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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 about z0:
```       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 z0, i.e., it can be written in the form f(z) = phi(z)/(z-z0)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.