# Schwarz-Christoffel mapping of the upper half-plane

I am given the function $w(z)=\int_0^z \frac1{\sqrt{1-t^2} \sqrt{1-k^2t^2}}dz$ and shall show that this is mapping the upper half-plane onto a rectangle. We just discussed the Schwarz-Christoffel integral, and we can rewrite this as $\int_0^z \frac1{\sqrt{t-1} \sqrt{t+1} \sqrt{kt-1} \sqrt{kt+1}}dt$, and since the exponents of the four factors are all $-1/2$, we have $\alpha_i-1=-1/2$ which tells us that we are dealing with four right angles.

But, isn't the map actually mapping the unit circle to the rectangle, and not the upper half-plane onto the rectangle? Where is my mistake?

Also, I want to show that the inverse function extends to a meromorphic function on $\mathbb C$. What is the trick here? I don't have any idea on it.

Best regards,

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Note: $w(z)$ is one of the canonical examples of an elliptic integral, and the inverse function of $w(z)$ is in fact one of Jacobi's elliptic functions. There should be a discussion of this particular Schwarz-Christoffel mapping in McKean and Moll's book. –  Guess who it is. Dec 2 '11 at 0:14
See also the discussion here. –  Guess who it is. Dec 5 '11 at 2:23
Marie, don't forget to accept answers you are satisfied with! :) –  Bruno Joyal Sep 28 '13 at 1:18
Hi @Guesswhoitis., I studied the section in McKean's book just now, and I understand how the real line, traces out the perimeter of a rectangle. However, how do we know that the UHP is necessarily mapped to the interior of the rectangle? This doesn't seem obvious to me, for some reason. Is it just simply testing a point from the UHP, e.g., integrating the above integral from 0 to +i, and try to show that +i actually gets mapped to the interior of the rectangle? Thanks, –  Lebron James Jul 18 at 1:23

As pointed out by J.M., this is an elliptic integral, whose inverse function is Jacobi's elliptic function $\text{sn}$.
As $z$ travels along the real line, the argument of $\sqrt{(1-t^2)(1-k^2t^2)}$ undergoes sudden $\pi/2$ changes as $z$ crosses the points $1,-1,1/k,-1/k$, provided $k$ is real. Hence the real line is mapped to the boundary of a rectangle of side lengths $\int_{-1}^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$ and $\int_1^{1/k}\frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}$ (provided here that $|1/k|>1$). Hence no, it is not the unit disc which is mapped to a rectangle. However, the upper-half plane and unit disc are conformally equivalent and by a change of variables (eg. $z \mapsto (z-i)/(z+i)$) you can obtain a conformal map which does take the unit disc to the rectangle.
Proving that the inverse function extends to a meromorphic function on $\mathbb{C}$ can be done using the Lagrange inversion theorem.
Conventionally, the modulus $k$ is taken to be $0 < k < 1$ in most applications, so in that case $1/k$ is definitely bigger than $1$. For other real values of $k$, things are a bit more complicated, but there are ways to reduce to the $0 < k < 1$ case. –  Guess who it is. Dec 5 '11 at 4:55