Page 126 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 12: Fundamental Theorem of Algebra

(iii) If f(z) has a pole of order m at z = a then we can write Notes

 (z)
f(z) = (1)
(z a) m

where (z) is analytic and (a)  0.

1 1  (z)
Now, Res (z = a) = b = f(z) dz = dz

1 2 i  C 2 i (z a) m



C
1 | m 1  (z)

=  m 1 1 dz
| m 1 2 i C (z a)  



1
=  (a) [By Cauchy’s integral formula for derivatives] (2)
m-1

| m 1
Using (1), formula (2) take the form
1 d m 1

Res (z = a) = [(z–a) f(z)] as z  a
m

| m 1 dz m 1


i.e. Res (z = a) = lim 1 d m 1 [(z-a) f(z)] (3)
m

|
z a m 1 dz m 1
Thus, for a pole of order m, we can use either formula (2) or (3).
(iv) If z = a is a pole of any order for f(z), then the residue of f(z) at z = a is the co-efficient of
1
z a in Laurent’s expansion of f(z)

1
(v) Res (z = ) = Negative of the co-efficient of in the expansion of f(z) in the neighbourhood
z
of z = .
z 4
Example: (a) Find the residue of at z = –ia
2
z  a 2
z 4
Solution. Let f(z) = .
2
z  a 2
Poles of f(z) are z = ± ia
Thus z = –ia is a simple pole, so

Res (z = –ia) = lim (z + ia) f(z)
z ia
z 4
= lim (z ia)



z ia (z ia)(z ia)

LOVELY PROFESSIONAL UNIVERSITY 119