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?
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