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MathGroup Archive 2005

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Getting simple answers from Reduce, ComplexExpand and FullSimplify

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57832] Getting simple answers from Reduce, ComplexExpand and FullSimplify
  • From: "Hugh Goyder" <h.g.d.goyder at cranfield.ac.uk>
  • Date: Thu, 9 Jun 2005 05:55:33 -0400 (EDT)
  • Organization: University of Warwick, UK
  • Sender: owner-wri-mathgroup at wolfram.com

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



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