# Conformal mapping from upper half plane onto unit square

Question:

Suppose that $f(z)$ is a conformal mapping of $\{\mathrm{Im}(z)>0\}$ onto the unit square $$\{0<\mathrm{Re}(z)<1,\;0<\mathrm{Im}(z)<1\}$$ such that the boundary points $0$, $1$, $\infty$ correspond to $0$, $1$, $1+i$ respectively.

a) Which point $x$ on the real axis corresponds to the vertical $i$?
b) Prove that $f^{-1}$ has an analytic continuation to a meromorphic function $F$.

Thanks.

I think it should be related to Schwartz Christoffel.

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Do you mean the unit square :$$\mid Re(z) \mid < 1 \space and \space \mid Re(z) \mid <1 ?$$ – Amir Alizadeh Dec 29 '12 at 16:22

$$f(z)=\int_0^z s^{-\frac{1}{2}}(s-x)^{-\frac{1}{2}}(1-s)^{-\frac{1}{2}}ds$$
for some $x<0$. Here the branch of the integrand is taken such that it is positive for $s \in (0,1)$. Then $f$ maps the upper half plane onto a rectangle. If you take $x=-1$ and continue the integrand to the interval $(-1,0)$ then it follows that $f(-1)=i\, f(1)$ and so $f$ maps to a square and $f$ can be properly scaled such that it maps onto the unit square. The entire complex plane can be tesselated by repeatedly reflecting this unit square in any of its sides. The Schwarz reflection principle shows that $f^{-1}$ extends to a meromorphic function (it extends holomorphically over three of the edges of the square and has a second order pole at the fourth). In fact $f^{-1}$ can be easily seen to be a doubly periodic function for the lattice $\mathbb{Z}\,2+\mathbb{Z}\, 2i$.