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Laurent Series — Definition, Formula & Examples

A Laurent series is a representation of a complex function as a series that includes both positive and negative powers of (z−z0)(z - z_0). It generalizes the Taylor series by allowing you to expand functions around points where they are not analytic, such as poles.

Let f(z)f(z) be analytic in an annular region r<∣z−z0∣<Rr < |z - z_0| < R. The Laurent series of ff about z0z_0 is the unique expansion f(z)=∑n=−∞∞an(z−z0)nf(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n, where the coefficients are given by an=12πi∮Cf(z)(z−z0)n+1 dza_n = \frac{1}{2\pi i} \oint_C \frac{f(z)}{(z - z_0)^{n+1}}\,dz and CC is any positively oriented simple closed contour in the annulus encircling z0z_0.

Key Formula

f(z)=∑n=−∞∞an(z−z0)n=⋯+a−2(z−z0)2+a−1z−z0+a0+a1(z−z0)+⋯f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n = \cdots + \frac{a_{-2}}{(z-z_0)^2} + \frac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0) + \cdots
Where:
  • z0z_0 = The center of the expansion (often a singularity of f)
  • ana_n = Laurent coefficients, determined by contour integration or algebraic methods
  • a−1a_{-1} = The residue of f at z_0

How It Works

The Laurent series splits into two parts: the analytic part (non-negative powers, ∑n=0∞an(z−z0)n\sum_{n=0}^{\infty} a_n(z-z_0)^n) and the principal part (negative powers, ∑n=1∞a−n(z−z0)−n\sum_{n=1}^{\infty} a_{-n}(z-z_0)^{-n}). The principal part captures the singular behavior of the function near z0z_0. If the principal part is finite (only finitely many negative-power terms), z0z_0 is a pole; if it is infinite, z0z_0 is an essential singularity. The coefficient a−1a_{-1} is especially important — it equals the residue of ff at z0z_0, which is the key quantity used in the residue theorem for evaluating contour integrals.

Worked Example

Problem: Find the Laurent series of f(z)=1z(1−z)f(z) = \frac{1}{z(1-z)} centered at z0=0z_0 = 0, valid for 0<∣z∣<10 < |z| < 1.
Partial fractions: Decompose the function into simpler terms.
1z(1−z)=1z+11−z\frac{1}{z(1-z)} = \frac{1}{z} + \frac{1}{1-z}
Expand the analytic part: For ∣z∣<1|z| < 1, expand 11−z\frac{1}{1-z} as a geometric series.
11−z=∑n=0∞zn=1+z+z2+⋯\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n = 1 + z + z^2 + \cdots
Combine: Add the 1z\frac{1}{z} term to get the full Laurent series.
f(z)=1z+1+z+z2+⋯=∑n=−1∞znf(z) = \frac{1}{z} + 1 + z + z^2 + \cdots = \sum_{n=-1}^{\infty} z^n
Answer: The Laurent series is ∑n=−1∞zn=1z+1+z+z2+⋯\displaystyle\sum_{n=-1}^{\infty} z^n = \frac{1}{z} + 1 + z + z^2 + \cdots for 0<∣z∣<10 < |z| < 1. The residue at z=0z=0 is a−1=1a_{-1} = 1.

Why It Matters

Laurent series are essential in complex analysis for classifying singularities and computing residues. The residue theorem — which relies on identifying a−1a_{-1} in a Laurent expansion — is one of the most powerful tools for evaluating real integrals that arise in physics, engineering, and applied mathematics.

Common Mistakes

Mistake: Using a Taylor series expansion at a point where the function has a singularity.
Correction: Taylor series require the function to be analytic at the center. At singularities, you must use a Laurent series, which accommodates negative powers of (z−z0)(z - z_0).