Complex Analysis - Maths KU

An alternative method for computing
a residue at a simple pole
☞
6.5 Residues and Residue Theorem 345
Now at the pole of order 2, the result in (2) gives
Res(f(z), 1) = 1
1! lim
d
z→1dz
(z − 1)2f(z) d 1
= lim
z→1dz
z − 3
−1
= lim = −1
z→1(z
− 3) 2 4 .
When f is not a rational function, calculating residues by means of (1)
or (2) can sometimes be tedious.It is possible to devise alternative residue
formulas.In particular, suppose a function f can be written as a quotient
f(z) =g(z)/h(z), where g and h are analytic at z = z0.If g(z0) �= 0 and if
the function h has a zero of order 1 at z0, then f has a simple pole at z = z0
and
Res(f(z),z0) = g(z0)
h ′ . (4)
(z0)
To derive this result we shall use the definition of a zero of order 1, the
definition of a derivative, and then (1).First, since the function h has a zero
of order 1 at z0, we must have h(z0) = 0 and h ′ (z0) �= 0.Second, by definition
of the derivative given in (12) of Section 3.1,
h ′ (z0) = lim
z→z0
0
� �� �
h(z) − h(z0) h(z)
= lim .
z − z0 z→z0 z − z0
We then combine the preceding two facts in the following manner in (1):
Res(f(z),z0) = lim (z − z0)
z→z0
g(z)
= lim
h(z) z→z0
g(z)
h(z)
z − z0
= g(z0)
h ′ (z0) .
There are several alternative ways of arriving at formula (4).For instance, it
can be obtained by a single application of L’Hôpital’s rule (page 147), but you
are asked in Problem 40 in Exercises 6.5 to derive (4) using (5) of Section 6.4.
Residue formulas for poles of order greater than 1 are far more complicated
than (4) and will not be presented here.Practicality aside, a derivation of
one of these higher-order formulas provides an opportunity to review and use
important concepts.See Problem 41 in Exercises 6.5.
EXAMPLE 3 Using (4) to Compute Residues
The polynomial z 4 + 1 can be factored as (z − z1)(z − z2)(z − z3)(z − z4),
where z1, z2, z3, and z4 are the four distinct roots of the equation z 4 +1=0