[Tutor] Little problem with math module

[Terry Carroll]

On Wed, 21 May 2008, Terry Carroll wrote:

> The following (barely-tested) routine should calculate all the Nth roots
> of a given x, even when x is negative and N is even:

I realize I'm probably talking to myself here, but for the benefit of the 
archives, I found a more elegant approach after reading 
http://everything2.com/node/1309141.  The following routine instead uses 
the formula at the end of the page (although I must admit I needed to read 
the entire page very carefully before I actually understood it):

def nthroots(x,n):
    """returns a list of all nth roots of x;
    see http://everything2.com/node/1309141 """
    from math import atan2, pi
    from cmath import exp
    i = 0+1j
    z=complex(x)
    # express the cartesian (x+yi) complex number z
    # as polar (modulus R & argument theta) 
    R = abs(z)
    theta = atan2(z.imag, z.real)
    coef = R**(1.0/n)
    return [coef * exp(i*(theta/n + 2*pi*k/n)) for k in range(0,n)]  

I'm not sure it's faster than the method from the prior email, but it 
certainly seems a bit less ham-handed.  It could probably be sped up, at 
the cost of some obfuscation, by pulling a few more things out of the list 
comprehension, e.g.,

    X = i*theta/n
    Y = (0+2j)*pi/n
    return [coeff * exp(X + Y*k) for k in range(0,n)]