Laurent's Theorem

Laurent's theorem guarantees that every holomorphic function on an annulus has a unique doubly-infinite series expansion, generalizing Taylor series to regions with singularities.

Statement

Theorem5.10Laurent's theorem

Let $f$ be holomorphic on the annulus $A = \{z : r_1 < |z-z_0| < r_2\}$ where $0 \leq r_1 < r_2 \leq \infty$ . Then $f$ has a unique representation

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

converging absolutely and uniformly on compact subsets of $A$ . The coefficients are

$a_n = \frac{1}{2\pi i}\oint_\gamma \frac{f(\zeta)}{(\zeta-z_0)^{n+1}}\,d\zeta$

for any simple closed positively oriented curve $\gamma$ in $A$ winding once around $z_0$ .

Proof Outline

Proof

By Cauchy's theorem for multiply connected domains:

$f(z) = \frac{1}{2\pi i}\oint_{C_2}\frac{f(\zeta)}{\zeta - z}\,d\zeta - \frac{1}{2\pi i}\oint_{C_1}\frac{f(\zeta)}{\zeta - z}\,d\zeta.$

For the outer integral (on $C_2$ , $|\zeta-z_0| > |z-z_0|$ ): expand as in the Taylor series proof:

$\frac{1}{\zeta - z} = \sum_{n=0}^{\infty}\frac{(z-z_0)^n}{(\zeta-z_0)^{n+1}} \implies \frac{1}{2\pi i}\oint_{C_2}\frac{f(\zeta)}{\zeta-z}\,d\zeta = \sum_{n=0}^{\infty}a_n(z-z_0)^n.$

For the inner integral (on $C_1$ , $|\zeta-z_0| < |z-z_0|$ ): expand the other way:

$\frac{-1}{\zeta - z} = \frac{1}{z-\zeta} = \frac{1}{z-z_0}\sum_{n=0}^{\infty}\frac{(\zeta-z_0)^n}{(z-z_0)^n}$

giving $\frac{-1}{2\pi i}\oint_{C_1}\frac{f(\zeta)}{\zeta-z}\,d\zeta = \sum_{n=1}^{\infty}a_{-n}(z-z_0)^{-n}$ with $a_{-n} = \frac{1}{2\pi i}\oint_{C_1}f(\zeta)(\zeta-z_0)^{n-1}\,d\zeta$ .

Combining both contributions yields the Laurent series. Uniqueness follows from the uniqueness of Fourier coefficients (since $a_n$ are the Fourier coefficients of $f(z_0 + \rho e^{i\theta})$ on $|z-z_0|=\rho$ ). $\blacksquare$

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Applications

ExampleLaurent series by algebraic manipulation

For $f(z) = \frac{z}{(z-1)(z-2)}$ in the annulus $1 < |z| < 2$ , use partial fractions:

$f(z) = \frac{-1}{z-1} + \frac{2}{z-2}.$

For $|z| > 1$ : $\frac{-1}{z-1} = \frac{-1}{z}\cdot\frac{1}{1-1/z} = -\sum_{n=0}^\infty z^{-n-1}$ .

For $|z| < 2$ : $\frac{2}{z-2} = \frac{-1}{1-z/2} = -\sum_{n=0}^\infty \frac{z^n}{2^n}$ .

Therefore: $f(z) = -\sum_{n=0}^\infty z^{-n-1} - \sum_{n=0}^\infty \frac{z^n}{2^n}$ .

RemarkDifferent annuli yield different series

The same function can have different Laurent expansions in different annuli. For $f(z) = 1/[(z-1)(z-2)]$ , there are three distinct Laurent series centered at $z = 0$ : one for $|z| < 1$ , one for $1 < |z| < 2$ , and one for $|z| > 2$ . Each expansion is valid only in its annulus.

ExampleConnection to Fourier series

Setting $z = z_0 + re^{i\theta}$ in the Laurent series:

$f(z_0 + re^{i\theta}) = \sum_{n=-\infty}^\infty a_n r^n e^{in\theta}.$

This is a Fourier series in $\theta$ , with $a_n r^n$ as the Fourier coefficients. This connection between Laurent series and Fourier series is fundamental and explains why the coefficients are given by integral formulas.