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