Page 412 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 16] FUNCTIONS OF A COMPLEX VARIABLE 403

z þ z z z z z
z .
This can also be accomplished by nothing that z ¼ x þ iy; z ¼ x iy so that x ¼ , y ¼
2 2i
z
The result is then obtained by substitution; the terms involving z drop out.
INTEGRALS, CAUCHY’S THEOREM, CAUCHY’S INTEGRAL FORMULAS

ð 2þ4i
2
16.10. Evaluate z dz
1þi
2
(a) along the parabola x ¼ t; y ¼ t where 1 @ t @ 2,
(b) along the straight line joining 1 þ i and 2 þ 4i,
(c) along straight lines from 1 þ i to 2 þ i and then to 2 þ 4i.
We have
ð 2þ4i ð ð2;4Þ ð ð2;4Þ
2 2 2 2
z dz ¼ ðx þ iyÞ ðdx þ idyÞ¼ ðx y þ 2ixyÞðdx þ idyÞ
1þi ð1;1Þ ð1;1Þ
ð ð
ð2;4Þ 2 2 ð2;4Þ 2 2
ðx y Þ dx 2xy dy þ i 2xy dx þðx y Þ dy
¼
ð1;1 ð1;1Þ
Method 1. (a) The points ð1; 1Þ and ð2; 4Þ correspond to t ¼ 1 and t ¼ 2, respectively. Then the above
line integrals become
ð 2 ð 2 86
4
2
2
2
4
2
fðt t Þ dt 2ðtÞðt Þ2tdtgþ i f2ðtÞðt Þ dt þðt t Þð2tÞ dtg¼ 6i
t¼1 t¼1 3
4 1
ðx 1Þ or y ¼ 3x 2. Then we find
(b) The line joining ð1; 1Þ and ð2; 4Þ has the equation y 1 ¼
2 1
2
ð
2 2
½x ð3x 2Þ Š dx 2xð3x 2Þ3 dx
x¼1
ð 2
2 2 86
þ i 2xð3x 2Þ dx þ½x ð3x 2Þ Š3 dx ¼ 6i
x¼1 3
(c) From 1 þ i to 2 þ i [or ð1; 1Þ to ð2; 1Þ], y ¼ 1; dy ¼ 0 and we have
ð 2 ð 2 4
2
ðx 1Þ dx þ i 2xdx ¼ þ 3i
x¼1 x¼1 3
From 2 þ i to 2 þ 4i [or ð2; 1Þ to ð2; 4Þ], x ¼ 2; dx ¼ 0 and we have
ð 4 ð 4
2
4ydy þ i ð4 y Þ dy ¼ 30 9i
y¼1 y¼1
4 86
3
Adding, ð þ 3iÞþð 30 91Þ¼ 3 6i.
Method 2. By the methods of Chapter 10 it is seen that the line integrals are independent of the path, thus
accounting for the same values obtained in (a), (b), and (c) above. In such case the integral can be evaluated
directly, as for real variables, as follows:
2þ4i 86

ð 3 2þ4i 3 3
2 z ð2 þ 4iÞ ð1 þ iÞ
6i
z dz ¼ ¼ ¼
1þi 3 1 i 3 3 3
16.11. (a) Prove Cauchy’s theorem: If f ðzÞ is analytic inside and on a simple closed curve C, then
þ
f ðzÞ dz ¼ 0.
C