If $u(z) = \frac{az+b}{cz+d}$ is bijective on the upper half complex plane, show that $a,b,c,d$ are real. The following question is from Greene-Krantz, Function Theory of One Complex Variable (Q 1.10)
Let $U = \{ z \in \mathbb{C} \colon \textrm{Im} \, z > 0 \}$. 
Prove that if $ u(z) =\displaystyle \frac{az+b}{cz+d}$ with $a,b,c,d \in \mathbb{C}$ and $u : U \to U$ one-to-one and onto, then $a,b,c,d$ are real (after multiplying numerator and denominator by a constant) and $ad-bc>0$.
Background:
I am trying to work through Greene-Krantz on my own. I have shown the converse (that if $a,b,c,d$ are real and $ad-bc>0$, then $u$ is one-to-one and onto. However, I am stuck as to how to even begin to tackle the part above. Hints will be very much appreciated.
 A: You don't tell us how much you already know about Moebius transformations. Therefore I shall give a very simpleminded proof.
The set $R:=\{x\in{\mathbb R}\>|\>cx+d\ne0\}$ is the real axis minus at most one point. Take an $x_0\in R$. Then $u$ is defined and continuous at $x_0$. Let $w_0:=u(x_0)$. I claim that $w_0$ is real.
Proof. If $v_0:={\rm Im}(w_0)<0$ then some points $z:=x_0+i\epsilon$ with $0<\epsilon\ll1$ would be mapped by $u$ onto the lower half-plane – contrary to assumption. If $v_0>0$ there would be a point $z_0\in U$ with $u(z_0)=w_0$. We now have $u(x_0)=u(z_0)$ for two different points $x_0$, $z_0$. A little computation shows that this is impossible, since $ad-bc\ne0$ by assumption.
Therefore we may choose three points $x_1$, $x_2$, $x_3\in R$ and are sure that the three values $u_k:=u(x_k)$ are real. But these data determine $a$, $b$, $c$, $d$ in a rational way up to a common factor, so that we now know that these coefficients can be taken to be real.
Finally we compute
$${\rm Im}\bigl( u(i)\bigr)={ad-bc\over|ci+d|^2}\ ,$$
from which it follows that necessarily $ad-bc>0$.
A: I think you need to use the symmetry principle to show that $u$ must map the real line to itself, then it will follow that $a,b,c,d$ are real. For example, if $u$ maps $\Bbb R$ onto a circle in the upper half plane, then $u^{-1}$ maps a pair of points which are symmetric to the circle and in the upper half plane to a pair of conjugates one of which will not be in the upper half plane. There's definitely more detail to fill in, but I believe this is the right idea.
