Express ${1\over 2\pi i}\int_{\partial \Omega}{g(z){f'(z)\over f(z)}}dz$ with poles and zeros of $f$ The question states, in fact, show ${1\over 2\pi i}\int_{\partial \Omega}{g(z){f'(z)\over f(z)}}dz=\sum_{a_k}g(a_k)-\sum_{b_k}g(b_k)$, where $a_k$ are the zeros of $f$ in $\Omega$ and $b_k$ are the poles of $f$ given $f\in M(G)$ (meromorphic in $G$), $g\in H(G)$, $G$ is a bounded domain with piecewise smooth boundary $\Omega, \overline{\Omega}\subset G$ and $f\ne 0,\infty$ on $\partial {\Omega}$. (zeros of high order will show up more than once).
I tried to use the Argument principle, as well as trying to analyze the functions of question, but I have a problem trying to compute the residue, as I don't know what order of pole a given zero of $f$ might give. I could use your help.

Hint 
If $f(z)$ has a zero of order $k$ at $a$ then $f'(z)$ has a zero of order $k-1$ at $a$. Therefore $\frac{f'(z)}{f(z)}$ has a simple pole at the zeroes of $f$.
If $f(z)$ has a pole of order $k$ at $a$ then $f'(z)$ has a pole of order $k-1$ at $a$. Therefore $\frac{f'(z)}{f(z)}$ has a simple pole at the zeroes of $f$.
Hint 2: At the zeroes of $f$ write $f(z)=(z-a)^n h(z)$ with $h(a) \neq 0$. Then
$$g(z)\frac{f'(z)}{f(z)} = g(z) \frac{n(z-a)^{n-1}h(z)+(z-a)^nh'(z)}{(z-a)^n h(z)} \\
= g(z) \frac{nh(z)+(z-a)h'(z)}{(z-a) h(z)}
$$
Therefore, the residue at $a$ is
$$Res=g(a) \frac{nh(a)}{ h(a)}=n g(a) $$
Hmm, there seems to be an $n$ missing here.
I think the problem is not true, try $f(z)=z^n$ and $g(z)=1$ to get a counterexample...
At the poles
If $f(z)$ has a pole of order $n$ at $a$ then, writing $f(z)=\frac{h(z)}{(z-a)^n}=h(z)(z-a)^{-n}$ a fast computation shows that 
$$\frac{f'(z)}{f(z)}=\frac{h'(z)(z-a)^{-n}-nh(z)(z-a)^{-n-1}}{h(z)(z-a)^{-n}}=\frac{h'(z)(z-a)-nh(z)}{h(z)(z-a)}$$
Therefore the residue at $z=a$ is
$$Res= -ng(a)$$
Again there is an $n$ missing.
Therefore by the residue Theorem you get
$${1\over 2\pi i}\int_{\partial \Omega}{g(z){f'(z)\over f(z)}}dz=\sum_{a_k}n_kg(a_k)-\sum_{b_k}m_kg(b_k) $$
where $n_k, m_k$ are the orders of the zeroes/poles.
