16 - Laurent's Theorem

Laurent's Theorem

Suppose $0<r<R$ and

$$A=A(z_0,r,R)=\{z:r<|z-z_0<R|\}$$

where $A$ is an annulus centred at $z_0$

If $f:U\to \mathbb{C}$ is holomorphic on an open set $U$ which contains $\overline{A}$ then exists $c_n\in \mathbb{C}$ such that

$$f(z)=\sum _{n=-\infty }^{\infty }{c}_n(z-z_0{)}^n\forall z\in A$$

Series is known as the Laurent Series of $f$ and

1. Converges for all $z\in A$

2. Converges uniformly for all $z\in A(x_0,r^{\prime },R^{\prime })$ where $r<r^{\prime }<R^{\prime }<R$

Moreover $c_n$ is unique and given by

$$c_n=\frac{1}{2\pi i}{\int }_{{\gamma }_s}\frac{f(z)}{(z-z_0{)}^{n+1}}dz$$

where $s\in [r,R]$ and for any $s>0$ then ${\gamma }_s(t):=z_0+s{e}^{2\pi it}$

Proof

By Cauchy formula for multiple curves for any $w\in A$ then

$$f(w)=\frac{1}{2\pi i}{\int }_{{\gamma }_R}\frac{f(z)}{z-w}dz-\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)}{z-w}dz$$

Note that both boundary components are counter-clockwise oriented
However in Cauchy formula for multiple curves, inner component is clockwise oriented
So it is compensated by the minus sign in front of second integral

Fixing $w$ then for $|w|<R=|z|$ then

$$\frac{1}{z-w}=\sum _{n=0}^{\infty }\frac{w^n}{z^{n+1}}$$

and for $|w|<R-ϵ=|z|-ϵ$ then series converges uniformly $z$ for any $ϵ>0$ thus

$${\int }_{{\gamma }_R}\frac{f(z)}{z-w}={\int }_{{\gamma }_R}\sum _{n=0}^{\infty }\frac{f(z)w^n}{z^{n+1}}=\sum _{n=0}^{\infty }\left({\int }_{{\gamma }_R}\frac{f(z)}{z^{n+1}}dz\right)w^n$$

for all $w\in A$

Similarly for $|z|=r<|w|$ then

$$\frac{1}{w-z}=\sum _{n=0}^{\infty }\frac{z^n}{w^{n+1}}=\sum _{n=-1}^{-\infty }\frac{w^n}{z^{n+1}}$$

which again converges uniformly when $|z|<r<|w|-ϵ$ for $ϵ>0$ then

$${\int }_{{\gamma }_r}\frac{f(z)}{w-z}dz={\int }_{{\gamma }_r}\sum _{n=-1}^{-\infty }\frac{f(z)w^n}{z^{n+1}}dz=\sum _{n=-1}^{-\infty }\left({\int }_{{\gamma }_r}\frac{f(z)}{z^{n+1}}dz\right)w^n$$

Thus by taking $(c_n{)}_{n\in \mathbb{Z}}$ to be the statement in the theorem then

$$f(w)=\frac{1}{2\pi i}{\int }_{{\gamma }_R}\frac{f(z)}{z-w}dz-\frac{1}{2\pi i}{\int }_{{\gamma }_r}\frac{f(z)}{z-w}dz=\sum _{n=-\infty }^{\infty }{c}_n{w}^n$$

as required