For a complex function
with an isolated singularity
at a point z0
, the residue at z0
is the a-1
coefficient of the function's Laurent series
f(z) = > a_n (z - z_0)
Res( f(z); z_0 ) = a_-1
Residues are used with the Residue theorem to calculate contour integrals of functions in the complex plane more easily. The value of an integral over a closed curve C is simply 2*pi*i times the sum of the residues of the singularities within it.
| f(z) dz = 2*pi*i > Res( f(z); z_0 )
Calculating residues at poles is easy. Suppose f(z) has an nth order pole at z0, i.e., it can be written in the form f(z) = phi(z)/(z-z0)n. Then
Res( f(z); z_0 ) = ------------
(Superscript here indicating the (n-1)st derivative at z_0).
Example: Find the residue of f(z) = (e^z)/z at z=0.
The Laurent series about the origin is
1 / z^2 z^3 \
f(z) = - * | 1 + z + --- + --- + ... |
z \ 2 3! /
= - + 1 + - + ...
So by examination of the first term, the residue is just 1. Therefore any contour integral of f(z) around a closed curve containing the origin will be 2*pi*i.