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I'm studying complex calculus and one conclusion of the Riemann theorem is that the domain where the complex natural logarithm is holomorphic, meaning $\mathbb{C}\setminus\mathbb{R}_-$, can be mapped onto the unit disk with some Mobius transform since it is a simply-connected domain different from $\mathbb{C}$. I've tried to find this using methods we saw in class (picking 3 points on the edge of the starting domain and using the formula for $a$, $b$, $c$, and $d$) and the only this I've managed to map is the top half of the plane onto the unit disk. What would the Möbius transformation be in this special case (of a domain that has a one-dimensional outside)?

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  • $\begingroup$ Can you show us what you tried? $\endgroup$ Jul 27, 2020 at 12:12
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    $\begingroup$ Hint: you can take log and square-roots on $\mathbb{C}-\mathbb{R}_-$ since you no longer has the problem of winding number about 0. Note you can't do it with Mobius alone $\endgroup$ Jul 27, 2020 at 12:30
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    $\begingroup$ There is a conformal map from $\mathbb{C}\setminus\mathbb{R}_-$ to the unit disk, but not a Möbius transformation. $\endgroup$
    – Martin R
    Jul 27, 2020 at 13:49
  • $\begingroup$ But then, wouldn't you be able to map $\mathbb{C}$ to the unit disk with a function like $$ f(z)=(1-e^{-|z|})(e^{i\cdot \arg(z)})$$ or would this mapping not qualify as conformal? $\endgroup$ Jul 28, 2020 at 12:04

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You need two steps to perform the task.

In the first step you map the given domain $\Omega$ to the right half plane $H$, using the principal value of the square root function:

$${\rm pv}\sqrt{\cdot}:\quad\Omega\to H,\qquad z=re^{i\phi}\to w:=\sqrt{r}\,e^{i\phi/2}\qquad(r>0, \ -\pi<\phi<\pi)\ .$$ You could also write $${\rm pv}\sqrt{z}:=e^{{\rm Log}(z)/2}\qquad(z\in\Omega)\ .$$ In the second step you map the $w$-halfplane $H$ via a Moebius transformation $T$ to the unit circle. You could, e.g., require that $$T(0)=-1,\quad T(\infty)=1,\quad T(i)=i\ .$$

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