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UNIMODULARITY OF ZEROS OF SELF-INVERSIVE POLYNOMIALS MATILDE N. LALÍN AND CHRIS J. SMYTH Abstract. We generalisea necessaryand sufficient conditiongivenby 2 Cohn for all the zeros of a self-inversive polynomial to be on the unit 1 0 circle.OurtheoremimpliessomesufficientconditionsfoundbyLakatos, 2 LosoncziandSchinzel.We applyourresulttothestudy ofapolynomial family closely related to Ramanujan polynomials, recently introduced n a by Gun, Murty and Rath, and studied by Murty, Smyth and Wang and J Lal´ın and Rogers. We prove that all polynomials in this family have 3 their zeros on the unit circle, a result conjectured by Lal´ın and Rogers on computational evidence. ] T N . h t 1. Introduction a m A self-inversive polynomial of degree d is a nonzero complex polynomial [ P(z) that satisfies 1 v P(z) = εzdP(1/z) (1) 4 7 for some constant ε. So if P(z) = d A zj is self-inversive, then A = 7 j=0 j j 0 εA for j = 0,...,d. In particular A = εA and A = ε¯A , so that ε . d−j P 0 d d 0 1 is necessarily of modulus 1. It is easy to check that if a polynomial has 0 2 all its zeros on the unit circle then it is self-inversive. In fact, Cohn [5] 1 : proved that a polynomial P(z) has all its zeros on the unit circle if and v i only if it is self-inversive and its derivative P′(z) has all its zeros in the X closed unit disk |z| 6 1. One might think that Cohn’s result completely r a settles the matter. Indeed, Cohn’s condition on P′ can be verified for a specific self-inversive polynomial, for instance by finding the zeros of P′, or checking thattheMahlermeasureofP′ isequal tothemodulusofitsleading coefficient. However, it may not be possible to use these methods for whole parametrized families of polynomials. We prove here an extension of Cohn’s result – see Theorem 1 below – which turns out to be more flexible than Cohn’s Theorem for applications. We apply it to some polynomial families, 2010 Mathematics Subject Classification. Primary 26C10; Secondary 11B68. Key words and phrases. reciprocalpolynomials,self-inversivepolynomials,unitcircle, Ramanujan polynomials. M.L. is supported by NSERC Discovery Grant 355412-2008, FQRNT Subvention établissement de nouveaux chercheurs 144987, and a start-up grant from the Univer- sité de Montréal. 1 2 M. N.LALÍN ANDC. J. SMYTH including the polynomial family P given by k (2π)2k−1 k 2k P (z) = (−1)jB B z2j+ζ(2k−1) z2k−1 +(−1)kz . k 2j 2k−2j (2k)! 2j j=0 (cid:18) (cid:19) X (cid:0) (cid:1) (2) Here, as usual, the B are Bernoulli numbers. It was known from work of 2j Lal´ın and Rogers [11] that the polynomials in this family had all their zeros onthe unit circle for k 6 1000.They conjectured that this was true forall k. However, this conjecture had resisted previous attempts to prove it. We do this in Theorem 8 below, as an application of our main theorem (Theorem 1). There