How to prove $g(z) = \frac {a \bar z + b } {c \bar z + d}$ is a bijective mapping of UHP to UHP?

Question

I saw the answer to the question where it is $z$ instead of $\bar z $ here When does a Möbius transformation map $\Im(z)>0$ to itself?

Here's the complete question I'm wondering how having $\bar z$ would change the proof ?

Answer 1

Generally it is not, for example $a=d=1, b=c=0$, $g(z)=\bar z$ maps a point in UHP to LHP.

But when $ad-bc<0$, it is true. We can write $g=f\circ h$, where $h(z)=\bar z$.

First, you can show that for $z=x+iy$, $\Im f(z)={(ad-bc)y\over |cz+d|^2}$, so $f(z)$ gives a map from UHP into LHP. Then you can find the inverse of $f(z)$ similar to the case where $ad-bc>0$, so to prove that $f(z)$ is a bijection from UHP to LHP.

Then, it can be easily shown that $h(z)=\bar z$ gives a bijection from LHP to UHP.

Answer 2

Hint: To prove it is one to one, just use the definition

$$ f({z_1})=f({z_2}) \implies {z_1}={z_2}. $$