1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

292 CHAPTER 8 • RESIDUE THEORY

• EXAMPLE 8.1 If f (z) = exp(~), then the Laurent series off about the

point 0 has the form

(

2) 2 ~ ~

f(z)= exp - =l+-+--+--+ ··-,

z z 21z^2 3!z^3

and Res(!, OJ = a _ 1 = 2.

• EXAMPLE 8.2 Find Res[g, OJ if g (z) = 2 >+;,_,,.

Solution Using Example 7.7, we find that g has three Laurent series repre-

sentations involving powers of z. The Laurent series valid in the punctured disk

00

Di (O) is g (z) = I::. [(-lt + 2 Ai] zn-^1. Computing the first few coefficients,

n =O

we obtain

3 1 3 9 15 2

g(z)=----+-z-- z +···.

2 z 4 8 16

Therefore, Res(g,O] = a-1 = !·

Recall that, for a function f analytic in Dn (zo) and for any r with 0 < r < R,

the Laurent series coefficients of f are given by

1 (e) de

(e -zot+

1 for n = 0, ±1, ±2, ... , (8-1)

where c,: (zo) denotes the circle {z: [z - zol = r} with positive orientation. This

result gives us an important fact conceming Res[/, z 0 ]. If we set n = - 1 in

Equation (8-1) and replace c: (zo) with any positively oriented simple closed

contour C containing ZQ, provided zo is the still only singularity of f that lies

inside C, then we obtain

j J(e)~ = 2nia_ 1 =2ni Reslf,zol ·

c

(8-2)

If we are able to find the residue of f at zo, then Equation (8-2) gives us an

important tool for evaluating contour integrals.