The Residue Theorem

The residue theorem is the culmination of the theory of complex integration, providing a powerful method to evaluate contour integrals by reducing them to algebraic computations at isolated singularities.

Statement

Theorem6.6Residue theorem

Let $D$ be a simply connected domain and let $f$ be holomorphic on $D$ except at finitely many isolated singularities $z_1, \ldots, z_n \in D$ . If $\gamma$ is a positively oriented simple closed contour in $D$ enclosing all singularities, then

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

Proof

Surround each singularity $z_k$ by a small circle $C_k$ of radius $\varepsilon_k$ (chosen small enough so the circles are disjoint and contained inside $\gamma$ ). By Cauchy's theorem for multiply connected domains, applied to the region between $\gamma$ and the circles $C_k$ :

$\oint_\gamma f\,dz = \sum_{k=1}^n \oint_{C_k} f\,dz.$

By definition of residue, $\oint_{C_k} f\,dz = 2\pi i \cdot \text{Res}(f, z_k)$ . Therefore:

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

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Systematic Methods for Computing Residues

RemarkResidue formulas catalog

Let $f$ have an isolated singularity at $z_0$ .

1. Simple pole ( $m=1$ ): $\text{Res}(f, z_0) = \lim_{z \to z_0}(z-z_0)f(z).$

2. Simple pole of $p(z)/q(z)$ where $q(z_0) = 0$ , $q'(z_0) \neq 0$ : $\text{Res}(p/q, z_0) = \frac{p(z_0)}{q'(z_0)}.$

3. Pole of order $m$ : $\text{Res}(f, z_0) = \frac{1}{(m-1)!}\lim_{z \to z_0}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)].$

4. Essential singularity: Compute the Laurent series and read off $a_{-1}$ .

ExampleComplete evaluation example

Evaluate $I = \oint_{|z|=3} \frac{z^2 e^z}{(z-1)^2(z+2)}\,dz$ .

Singularities inside $|z|=3$ : $z=1$ (pole of order 2) and $z=-2$ (simple pole).

At $z = 1$ : $\text{Res} = \frac{d}{dz}\left[\frac{z^2 e^z}{z+2}\right]_{z=1} = \frac{d}{dz}\left[\frac{z^2 e^z}{z+2}\right]_{z=1}$ .

Computing: $\frac{d}{dz}\frac{z^2 e^z}{z+2} = \frac{(2z+z^2)e^z(z+2) - z^2 e^z}{(z+2)^2}$ . At $z=1$ : $\frac{3e \cdot 3 - e}{9} = \frac{8e}{9}$ .

At $z = -2$ : $\text{Res} = \frac{4e^{-2}}{(-3)^2} = \frac{4e^{-2}}{9}$ .

Therefore $I = 2\pi i\left(\frac{8e}{9} + \frac{4e^{-2}}{9}\right) = \frac{2\pi i(8e + 4e^{-2})}{9}$ .

The Residue Theorem on the Riemann Sphere

Theorem6.7Global residue theorem

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

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

ExampleVerification of global residue theorem

For $f(z) = \frac{1}{z(z-1)}$ : residues at $z=0$ : $-1$ ; at $z=1$ : $1$ . At $\infty$ : $\text{Res}(f, \infty) = -\text{Res}(\frac{1}{w^2}\cdot\frac{1}{(1/w)(1/w-1)}, 0) = -\text{Res}(\frac{-1}{w(1-w)}, 0) = -(-1) = 1$ . Wait: sum should be $0$ . Rechecking: $\text{Res}(f, \infty) = 0$ , and indeed $-1 + 1 + 0 = 0$ .

RemarkPower of the residue theorem

The residue theorem unifies many seemingly different results: the Cauchy integral formula (residue of $f(z)/(z-z_0)$ at $z_0$ is $f(z_0)$ ), the argument principle (residue of $f'/f$ ), and the evaluation of definite integrals. It is one of the most versatile tools in all of mathematics.