Analytic continuation of Schwarz-Christoffel mappings? I have not studied carefully the topic of analytic continuation. Here is a question that popped up in my mind.
Let $f: \mathbb{D} \to \Omega$, where $\mathbb{D}$ is the open unit disk, and $\Omega$ is the interior of a polygon be a Schwarz-Christoffel mapping. Then $f$ extends continuously from $\overline{\mathbb{D}}$ into $\overline{\Omega}$, and maps the unit circle continuously onto the polygon $\partial \Omega$.
My question is this. Can one analytically continue $f$? If so, what would be its maximal domain of definition? What type of singularities would lie at the preimages of the polygon under $f$?
 A: My other answer is a little sloppy. I will probably delete it once thisone is completed.
I'll look at a biholomophic map from the upper half-plane to a polygon. 


*

*Let $a_1,\ldots,a_n \in (0,1)^n, \sum_m a_m < n-1$ and $b_1<\ldots < b_n\in \mathbb{R}$
$$f'(z) =\prod_{m=1}^n (z-b_m)^{a_m-1}, \qquad f(z) = \int_0^z f'(s)ds$$
In $\int_0^z f'(s)ds$ we need to choose a curve $0 \to z$. When integrating $f'$ we continue it analytically along the curve.
All the analytic continuations of $f'$ and $f$ are locally analytic away from $b_1,\ldots,b_n$.


*

*As $|z| \to \infty$, $f'(z) = O(z^{-1-\varepsilon})$ thus $\lim_{R \to \infty} \int_0^t f'(R e^{it}) d(R e^{it}) = 0$ and $f$ is continuous at $\infty$.
Look at the upper semi-disks $D_R = \{ z, Im(z) > 0, |z|< R\}$ and the branch of $f$ analytic on $Im(z) >0$ and continuous on $Im(z) \ge 0$. Then $f(\partial D_R)$ is a closed curve and $\lim_{R \to \infty}  f(\partial D_R) = f(\mathbb{R})$ is the boundary of the polygon $P$ with vertices $f(b_0),f(b_1),\ldots,f(b_m)$ where $f(b_0)= f(+\infty) = f(i\infty) = f(-\infty) $. Also the $\pi a_m$ are the angles at those vertices.

*Since $f'$ doesn't vanish on $Im(z) > 0$ then $f$ maps biholomorphically $Im(z) > 0$ to the interior of $P$.

*Rotation around a branch point : given a branch of $f$ analytic around $b_m+z$, continue analytically $f(b_m+ze^{2i\pi t})$ from $t=0$ to $t=1$ and set $f(b_m+ze^{2i\pi}) =\lim_{t \to 1} f(b_m+ze^{2i\pi t})$.

*For $|z| < \min_{l \ne m} |b_l-b_m|$ then $f'(b_m+z e^{2i \pi}) = f'(b_m+z ) e^{2i \pi a_m}$ thus $$f(b_m+ze^{2i\pi}) = f(b_m+z)+\lim_{\epsilon \to 0} \int_z^{\epsilon z} f'(b_m+s)ds\qquad\qquad\qquad\qquad \\ \qquad\qquad\qquad\qquad + \int_0^{2\pi} f'(b_m+\epsilon z e^{it}) d(\epsilon e^{it})+\int_{\epsilon z}^{z} f'(b_m+se^{2\pi})ds$$
$$ = f(b_m+z)+ \int_z^0 f'(b_m+s)ds+\int_0^{z} f'(b_m+se^{2\pi})ds$$
$$ = f(b_m)+e^{2i \pi a_m} (f(b_m+z)-f(b_m))$$

*Fix some branch $\tilde{f}$ of $f$ and let $G$ be the group generated by the affine transformations $g_m(f(z))= e^{2i \pi a_m} f(z) + (1-e^{2i \pi a_m})\tilde{f}(b_m)$.

Then the monodromy $M$ group of $f$ is the group generated by the affine transformations $\sigma_{m,s}(f(z)) = e^{2i \pi a_m} f(z) + (1-e^{2i \pi a_m})s(\tilde{f}(b_m))$ for each $m$ and each $s \in G$.


*How do we get that for some choices of $a_m,b_m$ then $f^{-1}$ is periodic, or doubly-periodic, or is some sort of tilling of the plane ?
For example with $n=2,a_1 =a_2=1/2, b_1=1,b_2 = -1$ then $f(z) = \text{arcosh}(z)$ and $f^{-1}(z) = \cosh(z)$ is $2i\pi$ periodic because rotating around $z=1$ then rotating in the other direction around $z=-1$ yields $f(z) \mapsto -f(z)+2 f(1) \mapsto -(-f(z)+2 f(1))-2 (-f(-1)+2 f(-1))= f(z)+C$ where $C=-2 f(1)-2f(-1)=2i\pi$ 
so $f^{-1}(f(z)) = f^{-1}(f(z)+C) $ and $f^{-1}$ is $C= 2i\pi$ periodic.
