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MathGroup Archive 2007

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Re: Weird result in Mathematica 6

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76455] Re: [mg76393] Weird result in Mathematica 6
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Tue, 22 May 2007 03:00:09 -0400 (EDT)
  • References: <200705211001.GAA10071@smc.vnet.net>

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On 21 May 2007, at 19:01, Sebastian Meznaric wrote:

> I was playing around with Mathematica 6 a bit and ran this command to
> solve for the inverse of the Moebius transformation
>
> FullSimplify[
>  Reduce[(z - a)/(1 - a\[Conjugate] z) == w && a a\[Conjugate] < 1 &&
>    w w\[Conjugate] < 1, z]]
>
> This is what I got as a result:
> -1 < w < 1 && -1 < a < 1 && z == (a + w)/(1 + w Conjugate[a])
>
> Why is Mathematica assuming a and w are real? The Moebius
> transformation is invertible in the unit disc regardless of whether a
> and w are real or not. Any thoughts?
>
>


Reduce and FullSimplify will usually deduce form the presence of  
inequalities in an expression like the above that the variables  
involved in the inequalites are real. In your case it "sees"  
a*Conjugate[a]<1 and "deduces" that you wanted a to be real. This was  
of coruse not your intention but you can get the correct behaviour by  
using:

  FullSimplify[
  Reduce[(z - a)/(1 - Conjugate[a]*z) == w && Abs[a]^2 < 1 && Abs[w] 
^2 < 1, z]]


  -1 < Re[w] < 1 && -Sqrt[1 - Re[w]^2] < Im[w] < Sqrt[1 - Re[w]^2] &&  
-1 <
   Re[a] < 1 &&
    -Sqrt[1 - Re[a]^2] < Im[a] < Sqrt[1 - Re[a]^2] &&
  z == (a + w)/(w*Conjugate[a] + 1)

Mathematica knows that the fact that an inequality involves Abs[a]  
does not imply that a is real but it does not "know" the same thing  
about a*Conjugate[a]. This is clearly dictated by considerations of  
performance than a straight forward bug.
Andrzej Kozlowski


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