1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$T(z) = \frac{az+b}{cz+d} = \frac{az+\beta a}{\gamma az + \delta a} = \frac{z+\beta}{\gamma z + \delta}$$

$\endgroup$
1
  • $\begingroup$ Very nice approach. Thanks. $\endgroup$ – user153009 Feb 9 '15 at 0:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.