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 =-