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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 enter image description here I'm wondering how having $\bar z$ would change the proof ?

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

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

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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.

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You have me confused. The question specifically asks to prove this - So according to your answer should I be answering that is false ? – devcoder Apr 9 '13 at 3:45
@devcoder you should add the condition $ad-bc<0$ in your question – Ziqian Xie Apr 9 '13 at 3:50
Added!, thanks. – devcoder Apr 9 '13 at 4:03

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