Residue

Consider a function $f$ holomorphic in $B (z_{0}, r) - {z_{0}}$ , the point $z_{0}$ being a (possibly removable) isolated singularity. We call residue of $f$ at $z_{0}$ , denoted by $Res (f, z_{0})$ , to the term $b_{1}$ in the corresponding Laurent series.

Residue Calculation for Poles

For a pole of order $m$ :

f (z) = \frac{b_{m}}{(z - z_{0})^{m}} + \dots + \frac{b_{2}}{(z - z_{0})^{2}} + \frac{b_{1}}{z - z_{0}} + a_{0} + \dots

Multiply to cancel the singularity:

(z - z_{0})^{m} f (z) = b_{m} + b_{m - 1} (z - z_{0}) + \dots + b_{1} (z - z_{0})^{m - 1} + a_{0} (z - z_{0})^{m} + \dots

Differentiate $(m - 1)$ times:

\frac{d^{m - 1}}{d z^{m - 1}} [(z - z_{0})^{m} f (z)] = (m - 1)! b_{1} + terms with (z - z_{0})

Evaluate at $z = z_{0}$ :

b_{1} = \frac{1}{(m - 1)!} lim_{z \to z_{0}} \frac{d^{m - 1}}{d z^{m - 1}} [(z - z_{0})^{m} f (z)]

Since $Res (f, z_{0}) = b_{1}$ :

Res (f, z_{0}) = \frac{1}{(m - 1)!} lim_{z \to z_{0}} \frac{d^{m - 1}}{d z^{m - 1}} [(z - z_{0})^{m} f (z)]

Special Cases

Simple Pole ( $m = 1$ )

Res (f, z_{0}) = lim_{z \to z_{0}} (z - z_{0}) f (z)

Double Pole ( $m = 2$ )

Res (f, z_{0}) = lim_{z \to z_{0}} \frac{d}{d z} [(z - z_{0})^{2} f (z)]