Page 415 - Schaum's Outline of Theory and Problems of Advanced Calculus

406 FUNCTIONS OF A COMPLEX VARIABLE [CHAP. 16

1 þ cos z
(a)Since z ¼ lies within C, dz ¼ cos ¼ 1byProblem 16.15 with f ðzÞ¼ cos z, a ¼ .
cos z
þ 2 i C z
Then dz ¼ 2 i.
C z
e z 1 1 e e
þ z þ þ z þ z
e dz
ðbÞ dz ¼ z z þ 1 dz ¼ dz C z þ 1
C zðz þ 1Þ C C z
1
0
¼ 2 ie 2 ie 1 ¼ 2 ið1 e Þ
by Problem 16.15, since z ¼ 0 and z ¼ 1are both interior to C.
5z 3z þ 2
þ 2
16.17. Evaluate dz where C is any simple closed curve enclosing z ¼ 1.
3
C ðz 1Þ
n! þ
Method 1. By Cauchy’s integral formula, f ðnÞ f ðzÞ dz.
nþ1
ðaÞ¼
2 i C ðz aÞ
2
If n ¼ 2 and f ðzÞ¼ 5z 3z þ 2, then f ð1Þ¼ 10. Hence,
00
2
2
2! þ 5z 3z þ 2 þ 5z 3z þ 2
dz or dz ¼ 10 i
2 i C ðz 1Þ C ðz 1Þ
10 ¼ 3 3
2
2
Method 2. 5z 3z þ 2 ¼ 5ðz 1Þ þ 7ðz 1Þþ 4. Then
5z 3z þ 2 5ðz 1Þ þ 7ðz 1Þþ 4
þ 2 þ 2
dz
3 dz ¼ 3
C ðz 1Þ C ðz 1Þ
d dz dz
þ þ þ
¼ 5 þ 7 þ 4 ¼ 5ð2 iÞþ 7ð0Þþ 4ð0Þ
C z 1 C ðz 1Þ 2 C ðz 1Þ 3
¼ 10 i
by Problem 16.13.
SERIES AND SINGULARITIES
16.18. For what values of z does each series converge?
1 n n
X z z
: : Then
n 2 n 2
ðaÞ 2 n The nth term ¼ u n ¼ 2 n
n¼1

nþ1 2
z n
jzj
lim u nþ1 n 2
¼ lim 2 nþ1 n ¼
n!1 u n n!1 ðn þ 1Þ 2 2
z

By the ratio test the series converges if jzj < 2 and diverges if jzj > 2. If jzj¼ 2the ratio test fails.
n n

1 1 1
X jzj X 1
X
z
However, the series of absolute values ¼ converges if jzj¼ 2, since
2
2
n 2 n n 2 n n 2
n¼1 n¼1 n¼1
converges.
Thus, the series converges (absolutely) for jzj @ 2, i.e., at all points inside and on the circle jzj¼ 2.
n 1 2n 1 3 5
1
X z z z
: We have
ð 1Þ
ð2n 1Þ! 3! 5!
ðbÞ ¼ z þ
n¼1

n 2nþ1 2
ð 1Þ z z

lim u nþ1 ð2n 1Þ! ¼ 0

n!1 ð2n þ 1Þ!
z
¼ lim ¼ lim
n!1 u n n 1 2n 1 n!1 2nð2n þ 1Þ

ð 1Þ
Then the series, which represents sin z,converges for all values of z.