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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 meet at a right angle. And we know that the power 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 straighten the right angle at $0$ of the boundary of upper half plane , I need $$k=1/2$$

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

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

ViktorStein
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Nrsnr
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  • Yep, that's right. Seems you understood. (Except, we don't straighten the angle here, we produce it with the square root.) – Daniel Fischer Oct 28 '13 at 12:08
  • 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 "$T$"? – Nrsnr Oct 28 '13 at 12:16
  • I am grateful of you to look at my answer. Thanks Dear @DanielFischer – Nrsnr Oct 28 '13 at 12:17
  • 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"? @DanielFischer – Nrsnr Oct 28 '13 at 12:25
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    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. – Daniel Fischer Oct 28 '13 at 12:35
  • @DanielFischer suppose I wanted to find a function $f$ which maps the open unit disk in the complex plane one-to-one conformally onto the quarter plane. Taking $f_1(z)=-i\frac{z+1}{z-1}$ (unit disk to upper half plane), then $f_2(z)=\sqrt{z}$ (upper half plane to first quadrant), would the map then be $\sqrt{-i\frac{z+1}{z-1}}$? – User69127 Feb 26 '14 at 12:21
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    @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. – Daniel Fischer Feb 26 '14 at 12:29
  • @DanielFischer thanks a lot. Only reason I ask is because, when we go the other way around (quarter to disk), why is there a need to multiply the map by $-i$? Going from quarter to disk, using $f_1=z^2$ (quarter to upper half) and $f_2=\frac{z-i}{z+i}$ (upper half to disk), I obtain a simple $\frac{z^2-i}{z^2+i}$. Why is this wrong though? – User69127 Feb 26 '14 at 12:38
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    It isn't wrong. What makes you think it is? – Daniel Fischer Feb 26 '14 at 12:44
  • @DanielFischer http://math.stackexchange.com/questions/298075/find-conformal-mapping-from-sector-to-unit-disc – User69127 Feb 26 '14 at 20:13
  • second answer on the bottom – User69127 Feb 26 '14 at 20:14
  • @User69127 $$\frac{iw^2+1}{-w^2-i} = \frac{i(w^2-i)}{-(w^2+i)} = (-i)\frac{w^2-i}{w^2+i},$$ so that is your map followed by a rotation of the unit disk. The difference comes because he chose a different map to map the unit disk to the upper half-plane (yours, preceded by a rotation), so when inverting, he got a rotation at the end. – Daniel Fischer Feb 26 '14 at 20:31
  • @DanielFischer thank you, God's missionary! – User69127 Feb 26 '14 at 21:10

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