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Find all the Möbius transformations that maps the right half plane $\{z: \operatorname{Re}z>0\}$ to the unit disc $\{z: \, |z|<1\}$.

I have no clue how to do this. I know how to find a transformation when you map points to points but not like this. I notice however that the Imaginary axis must map to $|z|=1$ and that one point on the right half plane must map into some interior point of this. But that is pretty much how far I got. If anyone could take me through a full explanation or send to some website with information about this i would be so thankful.

Maybe I should add that I know that all the Möbius transformations can be written on the form

$$w=f(z)=\frac{az+b}{cz+d}$$

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    $\begingroup$ There is also a restriction on $ad-bc$ for Möbius transformations. $\endgroup$ – Marconius Oct 21 '15 at 19:09
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Suppose $f(z_0)=0, \:\Re{(z_0)}>0$. Then the symmetric point of $z_0$ with respect to $y$ axis is $-\bar{z_0}$, which must be mapped to $\infty$. So $$ f(z)=\frac{a}{c}\frac{z-z_0}{z+\bar{z_0}} $$ If $z'=0$, then $|f(z')|=1$ (boundary $y=0$ must be mapped to boundary $|\omega|=1$). So $|\frac{a}{c}|=1$ and we have $$ f(z)=e^{i\theta}\frac{z-z_0}{z+\bar{z_0}} $$ You need one more point to decide $\theta$.

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  • $\begingroup$ Okay! I got it so far $$ f(z)=\frac{a}{c}\frac{z-z_0}{z+\bar{z_0}} $$ and this I understand clearly, I think. Is it always like this, symmetric points respect to the boundary (in this case the Im-axis) mapps to opposite points? What about the latter statement you did, what is $z'$ here and how do you know that $\left| \frac{a}{c} \right|=1$? The point $z_0$ is still unknown in the final expression, right? $\endgroup$ – user269620 Oct 22 '15 at 6:59
  • $\begingroup$ So if I leave $\theta$ unknown I have found all the transformations and for each $\theta$ I have a unique transformation? $\endgroup$ – user269620 Oct 22 '15 at 7:16
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    $\begingroup$ Yes, you are right. $\endgroup$ – Math Wizard Oct 22 '15 at 7:17
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Hint: It's enough to find one such function, assuming you know the automorphism group of the unit disc. Every conformal mapping from the disc onto itself is of the form $$ z \mapsto e^{i\theta}\,\frac{a-z}{1-\bar a z} $$ for some $a$ in the unit disc and some real $\theta$, so if you can find one conformal map from the right half plane to the unit disc, you get all others by composing with maps of the above type.

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  • $\begingroup$ Ok, this adds some useful information but its still a bit unclear. Maybe you can see my comment on hermes post and help out with the explanation of that last part? By the way, greetings from KTH! :) $\endgroup$ – user269620 Oct 22 '15 at 7:08

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