Kerry, When n is a Gaussian integer things are more complicated since then z^n is a multi-valued function. Proceed as follows: let z = r*exp(2 pi * it), where r > 0 and t is in [0,1]. Then log(z) = log(r) + 2*pi*i (t + N) for any integer N. Let n = a + ib, where a,b are integers. So in order for z^n = 1 we must have: (log(r) + 2*pi*i*(t+N))*(a + i b) = 2*pi*i*M, for some integer M. Multiplying out and equating real and imaginary parts you find that (a^2 + b^2)(log(r)) = 2*pi*M*b and (a^2 + b^2)* t = 2*pi(-b^2*N + a*M - a^2*N) Victor On Thu, May 9, 2013 at 2:25 AM, Kerry Mitchell <lkmitch@gmail.com> wrote:
Hi all,
I am playing with a problem that has boiled down to this: for positive integer n, there are n complex roots of 1 (or any complex number). What happens when n is a Gaussian integer? How many roots are there and what are they like?
I've done some preliminary work on this; can someone point me to a reference so I can see if I'm on the right track?
Thanks, Kerry _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun