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How can I use Mathematica to check if a statement like $$\left| \frac{\sqrt{1+2z}-2}{2z-3} \right| \le 1$$ is true for all complex numbers $z$?

Also, how can I use Mathematica to check such a statement for a subset of the complex numbers (for example, all complex numbers with imaginary part less than zero)?

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For z==3/2 it's undefined –  belisarius Nov 23 '13 at 13:50
    
@Belisarius supposedly the OP would consider the limit of the 0/0 expression, which is 1/4. For the OP: by simplifying (Sqrt[1 + 2z] - 2)/(2z - 3) /. z -> x + I y and going back to z you'll get Abs[Sqrt[1 + 2z] + 2] ≥ 1. Is this easier fro you to tackle? –  Peltio Nov 23 '13 at 14:09
    
Oh, the above expression is just an arbitrary example of a relational expression. –  David Zhang Nov 23 '13 at 14:17
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1 Answer

up vote 4 down vote accepted
Resolve[Exists[z, Abs[(Sqrt[1 + 2 z] - 2)/(2 z - 3)] > 1], Complexes]
(*
False ... after a few looong minutes
*)

Edit

Slightly faster:

Reduce[FullSimplify[Abs[(Sqrt[1 + 2 z] - 2)/(2 z - 3)]] <= 1, z, Complexes]

Edit 2

Much faster:

See that:

FullSimplify[Abs[(Sqrt[1 + 2 z] - 2)/(2 z - 3)]]
(*
 1/Abs[2 + Sqrt[1 + 2 z]]
*)

So:

Reduce[FullSimplify[(2 + Sqrt[1 + 2 z]) Conjugate[2 + Sqrt[1 + 2 z]]] > 1, z, Complexes]
(*
  True
*)
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Whoa, that takes quite a bit longer than what I was hoping for. Is there any faster way? –  David Zhang Nov 23 '13 at 14:09
    
@DavidZhang See my edit 2 –  belisarius Nov 23 '13 at 14:26
    
FullSimplify[(2 + Sqrt[1 + 2 z]) Conjugate[2 + Sqrt[1 + 2 z]] > 1] evalutes to true as well, but much faster –  Coolwater Nov 23 '13 at 16:11
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