Practice Exercise Set I * – Problems may be diﬃcult. 1makyli/3043/exer.pdf · Practice Exercise...

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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
αk2n
Σk=1
βk2 − Σ1≤k<j≤n
∣∣αkβj − αjβk
∣∣2 for the case n = 2.
(This implies the CauchySchwarz inequality∣∣∣∣
n
Σk=1
αkβk
∣∣∣∣2
≤√
n
Σk=1
αk2√
n
Σk=1
βk2.)
(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 xaxis. 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 CauchyRiemann 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
11 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 − 1dz, 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 realvalued 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
2112 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 onetoone 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=1ncn2.
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 forz = 2. Prove that f(z) ≤ z2 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 realvalued 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 orientation into the unit circle with clockwise orientation?
*(4) If f is analytic on and inside a simple closed curve Γ, and f is onetoone on Γ, then f is onetooneinside 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 semicircle. ]
(3) Find∫ ∞
0
ln x
x2 + 1dx. [ Hint: Use the contour
r r RR
. ]
(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 − 12 = 4z + 12 = 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−1z +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 figureeight 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.
21/21/22
 45
o45
o
Alternatively, for z = r cis θ, (r2 cos 2θ−1)2+(r2 sin 2θ)2 = z2−12 = 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 − γ2)
=Im(αβ − γβ − αγ + γ2)
β − γ2 = 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
∣∣∣∣ ≤ zn
1 − z2n=
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
∣∣∣∣ ≥zn
1 + z2n=
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+α2β22 = (α1β1+α2β2)(α1β1+α2β2) = α12β12+α1β2α2β1+α2β1α1β2+ α22β22.R.H.S. = (α12+ α22)(β12+ β22)−α1β2−α2β12 = α12β12+ α12β22+ α22β12+ α22β22−(α1β2 −α2β1)(α1β2 −α2β1) = α12β12 + α12β22 + α22β12 + α22β22− (α12β22−α1β2α2β1 −α2β1α1β2 + α22β22).
Alternative solution for n ∈ N, n ≥ 2 by Chow ChakOn.
αkβj − αjβk2 = (αkβj − αjβk)(αkβj − αjβk) = αk2βj2 − αkβkαjβj − αjβjαkβk + αj2βk2.nΣ
k=1
nΣ
j=1αkβj − αjβk2 =
nΣ
k=1
nΣ
j=1(αk2βj 2 − αkβkαjβj − αjβjαkβk + αj2βk2).
Sincen
Σk=1
n
Σj=1
αkβj − αjβk2 = Σ1≤k=j≤n
αkβj − αjβk2 + Σ1≤k<j≤n
αkβj − αjβk2 + Σ1≤j<k≤n
αkβj − αjβk2
= 2 Σi≤k<j≤n
αkβj − αjβk2
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 −αjβk2 = 2n
Σk=1
αk2n
Σk=1
βk2−2∣∣∣∣
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 − [a0zn+1 + (a1 − a0)zn + · · ·+ (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 z9 = 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, forcingz9 > 1, a contradiction.If z > 1, then x2 + y2 > 1 and 121 + 220y + 100y2 + 100x2 < 121x2 + 121y2 + 220y + 100, forcingz9 < 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 − r1i − 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 − ri − r). So there must be a pair of complex roots
r = x+iy and r = x−iy such that i−ri−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 Mtest,
∞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)
3113
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 CauchyRiemann 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 − 1dz =∫ 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
2112 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
CauchyRiemann 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=1n2cn2z2n−2 + Σ
m 6=nmncmcnzm−1zn−1
)dA
(b)=
∞Σ
n=1n2cn2
∫ 2π
0
∫ 1
0
r2n−2r dr dθ =∞Σ
n=1n2cn22π
12n
= π∞Σ
n=1ncn2.
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)z2 = g(z) ≤ max
w=1,2g(w) = max
w=1,2
f(w)w2 = 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≤1f(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 realvalued 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 onetoone 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) ≤ maxw=1
g(w) = maxw=1
f(w)∣∣∣∣n
Πj=1
w − αj
1 − αjz
∣∣∣∣= max
w=1f(w) = 1.
Since g has no root in D =z: z < 1
, the minimum modulus theorem implies, for z ≤ 1, g(z) ≥
minw=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 crosscuts as shown below and applying Cauchy’s theoremto the upper and lower simple closed curves, we get (after cancelling theintegrals over the crosscuts)
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)zk+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 CasoratiWeierstrass 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 onetoone on Γ, f Γ has no selfintersections, 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 onetoone 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(iβ 1
n ) = iα2β2n−1
n , f(0) = β2, f(−iβ 1n ) = −iα2β
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)
RR
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 RR
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.