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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](http://books.google.com/books?id=CDP1zxFJLucC&pg=PA55).2011-12-02
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    See also the discussion [here](http://books.google.com/books?id=DYCOCBCBwoIC&pg=PA124).2011-12-05
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    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,2015-07-18

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