help for question on order of pole of function

****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! :D

Re: help for question on order of pole of function

$\displaystyle \text{Define } \alpha(z) = \phi(z)^2, \text{ where } \alpha \text{ has the same domain } U \text{ as } \phi.$

$\displaystyle \text{Then } g(z) = \frac{\alpha(z)}{(z-z_0)^2} \text{ on } U\backslash \{z_0\} \subset D\backslash \{z_0\}.$

$\displaystyle \text{Is } \alpha \text{ holomorphic on the neighborhood } U \text{ of }z_0? \text{ Does } \alpha(z_0) = 0?$

$\displaystyle \text{If you answered yes to both questions, then isn't that exactly the definition}$

$\displaystyle \text{of } g \ (=f^2) \text{ having a pole of order 2 at }z_0?$

Re: help for question on order of pole of function

I'm sorry - I phrased that wrong.

The point is that $\displaystyle \alpha(z_0) \ne 0.$