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I tried to find the Moebius transformation which maps unit-disk to unit-disk and maps 2 points: $z_1=1$ to $w_1=1$ and $z_2=1+i$ to $w_2=\infty$.

What I have:

  1. $w_1= \dfrac{a+b}{c+d} = 1$
  2. $w_2= \dfrac{a(1+i)+b}{c(1+i)+d} = \infty$

then $c(1+i)+d=0$ and $d = -c(1+i)$

What is the next step? It is not enough equations to find all a,b,c,d.

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Every Möbius transformation of the type$$z\mapsto\omega\frac{z-a}{\overline az-1},$$with $|\omega|=1$, maps the unit disk into itself. If you want it to map $1+i$ into $\infty$, take $a$ such that $\overline a(1+i)-1=0$; in other words, take $a=\frac12+\frac i2$. So, $f(1)=\omega i$. Therefore, take $\omega=-i$, and your Möbius transformation will be$$z\mapsto\frac{(-1+i)z+1}{-z+1+i}.$$

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  • $\begingroup$ thank you very much. I can't get your Moebius transformation result. I put $\omega=-i$ and $a=\frac12+\frac i2$ to the $z\mapsto\omega\frac{z-a}{\overline az-1}$. $\endgroup$ – Metso Jun 17 '18 at 9:45
  • $\begingroup$ @Metso And what do you get when you do that? What I get is what I wrote:$$\frac{(-1+i)z+1}{-z+1+i}.$$ $\endgroup$ – José Carlos Santos Jun 17 '18 at 9:48
  • $\begingroup$ $\omega\frac{z-a}{\overline az-1} = i\frac{z-(0.5)(1+i)}{0.5(1-i)z-1}$ and then $\frac{2zi-i+1}{z-zi-2} $ $\endgroup$ – Metso Jun 17 '18 at 10:09
  • $\begingroup$ @Metso I wrote that $\omega=-i$, not that $\omega=i$. $\endgroup$ – José Carlos Santos Jun 17 '18 at 10:17
  • $\begingroup$ sorry, $\omega\frac{z-a}{\overline az-1} = -i\frac{z-(0.5)(1+i)}{0.5(1-i)z-1}$ is not same as your result. I have got $\frac{-2zi+i-1}{z-zi-2}$ $\endgroup$ – Metso Jun 17 '18 at 10:30

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