# How to show $\mu(\overline{\mathbb{R}}) = \overline{\mathbb{R}}$ iff $a, b, c, d \in \mathbb{R}$ for $\mu(z) = \frac{az+b}{cz+d}$?

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.

$$\mu(\overline{\mathbb{R}}) = \overline{\mathbb{R}} \iff a, b, c, d \in \lambda \mathbb{R}$$ for some constant $\lambda \in \mathbb{C}$.
The claim is false. E.g. take $\mu(z)=\frac{iz+0}{0z+i}$. – Chris Eagle Dec 21 '11 at 18:55
what you state isn't exactly true, eg multiply $a,b,c,d$ all by $i$ to get the same function, but there will be a constant $\lambda$ st $a,b,c,d\in\lambda\mathbb{R}$. – yoyo Dec 21 '11 at 19:00
Note that the Moebious transform is uniquely determined by its value on three points. In particular, $$\tilde{\mu} = \frac{(z - z_0)(z_1 - z_{\infty})}{(z - z_{\infty})(z_1 - z_0)}$$ is the Moebious transform which maps $z_0, z_1$ and $z_{\infty}$ to $0, 1$ and $\infty$, respectively. Now let $z_0 = \mu(0)$, $z_1 = \mu(1)$ and $z_{\infty} = \mu(\infty)$. These are elements of $\bar{\mathbb{R}}$, so $\tilde{\mu}$ defines a Moebious transform associated to a real matrix such that $\mu^{-1} = \tilde{\mu}$. Hence $\mu$ itself is also associated to a real matrix. – sos440 Dec 21 '11 at 19:36