Residue (complex analysis)

The residue of a function $f$ at an isolated singularity $z_0$ is the coefficient $a_{-1}$ of $(z-z_0{)}^{-1}$ in the Laurent series of $f$ at $z_0$:

$\mathrm{Res}(f,z_0)=a_{-1}.$

The residue is the only Laurent coefficient that contributes to a closed-contour integral around $z_0$ — the rest integrate to zero. The Residue theorem makes this precise:

${\oint }_{C}f(z)dz=2\pi i{\sum }_k\mathrm{Res}(f,z_k)$

over singularities $z_k$ inside the positively oriented closed contour $C$.

Computing residues

The formula depends on the type of singularity.

Removable singularity

$\mathrm{Res}=0$. Removable singularities contribute nothing to contour integrals.

Simple pole (order 1)

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

If $f(z)=g(z)/(z-z_0)$ with $g$ analytic at $z_0$ and $g(z_0)\ne 0$: $\mathrm{Res}=g(z_0)$.

If $f(z)=p(z)/q(z)$ with $p(z_0)\ne 0$, $q(z_0)=0$, and $q^{\prime }(z_0)\ne 0$ — i.e., a simple zero of $q$ at $z_0$ that isn’t cancelled by $p$: the $p/q^{\prime }$ formula:

$\mathrm{Res}(f,z_0)=\frac{p(z_0)}{q^{\prime }(z_0)}.$

Proof: near $z_0$, $q(z)\approx q^{\prime }(z_0)(z-z_0)$, so $f(z)\approx p(z_0)/(q^{\prime }(z_0)(z-z_0))$. Residue is $p(z_0)/q^{\prime }(z_0)$.

Pole of order $m$

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

The factor $(z-z_0{)}^m$ “clears” the pole, then differentiating $m-1$ times and dividing by $(m-1)!$ extracts the $a_{-1}$ coefficient.

For $m=2$: $\mathrm{Res}={\lim }_{z\to z_0}\frac{d}{dz}[(z-z_0{)}^{2}f(z)]$.

For $m=1$: reduces to the simple-pole formula.

Essential singularity

Expand the Laurent series explicitly and read off $a_{-1}$. No formula.

Worked computations

Simple pole, direct. $f(z)=(3z+1)/(z-2)$ at $z_0=2$. $\mathrm{Res}={\lim }_{z\to 2}(z-2)f(z)={\lim }_{z\to 2}(3z+1)=7$.

Simple pole via $p/q^{\prime }$. $f(z)=e^z/(z^2+1)$ at $z_0=i$. $p(z)=e^z$, $q(z)=z^2+1$, $q^{\prime }(z)=2z$. $\mathrm{Res}=p(i)/q^{\prime }(i)=e^i/(2i)=e^i/(2i)$. Multiply numerator and denominator by $-i$: $-i{e}^i/2$. So $\mathrm{Res}=-i{e}^i/2$.

Pole of order 2. $f(z)=e^z/z^2$ at $z_0=0$. By formula with $m=2$:

$\mathrm{Res}={\lim }_{z\to 0}\frac{d}{dz}[z^2\cdot e^z/z^2]={\lim }_{z\to 0}\frac{d}{dz}[e^z]={\lim }_{z\to 0}{e}^z=1.$

Verify with Laurent: $e^z/z^2=1/z^2+1/z+1/2!+z/3!+\cdots$. Coefficient of $1/z$: $1$. ✓

Pole of order 3 via Laurent expansion. $f(z)=\sin z/z^4$ at $z_0=0$. Laurent: $\sin z=z-z^3/6+z^5/120-\cdots$, so $f=1/z^3-1/(6z)+z/120-\cdots$. Coefficient of $1/z$: $-1/6$. So $\mathrm{Res}=-1/6$.

(Using the formula with $m=3$ would require taking two derivatives of $z^3\sin z/z^4=\sin z/z$; the Laurent approach is faster.)

Essential singularity. $f(z)=e^{1/z}$ at $z_0=0$. Laurent: $1+1/z+1/(2!z^2)+\cdots$. Coefficient of $1/z$: $1$. So $\mathrm{Res}=1$.

In context

Residue computation is the heart of practical contour-integral calculation. The Residue theorem reduces any closed-contour integral with isolated singularities to a finite sum of residues — and residues themselves are computed by short formulas or direct Laurent expansion.

This collapses calculations that would be intractable by real methods. Real improper integrals like ${\int }_{-\infty }^{\infty }1/(x^4+1)dx$ become trivial via contour integration plus residues.

The full toolkit lives in the closed-contour decision tree: identify singularities inside $C$, classify each, compute residue by the right formula, sum, multiply by $2\pi i$. See Residue theorem for the master template.