Mathematical Methods for Physicists: A concise introduction - Site Map

FUNCTIONS OF A COMPLEX VARIABLE
proof. Let us go back to the de®nition:
Z
X n
f …z†dz ˆ lim f …w k †z k :
C
n!1
kˆ1
Now
X n
kˆ1
f …w k †z k
Xn
kˆ1
jf …w k †
jjz k j M Xn
kˆ1
jz k j ML;
where we have used the fact that j f …z†j M for all points z on C and that
P jzk j represents the sum of all the chord lengths joining z k1 and z k , and
that this sum is not greater than the length L of C. Now taking the limit of
both sides, and property (5) follows. It is possible to show, more generally, that
Z
Z
f …z†dz
jf …z† jjdzj: …6:24†
C
C
Example 6.11
Evaluate the integral R C …z*†2 dz, where C is a straight line joining the points z ˆ 0
and z ˆ 1 ‡ 2i.
Solution:
Since
we have
Z
C
Z
…z*† 2 dz ˆ
C
…z*† 2 ˆ…x iy† 2 ˆ x 2 y 2 2xyi;
Z
‰…x 2 y 2 †dx ‡ 2xydyŠ‡i ‰2xydx ‡…x 2 y 2 †dyŠ:
C
But the Cartesian equation of C is y ˆ 2x, and the above integral therefore
becomes
Z Z 1 Z 1
…z*† 2 dz ˆ 5x 2 dx ‡ i …10x 2 †dx ˆ 5=3 i10=3:
C
0
0
Example 6.12
Evaluate the integral
Z
C
dz
…z z 0 † n‡1;
where C is a circle of radius r and center at z 0 , and n is an integer.
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