From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Polynomials with roots of norm one (partial answer)
Date: 25 Nov 1995 20:25:41 GMT

It must have rolled off my system already, but someone was asking
about a condition on the coefficients of a polynomial which expresses
the condition that the roots all have magnitude 1. This certainly
seems like the kind of thing someone ought to know already but I
couldn't think of a reference. I do have a partial description of the
situation.

There is a method of estimating the roots of a polynomial which is based on
successive squaring of the roots until the ones of smaller magnitude
become insignificant. (If I remember correctly, this is the one attributed
to a Frenchman known only as "M. Vincent".) This method can be used for
the problem at hand.

Indeed, as the original poster noted, there are some easy bounds on
the coefficients of the polynomial: the constant term is of magnitude  1
and the coefficient of  x^k  is in magnitude at most  (n choose k),
where  n  is the degree of the polynomial.

But more is true: if all the roots are of magnitude  1, then the same
inequalities are true of the polynomial whose roots are the squares of
those of the original polynomial. This gives more inequalities, and
of course the process may be iterated.

Moreover, note that if there are roots of magnitude _not_ equal to 1, then
since the product of all the roots does have magnitude 1, there must be
some of larger magnitude. If there are  j  roots of maximal magnitude  r,
then when the roots are squared sufficiently often, the coefficient of
x^(n-j) will eventually exceed any preset bound, including  (n choose  n-j).

Thus, if  P  is the set of all monic polynomials of degree  n  having a 
constant term of magnitude  1, and if  F : P --> P  is the function which 
assigns to each polynomial  p  the polynomial  F(p)  whose roots are squares 
of those of  p, then a polynomial  p  has all roots of magnitude  1
iff all iterates  F^s(p)  have all coefficients bounded by the
appropriate binomial coefficient. In other words, if  S  is the subset of
P  determined by the conditions
	S = { Sum a_k x^k : a_k <= (n choose k)  for each k}
then the polynomials whose roots all have magnitude 1  are precisely the
intersection  
	T = \intersection  (F^s)^(-1)(S), s >=1.
(It is not critical to select  S  in the manner indicated. Really, the
only point is to create this  F-invariant  subset  T.)

This description of the polynomials may not be particularly useful, but
it is not completely without merit, either. First, I should note that  F
is not hard to compute.  If  p  has roots  r_i, then we want  F(p)  to 
have roots  r_i^2. Well, if we write  p(x) = p1(x^2) + x p2(x^2)  as the
sum of its even and odd degree terms, then we know  p1(r_i^2) = -r_i p2(r_i^2);
squaring, we see that  q(x)=p1(x)^2-x p2(x)^2 vanishes at each  r_i^2.
q  is also a polynomial of the right degree, so that  q=F(s) (up to sign).
So for the lowest-degree polynomials we have
	F(x^2+ax+b)=(x+b)^2-x(a^2) = x^2 + (2b-a^2) + b^2
	F(x^3+ax^2+bx+c)=(ax+c)^2-x(x+b)^2= -x^3-(2b-a^2)x^2-(b^2-2ac)x+c^2
and in general, F(xp+d)=x F(p) + d (2 x p2 + d).

Note that  F: P --> P  is an algebraic map, and almost everywhere
2-to-1. The only elements of  P  fixed by  F   are those polynomials whose
roots are certain sets of roots of unity.

For example, suppose you seek the real cubic polynomials with all three roots 
on the unit circle. The product of the roots is of magnitude  1  and real;
after squaring once, this product will be +1, so that the cubic has the form
x^3+ax^2+bx-1. We look for the pairs (a,b)  in the plane which yield 
such a cubic. We begin with the set  S = { (a,b) : |a|<3, |b|<1 }, and the 
function  F( (a,b) ) = (2b-a^2, b^2+2a). This  F  is two-to-one except
at the fixed points  (-3,3) and (0,0). Our  T  will be an  F-invariant subset
of the plane including these points. I'm not sure if there is a reasonable
way to compute this fixed-point set. If the boundary of  T  is an algebraic
curve, then it can be found by elimination if we know the degree of the
curve, but my brief attempt to find such an invariant curve yielded only the
line y=-x.

Of course the "proper" context for examining  F  is with complex 
coefficients (indeed, one should arguably define  F  as a mapping from
projective space to itself so as not to have to specialize to monic
polynomials) but my geometric visualize peters out pretty quickly in
high complex dimensions.

dave