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