Thanks Andy, Charles and Henry, but I guess I wasn't clear. I'm aware of the Fundamental Theorem of Algebra and what happens when n is a real integer. My question is about what happens when n is a complex number with integer parts. I've determined (and had confirmed in offline discussion) that the number of roots depends on which branch one uses for the angle (t or theta). I'm wondering if there's any standard definition of the roots of 1 when n is complex (not purely real). Kerry On Thu, May 9, 2013 at 6:28 AM, Andy Latto <andy.latto@pobox.com> wrote:

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?

Write your complex number z as r e^(i * t), where r and t are real. (I use t instead of the standard theta for ease of expressing in ASCII)

Then the nth roots of z are r^(1/n) e^(i (2pi k + t)/n), for k = 0, 1, 2, ... n - 1.

Geometrically, the n roots are evenly spaced on a circle centered at the origin.

Andy

...

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

-- Andy.Latto@pobox.com

_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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

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

Kerry, I get that, for z = x+iy nonzero, the set of values {1^1/z)} is all numbers of the form 1^(1/z) = exp((1/z)log(1)) = exp(2pi*i* (x-iy)/(x^2+y^2)) ^ K = exp(2pi*y/(x^2+y^2)) * exp(i * 2pi*x/(x^2+y^2)) ^ K , as Victor said (and except for an erroneous minus sign, I almost said). All powers of a nonzero complex number form a finite set if and only if a) its absolute value (modulus) is 1 and b) its angle is a rational multiple of 2pi. For z a Gaussian integer -- i.e., when x and y are integers -- the modulus exp(2pi*y/(x^2+y^2)) = 1 if and only if y = 0. (For, the function exp is monotonically increasing on the real numbers and so takes the value 1 only at 0.) In that case the (or an) angle is 2pi/x, clearly a rational multiple of 2pi. Then it's clear that this determination of 1^(1/z) will have exactly |x| distinct powers before they repeat: Just the |x| |x|th roots of unity. --Dan P.S. Minor stylistic suggestions: the letter n is best reserved for actual integers. And bullets before mathematical expressions might work better as asterisks or something else that's not an arithmetical operation. On 2013-05-10, at 7:20 AM, Kerry Mitchell wrote:

Thanks for all the responses to my question. Based on those, my own analyses, and lots of numerical work, this is what I’ve come up with.

Let z be an ordinary complex number, z = x+iy = r exp(i t), where r and t are defined in the standard ways. I limit t to the branch [-pi, pi). Let n = a Gaussian integer; n = a+b i, where a and b are both integers and not both 0.

For z to be an nth root of 1, then z^n = 1 = exp(i 2k pi), where k is an integer. And z^n = exp[n ln(z)], where ln(z) = ln(r)+i t, using only the principal value of the ln() function. So, assembling the parts,

- ln(r) = 2k pi b / (a^2 + b^2), and - t = 2k pi a / (a^2 + b^2).

Looking at the t equation, with the restriction that t is in [-pi,pi),

- -pi <= 2k pi a / (a^2 + b^2) < pi, or - -(a^2 + b^2) <= 2k a < (a^2 + b^2).

So, finding the number of roots for a given a+b i is just counting how many k’s will satisfy the inequality. Since everything in the last inequality is an integer, this can be performed exactly numerically. Here’s what I found:

- For b = 0, a common factor of a drops out of the inequality and it reduces to the standard roots of 1 for integer n. - For a = 0, t = 0 and there are infinite real roots: z = exp(2k pi / b). - For |a| = |b|, the inequality reduces to: -|a| <= k < |a|, and there are 2|a| roots. - For a != 0, |a| can be factored out of each term, leaving -(a^2+b^2)/|a| <= 2k < (a^2+b^2)/|a|. Each side would contribute about (a^2+b^2)/|a| roots, but the 2k term cuts that in half, so the approximate number of roots is (a^2+b^2)/|a|. This result is exact for the previous three cases (assuming 1/0 is infinity).

I've constructed a table of the numbers of roots for values of a & b. Is that of sufficient interest to be added to the OEIS?

Kerry _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun