Extension of a conformal mapping to an elliptic function Let $\rho$ be the conformal mapping from the interior of the triangle with vertices $-1,i\sqrt{3},1$ onto the upper half-plane. Show that $\rho$ has an elliptic extension. 
The hypothesis makes sense by invoking the Riemann Mapping Theorem. Furthermore, $\rho$ has an extension to a homeomorphism on the boundary of the triangle to the real line due to Caratheodory extension. In fact, this extension of $\rho$, denoted as $P$, takes the real value on the boundary of the triangle except that one of the vertices is mapped to the point of infinity under compactification. Thereby, $P$ has an analytic continuation across the boundary of the triangle and to a meromorphic function on $\mathbb{C}$ and the vertices are invariant under $P$ by rotations of $\pm \frac{2\pi}{3}$. At this point, I do not know how to continue. 
Any help is sincerely valued. 
 A: Extension across the boundaries of triangles results in a triangular grid. Mark its vertices according to where they are mapped (say, to -1, 1, or $\infty$, so that the two bottom vertices stay fixed and the top vertex goes to infinity). You will see identically labeled triangles appearing. Since a conformal map is determined by the images of three boundary points, this means it's identical in those triangles; the periodicity follows. Refer to the picture below (borrowed from Wikipedia): 
Pick any blue triangle here and say it's the original triangle; label its vertices $-1, 1, \infty$. This determines the labels of all other vertices, because reflection across a side corresponds to conjugation in the image plane, and all the vertex images are real. For example, reflecting across the right leg of a blue triangle shows that the bottom-left vertex of another blue triangle also goes to -1. Continuing this, you will see all blue triangles are labeled the same way.  Hence, the function had two periods: one vertical, $2\sqrt{3} i$, and another slanted, $3+i\sqrt{3}$. 

