If a function fails to be analytic at a point $z_0$, we cannot apply Taylor’s theorem at that point. However, it may be possible to find a series representation for $f(z)$ involving both positive and negative powers of $z-z_0$.

*Theorem* [Laurent's Theorem]. Suppose that a function $f$ is analytic throughout an annular domain $R_1<|z-z_0|<R_2$, 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

$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\sum_{n=1}^\infty\frac{b_n}{(z-z_0)^n}\ (R_1<|z-z_0|<R_2),$$

where

\begin{align*}

a_n&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz\ (n=0,1,2,\cdots),\\

b_n&=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{-n+1}}dz\ (n=1,2,\cdots).

\end{align*}

The expansion can be also written as

$$f(z)=\sum_{n=-\infty}^\infty c_n(z-z_0)^n\ (R_1<|z-z_0|<R_2),$$

where

$$c_n=\frac{1}{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^{n+1}}dz\ (n=0,\pm 1,\pm 2,\cdots).$$

*Example*. From the Maclaurin series expansion $e^z=\sum_{n=0}^\infty\frac{z^n}{n!}$, we obtain Laurent series expansion for $e^{\frac{1}{z}}$

$$e^{\frac{1}{z}}=\sum_{n=0}^\infty\frac{1}{n!z^n}=1+\frac{1}{1!z}+\frac{1}{2!z^2}+\cdots\ (0<|z|<\infty).$$

The coefficient $b_1$ is

$$b_1=\frac{1}{2\pi i}\oint_C e^{\frac{1}{z}}dz=1$$

for any positively oriented simple closed contour $C$ around $0$ and lying in the domain $0<|z|<\infty$. Hence, we obtain the integral

$$\oint_C e^{\frac{1}{z}}dz=2\pi i$$

for any positively oriented simple closed contour $C$ around $0$ and lying in the domain $0<|z|<\infty$.

*Example*. $f(z)=\frac{1}{(z-i)^2}$ is already in the form of a Laurent series, where $z_0=i$. From

$$f(z)=\sum_{n=-\infty}^\infty c_n(z-i)^n\ (0<|z-i|<\infty),$$

we find that $c_{-2}=1$ and all other coefficients are zero. Hence, we have

$$\oint_C\frac{dz}{(z-i)^{n+3}}=\left\{\begin{array}{ccc}

0 &\mbox{if}&n\ne -2,\\

2\pi i &\mbox{if}& n=-2

\end{array}

\right.$$

for any positively oriented simple closed contour $C$ around $i$ and lying in the domain $0<|z-i|<\infty$.

*Example*. Let $f(z)$ be the function

$$f(z)=\frac{-1}{(z-1)(z-2)}=\frac{1}{z-1}-\frac{1}{z-2}.$$

Since $f(z)$ has two singularities $z=1$ and $z=2$, we may consider the following three different domains to obtain Laurent series expansion in each domain

$$D_1: |z|<1,\ D_2: 1<|z|<2,\ D_3: |z|>2.$$

For $D_1: |z|<1$,

$$\frac{1}{z-1}=-\frac{1}{1-z}=-\sum_{n=0}^\infty z^n$$

and

$$-\frac{1}{z-2}=\frac{1}{2-z}=\frac{1}{2}\frac{1}{1-\frac{z}{2}}.$$

Since $|z|<1$, $|z|<2$ and so $\frac{1}{1-\frac{z}{2}}=\sum_{n=0}^\infty\left(\frac{z}{2}\right)^n$. Hence, we obtain the Taylor series expansion

\begin{align*}

f(z)&=-\sum_{n=0}^\infty z^n+\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}\\

&=\sum_{n=0}^\infty(2^{-n-1}-1)z^n\ (|z|<1).

\end{align*}

For $D_2: 1<|z|<2$,

$$\frac{1}{z-1}=\frac{1}{z}\frac{1}{1-\frac{1}{z}}=\frac{1}{z}\sum_{n=0}^\infty\left(\frac{1}{z}\right)^n.$$

Hence, we obtain the Laurent series expansion

\begin{align*}

f(z)&=\frac{1}{z}\sum_{n=0}^\infty\left(\frac{1}{z}\right)^n+\frac{1}{2}\sum_{n=0}^\infty\left(\frac{z}{2}\right)^n\\

&=\sum_{n=0}^\infty\frac{z^n}{2^{n+1}}+\sum_{n=1}\frac{1}{z^n}\ (1<|z|<2).

\end{align*}

For $D_3: |z|>2$,

\begin{align*}

f(z)&=\frac{1}{z-1}-\frac{1}{z-2}\\

&=\frac{1}{z}\sum_{n=0}^\infty\left(\frac{1}{z}\right)^n-\frac{1}{z}\sum_{n=0}^\infty\left(\frac{2}{z}\right)^n\\

&=\sum_{n=1}^\infty\frac{1-2^{n-1}}{z^n}\ (2<|z|<\infty).

\end{align*}

Ukwule Martins AdimaWonderful! What a credible, substantive guide and meticulous adherence to technicality of solving Laurent series problem? Remain bless, more Greece to us elbow for enlightening me on Laurent series.