COMPLEX ANALYSIS: Mathematica 4.1 Notebooks
(c) John H. Mathews, and
Russell W. Howell, 2002
Complimentary software to accompany our textbook:

COMPLEX ANALYSIS: for Mathematics and Engineering, 
4th Edition,  2001,  ISBN: 0-7637-1425-9
Jones & Bartlett Publishers, Inc.  
40 Tall Pine Drive, Sudbury, MA 01776  
Tele.  (800) 832-0034, FAX:  (508)  443-8000  
E-mail:  mkt@jbpub.com  
Internet:  http://www.jbpub.com/  
This free software is compliments of the authors.
John H. Mathews,   mathews@fullerton.edu 
Russell W. Howell,  howell@westmont.edu

CHAPTER 8   RESIDUE THEORY

a = t = x = y = z = Z = 
                  0; 
Clear[A, center, centers, dot, dots, f, f0, F, g, G, i, p, R, Res, 
        Res1, Res2, Res3, Res4, Res5, Res6, solset, sub, S, t, θ, val, 
        x, y, z, z1, z2, z3, z4, z5, z6, Z]; 
Needs["Graphics`Colors`"]; 
Needs["Graphics`FilledPlot`"];

Section 8.1  The Residue Theorem

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  ((f (z))/(((z  -  (Subscript[z, 0])))^k))  and  f  is an analytic function.  In this case, the singularity of the integrand is at worst a pole of order  k  at  (Subscript[z, 0]).  In this section we extend this result to integrals that have a finite number of isolated singularities that lie inside the contour  C.  This new method can be used in cases where the integrand has an essential singularity at  (Subscript[z, 0])  and is an important extension of the previous method.

Definition 8.1 (Residue),  Page 307.  Let f(z) have a nonremovable isolated singularity at the point  (Subscript[z, 0]).  Then f(z) has the Laurent series representation for all z in some disk \(\(D\_R\%*\) \((z\_0)\)\)given by  

			f (z) = (∑_(n = -∞)^∞) 
                (Subscript[a, n]) (((z - (Subscript[z, 0])))^n).  

The coefficient  (Subscript[a, -1])  of  (1/(z  -  (Subscript[z, 0])))  is called the residue of  f  at  (Subscript[z, 0]) and we use the notation  Res[f, (Subscript[z, 0])] = (Subscript[a, -1]).

Example 8.1,  Page 307.  Find the residue at z = 0 for the function  f[z] = (e^((2/z))).

Solution 8.1.

Enter the function  f[z] = (e^((2/z)))  and construct the Laurent series about  z = 0.

Remove[z]; 
Clear[f, Res, S]; 
f[z_]  =  (E^((2/z))); 
S[z_]  =  Series[f[z], {z, ∞, 5}]; 
Print["f[z] = ", f[z]]; 
Print["S[z] = ", S[z]];

The coefficient of  (1/z)  is  2  so the residue is Res[f, 0] = 2.

Res  =  (Subscript[(S[z]), [[3, 2]]]); 
Print["Res[f,0] = Res[", f[z], ",0] = ", 
        Res];

Aside. We can verify this computation with Mathematica's built in "Residue" procedure.

Res = 
        Residue[(E^((2/z))), {z, 0}]; 
Print["Res[f,0] = Res[", f[z], ",0] = ", 
        Res];

Example 8.2,  Page 308.  Find the residue at  z = 0  for the function  g[z] = (3/(2  z  +  (z^2)  -  (z^3))).

Solution 8.2.

Enter the function  g[z] = (3/(2  z  +  (z^2)  -  (z^3)))  and construct the Laurent series about  z = 0.

Remove[z]; 
Clear[g, Res, S, z]; 
g[z_]  =  (3/(2  z  +  (z^2)  -  (z^3))); 
S[z_]  =  Series[g[z], {z, 0, 4}]; 
Print["g[z] = ", g[z]]; 
Print["S[z] = ", S[z]];

The coefficient of  (1/z)  is  (3/2)  so the residue is  Res[g, 0] = (3/2).

Res  =  (Subscript[(S[z]), [[3, 1]]]); 
Print["Res[g,0] = Res[", g[z], ",0] = ", 
        Res];

Aside. We can verify this computation with Mathematica's built in "Residue" procedure.

