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Re: Trouble computing conjugates

  • To: mathgroup at smc.vnet.net
  • Subject: [mg88561] Re: Trouble computing conjugates
  • From: "David Park" <djmpark at comcast.net>
  • Date: Thu, 8 May 2008 04:15:31 -0400 (EDT)
  • References: <fvs2k5$enp$1@smc.vnet.net>

I don't think you need to use $Assumptions or Refine. Just use 
ComplexExpand.

u1 = f + I g;
u2 = f - I g;

Conjugate[c1 u1]
ComplexExpand[%]
Conjugate[c1] (Conjugate[f] - I Conjugate[g])
c1 f - I c1 g

Conjugate[c1 u1 + u2]
ComplexExpand[%]
Conjugate[f + c1 (f + I g)] + I Conjugate[g]
f + c1 f + I (g - c1 g)

-- 
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/


"Roy" <sarahroy at earthlink.net> wrote in message 
news:fvs2k5$enp$1 at smc.vnet.net...
> I'm having trouble getting mathematica to compute complex conjugates
> of some fairly simple expressions:
>
> If I type the following:
>
> $Assumptions = {g \[Element] Reals, f \[Element] Reals}
> u1 = f + \[ImaginaryI] g
> u2 = f - \[ImaginaryI] g
>
> Then the command:
>
> Refine[Conjugate [c1 u1]]
>
> returns:
>
> f - \[ImaginaryI] g) Conjugate[c1]
>
> and the command:
>
> Refine[Conjugate[c1 u1 + u2]]
>
> returns:
>
> f + \[ImaginaryI] g + (f - \[ImaginaryI] g) Conjugate[c1]
>
> as I would expect.  But the command:
>
> Refine[Conjugate[c1 u1 + c2 u2]]
>
> returns:
>
> Conjugate[c2 (f - \[ImaginaryI] g) + c1 (f + \[ImaginaryI] g)]
>
> i.e. it refuses to distribute the complex conjugate throughout the
> expression.  What I would like it to tell me is:
>
> (f + \[ImaginaryI] g) Conjugate[c2] + (f - \[ImaginaryI] g)
> Conjugate[c1]
>
> The closest I have been able to come to getting what I want is by
> using:
>
> ComplexExpand[Refine[Conjugate[ c1 u1 + c2 u2]], {c1, c2}]
>
> but this separates c1 and c2 into their real and imaginary parts.  The
> above expressions are much simpler than the ones I REALLY want
> Mathematica's help in simplifying.  If I use this ComplexExpand
> command, then I'm going to have to recombine them into complex numbers
> again, which would be very very bad.
>
> Thanks,
> Roy
> 



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