![]()
![]()
for
Chapter 8 Residue Theory
8.1 The Residue Theorem
Overview
We now have the necessary machinery to see
some amazing applications of the tools we developed in the last few
chapters. You will learn how Laurent expansions can give
useful information concerning seemingly unrelated properties of
complex functions. You will also learn how the ideas of
complex analysis make the solution of very complicated integrals of
real-valued functions as easy - literally - as the computation of
residues. We begin with a theorem relating residues to the
evaluation of complex integrals.
The Cauchy
integral
formulae in Section
6.5 are useful in evaluating contour integrals over a simple
closed contour C where the integrand
has the form
and f is
an analytic function. In this case, the singularity of the
integrand is at worst a pole of order k
at
. We
begin this section by extending this result to integrals that have a
finite number of isolated singularities inside the contour
C. This new method can be
used in cases where the integrand has an essential singularity at
and is an important extension of the previous method.
Definition 8.1 (Residue). Let
f(z) have a nonremovable isolated
singularity at the point
. Then
f(z) has the Laurent series
representation for all z in some disk
given
by
.
The coefficient
of
is
called the residue of f(z) at
and we use the notation
.
Example
8.1. If
, then
the Laurent series of f about the point
has the form
, and
.
Example
8.2. Find
if
.
Solution. Using Example 7.7, we find that g(z)
has three Laurent series representations involving powers of
z. The Laurent series
valid in the punctured disk
is
.
Computing the first few coefficients, we obtain
![[Graphics:Images/ResidueCalcMod_gr_27.gif]](residuecalculus/ResidueCalcMod/Images/ResidueCalcMod_gr_27.gif)
Therefore,
.
Recall that, for a function f(z)
analytic in
and for any r with
,
the Laurent series coefficients of f(z)
are given by
(8-1)
for
,
where
denotes the circle
with positive orientation. This gives us an important fact
concerning
. If
we set
in
Equation (8-1) and replace
with any positively oriented simple closed contour C
containing
,
provided
is the still only singularity of f(z)
that lies inside C, then we
obtain
(8-2)
.
If we are able to find the Laurent series expansion for f(z),
then above equation gives us an important tool for evaluating contour
integrals.
Example
8.3. Evaluate
where
denotes the circle
with positive orientation.
![[Graphics:Images/ResidueCalcMod_gr_55.gif]](residuecalculus/ResidueCalcMod/Images/ResidueCalcMod_gr_55.gif)
Solution. In Example 8.1 we showed that the residue
of
at
is
. Using
Equation (8-2), we get
.
Theorem 8.1 (Cauchy's
Residue
Theorem). Let
D be a simply connected domain, and
let C be a simple closed
positively oriented contour that lies in D. If
f(z) is analytic
inside C and on C, except
at the points
that
lie inside C, then
.
The situation is illustrated in Figure 8.1.
![]()
Figure 8.1 The domain D and contour C and the singular points
in the statement of Cauchy's residue theorem.
Proof of Theorem 8.1 is in the book.
Complex
Analysis for Mathematics and Engineering
The calculation of a Laurent
series
expansion is tedious in most circumstances. Since the
residue at
involves only the coefficient
in the Laurent expansion, we seek a method to calculate the residue
from special information about the nature of the singularity at
.
If f(z) has a
removable singularity at
,
then
for
. Therefore,
. Theorem
8.2 gives methods for evaluating residues at poles.
Theorem 8.2 (Residues at
Poles).
(i) If
f(z) has a simple pole
at
, then
.
(ii) If
f(z) has a pole of order 2
at
, then
.
(iii) If
f(z) has a pole of order 3
at
, then
.
(v) If
f(z) has a pole of order k
at
, then
.
Proof of Theorem 8.2 is in the book.
Complex
Analysis for Mathematics and Engineering
Example 8.4. Find
the residue of
at
.
Solution. We write
. Because
has
a zero of order 3 at
and
. Thus
f(z) has a pole of order 3
at
. By
part (iii) of Theorem 8.2, we
have
This last limit involves an indeterminate form, which we evaluate
by using L'Hôpital's rule:
Example
8.5. Find
where
denotes the circle
with positive orientation.
![[Graphics:Images/ResidueCalcMod_gr_107.gif]](residuecalculus/ResidueCalcMod/Images/ResidueCalcMod_gr_107.gif)
Solution. We write the integrand
as
.
The singularities of f(z) that lie
inside
are simple poles at the points
and
,
and a pole of order 2 at the
origin. We compute the residues as
follows:
Finally, the residue theorem yields
![[Graphics:Images/ResidueCalcMod_gr_115.gif]](residuecalculus/ResidueCalcMod/Images/ResidueCalcMod_gr_115.gif)
The answer,
, is
not at all obvious, and all the preceding calculations are required
to get it.
Example
8.6. Find
where
denotes the circle
with positive orientation.
![[Graphics:Images/ResidueCalcMod_gr_134.gif]](residuecalculus/ResidueCalcMod/Images/ResidueCalcMod_gr_134.gif)
Solution. The singularities of the
integrand
that
lie inside
are simple poles occurring at the points
, as
the points
, lie
outside
. Factoring
the denominator is tedious, so we use a different
approach. If
is any one of the singularities of f(z)
, then we can use L'Hôpital's rule to compute
:
![[Graphics:Images/ResidueCalcMod_gr_142.gif]](residuecalculus/ResidueCalcMod/Images/ResidueCalcMod_gr_142.gif)
Since
, we
can simplify this expression further to yield
![]()
We now use the residue theorem to get
The theory of residues can be used to expand the quotient of two polynomials into its partial fraction representation.
Example 8.7. Let
P(z) be a polynomial of degree at
most 2. If a
, b and c
are distinct complex numbers, then
,
where
Solution. It will suffice to prove
that
. We
expand f(z) in its Laurent series
about the point
by writing the three terms
,
,
and
in their Laurent series about the point
and adding them. The term
is itself a one-term Laurent series about the point
. The
term
is analytic at the point
,
and its Laurent series is actually a Taylor series given
by
which is valid for
.
Likewise, the Laurent expansion of the term
is
![]()
,
which is valid for
. Thus
the Laurent series of f(z) about the
point
is
,
which is valid for
,
where
. Therefore
, and
calculation reveals that
Example
8.8. Express
in
partial fractions.
Solution. In Example 8.7 use
and
. Computing
the residues, we obtain
![[Graphics:Images/ResidueCalcMod_gr_193.gif]](residuecalculus/ResidueCalcMod/Images/ResidueCalcMod_gr_193.gif)
The formula for f(z) in Example 8.7
gives us
Remark 8.1. If a
repeated root occurs, then the process is similar, and it is easy to
show that if P(z) has degree of at
most 2, then
,
where
Example
8.9. Express
in
partial fractions.
Solution. Using the Remark 8.1
and
and
, we
have
![]()
where
Thus,
Extra Example
1. Express of
in
partial fractions.
Explore Solution for Extra Example 1.
Exercises for Section 8.1. The Residue Theorem
Download This Mathematica Notebook
The Next Module is
Trigonometric Integrals via Contour Integrals
Return to the Complex Analysis Modules
Return to the Complex Analysis Project