Mathematics.

singularities

Laurent Series

Complex Analysis35 minDifficulty6 out of 10

You should know: taylor series, analytic functions

Overview

A Laurent series generalizes the Taylor series to functions that are analytic on an annulus (a ring-shaped region) rather than a full disk, allowing negative powers of (z − z0) in addition to nonnegative ones. If f is analytic in the annulus r < |z − z0| < R, it has a unique expansion f(z) = Σ_{n=−∞}^{∞} a_n (z − z0)^n valid throughout the annulus, where the coefficients are given by contour integrals over any circle inside the annulus. The part with negative powers, Σ_{n=1}^{∞} a_{-n}(z-z0)^{-n}, is called the principal part and encodes the singular behavior of f at z0; the coefficient a_{-1} is the residue, central to the residue theorem. Laurent series are the essential tool for classifying isolated singularities as removable, poles, or essential.

Intuition

A Taylor series only works where a function is perfectly smooth (analytic) all the way down to the center point; if there's a singularity right at z0, you cannot expand in nonnegative powers alone. Laurent series fix this by allowing negative powers, which can blow up as z → z0 and thereby capture exactly the singular behavior that a Taylor series cannot represent. Splitting the series into its 'analytic part' (nonnegative powers, well-behaved even at z0) and 'principal part' (negative powers, capturing the singularity) is the key move that lets Laurent series classify how badly a function misbehaves at an isolated singular point.

Formal Definition

Definition

If f is analytic on the annulus r < |z − z0| < R, then for every z in the annulus:

f(z)=∑n=−∞∞an(z−z0)n=∑n=0∞an(z−z0)n⏟analytic part+∑n=1∞a−n(z−z0)−n⏟principal partf(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n = \underbrace{\sum_{n=0}^{\infty} a_n(z-z_0)^n}_{\text{analytic part}} + \underbrace{\sum_{n=1}^{\infty} a_{-n}(z-z_0)^{-n}}_{\text{principal part}}
Laurent expansion
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
Coefficient formula (C any circle in the annulus)
a−1=Res⁡(f,z0)a_{-1} = \operatorname{Res}(f, z_0)
The residue is the coefficient of (z-z0)^{-1}

Worked Examples

  1. The function is already expressed as a sum of a negative power and a nonnegative power.

    f(z)=z−1+z1f(z) = z^{-1} + z^1

  2. Read off the coefficients: a_{-1}=1 (coefficient of 1/z), a_1 = 1, all other a_n = 0.

    a−1=1,a1=1a_{-1} = 1, \quad a_1 = 1

  3. The residue at z0=0 is by definition the coefficient a_{-1}.

    Res⁡(f,0)=1\operatorname{Res}(f,0) = 1

Answer: Laurent series: f(z) = z^{-1} + z; Res(f,0) = 1.

Practice Problems

Difficulty 5/10

Find the Laurent series of f(z) = sin(z)/z² about z0=0 up to the term in z, and give Res(f,0).

Difficulty 6/10

Find the Laurent series of f(z) = 1/(z(z-1)) valid on the annulus 0 < |z| < 1, and identify Res(f,0).

Difficulty 7/10

Find the Laurent series of f(z) = e^{1/z} about z0=0, and state what type of singularity this represents.

Quiz

A Laurent series differs from a Taylor series in that it allows:
The residue of f at z0 is defined as:
For f(z) = e^z/z², the residue at z=0 is:

Summary

  • A Laurent series f(z) = Σ_{n=-∞}^{∞} a_n(z-z0)^n represents f on an annulus, extending Taylor series with a principal part of negative powers.
  • The coefficient a_{-1} is the residue, central to evaluating contour integrals via the residue theorem.
  • Laurent series classify isolated singularities by the structure of the principal part: no negative terms (removable), finitely many (pole), infinitely many (essential).

References

  1. WebsiteWikipedia — Laurent series