help for question on order of pole of function

**alphabeta89** · October 26th 2012, 12:35 AM

Let $D$ be a domain in $\mathbb{C}$ , and let $z_0 \in D$ . It is given that a function
$f : D\backslash \{z_0\}\rightarrow \mathbb{C}$ has a simple pole at $z_0$ . Consider the function $g$ given by

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

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

This is my attempt.

Since f has a pole of order 1 at $z_0$ . Then
$f(z) = \frac{\phi(z)}{z-z_0}$ where $\phi(z)$ is analytic at $z_0$ and $\phi(z_0) \neq 0$ .

Since $g(z) = [f(z)]^2$ .
$g(z) = \frac{(\phi(z))^2}{(z-z_0)^2}$ .

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

**johnsomeone** · October 26th 2012, 04:35 PM

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

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

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

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

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

**johnsomeone** · October 30th 2012, 12:48 PM

I'm sorry - I phrased that wrong.
The point is that $\alpha(z_0) \ne 0.$

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