Solving for the Moebius transform that sends $(x_0, y_0)$ to $(x_1, y_1)$ with two known fixed points Suppose $M$ is a hyperbolic Moebius transformation with fixed points at $(0, 0), (1, 0)$ which, when applied to the complex $(x_0, y_0)$, yields the result $(x_1, y_1)$.
How do I solve for $M$ given $x_0, y_0, x_1$, and $y_1$?
 A: Hint: You can take your matrix representation of the Möbius transformation to be an element of $PSL(2,\mathbb{R})$, or your transformation to be $T(z)= \frac{az+b}{cz+d}$ with $ad-bc=1$. Then with the conditions $T(0)=0, T(1)=1 $ and $ T(z_0)=z_1$ you should be able to solve for $a, b , c$ and $d$.
A: The transformation we are looking for is 
$$\frac{w}{w-1}=k~\frac{z}{z-1}$$
with $0<k$.
Let $z_0=x_0+y_0~i$ and $z_1=x_1+y_1~i$. We can suposse $z_0\neq z_1$ because if $z_0=z_1$ then the function is the identity.
Then
$$k=\frac{z_1(z_0-1)}{z_0(z_1-1)}.$$
The inverse of $\displaystyle f(z)=\frac{z}{z-1}$ is itself. Therefore
$$w=f(k\cdot f(z))=\frac{k~z}{(k-1)z+1}$$
It remains the question that $k$ must be a positive real number. 
As $0$ and $1$ are fixed points, then $w$ is a traslation in $\mathbb{H}^2$ along the geodesic with infinity points $0$ and $1$. How the equidistant through $z_0$ to this geodesic is invariant, then $z_1$ is in the same equidistant. Then is necessary for $0<k$ that $z_1$ lives in the circumference through $0$, $1$ and $z_0$. 
