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MathGroup Archive 2005

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Re: Solutions of an equation under complex form

  • To: mathgroup at smc.vnet.net
  • Subject: [mg54570] Re: Solutions of an equation under complex form
  • From: bghiggins at ucdavis.edu
  • Date: Wed, 23 Feb 2005 03:11:40 -0500 (EST)
  • References: <cvc7un$qqr$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

 Michaël,

Try the following:

Solve [x^2 +
    x + 1 == 0, x] /. Rule[x_, y_] :> Rule[x, ComplexExpand[y ,
       TargetFunctions -> {Re, Im} ]]


{{x -> -(1/2) - (I*Sqrt[3])/2},  {x -> -(1/2) + (I*Sqrt[3])/2}}

Hope that helps,

Brian


Michaël Monerau wrote:
> Hello,
>
> I'm running into a little problem under Mathematica 5.0 but I'm sure
> people here will just take it as "too easy", but, well :) I just want

> the solutions of the equation :
>
> x^2 + x + 1 == 0
>
> under their complex form.
>
> So, I type :
>
> Solve [x^2 + x + 1 == 0, x]
>
> But I unfortunately get :
> { { {x -> -(-1)^(1/3) }, { x -> (-1)^(2/3) } } }
>
> And I'd prefer to obtain the more "readable" form :
> -1/2 + I*1/2*Sqrt[3], -1/2 - I*1/2*Sqrt[3]
>
> that I would get under another system for instance. What special
function
> should I call to get this form under Mathematica ?
> 
> Thanks for any help
> -- 
> Michaël Monerau
> -= JJG =-


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