For a

complex function with an isolated

singularity at a point z

_{0}, the residue at z

_{0} is the a

_{-1} coefficient of the function's

Laurent series about z

_{0}:

inf
-- n
f(z) = > a_n (z - z_0)
--
n=-inf
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 )
/c --
z_0

Calculating residues at poles is easy. Suppose f(z) has an nth order pole at z_{0}, i.e., it can be written in the form f(z) = phi(z)/(z-z_{0})^{n}. Then

(n-1)
phi (z_0)
Res( f(z); z_0 ) = ------------
(n-1)!
(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 z
= - + 1 + - + ...
z 2

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.