two questions
- To: mathgroup at smc.vnet.net
- Subject: [mg71271] two questions
- From: "dimitris" <dimmechan at yahoo.com>
- Date: Sun, 12 Nov 2006 06:48:33 -0500 (EST)
Supose the following
Solve[x^12 == 1, x]
{{x -> -1}, {x -> -I}, {x -> I}, {x -> 1}, {x -> -(-1)^(1/6)}, {x ->
(-1)^(1/6)}, {x -> -(-1)^(1/3)}, {x -> (-1)^(1/3)}, {x -> -(-1)^(2/3)},
{x -> (-1)^(2/3)}, {x -> -(-1)^(5/6)}, {x -> (-1)^(5/6)}}
% /. (a_ -> b_) :> a -> ComplexExpand[b]
{{x -> -1}, {x -> -I}, {x -> I}, {x -> 1}, {x -> -(I/2) - Sqrt[3]/2},
{x -> I/2 + Sqrt[3]/2}, {x -> -(1/2) - (I*Sqrt[3])/2}, {x -> 1/2 +
(I*Sqrt[3])/2}, {x -> 1/2 - (I*Sqrt[3])/2}, {x -> -(1/2) +
(I*Sqrt[3])/2}, {x -> -(I/2) + Sqrt[3]/2}, {x -> I/2 - Sqrt[3]/2}}
How is it possible to go "bacwards"; that is to get {x -> (-1)^(5/6)}
instead of {x -> I/2 - Sqrt[3]/2}}?
Also
(Cos[#1*(Pi/3)] + Sin[#1*(Pi/3)] & ) /@ Range[3]
{1/2 + Sqrt[3]/2, -(1/2) + Sqrt[3]/2, -1}
Is it possible to go "backwards"?
Regards
Dimitris