Re: [math-fun] Cube root of a complex number

Re: [math-fun] Cube root of a complex number somewhat tangential to the original question, Möbius transformations map three points into three points. So, why not map the three roots of the cubic into the three complex roots of unity? You only get to use the coefficients of the polynomial. I recall a paper in an ACM journal humorously dated March 32 some years ago where someond did that; I don't remember if the solution was relevant to the present inquiry. ------------------------------------------------- www.correo.unam.mx UNAMonos Comunicándonos