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 neighborhood 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 neighborhood 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:

Suppose that a function $f$ is analytic throughout an annular domain $R_1 \lt |z - z_0| \lt R_2,$ centered 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}

Annulus — $f(z)$ is analytic throughout an annular domain $R_1 \lt |z - z_0| \lt R_2.$

As a very simple example of (\ref{laurentfunction}) let us consider the function \[ f(z) = \frac{1}{z-1}. \] As you can seen, the point $z = 1$ is an isolated singularity of $f$ and consequently the function cannot be expanded in a Taylor series centered at that point. Nevertheless, $f$ can expanded in a series of the form given in (\ref{laurentfunction}) that is valid for all $z$ near $1:$

\begin{eqnarray}\label{simple-example} f(z) = \cdots + 0 \cdot (z-1)^2 + 0 \cdot (z-1) + 0 + \frac{1}{(z-1)} + \frac{0}{(z-1)^2} + \cdots \end{eqnarray}

This series representation is valid for $0 \lt |z - 1| \lt \infty.$

The series on the right-hand side of (\ref{laurentfunction}) are given special names. The part consisting of the nonnegative powers of $z - z_0$, \begin{eqnarray}\label{analytic-part} \sum_{n=0}^{\infty} a_n(z-z_0)^n \end{eqnarray} is called the analytic part of the series (\ref{laurentfunction}) and will converge for $|z - z_0| \lt R_2.$ The part with negative powers of $z - z_0,$ that is, \begin{eqnarray}\label{principal-part} \sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n} \end{eqnarray} is called the principal part of the series (\ref{laurentfunction}) and will converge for

$$\abs{\frac{1}{z-z_0}}\lt \frac{1}{R_1} \quad \text{or equivalent }\quad R_1 \lt \abs{z-z_0}.$$

Hence, the sum of (\ref{analytic-part}) and (\ref{principal-part}) converges when $z$ satisfies both $R_1 \lt \abs{z-z_0}$ and $\abs{z-z_0}\lt R_2$, that is, when $z$ is a point in an annular domain defined by $R_1 \lt |z - z_0| \lt R_2.$

The principal part of the series (\ref{simple-example}) consists of exactly one nonzero term, whereas its analytic part consists of all zero terms. The following example illustrates a series of the form (\ref{laurentfunction}) in which the principal part of the series also consists of a finite number of nonzero terms, but this time the analytic part consists of an infinite number of nonzero terms.

Example 1: The function $f(z) = \dfrac{\sin z}{z^4}$ is not analytic at the isolated singularity $z = 0 $ and hence cannot be expanded in a Maclaurin series. However, $\sin z$ is an entire function, and we know that its Maclaurin series,

$$\sin z = z - \frac{z^3}{3!}+ \frac{z^5}{5!} - \frac{z^7}{7!} + \frac{z^9}{9!}- \cdots,$$

converges for $|z|\lt \infty.$ By dividing this power series by $z^4$ we obtain a series for $f$ with negative and positive integer powers of $z:$

\begin{eqnarray}\label{example-1} f(z) = \dfrac{\sin z}{z^4} = \overbrace{\frac{1}{z^3} - \frac{1}{3!z} }^{\large \textbf{Principal part}} \underbrace{+ \frac{z}{5!} - \frac{z^3}{7!} + \frac{z^5}{9!}- \cdots}_{\large \textbf{Analytic part}} \end{eqnarray}

The analytic part of the series in (\ref{example-1}) converges for $|z|\lt \infty.$ (Verify!) The principal part is valid for $|z|>0.$ Thus (\ref{example-1}) converges for all $z$ except at $z=0;$ that is, the series representation is valid for $0 \lt |z| \lt \infty.$

In practice, the above integral formulae (\ref{non-principalpart}) and (\ref{principalpart}) provided in Theorem 1 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, as we did in Example 1. 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 $f(z).$

Example 2: Consider the function $$f(z)=\frac{1}{z(1+z^2)}$$ which has isolated singularities at $z=0$ and $z=\pm i.$ 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: Expand $f(z)=e^{3/z}$ in a Laurent series valid for $0\lt |z| \lt \infty.$