Re: Solutions of an equation under complex form

*To*: mathgroup at smc.vnet.net*Subject*: [mg54516] Re: [mg54478] Solutions of an equation under complex form*From*: "David Park" <djmp at earthlink.net>*Date*: Tue, 22 Feb 2005 04:23:23 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

Michael, A nice question. You can obtain the form you want by using RootReduce. sols = Solve [x^2 + x + 1 == 0, x] % // RootReduce {{x -> -(-1)^(1/3)}, {x -> (-1)^(2/3)}} {{x -> (1/2)*(-1 - I*Sqrt[3])}, {x -> (1/2)*(-1 + I*Sqrt[3])}} You can also obtain the individual roots by using the polynomial as a pure function in Root. Root[#1^2 + #1 + 1 & , 1] (1/2)*(-1 - I*Sqrt[3]) Root[#1^2 + #1 + 1 & , 2] (1/2)*(-1 + I*Sqrt[3]) David Park djmp at earthlink.net http://home.earthlink.net/~djmp/ From: Michaël Monerau [mailto:mmonerau at gmail.com] To: mathgroup at smc.vnet.net 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 =-