p'(z) = nl[(z-cj), \Cj\<l (1 <;<"-!). 7 = 1

Transcription

1 proceedings of the american mathematical society Volume 3, Number 4, December 99 ON THE SENDOV CONJECTURE FOR SIXTH DEGREE POLYNOMIALS JOHNNY E. BROWN (Communicated by Clifford J. Earle, Jr.) Abstract. The Sendov conjecture asserts that if p{z) = nk=l{z - zk) is a polynomial with zeros \zk\ <, then each disk \z - zk\ < I, ( < k < n) contains a zero of p (z). This conjecture has been verified in general only for polynomials of degree n = 2, 3, 4, 5. If p{z) is an extremal polynomial for this conjecture when n = 6, it is known that if a zero \z,\ < A6 = then \z - z \ < contains a zero of p (z). (The conjecture for n = 6 would be proved if A6 =.) It is shown that À6 can be improved to A6 = 63/64 = The well-known conjecture of Sendov [6, Problem 4.5] posed in 962 states that if p(z) = n =i(z-z/fc) ' \zk\ î (^ < k < n), then each disk \z- zk\ < ( < k < n) contains a zero of p'(z). Surprisingly this conjecture has been proved only for polynomials of degree n < 5 and in a few special cases [2-4, 9-5]. (See Marden [8] for an excellent expository article on this conjecture.) The case n = 5 was proved in 969 [9], while the general case for n = 6 is still open. The Sendov conjecture can be viewed as an extremal problem over a compact family of functions as follows. Let «^ denote the family of all monic polynomials of the form n p(z) = Y[(z-zk), \zk\<l (l<k<n). k=\ Thus, by the classical Gauss-Lucas Theorem we have p'(z) = nl[(z-cj), \Cj\<l ( <;<"-!). 7 = Define I(zk) = min,^^,., \zk - C;, I(p) = maxx<k<ni(zk), and I(0>n) = sup e3 I(p). Phelps and Rodriguez [0] proved that there exists an extremal polynomial pt(z) = Yll=x(z - z*k) 3Pn such that I(pJ = I( &' ) (see Lemma Received by the editors November 5, 989; presented to the 96th Annual Meeting of the American Mathematical Society, January Mathematics Subject Classification (985 Revision). Primary 30C5; Secondary 30C American Mathematical Society /9 $.00+ $.25 per page

2 940 J. E. BROWN A). It thus suffices to prove the conjecture for pt i.e., show I(z*k) < for < k < n. They also proved a result which implies that if n > 5 and z*, for some j, satisfies \z*\ < Xn, where Xn is the unique root of (l+x2)(l+x)"~3 -n = 0, then I(z*) <. In the special case n 6 their result gives X6 = Of course the conjecture for n 6 would be proved if X6 =. The purpose of this paper is to improve the bound X6 : Theorem. If pt(z) = rxc=i(z ~ zl) '5 an extremal polynomial for I(â 6) and \Zj\ < 63/64, for some j =, 2,..., 6, then the disk \z - z*\ < contains a zero of p\(z). This result now reduces the conjecture for the case n = 6 to considering the zeros of pt close to z =. In this direction, we point out that Goodman, Rahman, and Ratti [5] have proved that if zq is a zero of p(z) and \zq\ =, then the disk \z - zq/2\ < \ contains a zero of p'(z). Thus for boundary zeros a result stronger than the Sendov conjecture holds. Our proof of the main result involves a blend of analytic and geometric ideas. The method presented here can be used to increase the bounds for Xn for all n > 6. We make use of the following known results: Lemma A (Phelps and Rodriguez [0]). There exists an extremal polynomial pt Pn such that I(pt) = I( P ) Moreover, pt(z) has a zero on each subarc of \z\ of length n. Lemma B (Bojanov, Rahman, and Szynal [2]). If Q(z) is a monk polynomial of degree n, ß(0) = 0,and Q'(z) 0 in \z\<r, then \Q(z)\ >R" -(R-X)" We also need the following estimates: Lemma. Let p(z) = ( «)nli(z»n i!(*-cy). If (a) > I, then Cj açj - > l+a-a for\z\=x<rsinn/n. - zk)e &n' <a < l and p'w 7 =,2-«-. Lemma 2. If p(z) = (z -a) Yik=\ (z _ zk> is an extremal polynomial for I(3 n), 0 < a < and p'(z) = n Yl]Z\i.z ~,) men n7 = Ci «j -l < n - (n - 3)a 4az l+a Lemma 3. Let p(z) = (z - a)t[k=x(z - zk) be an extremal polynomial for (i) // 0 < a < I, then (a) <.

