Abstract
We study the root distribution of a sequence of polynomials , 2) and (4, 3) where A(z) and B(z) are arbitrary polynomials in z with complex coefficients. We show that the zeros of Pn(z) which satisfy A(z)B(z) = 0 lie on a real algebraic curve which we describe explicitly. P n (z) + B(z)P n−1 (z) + A(z)P n−k = 0 for n = 1, 2, . . .
2010 Mathematics Subject Classification. Primary 12D10, Secondary 26C10, 30C15.
Introduction
The study of zeros of sequences of polynomials plays an important role in many areas of mathematics such as analysis, probability theory, combinatorics and geometry. In this article, we study the distribution of zeros of polynomials in a polynomial sequence {P n } ∞ n=0 generated by a certain type of three-term recurrence relation. Such polynomials are of interest due to several remarkable properties they possess. For example, three-term recurrence relations are useful in numerical mathematics for producing sequences of orthogonal polynomials. Let Q 1 (z), Q 2 (z) . . . , Q k (z) be fixed complex polynomials and {P n (z)} ∞ n=0 be the sequence of polynomials generated by recurrence relation of the form: P n (z)+Q 1 (z)P n−1 (z)+Q 2 (z)P n−2 (z)+· · ·+Q k (z)P n−k (z) = 0, n = 1, 2, . . . (1.1) subject to certain initial conditions. The problem of describing the location of the zeros of polynomials in the sequence {P n (z)} ∞ n=0 might have two different versions. The first is asymptotic and it aims at finding the limiting curve for the zeros of P n (z) as n → ∞. This is the approach taken in the papers [3, 4, 6, 12] . The second is of exact type; it aims at finding the curve where all the zeros of P n (z) lie for all n (or at least for all large n), for example [1, 2] . For general recurrence relations, such curves do not exist. However, for three-term recurrences with 2 k 5 and appropriate initial conditions, such a curve containing all the zeros of P n (z) exists and is given in [1] . More generally in [1, Conjecture 6] , K. Tran conjectured existence of such a curve for all k 6. Below we reformulate Tran's Conjecture.
Conjecture A. For an arbitrary pair of complex polynomials A(z) and B(z), every zero of every polynomial in the sequence {P n (z)} ∞ n=1 satisfying the three-term recurrence relation of length k with the standard initial conditions P 0 (z) = 1, P −1 (z) = · · · = P 1−k (z) = 0 which is not a zero of A(z) lies on the portion of the real algebraic curve C ⊂ C given by
The first part of the above conjecture explicitly defines the real algebraic curve on which all the zeros of the polynomials P n (z) (except the zeros shared with A(z)) are located. The second part describes the exact portion on this curve where these zeros lie. In [1] , Conjecture A was settled for k ≤ 5. In a subsequent paper [2] , K. Tran proved existence of such a curve containing all the zeros of P n (z) for sufficiently large n and arbitrary k. More supporting results can be found in [7] . Additionally, a criterion for the reality of all the zeros of every polynomial in the sequence {P n (z)} ∞ n=0 for k = 2 can be found in [10] . Based on numerical experiments, B. Shapiro has formulated the following generalization of Conjecture A (private communication).
Conjecture 1.
For an arbitrary pair of complex polynomials A(z) and B(z), every zero of every polynomial in the sequence {P n (z)} ∞ n=0 satisfying the three-term recurrence relation of length k P n (z) + B(z)P n− (z) + A(z)P n−k (z) = 0 for n = 1, 2, . . . (1.2) with the standard initial conditions P 0 (z) = 1, P −1 (z) = · · · = P 1−k (z) = 0 and coprime 1 ≤ < k which is not a zero of A(z) or B(z) lies on the real algebraic curve C ⊂ C given by
The purpose of this paper is to prove some specific cases of Conjecture 1 and make them more precise. It is important to note that the rational function B k (z)/A (z) can be written in terms of the discriminant of the denominator of the generating function of the polynomial sequence generated by (1.2) . Our approach to the proof of the specific cases of Conjecture 1 uses the q-analogue of the discriminant of a polynomial, a concept introduced by Ismail in [9] and used in [1] . In particular, the ratios of zeros of the denominator of the generating function with equal modulus will play a fundamental role.
