8.1 Laurent Series

Home | Assessment | Notes | Worksheets | Blackboard

Over the previous week we developed an understanding of what is called the local picture of a holomorphic function: if $f:\mathsf{B}(b,R)\to \mathbb{C}$ is holomorphic then $f$ can be expanded as a power series centered at $b$ with radius of convergence at least $R$. This is an incredible amount of regularity to get from merely the assumption that $f$ is differentiable at every point in $\mathsf{B}(b,R)$.

Our next goal is to develop the theory of functions holmorphic on an annulus. Given $b\in \mathbb{C}$ and $0\le r<R$ we write $$\mathsf{Ann}(b,r,R)=\{z\in \mathbb{C}:r<|z-b|<R\}$$ for the annulus of inner radius $r$ and outer radius $R$.

Figure 1: The annulus $\mathsf{Ann}(b,r,R)$ centered at $b$ of inner radius $r$ and outer radius $R$.

Example

The function $f(z)=1/z$ is holomorphic on the annulus $\mathsf{Ann}(0,0,R)$. However, there is no power series centered at 0 equal to $f$ throughout $\mathsf{Ann}(0,0,R)$ because any such power series would be holomorphic on $\mathsf{B}(0,R)$ and would therefore have a contour integral of zero over the circle of radius $R/2$ centered at 0, whereas the contour integral of $f$ over the same contour is $2\pi i$. We will therefore need a broader type of expansion for functions on annuli.

Although we cannot expand holomorphic functions on annuli in terms of power series, we will see that a more general expansion in terms of Laurent series is always possible.

Definition (Laurent Series)

A Laurent series centered at $b$ with coefficients $a_n$ is an expression of the form $$\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{a}_n(z-b{)}^n=\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{\displaystyle \frac{1}{(z-b{)}^n}}+a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ and we say that a Laurent series converges at a specific $z\in \mathbb{C}$ if both of the series $$\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{\displaystyle \frac{1}{(z-b{)}^n}}\sum _{n=1}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ converge.

The coefficients of a Laurent series are indexed by integers and therefore corerspond to a map from $\mathbb{Z}$ to $\mathbb{C}$.

Fix a map $a:\mathbb{Z}\to \mathbb{C}$. The power series $$\sum _{n=1}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ has (as all power series do) some radius of convergence $R\ge 0$. The power series $$\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{w}^n$$ has radius of convergence $\rho \ge 0$. Taking $w=1/(z-b)$ we conclude that $$\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{\displaystyle \frac{1}{(z-b{)}^n}}$$ converges whenever $|z-b|>1/\rho$. Putting $r=1/\rho$ we have proved that the Laurent series $$\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{\displaystyle \frac{1}{(z-b{)}^n}}+a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ converges whenever $|z-b|>r$ and $|z-b|<R$. If $r<R$ then the Laurent series converges on the annulus $\mathsf{Ann}(b,r,R)$. If $r>R$ the Laurent series does not converge anywhere.

Our main theorem is that every holomorphic function on an annulus can be represented by a Laurent series.

Theorem (Laurent's Theorem)

If $f$ is holomorphic on the annulus $\mathsf{Ann}(b,r,R)$ then $f$ can be represented by a Laurent series $$f(z)=\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{\displaystyle \frac{1}{(z-b{)}^n}}+a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n(z-b{)}^n$$ on all of $\mathsf{Ann}(b,r,R)$. Moreover $$a_n=\frac{1}{2\pi i}\underset{\gamma }{\int }{\displaystyle \frac{f(z)}{(z-b{)}^{n+1}}}\mathsf{d}z$$ for all $r<s<R$ and all $n\in \mathbb{Z}$ where $\gamma (t)=b+s{e}^{it}$ on $[0,2\pi ]$.

Definition (Laurent Series of a Function)

We call the Laurent series representing $f$ on $\mathsf{Ann}(b,r,R)$ the Laurent series of $f$. The portion $${\displaystyle \frac{a_{-1}}{(z-b)}}+{\displaystyle \frac{a_{-2}}{(z-b{)}^2}}+\cdots =\sum _{n=1}^{\mathrm{\infty }}{a}_{-n}{\displaystyle \frac{1}{(z-b{)}^n}}$$ is called the principal part of $f$ on $\mathsf{Ann}(b,r,R)$.

We will not prove Laurent's theorem. The proof is somewhat similar to the proof of Taylor's theorem. Note, however, that we cannot conclude that $a_n=f^{(n)}(b)/n!$ in this case. Indeed $f$ may not even be defined at $b$!