On page 173 of Complex Analysis, Gamelin, the author states the following theorem:
Let $z_0$ be an isolated singularity of $f(z)$. Then $z_0$ is a pole of $f(z)$ of order $N$ if and only if $1/f(z)$ is analytic at $z_0$ and has a zero of order $N$.
How can $1/f(z)$ have a zero, if $1/f(z) \neq 0$ for all $z$?
If $z_0$ is the n-th order pole, then $f(z)$ has the Laurent series at the neighborhood of $z_0$, which is
$$f(z)=\sum_{k=-n}^\infty a_n (z-z_0)^k$$
where $\color{red}{a_{-n}\neq0}$, hence we have
$$\frac1{f(z)}=\frac1{\displaystyle\sum_{k=-n}^\infty a_k (z-z_0)^k}=\frac{(z-z_0)^n}{\displaystyle\sum_{k=-n}^\infty a_k (z-z_0)^{k+n}}$$
Let $m=k+n$
$$\frac1{f(z)}=\frac{(z-z_0)^n}{\displaystyle\sum_{m=0}^\infty a_{m-n} (z-z_0)^{m}}=\frac{(z-z_0)^n}{a_{-n}+\displaystyle\sum_{m=1}^\infty a_{m-n} (z-z_0)^{m}}$$
Since $\color{red}{a_{-n}\neq0}$, the denominator is analytic and non-vanish at $z=z_0$, therefore, $z_0$ is the n-th order zero of $1/f(z)$.