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 $z_0$

$\;z_0\;$ is a simple pole of $\;f(z)\;$ iff there's a holomorphic $\;h\;$ in some domain $\;D_0-\{z_0\}\;$ s.t.
$$f(z)=\frac{h(z)}{(z-z_0)}\;\;,\;\;\forall\,z\in D_0\;\;\;\text{and}\;\;\;h(z_0)\neq 0$$
From here, we have that
$$\forall\,z\in D_0-\{z_0\}\;\;,\;\;\;g(z):=f(z)^2=\frac{h^2(z)}{(z-z_0)^2}$$
Prove now $\;h(z)^2\;$ is holomorphic in $\;D_0-\{z_0\}\;$ and $\;h(z_0)^2\neq 0\;$  and you're done...
