Recall that if $f(z)$ has an isolated singularity at $z=z_0$, it may be represented by a Laurent series
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_n}{(z-z_0)^n}+\cdots$$
in a puctured disk $0<|z-z_0|<R$. The part of series that contains negative powers of $z-z_0$
$$\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_n}{(z-z_0)^n}+\cdots$$
is called the principal part of $f(z)$ at $z_0$.
Suppose that there exists $m\in\mathbb{N}$ such that $b_m\ne 0$ and $b_{m+1}=b_{m+2}=\cdots=0$. Then
$$f(z)=\sum_{n=0}^\infty a_n(z-z_0)^n+\frac{b_1}{z-z_0}+\frac{b_2}{(z-z_0)^2}+\cdots+\frac{b_m}{(z-z_0)^m},$$
where $0<|z-z_0|<R$. In this case, the isolated singularity $z_0$ is called a pole of order $m$. A pole of order 1 is usually called a simple pole.
Example.
\begin{align*}
\frac{z^2-2z+3}{z-2}&=z+\frac{3}{z-2}\\
&=2+(z-2)+\frac{3}{z-2},\ 0<|z-2|<\infty
\end{align*}
has a simple pole at $z_0=2$. The residue at $z_0=2$ is 3.
Example.
\begin{align*}
\frac{\sinh z}{z^4}&=\frac{1}{z^4}\left(z+\frac{z^3}{3!}+\frac{z^5}{5!}+\cdots\right)\\
&=\frac{1}{z^3}+\frac{1}{3!z}+\frac{z}{5!}+\frac{z^3}{7!}+\cdots,\ 0<|z|<\infty
\end{align*}
has a pole of order $m=3$ at $z_0=0$. The residue at $z_0=0$ is $\frac{1}{6}$.
When $b_n=0$ for all $n\geq 1$, so that
$$f(z)=\sum_{n=0}^\infty (z-z_0)^n$$
the point $z_0$ is called a removable singularity.
Example.
\begin{align*}
f(z)&=\frac{1-\cos z}{z^2}\\
&=\frac{1}{z^2}\left[1-\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+\cdots\right)\right]\\
&=\frac{1}{2!}-\frac{z^2}{4!}+\frac{z^4}{6!}-\cdots,\ 0<|z|<\infty.
\end{align*}
Thus $z_0=0$ is a removable singularity. Define
$$g(z)=\left\{\begin{array}{ccc}
f(z) & \mbox{if} & z\ne 0,\\
\frac{1}{2!} & \mbox{if} & z=0.
\end{array}\right.$$
Then $g(z)$ is entire.
When an infinite number of the $b_n$ are nonzero, $z_0$ is called an essential singularity.
Example.
\begin{align*}
\exp\left(\frac{1}{z}\right)&=\sum_{n=0}^\infty\frac{1}{n! z^n}\\
&=1+\frac{1}{1!z}+\frac{1}{2!z^2}+\cdots,\ 0<|z|<\infty
\end{align*}
has an essential singularity at $z_0=0$.
Example. $e^z=-1$ when $z=(2n+1)\pi i$ $(n=0,\pm 1, \pm 2,\cdots)$. So $e^{\frac{1}{z}}=-1$ when $\frac{1}{z}=(2n+1)\pi i$ or $z=-\frac{i}{(2n+1)\pi}$ $(n=0,\pm 1,\pm 2,\cdots)$and an infinite number of these points lie in any neighbourhood of the essential singularity $z_0=0$.
Picard’s Theorem. In each neighbourhood of an essential singularity, a function assumes every finite value, with one possible exception, an infinite number of time.
In the above example, since $e^{\frac{1}{z}}\ne 0$ for all $z$, $z=0$ is the exceptional value in Picard’s theorem.