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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.

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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.

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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.

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  • $\begingroup$ So you are saying $T(\infty) \in \mathbb{R}$? Why? $\endgroup$ Sep 10 '17 at 23:55
  • $\begingroup$ @CamiloEscobar By continuity. $\endgroup$
    – Wojowu
    Sep 11 '17 at 6:55
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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?

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  • $\begingroup$ You are assuming $\frac{-d}{c}\in\mathbb R$. $\endgroup$
    – Wojowu
    Sep 10 '17 at 6:59
  • $\begingroup$ @Wojowu yes, to get a contradiction. $\endgroup$ Sep 10 '17 at 7:58
  • $\begingroup$ You are supposed to show that it's not possible to have $T(\mathbb R)=\mathbb R$ if they are neither all real nor all imaginary. That's what (the contrapositive of) the question is. You show that $T(\mathbb R)=\mathbb R$ is impossible if they are all real or all imaginary. $\endgroup$
    – Wojowu
    Sep 10 '17 at 8:01
  • $\begingroup$ @Wojowu The question says that if we have a $T$ that has $[\mathbb{R}] = \mathbb{R}$ and $ad-bc=1$, then the coefficients are all real or all imaginary.(say, $\phi \implies \psi$). I showed that if $T$ obeys the left hand side, the right hand side cannot hold. So I showed $\lnot (\phi \land \psi) = \lnot \phi \lor \lnot \psi$, while the implication is equivalent to $\lnot \phi \lor \psi)$. So I think the question is wrong. $\endgroup$ Sep 10 '17 at 8:06
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    $\begingroup$ By the way, it's possible to have $T(\mathbb R)=\mathbb R$ -- take $a=d=1,b=c=0$. You are not taking $c=0$ into consideration. $\endgroup$
    – Wojowu
    Sep 10 '17 at 8:10

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