670 Chapter9
(b) Residue at a finite pole of order m (m > 1). If a# oo is a pole of
order m of f(z), we have the Laurent representation
f(z)= A_m ·+ .. ·+ A_^1 +Ao+A1(z-a)+ ...
(z-a)m z-awhere A--m # 0, valid in 0 < lz -al < 6 for some 6 > 0. Hence
g(z) = (z -ar J(z) = A_m + · · · + A-1(z -ar-^1 + Ao(z - ar + · · ·
is regular at z = a, so by Taylor's formula we get
R z~: !( z ) = A -1 = (m -1)!^1 g m-lc a )
= (1
)' lim nm-^1 [(z - ar f(z)]
m -1. z-+aIn particular, suppose that f(z) is given in the form
f(z) = g(z)
h(z)(9.8-3)where both g and h are regular at a, and g(a) # 0, h(a) = h'(a) = 0,
h"(a) # 0. In this case a is a double pole of f(z), since h(z) = (z-a)^2 h 1 (z),
with hl(a) # 0, so that
f(z) - g(z) = ,,P(z)- (z - a)^2 h 1 (z) (z - a)^2
where ,,P(z) = g(z)/h 1 (z) is regular at a, and ,,P(a) = g(a)/h 1 (a) # 0.
If we let
a2 alf(z) = + --+ ao + a1(z - a)+···
(z-a)^2 z-a
g(z) = bo + b1(z -a)+ .. ·
h(z) = c2(z - a)^2 + c 3 (z ·_ a)^3 + · · ·
from h(z)f(z) = g(z), or[c2(z-a)2+ca(z-a)^3 +· · ·] [ a
2
+ ~ + · .. J = bo+b1(z-a)+· · ·
(z - a)^2 z -a
we get, equating coefficients,c2a2 = bo, c3a2 + c2a1 = b1,
and solving for ai,