I am given this problem from a past test that I am trying to figure out, I have tried finding the conjugate and going about it. But i am not getting the right transformation.
Please help out.
Show that a linear fractional transformation $f$ sends the upper half-plane
$$H:= \{z \in \mathbb{C} \mid \operatorname{Im}(z) > 0 \}$$
to itself if and only if it is of the form
$$f(z) = \frac{az+b}{cz+d}$$
where $a, b, c, d$ are real numbers and $ad-bc >0$. You may assume that such a transformation must also send $\hat{\mathbb{R}} = \mathbb{R} \cup \{\infty\ \} $ to itself.
Thank you
A linear fractional transformation $f$ mapping $H$ to itself maps $\bar{\mathbb R}$ to itself. Take any three real numbers $x_k$ such that $y_k:=f(x_k)\in\Bbb R$. Then the data $x_k$, $y_k$ determine $a$, $b$, $c$ as real numbers (up to a common factor); furthermore $ad-bc\ne0$. Since $f$ maps $H$ to $H$ it has to be monotonically increasing on $\Bbb R$; therefore $$f'(x)={ad-bc\over (cx+d)^2}>0\ \qquad(x\in\Bbb R),$$ which in turn implies $ad-bc>0$. (You could also argue about ${\rm Im}f(i)$, which should be positive.)
Conversely, a function of the form $f(z):={az+b\over cz+d}$ sends the real axis to itself. Therefore it maps $H$ onto $H$ or onto $\bar H$. Which of the two is the case depends solely on the sign of $ad-bc$.