Suppose that $\alpha$ is a regular closed contour. $f$, our function, lacks zeros and poles on $\beta$ and if A=the number of zeros of f inside $\beta$ (a zero of order n is counted n times) and B= the number of poles of f inside $\beta$ (same idea with counting multiplicity). Prove that the line integral on $\beta$ of $f'(z)/f(z) = (A-B)/2 \pi i$.
I'm not really sure how to proceed with this proof. If someone can give me a hint or some direction that would be really appreciated.
Let $\;Z_i\,,\,P_j\;$ be little, non-overlapping nor intersecting circles around the zeros and poles, resp. of $\;f(z)\;$ in the domain determined by $\;\alpha\;$ . then
$$\oint_\alpha\frac{f'}fdz=\sum_{i=1}^n\oint_{Z_i}\frac{f'}fdz+\sum_{j=1}^m\oint_{P_i}\frac{f'}fdz$$
Now, if the zeros are $\;\{z_1,...,z_k\}\;$ with respectives multiplicities $\;n_1,...,n_k\;$ , and the poles are $\;\{p_1,...,p_m\}\;$ with respective multiplicities $\;t_1,...,t_m\;$ (and thus we have that $\;A=\sum\limits_{i=1}^kn_i\,,\,\,B=\sum\limits_{j=1}^mt_j\;$), we can then write
$$\begin{align}&\text{Around}\;z_i\;:\;\;f(z)=(z-z_i)^{n_i}g_i(z)\implies f'(z)=n_i(z-z_i)^{n_i-1}g(z)+(z-z_i)^{n_i}g'(z)\\{}\\&\text{Around}\;p_j\;:\;\;f(z)=\frac{h(z)}{(z-p_j)^{t_j}}\implies f'(z)=\frac{h'(z)(z-p_j)-t_jh(z)}{(z-p_j)^{t_j+1}}\end{align}$$
with $\;g_i\,,\,h_j\;$ holomorphic functions without zeros, poles resp. in the region limited by $\;Z_i\,,\,\,P_i\;$ resp.
Thus, for each $\;i\,,\,j\;$ we get
$$\oint_{Z_i}\frac{f'}fdz=\int_{Z_i}\frac{n_i}{z-z_i}dz+\oint_{Z_i}\frac{g'}gdz=2\pi i n_i+0$$
and likewise for the other integrals, using Cauchy's theorem and etc. Complete and fill in details in the argument now.