Dick Mores wrote: <snip> My trusty $10 Casio calculator tells me that the 3 cube roots of 1 are: 1, (-.5 +0.866025403j), and (-.5 -0.866025403j), or thereabouts. Is there a way to do this in Python?
<snip> I think the neatest approach might be to consider that the n-complex roots form at equal angle around the origin on an Argand diagram (basically a cartesian plane) - here the points (in "standard") cartesian notation are at (1,0), (-0.5, 0.86) and (-0.5, -0.86). The whole effect looks a bit like a Mercedes-Benz symbol rotated clockwise by 90 degrees. The points, in this case, lie on the unit circle. As a result, I think you could just divide 360 by n (for the n-roots), set the 1st root at (1,0) and then position the others around the circle, incrementing by the required number of degrees. If I've remembered correctly, for c where |c| <>1, the argument is the same, but you need to take the nth root of the absolute value of c (this can be ignored when we're doing it for 1, as of course the nth root of 1 is 1). Obviously I haven't included any code....but I hope this helps a bit. Matt P.S. Thrilled to be able to answer something on the tutor list, instead of just asking dumb questions! -- Dr. M. Williams MRCP(UK) Clinical Research Fellow,Cancer Research UK +44 (0)207 269 2953 +44 (0)7834 899570 The views, opinions and judgements expressed in this message are solely those of the author. The message may not have been reviewed or approved by Cancer Research UK _______________________________________________ Tutor maillist - [EMAIL PROTECTED] http://mail.python.org/mailman/listinfo/tutor
