COMPLEX ANALYSIS: Maple Worksheets, 2001
(c) John H. Mathews Russell W. Howell
mathews@fullerton.edu howell@westmont.edu

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COMPLEX ANALYSIS: for Mathematics & Engineering, 4th Ed, 2001, ISBN: 0-7637-1425-9
Jones and Bartlett Publishers, Inc., 40 Tall Pine Drive, Sudbury, MA 01776
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CHAPTER 8 RESIDUE THEORY

Section 8.1 The Residue Theorem

The Cauchy-integral formulas 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-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 z[0] . In this section we extend this result to integrals that have a finite number of isolated singularities and lie inside the contour C . This new method can be used in cases where the integrand has an essential singularity at z[0] and is an important extension of the previous method.

 

Definition 8.1: Residue

 

Let f(z) have a nonremovable isolated singularity at the point z[0] . Then f(z) has
the Laurent series representation
f(z) = sum(a[n]*(z-z0)^n,n = -infinity .. infinity)... . The coefficient a[-1] of 1/(z-z[0]) is
called the
residue of f(z) at z[0] and we use the notation Res[f , z[0] ] = a[-1] .

 

Load Maple's "residue" procedure.
Make sure this is done only ONCE during a Maple session.

> readlib(residue):

Example 8.1, Page 307. Use Laurent series to find the residue at z = 0 for the function f(z) = exp(2/z) .

> f:='f': s:='s': z:='z':
f := z -> exp(2/z):
s := series(f(z), z=infinity, 7):
`f(z) ` = f(z);
`f(z) ` = s;

`f(z) ` = exp(2*1/z)

`f(z) ` = 1+2/z+2/z^2+4/3/z^3+2/3/z^4+4/15/z^5+4/45...

 

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


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

> g:='g': s:='s': z:='z':
g := z -> 3/(2*z + z^2 - z^3):
s := series(g(z), z=0, 5):
`g(z) ` = g(z);
`g(z) ` = s;

`g(z) ` = 3*1/(2*z+z^2-z^3)

`g(z) ` = series(3/2*z^(-1)-3/4+9/8*z-15/16*z^2+33/...

 

The coefficient of 1/z is 3/2 so the residue is `Res[g,0]` = 3/2 .
We compare this with Maple's residue procedure for computing residues.

> `g(z) ` = g(z);
`g(z) ` = s;
`Res[g,0] ` = residue(g(z), z=0);

`g(z) ` = 3*1/(2*z+z^2-z^3)

`g(z) ` = series(3/2*z^(-1)-3/4+9/8*z-15/16*z^2+33/...

`Res[g,0] ` = 3/2

 

Example 8.3, Page 308. Use residues to integrate int(exp(2/z),z = C .. `   `) around C : abs(z) = 1 .
From Example 8.1
`Res[f,0]` = 2 . Thus the value of the integral is 2*pi*i*`Res[f,0]` .

> f:='f': F:='F': s:='s': z:='z':
f := z -> exp(2/z):
`F(z) ` = f(z);
s := series(f(z), z=infinity, 5):
res := 2:
`Res[F,0] ` = res;
print(int(F(z),z=C..``) = `2*Pi*I*Res[f,0])`);
print(int(F(z),z=C..``) = 2*Pi*I*res);

`F(z) ` = exp(2*1/z)

`Res[F,0] ` = 2

int(F(z),z = C .. ``) = `2*Pi*I*Res[f,0])`

int(F(z),z = C .. ``) = 4*I*Pi

 

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 is analytic inside C and on C , except at the points z[1] , z[2] , ..., z[n] that lie inside C , then

 

Int(f(z),z = C .. ``) = 2*pi*i Sum(`Res[ f , `*z[k]*`]`,k = 1 .. n) .

 

End of Section 8.1.