Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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,

share|cite|improve this question
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. – J. M. Dec 2 '11 at 0:14
See also the discussion here. – J. M. 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, – User001 Jul 18 '15 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.

A very detailed study of this integral can be found in Markushevish's centennial Theory of functions.

share|cite|improve this answer
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. – J. M. Dec 5 '11 at 4:55
This shows the meromorphicity and doubly periodic character of the inverse of the elliptic integral, and more. – J. M. Dec 5 '11 at 5:00
Thank you for the added precision J.M. :) – Bruno Joyal Dec 5 '11 at 5:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.