Res  =  Residue[g[z], {z, 0}]; 
Print["Res[g,0] = Res[", g[z], ",0] = ", 
        Res];

Let us recall that, for a function f analytic in \(\(D\_R\%*\) \((z\_0)\)\) and for any r with 0 < r < R, the Laurent series coefficients of f are  

			  \(a\_n = \(1\/\(2  π\ i\)\) \(∫\_\(\(C\_r\%+\) \((z\_0)\)\)\(\(f \((ξ)\)\)\/\((ξ\  - \ z\_0)\)\^\(n + 1\)\) dξ\)\)  for  n = 0, ±1, ±2,  ... ,  
  
  where \(\(C\_r\%+\) \((z\_0)\)\) denotes the circle \(\(C\_r\%+\) \((z\_0)\) = {z : \ \(\(|\)\(z - z\_0\)\(|\)\) = r}\) with positive orientation.  

This gives us an important fact concerning  Res[f, (Subscript[z, 0])].  If we set  n = -1  in the integrand and replace \(\(C\_r\%+\) \((z\_0)\)\) with any positively oriented simple closed contour C containing (Subscript[z, 0]), provided (Subscript[z, 0]) is the still only singularity of f that lies inside C, then we obtain

			\(∫\_C\ 
          f \((z)\) dz\  = \ \(2  π\ i\ a\_\(-1\)\  = \ 2  π\ i\ Res[f, z\_0]\)\).  

If we are able to find the Laurent series expansion for f, then above equation gives us an important tool for evaluating contour integrals.

Example 8.3,  Page 308.  Use residues to integrate  (∫_C) (e^(2/z)) dz  around the circle  C : |z| = 1.

Solution 8.3.

From Example 8.1  we have  Res[f,0]  =  2.

Remove[z]; 
Clear[f, Res, val]; 
f[z_]  =  (E^((2/z))); 
Res  =  Residue[f[z], {z, 0}]; 
Print["Res[f,0] = Res[", f[z], ",0] = ", 
        Res];

Thus the value of the contour integral is given by the following computation.

val  =  
        2  π I Res; 
(Subscript[Z, 1]) = 0; 
Print["Res[f,0] = Res[", f[z], ",0] = ", 
        Res]; 
dot = 
        Graphics[{{Red, PointSize[0.03], 
              Point[{Re[(Subscript[Z, 1])], Im[(Subscript[Z, 1])]}]}}]; 
\(\(center = 
        Graphics[{Text[\*""\!\(z\_1\)"", {Re[Z\_1], 
                Im[Z\_1] + 0.05}, {\(-1\), 
                0}]}];\)\ \)
contour  =  Graphics[{Magenta, Circle[{0, 0}, 1]}]; 
Show[dot, center, contour, 
        PlotRange -> {{-1.3, 1.3}, {-1.3, 1.3}}, 
        Ticks -> {Range[-1, 1, (1/2)], 
            Range[-1, 1, (1/2)]}, 
AspectRatio -> 
          1, Axes -> True, 
        AxesLabel -> {"x", "y"}]; 
\(Print[\*""\!\(∮\+C\) "", 
      f[z], " \[DifferentialD]z = 2πI Res[", 
      f[z], ",0] = ", val]\)
Print["The contour is the unit circle  C:|z| = 1,"]; 
Print["with positive orientation."];

Theorem 8.1,  Page 309.  (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  (Subscript[z, 1]), (Subscript[z, 2]),   ... , (Subscript[z, n])  that lie inside  C,  then  

			(∫_C) f (z) dz = 
        2  π i (∑_(k = 1)^n) Res[f, (Subscript[z, k])].

Proof of Theorem 8.1, see text Page 309.

Section 8.2	Calculation of Residues

Section 8.3	Trigonometric Integrals

Section 8.4	Improper Integrals of Rational Functions

Section 8.5	Improper Integrals Involving Trigonometric Functions

Section 8.6	Indented Contour Integrals

Section 8.7	Integrands with Branch Points

Section 8.8	The Argument Principle and Rouche's Theorem

GO TO  THE  Table of Contents