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Residue Theorem - Residue Calculus

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

.

Explore Solution 8.1.

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

Therefore,   .

Explore Solution 8.2.

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.

Solution.  In Example 8.1 we showed that the residue of    at    is  .  Using Equation (8-2), we get

.

Explore Solution 8.3.

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.

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.

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:

Explore Solution 8.4.

Example 8.5.  Find    where denotes the circle with positive orientation.

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

The answer,  ,  is not at all obvious, and all the preceding calculations are required to get it.

Explore Solution 8.5.

Example 8.6.  Find    where denotes the circle with positive orientation.

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 :

Since ,  we can simplify this expression further to yield

We now use the residue theorem to get

Explore Solution 8.6.

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

Explore Solution 8.7.

Example 8.8.  Express    in partial fractions.

Solution.  In Example 8.7 use    and  .  Computing the residues, we obtain

The formula for f(z) in Example 8.7 gives us

Explore Solution 8.8.

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,

Explore Solution 8.9.

Extra Example 1.  Express  of    in partial fractions.

Explore Solution for Extra Example 1.

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(c) 2012 John H. Mathews, Russell W. Howell