From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Distribution of roots Date: 19 Oct 1995 04:24:17 GMT In article , Nick Halloway wrote: >Some questions I have been wondering about roots: > >a) Are roots of polynomials over Q dense in C^n? That is for arbitrary > c_1, ..., c_n in C, and e > 0 in R, is there an irreducible polynomial > in Q[x] with roots z_1, ..., z_n, conj (z_1), ...., conj (z_n), with > |z_i - c_i| < e for i = 1, n? Yes. If P(z) = Prod(z-c_i), then P1(z) = P(z)*(conj P(z)) is a polynomial with real coefficients and roots c_i and conj(c_i). You can choose rational polynomials P2 as close as you like to this P1; as you do so, you will get as roots collections of n pairs of complex conjugates, which approach the c_i as P2 approaches P1 -- in particular, you can keep |z_i - c_i| < e/2. Likewise, all the other rational polynomials P3 in a neighborood of P2 have roots within e/2 of the roots of P2, and thus within e of the roots of P1. Now you need the Hilbert Irreducibility Theorem: that the rational polynomials which are irreducible over Q are dense in the collection of all rational polynomials. Thus there will be choices for P3 which meet your additional requirement of being irreducible. >b) Suppose that f is complex analytic over the whole complex plane. > Is there anything you can say about the distribution of its roots? > i.e. that the distance between them has a positive lower bound? > Or that the roots have no accumulation point? The distance between consecutive zeros of f(z) = sin(pi * z^2) is sqrt(n+1)-sqrt(n), which is roughly 1/(2 sqrt(n)) for large n. So, no, there is no positive lower bound on the distance between roots. (By the way, for non-polynomials, they're usually called zeros of f.) Let Z = { x : f(x)=0 }. Accumulation points lie in closures, but Z = f^(-1)(0) is already closed in C as f is continuous on all of C. Thus if the roots have an accumulation point, that accumulation point is also a root. (You need to consider e.g. f(z) = sin(1/z) to see why I bother with this paragraph.) But if f(z_0)=0 and f(z_n)=0 for a sequence of z_n approaching z_0, then the Taylor series of f at z_0 vanishes. If f is entire, then f must be identically zero. So, yes, you can rule out zero-sets with accumulation points. You ought to look at factorizations of entire functions. Roughly, an entire function f is a product f(z) = Prod(z - c_i) * exp(g(z)) for some other entire function g; then c_i are the zeros of f. Of course there's a lot to worry about when it comes to convergence and analyticity. As I recall there's a readable exposition of this topic towards the end of Conway's book on complex analysis. dave