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Re: Getting simple answers from Reduce, ComplexExpand and FullSimplify


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

-- 
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