54  Series and residues

Laurent series, isolated singularities, and the residue theorem

54.1 What this chapter helps you do

Symbols to keep handy

These are the bits of notation you'll see a lot. If a line of symbols feels like a fence, read it out loud once, then keep going.

  • [f, z_0]: residue of f at z-zero

  • _C f,dz = 2i : residue theorem

Definitions to keep handy

These are the words we keep coming back to. If one feels slippery, come back here and steady it before you push on.

  • Taylor series: A power series that represents an analytic function away from singularities.

  • Laurent series: A power series that can include negative powers, so it still works near a singularity.

  • singularity: A point where the function stops being analytic (it blows up or misbehaves).

  • residue: A single coefficient that captures the local contribution of a singularity to a contour integral.

  • residue theorem: A rule that turns a contour integral into a sum of residues.

Here is the main move this chapter is making, in plain terms. You do not need to be fast. You just need to keep the thread.

  • Coming in: An analytic function has a Taylor series that converges in any disk that avoids singularities. At a singularity, the Taylor series breaks down. The Laurent series extends the power series to include negative powers of (z−z₀) and converges in an annulus around the singularity. The coefficient of the (z−z₀)^{−1} term — the residue — encodes everything needed to compute contour integrals.

  • Leaving with: The Laurent series of f around an isolated singularity z₀ splits into an analytic part (non-negative powers) and a principal part (negative powers). The classification — removable, pole of order m, or essential singularity — is determined by how many negative-power terms are present. The residue Res[f, z₀] is the coefficient of (z−z₀)^{−1}. The residue theorem states that the integral of f around any closed contour equals 2πi times the sum of residues at the enclosed singularities. This converts many difficult real integrals to simple residue calculations.

54.2 Taylor series for analytic functions

If f is analytic in a disk |z - z_0| < R, it has a Taylor series that converges in that disk:

f(z) = \sum_{n=0}^{\infty} c_n(z-z_0)^n, \qquad c_n = \frac{f^{(n)}(z_0)}{n!} = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz

The series converges absolutely for |z - z_0| < R and uniformly on any closed subdisk. The radius of convergence R equals the distance from z_0 to the nearest singularity of f.

Key Taylor series (centred at 0, valid for all z):

e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!}, \qquad \sin z = \sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{(2n+1)!}, \qquad \cos z = \sum_{n=0}^{\infty}\frac{(-1)^n z^{2n}}{(2n)!}

\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n \quad (|z| < 1)

These are exactly the real Taylor series with x replaced by z.


54.3 Laurent series

At a point z_0 where f is not analytic, the Taylor series does not converge. The Laurent series extends the power series to include negative powers:

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

where the coefficients are:

c_n = \frac{1}{2\pi i}\oint_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz

This converges in an annulus r < |z - z_0| < R (with r possibly 0 and R possibly \infty).

The principal part consists of the negative-power terms \sum_{n=1}^{\infty} c_{-n}(z-z_0)^{-n}; the analytic part is the non-negative power series \sum_{n=0}^{\infty}c_n(z-z_0)^n. Whether the principal part is absent, has finitely many terms, or has infinitely many terms classifies the type of singularity at z_0 — which is the point of the next section.

Computing Laurent series in practice. For functions that are products, quotients, or compositions of known series, it is usually more efficient to manipulate the known Taylor series directly rather than computing coefficients by integration.

Example. \sin z / z^3 near z = 0:

\frac{\sin z}{z^3} = \frac{1}{z^3}\left(z - \frac{z^3}{6} + \frac{z^5}{120} - \cdots\right) = \frac{1}{z^2} - \frac{1}{6} + \frac{z^2}{120} - \cdots

The principal part has only one negative-power term, 1/z^2.


54.4 Classification of isolated singularities

An isolated singularity of f at z_0 is a point where f is not analytic but is analytic in some punctured disk 0 < |z - z_0| < \delta. The Laurent series reveals the type:

Removable singularity. The principal part is absent (no negative-power terms). f can be assigned a value at z_0 to make it analytic. Example: \sin z / z at z = 0 — the Laurent series is 1 - z^2/6 + \cdots, which extends continuously to z = 0 with value 1.

Pole of order m. The principal part has exactly m terms: f(z) = \frac{c_{-m}}{(z-z_0)^m} + \cdots + \frac{c_{-1}}{z-z_0} + c_0 + c_1(z-z_0) + \cdots, \quad c_{-m} \neq 0 |f(z)| \to \infty as z \to z_0. A simple pole is a pole of order 1.

Essential singularity. The principal part has infinitely many terms. The behaviour near z_0 has no limit — finite or infinite. Example: e^{1/z} at z = 0. Approach the origin along the positive real axis and e^{1/z} \to +\infty; along the negative real axis, e^{1/z} \to 0; along the imaginary axis, e^{1/z} spirals and oscillates.

The takeaway for engineering work is simple: near an essential singularity, the function does not settle into any single kind of behaviour (not “tends to a value” and not “blows up like a fixed power”). It can do wildly different things depending on how you approach the point.


