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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 $\lambda \in \mathbb{C}$.

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    @sos440: I second that.2011-12-21

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One direction is clear: if there exists $\lambda\in\mathbb C$ such that $a,b,c,d\in\lambda\mathbb R$, then after canceling $\lambda$ we have all real coefficients. The converse was proved by sos440 in the comments: I copy the comment with improved formatting.

Note that the Möbius transform is uniquely determined by its value on three points. In particular, $\tilde \mu(z)=\frac{(z−z_0)(z_1−z_\infty)}{(z−z_\infty)(z_1−z_0)} \tag{*}$ is the Möbius transform that maps $z_0$, $z_1$ and $z_\infty$ to $0$, $1$ and $\infty$, respectively. Given a transform $\mu$ that fixes the extended real line, let $z_0=\mu (0)$, $z_1=\mu(1)$ and $z_\infty=\mu(\infty)$ in (*). These are elements of $\overline{\mathbb R}$, so $\tilde \mu$ defines a Möbius transform associated to a real matrix. Since $\mu^{-1}=\tilde \mu$, it follows that $\mu$ itself is associated to a real matrix.