24.11 TAYLOR AND LAURENT SERIES
denominator take the form (1−αz), whereαis some constant, and thus obtain
f(z)=−1
8 z(1−z/2)^3=−
1
8 z[
1+(−3)
(
−
z
2)
+
(−3)(−4)
2!
(
−
z
2) 2
+
(−3)(−4)(−5)
3!
(
−
z
2) 3
+···
]
=−
1
8 z−
3
16
−
3 z
16−
5 z^2
32−···.
Since the lowest power ofzis−1, the pointz= 0 is a pole of order 1. The residue off(z)
atz= 0 is simply the coefficient ofz−^1 in the Laurent expansion about that point and is
equal to− 1 /8.
The Laurent series aboutz= 2 is most easily found by lettingz=2+ξ(orz−2=ξ)
and substituting into the expression forf(z)toobtain
f(z)=1
(2 +ξ)ξ^3=
1
2 ξ^3 (1 +ξ/2)=
1
2 ξ^3[
1 −
(
ξ
2)
+
(
ξ
2) 2
−
(
ξ
2) 3
+
(
ξ
2) 4
−···
]
=
1
2 ξ^3−
1
4 ξ^2+
1
8 ξ−
1
16
+
ξ
32−···
=
1
2(z−2)^3−
1
4(z−2)^2+
1
8(z−2)−
1
16
+
z− 2
32−···.
From this series we see thatz= 2 is a pole of order 3 and that the residue off(z)atz=2
is 1/8.
As we shall see in the next few sections, finding the residue of a functionat a singularity is of crucial importance in the evaluation of complex integrals.
Specifically, formulae exist for calculating the residue of a function at a particular
(singular) pointz=z 0 without having to expand the function explicitly as a
Laurent series aboutz 0 and identify the coefficient of (z−z 0 )−^1. The type of
formula generally depends on the nature of the singularity at which the residue
is required.
Suppose thatf(z)has a pole of ordermat the pointz=z 0. By considering the Laurent
series off(z)aboutz 0 , derive a general expression for the residueR(z 0 )off(z)atz=z 0.
Hence evaluate the residue of the functionf(z)=expiz
(z^2 +1)^2
at the pointz=i.Iff(z) has a pole of ordermatz=z 0 , then its Laurent series about this point has the
form
f(z)=a−m
(z−z 0 )m+···+
a− 1
(z−z 0 )+a 0 +a 1 (z−z 0 )+a 2 (z−z 0 )^2 +···,which, on multiplying both sides of the equation by (z−z 0 )m,gives
(z−z 0 )mf(z)=a−m+a−m+1(z−z 0 )+···+a− 1 (z−z 0 )m−^1 +···.