1550251515-Classical_Complex_Analysis__Gonzalez_

734 Chapter^9

provided that mis large enough so that m >Mand b/m < 1. Thus if we

take limits in (9.12-1) as m -t oo, we get

0 = L ,8ng( a + n) + S

or

(9.12-2)

where the summation on the right extends now to all poles of f where g

is analytic.

Examples 1. Let f(z) = 7rcot7l"Z and g(z) = 1/((^2 - z^2 ), where ( f= n

(n = 0,±1,±2, ... ). The function f(z) = 7rcot7rz has simple poles at the

points Zn = n with residues

,8

_

1

. ( ) COS 7l"Z 7r COS n7r l
n - 1m z-n 71"-.--- _
z-m sin 7rZ 7r cos n7r
while g(z) = 1/((^2 - z^2 ) has simple poles at z = ( and z = -(. The
residues of f(z)g(z) at those two poles are
7r cot 7rZ 7r
Res ( 2 2 = -
2
1" cot 7r(
z=( - Z '>

R

7rCOt7rz 7r · 7r

es =

2

1" cot( -7r() = -

2

1" cot 7r(

z=-( (2 - z2 '> '>

Also, for lzl > ICI,

lz]

lzg(z)I:::; lzl2~1c12 -to as lzl -too

In this case a = 0 and we may choose b = %, thus making c;;; the boundary
of the rectangle with vertices (1/ 2 + m, -m ), (% + m, m) (-%-m, m ), and

(-% - m,-m). On the sides x = ±(% + m), we have

Jcos^2 7r(% + m) + sinh

2

7rY I sinh7ryl

I cot 7rzl = --'-;:========= :=:; :=:; 1

Jsin^2 7r(% + m) + sinh^2 7ry J1+sinh

2

7ry

and on the sides y = ±m,

I I

Vl + sinh^2 7rm <

1

1 1

cot 7rz ::=; I sin. h 7rm I _ +. h sin 7rm ::=; 1 + -7rm < 2

so that lf(z)I = 17rcot 7rzl < 271" for all z E c;;;. Hence all conditions for

the validity of (9.12-2) are met, and we obtain

7r +oo 1

( cot 7r( = L (2 - n2

n=-oo