Laurent Series in Complex Analysis: Explained with Examples

A Laurent series generalises the Taylor series by allowing negative powers: f(z) = Σ (n = −∞ to ∞) aₙ(z − a)ⁿ = … + a₋₂/(z − a)² + a₋₁/(z − a) + a₀ + a₁(z − a) + …. It represents a function that is analytic on an annulus r < |z − a| < R — a ring-shaped region — where the function may have a singularity at the centre a.

This matters because Taylor series cannot exist at a singularity, yet most interesting complex analysis happens exactly there. The Laurent expansion exposes what kind of singularity sits at a, and its coefficient a₋₁ — called the residue — is the key that unlocks contour integration of the function.

Key Concepts

• Annulus of convergence: the ring r < |z − a| < R on which the series converges; r can be 0 and R can be ∞.
• Principal part: the negative-power terms … + a₋₂/(z − a)² + a₋₁/(z − a), which capture the singular behaviour at a.
• Analytic part: the non-negative-power terms a₀ + a₁(z − a) + …, an ordinary Taylor-like series.
• Uniqueness: on a given annulus, the Laurent coefficients are unique; different annuli give different expansions of the same function.
• Coefficient formula: aₙ = (1/2πi) ∮ f(z)/(z − a)ⁿ⁺¹ dz around any positively oriented circle inside the annulus.

Formulas

• General form: f(z) = Σ (n = −∞ to ∞) aₙ(z − a)ⁿ, convergent for r < |z − a| < R.
• Same function, two annuli: 1/(1 − z) = Σ zⁿ for |z| < 1, but for |z| > 1 it equals −Σ 1/zⁿ⁺¹ = −(1/z + 1/z² + 1/z³ + …).
• Essential example: e^(1/z) = Σ 1/(n!zⁿ) for n = 0, 1, 2, …, valid for 0 < |z| < ∞, with infinitely many negative powers.
• Assumptions: f must be analytic at every point of the annulus; the expansion is tied to the chosen annulus, not just the function.

Worked Example

Find the Laurent expansion of f(z) = 1/(z(z − 1)) valid on the annulus 0 < |z| < 1.

1. Factor and split: 1/(z(z − 1)) = −(1/z) · 1/(1 − z).
2. On |z| < 1, the geometric series gives 1/(1 − z) = 1 + z + z² + z³ + ….
3. Multiply by −1/z: f(z) = −1/z − 1 − z − z² − z³ − ….
4. The principal part is just −1/z, so z = 0 is a simple pole with residue −1.
5. Numerical check at z = 0.5: the function gives 1/(0.5 × −0.5) = −4, while the series gives −2 − 1 − 0.5 − 0.25 − 0.125 − … = −4. They agree.

Common Misconceptions

• "Negative powers mean the series diverges." No — the negative powers converge for |z − a| > r, the positive powers for |z − a| < R; together they converge exactly on the annulus.
• "A function has one Laurent series." No — 1/(1 − z) has a Taylor series on |z| < 1 and a purely principal-part series on |z| > 1. The annulus is part of the specification.
• "The principal part is just a shifted Taylor series." No — infinitely many negative powers signal an essential singularity, qualitatively different from any pole.

Real-World Applications

• Residue computation: reading a₋₁ off the Laurent series is the fastest way to find residues for contour integrals.
• Singularity classification: the principal part tells you instantly whether a singularity is removable, a pole, or essential.
• Signal processing: Laurent expansions on annuli are the analytic backbone of z-transforms for discrete-time signals.

A Laurent series is a Taylor series freed from the requirement of analyticity at the centre: negative powers describe the singularity, non-negative powers describe the smooth part. Because the expansion depends on the annulus, always state the region — and the a₋₁ coefficient you read off is the residue that powers integral computations.