1. NCU Math, Spring 2014: Complex Analysis Homework Solution 3
Text Book: An Introduction to Complex Analysis
Problem. 21.2
Sol:
(a)
Since | ij
(j+1)2 | ≤ 1
(j+1)2 for all j ∈ N ∪ {0} and
∑∞
j=0
1
(j+1)2 < ∞, we can get
∑∞
j=0
ij
(j+1)2 converges by Q9 in
Chapter 21.
(b)
Since
(1 + 3i)j
5j
=
|1 + 3i|
j
5j
≤
(1 + |3i|)
j
5j
=
(
4
5
)j
for all j ∈ N ∪ {0} and
∑∞
j=0
(4
5
)j
< ∞, we can get
∑∞
j=0
(1+3i)j
5j converges by Q9 in Chapter 21.
(c)
Since
Sm ≡
m∑
j=0
(cos
jπ
5
+ i sin
jπ
5
)
=
m∑
j=0
ei jπ
5
=
1 − e
i(m+1)π
5
1 − e
iπ
5
for all m ∈ N and limm→∞ e
i(m+1)π
5 doesn't exist, we can get that
∑∞
j=0(cos jπ
5 + i sin jπ
5 ) diverges.
(d)
Since
2ij
5 + ij2
=
2
√
25 + j4
≤
2
j2
1

2. for all j ∈ N and
∑∞
j=1
2
j2 < ∞, we can get
∑∞
j=1
2ij
5+ij2 converges by Q9 in Chapter 21. Thus
∑∞
j=0
2ij
5+ij2 =
2
5 +
∑∞
j=1
2ij
5+ij2 converges.
Problem. 21.4
Sol:
Because
Sm ≡
m∑
j=0
rj
(cos jθ + i sin jθ)
=
m∑
j=0
rj
eijθ
=
{
1−r(m+1)
ei(m+1)θ
1−reiθ if 0 < r < 1,
1 if r = 0,
for all m ∈ N and
lim
m→∞
r(m+1)
ei(m+1)θ
= 0
for all r ∈(0,1), we have
∑∞
j=0 rj
(cos jθ + i sin jθ) converges for all 0 ≤ r < 1 and
∞∑
j=0
rj
(cos jθ + i sin jθ) =
{
1
1−reiθ if 0 < r < 1,
1 if r = 0.
(a)
Since for 0 < r < 1
1
1 − reiθ
=
1 − re−iθ
(1 − reiθ) · (1 − re−iθ)
=
1 − r cos θ + ir sin θ
1 − 2r cos θ + r2
=
1 − r cos θ
1 − 2r cos θ + r2
+ i
r sin θ
1 − 2r cos θ + r2
and
rj
(cos jθ + i sin jθ) = rj
cos jθ + irj
sin jθ,
we can get that
∑∞
j=0 rj
cos jθ = 1−r cos θ
1−2r cos θ+r2 and
∑∞
j=0 rj
sin jθ = r sin θ
1−2r cos θ+r2 by virtue of Q5 in Chapter 21. For
r = 0, we have
∑∞
j=0 rj
cos jθ = 1 = 1−r cos θ
1−2r cos θ+r2 . Thus
∑∞
j=0 rj
cos jθ = 1−r cos θ
1+r2−2r cos θ for all 0 ≤ r < 1.
2

3. Problem. 21.5
Sol:
Since limn→∞ zn = z, for all ϵ > 0, we can nd m ∈ N so that when n ≥ m, |zn − z| ϵ
2 . For k = 0, · · · , m − 1,
we have limn→∞
zk−z
n+1 = 0. So we can nd l ∈ N such that when n ≥ l, |zk−z
n+1 | ϵ
2m for all k = 0, · · · , m − 1.
We dene N = max{m.l}. Then when n ≥ N = max{m.l}, we have
|
z0 + z1 + · · · + zn
n + 1
− z|
≤
n∑
k=0
|
zk − z
n + 1
|
=
m−1∑
k=0
|
zk − z
n + 1
| +
n∑
k=m
|zk − z|
n + 1
m−1∑
k=0
ϵ
2m
+
n∑
k=m
ϵ
2(n + 1)
ϵ
2
+
ϵ
2
=ϵ.
Thus we can get that limn→∞
(z0+z1+···+zn)/(n+1) = z.
Problem. 22.1
Sol:
(a)
Since
zj
(j + 1)(j + 2)
≤
1
(j + 1)(j + 2)
for |z| ≤ 1, j ∈ N ∪ {0} and
∑∞
j=0
1
(j+1)(j+2) = 1 converges, we have
∑∞
j=0 fj(z) converges uniformly on |z| ≤ 1 by
Theorem 22.4. So
∑∞
j=0 fj(z) converges pointwise on |z| ≤ 1.
(b)
Since
1
(z + j + 1)2
=
1
|z + j + 1|
2
=
1
(Re(z) + j + 1)2 + (Im(z))2
≤
1
(j + 1)2
3