have also been a number of results in recent years that provide sufficient, easier-to-verify conditions for a self-inversive polynomial to have its zeros on the unit circle. Lakatos [7] proved that a reciprocal polynomial d A zj with real coefficients that satisfies j=0 j P d |A | > |A −A |, (3) d j d j=0 X has its zeros on the unit circle. Moreover, if the inequality is strict, the zeros are simple. Schinzel [18] improved Lakatos’ result. His result – see Corollary 5 and Remark 6 below – follows from Theorem 4. On the other hand, Lakatos and Losonczi [9] proved that a reciprocal polynomial satisfying d−1 |(1+α)A | > |A −(1−α)A | (4) d j d j=1 X for some 0 6 α 6 1, has all its zeros on the unit circle. We have not been able to deduce their result using our approach, except in the case α = 0, see (3) and Remark 6 below, and in the case α = 1, which is Corollary 3. Their result was combined with Schinzel’s into a more general statement, proved in [10]. We will prove another result (Theorem 4) of the same general type that extends Schinzel’s criterion in a different direction. Forotherresultsconcerning self-inversive andreciprocalpolynomials, see the books by Marden [12, pp. 201–206], Rahman and Schmeisser [15] and Schinzel [17], as well as the papers by Ancochea [1], Bonsall and Marden [3, 4] and O’Hara and Rodriguez [14]. In Section 2 we state and prove our main result (Theorem 1), anddeduce various consequences of it, including Theorem 4 and Lemma 7. In Section 3 we apply Lemma 7 to prove that the polynomial family P given by (2) have k UNIMODULARITY OF ZEROS OF SELF-INVERSIVE POLYNOMIALS 3 all their zeros on the unit circle. Then in Section 4 we give some details of how the same result, due originally to Lal´ın and Rogers [11], can be proved in a similar way for two other polynomial families Q and W . k k 2. Results Our main theorem is the following. Theorem 1. Let h(z) be a nonzero complex polynomial of degree n having all its zeros in the closed unit disk |z| 6 1. Then for d > n and any λ on the unit circle, the self-inversive polynomial P{λ}(z) = zd−nh(z)+λh∗(z) (5) has all its zeros on the unit circle. Conversely, given a self-inversive polynomial P(z) having all its zeros on the unit circle, there is a polynomial h having all its zeros in |z| 6 1 such that P has a representation (5). In particular, we can take h(z) = 1P′(z). d Here h∗(z) = znh(1/z). Proof. Assume first that the polynomial h, of degree n, has all its zeros in |z| < 1. Then zd−nh(z) has all of its zeros in the open unit disk while h∗(z) has all of its zeros with absolute value greater than 1. Now take z such that |z| = 1. We have |h∗(z)| = |h(1/z)| = |h(z)| = |zd−nh(z)|. Assume for the time being that λ has absolute value greater (respectively less) than 1. Then |zd−nh(z)| is less (respectively greater) than |λh∗(z)|. Hence, by Rouché’s Theorem, P{λ}(z) has all of its zeros in |z| > 1 (respectively all of its zeros in |z| < 1). As the zeros of P{λ} are continuous functions of λ, we see that when |λ| = 1 then P{λ} must have all its zeros on the unit circle. The result under the weaker assumption that h has all its zeros in the closed unit disc |z| 6 1 then follows by continuity. Conversely, given P self-inversive of degree d with all its zeros on the unit circle, we note that differentiating (1) gives z zd−1 P(z) = P′(z)+ε P′(1/z), (6) d d which is of the form (5) with h(z) = 1P′(z), λ = ε and n = d−1. Further, d the zeros of P′ certainly all lie in |z| 6 1. This is because the zeros of P′ lie within the convex hull of the zeros of P, a result due originally to Gauss and Lucas — see [12, pp. 23–24] and [15, pp. 72–73, pp. 92–93] for relevant 4 M. N.LALÍN ANDC. J. SMYTH references. For completeness, and because of its elegance, we now reproduce a proof, by Cesa`ro, of this latter result, taken from [15]. Let P(z) have zeros α ,...,α , and suppose that P′(β) = 0. If β equals 1 d some α then β is clearly in the convex hull of all the α . So we can assume j i thatP(β) 6= 0,andthenonlogarithmicdifferentationwehaveP′(z)/P(z) = 1 , and hence, putting z = β and taking complex conjugates, that i z−αi 1 = 0. Then β−αi = 0, giving β = λ α , where Pi β−αi i |β−αi|2 i i i P P |β −α |−2 P i λ = . i |β −α |−2 j j Thus the λi are all positive and sPum to 1. (cid:3) It is not the case in general that if (5) holds for a particular h, and P has all its zeros on the unit circle then h must have all its zeros in |z| 6 1. For example, it is known (see e.g., [16, p. 9]) that the polynomial P(z) = zk(z3 −z −1)+(z3 +z2 −1) has all its zeros on the unit circle for k = 0,1,...,7, while h(z) = z3 −z −1 has a zero 1.3247... > 1. However, in that direction we can say the following. Observation 2. For any d > n, let P{λ}(z) = zd−nh(z)+λh∗(z). d {λ} If there is a K > 0 such that for every d > K, P (z) has all its zeros on d the unit circle, then h(z) has all its zeros in the unit circle |z| 6 1. Proof. Assume that h(z) has a zero z with |z | > 1. Take δ < |z |−1, so 0 0 0 that |z | − δ > 1. Then for z on the circle |z − z | = δ and d sufficiently 0 0 large we have |zd−nh(z)| > (|z |−δ)d−n|h(z)| > |λh∗(z)|. 0 Hence, by Rouché’s Theorem, P(λ)(z) has the same number of zeros in the disc |z−z | < δ aszd−nh(z) has, namely at least one. This disc is completely 0 outside the unit circle. (cid:3) As a simple consequence of Theorem 1, we obtain the following known result. Corollary 3 (LakatosandLosonczi[8]). A self-inversivepolynomialP(z) = d A zj satisfying j=0 j d−1 P |A | > 1 |A | (7) d 2 j j=1 X has all its zeros on the unit circle. UNIMODULARITY OF ZEROS OF SELF-INVERSIVE POLYNOMIALS 5 Proof. We take Adzd2 +Ad−1zd2−1 +...+Ad+1z + 12Ad, d even, 2 2 h(z) =  d−1 d−3  Adz 2 +Ad−1z 2 +...