This is an exercise question about Schwarz-Christoffel Formula.

Find a conformal mapping from upper half plane to the domain $\mathbb{C}\setminus\left(\{x+i \mid x\in(-\infty,0)\}\cup\{x-i \mid x\in(\infty,0)\}\right)$

I know that S-C formula is to find the conformal mapping from upper half plane to any $n$-gon. But I have no idea what $n$-gon this domain is. Any help is appreciated.


If I recall correctly (please check everything carefully before believing me!), you need to squint at the desired target domain until it looks like a quadrilateral with sides going from $-\infty+i$ to $i$, then from $i$ to $-\infty+i$ (think of those two sides as the "top" and "bottom" parts of the ray), then from $\infty-i$ to $-i$, and finally to $-i$ to $\infty-i$. The interior angles are $2\pi$ as it whips around the vertex $i$ from the first side to the second, then $-\pi$ (because $-\infty+i$ is the same point as $\infty-i$, but the changed direction reveals the angle "at infinity"), then $2\pi$ again at the vertex $-i$, then $-\pi$ again.

  • $\begingroup$ Thank you, but there are two things I dont understand. First, why the "angle at infinity" is $\pi$. I think it should be $0$ if it changes direction. Also, This quadrilateral has two vertices at infinity, which is essentially the same point. So should they be mapped to the same point or different points? $\endgroup$ – NECing Mar 24 '13 at 21:50
  • $\begingroup$ The interior angles of a quadrilateral should add to $2\pi$ - that's one fact supporting my belief that the angles at infinity should be $\pi$. And yes, the quadrilateral has two vertices at infinity, but they should be treated separately (as would be the case if you had a non-convex actual polygon with two equal vertices). The conformal map doesn't have to be one-to-one on the boundary. $\endgroup$ – Greg Martin Mar 24 '13 at 22:15
  • $\begingroup$ Ok, the interior angle add to $2\pi$ makes sense. But could the angles be $2\pi, 0, 2\pi,2\pi$? Imaging the boundary to be two isoceles triangle with same base $i,-i$,and two vertices at infinity. $\endgroup$ – NECing Mar 24 '13 at 22:40
  • $\begingroup$ Right. The reason I think both angles should be $\pi$ is because of the specific directions that the rays go off to $\infty$. Look at it this way: if you applied $1/z$ to the entire target domain, then the two rays going to infinity would be mapped to two circular arcs meeting at $0$, and the angle between those arcs (as you can check) would be $\pi$. $\endgroup$ – Greg Martin Mar 25 '13 at 3:00

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