Root Sets of Polynomials and Power Series with Finite Choices of Coefficients

Given H⊆C\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H\subseteq \mathbb {C}$$\end{document} two natural objects to study are the set of zeros of polynomials with coefficients in H, z∈C:∃k>0,∃(an)∈Hk+1,∑n=0kanzn=0,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \left\{ z\in \mathbb {C}: \exists k>0,\, \exists (a_n)\in H^{k+1}, \sum _{n=0}^{k}a_{n}z^n=0\right\} , \end{aligned}$$\end{document}and the set of zeros of a power series with coefficients in H, z∈C:∃(an)∈HN,∑n=0∞anzn=0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\begin{aligned} \left\{ z\in \mathbb {C}: \exists (a_n)\in H^{\mathbb {N}}, \sum _{n=0}^{\infty } a_nz^n=0\right\} . \end{aligned}$$\end{document}In this paper, we consider the case where each element of H has modulus 1. The main result of this paper states that for any r∈(1/2,1),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$r\in (1/2,1),$$\end{document} if H is 2cos-1(5-4|r|24)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$2\cos ^{-1}(\frac{5-4|r|^2}{4})$$\end{document}-dense in S1,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$S^1,$$\end{document} then the set of zeros of polynomials with coefficients in H is dense in {z∈C:|z|∈[r,r-1]},\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{z\in {\mathbb {C}}: |z|\in [r,r^{-1}]\},$$\end{document} and the set of zeros of power series with coefficients in H contains the annulus {z∈C:|z|∈[r,1)}\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\{z\in \mathbb {C}: |z|\in [r,1)\}$$\end{document}. These two statements demonstrate quantitatively how the set of polynomial zeros/power series zeros fill out the natural annulus containing them as H becomes progressively more dense.


Introduction
Let H ⊆ C be a finite set. Given such a H we define the root set of polynomials with coefficients in H to be: R(H ) := z ∈ C : ∃k > 0, ∃(a n ) ∈ H k+1 , k n=0 a n z n = 0 .
Similarly, we define the root set of power series with coefficients in H to be R * (H ) := z ∈ C : ∃(a n ) ∈ H N , ∞ n=0 a n z n = 0 .
The study of the sets R(H ) and R * (H ) can be dated back to Littlewood [6] who studied the case where H = {−1, 1}. Since then many related works have appeared, most notable amongst these are the number theoretic results of Beaucoup, Borwein, Boyd and Pinner [3], and Borwein, Erdélyi and Littmann [4], who studied the distribution of roots and multiple roots. Related work also appeared in Bousch [5], where it was shown that R({−1, 1}) is dense in {z : |z| 4 ∈ [1/2, 2]}. In Shmerkin and Solomyak [8] some measure theoretic and topological properties of R({−1, 0, 1}) are studied in detail.
In what follows, we will adopt the following notational conventions: S r := z ∈ C : |z| = r }, B(z, r ) := {z ∈ C : |z − z| < r , and given some interval I in R let In this paper, we focus on the case where H is a subset of the unit circle S 1 . Under this assumption it is straightforward to show that Intuitively, one might expect that if we allowed H to become a progressively more dense subset of S 1 , then R(H ) and R * (H ) would begin to fill out their respective annuli. The main result of this paper shows that this intuition is correct.
Before stating this result we need to define a metric on S 1 to properly quantify the density of H . Given e iθ , e iθ ∈ S 1 let d(e iθ , e iθ ) = min{|θ − θ |, |2π − (θ − θ )|}. This metric measures the interior angle of the sector of S 1 determined by the two radii e iθ and e iθ .
The sets R(H ) and R * (H ) are related by the following formula.

Proposition 1.2
Let H ⊆ C be any finite set, then the following relations hold: In the statement of Proposition 1.2, A denotes the closure of a set A, and 1 A denotes the set {z ∈ C : z −1 ∈ A}.
Proof Given z ∈ C suppose there is polynomial P ∈ H [x] such that, P(z) = k n=0 a n z n = 0.
We can construct another polynomial Q ∈ H [x] such that Q(1/z) = 0. Just consider Q(x) = x k P( 1 x ) with k = deg P. Therefore, whenever z ∈ R(H ) we also have 1/z ∈ R(H ). This proves our first relation. Now we shall show that, Without loss of generality we can assume that H ⊆ {z : |z| ≤ 1}. If z * ∈ R(H ) ∩ B(0, 1) then we can find a sequence (z i ) ∈ R(H ) N with: Moreover, since z * ∈ B(0, 1) there exists a positive number M such that 1 < M < 1 |z * | .

Now let us consider any polynomial in H [x]
P(x) = k n=0 a n x n .

