5.3: Laurent Series

Recall that a function $f$ of the complex variable $z$ is analytic at a point $z_0$ if it has a derivative at each point in some neighbourhood of $z_0$. An entire function is a function that is analytic at each point in the entire finite plane. If a function $f$ fails to be analytic at a point $z_0$ but is analytic at some point in every neighbourhood of $z_0$, then $z_0$ is called a singular point, or singularity, of $f$.

Suppose that $f(z)$, or any single valued branch of $f(z)$, if $f(z)$ is multivalued, is analytic in the region $0\lt|z-z_0|\lt R$ and not at the point $z_0$. Then the point $z_0$ is called an isolated singular point of $f(z)$.

Now, recall also that any function which is analytic throughout a disk $|z -z_0|\lt R_0$ must have a Taylor series about $z_0$. If the function fails to be analytic at a point $z_0$, it is often possible to find a series representation for $f(z)$ involving both positive and negative powers of $z−z_0$. Formally speaking we have the following result:

Theorem $\PageIndex{1}$

Suppose that a function $f$ is analytic throughout an annular domain $R_1 \lt |z - z_0| \lt R_2$, centred at $z_0$, and let $C$ denote any positively oriented simple closed contour around $z_0$ and lying in that domain. Then, at each point in the domain, $f(z)$ has the series representation

$\begin{eqnarray}\label{laurentfunction}                 f(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n},                 \end{eqnarray}$

where

$\begin{eqnarray}\label{non-principalpart}                 a_n=\frac{1}{2\pi i}\oint_C \frac{f(z)dz}{(z-z_0)^{n+1}},\quad n=0,1,2,\ldots                 \end{eqnarray}$

and

$\begin{eqnarray}\label{principalpart}                 b_n=\frac{1}{2\pi i}\oint_C \frac{f(z)dz}{(z-z_0)^{-n+1}},\quad n=1,2,\ldots                 \end{eqnarray}$

Figure 1: $f(z)$ is analytic throughout an annular domain $R_1 \lt |z - z_0| \lt R_2$.

In practice, the above integral formulae (2) and (3) may not offer the most practical method for computing the coefficients $a_n$ and $b_n$ for a given function $f(z)$; instead, one often pieces together the Laurent series by combining known Taylor expansions. Because the Laurent expansion of a function is unique whenever it exists, any expression of this form that actually equals the given function $f(z)$ in some annulus must actually be the Laurent expansion of $f(z)$.

For example, consider the function

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

which has isolated singularities at $z=0$ and $z=±i$. In this case, there is a Laurent series representation for the domain $0\lt |z|\lt 1$ and also one for the domain $1\lt |z|\lt \infty$, which is exterior to the circle $|z|=1$. To find each of these Laurent series, we recall the Maclaurin series representation

$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n,\quad |z|\lt 1.$

For the domain $0\lt |z|\lt 1$, we have

$\begin{eqnarray*}             f(z)&=&\frac{1}{z}\frac{1}{1+z^2}=\frac{1}{z}\sum_{n=0}^{\infty}\left(-z^2\right)^n \\             &=&\sum_{n=0}^{\infty}(-1)^nz^{2n-1}\\             &=&\frac{1}{z}+\sum_{n=1}^{\infty}(-1)^nz^{2n-1} \\             &=& \sum_{n=0}^{\infty}(-1)^{n+1}z^{2n+1}+\frac{1}{z}.             \end{eqnarray*}$

On the other hand, when $1\lt |z|\lt \infty$,

$\begin{eqnarray*}             f(z)&=&\frac{1}{z^3}\frac{1}{1+\frac{1}{z^2}}=\frac{1}{z^3}\sum_{n=0}^{\infty} \left(-\frac{1}{z^2}             \right)^n\\             & =&\sum_{n=0}^{\infty}\frac{(-1)^n}{z^{2n+3}}\\             &=&\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{z^{2n+1}}             \end{eqnarray*}$

In this last part we use the fact that $(-1)^{n-1}=(-1)^{n-1}(-1)^2=(-1)^{n+1}$.

Exercise $\PageIndex{1}$

Expand $f(z)=e^{3/z}$ in a Laurent series valid for $0\lt |z| \lt \infty$.