+Ad+1, d odd. 2 Then, since by(7) the leading coefficient of h(z) is at least as big as the sum of the moduli of the other coefficients, h(z) has no zeros in |z| > 1. Now P(z)/Ad = z⌊d+21⌋h(z) + εh∗(z), where ε = A0/Ad. Therefore, by Theorem 1, P(z) has all its zeros on the unit circle. (cid:3) We can also deduce the next result from Theorem 1. Theorem 4. A self-inversive polynomial P(z) = d A zj satisfying j=0 j d−1 P |A | > 1 inf |A −µA | (8) d 2 µ∈C j j+1 |µ|=1Xj=0 has all its zeros on the unit circle. Proof. We first consider the case µ = 1, and look at P(z)(z −1). We take d d Adz2 +(Ad−1 −Ad)z2−1 +... + A −A , d even,  d d+1 2 2 h(z) =  (cid:16) (cid:17)   d+1 d−1  Adz 2 +(Ad−1 −Ad)z 2 +...  + Ad+21 −Ad+23 z + 21 Ad−21 −Ad+21 , d odd.   (cid:16) (cid:17) (cid:16) (cid:17) Again, the absolute value of the leading coefficient of h(z) is at least as big as the sum of the moduli of the other coefficients. This implies that h(z) has no zeros in |z| > 1 and we can apply the Theorem 1 with λ = −ε to conclude that P(z)(z −1), and therefore P(z), has all its zeros on the unit circle. To obtain the result in general, let Q(z) := P(µz) and apply what has been proved to Q(z), using the fact that |µjA −µj+1A | = |A −µA |. j j+1 j j+1 (cid:3) Corollary 5 (Schinzel [18]). A self-inversive polynomial P(z) = d A zj j=0 j satisfying P d |A | > inf cA −µd−jA , (9) d j d c,µ∈C |µ|=1Xj=0 (cid:12) (cid:12) (cid:12) (cid:12) must have all of its zeros on the unit circle. 6 M. N.LALÍN ANDC. J. SMYTH Proof. Considertheself-inversive polynomialP(z) = d A zj andassume j=0 j that (9) is satisfied. Then we claim that (8) holds. We first check this for P µ = 1. Indeed, by applying twice the triangle inequality, d−1 d−1 d−1 |A −A | 6 |A −1/cA |+ |1/cA −A | j j+1 j d d j+1 j=0 j=0 j=0 X X X d = 2 |A −1/cA |−2|1−1/c||A | j d d j=0 X 6 2|1/cA |−2|1−1/c||A | d d 6 2|A |. d Then, again applying the result to P(µz) for general µ on the unit circle gives the full result. (cid:3) Remark 6. The condition (3) of Lakatos is the special case c = µ = 1 of (9). We next show that the result of Theorem 1 still holds if P{λ}(z) is per- turbed by a small self-inversive ‘error’ polynomial. Lemma 7. Let h and λ be as in Theorem 1, with |h(z)| > c > 0 for |z| = 1. Let e(z) be a polynomial of degree m such that |e(z)| 6 c for |z| = 1. Then for k > max{m,n}, the self-inversive polynomial z2k−nh(z)+zke(z)+λ(h∗(z)+zk−me∗(z)) has all its zeros on the unit circle. Proof. We first assume that for some positive c′ < c we have |e(z)| 6 c′ < c for all z with |z| = 1. Now h (z) = zk−nh(z)+e(z) is a polynomial of degree e k. Because |zk−nh(z)| > c > c′ > |e(z)| for |z| = 1, Rouché’s Theorem tells us that h has all its zeros in the open unit disk |z| < 1. Also h∗(z) = e e zkh (1/z) = h∗(z) +zk−me∗(z). Now apply Theorem 1 with h replaced by e h and d replaced by k. e The general case, where we assume only that |e(z)| 6 c for |z| = 1, then follows by continuity. (cid:3) 