4. for Re(z) 0, j ∈ N ∪ {0} and
∑∞
j=0
1
(j+1)2 converges, we have
∑∞
j=0 fj(z) converges uniformly on Re(z) 0 by
Theorem 22.4. So
∑∞
j=0 fj(z) converges pointwise on Re(z) 0.
(c)
Since
1
(1 + j2 + z2)
=
1
|1 + j2 + z2|
≤
1
||1 + j2| − |z2||
≤
1
1 + j2 − 22
=
1
j2 − 3
for 1 |z| 2, j ∈ N − {1} and
∑∞
j=2
1
j2−3 converges, we have
∑∞
j=2 fj(z) converges uniformly on 1 |z| 2 by
Theorem 22.4. This implies that
∑∞
j=0 fj(z) = f0(z) + f1(z) +
∑∞
j=2 fj(z) converges uniformly on 1 |z| 2. So
∑∞
j=0 fj(z) converges pointwise on 1 |z| 2.
(d)
Fix z ∈ C with 0 |z| ≤ 1. Then for j 1
|z|2 , we have
1
(1 + j2z2)
=
1
|1 + j2z2|
≤
1
||j2z2| − 1|
=
1
j2|z|2 − 1
.
It is easy to see that
∑
j 1
|z|2
1
j2|z|2−1 converges. This implies that
∑
j 1
|z|2
1
(1+j2z2) converges by Q9 in Chapter 21.
So
∑∞
j=0
1
(1+j2z2) converges. If z = 0, fj(z) = 1 for all j ∈ N ∪ {0}. Then
∑∞
j=0 fj(0) diverges. Thus
∑∞
j=0 fj(z)
converges pointwise on 0 |z| ≤ 1 and diverges at z = 0.
Because
∑∞
j=0 fj(z) converges pointwise on 0 |z| ≤ 1 and diverges at z = 0, we can get that
∑∞
j=0 fj(z) does
not converge uniformly on |z| ≤ 1.
Problem. 22.2
Sol:
No! fn does not converge uniformly on the unit disk B(0, 1).
First, for xed z ∈ C with 0 |z| ≤ 1, we have
fn(z) =
{
n|z| if n ≤ 1
|z| ,
1 if n 1
|z| ,
4

5. by the denition of fn. This implies that for all z ∈ C with 0 |z| ≤ 1, limn→∞ fn(z) = 1. For z = 0, fn(z) = 0
for all n ∈ N. So limn→∞ fn(0) = 0. Thus fn(z) converges pointwise to f(z) in B(0, 1) where we dene
f(z) =
{
1 if 0 |z| ≤ 1,
0 if z = 0.
Next, we want to show that fn(z) doesn't converge uniformly to f(z) in B(0, 1). Suppose fn(z) converges
uniformly to f(z) in B(0, 1). It is easy to see that fn is continuous in B(0, 1) for all n ∈ N. So by Theorem 22.2,
we can get that f is also continuous in B(0, 1). But it is obvious that f is not continuous at z = 0. We get a
contradiction. Thus fn(z) doesn't converge uniformly to f(z) in B(0, 1).
Problem. 22.4
Sol:
(⇒)
Since {fn(z)} converges to f(z) uniformly on S, for all ϵ 0, we can nd N ∈ N such that n ≥ N implies
|fn(z) − f(z)| ϵ for all z ∈ S. So n ≥ N implies that
Mn = sup{|fn(z) − f(z)| : z ∈ S} ≤ ϵ.
From above, we have for all ϵ 0, there exists N ∈ N such that when n ≥ N, |Mn| = Mn ≤ ϵ. Thus limn→∞ Mn = 0.
(⇐)
From limn→∞ Mn = 0, we have for all ϵ 0, there exists N ∈ N such that when n ≥ N, |Mn| ≤ ϵ
2 . So when
n ≥ N ,
|fn(z) − f(z)| ≤ sup{|fn(z) − f(z)| : z ∈ S}
= Mn
= |Mn|
≤
ϵ
2
ϵ for all z ∈ S.
From above, we can get that for all ϵ 0, there exists N ∈ N such that n ≥ N implies |fn(z) − f(z)| ϵ for all
z ∈ S. Thus {fn(z)} converges to f(z) uniformly on S.
Problem. 23.1
Sol:
5

6. (a)
Since
1/(1+3i)j+2
1/(1+3i)j+1
=
1
1 + 3i
=
1
√
10
→
1
√
10
as j → ∞,
we can get that the radius of convergence is
√
10 by the rario test.
(b)
Since
((j+1)!)2
/(2(j+1))!
(j!)2
/(2j)!
=
j2
(2j + 2)(2j + 1)
=
j2
2(j + 1)(2j + 1)
→
1
4
as j → ∞,
we can get that the radius of convergence is 4 by the rario test.
Problem. 23.3
Sol:
(a)
Since
|aj|
1
j
=
j∑
k=0
1
k!
→ e as j → ∞,
we can get that lim supj→∞ |aj|
1
j
= e. Thus the radius of convergence is 1
e by the Cauchy-Hadamard formula.
(b)
6

7. Since
|aj|
1
j
= 8 − (−3)j
=
{
3j
+ 8 if j ∈ Odd,
3j
− 8 if j ∈ Even,
→ ∞ as j → ∞,
we can get that lim supj→∞ |aj|
1
j
= ∞. Thus the radius of convergence is 0 by the Cauchy-Hadamard formula.
Problem. 23.5
Sol:
From Stirling's formula, we have
lim
n→∞
n!
√
2πn
(n
e
)n = 1.
So we can get that
(
n!
nn
) 1
n
= exp
(
log
( n!
nn
)
n
)
= exp




log
(
n!√
2πn(n
e )
n
)
+ log
√
2πn
en
n




= exp




log
(
n!√
2πn(n
e )
n
)
+ 1
2 (log 2π + log n) − n
n




→ e−1
as n → ∞
by virtue of limn→∞
n!√
2πn(n
e )
n = 1 and limn→∞
log n
n = 0.
Then from
lim
n→∞
jj
j!
1
j
=
1
e
,
we can get that the radius of convergence is e by the Cauchy-Hadamard formula.
7