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Verify that the Schwarz-Christoffel mapping of $\mathbb H$ onto the infinite half strip described by $|\Re(z)|<\frac \pi 2$ and $\Im(z)>0$ is given by the arcsine function.

What does that mean and how do I do it?

Schwarz-Christoffel formula for the half-plane $\mathbb H$ to the polygon with exterior angles described by coefficients $\beta_k$ is $$f(z)=A_1\int_0^z \frac 1{(w-x_1)^{\beta_1}(w-x_2)^{\beta_2}\cdots(w-x_n)^{\beta_n}}\ dw+A_2,\quad (z \in \mathbb H).$$

I realize this isn't the best question, but I'm not even sure what to ask.

Edit: An attempt to add more specific questions:

  1. The $x_n$ in the integral are supposed to be vertex points of the pre-image. If the pre-image is the upper half plane, how do I find vertex points?
  2. How do I get the integral to map to the described half plane?
  3. Where does the arcsine fit in? What does it mean for the mapping created by an integral to be "given by the arcsine function"?
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    So, you're asking why you're supposed to be interested in the mapping $$\int_0^z \frac1{\sqrt{1-w^2}} \mathrm dw$$ I take it?2012-05-03
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    No. I'm wondering how to verify that the mapping of $\mathbb H$ onto the given half strip is given by the arcsine function.2012-05-03
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    Yes, and the integral I gave is supposed to be the result of applying Schwarz-Christoffel to the region you're studying...2012-05-03
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    How did you get that? Note that I edited my question, too.2012-05-03
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    An infinite half-strip is a triangle on the Riemann sphere (with a vertex located at the "point at infinity" $\hat{\infty}$). Perhaps a sequence of triangles that converges to the strip in the limit will help you?2012-05-03
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    J.M. The target plane has vertices at $(-\pi/2,0)$ and $(\pi/2,0)$, both of which have external angles $\beta_1=\beta_2=1/2$. The arcsine of $\pm1$ maps to $\pm \pi/2$, so you plug $\pm1$ into SC for $x_1, x_2$. But if we didn't already know the mapping was an arcsine, how would we find that (or is that beyond the scope of this question)? Also, when I plug it in I get $$\int_0^z \frac {1}{\sqrt{w^2-1}}$$. How come your denominator is $1-w^2$ (which doesn't integrate to arcsine)?2012-05-03
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    Hmm, what you have seems to come from a different definition of the SC map. From what book did your definition come from?2012-05-03
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    An online book called "Exploring Complex Analysis" (here: http://www.jimrolf.com/explorationsInComplexVariables.html). See Ch. 5 (http://www.jimrolf.com/explorationsInComplexVariables/bookChapters/Ch5.pdf), equation 67 on page 324. I think I get to your equation by factoring out $-1$ from the square root in the denominator. Then my equation becomes $$\frac{-Ai}{\sqrt{w^2-1}}$$ and the integral returns $-Ai\arcsin(1-w^2)$. Is that right?2012-05-03
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    BTW: you'll want to see [this](http://books.google.com/books?hl=en&id=CDP1zxFJLucC&pg=PA71).2012-05-03
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    Thanks. I don't think I'll need to learn elliptical functions for my immediate needs, but I do want to.2012-05-03

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