Find mobius transformation that maps open half annulus onto itself

Question

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?

Answer 1

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.

Answer 2

Alternatively, you may use the fact that the cross-ratio is preserved for a billinear mapping i.e.

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