1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/Applications 671

or

_ (1/2!)h"(a)g'(a)-(1/3!)h'"(a)g(a)

¥-~:J(z) - [h"(a)/2!]2

g'(a) 2 h"'(a)g(a)

=

2

h"(a) - 3 [h"(a)]2

(9.8-4)

More generally, if both g and h are regular at a, and g( a) =/: O, h( a) =

h'(a) = · · · = Mm-l)(a) = 0 h(m>(a) =/: O, then a is a pole of order m of

f, and if we let

am ai

f(z)= +"·+--+ao+···

(z - a)m z - a

g(z) = bo + bi(z -a)+ ... + bm-1(z -ar-^1 + ...

h(z) = Cm(z -ar + Cm+i(Z -ar+l + ''' + C2m-1(z - a)^2 m-l + • • •

we have

[cm(z-a)m+cm+i(z-ar+i+···] [ am + am-1 - +···]

(z-a)m (z-ar^1

= bo + bi(z -a)+ ... + bm-1(z - ar-^1 + ...

Multiplying out and equating coefficients of the same powers of z - a, we

obtain the system

Cm+1am + Cmam-1

Cm+2am + Cm+lam-1 + Cmam-2 (9.8-5)

Clearly, the determinant of the system (9.8-5) equals ( cmr. Hence solving

for a 1 we have

Cm 0 0 bo

Cm+l Cm·^0 bi

(^1) b2
Resf(z) z=a = ai = -( Cm -)-m Cm+2 Cm+l Cm
C2m-l C2m-2 C2m-3 bm-1

where bk = (1/k!)g(k)(a) and ck = (1/k!)h(k)(a).

Example To find Resz=O ez /(1-cosz). Let g(z) = ez, h(z) = 1-cosz.

We have g'(z) = ez, h'(z) = sinz, h"(z) = cosz, h"'(z) = -sinz, so