i'm triyin to solve this:
Find a moebius transformation that applies the exterior of the circle $\{z:|z-1|=1\}$ to the right half plane. $\{z: Re(z)>0\}$.
So i fixed $a=1$, then
$T(2) \to \infty$ $\implies \frac{2\cdot 1+b}{2c+d}=\infty$
$T(1-i) \to 1 \implies \frac{(1-i)+b}{c(1-i)+d}=1$
$T(0) \to 0 \implies \frac{0 \cdot 1 +b}{c\cdot 0 + d}=0$
with this I get the following equation system
$a=1$
$2c+d=0$
$1-i+b=c-ci+d$
$b=0$
and i get $a=1, b=0, c=i$ and $d=-2i$ so i get the next moebius transformation $T(z)=\frac{z}{iz-2i}$ wich verifies for $2,1-i$ and $0$ but if i consider $-1$ in the exterior of the circle then $T(-1)=\frac{-1}{-i-2i}=\frac{-1}{-3i}=\frac{-i}{3}$ and $\frac{-i}{3}$ is not into the right half plane. $\{z: Re(z)>0\}$.
I don't know where I went wrong
Doing this by solving equations for the coefficients may not be the easiest way. There are "geometric" options. First, translate the circle $|z-1|=1$ to the circle $|z|=1$. Then invert: $z\to 1/z$. Now, true, this sends $0$ to the point at infinity, but this is inescapable: if the exterior of a circle is not taken to include the point at infinity, then it has the homotopy type of a circle, which is not the same as the homotopy type of a half-plane... Then map the unit disk to a half-plane by a "Cayley map"... (for example, sending $-1\to 0$, $0\to 1$, and $1\to\infty$).