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Find a conformal mapping from the quarter-disk $Q=\{ |z|<1 : rez>0,im z>0 \}$ onto the upper half plane set $U=\{im z>0\}$

I'm guided through this problem:

First I need to find the image of the quarter disk $Q=\{ |z|<1 : rez>0,im z>0 \}$ by $f(z)=z^2$?

My answer to this is a semi-circle (edit:disk), with unit radius.

Now I need to find the map of the halfdisk $D=\{ |z-1|<1 :im z>0 \}$ onto the the first quadrant $\{im z>0, rez>0\}$.

I'm stuck on this bit:

I tried: rewriting the halfdisk as $1+e^{it}$ for $0\leq t \leq 2\pi$ , but I didn't get far.

Lastly, I need to somehow put the first two together, to get a map from the quarter disk onto the upper halfplane

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  • $\begingroup$ Your words "semi-circle" are less appropriate that "semi-disc" (in analogy to "quarter-disc"). $\endgroup$ – Lee Mosher Apr 23 '16 at 21:35
  • $\begingroup$ Yes, sorry, I meant disk $\endgroup$ – GRS Apr 23 '16 at 21:40
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Consider the map $f_1=z^2$, then the map $f_2=-1/2(z+z^{-1})$.

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  • $\begingroup$ I'm assuming the second map is for the mapping of halfdisk onto the 1st quadrant. Then I'll need to take the composition. Is that right? $\endgroup$ – GRS Apr 23 '16 at 22:07
  • $\begingroup$ First map takes the quarter-disk into upper half-disk. The second takes the upper half-disk into the upper half-space. $\endgroup$ – Ruzayqat Apr 23 '16 at 22:09
  • $\begingroup$ @Ruzayqat did you mean $f_2=-\frac{1}{2} (z+z^{-1})$ or $f_2=\frac{-1}{2 (z+z^{-1})}$, at first I thought it is the first one but I couldn't show that it maps the upper half-disk into the upper half-space. $\endgroup$ – thehardyreader Jun 19 '18 at 20:54

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