# 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: 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}

In practice, the above integral formulae (\ref{non-principalpart}) and (\ref{principalpart}) 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 $z=\pm i$. 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 $0\lt |z| \lt \infty$.

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