* Henry Baker <hbaker1@pipeline.com> [Nov 14. 2009 18:50]:
Ok, so if we are given c+is, c,s real, cc+ss=1, then we want to compute cos(acos(c)/3) for the real part of one of the cube roots of c+is.
cos(acos(c)/3) = T_(1/3)(c) = d, so
cos(3acos(d)) = T_3(d) = c, i.e.,
4d^3-3d=c
So we have reduced the problem to one of finding a real root of a real cubic 4d^3-3d-c=0. Once we have d, we can easily compute the imaginary part, since we know that the absolute value is 1.
Question:
Since we know that the real parts of all three of the complex cube roots of c+is are real (duh!), this would seem to imply that the equation 4d^3-3d-c=0 always has 3 real roots. Is this easy to see?
Also, all three roots should lie within the range [-1,1]. Is this easy to see?
[...]
Taking "see" literally: (using pari/gp) ploth(x=-1,1,poltchebi(3)) Plot also at http://mathworld.wolfram.com/ChebyshevPolynomialoftheFirstKind.html