Analysis

Residues and the residue theorem

A technique for evaluating complex contour integrals by summing residues at enclosed poles.

IntuitionFrom an isolated pole to an infinite sum

Imagine walking a closed loop $\gamma$ in the complex plane around a function $f$ that blows up at a few isolated points. Near each such point, $f$ behaves like a spiral staircase: the closer you get, the faster the value spins and grows. The residue theorem says something remarkable: the value of the whole loop integral depends only on a single number extracted at each trapped point — its residue — never on the shape of the loop itself, as long as the loop does not cross a singularity.

Domain-coloring plot of a function with an isolated pole; the color swirls around the pole, and its winding encodes the residue there.

UndergraduateIsolated singularities and the Laurent series

Definition: Isolated singularity and its residue

A point $z_0$ is an isolated singularity of $f$ if $f$ is holomorphic on some punctured disk around $z_0$ but not at $z_0$ itself. On that punctured disk, $f$ has a unique Laurent expansion $f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$ , and the coefficient $a_{-1}$ of the term $(z-z_0)^{-1}$ is called the residue of $f$ at $z_0$ , written $\mathrm{Res}(f, z_0) = a_{-1}$ . Singularities are classified by how many negative powers appear: none is removable ( $\lim_{z \to z_0} f(z)$ exists), finitely many of order $m$ is a pole, and infinitely many (as in $\sin(1/z)$ at $z=0$ ) is essential.

f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n

Practically, no one expands a full Laurent series just to find a residue at a pole. For a simple pole (order $m=1$ ), $\mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0) f(z)$ . For a pole of higher order $m$ , the formula generalizes to $\mathrm{Res}(f, z_0) = \dfrac{1}{(m-1)!} \lim_{z \to z_0} \dfrac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]$ , which differentiates away the singular part before evaluating.

\mathrm{Res}(f, z_0) = \dfrac{1}{(m-1)!} \lim_{z \to z_0} \dfrac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]

Classifying an isolated singularity and computing its residue
Type of singularity	Laurent series	Residue formula
Removable	No negative powers	$\mathrm{Res}(f, z_0) = 0$
Simple pole ( $m=1$ )	One negative power	$\mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0) f(z)$
Pole of order $m$	Finitely many negative powers	$\mathrm{Res}(f, z_0) = \dfrac{1}{(m-1)!} \lim_{z \to z_0} \dfrac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]$
Essential	Infinitely many negative powers	Read off $a_{-1}$ directly

UndergraduateKey theorems

Theorem: The residue theorem

Let $f$ be holomorphic on a simply connected domain except at finitely many isolated singularities $z_1, \dots, z_n$ inside a positively oriented simple closed contour $\gamma$ . Then $\oint_\gamma f(z)\,dz = 2\pi i \sum_{k} \mathrm{Res}(f, z_k)$ .

Why is it true?

It reduces a hard geometric problem (integrating along a curve) to an easy algebraic one (adding up finitely many numbers), because deforming the contour around each pole shrinks it to a tiny circle where the Laurent series does all the work.

Proof

Step 1 (Deform the contour). By Cauchy's integral theorem, $\oint_\gamma f(z)\,dz = 0$ for any closed curve bounding a region where $f$ is holomorphic. Since $f$ fails to be holomorphic only at $z_1,\dots,z_n$ , surround each $z_k$ with a tiny positively oriented circle $C_k$ of radius $\varepsilon$ small enough that the circles are disjoint and lie inside $\gamma$ . Cutting slits from $\gamma$ to each $C_k$ turns the region between $\gamma$ and the $C_k$ into a simply connected domain where $f$ is holomorphic, so the integral over the boundary of that region is $0$ ; the slit contributions cancel in pairs, leaving $\oint_\gamma f(z)\,dz = \sum_{k} \oint_{C_k} f(z)\,dz$ .

