Consider the following fragment on p99 of Conway's book "Functions of one complex variable":
I don't fully understand this proof.
(1) Firstly, what are the $z_k(\zeta)$ for $1 \le k \le p$? Are they the zeros of $f(z)-\alpha$ inside $B(a, \epsilon)?$ If so, why are there only finitely many such zeros?
Maybe because otherwise $\overline{B}(a,\epsilon)$ has infinitely many zeros, and has therefore a limit point by compactness. But I don't see how to proceed then.
(2) How exactly does the equality $n(\sigma, \zeta) = \sum_{k=1}^p n(\gamma, z_k(\zeta))$ ensure that there are exactly $m$ solutions of $f(z) = \zeta$ inside $B(a, \epsilon)$?
On page 98, Conway defines the object $z_k(\zeta)$ to be the points satisfying $f(z)=\zeta$. On the other hand,
$$ n(\sigma;\alpha) = \frac{1}{2\pi i}\int_\sigma \frac{\mathrm dw}{w-\alpha}$$
which is
$$\frac1{2\pi i}\int_{\gamma}\frac{f’(z)}{f(z)-\alpha}.$$
This counts the amount of zeroes of $f(z)-\alpha$ inside $B(a;\varepsilon)$. But remember—we explicitly made sure that $f(z)-\alpha$ has no solution with $0<|z-\alpha|<2\varepsilon$. Therefore, the only zero it will count is the one with order $m$ at $a$, whence $n(\sigma;\alpha) =m$.
But also, we specifically chose $\gamma$ such that $\alpha\notin \{\sigma\}$. Therefore, there exists some $\delta>0$ such that the entirety of $B(\alpha;\delta)$ belongs to the same component of $\mathbb C-\{\sigma\}$. Intuitively, the winding number of $\sigma$ about $\alpha$ counts the amount of times $\sigma$ wraps around $\alpha$, so for any $\zeta$ such that $|\alpha-\zeta|<\delta$, $n(\sigma;\alpha) = n(\sigma;\zeta)$.
But by the same integral that we did above,
$$n(\sigma;\zeta)=\sum_{k=1}^p n(\gamma;z_k(\zeta))$$
where $z_k(\zeta)$ are the points in $B(a;\varepsilon)$ such that $f(z)=\zeta$ (remember, this is the same thing as counting the zeroes of $f(z)-\zeta$). This implies that there are $m$ of these solutions because $n(\gamma;z)$ can only be zero or one (intuitively, $\gamma$ can wrap around a point at most once). All of them must be simple because we chose $\varepsilon$ such that $f’(z)\ne 0$ on $0<|z-a|<\varepsilon$.