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(b) Consider any two paths joining points P1 and P2 (see Fig. 16-5). f z dz 0
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P1 AP2 BP1
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By Cauchy s theorem,
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Path 1 A
or
P1 AP2
f z dz
f z dz
P1 Path 2
i.e., the integral along P1 AP2 (path 1) integral along P1 BP2 (path 2), and so the integral is independent of the path joining P1 and P2 . This explains the results of Problem 16.10, since f z z2 is analytic.
Fig. 16-5
16.12. If f z is analytic within and on the boundary of a region bounded by two closed curves C1 and C2 (see Fig. 16-6), prove that f z dz f z dz
C1 C2
As in Fig. 16-6, construct line AB (called a cross-cut) connecting any point on C2 and a point on C1 . By Cauchy s theorem (Problem 16.11), f z dz 0
AQPABRSTBA
since f z is analytic within the region shaded and also on the boundary. Then f z dz f z dz f z dz f z dz 0 1
AQPA AB BRSTB BA
But
f z dz
f z dz.
Hence, (1) gives Fig. 16-6 f z dz
AQPA BRSTB
f z dz
f z dz
BTSRB
CHAP. 16]
FUNCTIONS OF A COMPLEX VARIABLE f z dz
C1 C2
i.e.,
f z dz
Note that f z need not be analytic within curve C2 .
& dz 2i if n 1 16.13. (a) Prove that , where C is a simple closed curve bounding 0 if n 2; 3; 4; . . . z a n C a region having z a as interior point. (b) What is the value of the integral if n 0; 1; 2; 3; . . .
(a) Let C1 be a circle of radius  having center at z a (see Fig. 16-7). Since z a n is analytic within and on the boundary of the region bounded by C and C1 , we have by Problem 16.12, dz dz n n C z a C1 z a
Fig. 16-7
To evaluate this last integral, note that on C1 , jz aj  or z a ei and dz iei d. integral equals  2 i ie d i 2 1 n i i e 1 n i 2  0 n 1 e d n 1 if n 6 1 n in 1 n i 0   0  e 0 If n 1, the integral equals i 2
d 2i.
(b) For n 0; 1; 2; . . . the integrand is 1; z a ; z a 2 ; . . . and is analytic everywhere inside C1 , including z a. Hence, by Cauchy s theorem the integral is zero.
16.14. Evaluate
dz , where C is z 3
(a) the circle jzj 1;
b the circle jz ij 4.
(a) Since z 3 is not interior to jzj 1, the integral equals zero (Problem 16.11). (b) Since z 3 is interior to jz ij 4, the integral equals 2i (Problem 16.13).
16.15. If f z is analytic inside and on a simple closed curve C, and a is any point within C, prove that 1 f z dz f a 2i C z a
Referring to Problem 16.12 and the gure of Problem 16.13, we have f z f z dz dz z a z a C C1 2 f a ei d. But since f z is analytic, it is Letting z a ei , the last integral becomes i 0 continuous. Hence, 2 2 2 lim i f a ei d i lim f a ei d i f a d 2i f a
!0 0 0 !0 0
and the required result follows.
16.16. Evaluate (a)
cos z dz; z  C
b
ex dz, where C is the circle jz 1j 3. C z z 1
FUNCTIONS OF A COMPLEX VARIABLE 1 (a) Since  lies within C, z 2i cos z Then dz 2i. Cz  b
[CHAP. 16
cos z dz cos  1 by Problem 16.15 with f z cos z, a . z  C ez
ez dz z z 1 C
  z 1 1 e ez dz dz dz z z 1 z z 1 C C
2ie0 2ie 1 2i 1 e 1 by Problem 16.15, since z 0 and z 1 are both interior to C.
16.17. Evaluate
Method 1.
5z2 3z 2 dz where C is any simple closed curve enclosing z 1. z 1 3 C
By Cauchy s integral formula, f n a n! 2i f z dz. z a n 1 C
If n 2 and f z 5z2 3z 2, then f 00 1 10. Hence, 2! 5z2 3z 2 5z2 3z 2 dz or dz 10i 10 3 2i C z 1 z 1 3 C Method 2. 5z2 3z 2 5 z 1 2 7 z 1 4. Then 5z2 3z 2 5 z 1 2 7 z 1 4 dz dz 3 z 1 z 1 3 C C d dz dz 7 4 5 2i 7 0 4 0 5 2 3 z 1 C C z 1 C z 1 10i
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