# A homeomorphism from a semidisk to the unit disk

In Bak and Newman's Complex Analysis Chapter 14, Problem 12, the reader is asked to find a conformal mapping from the upper semidisk (with norm 1) $S$ to the unit disk $U$. Then, they ask to show that this extends to a homeomorphism from the closure of $S$ to the closure of $U$. Finally, they ask to show that this map is analytic on the closure of $S$ but its inverse is not analytic on the closure of $U$.

The first part (I think) is easy: I know that the function $$f_1=\frac{-(z-1)^2}{4(z+1)^2}$$ maps $S$ onto $H$ (the upper half-plane) and further I know that the simplest mapping from $H$ onto $U$ is given by my old friend $$f_2=\frac{z-i}{z+i}$$ so that $f_2(f_1(z))$ gives the desired conformal mapping from $S$ onto $U$. I'm not sure how to proceed for the next part, though. I assume that I will somehow need to show that $f_2(f_1(z))$ maps $\partial S$ onto $\partial U$, so if $z_0 \in S$, we have argument in $[0, \pi)$ with norm 1 and our mapping takes this to something with norm 1 but any argument in $[0, 2\pi)$. The problem I'm encountering is that I do not know how to show this. Trying to algebraically simplify the mapping I described has not been helpful and I am not sure how to proceed - perhaps going about this so explicitly is the wrong approach. Then, I assume once I show that I can map boundary to boundary, and that the function and its inverse are continuous, I will need to show that the inverse is not analytic, which I am not sure how to do. (I know how to show some maps are analytic, but I am not so sure on how to show they are not).

Edit: If I can show that the map is continuous, does this imply that since it maps $S$ to $U$, it maps $\partial S$ to $\partial U$?

Finally, please avoid any detailed topological arguments as much as you can in your answers as I have not studied topology and so complex analysis is as close as I have gotten to the subject. Thanks!

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The map can be extended by continuity to $\partial S$. In general, a continuous map doesn't have to map $\partial S$ to $\partial U$. E.g. $\sin x$ maps $(0, 2\pi)$ to $[-1,1]$ but it doesn't map $\{0, 2\pi\}$ (the boundary of $(0,2\pi)$) to $\{-1, 1\}$ (the boundary of $[-1,1]$). However, in your problem, the map is conformal, so its extension must map $\partial S$ to $\partial U$. The inverse map is not analytic at points $f_2(f_1(1))$ and $f_2(f_1(-1))$ (and analytic at all other points). To prove that, find the Taylor expansion of $f_2(f_1(z))$ at $z=1$ (compute just the first 3 terms). –  Yury Nov 25 '12 at 6:21
I don't know what you know, what theorems and methods you can use in the proof. One way to prove that $f_2(f_1(z))$ is analytic is to use the Schwarz reflection principle (for all boundary points except for $\{-1,1\}$); another is to directly prove that the explicit formula for $f_2(f_1(z))$ defines an analytic function in a neighborhood of $\bar S$. –  Yury Nov 25 '12 at 6:30