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Find a conformal mapping of the disk $x^2+(y-1)^2\lt 1$ onto the first quadrant $x, y \gt 0$


I did something, which may be false or not, I cannot exactly say anything.

I used the composition of a conformal map, which is conformal.

Firstly, let's get a conformal map from the disk to the unit disk, say $f_1(z)$. And let's get a conformal map from the unit disk to half plane, say $f_2(z)$. And let's other conformal map from the half plane to the first quadrant, say $f_3(z)$

First of all, $f_1$ maps the given disc $\{z\in \Bbb C : x^2+(y-1)^2\lt 1\}$ conformally to the unit disc $\{ z\in \Bbb C : x^2+ y^2\lt 1\}$ by $$f_1(z)= z-i$$

Secondly, $f_2$ maps from the unit disk $\{ z\in \Bbb C : x^2+ y^2\lt 1\}$ conformally to the upper half plane $\{z\in \Bbb C : Im(z)\gt 0\}$ by a möbius transformation $T=\frac{z-1}{z+1}$ and $T(0)=-1$

So, $$f_2(z)=-i\frac{z-1}{z+1}$$

Thirdly, $f_3$ maps from the upper half plane to the first quadrant $\{z\in \Bbb C :x,y\gt 0\}$ We know that the boundary of a half plane has No vertex in the plane, in the straightforward line, in the sphere and the boundary of first quadrant has a vertex where two straightforward half linear making up the boundary Mete at the right angle. And we know that the powe maps $z\to z^k$ multiple angles at $0$ by k-writing $z=p.e^{i\theta}$so we have $z^k=p^ke^{ik\theta}$

In order to straigthen the right angle at $0$ of the boundary of upper half plane , I need $$k=1/2$$

Then, $$f_3(z)=\sqrt{z}$$

Then $$f(z)= f_3 \circ f_2 \circ f_1= \sqrt{-i{\frac{z-i-1}{z-i+1}}}$$

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  • $\begingroup$ Yep, that's right. Seems you understood. (Except, we don't straighten the angle here, we produce it with the square root.) $\endgroup$ – Daniel Fischer Oct 28 '13 at 12:08
  • $\begingroup$ Okay thank you a lot:) but my teacher said a solution hint as disk to unit disk to left half plane to upper half plane to first quadrant. But I skiped the left half plane. Hopefully, No problem! In addition to this, I taked $T=(z-1)/(z+1)$ accourding to my feelings:) but I dont understand clearly how to write $1$ in the "$T$"? $\endgroup$ – Nrsnr Oct 28 '13 at 12:16
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    $\begingroup$ You didn't skip the left half plane, $z \mapsto \frac{z-1}{z+1}$ takes the unit disk to the left half plane. The rotation by multiplication with $-i$ that takes the left half plane to the upper half plane is so trivial that you looked at it as one step, but in the plan the teacher laid out those are two steps. $\endgroup$ – Daniel Fischer Oct 28 '13 at 12:35
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    $\begingroup$ @User69127 Yes. Of course the map is not unique, you can compose it with an automorphism to get different maps. But that one is (one of) the simplest. $\endgroup$ – Daniel Fischer Feb 26 '14 at 12:29
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    $\begingroup$ It isn't wrong. What makes you think it is? $\endgroup$ – Daniel Fischer Feb 26 '14 at 12:44

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