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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  [Graphics:Images/ResidueCalcMod_gr_1.gif]  and  f  is an analytic function.  In this case, the singularity of the integrand is at worst a pole of order k at [Graphics:Images/ResidueCalcMod_gr_2.gif].  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 [Graphics:Images/ResidueCalcMod_gr_3.gif] and is an important extension of the previous method.

 

Definition 8.1 (Residue).  Let f(z) have a nonremovable isolated singularity at the point [Graphics:Images/ResidueCalcMod_gr_4.gif].  Then f(z) has the Laurent series representation for all z in some disk [Graphics:Images/ResidueCalcMod_gr_5.gif]given by  

            [Graphics:Images/ResidueCalcMod_gr_6.gif].  

The coefficient  [Graphics:Images/ResidueCalcMod_gr_7.gif]  of  [Graphics:Images/ResidueCalcMod_gr_8.gif]  is called the residue of f(z) at [Graphics:Images/ResidueCalcMod_gr_9.gif] and we use the notation  

            [Graphics:Images/ResidueCalcMod_gr_10.gif].  

 

Example 8.1.  If  [Graphics:Images/ResidueCalcMod_gr_11.gif],  then the Laurent series of f about the point [Graphics:Images/ResidueCalcMod_gr_12.gif] has the form   

            [Graphics:Images/ResidueCalcMod_gr_13.gif],  and

            [Graphics:Images/ResidueCalcMod_gr_14.gif].  

Explore Solution 8.1.

 

Example 8.2.  Find  [Graphics:Images/ResidueCalcMod_gr_23.gif]  if  [Graphics:Images/ResidueCalcMod_gr_24.gif].  

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  [Graphics:Images/ResidueCalcMod_gr_25.gif]  is  

            [Graphics:Images/ResidueCalcMod_gr_26.gif].  

Computing the first few coefficients, we obtain  

            [Graphics:Images/ResidueCalcMod_gr_27.gif]

Therefore,   [Graphics:Images/ResidueCalcMod_gr_28.gif].

Explore Solution 8.2.

 

    Recall that, for a function f(z) analytic in [Graphics:Images/ResidueCalcMod_gr_40.gif] and for any r with [Graphics:Images/ResidueCalcMod_gr_41.gif], the Laurent series coefficients of f(z) are given by  

(8-1)              [Graphics:Images/ResidueCalcMod_gr_42.gif]    for    [Graphics:Images/ResidueCalcMod_gr_43.gif],  
  
  where [Graphics:Images/ResidueCalcMod_gr_44.gif] denotes the circle [Graphics:Images/ResidueCalcMod_gr_45.gif] with positive orientation.  This gives us an important fact concerning  [Graphics:Images/ResidueCalcMod_gr_46.gif].  If we set  [Graphics:Images/ResidueCalcMod_gr_47.gif]  in Equation (8-1) and replace [Graphics:Images/ResidueCalcMod_gr_48.gif] with any positively oriented simple closed contour C containing [Graphics:Images/ResidueCalcMod_gr_49.gif], provided [Graphics:Images/ResidueCalcMod_gr_50.gif] is the still only singularity of f(z) that lies inside C, then we obtain  

(8-2)            [Graphics:Images/ResidueCalcMod_gr_51.gif].  

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  [Graphics:Images/ResidueCalcMod_gr_52.gif]  where [Graphics:Images/ResidueCalcMod_gr_53.gif] denotes the circle [Graphics:Images/ResidueCalcMod_gr_54.gif] with positive orientation.  

[Graphics:Images/ResidueCalcMod_gr_55.gif]

Solution.  In Example 8.1 we showed that the residue of  [Graphics:Images/ResidueCalcMod_gr_56.gif]  at  [Graphics:Images/ResidueCalcMod_gr_57.gif]  is  [Graphics:Images/ResidueCalcMod_gr_58.gif].  Using Equation (8-2), we get  

            [Graphics:Images/ResidueCalcMod_gr_59.gif].  

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  [Graphics:Images/ResidueCalcMod_gr_66.gif]  that lie inside C, then  

            [Graphics:Images/ResidueCalcMod_gr_67.gif].  

The situation is illustrated in Figure 8.1.

Figure 8.1 The domain D and contour C and the singular points [Graphics:Images/ResidueCalcMod_gr_68.gif] 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 [Graphics:Images/ResidueCalcMod_gr_69.gif] involves only the coefficient [Graphics:Images/ResidueCalcMod_gr_70.gif] in the Laurent expansion, we seek a method to calculate the residue from special information about the nature of the singularity at [Graphics:Images/ResidueCalcMod_gr_71.gif].

