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 =-