Residues and the Residue Theorem

The residue theorem converts contour integrals into algebraic computations involving the singularities of the integrand. It is the central tool for evaluating integrals in complex analysis.

Definition of Residue

Definition6.1Residue

Let $f$ have an isolated singularity at $z_0$ with Laurent expansion $f(z) = \sum_{n=-\infty}^\infty a_n(z-z_0)^n$ in $0 < |z-z_0| < R$ . The residue of $f$ at $z_0$ is

$\text{Res}(f, z_0) = a_{-1} = \frac{1}{2\pi i}\oint_\gamma f(z)\,dz$

where $\gamma$ is any small positively oriented circle around $z_0$ .

RemarkComputing residues

For practical computation:

Simple pole ( $m = 1$ ): $\text{Res}(f, z_0) = \lim_{z \to z_0}(z - z_0)f(z)$ .
If $f = p/q$ with simple zero of $q$ : $\text{Res}(f, z_0) = p(z_0)/q'(z_0)$ .
Pole of order $m$ : $\text{Res}(f, z_0) = \frac{1}{(m-1)!}\lim_{z \to z_0}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)]$ .
Essential singularity: Extract $a_{-1}$ from the Laurent series directly.

ExampleResidue calculations

$f(z) = \frac{e^z}{z^2}$ at $z=0$ (pole of order 2): $\text{Res} = \frac{d}{dz}e^z\big|_{z=0} = 1$ .
$f(z) = \frac{z}{z^2+1}$ at $z=i$ (simple pole): $\text{Res} = \frac{i}{2i} = \frac{1}{2}$ .
$f(z) = \frac{\cos z}{z}$ at $z=0$ : Since $\cos z/z = 1/z - z/2 + \cdots$ , $\text{Res} = 1$ .

The Residue Theorem

Theorem6.1Residue theorem

Let $f$ be holomorphic on a domain $D$ except for isolated singularities $z_1, \ldots, z_n$ . If $\gamma$ is a positively oriented simple closed contour in $D$ with $z_1, \ldots, z_n$ in its interior, then

$\oint_\gamma f(z)\,dz = 2\pi i \sum_{k=1}^n \text{Res}(f, z_k).$

ExampleApplication of the residue theorem

Evaluate $\oint_{|z|=2} \frac{z^2}{(z-1)(z^2+1)}\,dz$ .

The singularities inside $|z|=2$ are $z = 1, i, -i$ . Computing residues:

$\text{Res}(f, 1) = \frac{1}{1+1} = \frac{1}{2}, \quad \text{Res}(f, i) = \frac{-1}{(i-1)(2i)} = \frac{1+i}{4}$ $\text{Res}(f, -i) = \frac{-1}{(-i-1)(-2i)} = \frac{1-i}{4}.$

Sum: $\frac{1}{2} + \frac{1+i}{4} + \frac{1-i}{4} = \frac{1}{2} + \frac{1}{2} = 1$ . Therefore the integral is $2\pi i$ .

Residue at Infinity

Definition6.2Residue at infinity

For $f$ holomorphic in $|z| > R$ , the residue at infinity is

$\text{Res}(f, \infty) = -\frac{1}{2\pi i}\oint_{|z|=\rho} f(z)\,dz = -a_{-1}$

where the integral is over a large circle (positively oriented) and $a_{-1}$ is the coefficient of $z^{-1}$ in the Laurent expansion at $\infty$ . Equivalently, $\text{Res}(f, \infty) = -\text{Res}\left(\frac{1}{z^2}f(1/z), 0\right)$ .

RemarkGlobal residue theorem

If $f$ is meromorphic on $\hat{\mathbb{C}}$ (the Riemann sphere) with singularities $z_1, \ldots, z_n$ (including possibly $\infty$ ), then

$\sum_{k=1}^n \text{Res}(f, z_k) + \text{Res}(f, \infty) = 0.$

The sum of all residues on the Riemann sphere is always zero.