Mathematical Methods for Physics and Engineering : A Comprehensive Guide

COMPLEX VARIABLES

Differentiating both sidesm−1times,weobtain

dm−^1

dzm−^1

[(z−z 0 )mf(z)] = (m−1)!a− 1 +

∑∞

n=1

bn(z−z 0 )n,

for some coefficientsbn. In the limitz→z 0 , however, the terms in the sum disappear, and
after rearranging we obtain the formula

R(z 0 )=a− 1 = lim

z→z 0

{

1

(m−1)!

dm−^1

dzm−^1

[(z−z 0 )mf(z)]

}

, (24.56)

which gives the value of the residue off(z) at the pointz=z 0.
If we now consider the function

f(z)=

expiz

(z^2 +1)^2

=

expiz

(z+i)^2 (z−i)^2

,

we see immediately that it has poles of order 2 (doublepoles) atz=iandz=−i.To
calculate the residue at (for example)z=i, we may apply the formula (24.56) withm=2.
Performing the required differentiation, we obtain

d

dz

[(z−i)^2 f(z)] =

d

dz

[

expiz

(z+i)^2

]

=

1

(z+i)^4

[(z+i)^2 iexpiz−2(expiz)(z+i)].

Settingz=i, we find the residue is given by

R(i)=

1

1!

1

16

(

− 4 ie−^1 − 4 ie−^1

)

=−

i

2 e

.

An important special case of (24.56) occurs whenf(z) has asimple pole(a pole

of order 1) atz=z 0. Then the residue atz 0 is given by

R(z 0 ) = lim

z→z 0

[(z−z 0 )f(z)]. (24.57)

Iff(z) has a simple pole atz=z 0 and, as is often the case, has the form

g(z)/h(z), whereg(z) is analytic and non-zero atz 0 andh(z 0 ) = 0, then (24.57)

becomes

R(z 0 ) = lim

z→z 0

(z−z 0 )g(z)

h(z)

=g(z 0 ) lim

z→z 0

(z−z 0 )

h(z)

=g(z 0 ) lim

z→z 0

1

h′(z)

=

g(z 0 )

h′(z 0 )

, (24.58)

where we have used l’Hopital’s rule. This result often provides the simplest wayˆ

of determining the residue at a simple pole.

24.12 Residue theorem

Having seen from Cauchy’s theorem that the value of an integral round a closed

contourCis zero if the integrand is analytic inside the contour, it is natural to

ask what value it takes when the integrand is not analytic insideC.Theanswer

to this is contained in the residue theorem, which we now discuss.