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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

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2 Answers 2

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

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  • $\begingroup$ Do you mean into the equation I posted and try to prove that LHS not equal RHS ? $\endgroup$
    – devcoder
    Apr 9, 2013 at 2:11
  • $\begingroup$ @devcoder, No, you need to show that something with positive imaginary part maps to something with negative imaginary part. $\endgroup$ Apr 10, 2013 at 0:52
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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 $\Im{w} > 0$?
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