Residue at an Isolated Complex Singularity

Analysis, Measure & PDE · Undergraduate · Mathematics Resources editorial compilation (AI-assisted)

The residue is the coefficient of the inverse first power in a local Laurent expansion.

Overview

A residue extracts one coefficient from the behavior of a holomorphic function around an isolated singularity. That coefficient measures the function's contribution to a contour integral. It is a local quantity with a global use: contributions from finitely many enclosed singularities can be added to evaluate an integral.

A residue does not classify a singularity by itself. Different singularities can have the same residue, including zero, while having very different growth near the exceptional point.

Prerequisites

Understand holomorphic functions, oriented contour integration, and Laurent series. A Laurent expansion permits negative as well as nonnegative powers, and represents a holomorphic function on an annulus. Positive orientation around a disk means counterclockwise traversal.

Definition and conventions

Suppose $f$ is holomorphic on $0<|z-a|<r$ . In its Laurent expansion

f(z)=\sum_{k=-\infty}^{\infty}c_k(z-a)^k,

the residue is $\operatorname{Res}(f,a)=c_{-1}$ . Equivalently, for $0<\rho<r$ ,

\operatorname{Res}(f,a)=\frac1{2\pi i}\int_{|z-a|=\rho}f(z)\,dz.

The circle has positive orientation. The residue at infinity is a separate convention, defined through the coordinate $w=1/z$ with the differential's sign taken into account.

Intuition

Integrating a Laurent term around a circle makes every power cancel except $(z-a)^{-1}$ . That exceptional term records one complete winding around the missing point. The residue is therefore the coefficient visible to circular integration, rather than the whole singular behavior.

Examples and non-examples

For

f(z)=\frac{z+2}{(z-1)(z+1)},

the simple-pole formula gives residue $3/2$ at $1$ . At $-1$ , canceling the factor $z+1$ gives $(z+2)/(z-1)$ , whose value there is $-1/2$ . The enclosed residues sum to one, so on the positively oriented circle $|z|=2$ the integral is $2\pi i(3/2-1/2)=2\pi i$ .

The function $1/z^2$ has a pole of order two at zero but residue zero. The function $e^{1/z}$ has an essential singularity and residue one, since its expansion contains $1/z$ with that coefficient. Thus the residue need not be nonzero at a pole, and it need not require a pole at all.

Important properties

At a simple pole, $\operatorname{Res}(f,a)=\lim_{z\to a}(z-a)f(z)$ . More generally, if $f=g/h$ with holomorphic $g,h$ , $h(a)=0$ , and $h'(a)\ne0$ , then the residue is $g(a)/h'(a)$ .

For a bounded region with suitable piecewise smooth boundary, if $f$ is holomorphic near its closure except at finitely many isolated interior singularities and none lies on the boundary, the contour integral is $2\pi i$ times the sum of residues. The proof removes small disks around the exceptional points, applies Cauchy's theorem to the remaining region, and integrates the Laurent expansions on the small circles.

Connections and further reading

Applying the residue theorem to $f'/f$ yields the argument principle, counting zeros against poles with multiplicity. Real improper integrals require additional estimates on any closing arcs; the residue calculation alone does not justify their disappearance. Taylor, §16 states the local coefficient convention and contour formula.