1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

7.3 • LAURENT SERIES REPRESENTATIONS 273

Equation (7-23). The following examples illustrate some methods for finding

Laurent series coefficients.

• EXAMPLE 7.7 Find three different Laurent series representations for the
  function f (z) = 2 +Lz> involving powers of z.

Solut ion The function f has singularities at· z = - 1, 2 and is analytic in the

disk D : lzl < 1, in the annulus A : 1 < lzl < 2, and in the region R: lzl > 2.

We want to find a different Laurent series for f in each of the three domains D,

A, and R. We start by writing f in its partial fraction form:

(7-30)

We use Theorem 4.12 and Corollary 4 .2 to obtain the following representa-

tions for the terms on the right side of Equation (7-30):

00

_ l_ = L (- 1)" Zn (valid for lzl < 1), {7-31)

1 +z n=O

1 00 (-1)"+1

l+z=L zn (valid for lzl > 1) , (7-32)

n=l

1 ( 1 )

00

zn

2 1 - ~ = L 2 n+l (valid for lzl < 2), and (7-33)

2 n=O

~ ( - 1 ) =

00

-2n- l (valid for lzl > 2). (7-34)

2 1- ~ L zn

2 n=l

Representations (7-31) and (7-33) are both valid in the disk D, and thus we have

(valid for lzl < 1) ,

which is a Laurent series that reduces to a Maclaurin series. In the annulus A ,

Representations -(7-32) and (7-33) are valid; hence we get

oo ( - 1 r+l oo Zn

I (z) = L zn + L zn+I

n=l n=O

(valid for 1 < lzl < 2).

Finally, in the region R we use Representations (7-32) and (7-34) to obtain

oo {-l)"+l _ 2n-l

f (z) = L zn

n=l

{valid for lzl > 2).