54.5 Residues

The residue of f at an isolated singularity z_0 is the Laurent coefficient c_{-1}:

\text{Res}[f, z_0] = c_{-1}

Why it matters. Chapter 2 proved by direct parametrisation that \oint_{|z-z_0|=r}(z-z_0)^n\,dz = 2\pi i when n = -1 and = 0 for every other integer n. (The same substitution z = z_0 + re^{i\theta} works for all integers, not just n = 0 as computed there.) So when the Laurent series is integrated term by term around a small circle enclosing z_0, every term vanishes except the c_{-1} term: \oint_C \sum_{n=-\infty}^{\infty} c_n(z-z_0)^n\,dz = c_{-1}\cdot 2\pi i

54.5.1 Computing residues

At a simple pole: \text{Res}[f, z_0] = \lim_{z \to z_0}(z-z_0)f(z).

At a simple pole of p/q where q(z_0)=0, q'(z_0)\neq 0: \text{Res}\!\left[\frac{p}{q}, z_0\right] = \frac{p(z_0)}{q'(z_0)}

This is just L’Hôpital applied to the simple-pole limit. Quick check: \text{Res}[1/(z^2-1),\,1] — here p = 1, q = z^2-1, q'(z) = 2z, so the residue is 1/(2\cdot 1) = 1/2.

At a 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}}\left[(z-z_0)^m f(z)\right]

Quick check: \text{Res}[e^z/(z-1)^2,\, 1] — multiply by (z-1)^2 to get e^z, differentiate once (since m=2, so m-1=1), divide by 1! = 1: d(e^z)/dz\big|_{z=1} = e.


54.6 The residue theorem

Residue theorem. Let f be analytic on and inside a simple closed contour C (counterclockwise) except for isolated singularities z_1, \ldots, z_k inside C. Then:

\oint_C f(z)\,dz = 2\pi i\sum_{j=1}^{k}\text{Res}[f, z_j]

This is the master formula of complex analysis. Cauchy’s integral formula is the special case where f has only simple poles.


54.7 Evaluating real integrals

54.7.1 Type 1: trigonometric integrals \int_0^{2\pi} R(\cos\theta, \sin\theta)\,d\theta

Substitute z = e^{i\theta}: \cos\theta = (z+z^{-1})/2, \sin\theta = (z-z^{-1})/(2i), d\theta = dz/(iz). The integral becomes a contour integral on |z| = 1.

Example. \displaystyle\int_0^{2\pi}\frac{d\theta}{3 + 2\cos\theta}. Substituting gives \oint_{|z|=1}\frac{dz}{iz(3 + z + z^{-1})} = \oint_{|z|=1}\frac{dz}{i(z^2+3z+1)}. The roots of z^2+3z+1 are z = (-3\pm\sqrt{5})/2; only z_1 = (-3+\sqrt{5})/2 \approx -0.38 lies inside |z|=1. Residue: 1/(i(2z_1+3)) = 1/(i\sqrt{5}). Integral: 2\pi i \cdot 1/(i\sqrt{5}) = 2\pi/\sqrt{5}. Exercise 5 follows the same pattern.

54.7.2 Type 2: improper integrals \int_{-\infty}^{\infty} R(x)\,dx

Close the contour with a large semicircle in the upper (or lower) half-plane. The semicircular arc’s contribution vanishes as the radius grows — but the condition depends on the form of the integrand. For a plain rational R(x) with no oscillatory factor, the ML estimation lemma guarantees the arc vanishes when the degree of the denominator exceeds the degree of the numerator by at least 2. When the integrand takes the form e^{iax}R(x) with a > 0 (for example, e^{ix}/(1+x^2), whose real part gives the hook integral), Jordan’s lemma applies and degree excess of at least 1 suffices. The residue theorem then gives the real integral as 2\pi i times the sum of residues in the upper half-plane.

Example. \displaystyle\int_{-\infty}^{\infty}\frac{dx}{1+x^2}. Poles of 1/(1+z^2): z = \pm i. Only z = i is in the upper half-plane. Residue: 1/(2i). Integral: 2\pi i \cdot 1/(2i) = \pi.

The hook integral \int_{-\infty}^{\infty}\cos x/(1+x^2)\,dx uses the same idea with one extra step: write \cos x = \text{Re}(e^{ix}), so the integral equals \text{Re}\!\int_{-\infty}^{\infty} e^{ix}/(1+x^2)\,dx. The integrand is now of the form e^{iax}R(x) with a=1, so Jordan’s lemma applies. The pole at z=i gives residue e^{i\cdot i}/(2i) = e^{-1}/(2i), and the integral equals \text{Re}(2\pi i \cdot e^{-1}/(2i)) = \pi/e.

The residue theorem and Laurent series complete the core tools of complex integration. The next chapter takes a different direction: conformal mapping uses analytic functions to deform geometry — turning complicated domains into simple ones where Laplace’s equation can be solved directly.


54.8 Exercises


54.8.1 Exercise 1: Laurent series around a pole

Find the Laurent series of f(z) = \dfrac{1}{z(z-1)} valid for 0 < |z| < 1, and classify the singularity at z = 0.


54.8.2 Exercise 2: Classifying \sin z / z^2 at z = 0


54.8.3 Exercise 3: Residue at a simple pole

Find \text{Res}\!\left[\dfrac{z^2+1}{(z-2)(z+3)},\; z=2\right].


54.8.4 Exercise 4: Residue at a double pole

Find \text{Res}\!\left[\dfrac{e^z}{(z-1)^2},\; z=1\right].


54.8.5 Exercise 5: Trigonometric integral

Evaluate \displaystyle\int_0^{2\pi}\frac{d\theta}{2 + \cos\theta}.


54.8.6 Exercise 6: Improper real integral via residues

Evaluate \displaystyle\int_{-\infty}^{\infty}\frac{dx}{(1+x^2)^2}.