The Residue Theorem

In this lesson you’ll state the residue theorem, learn the formulas for computing residues at simple and higher-order poles, and use them to evaluate contour integrals that would be hopeless by parametrisation.

Residue theorem. Let $C$ be a positively oriented simple closed contour, and let $f$ be analytic on and inside $C$ except at finitely many isolated singularities $z_1,\ldots,z_n$ inside $C$. Then

$$\oint_C f(z)\,dz = 2\pi i\sum_{k=1}^{n}\operatorname{Res}_{z=z_k}f$$

The residue at $z_k$ is the Laurent coefficient $c_{-1}$ of the expansion about that point.

This is the culmination of the section, and everything in it has been preparation. From Cauchy’s theorem we know the contour can be shrunk onto small circles around each singularity. From Laurent series we know each such circle contributes $2\pi i c_{-1}$. Put those together and the theorem is immediate.

A global integral has become a finite sum of local data. Note what has been discarded: the shape of the contour, the behaviour of $f$ away from the singularities, everything except one number per enclosed pole.

The figure checks it. A circle of radius 2.4 encloses two of the three poles of $\frac{1}{(z-1)(z+1)(z-4)}$; the residues at those two are computed from the formula, and the contour integral is computed numerically. They agree to numerical precision, and the excluded pole contributes nothing.

You almost never build the Laurent series. Use these instead.

Simple pole.

$$\operatorname{Res}_{z=z_0}f = \lim_{z\to z_0}(z-z_0)f(z)$$

Simple pole of a quotient. If $f = \frac{p}{q}$ with $p(z_0)\ne0$ and $q$ having a simple zero at $z_0$, then

$$\operatorname{Res}_{z=z_0}\frac{p}{q} = \frac{p(z_0)}{q'(z_0)}$$

This is the most useful formula in the lesson. It avoids all factoring, and it works beautifully for things like $\frac{1}{\sin z}$ where you cannot factor at all.

Pole of order $m$.

$$\operatorname{Res}_{z=z_0}f = \frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}\Big[(z-z_0)^mf(z)\Big]$$

The differentiation gets unpleasant for large $m$; for $m \ge 3$ it is often faster to expand the series directly.

Essential singularity. No formula exists. Expand the Laurent series and read off $c_{-1}$.

Order and residue are independent. A pole of order 5 can have residue 0, as $\frac{\sin z}{z^3}$ showed last lesson with order 2 and residue 0.

Removable singularities have residue 0, so they can be ignored entirely.

Residues can cancel. $\oint_{|z|=2}\frac{dz}{z^2-1} = 0$ because the two residues are $\frac12$ and $-\frac12$. Zero does not imply no poles.

Only enclosed singularities matter. This bears repeating because it is the most common source of error: locate every singularity, then check each one against the contour.

Sometimes it is faster to work from outside. Define

$$\operatorname{Res}_{z=\infty}f = -\operatorname{Res}_{w=0}\left[\frac{1}{w^2}f\!\left(\frac1w\right)\right]$$

Then the sum of all residues, including the one at infinity, is zero for any rational function. So a contour enclosing many poles can be evaluated by computing the single residue at infinity, which is often much less work.

Example 1: Find the residue of $\frac{e^z}{z-2}$ at $z=2$.

Solution. Simple pole, so multiply and take the limit:

$$\operatorname{Res} = \lim_{z\to2}(z-2)\cdot\frac{e^z}{z-2} = e^2$$

∎

Example 2: Evaluate $\displaystyle\oint_{|z|=3}\frac{dz}{z^2+1}$.

Solution. Poles at $\pm i$, both inside $|z|=3$. Use the quotient formula with $p=1$, $q=z^2+1$, $q'=2z$:

$$\operatorname{Res}_{z=i} = \frac{1}{2i}, \qquad \operatorname{Res}_{z=-i} = \frac{1}{-2i}$$

Sum: $\frac{1}{2i}-\frac{1}{2i} = 0$, so

$$\oint_{|z|=3}\frac{dz}{z^2+1} = 0$$

∎

Zero by cancellation, not by absence of poles.

