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I need to find a mobius transformation that maps the open half annulus $1<|z|<2$ with $Im(z)>0$ onto itself, while mapping the points $z_1=1, z_2=i,$ and $z_3=-1$ to $w_1=-2, w_2=2i,$ and $w_3=2$.

I came up with three equations:

$$-2=\frac{a+b}{c+d}$$

$$2i=\frac{ai+b}{ci+d}$$

and $$2=\frac{-a+b}{-c+d}$$

I think I need one more piece of information but I don't know what. Can somebody please help me out?

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  • $\begingroup$ It is sufficient to assume $w(z)=\frac{z+b}{cz+d}$. $\endgroup$ – Nitin Uniyal Nov 14 '16 at 6:45
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    $\begingroup$ @mathlover Not necessarily. What if the transformation he's after is $\frac{-2}{z}$ or something along those lines? $\endgroup$ – Arthur Nov 14 '16 at 7:13
  • $\begingroup$ @Arthur...thanks. $\endgroup$ – Nitin Uniyal Nov 14 '16 at 7:49
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The "one more information" you need is that expanding or simplifying the fraction gives the same transformation (e.g. $\frac{az+b}{cz+d}=\frac{2az+2b}{2cz+2d}$). That removes one degree of freedom, and your three equations fix the rest.

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  • $\begingroup$ I'm sorry I still don't fully understand how that helps, would you mind elaborating a little? $\endgroup$ – user342661 Nov 14 '16 at 10:02
  • $\begingroup$ @user326441 What I mean is that if you try to solve these three equations, you will probably end up with something looking like $a=2b$ or $c=-3d$. At that point you're free to choose your favourite non-zero value for one of those variables, say $a=1$ in the first example or $d=6$ in the second. Or $e^{5\pi+\sqrt2}$, if you like that number better. $\endgroup$ – Arthur Nov 14 '16 at 10:10
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Alternatively, you may use the fact that the cross-ratio is preserved for a billinear mapping i.e.

$\frac{w-w_1}{w_1-w_2}.\frac{w_2-w_3}{w_3-w}=\frac{z-z_1}{z_1-z_2}.\frac{z_2-z_3}{z_3-z}$

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