Laurent Series

Core Idea

A Laurent series is Σ_(n=-∞)^∞ aₙ(z - z₀)ⁿ. It converges on an annulus r < |z - z₀| < R. Any holomorphic function on an annulus has a unique Laurent expansion. The coefficient a₋₁ (the residue) plays a special role. The principal part Σ_(n=-∞)^-1 aₙ(z - z₀)ⁿ captures the behavior near the singularity at z₀.

How It's Best Learned

Expand f(z) = 1/(z(z-1)) as a Laurent series around z = 0 in the annulus 0 < |z| < 1. Notice the negative powers appear and identify the residue (a₋₁).

Common Misconceptions

Thinking Laurent series are like Taylor series but with negative powers; they describe behavior near singularities. Confusing the principal part with the regular part; they represent different aspects of the singularity.

Explainer

You have worked with power series in the complex plane — series of the form Σ aₙ(z − z₀)ⁿ with non-negative exponents, which converge on a disk centered at z₀. A Laurent series extends this idea by allowing negative exponents: Σ_{n=−∞}^{∞} aₙ(z − z₀)ⁿ. This extension is not just notational generosity — it is forced on you by the existence of singularities. If f(z) has a singularity at z₀, it cannot have a Taylor series there, but it may still have a perfectly convergent Laurent expansion on the punctured neighborhood 0 < |z − z₀| < R. The negative-power terms encode exactly how badly f blows up as z → z₀.

The natural domain of a Laurent series is an annulus r < |z − z₀| < R, not a disk. The outer radius R is determined by the nearest singularity outside z₀, exactly as for a Taylor series. The inner radius r accounts for the singularity at z₀ itself — the series breaks down at z₀, so we exclude it. A Taylor series is the special case r = 0 and no negative-power terms, where the disk is non-punctured. A concrete example: f(z) = 1/(z(z−1)) around z₀ = 0 has a singularity at z = 0 and at z = 1. The annulus 0 < |z| < 1 avoids both. Expanding 1/(z−1) = −1/(1−z) = −Σzⁿ for |z| < 1 and dividing by z gives f(z) = −Σzⁿ⁻¹ = −z⁻¹ − 1 − z − z² − ⋯. The negative-power term is −z⁻¹.

The Laurent series splits into two parts. The regular part Σ_{n=0}^{∞} aₙ(z − z₀)ⁿ behaves like a Taylor series; it converges inside the outer circle. The principal part Σ_{n=−∞}^{−1} aₙ(z − z₀)ⁿ consists of all the negative-power terms; it captures the singularity's character. The nature of the principal part classifies the singularity: if it has finitely many terms (stopping at (z−z₀)^−m), the singularity is a pole of order m. If it has infinitely many terms, the singularity is essential. If the principal part is empty entirely, there is no singularity — the Laurent series is just a Taylor series.

The coefficient a₋₁ — the coefficient of (z − z₀)⁻¹ — is called the residue and occupies a special role. By Cauchy's integral formula, integrating f(z) around a small loop encircling z₀ picks out exactly this coefficient: (1/2πi) ∮ f(z) dz = a₋₁. This is why the residue appears as the output of contour integrals. No other Laurent coefficient contributes to the integral — all the (z − z₀)ⁿ terms with n ≠ −1 integrate to zero around a closed loop. The residue theorem, which you will study next, turns this observation into a powerful computational machine: to evaluate complicated real integrals, encode them as contour integrals in ℂ, locate the singularities inside the contour, read off the residues from the Laurent expansions, and sum them.