# 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:

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 $1:$

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

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,

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 $z:$

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
$0\lt |z| \lt \infty.$