The paper is organized as follows. In Section 2, we review the most important notions and basic results. In Section 3, we prove some relevant general results. In Section 4, we prove the main result of the paper. In Section 5, we give some numerical examples.
Discriminants
This paper uses discriminants, and therefore we present below a few key results about them from the literature. Our initial step is to recall the concept of discriminants of polynomials, both ordinary and q-discriminants.
Definition 1 (see [11] ). Let P (x) be a univariate polynomial of degree n with zeros x 1 , . . . , x n and leading coefficient a n . The ordinary discriminant of P (x) is defined as Disc x (P (x)) = a 2n−2 n 1≤i<j≤n
This ordinary discriminant of P (x) can also be expressed in terms of the resultant of P (x) and its derivative. Let P (x), Q(x) be polynomials of degrees n, m and leading coefficients a n , b m respectively. The resultant of P (x) and Q(x) is defined by
The discriminant of P (x) is then computed as follows Disc x (P (x)) = (−1) n(n−1) 2 1 a n Res(P (x), P (x)).
For more details on the ordinary discriminants and resultants, see [5, 11] .
Example 1. If f (x) and g(x) are fixed complex polynomials in x with f ≡ 0, then the discriminant of P (x, t) = f (x)t 3 + g(x)t 2 + 1 as a polynomial in t is given by
The proof/ verification of the above example follows directly from equation (2.2).
Generally, the discriminant of a polynomial connects with the ratio of its zeros in the sense that the discriminant is zero if and only if the polynomial has a multiple zero. In particular, the discriminant of a polynomial vanishes whenever there exist at least two zeros that are equal, i.e., zeros whose ratio is 1.
Definition 2 (see [9] ). Let P (t) be a polynomial of degree n with zeros x 1 , . . . , x n and leading coefficient a n . The q-discriminant of P (t) is Disc t (P (t); q) = q n(n−1)/2 a 2n−2 n 1≤i<j≤n
In other words, Disc t (P (t); q) = q n(n−1)/2 a 2n−2 n 1≤i<j≤n
It is clear from Definition 2 that Disc t (P (t); 1) = Disc t (P (t)), the ordinary discriminant of the polynomial. Additionally, the q-discriminant vanishes whenever there is a pair of zeros with ratio q. Below we give some examples of q-discriminants.
Example 2. Let P (t) = at 2 + bt + c be a quadratic polynomial with complex coefficients. The q-discriminant of P (t) is
This case can easily be computed from Equation (2.3). A similar but tedious calculation give the q-discriminant of cubic polynomial P (t) = at 3 + bt 2 + ct + d as
where Φ = 1 + 5q + 6q 2 + 5q 3 + q 4 .
In general computing q-discriminants using Definition 2 can be a tedious exercise. However, the following proposition proved in [9] is often used. Proposition 1. Let P (t) be a polynomial of degree n with zeros x 1 , . . . , x n and leading coefficient a n . The q-discriminant of P (t) is given by:
Proposition 1 is used in [2, 14] to derive the expression of the q-discriminant of the polynomial D(t, x) = A(x)t n + B(x)t + 1. This is stated in Proposition 2, (we suppress the parameter x).
Proposition 2 (K. Tran [2] ). Let D(t) = At n + Bt + 1 be a polynomial in t of degree n where A and B are arbitrary fixed complex functions. The q-discriminant of D(t) = At n + Bt + 1 is given by
For completeness let us review some definitions (also obtained in [8, 12] ) about the root distribution of a sequence of functions
where α i (z) and β i (z) are analytic in a domain D. Let us call an index i dominant at z if |β i (z)| ≥ |β j (z)| for all j(1 ≤ j ≤ n). Let
Denote Z(f m ) the set of zeros of f m (z). Definition 3. The set of all z ∈ D such that every neighborhood U of z has a nonempty intersection with infinitely many of the sets Z(f m ) is called limit superior of Z(f m ) and is denoted by lim sup Z(f m ). On the other hand, the set of all z ∈ D such that every neighborhood U of z has a non-empty intersection with all but finitely many of the sets Z(f m ) is called limit inferior of Z(f m ) and is denoted by lim inf Z(f m ).
