Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify

*To*: mathgroup at smc.vnet.net*Subject*: [mg57871] Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify*From*: Paul Abbott <paul at physics.uwa.edu.au>*Date*: Sat, 11 Jun 2005 03:35:25 -0400 (EDT)*Organization*: The University of Western Australia*References*: <d893st$t26$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

In article <d893st$t26$1 at smc.vnet.net>, "Hugh Goyder" <h.g.d.goyder at cranfield.ac.uk> wrote: > Below the expression ex has roots, rts, where n in an integer and the > constant a is in the interval 0 < a <1. We may check this using FullSimplify > as follows. > > ex=2 Cos[z]+I a Sin[z]; > > rts=z ->Pi(2n-1)/2 + I Log[(2+a)/(2-a)]/2; > > FullSimplify[0 == ex/.rts,Element[n,Integers]] The following works for any a: ex = 2 Cos[z]+I a Sin[z]; rts = z -> Pi(2n-1)/2 + I Log[(2+a)/(2-a)]/2; Simplify[ex /. rts, Element[n,Integers]]; FullSimplify[% == 0] so this result holds for arbitrary complex a. Cheers, Paul -- Paul Abbott Phone: +61 8 6488 2734 School of Physics, M013 Fax: +61 8 6488 1014 The University of Western Australia (CRICOS Provider No 00126G) AUSTRALIA http://physics.uwa.edu.au/~paul http://InternationalMathematicaSymposium.org/IMS2005/