Page 414 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 405

þ þ
i.e., f ðzÞ dz ¼ f ðzÞ dz
C 1 C 2
Note that f ðzÞ need not be analytic within curve C 2 .

dz 2 i if n ¼ 1
þ
16.13. (a) Prove that n ¼ 0 if n ¼ 2; 3; 4; ... , where C is a simple closed curve bounding
C ðz aÞ
a region having z ¼ a as interior point.
(b)What is the value of the integral if n ¼ 0;
1; 2; 3; ... ?
(a)Let C 1 be a circle of radius having center at z ¼ a (see Fig.
n
16-7). Since ðz aÞ is analytic within and on the boundary
of the region bounded by C and C 1 ,we have by Problem
16.12,
Fig. 16-7
dz dz
þ þ
n ¼ n
C ðz aÞ C 1 ðz aÞ
i i
To evaluate this last integral, note that on C 1 , jz aj¼ or z a ¼ e and dz ¼ i e d . The
integral equals
i
ð 2 i e d i ð 2 ð1 nÞi i e ð1 nÞi 2
n in
0 e ¼ n 1 0 e d ¼ n 1 ð1 nÞi 0 ¼ 0 if n 6¼ 1
ð 2
If n ¼ 1, the integral equals i d ¼ 2 i.
0
2
(b) For n ¼ 0; 1; 2; .. . the integrand is 1; ðz aÞ; ðz aÞ ; ... and is analytic everywhere inside C 1 ,
including z ¼ a. Hence, by Cauchy’s theorem the integral is zero.
dz
þ
16.14. Evaluate , where C is (a) the circle jzj¼ 1; ðbÞ the circle jz þ ij¼ 4.
C z 3
(a)Since z ¼ 3is not interior to jzj¼ 1, the integral equals zero (Problem 16.11).
(b)Since z ¼ 3isinterior to jz þ ij¼ 4, the integral equals 2 i (Problem 16.13).

16.15. If f ðzÞ is analytic inside and on a simple closed curve C,and a is any point within C, prove that
1 þ f ðzÞ
dz
f ðaÞ¼
2 i C z a
Referring to Problem 16.12 and the figure of Problem 16.13, we have
þ þ
f ðzÞ f ðzÞ
dz
C z a C 1 z a
dz ¼
ð 2
i
i
Letting z a ¼ e ,the last integral becomes i f ða þ e Þ d . But since f ðzÞ is analytic, it is
continuous. Hence, 0
ð 2 ð 2 ð 2
i
i
lim i f ða þ e Þ d ¼ i lim f ða þ e Þ d ¼ i f ðaÞ d ¼ 2 if ðaÞ
!0 0 0 !0 0
and the required result follows.
cos z e
þ þ x
16.16. Evaluate (a) dz; ðbÞ dz, where C is the circle jz 1j¼ 3.
C z C zðz þ 1Þ