We note that both lim inf Z(f m ) and lim sup Z(f m ) always exist. In addition, it always holds that lim inf Z(f m ) ⊂ lim sup Z(f m ). Moreover, in the event that both the limit superior and the limit inferior of Z(f m ) coincide, then we call this set, the limit of the Z(f m ), namely,
We now state the following theorem, see ([9, Theorem 1.5]) which will be useful in this work.
Theorem 3. Let D be a domain in C, and let α 1 , . . . , α n , β 1 , . . . , β n , (n ≥ 2) be analytic functions on D, none of which is identically zero. Let us further assume a "no-degenerate-dominance" condition, that is, there do not exist indices i = i such that β i ≡ ωβ i for some constant ω with |ω| = 1 and such that D i (= D i ) has nonempty interior. For each integer m ≥ 0, define f m by
Then lim inf Z(f m ) = lim sup Z(f m ), and a point z lies in this set if and only if either (a) there is a unique dominant index i at z, and α i (z) = 0, or (b) there are two or more dominant indices at z.
Remark 1.
If z * ∈ C is a fixed complex number such that the zeros in t of D(t, z * ) = A(z * )t k +B(z * )t +1 are distinct, then by use of partial fraction decomposition and Theorem 3, z * belongs to lim inf Z(f m ) when the two smallest zeros (in modulus) of D(t, z * ) have the same modulus. Note that the functions t 0 (z), t 1 (z), . . . , t k (z) are analytic in a neighborhood of z * by the implicit function theorem [15] . Now with our setup, if we can obtain a point z * ∈ C so that the zeros of D(t, z * ) are distinct and the two smallest (in modulus) zeros of D(t, z * ) have the same modulus, then for such a point z * , we have that z * ∈ lim inf Z(f m ) = lim sup Z(f m ). This implies that on a small neighborhood of z * , there is a zero of f m for all large m by the Definition 3 of lim inf Z(f m ). For details see [15] .
It is important to note that this elegant theorem provides a description of the asymptotic behaviour of the zeros of {P n (x)} in the general case. On the other hand, Conjectures A and 1 describe specific cases where all the zeros of P n (x) (different from those of A(x)) actually lie on the curve C for all n or for all sufficiently large n.
GENERAL RESULTS
In order to prove some specific cases of Conjecture 1, we first prove some general auxiliary lemmas. The results for cases when (k, ) = (3, 2) and (k, ) = (4, 3) then fit in as specific cases. In this section among other things, we compute the q-discriminant of a special polynomial D(t, z) = A(z)t k + B(z)t + 1 for coprime 1 ≤ < k. We begin as follows. 
with the standard initial conditions P 0 (z) = 1,
Proof. Let G(t) = n≥0 P n t n be the generating function of {P n } ∞ i=0 . Multiplying (3.1) by t n and summing over from n = k to infinity, we obtain
Simplifying gives
From Equation (3.1), we have P n = 0 for 0 < n ≤ − 1 and P n + BP n− = 0 for ≤ n ≤ k − 1. It follows that 1 + −1 n=1 P n t n + k−1 n= (P n + BP n− )t n = 1, which proves the required result.
Let us now consider the q-discriminant of D(t, z). In the following proposition, we shall suppress the variable z.
Proof. By Proposition 1, we have
This implies that
Now, D(t) and D q D(t) are related by the equation
.
implies that 0 is a root that occurs with multiplicity − 1. The other zeros satisfy the relation
Since the integers k and are coprime which is equivalent k − and being coprime, we can make the change of variables ξ = t . We observe that t
The zero t * i = 0 for 1 ≤ i ≤ − 1 contributes the following factor to the product of the discriminant
The expression for Disc t (D(t); q) now has the form
.
