Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify

*To*: mathgroup at smc.vnet.net*Subject*: [mg57842] Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify*From*: Andrzej Kozlowski <andrzej at akikoz.net>*Date*: Fri, 10 Jun 2005 02:29:06 -0400 (EDT)*References*: <200506090955.FAA29632@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 9 Jun 2005, at 18:55, Hugh Goyder 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]] > > > > My problem is to try and deduce the roots using computer algebra > rather than > by hand. (I also have other expressions I would like to work on.) > My attempt > at using Reduce is partially successful but leads to unfamiliar, and > difficult to interpret, ArcTanh functions > > Reduce[{ex == 0,0<a<1},z] > > An attempt with FullSimplify and ComplexExpand to crack the ArcTanh > function > is again partially successful in giving the imaginary part I require. > However, I am stuck with Arg functions with complex arguments. The Arg > Functions are just equal to -Pi/2 but I cannot crack them further. Any > suggestions for simplifying the output from Reduce so that I get > the simple > form I guessed at the start? Alternativly, are there more methods for > simplifing the complex output? > > > > FullSimplify[ComplexExpand[(2*I)*ArcTanh[a/2+(I/2)*Sqrt[4-a^2]]], > {0<a<1}] > > FullSimplify[ComplexExpand[Arg[2-a-I Sqrt[4-a^2]]-Arg[ > > 2+a+I Sqrt[4-a^2]]],{0<a<1}] > > > > Thanks > > Hugh Goyder > > > Reduce[{ComplexExpand[ex, {z}] == 0, 0 < a < 1}, z] 2*Pi*Im[C[1]] == 0 && 0 < a < 1 && C[1] â?? Integers && (z == (1/2)*(4*Pi*C[1] - Pi) + I*Log[Sqrt[(-a - 2)/(a - 2)]] || z == (1/2)*(4*Pi*C[1] + Pi) + I*Log[Sqrt[(-a - 2)/(a - 2)]]) Is this what you wanted? Andrzej Kozlowski Chiba, Japan

**References**:**Getting simple answers from Reduce, ComplexExpand and FullSimplify***From:*"Hugh Goyder" <h.g.d.goyder@cranfield.ac.uk>