Step 2 (Evaluate each small circle). Fix $k$ and expand $f$ as its Laurent series $f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$ around $z_k$ , valid on the punctured disk containing $C_k$ . Every term $a_n(z-z_k)^n$ with $n \neq -1$ has an antiderivative single-valued on the punctured disk, so it integrates to $0$ around the closed circle $C_k$ ; only the term $a_{-1}(z-z_k)^{-1}$ survives.

Step 3 (Compute the surviving integral). Parametrize $C_k$ by $z = z_k + \varepsilon e^{i\theta}$ for $\theta \in [0, 2\pi]$ , so $dz = i\varepsilon e^{i\theta}\,d\theta$ and $\dfrac{dz}{z-z_k} = i\,d\theta$ . Then $\oint_{C_k} \dfrac{a_{-1}}{z-z_k}\,dz = a_{-1} \int_0^{2\pi} i\,d\theta = 2\pi i\, a_{-1} = 2\pi i\, \mathrm{Res}(f, z_k)$ .

Step 4 (Sum up). Substituting Step 3 into the identity from Step 1 gives $\oint_\gamma f(z)\,dz = 2\pi i \sum_{k} \mathrm{Res}(f, z_k)$ , which is exactly the residue theorem.

Theorem: Residue formula at a pole of order

m

If $f$ has a pole of order $m$ at $z_0$ , meaning $g(z) = (z-z_0)^m f(z)$ extends holomorphically to $z_0$ with $g(z_0) \neq 0$ , then $\mathrm{Res}(f, z_0) = \dfrac{1}{(m-1)!} \lim_{z \to z_0} \dfrac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]$ . In particular, for a simple pole ( $m=1$ ), $\mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0) f(z)$ .

Why is it true?

Multiplying by $(z-z_0)^m$ kills the singular part of the Laurent series, turning it into an ordinary Taylor series whose coefficients are just derivatives — so extracting $a_{-1}$ becomes an ordinary calculus computation.

Proof

Step 1 (Write the Laurent series explicitly). Since $f$ has a pole of order $m$ at $z_0$ , its Laurent series has the form $f(z) = \dfrac{a_{-m}}{(z-z_0)^m} + \cdots + \dfrac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0) + \cdots$ with $a_{-m} \neq 0$ .

Step 2 (Clear the pole). Multiply both sides by $(z-z_0)^m$ : $g(z) := (z-z_0)^m f(z) = a_{-m} + a_{-m+1}(z-z_0) + \cdots + a_{-1}(z-z_0)^{m-1} + a_0(z-z_0)^m + \cdots$ , which is an ordinary power series, so $g$ is holomorphic at $z_0$ .

Step 3 (Recognize the coefficient as a derivative). In this power series, $a_{-1}$ is the coefficient of $(z-z_0)^{m-1}$ . For any holomorphic $g(z) = \sum_j c_j (z-z_0)^j$ , Taylor's formula gives $c_j = \dfrac{g^{(j)}(z_0)}{j!}$ , so $a_{-1} = \dfrac{g^{(m-1)}(z_0)}{(m-1)!}$ .

Step 4 (Conclude). Since $\mathrm{Res}(f,z_0) = a_{-1}$ and $g^{(m-1)}(z_0) = \lim_{z\to z_0} \dfrac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)]$ , substituting gives $\mathrm{Res}(f, z_0) = \dfrac{1}{(m-1)!} \lim_{z \to z_0} \dfrac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]$ . Setting $m=1$ collapses the derivative and factorial to give $\mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0) f(z)$ .

UndergraduateReal-World Applications and Worked Examples

Residues turn otherwise intractable definite real integrals into arithmetic, which is why they appear throughout physics and engineering: evaluating oscillatory Fourier and Laplace transforms in signal processing, computing scattering amplitudes and propagators in quantum field theory, inverting Laplace transforms to solve control-system differential equations, and estimating asymptotics of counting functions in analytic number theory and combinatorics.

