Residue

The residue of a complex function $f$ at an isolated singularity $z_0$ is the coefficient $A_{-1}$ of $(z-z_0{)}^{-1}$ in its Laurent Series expansion. It is the unique part of the singularity that survives a contour integral.

When you integrate $(z-z_0{)}^n$ around a closed contour, every $n\ne -1$ gives zero because $(z-z_0{)}^n$ has an antiderivative. The $n=-1$ term is the exception: its antiderivative $\log (z-z_0)$ picks up a branch contribution of $2\pi i$ each time the contour winds around $z_0$:

$${\oint }_{\gamma }(z-z_0{)}^{n}dz=\{\begin{array}{ll}2\pi i& n=-1\\ 0& n\ne -1\end{array}$$

Integrating the Laurent series term by term, all coefficients cancel except $A_{-1}$, giving:

$${\oint }_{\gamma }f(z)dz=2\pi i\cdot A_{-1}$$

So via the Cauchy Integral Formula:

$$\mathrm{Res}(f,z_0)=A_{-1}=\frac{1}{2\pi i}{\oint }_{\gamma }f(z)dz$$

where $\gamma$ is any simple positively-oriented contour encircling $z_0$ and no other singularity. A function can have arbitrarily bad singular behavior at $z_0$, but none of it affects the contour integral. Only the $-1$ power does.

Computing Residues

For a simple pole (order 1):

$$\mathrm{Res}(f,z_0)=\underset{z\to z_0}{\lim }(z-z_0)f(z)$$

For a pole of order $m$:

$$\mathrm{Res}(f,z_0)=\frac{1}{(m-1)!}\underset{z\to z_0}{\lim }\frac{d^{m-1}}{d{z}^{m-1}}\left[(z-z_0{)}^{m}f(z)\right]$$