    If f(z) has a removable singularity at [Graphics:Images/ResidueCalcMod_gr_72.gif], then  [Graphics:Images/ResidueCalcMod_gr_73.gif]  for  [Graphics:Images/ResidueCalcMod_gr_74.gif].  Therefore,  [Graphics:Images/ResidueCalcMod_gr_75.gif].  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  [Graphics:Images/ResidueCalcMod_gr_76.gif],  then  [Graphics:Images/ResidueCalcMod_gr_77.gif].  

(ii)     If f(z) has a pole of order 2 at  [Graphics:Images/ResidueCalcMod_gr_78.gif],  then  [Graphics:Images/ResidueCalcMod_gr_79.gif].  

(iii)     If f(z) has a pole of order 3 at  [Graphics:Images/ResidueCalcMod_gr_80.gif],  then  [Graphics:Images/ResidueCalcMod_gr_81.gif].  

(v)     If f(z) has a pole of order k at  [Graphics:Images/ResidueCalcMod_gr_82.gif],  then  [Graphics:Images/ResidueCalcMod_gr_83.gif].  

Proof.

Proof of Theorem 8.2 is in the book.
Complex Analysis for Mathematics and Engineering

 

Example 8.4.  Find the residue of  [Graphics:Images/ResidueCalcMod_gr_84.gif]  at  [Graphics:Images/ResidueCalcMod_gr_85.gif].  

Solution.  We write  [Graphics:Images/ResidueCalcMod_gr_86.gif].  Because  [Graphics:Images/ResidueCalcMod_gr_87.gif]  has a zero of order 3 at [Graphics:Images/ResidueCalcMod_gr_88.gif] and  [Graphics:Images/ResidueCalcMod_gr_89.gif].  Thus f(z) has a pole of order 3 at [Graphics:Images/ResidueCalcMod_gr_90.gif].  By part (iii) of Theorem 8.2, we have  

            [Graphics:Images/ResidueCalcMod_gr_91.gif]  

This last limit involves an indeterminate form, which we evaluate by using L'Hôpital's rule:  

            [Graphics:Images/ResidueCalcMod_gr_92.gif]  

Explore Solution 8.4.

 

Example 8.5.  Find  [Graphics:Images/ResidueCalcMod_gr_104.gif]  where [Graphics:Images/ResidueCalcMod_gr_105.gif] denotes the circle [Graphics:Images/ResidueCalcMod_gr_106.gif] with positive orientation.  

[Graphics:Images/ResidueCalcMod_gr_107.gif]

Solution.  We write the integrand as  [Graphics:Images/ResidueCalcMod_gr_108.gif].  
The singularities of f(z) that lie inside [Graphics:Images/ResidueCalcMod_gr_109.gif] are simple poles at the points [Graphics:Images/ResidueCalcMod_gr_110.gif] and [Graphics:Images/ResidueCalcMod_gr_111.gif], and a pole of order 2 at the origin.  We compute the residues as follows:  

            [Graphics:Images/ResidueCalcMod_gr_112.gif]    
        
            [Graphics:Images/ResidueCalcMod_gr_113.gif]  
        
            [Graphics:Images/ResidueCalcMod_gr_114.gif]  

Finally, the residue theorem yields

            [Graphics:Images/ResidueCalcMod_gr_115.gif]

The answer,  [Graphics:Images/ResidueCalcMod_gr_116.gif],  is not at all obvious, and all the preceding calculations are required to get it.

Explore Solution 8.5.

 

Example 8.6.  Find  [Graphics:Images/ResidueCalcMod_gr_131.gif]  where [Graphics:Images/ResidueCalcMod_gr_132.gif] denotes the circle [Graphics:Images/ResidueCalcMod_gr_133.gif] with positive orientation.  