Example: A real integral with no elementary antiderivative shortcut

Evaluate $\displaystyle\int_{-\infty}^{\infty} \frac{dx}{x^2+1}$ using the residue theorem. (This integral can also be done by the substitution $x=\tan\theta$ , but residues generalize to integrals that substitution cannot touch.)

Solution

Step 1 (Set up the contour). Consider $f(z) = \dfrac{1}{z^2+1}$ and the closed contour $\gamma_R$ made of the segment $[-R,R]$ on the real axis together with the upper semicircle $|z|=R$ , traversed counterclockwise, for large $R$ .

Step 2 (Bound the arc). On the semicircular arc, $|f(z)| \le \dfrac{1}{R^2-1}$ , and the arc has length $\pi R$ , so the arc's contribution is at most $\dfrac{\pi R}{R^2-1} \to 0$ as $R \to \infty$ .

Step 3 (Find the enclosed pole). $f(z) = \dfrac{1}{z^2+1}$ has simple poles at $z = i$ and $z=-i$ ; only $z = i$ lies inside the upper semicircle. By the simple-pole formula, $\mathrm{Res}(f,i) = \lim_{z\to i}(z-i)\dfrac{1}{(z-i)(z+i)} = \dfrac{1}{2i}$ .

Step 4 (Apply the residue theorem and take the limit). $\oint_{\gamma_R} f(z)\,dz = 2\pi i \cdot \dfrac{1}{2i} = \pi$ for every $R>1$ , and this equals the real integral plus the vanishing arc term, so letting $R \to \infty$ gives $\displaystyle\int_{-\infty}^{\infty} \frac{dx}{x^2+1} = \pi$ .

Example: Inverting a Laplace transform in control theory

A control engineer models a damped system with transfer function $F(s) = \dfrac{1}{(s+1)(s+2)^2}$ and needs its impulse response $y(t)$ , the inverse Laplace transform, to check whether the system settles without oscillating.

Solution

Step 1 (Recall the inversion integral). The inverse Laplace transform is $y(t) = \dfrac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} F(s) e^{st}\,ds$ , which for a rational $F(s)$ decaying at infinity equals $2\pi i$ times the sum of residues of $F(s)e^{st}$ at all poles of $F(s) = \dfrac{1}{(s+1)(s+2)^2}$ .

Step 2 (Locate the poles). $F(s) = \dfrac{1}{(s+1)(s+2)^2}$ has a simple pole at $s=-1$ and a pole of order $2$ at $s=-2$ .

Step 3 (Residue at the simple pole). $\mathrm{Res}_{s=-1}\, F(s)e^{st} = \lim_{s\to -1}(s+1)\dfrac{e^{st}}{(s+1)(s+2)^2} = \dfrac{e^{-t}}{(-1)^2} = e^{-t}$ .

Step 4 (Residue at the order-2 pole). $\mathrm{Res}_{s=-2}\, F(s)e^{st} = \lim_{s\to -2}\dfrac{d}{ds}\left[\dfrac{e^{st}}{s+1}\right] = \lim_{s\to -2}\dfrac{t e^{st}(s+1) - e^{st}}{(s+1)^2} = -te^{-2t} - e^{-2t}$ .

Step 5 (Assemble the response). Summing residues gives $y(t) = e^{-t} - (1+t)e^{-2t}$ for $t \ge 0$ ; both exponentials decay and neither term oscillates, so the engineer confirms the system is overdamped.

If $f$ has a single simple pole at $z_0$ inside a positively oriented simple closed contour $\gamma$ , and $\mathrm{Res}(f,z_0)=3$ , what is $\oint_\gamma f(z)\,dz$ ?

What is the order of the pole of $f(z) = \dfrac{1}{z^3}$ at $z=0$ ?

What is $\mathrm{Res}_{z=0}\, \dfrac{e^z}{z}$ ?

An electrical engineer needs $\displaystyle\int_{-\infty}^{\infty} \dfrac{dx}{x^2+1}$ to normalize a Lorentzian frequency response. Using the residue at $z=i$ , what value should they get?