I guess you mean that 1 has n complex *nth* roots. Or, put another way, 1 has n complex (1/n)th powers. There's a general definition of how to take any nonzero complex number z to any complex power w: z^w := {exp(w log(z))} where exp() is the usual power series for e^z, and log is its multivalued inverse function, which is why you get a set of values in general. If c is any one value of log(z), i.e., if exp(L) = z, then all the values form the set {c + 2Kpi*i | K in Z}, where Z is the integers. So, z^w = {exp(w * (c + 2Kpi*i))} (***) = {exp(w*c) * exp(2pi*i*w)^K | K in Z}. From the last expression, it's clear that the only case where there's only one value of z^w is when w is an integer. For most combinations of z and w we get a discrete log spiral of the form {P*(Q^K) | K in Z}. ----------------------------------------- Now we specialize to z = 1 and w = 1/(M + Ni) with M, N (not both 0) in Z: From (***) we get 1^(1/(M+Ni) = {exp((1/(M+Ni)*0) * exp(2pi*i*(1/M+Ni))^K | K in Z} = {exp(2pi*i*(M-Ni)/(M^2+N^2))^K} = {exp(2pi*(N + iM)/(M^2+N^2)) ^ K | K in Z} --Dan On 2013-05-08, at 11:25 PM, Kerry Mitchell wrote:
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?