Property of Mobius transformations which send $\mathbb{R}$ in $\mathbb{R}$ and that $ad-bc=1$

Question

I am asked to show that if $T(z) = \dfrac{az+b}{cz+d}$ is a mobius transformation such that $T(\mathbb{R})=\mathbb{R}$ and that $ad-bc=1$ then $a,b,c,d$ are all real numbers or they all are purely imaginary numbers.

So far I've tried multiplying by the conjugate of $cz+d$ numerator and denominator and see if I get some information about $a,b,c,d$ considering that $T(z) \in \mathbb{R}$ whenever $z \in \mathbb{R}$ but this doesn't really work. Also I've considered $SL(2,\mathbb{C}) / \{ \pm I\}$ which is isomorphic to the group of Mobius transformations but this doesn't really help either.

Answer 1

Prove first that $$T(z)=\frac{a'z+b'}{c'z+d'}$$ for real $a'$, $b'$, $c'$ and $d'$ and that then $a=\pm a'/\sqrt{a'd'-b'c'}$ etc.

Answer 2

Hints: if we assume $c\neq 0$, $T(z)=\frac{a}{c}-\frac{1}{c(cz+d)}$. Letting $z$ go to infinity shows $\frac{a}{c}$ is real, hence $c(cz+d)$ is real for all $z\in\mathbb R$. From there conclude $cd$ and $c^2$ are real. The rest should be easy. The case $c=0$ is also not difficult.

Answer 3

Suppose $T$ is as asked for.

• Suppose $c\neq 0$. Then if $\frac{-d}{c} \in \mathbb{R}$ and we know that $T(\frac{-d}{c}) = \infty \notin \mathbb{R}$ unless the numerator is $0$ as well, in which $a(\frac{-d}{c}) = b = 0$ which means $\frac{-ad}{c} = -b$ or $-ad = -bc$ or $ad=bc$ and we have a contradiction with $ad-bc = 1$.

So $c=0$ and so $ad=1$. What more can you do now?