Mobius transformation maps the real axis to the unit circle 
Show that any Mobius transformation which takes the real axis (with
  $\infty$) to the unit circle can be written in the form
$$M(z)= \alpha \dfrac{z-\beta}{z-\overline{\beta}}$$ where
  $|\alpha|=1$.

I attempt to solve this: 
If $w=f(z)$ is a linear fractional transformation that transforms the real axis into unit circle, setting $z_1=1, z_2=0, z_3=-1$, we know $w_1=f(z_1), w_2=f(z_2), w_3=f(z_3)$ have module equal to $1$ . Using the cross ratio, we have
$$\frac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\frac{-(z-1)}{z+1}.$$
We can rewrite
$$\frac{w-w_1}{w-w_3}=\alpha\frac{z-1}{z+1}$$
Solving for $w$, we have
$$w=\frac{(w_1-\alpha w_3)z+(w_1+\alpha w_3)}{(1-\alpha)z+(1+\alpha)}.$$
But I can't continue and use the condition $|w_1|=|w_2|=|w_3|=1$.
 A: Let $$f(z) = \frac{az + b}{cz + d}.$$
First by normalizing $c=1$ ($c = 0$ is easily seen to be impossible), and then collecting a factor at the numerator, we may assume that $$f(z) = \alpha \frac{z + w_1}{z + w_2}.$$
The condition $|f(\infty)| = 1$ gives immediately that $|\alpha|=1$. Now the condition $|f(0)| = 1$ gives $|w_1|=|w_2|$, let's say $|w_1|=|w_2|=r$. Let now $t \in \mathbb{R}$. Then
$$
|t+w_j|^2 = t^2 + 2t\mathfrak{Re}(w_j) + r^2
$$
for $j=1,2$. Hence, $|f(t)|=1$ if and only if
$$
t^2 + 2t\mathfrak{Re}(w_1) + r^2 = t^2 + 2t\mathfrak{Re}(w_2) + r^2,
$$
i.e., since $t$ is arbitrary,
$$
\mathfrak{Re}(w_1) = \mathfrak{Re}(w_2).
$$
Now $w_1$ and $w_2$ share the same modulus and the same real part, so either $w_1 = w_2$ and the function is constant equal to $\alpha$ (you can choose $\beta = 0$ in your formula), or $w_1 = \overline{w_2}$ as wanted.
A: First we move unit circle for $1$ up, that is we act on it with the function $h(z)=z+i$.
Then we perform a stereografic projection on this new circle, so we map $2i\mapsto \infty$, $0\mapsto 0$ and $1+i\mapsto 2$. This map is of form $$g(z)={-2iz\over z-2i}$$
Now $$f(z) = g(h(z)) = {-2i(z+i)\over z-i}$$ is a Mobius transformation with that takes unit circle to extended real axsis. Now $$M(z) = f^{-1}(z) =i{z-2i\over z+2i}$$ is a desired Mobius transformation.
A: The required transform maps points that are symmetric wrt the real line to points symmetric (conjugate) wrt the unit circle. If $\beta$ is not on the real line and is mapped to $0$, then $\overline \beta$ is mapped to $\infty$. Therefore the transform necessarily has the form $\alpha (z - \beta)/(z - \overline \beta)$. The requirement that the image of $\infty$ is on the unit circle gives $|\alpha| = 1$.