We now simplify the remaining product. Consider the following equation:
Using the change of variable
Using Binomial Theorem, we get
For i = 1, 2, . . . , k − , let y i are the solutions to g(Y ) = 0. Then
. Now the product of solutions to g(Y ) = 0 is given by
Plugging equation (3.10) into the expression for Disc t (D(t); q) gives
Simplifying equation (3.11) gives
The proof is complete.
Proof. Set k = 3 and = 2 in the Theorem 5 above and obtain the result.
The following lemma will be used in the proof of the main results. Lemma 6. Let k, be integers and q ∈ C with |q| = 1. The function defined by
Proof. Since |q| = 1, we have q s − q −s = 2i (q s ), for any integer s, (here i 2 = −1). It therefore follows that ( 
Since the expressions (q = 0. We also provide the relevant inequality constraint.
For the case (k, ) = (3, 2), we begin with the following lemma. 
for some positive integer n. Then ξ 2 and ξ 3 lie on the curve Γ = C 1 ∪ C 2 ∪ C 3 and are dense there as n → ∞. Here the equations of C 1 , C 2 and C 3 (see Fig.1 ) are given by
The proof is similar to the one given in [2, Lemma 2] by letting ξ → 1 ξ . Remark 2.
Note that t 1 , t 2 and t 3 are zeros of D(t, z) = A(z)t 3 + B(z)t 2 + 1. By Vieta's formulae we have t 1 t 2 + t 1 t 3 + t 2 t 3 = 0. Dividing by t 2 1 and letting q i = t i /t 1 and q i = 1/ξ i , i = 2, 3 gives q 2 q 3 + q 2 + q 3 = 0 if and only if ξ 2 + ξ 3 + 1 = 0, the first condition in Lemma 7.
Theorem 8. Let {P n (z)} be a polynomial sequence whose generating function is
where B(z) and A(z) are polynomials in z with complex coefficients. Given any n ∈ N, those zeros z of P n (z) for which A(z)B(z) = 0 lie on the curve defined by
Proof. Let z 0 be a zero of P n (z) such that A(z 0 )B(z 0 ) = 0 and let
There are two cases we shall consider; namely, the case of repeated zeros and the case of distinct zeros of D. Case 1: Suppose D(t, z 0 ) has repeated zeros. The ordinary discriminant of
For repeated zeros, Disc t (D(t, z 0 )) = 0 which is equivalent to
Therefore, the point z 0 lies on the curve given by
Case 2: Suppose all the zeros of D(t, z 0 ) are distinct, i.e., Disc t (D(t, z 0 )) = 0. In this situation, let t 1 , t 2 and t 3 be the three distinct zeros of D(t, z). We first consider the distribution of quotients of zeros q i = t i /t 1 , 1 ≤ i ≤ 3. Then we examine the root distribution of P n (z) using q-discriminants. We show that these quotients q i = t i /t 1 , 1 ≤ i ≤ 3 lie on the curve in Fig. 1 . Let z 0 be a root of P n (z) which satisfies A(z)B(z) = 0 and let t 1 = t 1 (z 0 ), t 2 = t 2 (z 0 ), t 3 = t 3 (z 0 ) be the zeros of D(t, z 0 ) = A(z 0 )t 3 + B(z 0 )t 2 + 1. By partial fraction decomposition, we have
By using geometric progression, we get
Hence we obtain
(4.1)
Comparing coefficients of t n in equations (3.2) and (4.1) we obtain
Let q i = t i /t 1 , i = 2, 3. Now for any n ≥ 0, the equation P n (z 0 ) = 0 is equivalent to
. Therefore equation (4.2) can be rewritten as q n 2 q n 3 (q 2 − 1) − q n 2 q n 3 (q 3 − 1) = q n 2 (q 2 − 1) − q n 3 (q 3 − 1) and consequently since q 2 , q 3 = 0, 1 we obtain
.
(4.3)
If we add 1 on both sides of the equation (4.3) and set q i = 1/ξ i , i = 2, 3 we obtain [one side] q n 2 − 1 q n 2 (q 2 − 1)
Similarly on the other side of equation (4.3) we obtain
the second condition in Lemma 7. Next, we need show that ξ i , i = 2, 3 being on the curve Γ implies that q i ∈ Γ as well. In particular, if ξ i ∈ C 1 then the corresponding q i ∈ C 1 . If ξ i ∈ C 2 then the corresponding q i ∈ C 3 and for ξ i ∈ C 3 , the corresponding q i ∈ C 2 . To do this, it is enough to show that Γ is invariant under the Möbius inversion z → 1 z . This is shown as follows.