3 THE SENDOV CONJECTURE 94 (ii) If \ < a < and (a) >, then there exists a zero Ç0 = a + p0e'e of p'(z) such that p0 > and cos0o > /25 - a. Proof of Theorem. Suppose pt(z) = Ytk=\(z~z*k> is an extremal polynomial for 7(^6) and suppose that \z*\ < 63/64 for some j. By a rotation, if necessary, we may suppose that z* = a, 0<a<63/64. Hence 5 pt(z) = (z-a)l[(z-zk), \zk\<l k = l,2,...,5. k=l Assume by way of contradiction that (a) >. By Lemma 3(i) we must then have \ < a <. Thus, making use of Lemma 3(ii) we can assert that there in exists a critical point C0 = a + Poe w^ta Po > anc* cos0o > /25 - a. It follows that Co flco- [( - a2)2 + a2-2a(l- a2)cosdq]x/2 >,2x2 [(I - ay + a2-2a(l - a2)(l/25 - a)]x/2' Assuming as we may that (, - a)/(açx - ) < \(Çj - a)/(açj - l)\ for the critical points Ç, ^ Ç0 > we appty Lemma 2 with n = 6 to get (2) C al-l C0-a aco-l Using () and (2) we obtain 7= liât i - < 6-3fl-4a2/(l+a2)' < [(l-ay + a -2a(l-aí)(l/25-a)V'í = (3) < 6-3a-4a2/(l+a2) Simple numerical calculations show that <t>(a)< f for3-*-64-2 ^ ^ 63 I +a-az) From (3) we conclude that (, - a)/(açx -l) <l/(l+a-a Lemma. D We now turn to the proofs of the lemmas. contradicting Proof of Lemma. Assume by way of contradiction that \(ÇX- a)/(açx - ) < /( + a - a2). Hence we get (Ç, - a)/(a x - ) = reie, where 0 < r < /( + a - a2). This gives (, = (a - re'e)/(l - are'6) and so which contradicts (a) >. \l-arel D - ar

4 942 J. E. BROWN Proof of Lemma 2. Suppose that p(z) is extremal and (4) p(z) = (z-a)y[(z-zk), \zk\<l (l<k<), k=\ (5) p'(z) = nh(z-cj), \Cj\<l (l<j<), 7= and 0 < a < I. Let z = T(w) = (w - a)/(aw - ) and note that where p(t(w)) = p0(w)(aw - l)~", p0(w) = Aw(w"~ +bn_xw"~ +-\-bx). The zeros of p0(w) are 0, wx,w2,..., wn_x, where wk - T(zk), < k < n -. Hence we see that (6) and (7) Differentiating p(t(w)) where gives *,=(-ir'n w. k=l K---E w, k=l dp(t(w)) dp(t(w)) dz dw -^dz-ji-d^=a^w)[-a{aw-l) (8) Ai/aPo(w) = np0(w) +Í--WJ p'0(w) = B ]\(w - yß is the polar derivative of p0(w) with respect to I/a (see Marden [7, p. 44]). Thus we get (9) p (T(w)) = Ax/ap0(w)[-a(aw -«-i - ) dw/dz] (where ' denotes differentiation with respect to z). It follows from (5), (8), and (9) that the zeros of p'(z) and Ax/apa(w) are related by -, l ;io) Vj = acj - ' I <j <n- I. Next, we see after a simple check that Ai/aPo(w) = B w"-x n + ab 7 =