The following result holds for |z
Since |a n | ≤ 1 and M|z * | < 1 the latter summation can be bounded uniformly with respect to k, namely: Each z i is the root of some polynomial P i ∈ H [x], in which case by the above, for i sufficiently large we have For the sequence (P i ) there is either a uniform upper bound for the degrees of the P i , or there exists a subsequence along which the degrees tend to infinity. In the first case, there must exist a polynomial Q ∈ H [x] and a subsequence (P i j ) such that P i j = Q for all i j . By (1) we must then have Q(z * ) = 0. Suppose deg Q = L , then is a power series with digits in H . For this particular power series we clearly have T (z * ) = 0. Therefore, in the first case we have z * ∈ R * (H ). Now suppose there exists a subsequence (P i j ) such that deg P i j → ∞. Via a diagonalisation argument, one can assume without loss of generality that there exists a sequence (a n ) ∈ H N and an increasing sequence of natural numbers (l n ), such that for all i j ≥ l n the coefficient of the degree n term of P i j is a n . In other words, as the i j become sufficiently large the lower order terms of the P i j 's start to coincide. It follows from (1) then that for this sequence (a n ) we must have ∞ n=0 a n (z * ) n = 0. Therefore, z * ∈ R * (H ) and R(H ) ∩ B(0, 1) ⊆ R * (H ) ∩ B(0, 1). Now suppose z * ∈ R * (H ) ∩ B(0, 1). Then there is a sequence (a n ) ∈ H N such that ∞ n=0 a n (z * ) n = 0.
This series is absolutely and uniformly convergent in B(0, c) for any 0 < c < 1. Since z * ∈ B(0, 1) it is contained in one of these sets for c sufficiently close to 1. We see that the function: a n x n is holomorphic on the interior of the unit disc, and therefore, the roots of P must form a discrete set. Since z * belongs to the root set of P there must exist r > 0 such that We can then consider the following integral with P N (x) = N n=0 a n x n By our conditions on r we see that P N has no zeros in B(z * , r ) for all N ∈ N. Therefore, by the argument principle (see [1, p. 152]) we must have I N ≡ 0. One can also assume that r is sufficiently small so that P N converges to P absolutely and uniformly. Therefore, However, it follows from another application of the argument principle, and the fact that P(x) has a single zero in B(z * , r ) at z * , that the above integral cannot be 0. This contradiction implies z * ∈ R(H ) and our proof is complete.
It is natural to wonder whether there exists a set H such that the sets R(H ) and R * (H ) fill up their ambient annuli, that is A [1/2,2] and A [1/2,1) respectively. In fact such a H cannot exist. For any H ⊆ S 1 there exists z ∈ C with modulus 1/2 and δ > 0, such that R(H ) ∩ B(z, δ) = ∅ and R * (H ) ∩ B(z, δ) = ∅. This is because of the following simple reasoning. Since H is a finite set there exists z ∈ C such that |z| = 1/2 and |a i + a j z| > 1/2 for all a i , a j ∈ H . Equivalently for all a i , a j ∈ H . By continuity, equation (2) holds under small perturbations of z. Therefore, there must exist δ > 0, such that for all z ∈ B(z, δ) we have Since k n=2 a n (z ) n ≤ |z | 2 1 − |z | for all (a n ) ∈ H k and k ∈ N, it follows that z cannot be the zero of a power series or a polynomial. Therefore we must have R(H ) ∩ B(z, δ) = ∅ and R * (H ) ∩ B(z, δ) = ∅.

Proof Let us start by fixing
be the circle centered at z z −1 with radius |z| −1 .
Since z ∈ A (1/2,1) we must have S(z z −1 , |z| −1 ) ∩ B(0, 2) = ∅. In fact this intersection must contain an arc of S(z z −1 , |z| −1 ). This arc is parameterised by two radii of S(z z −1 , |z| −1 ) with interior angle θ . See Fig. 1 for a diagram describing the intersection of S(z z −1 , |z| −1 ) with B(0, 2). It is easy to see that the angle θ is minimised when z z −1 is as far from the origin as possible, i.e., when z has modulus 2. Employing elementary techniques from geometry we can see that the angle θ is at least twice the size of a particular angle of the triangle whose sides have length |z| −1 , 2, and 2|z| −1 (see Fig. 1). Therefore, we can use the well-known cosine rule from trigonometry to show that θ is always bounded below by 2 cos −1 5 − 4|z| 2 4 .
Since H is 2 cos −1 ( 5−4|z| 2   4 )-dense as a subset of S 1 , there must exist a ∈ H such that z z −1 − az −1 is contained in the arc of S(z z −1 , |z| −1 ) which intersects B(0, 2). In particular, for this choice of a we have z −1 (z − a) ∈ B(0, 2). Applying Proposition 2.1 again with x 0 in the place of z yields a 1 and x 1 := z −1 (x 0 − a 1 ), such that x 1 ∈ B(0, 2) and One can then apply Proposition 2.1 with z = x 1 . Repeating this procedure indefinitely yields a sequence (a n ) and (x n ) such that x n+1 = z −1 (x n − a n+1 ) for all n ∈ N.
The terms in (x n ) remain in B(0, 2). Therefore, we are able to repeatedly apply the substitution x n+1 = z −1 (x n − a n+1 ) in (3) and we obtain 0 = ∞ n=0 a n z n .
The proof of Theorem 1.1 was based upon ideas from β-expansions. The argument given relied upon adapting methods from [2,7]. The proof can easily be adapted to show that under the hypothesis of the theorem, for every z ∈ B(0, 2) there exists (a n ) ∈ H N such that ∞ n=0 a n z n = z .

Some Further Problems
There are some more challenging problems related to root sets R(H ), R * (H ). We mentioned in the beginning of this paper that there exist various results of multiple roots [3,4]. We say that a z ∈ C is a multiple root of a holomorphic function f of order k if for all integers i = 0, 1, 2, . . . , k f (i) (z) = 0.
Adopting the notation in this paper, we can define for any integer k ≥ 0: R k (H ) : = z ∈ C : ∃k > 0, ∃(a n ) ∈ H k+1 , P(w) = k n=0 a n w n , z is a k-th order root of P(w) . R * k (H ) : = z ∈ C : ∃(a n ) ∈ H N , P(w) = ∞ n=0 a n w n , z is a k-th order root of P(w) .
Not so much has been studied about the above multiple root set, some partial results can be found in [8]. We can, for example, consider the following questions: • Are R k (H ), R * k (H ) dense in any non-degenerate annulus? • What about the connectedness and path-connectness of R k (H ), R * k (H )? • What can we say about the boundary of R k (H ), R * k (H )?