(i) For any z = x + iy ∈ C 1 , we have |z| = 1 with (z) ≤ − 1 2 . In addition, z = 0 and so the image of z under inversion w = 1
With these ranges on x and y, we obtain | (w)| = | y 2x | ≥ In addition, since (w) = −y 1/4+y 2 and |y| ≥
The conclusion that Γ is invariant under the Möbius inversion z → 1 z thus follows. Therefore q 2 and q 3 lie on the curve Γ given in Lemma 7. The proof is complete. Now we know that q 2 and q 3 are given by the ratio of zeros of D and are therefore the zeros of the discriminant. (Recall from Ismail [9] that the q-discriminant equals zero if there is a pair of zeros with the ratio q). From Corollary 1, we know that the q-discriminant of D(t, z) = 1 + B(z)t 2 + A(z)t 3 is given by
Since q 2 and q 3 are the zeros of the discriminant then
with i = 2, 3. Consequently since q i = 0, 1 we have
maps the curve Γ to a real interval so that we conclude (f ) = 0. Now since q lies on the curve in Fig. 1 , we have three possibilities below. (a) |q| = 1, and this corresponds to the points ξ on the curve C 1 . (b) q +q = −1, and this corresponds to the points ξ on the curve C 2 . (c) | − 1 − q| = 1, and this corresponds to the points ξ on the curve C 3 .
For part (a)
For part (b), note thatq = −1 − q. Therefore
Consequently f (q) = f (q) = f (q), and we conclude that (f ) = 0.
Hence f (q) = f (q) = f (q), and we conclude that (f ) = 0.
The conclusion (f ) = 0 follows. Therefore, the point z 0 lies on the curve given by
Next we prove the inequality constraint. From the recurrence relation with = 2, k = 3 we have
Let t 1 , t 2 , t 3 be the zeros of D(t) labelled so that 0 < |t 1 | ≤ |t 2 | ≤ |t 3 | and set q i := t i /t 1 for i = 1, 2, 3. Note these are functions of z. Vieta's formulae give
If z 0 is a zero of P n (z) and A(z 0 )B(z 0 ) = 0, then by Theorem 3(b) and Remark 1, q 1 = 1, |q 2 | = |u| = 1 and |q 3 | ≥ 1.
So, we search u such that the above conditions are satisfied. Let u = e iθ where θ ∈ [0, 2π]. It follows from |q 3 | ≥ 1 that 1 = |u| ≥ |1 + u| = 2(1 + cos θ). This implies that θ ∈ 2π 3 , 4π 3 . Let Ω = e iθ : 2π 3 ≤ θ ≤ 4π
3
. Since q ∈ Ω ⊂ Γ, it follows that
We now compute the range of f on Ω. To do this, we let q = e iθ where θ ∈ [ 2π 3 , 4π 3 ]. After this parametrization we obtain
It is clear that F is well-defined and continuous on the union [ 2π 3 , π) ∪ (π, 4π 3 ]. Observe that for θ ∈ [ 2π 3 , 4π 3 ], we have −1 2 cos(θ)+1 0 and 0 2 cos(θ)+2 1 hence F (θ) 0. Moreover, F attains its minimal values equal to 0 at θ = 2π 3 and at θ = 4π 3 . Since lim θ→π F (θ) = +∞, it follows from continuity and non-negativity of F that the range of f is R 0 . This proves the inequality condition that
Next we settle the specific case of Conjecture 1 where (k, ) = (4, 3). We begin with the following lemma. Lemma 9. Let z 0 be a zero of P n (z) and q = q(z 0 ) be a quotient of two zeros in t of D(t, z 0 ) = A(z 0 )t 4 + B(z 0 )t 3 + 1. Then the set of all such quotients belongs to the curve depicted in Fig. 2 where the equation of the quartic curve C 5 (see the left part of Fig. 2 ) is 1 + 2x + 2x 2 + 2x 3 + x 4 − 2y 2 + 2xy 2 + 2x 2 y 2 + y 4 = 0, and the other curve C 4 (see the right part of Fig. 2) is the segment of the unit circle with real part at least −1/3.
