Let $f$ be an analytic function on $\overline{B(z,R)}$ with only simple zeroes. Assume that f has no zeroes on the boundary of the ball. Let $\gamma=z_{0}+Re^{it}$ be the boundary path.
I want to show that $\intop_{\gamma}\frac{f'(z)}{f(z)}dz$ counts the number of zeroes of $f$ in $B(z_{0},R)$.
I don't know how to attack this problem.
What I am thinking is something along the lines of singularities since when $z$ is a root then $\frac{f'(z)}{f(z)}$ is undefined? Also, I am thinking the winding number has to come in play here, since it's the only thing that "counts"...
The only theorem I know that bring these two together is the following: If $f$ is analytic on $U$, except for poles $z_{1},...,z_{n}$ and $\gamma$ is homologous to 0 on $U$ then
$\intop_{\gamma}f(w)dw=2\pi i{\sum}W(\gamma;z_{i})Res(f,z_{i})$
But it doesn't seem to fit since nothing tells me $f$ has poles and nothing tells me that $\gamma$ is homologous to 0.
Any suggestions?
If $f(z)=0$ then $f'(z) \neq 0$ because zeros are of order 1. Thus $\frac {f'} f$ has a pole of order 1 at each zero of $f$ and no other poles. Just apply Residue Theorem to complete the proof. If $f(c)=0$ then the residue of $\frac {f'} f$ at $c$ is $\lim _{z \to c} \frac {(z-c)f'(z)} {f{(z)}}=\lim _{z \to c} \frac {(z-c)f'(z)} {f{(z)}-f(c)}=\frac {f'(c)} {f"(c)}=1$ by the definition of $f'(c)$.