Homework #9 Solutions

Math 128, Fall 2013

Instructor: Dr. Doreen De Leon

1 HW #9(a)

Investigate the convergence of

∞
X n(i − 1)n
1.
(z − 2i)n
n=2
Solution:
n(i − 1)n
zn =
(z − 2i)n
 n(i − 1)n 
 
=⇒ |zn | =  
(z − 2i)n 
√ n
n|i − 1|n n 2
= = .
|z − 2i|n |z − 2i|n
Apply the ratio test.
 
 = lim |zn+1 |
 zn+1 
lim 
n→∞ zn  n→∞ |zn |
 √ n+1 
 (n+1) n+1
2 
|z−2i|
= lim √ n
n→∞  n 2 
|z−2i|n
√ n+1
(n + 1) 2 |z − 2i|n
= lim · √ n
n→∞ |z − 2i|n+1 n 2
√
n+1 2
= lim ·
n→∞ n |z − 2i|

√
2 n+1
= lim
|z − 2i| n→∞ n
√
2
= .
|z − 2i|
For absolute convergence, we require
√
2 √
< 1 =⇒ |z − 2i| > 2.
|z − 2i|

1
 √
What if |z − 2i| = 2? Then, |zn | = n, and limn→∞ |zn | = ∞, so the series diverges.
√
Thus, the series converges absolutely if |z − 2i| > 2.
∞
X
2. (−1)j j2j+1 z 2j
j=1

Solution: zj = (−1)j j2j+1 z 2j , so |zj | = j2j+1 |z|2j . Apply the ratio test:
 
 = lim |zj+1 |
 zj+1 
lim 
j→∞  zj  j→∞ |zj |

(j + 1)2j+2 |z|2(j+1)
= lim
j→∞ j2j+1 |z|2j
j+1
= lim · 2 · |z|2
j→∞ j
 
2 j+1
= 2|z| lim
j→∞ j
= 2|z|2 .
For absolute convergence, we need
2|z|2 < 1
1
|z|2 <
2
1
|z| < √ .
2
1
What if |z| = √ ? Then,
2  2j
1
|zj | = j2j+1 √ = 2j.
2
Since limj→∞ |zj | = ∞, the series diverges.
1
Therefore, the series converges absolutely if |z| < √ .
2

2 HW #9(b)

Find the Taylor series expansion about z0 and determine the set on which it converges for each of the
following functions.

1
1. , z0 = i
z
Solution:
1 1
f (z) = =⇒ f (i) = = −i
z i
f 0 (z) = −z −2 =⇒ f 0 (i) = 1
f 00 (z) = 2z −3 =⇒ f 00 (i) = 2i−3 = −2i

2
 f 000 (z) = −3!z −4 =⇒ f 000 (i) = −3!
f (4) (z) = 4!z −5 =⇒ f (4) (i) = −4!i
..
.

Then,
1 00
f (z) = f (i) + f 0 (i)(z − i) + f (i)(z − i)2 + · · · .
2!
So, we obtain
1 1 1
f (z) = −i + (z − i) + (2i)(z − i)2 + (−3!)(z − i)3 + (−4!i)(z − i)4 + · · ·
2! 3! 4!
= −i + (z − i) + i(z − i)2 − (z − i)3 − i(z − i)4 + · · ·
= i−1 + i0 (z − i) + i1 (z − i)2 + i2 (z − i)3 + i3 (z − i)4 + · · ·
∞
1 X n−1
= i (z − i)n .
z
n=0

Convergence:

|i(n+1)−1 (z − i)n+1 |
lim = lim |i(z − i)|
n→∞ |in−1 (z − i)n | n→∞

= |z − i|.

For convergence, we require

|z − i| < 1.

We can show that if |z − i| = 1, the series diverges, so the series only converges on |z − i| < 1.

2. z i , z0 = 1
Solution:

f (z) = z i =⇒ f (1) = 1i = 1
f 0 (z) = iz i−1 =⇒ f 0 (1) = i
f 00 (z) = i(i − 1)z i−2 =⇒ f 00 (1) = i(i − 1)
f 000 (z) = i(i − 1)(i − 2)z i−3 =⇒ f 000 (1) = i(i − 1)(i − 2)
..
.

Then,
1 00
f (z) = f (1) + f 0 (1)(z − 1) + f (1)(z − 1)2 + · · · .
2!
So, we obtain
i(i − 1) i(i − 1)(i − 2)
f (z) = 1 + i(z − 1) + (z − 1)2 + (z − 1)3 + · · ·
2! 3!
∞  
i
X i
=⇒ z = (z − 1)n
n
n=0

3
      
α α(α − 1) · (α − n + 1) i i(i − 1) · · · (i − n + 1)
where we define = , so = .
n n! n n!

