How to prove $\int_{\gamma}f'(z)g(z)dz/f(z)=2\pi i \sum^N_{j=1}g(\alpha_j)$, where f and g are holomorphic functions and $\gamma$ is a simple closed curve such that no zero of f lies on $\gamma$. And suppose $\alpha_1, \alpha_2,... \alpha_N$ are the zeroes of f in $Int(\gamma)$.
I think this can be considered as the generalization of the argument principle, but don't know how to prove this problem. Thanks.
Hint: Around a zero of order $k$, $f$ can be represented as $$(z-z_0)^kh(z)$$ with $h$ holomorphic and $h(z_0)\neq 0$. Noe apply the product rule and Cauchy's theorem.
EDIT: let $f(z)=(z-\alpha_i)^kh(z)$ as in the hint. Then $f'(z)=k(z-\alpha_i)^{k-1}h(z)+z-\alpha_i)^kh'(z)$, and hence $$\frac{f'}{f}=\frac{k(z-\alpha_i)^{k-1}h(z)+z-\alpha_i)^kh'(z)}{(z-\alpha_i)^kh(z)}=\frac{k}{z-\alpha_i}+\frac{h'}{h}.$$ The first summand has a simple pole, the second is holomorphic. By Cauchy's theorem one has, for any holomorphic function $g$ and any simple closed path around $z_0$ $$\frac{1}{2\pi i}\int_{\gamma}\frac{g}{z-z_0}dz=g(z_0,)$$ And for any closed path $$\int_\gamma gdz=0.$$ Conclude