Find a conformal map from the disc to the first quadrant.

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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