Convergence:
  
 i n+1

 n + 1 (z − 1)
 
 |i(i − 1) · · · (i − n + 1)(i − n)| n!
lim    = lim · |z − 1|
n→∞  i 
n
n→∞ |i(i − 1) · · · (i − n + 1)| (n + 1)!

 n (z − 1) 
|z − 1||i − n|
= lim
n→∞ n+1
√
|z − 1| n2 + 1
= lim
n→∞ n+1
= |z − 1|.
For convergence, then, we require |z − 1| < 1. If |z − 1| = 1, then it can be shown that the series
does not converge absolutely.
3. iz , z0 = 0
Solution:
f (z) = iz =⇒ f (0) = 1
f 0 (z) = (log i)iz =?
Note that
π
log i = i (choose the principal branch),
2 
π z π
f 0 (z) = i i =⇒ f 0 (0) = i
2 2
 π 2  π 2
00 z 00
f (z) = i i =⇒ f (z) = i
2 2
..
..
Then,
1 00
f (z) = f (0) + f 0 (0)z + f (0)z 2 + · · · .
2!
So, we obtain
π 1  π 2 2
f (z) = 1 + i z + i z + ···
2 2! 2
∞
X 1  π n n
=⇒ iz = i z .
n! 2
n=0
Convergence:
 
 1 π n+1 n+1 


 (n+1)! i 2 z  1  π 
lim  1 π
n  = lim i  |z|
n→∞ 
n! i 2 zn n→∞ n + 1 2
π
= lim |z|
n→∞ 2(n + 1)

= 0.
Therefore, the series converges for all z (i.e., on |z| < ∞).

4
3 HW #9(c)

1. Find the Laurent series expansion of the following functions around z0 = 0 in the regions indicated.
 
1
(a) sin , 0 < |z| < ∞
z
Solution: The Taylor series for sin z is
∞ ∞
X z n+1 X z 2n−1
sin z = (−1)n = (−1)n+1 , |z| < ∞.
(2n + 1)! (2n − 1)!
n=0 n=1

Therefore,
  X∞  2n−1  
1 n+1 1 1 1
sin = (−1) , 0 <   < ∞ =⇒ 0 < |z| < ∞.
z (2n − 1)! z z
n=1

Or,
∞
(−1)n+1 1
  X
1
sin = , 0 < |z| < ∞ .
z (2n − 1)! z 2n−1
n=1

1
(b) , 0 < |z| < 1
z(z + 1)
Solution:
Step 1: Partial fractions
1 1 1
= − .
z(z + 1) z z+1
1
Step 2: Taylor series for
z+1
1 1
=
z+1 1 − (−z)
∞
X
= (−z)n
n=0
X∞
= (−1)n z n ,
n=0

convergent for | − z| < 1 =⇒ |z| < 1. So,

∞
1 1 X
= − (−1)n z n ,
z(z + 1) z
n=0

Or,
∞
1 X 1
= (−1)n+1 z n + , 0 < |z| < 1 .
z(z + 1) z
n=0

5
 Alternate solution:
1 1 1
= ·
z(z + 1) z 1+z
1 1
= ·
z 1 − (−z)
∞
1X
= (−z)n
z
n=0

convergent for | − z| < 1 =⇒ |z| < 1. So,

∞
1 1X
= (−1)n z n
z(z + 1) z
n=0
1
1 − z + z2 − z3 + · · ·


=
z
1
= − 1 + z − z2 + · · ·
z
∞
1 X
= + (−1)n+1 z n
z
n=0
∞
X 1
= (−1)n+1 z n + .
z
n=0

z
(c) , |z| < 1
z+1
Solution:
 
z 1
=z .
z+1 z+1
Then,
1 1
=
z+1 1 − (−z)
∞
X
= (−z)n
n=0
X∞
= (−1)n z n ,
n=0

convergent for | − z| < 1 =⇒ |z| < 1. So,

∞
z X
=z (−1)n z n
z+1
n=0
∞
X
= (−1)n z n+1 .
n=0

So,
∞
z X
= (−1)n z n+1 , |z| < 1 .
(z + 1)
n=0

6
 ez
(d) , 0 < |z| < ∞
z2
Solution:
∞
z
X zn
e = , |z| < ∞ (Taylor series)
n!
n=0
z2 z3 z4
=1+z+ + + + ···
2! 3! 4!
z2 z3 z4
 
1 z 1
=⇒ 2 e = 2 1 + z + + + + ···
z z 2! 3! 4!
1 1 1 z z2
= 2+ + + + + ···
z z 2! 3! 4!
∞
1 1 X zn
= 2+ + .
z z (n + 2)!
n=0

So,
∞
ez X zn 1 1
2
= + + 2 , 0 < |z| < ∞ .
z (n + 2)! z z
n=0

1
2. Find the Laurent series expansion of around z0 = 0 valid in the region 1 < |z| < ∞.
z(z + 1)
Solution: We know that
1 1 1
= − (from problem 1(a)).
z(z + 1) z z+1
1 1 1 1 1
Since the Taylor series for = is convergent for |z| < 1, we write as · .
1+z 1 − (−z) z+1 z 1 + z1
Then,
1 1 1 1
· 1 = ·
z 1 − − z1


z 1+ z
∞ 
1 n

1X
= − ,
z z
n=0

convergent for − z1  < 1, or |z| > 1. So,

 

∞ 
1 n

1 1 1X
= − − ,
z(z + 1) z z z
n=0
 
1 1 1 1 1
= − 1 − + 2 − 3 + ···
z z z z z
1 1 1
= 2 − 3 + 4 + ···
z z z
∞
X 1
= (−1)n n .
z
n=2

So,
∞
1 X 1
= (−1)n+1 n+1 , 1 < |z| < ∞ .
z(z + 1) z
n=1