1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

238 CHAPTER 6 • COMPLEX INTEGRATION

• EXAMPLE 6.24 Let z 0 denote a fixed complex value. Show that if C is a
  simple closed positively oriented contour s uch that zo lies interior to C, then for
  any integer n ~ 1,

r ....!!:=___ = 21Ti and r dz n+l = o. (6-50)

le z - zo l e (z - zo)

Solution We let f (z) = 1. Then J(n) (z) = 0 for n :2'. 1. Theorem 6.10 implies

that the value of the first integral in Equations (6-50) is

r ....!!:=___ = 21Tif (zo) = 21Ti,

le z - zo

and Theorem 6.12 further implies that

1

dz = 27Tif(n) ( ) = O

e (z - zo) n+l n. I zo ·

This result is the same as that proven earlier in Corollary 6.1. Obviously, though,

the technique of using Theorems 6.10 and 6.12 is easier.

• EXAMPLE 6.25 Show that f et(O) t:~n: dz= 3~" ·

Solution If we let f (z) = expz^2 , then a straightforward calculation shows

that f(^3 ) (z) = (12z + 8z^3 ) expz^2 • Using Cauchy's integral formulas with n = 3,

we conclude that

r exp z

2

4 dz = 21Ti !(3) ( i) = 21Ti 4i = - 41T.

l e (z -i) 3! 6 e 3e