Complex integrals and poles

Question

I have to show, if $ f(z)$ has a pole of order 1 at $ z=z_0 $, then $ w(z) = [f(z)]^2 $ must have a pole of second order at $ z_0 $.

so far all I have is that $f(z) $ will have a pole of order 1 if $ m=1 $

$ f(z)=\sum_{n=-m}^\infty a_n(z-z_0)^n $

pole should be of order 1 when the smallest $ z_0$ term in the Laurent series has power -1

$f(z) = \sum_{-1}^\infty a_{-1}(z-z_0)^{-1} $

next I have to show how the the $ \phi $-rule can be put into the form:

if $w(z)$ has a pole of order m as $ z_0 $, $ Res_{z=z_0}w(z) =\lim_{z\to z_0} \frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^mw(z)]$ I have tried gathering some information like:

The $ \phi $-rule is $ Res_{z=z_0}f(z)= \frac{\phi^{m-1}(z_0)}{(m-1)!} $

and that if $ f $ has a pole of any order at $ z_0 $ then

$ f(z) \to \infty $ as $ z \to z_0 $

I can't seem to put this information together to get any closer to the solution

Answer 1

An analytic function $f$ has a pole of order $n$ at $z_0$ if and only if the limit $\lim_{z\to z_0}(z-z_0)^nf(z)$ exists and it is a complex number different from $0$.

So, if $f$ has a pole of order $1$ at $z_0$, then $\lim_{z\to z_0}(z-z_0)^nf(z)=w$, for some $w\in\Bbb C\setminus\{0\}$. But then$$\lim_{z\to z_0}(z-z_0)^2f^2(z)=\lim_{z\to z_0}\left((z-z_0)f(z)\right)^2=w^2\ne0,$$and therefore $f^2$ has a pole of order $2$ at $z_0$.

Answer 2

Why not directly? The function has a pole of order $\;1\;$ at $\;z=z_0\;$ meשns we get a Laurent series for $\;f\;$ of the form

$$f(z)=\frac{a_{-1}}{z-z_0}+\sum_{n=1}^\infty a_n(z-z_0)^n\;,\;\;\text{for all points}\;z\;\text{ in some neighborhood of}\;z_0$$

Inside this neighborhood, we get that

$$(f(z))^2=\frac{a_{-1}^2}{(z-z_0)^2}+2\frac{a_{-1}a_0}{z-z_0}+\ldots$$

and thus $\;(f(z))^2\;$ has a pole of order two in $\;z=z_0\;$ ...