Whoops, Forgot to divide by 2*pi in the second equation. They should be: log(r) = 2*pi*M*b/(a^2+b^2) t = M*a/(a^2+b^2) Note that if we take M=1, we get a fundamental solution: z=exp(2*pi*b/(a^2+b^2)) exp(2*pi*a/(a^2+b^2)). The second factor is an ordinary root of unity. Every solution is an (ordinary) integral power of the fundamental one, just as in the ordinary case. So in the case n=1+i, the fundamental solution is -exp(pi). Victor On Thu, May 9, 2013 at 9:52 AM, Victor Miller <victorsmiller@gmail.com>wrote:
Continuing, we can divide both equations by a^2+b^2 to get
log(r) = 2*pi*M*b/(a^2+b^2) t = 2*pi*M*a/(a^2+b^2) + 2pi*N, so we see that N doesn't matter. If b=0 we get the normal n-th roots. If a+ib = 1+i, we get
z = - exp(pi*M) for an integer M.
Victor
On Thu, May 9, 2013 at 9:47 AM, Victor Miller <victorsmiller@gmail.com>wrote:
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