1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applications 735

or changing ( into z,

1 00 2z 7r cot7rz = - + '\"' 2 2 z n=l L..!z -n

1

00 --+ ( --+--^1 1 )

z ~ z-n z+n
(9.12-3)

whenever z # 0, ±1, ±2, ... The series on the right of (9.12-3) converges absolutely and uniformly

on every compact set K that contains none of the integers 0, ±1, ±2, ....

In fact, if A = maxzEK lzl and n > J2A, we have

< <- I

2z I 2A 4A z2 - n2 - n2 - A2 n2

and the stated property follows from the Weierstrass M-test. By Corollary
8.5 we obtain, by differentiation of (9.12-3),

. +oo
2 2. I: 1
7r CSC · 7rZ = ( z-n ) 2
n=-oo

for z # O, ±1, ±2,

Let f(z) = ?rcsc?rz and g(z) = 1/((^2 - z^2 ), ( # 0, ±1, ±2,
f(z) = ?rcsc?rz has simple poles at the points Zn= n. The corresponding
residues are

f3n.= lim(z-n)~ = 7r =(-lt Z-tn sin 7rZ 7r cos n?r

On the other hand, the residues of f(z)g(z) at the poles of g(z) are

Res 7r csc 7rZ = - ~ csc 7r( z=C (2 - z2 2(

R

7r CSC 7rZ 7r · 7r es ---= 2 ,.. csc( -7r() = - 2 ,.. csc 7r( z=-C (2 - z2 '> '>

We have already seen that zg(z) -t 0 as z -t oo, and we may use as
contours the same rectangles of Example 1. On the sides x = ±(% + m)
we have

1 1

icsc?rzl= ---= -;===================

I sin?rzl Jsin^2 7r(% + m) + sinh^2 7ry 1 = <1 Vl + sinh^2 ?rY -

icsc?rzl= ---= -;===================