Prove that a Möbius transformation $T(z)=\dfrac{az+b}{cz+d}$ maps $\overline{\mathbb R}$ to $\overline{\mathbb R}$ if and only if it can be written with real coefficients.

If it can be written with real coefficients $a,b,c,d$, then for any $r \in \mathbb R, T(r)=\dfrac{ar+b}{cr+d}$, and by the closure property of $\mathbb R$ under addition and multiplication, we have $T(r) \in \mathbb R \cup \{\infty\}=\overline{\mathbb R}$

I don't know what to do to prove the other implication. I've tried to show it by the absurd: suppose $T$ maps $\overline{\mathbb R}$ to $\overline{\mathbb R}$ but $T$ can't be expressed with real coefficients, how could I arrive to an absurd? I would have to find $r \in \mathbb R$ such that $T(r) \in \mathbb C$. Maybe there is a straight forward way to prove it.


Hint: $T(0)$, $T(1)$, $T(-1)$ and $T(\infty)$ are all real.

  • $\begingroup$ Let me see: by my hypothesis, $T(0)=\dfrac{c}{d},T(\infty)=\dfrac{a}{b},T(1)=\dfrac{a+b}{c+d}, T(-1)=\dfrac{-a+b}{-c+d} \in \mathbb R$. I suppose I must somehow deduce from here that $a,b,c,d \in \mathbb R$, now I don't see this has to be the case necessarily, for example $\dfrac{2}{3} \in \mathbb R$, but $2i,3i \in \mathbb C$. $\endgroup$ – user156441 Jun 23 '14 at 1:52
  • $\begingroup$ Note that your question didn't say $T$ is written with real coefficients, it said it can be written with real coefficients. And for example $T(z)=(iz+2i)/(3iz+4i)$ can be written with real coefficients by cancelling the $i$s. $\endgroup$ – David Jun 23 '14 at 2:00
  • $\begingroup$ BTW I think you have the coefficients mixed up a bit, $T(0)$ should be $b/d$. $\endgroup$ – David Jun 23 '14 at 2:01
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    $\begingroup$ Sounds good. Try writing $T(1)=x\in\Bbb R$ and $T(-1)=y\in\Bbb R$ then eliminating $b$ from the resulting equations. $\endgroup$ – David Jun 23 '14 at 2:23
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    $\begingroup$ I agree that you need to look carefully at $a=0$ or $c=0$. You could do $d/c$ as above, or for a short cut consider $d/c=(d/a)(a/c)$. $\endgroup$ – David Jun 23 '14 at 3:28

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