Help to understand the generalization of the Argument Principle I'm reading Conway's complex analysis book and I'm trying to prove this theorem left to the reader on page 124:

I tried to use integration by parts without success. I need some hint how prove this theorem.
 A: You can proceed similarly as in the proof of the "ordinary" argument
principle. Write
$$
 f(z) = \frac{\prod_{j=1}^n (z-z_j)}{\prod_{k=1}^m (z-p_k)} h(z)
$$
where $h$ is holomorphic and $\ne 0$ in $G$. Now take the logarithmic
derivative
$$
 \frac{f'(z)}{f(z)} = \sum_{j=1}^n \frac{1}{z-z_j} - \sum_{k=1}^m \frac{1}{z-p_k} + \frac{h'(z)}{h(z)}
$$
and multiply with $g(z)$
$$
 g(z) \frac{f'(z)}{f(z)} = \sum_{j=1}^n \frac{g(z)}{z-z_j} - \sum_{k=1}^m \frac{g(z)}{z-p_k} + g(z) \frac{h'(z)}{h(z)} \\
 = \sum_{j=1}^n \frac{g(z_j)}{z-z_j} - \sum_{k=1}^m \frac{g(p_k)}{z-p_k} + \left( g(z) \frac{h'(z)}{h(z)} + \sum_{j=1}^n \frac{g(z)-g(z_j)}{z-z_j} - \sum_{k=1}^m \frac{g(z)-g(p_k)}{z-p_k} \right) \, .
$$
The term in parentheses has only removable singularities in $G$, therefore
$$ \tag{*}
g(z) \frac{f'(z)}{f(z)} = \sum_{j=1}^n \frac{g(z_j)}{z-z_j} - \sum_{k=1}^m \frac{g(p_k)}{z-p_k} + \phi(z)
$$
with an holomorphic function $\phi$ in $G$.
The assertion now follows since for $a =z_1, \ldots, z_n, p_1, \ldots, p_m$
$$
 \frac{1}{2 \pi i} \int_\gamma  \frac{g(a)}{z-a}\, dz = g(a) \operatorname{n}(\gamma; a)
$$ and
$$
 \frac{1}{2 \pi i} \int_\gamma \phi(z) \, dz = 0
$$
due to the Cauchy integral theorem.
Remark: Another method to obtain $(*)$ would be to observe that both
$$
g(z) \frac{f'(z)}{f(z)} 
$$
and
$$ \sum_{j=1}^n \frac{g(z_j)}{z-z_j} - \sum_{k=1}^m \frac{g(p_k)}{z-p_k} 
$$
are holomorphic in $A$ with the only exception of (at most) simple poles
at the zeros and poles of $f$, and that the principle parts at those
points are the same. Therefore the difference has only removable
singularities.
