Consider the following scenario. Let $γ$ bound a polygon in $C$ , and let $b_{i}$ denote the vertices of this polygon. By Caratheodary Theorem, there exists a conformal map $f$ from $H$ onto the interior of the polygon. Consider the interior angles of the polygon, and rewrite them as $π α_{i}$ for $0 < α_{i} < 2$ . Finally, We notice that the preimage of the vertices of the polygon lie on the real line (possibly infinity). Below we illustrate the situation:

Now, the theorem states that for such a function, we have $$ \frac{f''(z)}{f'(z)}=\begin{cases} \sum_{k=1}^n \frac{\alpha_{k}-1}{z-a_{k}} & a_{n}\in \mathbb{R} \\ \sum_{k=1}^{n-1} \frac{\alpha_{k}-1}{z-a_{k}} & a_{n}=\infty \end{cases} $$ $$ \LARGE \text{PROOF} $$ 1. Use ==1;;Schwarz Reflection principle== on the sides of the polygon 2. Use the reflections are an involution to create a ==1;;relation between the different Schwarz reflections== 3. Show that these relations represent a Euclidean motion. That is, for two reflections $f_{k},f_{\ell}$, we have that $f_{k}=pf_{\ell}+q$ with $|p|=1$ and $q\in \mathbb{C}$. 4. Obtain a function $\phi:\mathbb{C}-\{ a_{1},\dots,a_{n} \}\to \mathbb{C}$ (QUESTION: Why is $\phi$ not necessarily defined here?) given by $$ \phi(z)=\begin{cases} \frac{f''(z)}{f'(z)} & z\in \mathbb{H} \cup \mathbb{R} \\ \frac{f_{k}''(z)}{f_{k}'(z)} & z\in \overline{\mathbb{H}} \end{cases} $$ 6. Now we observe what happens near the $a_{i}$ 7. Post compose with $(z-b_{k})^{1/\alpha_{k}}$ to ==1;;normalize the angle of the polygon to a half circle around the origin==. 8. We apply the schwarz reflection principal again, argue that $f$ has a simple zero here. 9. Use the simple zero to determine the behavior of $\phi(z)$ near the $a_{i}$. 10. Show that $\phi$ has the desired form using ==1;; Louivilles Theorem==. 11. Make a remark that all the $\alpha_{i}$ add up to $n-2$ by ==1;;observing behavior at infinity== 12. Address case where $\alpha_{n}=\infty$ seperately but similarly.