Let $f$ be analytic in $D_r(z_0)$ and has a zero of order $m$ at $z_0$. I need to show that $f'/f$ has a simple pole at $z_0$. This is part of attempt.
Since f has a zero of order $m$ there exists a analytic $g$ such that $$f(z)=g(z)(z-z_0)^m,g(z_0)\neq0$$.
Also for some $R_1>0$ such that $0<R_1\leq r$, $g(z)\neq 0$ in $D_{R_1}(z_0)$. Now there are possibilities. $f$ is identically zero or for some $0<R_2\leq r, $ $f(z)\neq 0$ in $D_{R_2}(z_0)$\ {$z_0$}. Assuming the latter is true let $R$=min{$R_1,R_2$}.
Let $z\in D_R(z_0)$\ {$z_0$}. Then $$f'(z_0)=mg(z)(z-z_0)^{m-1}+g'(z)(z-z_0)^m \implies \frac{f'(z)}{f(z)}=\frac{m}{z-z_0}+\frac{g'(z)}{g(z)}$$.
I am stuck here and don't know how to finish it off. And what about the case where f is identically zero? Any help is appreciated thanks
Best Answer
You're almost there. You just need to write $$ \dfrac{f'(z)}{f(z)}=\dfrac{h(z)}{z-z_0} $$ with $h$ holomorphic.
Alternatively, you just need to prove that $$ (z-z_0)\dfrac{f'(z)}{f(z)} $$ is holomorphic.
The case $f\equiv0$ is not relevant because then $z_0$ would be a zero of infinite order, not finite $m$.