Proof. Let z 0 be a zero of P n (z) such that A(z 0 )B(z 0 ) = 0 and let t 1 = t 1 (z 0 ), t 2 = t 2 (z 0 ), t 3 = t 3 (z 0 ) and t 4 = t 4 (z 0 ) be the zeros of D(t, z 0 ) = A(z 0 )t 4 + B(z 0 )t 3 + 1.
By partial fraction decomposition, we have
By using geometric progression, we get
Hence we obtain
(4.5)
Comparing coefficients of t n in Equation (3.2) and Equation (4.5) we obtain
Let q i = t i /t 1 , i = 2, 3, 4. Now for any n ≥ 0, the equation P n (z 0 ) = 0 is equivalent to
and
Divide Equation (4.6) by t 2 1 and (4.7) by t 3 1 and solve the resulting equations simultaneously to get q 3 and q 4 in terms of q 2 = u as follows. Either
, (4.8) or
. (4.9)
Observe that the product and sum of the roots are respectively Splitting the real and imaginary parts of Equation (4.12), using Mathematica we obtain that G maps the interval − 1 3 ≤ cos θ ≤ 1 to the quartic curve 1 + 2x + 2x 2 + 2x 3 + x 4 − 2y 2 + 2xy 2 + 2x 2 y 2 + y 4 = 0.
Theorem 10. Let {P n (z)} be a polynomial sequence whose generating function is
where B(z) and A(z) are polynomials in z with complex coefficients. The zeros of P n (z) for all n which satisfy A(z)B(z) = 0 lie on the real algebraic curve defined by
Proof. Let z 0 be a zero of P n (z) such that A(z 0 )B(z 0 ) = 0 and let t 1 = t 1 (z 0 ), t 2 = t 2 (z 0 ), t 3 = t 3 (z 0 ), t 4 = t 4 (z 0 ) be the zeros of D(t, z 0 ) = A(z 0 )t 4 + B(z 0 )t 3 + 1. As above, we consider two cases. Case 1: Suppose D(t, z 0 ) has repeated zeros. The ordinary discriminant of
For repeated zeros, Disc t (D(t, z 0 )) = 0 which is equivalent to
Therefore, the point z 0 lies on the curve given by
Case 2: Suppose all the zeros of D(t, z 0 ) are distinct, i.e., Disc t (D(t, z 0 )) = 0. Let t 1 , t 2 , t 3 and t 4 be the distinct zeros of D(t, z). As above, we consider the distribution of quotients of zeros q i = t i /t 1 , 1 ≤ i ≤ 4. Then we examine the root distribution of P n (z) using q-discriminants and show that the quotient of zeros q i = t i /t 1 , 1 ≤ i ≤ 4 lie on the curve given in Fig. 2 .
By Theorem 5, the q-discriminant of D(t, z) = 1 + B(z)t 3 + A(z)t 4 is given by
If q is a quotient of two distinct zeros of D(t, z 0 ) = 1 + B(z 0 )t 3 + A(z 0 )t 4 then Disc t (D(t); q) = 0 which implies
Since q = 0, 1, we have
It remains to show that for q on the curve depicted in Fig. 2 , we have (f (q)) = 0. To do this, let u = e iθ be a point on a unit circle with − 1 3 ≤ cos θ ≤ 1. Then u ∈ C 4 . To each u, there are two possible values of q. Moreover, u = q since this would imply that u = −1 ± i √ 2 contradicting that u is a point on the segment of the unit circle with real part at least −1/3. Now, let u ∈ C 4 with the property that q(u) ∈ C 5 . Then q and u are related by the equation
Multiplying Equation (4.14) by u − q, we obtain
From (4.13) and the identity (4. 15) we show that f (q) = f (u) as follows.
