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

```

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