Re: Simplify and Abs in version 6.0

*To*: mathgroup at smc.vnet.net*Subject*: [mg78093] Re: [mg78034] Simplify and Abs in version 6.0*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 22 Jun 2007 06:52:09 -0400 (EDT)*References*: <200706210958.FAA27636@smc.vnet.net>

It is a good think that mathemaica does not return the answer that you suggest: {-\[ImaginaryI] a, \[ImaginaryI] a} because it would be completely wrong as in Mathematica {a,b} is an ordered pair and not a set. Mathematica has no built in structures corresponding to sets and has no way of expressing the answer you (seem to) have in mind. You can get the answer with Sqrt[a^2] more directly by: ComplexExpand[{I*Abs[a], (-I)*Abs[a]}, TargetFunctions -> {Re, Im}] {I*Sqrt[a^2], (-I)*Sqrt[a^2]} This, of course, can't be reduced any further without additional information about a the sign of a. Andrzej Kozlowski On 21 Jun 2007, at 18:58, Michael wrote: > Hi, > > In Mathematica version 6.0, I'm having difficulty trying to coax the > input > > In[1]:= FullSimplify[{\[ImaginaryI] Abs[a], -\[ImaginaryI] Abs[a]}, > Element[a, Reals]] > > to produce the set {-\[ImaginaryI] a, \[ImaginaryI] a}. > Interestingly, this *does* result for > > In[2]:= FullSimplify[{\[ImaginaryI] Abs[a], -\[ImaginaryI] Abs[a]}, a >> = 0] > > and > > In[3]:= FullSimplify[{Sqrt[-1] Abs[a], -Sqrt[-1] Abs[a]}, a < 0] > > Am I missing something here? I've tried Allan Hayes' suggestion in > 2003, with > > In[4]:= FullSimplify[{\[ImaginaryI] Abs[a], -\[ImaginaryI] Abs[a]}, > Element[a,Reals],ComplexityFunction -> ((Count[#, _Abs, Infinity]) &)] > Out[4]= {\[ImaginaryI] Sqrt[a^2], -\[ImaginaryI] Sqrt[a^2]} > > to no avail; of course, I could just add a PowerExand@ to the above > expression, but this seems like a lot to do, especially when > Mathematica already has been explicitly told that "a" is a real > number. > > Any tricks or hints would be greatly appreciated! > > Regards, > > Michael > >

**References**:**Simplify and Abs in version 6.0***From:*Michael <mcauxeu@gmail.com>