I got following proof for "Search for all Möbius transformations that map the upper half plane to itself bijectively"
Proof:
Define \begin{align*} &M(z) = \frac{az+b}{cz+d} \\ &M(0) = \frac{b}{d}\in \mathbb{R}_{\infty} \\ &M(\infty) = \frac{a}{c}\in \mathbb{R}_{\infty} \\ &M^{-1}(0) = -\frac{b}{a}\in \mathbb{R}_{\infty} \\ &M^{-1}(\infty) = -\frac{a}{c}\in \mathbb{R}_{\infty} \end{align*}
Then by $$M(i) = \frac{ai+b}{ci+d}= \frac{(ad-bc)i+(ac+bd)}{c^2+d^2}$$ we see that $\Im(M(i)) > 0 \Rightarrow ad-bc >0$
So, $$M(z) = \frac{az+b}{cz+d}, \quad a,b,c,d\in \mathbb{R}, \quad ad-bc=1$$ and the set of Möbius transformations is $$Aut(\{z\in \mathbb{C}\mid \Im z >0\})= \{\pmatrix{a&b\\c&d}\in \mathbb{R}^{2\times 2}\mid \det(...)=1\}/\{\pm id\}$$
If I understand correctly we first show that we can map the real axis to itself? That's where we define the special cases for $0, \infty$.
Then check with $M(i)$ what other constraints we need s.t. $M(i)$ actually maps to the upper half plane. We see that $ad-bc>0$.
But after that I don't understand why $ad-bc=1$ and why we can so easily follow that $a,b,c,d$ are real. They could also all be complex?
Can someone help me understand?