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Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify
*To*: mathgroup at smc.vnet.net
*Subject*: [mg57871] Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify
*From*: Paul Abbott <paul at physics.uwa.edu.au>
*Date*: Sat, 11 Jun 2005 03:35:25 -0400 (EDT)
*Organization*: The University of Western Australia
*References*: <d893st$t26$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
In article <d893st$t26$1 at smc.vnet.net>,
"Hugh Goyder" <h.g.d.goyder at cranfield.ac.uk> 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]]
The following works for any a:
ex = 2 Cos[z]+I a Sin[z];
rts = z -> Pi(2n-1)/2 + I Log[(2+a)/(2-a)]/2;
Simplify[ex /. rts, Element[n,Integers]];
FullSimplify[% == 0]
so this result holds for arbitrary complex a.
Cheers,
Paul
--
Paul Abbott Phone: +61 8 6488 2734
School of Physics, M013 Fax: +61 8 6488 1014
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AUSTRALIA http://physics.uwa.edu.au/~paul
http://InternationalMathematicaSymposium.org/IMS2005/
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