Since u ∈ C 4 , it implies |u| = 1. Lemma 6 then gives (f (u)) = (f (q)) = 0. The conclusion (f ) = 0 thus follows. Therefore, the point z 0 lies on the curve given by
Next we prove the inequality constraint. From the recurrence relation with = 3, k = 4 we have
Let t 1 , t 2 , t 3 , t 4 be the zeros of D(t) labelled so that 0 < |t 1 | ≤ |t 2 | ≤ |t 3 | ≤ |t 4 | and set q i := t i /t 1 for i = 1, 2, 3, 4. Note these are functions of z.
From the Equations (4.8) and (4.9), it is enough to consider only one pair of solutions (q 3 , q 4 ) since the other pair follows similarly and give the same results.
If z 0 is a zero of P n (z) and A(z 0 )B(z 0 ) = 0, then q 1 = 1, |u| = 1, |q 3 | ≥ 1 and |q 4 | ≥ 1. So, we search u such that the above conditions are satisfied. Set u = e iθ where θ ∈ [0, 2π]. It follows from |q 3 
.
If q ∈ Ω, it follows that
We now compute the range of f on Ω. To do this, we set q = e iθ where θ ∈
. After this parametrization we obtain f (q) = F (θ) = (cos(4θ) − 1) 2 (cos(3θ) − 1)(cos(2θ) − cos θ) .
Its clear that F is well-defined and continuous on the union [ π 2 , 2π 3 ) ∪ ( 4π 3 , 3π 2 ]. For θ ∈ Ω, we obtain 0 (cos(4θ) − 1) 2 9 4 , −1 (cos(3θ) − 1) 0 and − 1 (cos(2θ) − cos θ) 0. Hence F (θ) 0. Moreover, F attains its minimum value of 0 at θ = π 2 and θ = 3π 2 . Since lim θ→ 2π 3 − F (θ) = +∞ = lim θ→ 4π 3 + f (θ), it follows from continuity and non-negativity of F that the range of f is R 0 . This proves the inequality condition that 0 ≤ B 4 (z) A 3 (z) < ∞.
Examples
In this section we present several concrete examples using numerical experiments. In these examples, we consider the sequence of polynomials {P n } ∞ n=0 generated by the rational function On each graph, we plot the zeros of one of the polynomials (of our choice) in the polynomial sequence described by the given parameters k, , A(z) and B(z).
Example 3. For k = 3, = 2, A(z) = z + 5 and B(z) = −z 2 + 2z + 5, we obtain Fig. 3 and Fig. 4 . In Fig. 3 , we plot the zeros of P 30 (z) while in Fig. 4 , we plot the zeros of P 70 (z).
Example 4. For k = 3, = 2, A(z) = z 3 − z + 6 and B(z) = −z 2 + 7z − 5, we obtain Fig. 5 and Fig. 6 . In Fig. 5 , we plot the zeros of P 120 (z) while in Fig. 6 , we plot the zeros of P 200 (z).
Example 5. For k = 4, = 3, A(z) = z 2 + 1 and B(z) = z 3 − 1, we obtain Fig. 7 and Fig. 8 . In Fig. 7 , we plot the zeros of P 40 (z) while in Fig. 8 , we plot the zeros of P 70 (z).
Example 6. For k = 4, = 3, A(z) = 7z 5 − 2z + i and B(z) = −z 2 − 2z + 5, we obtain Fig. 9 and Fig. 10 . In Fig. 9 , we plot the zeros of P 50 (z) while in Fig. 10 , we plot the zeros of P 150 (z).
We end this paper with the following conjecture which we obtain based on numerical experiments. Conjecture 2. In the notation of Conjecture 1 with the condition > 1, we have the following. The conjecture excludes the case = 1 because these were settled [1, 2] . It is important to note that in all those cases, the real part of rational function is bounded unlike in Conjecture 2 where the real part is unbounded.
Final Remarks
Problem: Give a complete proof of Conjecture 1 as formulated by B. Shapiro. This still remains unknown to the author and it should be a project for a future work. Additionally to give a complete proof of Conjecture 2 as formulated in this paper.

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