Let $\mu(z) = \frac{az+b}{cz+d}$ be a Möbius transformation. I want to show that $\mu(\overline{\mathbb{R}}) = \overline{\mathbb{R}} \iff a, b, c, d \in \mathbb{R}.$ What would be an elegant, and hopefully short way to prove this statement?
I have tried to show one implication first but then I arrive at a lot of different cases and I don't think the way to show the statement is to make an awkward case-by-case analysis. Other exercises given by the author of the lecture notes are much easier - e.g. I showed that $\mu(\mathbb{D}) = \mathbb{D}$ and $\mu(0)=0$ if and only if $\mu(z) = \zeta z$ for $\zeta \in S^1$, where $\mathbb{D}$ denotes the open unit disk in $\mathbb{C}$ and $S^1$ denotes the unit circle in $\mathbb{C}$ - which is why I believe there must be a more elegant way to approach this problem.
Thanks for any answers in advance.
EDIT: As Chris and yoyo have pointed out, the statement is not correct. The correct statement would probably be (correct me if I'm wrong again)
$\mu(\overline{\mathbb{R}}) = \overline{\mathbb{R}} \iff a, b, c, d \in \lambda \mathbb{R}$ for some constant $\lambda \in \mathbb{C}$.