3. Application to the polynomials P k Let k > 2, and, as in [11], define P (z) by (2). The study of this poly- k nomial is motivated by the fact that it appears in a formula by Ramanujan UNIMODULARITY OF ZEROS OF SELF-INVERSIVE POLYNOMIALS 7 [2, p. 276]: ∞ ∞ 1 1 1 P (z) = (−z)−(k−1) −zk−1 , 2zk k n2k−1(e2πnz −1) n2k−1(e2πn/z −1) n=1 n=1 X X (10) valid for z 6∈ iQ. A variant of the polynomial P (z) (without the term with k the ζ-value) was first considered by Gun, Murty, and Rath [6] in the context of expressing the special value of the ζ-function as an Eichler integral that could yield information about the algebraic nature of the number. Murty, Smyth, and Wang [13] studied this variant of the polynomial and found that all but four of its zeros lie on the unit circle. Finally, other variants of P (z) were considered in [11], and were shown to have all their zeros on k the unit circle. However, the methods from [11] were not sufficient to prove that P (z) itself has all its zeros on the unit circle. k Theorem 8. For all k ∈ N, the polynomial P has all its zeros on the unit k circle. We need the following straightforward bounds. Put q = ζ(2j)ζ(2k−2j) for j ζ(2k) j = 0,1,...,k andδ = q −ζ(2j).Notethatδ = 0andδ > 0for0 < j < k. j j 0 j Lemma 9. (i) For n > 2 we have n+1 1 < ζ(n) < 1+ ·2−n. n−1 (ii) For k > 2 and j = 1,2,...,k −1 we have ζ(2k−2j) 0 < −1 < 3·4j−k. ζ(2k) (iii) For k > 11 we have ζ(2k−1) 0 < −1 < 11 ·4−k. ζ(2k) 5 (iv) For k > 4 and j = 1,2,...,k −1 we have 21·4−k if j = 1; |δ −δ | < j−1 j 3·4−k 4j + 2j−1 if j > 2. ( 2j−3 (v) For k > 2 and 2 6 j 6 k/2 (cid:16)we have q(cid:17)= q and j k−j |q −q | < 3·4−k 4j + 2j−1 + 2j−1 ·41−j. j−1 j 2j−3 2j−3 (vi) For k > 4 and 4 6 r 6 k we(cid:16)have (cid:17) r |δ −δ | < 5·4r−k. j−1 j j=1 X 8 M. N.LALÍN ANDC. J. SMYTH (vii) For k > 10 and r > 4 we have k ⌊ ⌋ 2 |q −q | < 5·2−k + 12 ·4−r. j−1 j 7 j=r+1 X Proof. Part (i)-(iii) are easy – see [13, Lemmas 4.4 and 4.6] for (i) and (ii). For (iv), we have, using (i) and (ii), that δ = ζ(2j) ζ(2k−2j) −1 j ζ(2k) (cid:16) (cid:17) < 1+ 2j+1 ·4−j ·3·4j−k 2j−1 (cid:16) (cid:17) = 3·4−k 4j + 2j+1 . 2j−1 (cid:16) (cid:17) Hence |δ −δ | = δ < 21·4−k, 0 1 1 while in general |δ −δ | ≤ max(δ ,δ ), j−1 j j−1 j from which the result for j > 2 follows. For (v), we have |q −q | 6 |δ −δ |+ζ(2j −2)−ζ(2j), j−1 j j−1 j which gives the result using (ii) and (iv). For (vi), we have, using (iv), that r r |δ −δ | < 21·4−k +3·4−k 4j + 2j−1 j−1 j 2j−3 Xj=1 Xj=2 (cid:16) (cid:17) r < 4r−k 21·4−r +3 4j−r +3·3·(r −1)·4−r ! j=−∞ X < 4r−k 21/44+4+27/44 < 5·4r(cid:0)−k. (cid:1) For (vii), as we have j > 5 in the summand, and r > 4, we obtain, using (v), that k k ⌊ ⌋ ⌊ ⌋ 2 2 |q −q | 6 3·4−k 4j + 2j−1 + 2j−1 ·41−j j−1 j 2j−3 2j−3 jX=r+1 j=Xr+1(cid:16) (cid:16) (cid:17) (cid:17) k ⌊2⌋ ∞ < 3·4j−k +3·4−k · 9 k −r + 9 41−j 7 2 7 j=−∞ j=r+1 X (cid:0)(cid:4) (cid:5) (cid:1) X 6 3·4−k/2 · 4 +2−k · 27 k −4 2−k + 9 ·4−r · 4 3 7 2 7 3 6 2−k(4+1)+ 12 ·4−(cid:0)r. (cid:0) (cid:1) (cid:1) 7 UNIMODULARITY OF ZEROS OF SELF-INVERSIVE POLYNOMIALS 9 (cid:3) We also need the standard identity B 2ζ(2j) 2j = (−1)j+1 , (2j)! (2π)2j valid for all j > 0, since B = 1 and ζ(0) = −1/2. From this we see that 0 k 2 P (z) = (−1)k (−z2)jζ(2j)ζ(2k−2j)+ζ(2k−1)(z2k−1 +(−1)kz). k π j=0 X Thus P has leading coefficient −ζ(2k)/π. To show that P has all its ze- k k ros on the unit circle it is sufficient to show that the monic polynomial M (z) = − π (z2 + 1)P (z) has all its zeros on the unit circle. (Most of k ζ(2k) k the coefficients of M are very small, making it easier to work with than k − π P , whose coefficient of z2j is close to 2(−1)j+k+1 for most j.) We ζ(2k) k easily calculate that πζ(2k−1) M (z) = z2k+2 +(−1)k− (z2k+1 +z2k−1 +(−1)kz3 +(−1)kz) k ζ(2k) k +2 (−1)jz2k+2−2j(q −q ). j−1 j j=1 X Hence, using the fact that for k odd and j = (k+1)/2 we have q −q = 0, j−1 j we obtain πζ(2k−1) z−(k+1)M (z) = zk+1 +(−1)kz−(k+1) − (zk +zk−2 +(−1)k(z2−k +z−k)) k ζ(2k) k ⌊ ⌋ 2 +2 (q −q )(−1)j(zk+1−2j +(−1)kz−(k+1−2j)) j−1 j j=1 X = zk+1 +(−1)kz−(k+1) −π(zk +zk−2 +(−1)k(z2−k +z−k)) r +2 (ζ(2j −2)−ζ(2j))(−1)j(zk+1−2j +(−1)kz−(k+1−2j)) j=1 X k ⌊ ⌋ 2 +2 (q −q )(−1)j(zk+1−2j +(−1)kz−(k+1−2j)) j−1 j j=r+1 X r +2 (δ −δ )(−1)j(zk+1−2j +(−1)kz−(k+1−2j)) j−1 j j=1 X ζ(2k−1) −π −1 (zk +zk−2 +(−1)k(z2−k +z−k)) ζ(2k) (cid:18) (cid:19) = z−(k+1)H (z)+z−(k+1)E (z), r r 10 M. N.LALÍN ANDC. J. SMYTH where H (z) = z2k+2 +(−1)k −π(z2k+1 +z2k−1 +(−1)k(z3 +z)) r r +2 (ζ(2j −2)−ζ(2j))(−1)j(z2k+2−2j +(−1)kz2j) j=1 X ζ(2k−1) E (z) = −π −1 z2k+1 +z2k−1 +(−1)k(z3 +z) r ζ(2k) (cid:18) (cid:19) r (cid:0) (cid:1) +2 (δ −δ )(−1)j(z2k+2−2j +(−1)kz2j) j−1 j j=1 X k ⌊ ⌋ 2 +2 (q −q )(−1)j(z2k+2−2j +(−1)kz2j). j−1 j j=r+1 X Here H is the main polynomial, with E the error polynomial. r r We can rewrite H (z) as r H (z) = z2k+2−2rh (z)+(−1)kh∗(z), r r r where r h (z) = z2r −πz2r−1 −πz2r−3 +2 (ζ(2j −2)−ζ(2j))(−1)jz2r−2j, (11) r j=1 X and E (z) as r E (z) = zk+2e (z)+(−1)kze∗(z), r r r where r ζ(2k−1) e (z) = −π −1 (zk−1 +zk−3)+2 (δ −δ )(−1)jzk−2j r j−1 j ζ(2k) (cid:18) (cid:19) j=1 X k ⌊ ⌋ 2 +2 (q −q )(−1)jzk−2j. j−1 j j=r+1 X We now take r = 4. We then have the following bound. Lemma 10. For k > 11 and |z| = 1 we have |e (z)| 6 0.019. 4 Proof. Take z with |z| = 1. Applying Lemma 9 (iii), (vi) and (vii) we have r ζ(2k−1) |e (z)| 6 −π −1 (zk−1 +zk−3) + 2 (δ −δ )(−1)jzk−2j r j−1 j ζ(2k) (cid:12) (cid:12) (cid:12)(cid:12) (cid:18) (cid:19) (cid:12)(cid:12) (cid:12) Xj=1 (cid:12) (cid:12) (cid:12) (cid:12) k (cid:12) ⌊ ⌋ (cid:12) (cid:12) (cid:12) 2 (cid:12) (cid:12) (cid:12) + 2 (q −q )(−1)jzk−2j j−1 j (cid:12) (cid:12) (cid:12) j=r+1 (cid:12) (cid:12) X (cid:12) < 2π · (cid:12)11 ·4−k +2·5·4r−k +2·5·2(cid:12)−k +2· 12 ·4−r, (cid:12) (cid:12) 5 7 (cid:12) (cid:12) which is less than 0.019 for r = 4 and k > 11. (cid:3)