Proving whether a squared function has double pole at $z_0$ Let D be a domain in $\Bbb C$ and let $z_0 \in$ D. If a function $f:$ D \ {$z_0$} $\rightarrow \Bbb C$ has a simple pole at $z_0$, is it true that $g$ has a double pole at $z_0$, where $g(z) = [f(z)]^2 $ $\forall z \in$ D\ {$z_0$}
Definition of a pole: If there exists $m\in \Bbb Z^+$ such that $b_m \neq 0$ for all $n>m$ (i.e. the principal part has finitely many non-zero terms) so that 
$$f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n + \frac{b_1}{z-z_0} + \frac{b_2}{(z-z_0)^2} +...+ \frac{b_m}{(z-z_0)^m}$$
Please help me check my ans:
$z_0$ is simple pole of $f(z) \Rightarrow$  there exist a function h which is analytic (i.e. holomorphic) on some domain D - {$z_0$} s.t. $f(z) = \displaystyle\frac{h(z)}{z-z_0}$, $\forall z \in$ D and $h(z_0) \neq 0$
$g(z) = [f(z)]^2 = \displaystyle\frac{[ h(z)]^2}{(z-z_0)^2}$, $z \in$ D - {$z_0$}
Since the only singular point in $g(z)$ is $z_0$, $g(z)$ is analytic everywhere except at $z_0$ $\Rightarrow$ $g(z)$ is also analytic in D - {$z_0$} $\Rightarrow [h(z)]^2$ is also analytic in D - {$z_0$}, $(h(z_0))^2 \neq 0$ 
Therefore, $g$ has a double pole at $z_0$

Hint: Use the definition of a simple pole to prove that $f$
can be represented as $$f(z)=\frac{h(z)}{z-z_{0}}$$ where $$h(z_{0})\neq0$$
