Complex analysis question relating to closed rectifiable curves homologous to $0$. I came across this problem on Functions of One Complex Variable by John B. Conway. (Chapter IV section 7, page 99)
Let G be open and suppose $γ : [0, 1] \to G$ is a closed rectifiable curve in $G$ such that $γ$ is homologous to $0$. Set $r = d(γ([0, 1]), \partial G)$ and $H = \{z ∈ ℂ : n(γ; z) = 0\}$.
(a) Show that $\{z : d(z, \partial G) < \frac 12 r\} \subset H$.
(b) Use your answer to part (a) to show that if $f : G \to ℂ$ is analytic then $f(z) = \alpha $ has at most a finite number of solutions $z$ such that $n(γ; z) \neq 0$.
Here, $n(γ; z)$ is the winding number of $γ$ around $z$. Intuitively, the result makes sense but I'm not sure how to go about proving it. Any ideas?
 A: If $r=0$, then $\gamma$ intersects the boundary of $G$ nontrivially and so $\gamma$ is not in $G$. Thus, $r>0$. Let $z$ be such that $d(z,\partial G) < \frac{r}{2}$. If $z \notin G$, then by definition IV.6.18 $\gamma \approx 0$, $n(\gamma ; z)=0$. If $z \in G$ then note that $z \notin {\gamma}$ since otherwise we would have $d(z,\{\gamma\})=0$ and the contradiction that $r < \frac{r}{2}$ from $d(\{\gamma\},\partial G) \leq d(\{\gamma\},z)+d(z,\partial G) $. Now, if $z$ is in the component inside $\gamma$, then there exists $k$ such that $\frac{1}{n} < d(z,\gamma ) $ for all $n \geq k$ (this is because $d(z,\gamma )>0$ and again we run into a contradiction via $|d(z,\gamma) - d(\gamma,\partial G)| \leq d(z,\partial G)$ namely $\frac{r}{2}<\frac{1}{n}$ and so $r=0$. Thus, $z$ is in the other component of $G$ whereby $n(\gamma ; z)=0$
For (b), let $a \in G$ such that $f(a)=\alpha$ and $n(\gamma ; a ) \neq 0$. The just established result tells us that $d(a,\partial G) \geq \frac{r}{2}$. If there were infinitely many such solutions, all of which are isolated, then the setup in Page 98 tells us that $n(\sigma; \alpha) = \infty$, which basically says that $\sigma$ is not rectifiable, a contradiction.
