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*}