# 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 $$|w_1|=|w_2|=|w_3|=1$$.

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.

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.

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 $$|\alpha| = 1$$.