=

14

∫

|z|=1

−i

1 − 14(z − 1

z)2

dz

z=

∫

|z|=1

iz

(z2 − 2z − 1)(z2 + 2z − 1)dz

=i

4

( ∫

|z|=1

dz

z2 − 2z − 1−

∫

|z|=1

dz

z2 + 2z − 1

)= −π

2

(Res

z=1−√

2

1z2 − 2z − 1︸ ︷︷ ︸

= 1−2

√2

− Resz=−1+

√2

1z2 + 2z − 1︸ ︷︷ ︸

= 12√

2

)=

π

2√

2

(7) (a) Res(f,∞) =1

2πi

∫

|z|=R

f(z) dz = −n

Σj=1

Res(f, aj) (clockwise orientation), where the last equality fol-

lows from residue theorem.(b) Suppose f(z) = · · ·+

a−k

zk+ · · ·+

a−1

z+ a0 + a1z + · · · on r < |z| < ∞, then

− 1z2

f(1z) = − 1

z2(· · ·+ a−kzk + · · ·+ a−1z + a0 +

a1

z+ · · ·) on 0 < |z| < 1

r= · · · − a−kzk−2 − · · · − a−1

z− a0

z2− a1

z3− · · ·

.

So Res(− 1z2

f(1z), 0) = −a−1 = Res(f,∞).

(c) Let f(z) =1

sin(1z), then

∫

|z|=1

f(z) dz = −2πi Res(f,∞) = −2πi Res(− 1z2

f(1z), 0) (counterclockwise

orientation). Now − 1z2

f(1z) = − 1

z2(z − z3

6+ · · ·)

= − 1z3

( 1

1 − z2

6+ · · ·

)= − 1

z3(1 +

z2

6+ · · ·) =

− 1z3

− 16z

− · · ·. So∫

|z|=1

1

sin(1z)

dz = −2πi(−16) =

πi

3(counterclockwise orientation).

Alternatively,∫

|z|=1

1

sin(1z)

dz =∫

|z|=1

11z− 1

6z3+ · · ·

dz =∫

|z|=1

(z +16z

+ · · ·) dz =162πi =

πi

3,

where all orientations are counterclockwise and the last integrand converges uniformly on |z| = 1.