Example 3: Evaluate $\displaystyle\oint_{|z-i|=1}\frac{dz}{z^2+1}$.

Solution. This contour is a small circle about $i$, so only that pole is enclosed - $-i$ is at distance 2 from the centre.

$$2\pi i\cdot\frac{1}{2i} = \pi$$

∎

Same integrand, different contour, different answer. The contour is doing all the work.

Example 4: Find the residue of $\frac{e^z}{z^3}$ at 0.

Solution. Pole of order 3, so use $m=3$:

$$\operatorname{Res} = \frac{1}{2!}\lim_{z\to0}\frac{d^2}{dz^2}\left[z^3\cdot\frac{e^z}{z^3}\right] = \frac{1}{2}\lim_{z\to0}\frac{d^2}{dz^2}e^z = \frac12$$

Matching the series computation from the Laurent lesson. ✓

Example 5: Find the residue of $\frac{1}{\sin z}$ at $z=n\pi$.

Solution. Simple pole, so use the quotient formula with $p=1$ and $q=\sin z$, $q' = \cos z$:

$$\operatorname{Res}_{z=n\pi}\frac{1}{\sin z} = \frac{1}{\cos(n\pi)} = (-1)^n$$

∎

No factoring was possible here, which is exactly why the $p/q'$ formula matters.

Example 6: Evaluate $\displaystyle\oint_{|z|=1}\frac{\cos z}{z^2(z-3)}\,dz$.

Solution. Singularities at $z=0$ (order 2) and $z=3$ (simple). Only 0 is inside.

$$\operatorname{Res}_{z=0} = \lim_{z\to0}\frac{d}{dz}\left[\frac{\cos z}{z-3}\right]$$

By the quotient rule,

$$\frac{d}{dz}\frac{\cos z}{z-3} = \frac{-\sin z(z-3)-\cos z}{(z-3)^2}$$

At $z=0$ this is $\frac{0-1}{9} = -\frac19$. So

$$\oint = 2\pi i\left(-\frac19\right) = -\frac{2\pi i}{9}$$

∎

Example 7: Evaluate $\displaystyle\oint_{|z|=1}z^3e^{1/z}\,dz$.

Solution. Essential singularity at 0, so no formula applies - expand instead:

$$z^3e^{1/z} = z^3\left(1+\frac1z+\frac{1}{2z^2}+\frac{1}{6z^3}+\frac{1}{24z^4}+\cdots\right)$$ $$= z^3+z^2+\frac{z}{2}+\frac16+\frac{1}{24z}+\cdots$$

The coefficient of $\frac1z$ is $\frac{1}{24}$, so

$$\oint_{|z|=1}z^3e^{1/z}\,dz = \frac{2\pi i}{24} = \frac{\pi i}{12}$$

∎

You never had to understand the singularity, only to locate one coefficient in the series. That is the practical power of the whole method.

Inverse Laplace transforms. Recovering a time response from a transfer function is a contour integral evaluated by residues, with each pole contributing one exponential term. This is what produces the familiar sums of decaying exponentials in step responses.

Partial fraction expansion. The coefficients in a partial fraction decomposition are residues, so every time an engineer expands a transfer function they are computing residues, usually without calling them that.

Summing series. Sums like $\sum 1/n^2$ and $\sum 1/(n^2+a^2)$ are evaluated by applying the residue theorem to a function with poles at the integers, typically $\pi\cot(\pi z)$. This is a standard technique in analytic number theory.

Feynman integrals. Loop integrals in quantum field theory are performed by closing a contour and summing residues, with each residue corresponding to a particle going on shell. The physics interpretation of a residue is standard vocabulary in the field.

Modal analysis. In structural and acoustic engineering, a system’s response is decomposed into modes, and each mode’s contribution is the residue at the corresponding pole. Measured frequency-response data is fitted by estimating those residues.

Filter and antenna design. Residues at poles determine the weight of each resonance in a network’s response, so shaping a filter is largely choosing pole locations and residues.