How to prove when Möbius transformation determinant ad - bc < 0, the upper half plane does not map to itself?

Question

I proved the part where $$\operatorname{Im} f(z) = \frac {(ad - bc) \operatorname{Im}z} {|cz + d|^2} $$

But now I need to show that when $ ad - bc < 0 $, then the upper half plane does not map to itself

score 4 · Answer 1 · answered Apr 09 '13 at 01:45

4

Try plugging in something in the upper half plane ($z=i$, for example).

answered Apr 09 '13 at 01:45

Paul Gustafson

Do you mean into the equation I posted and try to prove that LHS not equal RHS ? – devcoder Apr 09 '13 at 02:11
@devcoder, No, you need to show that something with positive imaginary part maps to something with negative imaginary part. – Paul Gustafson Apr 10 '13 at 00:52

score 0 · Answer 2 · answered Apr 09 '13 at 02:29

So, you're asking how to prove, when $(ad-bc) = 0$, that at least one of:

There exists a $z$ such that $\Im{z} > 0$, and $(ad-bc) \Im{z} / |cz+d|^2 < 0$,
There exists a $z$ such that $\Im{z} > 0$ and the equation $f(w)=z$ has no solutions with $\Im{w} > 0$?

2 Answers2