RE: How Mathematica select one
- To: mathgroup@smc.vnet.net
- Subject: [mg12326] RE: [mg12237] How Mathematica select one
- From: Ersek_Ted%PAX1A@mr.nawcad.navy.mil
- Date: Thu, 7 May 1998 18:52:13 -0400
S. Molyavko wrote: | | I have some question about Mathematica 3.0. | | When Mathematica 3.0 calculate | (-1.)^(1/3) | it generate complex value result | 0.5 + 0.8660254037844385*I. | | But when I try to Solve equation | x^3+1==0 | it generates three different roots. | | So, here is the question: | | "How Mathematica 3.0 select one of the roots in the first example?" | Suppose you are doing (-2.0)^(1/3) . Then -2.0=2.0 E^(Pi*I) (* where I=Sqrt[-1] *) so (-2.0)^(1/3) gives (2.0^(1/3))*E^(Pi*I*(1/3)) You then get (2.0^(1/3)) (Cos[Pi/3] + I*Sin[Pi/3]) The root above is called the principal root. The other roots are: (2.0^(1/3)) (Cos[2*Pi/3] + I*Sin[2*Pi/3]) (2.0^(1/3)) (Cos[3*Pi/3] + I*Sin[3*Pi/3]) This works for any (x^a) when (x<0); (-1<a<1). Ted Ersek