Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify

*To*: mathgroup at smc.vnet.net*Subject*: [mg57868] Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify*From*: "Hugh Goyder" <h.g.d.goyder at cranfield.ac.uk>*Date*: Sat, 11 Jun 2005 03:35:23 -0400 (EDT)*Organization*: University of Warwick, UK*References*: <d893st$t26$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Many thanks for all the answers I received. The best answer seems to be ex = 2 Cos[z] + I a Sin[z]; InputForm[Reduce[{0 == ex // TrigToExp, 0 < a < 1}, z]] C[1] \[Element] Integers && Inequality[0, Less, a, Less, 1] && z == (-I/2)*((2*I)*Pi*C[1] + Log[(-2 + a)/(2 + a)]) Two points are interesting. 1. I used ComplexExpand on the output of Reduce and got a different and incomprehensible answer compared to Andrzej Kozlowski and Pratik Desai who used ComplexExpand and TrigToExp, respectively, on the input to Reduce and got a simple answer. Is this obvious from the way Reduce works or does one get this by trial-and-error and tinkering? (I don't like tinkering I don't think you can really learn and understand by tinkering with computation.) 2. I note that Reduce sometimes generates a constant, C[1], that is an element of the Integers and also has Im[C[1]] == 0. I know that there are Gaussian Integers but I would have thought that the definition of Integers in Mathematica would be restricted to real integers. Thanks Hugh Goyder "Hugh Goyder" <h.g.d.goyder at cranfield.ac.uk> wrote in message news:d893st$t26$1 at smc.vnet.net... > 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 > >

**Follow-Ups**:**Re: Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify***From:*Pratik Desai <pdesai1@umbc.edu>