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 <snowe.814068166@rain.org>, Nick Halloway <snowe@rain.org> 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