5 and so () THE SENDOV CONJECTURE 943 EN«n + ab 7 = By Lemma A there is a zero on each subarc of \z\ = of length n. Hence without loss of generality \zn_x=ei6\ (If Im zn_x < 0, simply consider p(z).) O<0o<7t/2 Zn-2 = e 60 + n<dx< 2n. It is easy to check that ae'a«- ae^ - 2a-( -a2)cos0o 2a + ( + a2)cos0o ~ (I+a2)-2acosd0 (I + a2)+ 2acosd0 _ 4fl(l+q2)(l-cos26i0) 4a " (l+a2)2-4a2cos20o " I + a2' Using (6), (7), (), and the above inequality, we obtain n7 = tj- «j i The proof of the lemma is complete. < < n-atrkl\t(zk)\ n-aelzl^nz,) i M-4a2/(l+a2)-a(«-3)' Note that () immediately gives rj ji IYj\ ^ l/(n-()a). estimate is not good enough for our purposes. Proof of Lemma 3. Let p(z) = (z - a) n^=i(z _ zk) ^e extremal. D However this (i) 0 < a <. Clearly 7(0) < by the classical Gauss-Lucas Theorem. Next, from Lemma 2 with n = 6 we see that d2) n 7= Í,-««i 6-3a-4a2/(l+a2)' Assuming (C, - a)/(açx - ) < (Ç, - a)/(a^j - ) for ; =, 2,..., 5 we have from (2) and an easy calculation that C,-«flC,-l Thus, by Lemma, we get (a) < (6-3a-4a7(l+ait)) 2^/5 < + a - a

6 944 J. E. BROWN (ii) \ < a < and (a) >. For fixed a we apply Lemma B with Q(z) =p(z + a), n = 6, R =, and X = - ( - a)x/6 to conclude that (3) p(z) >l-(l-a)6 = a> p(0) for \z - a\ = X = I - (I - a)x/6. Note that A < \. Since p'(z) ^ 0 in \z - a\ < and the degree of p(z) is six, we see that by Alexander's Theorem (See Marden [7, p. 0, exercise 2]) that p(z) is univalent in \z - a\ < j. Thus from (3) we see that there exists a unique zq such that \zq- a\ < X with p(z0) = p(0). We may assume that Imz0 > 0 (if not, consider p(~z)). Let ro be the perpendicular bisector of the segment from 0 to z0, and let H^ and Hq be the closed halfplanes bounded by ro. By a variant of the Grace- Heawood Theorem (see [, Lemma ] for example) we see that p'(z) has a zero in both Hq and HQ~. Let coq = p0(x0, y0) + iv0(x0, y0) be the intersection of T0 with the circle \z - a\ = with v0 > 0. It follows that p'(z) has a zero if? C0 = a + p0e with p0 > and cos#0 > ^0(^0, v0) - a. (See Figure.) It suffices to prove that p0(x0, y0) > /25. Figure

7 THE SENDOV CONJECTURE 945 Let z* = x* + iy* be the point on \z - a\ = X such that the line through 0 and z* is tangent to \z - a\ = X and y* > 0. Hence x* = (a2 - X2)/a. If z0 = r^e"0 then by letting z = re"0 = x + iy (r < r0) be the point on \z - a\ = X with y > 0 and x < x*, we conclude that p0(x0, y0) > p0(x, y). Thus, it suffices to prove that p0(x, y) > /25 where (x - a)2 + y2 = X2, a - X < x < x*, and y > 0. A calculation shows that x(x2 + 3a2) + 2aß - y/y2(&ax + 4ß- ß2) (4) *<* '>--2(2^T^-' where /? = A - a. Observe that since a - X < x < x* we get y = X - (x - a) < 2X(x - a + X) and 2ax + ß < -ß. From (4) we conclude that x(x2 + 3a2) + 2aß - \ 2X(x -a + X)(8ax + 4ß- ß2) Paix, y) >-~~^2ß-~ßv It is enough to show that px > /25. Now px > /25 if and only if (5) F(x) = cxx + c2x + c3 > 0, a-x<x<x*, where cx = (X +3a ) - l6ax, c2 = 2ß(2a + 0M)(X2 + 3a2) - 2A/J(4 - ß) - l6ax(x - a), c3 = ß2(2a )2-2A/?(4 - ß)(X - a), (ß = X -a). An easy check shows that cx > 0 and so (5) follows if A = c2-4cxc3 < 0. A brief calculation shows that A = 4X(X -a) A0, where A0 = [(X + a)(4 - ß) + 8a)[8aX + (X + a){x(4 - ß) - (4a + 0.6)(A2 + 3a2)}] + (X + a)[l6a(x + a)(2a )2 + 2(4 - ß)(X2 + 3a2)2-32aX(4 - ß)]. Finally, a computation shows that A0 < 0 for \ < a <. (It can easily be checked numerically that A0 < ) Thus (5) holds and hence p0 > /25. D The proof of the Sendov conjecture has been elusive for more than twentyfive years and only verified in a few special cases. The method of proof in the case n = 5 does not seem to be useful for n > 6. It is not surprising to see the different ideas used to prove our results.

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