[Graphics:Images/ResidueCalcMod_gr_134.gif]

Solution.  The singularities of the integrand  [Graphics:Images/ResidueCalcMod_gr_135.gif]  that lie inside [Graphics:Images/ResidueCalcMod_gr_136.gif] are simple poles occurring at the points  [Graphics:Images/ResidueCalcMod_gr_137.gif],  as the points  [Graphics:Images/ResidueCalcMod_gr_138.gif],  lie outside [Graphics:Images/ResidueCalcMod_gr_139.gif].  Factoring the denominator is tedious, so we use a different approach.  If [Graphics:Images/ResidueCalcMod_gr_140.gif] is any one of the singularities of f(z) , then we can use L'Hôpital's rule to compute [Graphics:Images/ResidueCalcMod_gr_141.gif]:

            [Graphics:Images/ResidueCalcMod_gr_142.gif]

Since [Graphics:Images/ResidueCalcMod_gr_143.gif],  we can simplify this expression further to yield

            [Graphics:Images/ResidueCalcMod_gr_144.gif]

We now use the residue theorem to get  

            [Graphics:Images/ResidueCalcMod_gr_145.gif]  

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  

            [Graphics:Images/ResidueCalcMod_gr_160.gif],  
where
            [Graphics:Images/ResidueCalcMod_gr_161.gif]   

Solution.  It will suffice to prove that  [Graphics:Images/ResidueCalcMod_gr_162.gif].  We expand f(z) in its Laurent series about the point [Graphics:Images/ResidueCalcMod_gr_163.gif] by writing the three terms [Graphics:Images/ResidueCalcMod_gr_164.gif], [Graphics:Images/ResidueCalcMod_gr_165.gif], and [Graphics:Images/ResidueCalcMod_gr_166.gif] in their Laurent series about the point [Graphics:Images/ResidueCalcMod_gr_167.gif] and adding them.  The term [Graphics:Images/ResidueCalcMod_gr_168.gif] is itself a one-term Laurent series about the point [Graphics:Images/ResidueCalcMod_gr_169.gif].  The term [Graphics:Images/ResidueCalcMod_gr_170.gif] is analytic at the point [Graphics:Images/ResidueCalcMod_gr_171.gif], and its Laurent series is actually a Taylor series given by  

            [Graphics:Images/ResidueCalcMod_gr_172.gif]  

which is valid for  [Graphics:Images/ResidueCalcMod_gr_173.gif].  

Likewise, the Laurent expansion of the term [Graphics:Images/ResidueCalcMod_gr_174.gif] is  

            [Graphics:Images/ResidueCalcMod_gr_175.gif][Graphics:Images/ResidueCalcMod_gr_176.gif],  
        
which is valid for  [Graphics:Images/ResidueCalcMod_gr_177.gif].  Thus the Laurent series of f(z) about the point [Graphics:Images/ResidueCalcMod_gr_178.gif] is  

            [Graphics:Images/ResidueCalcMod_gr_179.gif],

which is valid for  [Graphics:Images/ResidueCalcMod_gr_180.gif], where  [Graphics:Images/ResidueCalcMod_gr_181.gif].  Therefore  [Graphics:Images/ResidueCalcMod_gr_182.gif],  and calculation reveals that

        [Graphics:Images/ResidueCalcMod_gr_183.gif]  

Explore Solution 8.7.

 

Example 8.8.  Express  [Graphics:Images/ResidueCalcMod_gr_190.gif]  in partial fractions.  

Solution.  In Example 8.7 use  [Graphics:Images/ResidueCalcMod_gr_191.gif]  and  [Graphics:Images/ResidueCalcMod_gr_192.gif].  Computing the residues, we obtain  

            [Graphics:Images/ResidueCalcMod_gr_193.gif]

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

            [Graphics:Images/ResidueCalcMod_gr_194.gif]  

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

            [Graphics:Images/ResidueCalcMod_gr_202.gif],  
where
            [Graphics:Images/ResidueCalcMod_gr_203.gif]   

 

Example 8.9.  Express  [Graphics:Images/ResidueCalcMod_gr_204.gif]  in partial fractions.  

Solution.  Using the Remark 8.1 and  [Graphics:Images/ResidueCalcMod_gr_205.gif]  and  [Graphics:Images/ResidueCalcMod_gr_206.gif],  we have  

            [Graphics:Images/ResidueCalcMod_gr_207.gif]

where  
            [Graphics:Images/ResidueCalcMod_gr_208.gif]    
            
            [Graphics:Images/ResidueCalcMod_gr_209.gif]    
                        
            [Graphics:Images/ResidueCalcMod_gr_210.gif]    

Thus,  
        [Graphics:Images/ResidueCalcMod_gr_211.gif]  

Explore Solution 8.9.

 

Extra Example 1.  Express  of  [Graphics:Images/ResidueCalcMod_gr_217.gif]  in partial fractions.  

Explore Solution for Extra Example 1.

 

Exercises for Section 8.1.  The Residue Theorem

 

Library Research Experience for Undergraduates

Residue Calculus  

Contour Integrals  

 

 

 

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