Let $\displaystyle D$ be a domain in $\displaystyle \mathbb{C}$, and let $\displaystyle z_0 \in D$. It is given that a function

$\displaystyle f : D\backslash \{z_0\}\rightarrow \mathbb{C}$ has a simple pole at $\displaystyle z_0$. Consider the function $\displaystyle g$ given by

$\displaystyle g(z) = [f(z)]^2$ for all $\displaystyle z \in D \backslash \{z_0\}$.

Is it true that $\displaystyle g$ has a double pole at $\displaystyle z_0$?

This is my attempt.

Since f has a pole of order 1 at $\displaystyle z_0$. Then

$\displaystyle f(z) = \frac{\phi(z)}{z-z_0}$ where $\displaystyle \phi(z)$ is analytic at $\displaystyle z_0$ and $\displaystyle \phi(z_0) \neq 0$.

Since $\displaystyle g(z) = [f(z)]^2$.

$\displaystyle g(z) = \frac{(\phi(z))^2}{(z-z_0)^2}$.

How do I proceed from here? Any idea or suggestion is welcome. Thanks in advance!