Re: Trick for getting the comples conjugate in symbolic calculations?

*To*: mathgroup at smc.vnet.net*Subject*: [mg30320] Re: Trick for getting the comples conjugate in symbolic calculations?*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Sat, 11 Aug 2001 03:39:43 -0400 (EDT)*References*: <9kqj1m$4bo$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Norbert, By default, Mathematica assumes that variables denote complex numbers, so for example Conjugate[z(x+I y)] Conjugate[(x + I*y)*z] This could be expanded to (Conjugate[x] - I*Conjugate[y])*Conjugate[z] but it is left to the user to make choices. To deal with such situations we have the functions ComplexExpand, FunctionExpand and the package Algebra`ReIm`. These can be looked up in the HelpBrowser. Here are some examples ComplexExpand[Conjugate[(x+I y)z]] (*x,y,z are assumed real*) x*z - I*y*z ComplexExpand[Conjugate[(x+I y)z],{z}] (*z, only is assumed complex*) (-y)*Im[z] + x*Re[z] - I*(x*Im[z] + y*Re[z]) FunctionExpand[Conjugate[(x+I y)z]] (*x,y,z are assumed complex*) (Conjugate[x] - I*Conjugate[y])*Conjugate[z] <<Algebra`ReIm` This allows some expansion, under the default assumption that x,y,z are complex. Conjugate[z(x+I y)] (Conjugate[x] - I*Conjugate[y])*Conjugate[z] We can declare z to be real ( note z:/ --- this setting is tagged by z, not Im) z/:Im[z]=0; Conjugate[(x+I y)z] z*(Conjugate[x] - I*Conjugate[y]) This declaration overides the beviour of ComplexExpand ComplexExpand[Conjugate[(x+I y)z],{z}] x*z - I*y*z -- Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 <riefler at iwt.uni-bremen.de> wrote in message news:9kqj1m$4bo$1 at smc.vnet.net... > Hi > > I read in the older news that the numeric functions real / Re and imag > / Im for / Mathematica won't do its job because of some > ambiguity. The same holds for the complex conjugate conj / Conjugate. > > Is there a trick to get anyway the algebraic solution of for example > (x1+i*y1)*conj(x2+i*y2) / (x1+i*y1)*Conjugate(x2+i*y2) in Mathematica? > > > Thanks a lot > > Norbert >