Find all Mobius transformations that satisfies the following conditions Cosinder Mobius functions $$ T\left(z\right)=\frac{az+b}{cz+d} $$
I want to find all Mobius functions, which satisfies the following:
$T(-2)=0 $, $T(0)=i$ , $T(S(0;2)) =S(-1;1) $ Where $S$ denotes sphere.
I tried to solve it but stumbled upon something I cannot explain. Here's my attempt:
The obvious things:
$$ T\left(i\right)=0\Rightarrow b=di $$
and $$ T\left(-2\right)=0\Rightarrow b=2a $$
So also $d=-2ia$
So far we have $$ T\left(z\right)=\frac{az+2a}{cz-2ia} $$
Now consider the spheres:

Since $T(z) $ is a conformal map, it must conserve angles. Which means that $ T $ maps points where the tangent of $S(0,2) $ perpendicular to the real axis, to points where the tangent of $S(-1,1) $ perpendicular to the real axus. That is, it maps the point $2$ to either $-2 $ or $0$. But we know already that $T(-2)=0$, which means that $ T $ must map $2$ to $-2$.
That is, $$ T\left(2\right)=\frac{2a+2a}{2c-2ia}=-2 $$
and we get $ 4a=-4c+4ia\Rightarrow c=\left(i-1\right)a $
So we have  $$ T\left(z\right)=\frac{az+2a}{\left(i-1\right)az-2ia}=\frac{z+2}{\left(i-1\right)z-2i} $$
Which seems like a solution, but in order to check myself I tried to check what is the image of $2i$ under $ T $. If everything is fine, $2i$ must be mapped to either $-1+i$ or $-1-i$ since $T $ preserves angles.
Now $$ T\left(2i\right)=\frac{2i+2}{\left(i-1\right)2i-2i}=\frac{i+1}{\left(i-1\right)i-i}=\frac{i+1}{-1-i-i}=\frac{i+1}{-1-2i} $$
Which is not what I expected, because $ -\frac{1+i}{1+2i}=-\frac{3-i}{5} $.
What went wrong?
Thanks in advance.
 A: There is an error in your approach: At $z=2$, $S(0;2)$ intersects the real axis at a right angle. “$T$ is angle-preserving“ means that at $w=T(2)$, $T(S(0;2))$ intersects $T(\Bbb R)$ at a right angle. You assume that the image circle intersects the real axis at a right angle. That would only be true if $T$ maps the real axis onto itself, which is not the case here.
In order to find a third condition which determines $T$ we can use the fact that Möbius transformations preserve symmetry with respect to a circle.
If $S$ is a circle with center $a$ and radius $R$ then $z, z^*$ are symmetric with respect to $S$ if
$$
 (z^*-a)(\bar z - \bar a) = R^2 \, \text{ or } \, \{ z, z^* \} = \{ a, \infty \} \, .
$$
It then follows that $T(z)$ and $T(z^*)$ are symmetric with respect to $T(S)$.
In our case $0$ and $\infty$ are symmetric with respect to $S(0;2)$. It follows that $T(0) = i$ and $T(\infty)$ are symmetric with respect to $T(S(0; 2)) = S(-1;1)$. Using the above formula we get
$$ 
T(\infty) = -1 + \frac{1}{-i+1} = -\frac 12 + \frac i 2 \, .
$$
Together with $T(-2)=0$ and $T(0) = i$ this determines $T$ uniquely.
