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MathGroup Archive 2005

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Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify

  • To: mathgroup at smc.vnet.net
  • Subject: [mg57855] Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify
  • From: Pratik Desai <pdesai1 at umbc.edu>
  • Date: Fri, 10 Jun 2005 02:29:18 -0400 (EDT)
  • References: <200506090955.FAA29632@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

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
>
>
>  
>
Try using TrigToExp, some times CAS like humans needs a little help with 
hyperbolic trig expressions

Clear[ex, z, a, sol1]
TagSet[a, Conjugate[a], a]
TagSet[a, Re[a], a]
TagSet[a, Im[a], 0]
ex[z_] := 2 Cos[z] + I* a* Sin[z] // TrigToExp
Reduce[ex[z] == 0, z]


C[1] â?? Integers && 2 + a != 0 && -2 + a != 0 && z == 
(-I/2)*((2*I)*Pi*C[1] + Log[(-2 + a)/(2 + a)])

You can use simplify or full simplify to get what you want at the end
Best regards

Pratik

-- 
Pratik Desai
Graduate Student
UMBC
Department of Mechanical Engineering
Phone: 410 455 8134



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