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Re: Solutions of an equation under complex form


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 =-




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