I'm trying to prove the following:
Show that a Möbius transformation T maps the x-axis to itself if and only if T can be written $$T(z)=\frac{az+b}{cz+d}$$ where the constants $a,b,c,d\in\Bbb{R}$.
I have found the following proof, but I am confused/not convinced by it (in particular the iff-part). Perhaps you can convince me or provide a correct proof. My confusion lies in the lack of justification for $cz+d$ in the denominator.
1) Assume that $T(\infty)=\infty$. Then, $T(z)=\frac{az+b}{c}=a'z+b'$.
$T(0)\in\Bbb{R} \implies b'\in\Bbb{R}$
$T(1)\in\Bbb{R} \implies a'+b'\in\Bbb{R}$
So $a',b'\in\Bbb{R}$.
2) Assume the contrary, $T(\infty)=e\in\Bbb{R}$.
Let $s(w)=\frac{1}{w-e}$ and $u(z)=S\circ T$. Then, $u(\infty)=\infty$ so $u$ can be written with real coefficients according to (1). Therefore, $T=S^{-1}\circ U$ can also be written with real coefficients.
I apologize for my poor formatting, this is my first time on this site.
Best Answer
Suppose $T(z) = \frac{az+b}{cz+d}$, just by plugging in $z=0$ and $z=\infty$ we see that $\frac{b}{d}$ and $\frac{a}{c}$ are reals. Then plug in $z=\frac{a}{c}$ we see
$$T(\frac{a}{c})= \frac{a^2/c +b}{a+d}= \frac{a^2/cd + b/d}{a/d +1}\in \mathbb{R}$$ Simple manipulation shows that $\frac{a}{d}$ is also real, thus $\frac{b}{c}$ and $\frac{d}{c}$ are reals. Now
$$T(z)=\frac{az+b}{cz+d}= \frac{ \frac{a}{c}z + \frac{b}{c}}{z+\frac{d}{c}}$$ with coefficients being reals.