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MathGroup Archive 2006

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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


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