Prove that a Möbius Transformation maps the real axis to the real axis iff the coefficients of the Möbius Transformation are real. If I assume the coefficients are real and take the $3$ points on the real line and show that it maps to $3$ points on the real line I think I can show the converse.
So,
If we have the Möbius Transformation $T(z)=\frac{az+b}{cz+d}$ and I assume $a,b,c,d$ are real, then if I take the points, $-1,1,\infty$ and show it maps to $3$ real points via the transformation I would think that is sufficient. Noting $T(\infty)=\frac{a}{c}$ (which is real if coefficients are real).
How would I go about proving the $\to$ direction?
Sorry, not too familiar with Latex.
Best Answer
One way to do it is to look at specific points on the $x$-axis under $T$.
For example, $T(x) = 0$ for some $x \in \mathbb{R}$, which means that $ax+b=0$ for some $x \in \mathbb{R}$, so there is some $\beta\in\mathbb{R}$ s.t. $a=\beta b$ (here $\beta = -1/x$ so you need to consider that $x$ might be $0$ and handle that separately).
You can go on and show that there are also $\gamma,\delta\in\mathbb{R}$ such that $a=\gamma c, a=\delta d$ and then we get $$T(z) = \frac{az+b}{cz+d} = \frac{az+\beta a}{\gamma az + \delta a} = \frac{z+\beta}{\gamma z + \delta}$$