Do you want : the roots of 1 ? or something more general ? Because the nth-roots of 1 are defined by :
r = exp(2*i*pi*k/n) with k = {0,1,...n-1}If it's more complex ... then I don't know :)
Pierre
Dick Moores a écrit :
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?
Checking the Casio results with Python: >>> 1**3 1 >>> (-.5 + .866025403j)**3 (0.99999999796196859+1.1766579932626087e-009j) >>> (-.5 - .866025403j)**3 (0.99999999796196859-1.1766579932626087e-009j)
Thanks,
Dick Moores [EMAIL PROTECTED]
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