proceedings of the
american mathematical society
Volume 113, Number 4, December 1991
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\ < 1 , then each disk \z - zk\ < I , (1 < 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 = 0.626997...
then \z - z \ < 1 contains a zero of p (z). (The conjecture for n = 6 would
be proved if A6 = 1 .) It is shown that À6 can be improved to A6 = 63/64 =
0.984375.
The well-known conjecture of Sendov [6, Problem 4.5] posed in 1962 states
that if p(z) = n£=i(z-z/fc) ' \zk\ —î (^ < k < n), then each disk \z- zk\ < 1
(1 < 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,
9-15]. (See Marden [8] for an excellent expository article on this conjecture.)
The case n = 5 was proved in 1969 [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
(1 <;<"-!).
7= 1
Define I(zk) = min,^^,.,
\zk - C;|, I(p) = maxx<k<nI(zk), and I(0>n) =
sup e3¡¡I(p). Phelps and Rodriguez [10] 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, 1989; presented to the 96th Annual Meeting of the American Mathematical Society, January 1990.
1980 Mathematics Subject Classification (1985 Revision). Primary 30C15; Secondary 30C10.
© 1991 American Mathematical Society
0002-9939/91 $1.00+ $.25 per page
939
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
J. E. BROWN
940
A). It thus suffices to prove the conjecture for pt i.e., show I(z*k) < 1 for 1 <
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*) < 1. In the special case n —6 their result gives X6= 0.626997....
Of course the conjecture for n — 6 would be proved if X6 = 1. 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 = 1, 2, ... , 6, then the disk \z - z*\ < 1 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| = 1 . 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\ = 1,
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 [10]). There exists an extremal polynomial
pt £ ¿Pn such that I(pt) = I(¿P¿) ■ Moreover, pt(z) has a zero on each subarc
of \z\ —1 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,
\Q(z)\ >R" -(R-X)"
then
for\z\=X<Rsinn/n.
We also need the following estimates:
«)nLi(z - zk)e &n' ° <a < l and p'w
»n£i!(*-Cy). If 1(a) > I, then
Lemma 1. Let p(z) = (
Cj
aÇj - 1
>
l+a-a
7 = 1,2-«-
1.
Lemma 2. If p(z) = (z -a) Yik=\ (z _ zk> iS an extremal polynomial for I(3°n),
0 < a < 1 and p'(z) = n Yl]Z\i.z ~ £,) ■men
n-l
Ci
n «j
7= 1
-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 1(a) < 1.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
941
THE SENDOV CONJECTURE
(ii) If \ < a < || and 1(a) > 1, then there exists a zero Ç0= a + p0e'e° of
p'(z) such that p0 > 1 and
cos0o > 1/25 - 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 1(a) > 1 . 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 > 1 anc* cos0o > 1/25 - a. It
follows that
1
Co
flCo-1
[(1 - 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 - 1)| < \(Çj - a)/(aÇj - l)\ for the
critical points Ç, ^ Ç0>we aPPty Lemma 2 with n = 6 to get
(2)
C
C0-a
aCo-l
aL-l
li7=1
ât i - 1
<
6-3fl-4a2/(l+a2)'
Using (1) and (2) we obtain
(3)
< [(l-ay
<
1
+ a¿-2a(l-aí)(l/25-a)V'í
=
6-3a-4a2/(l+a2)
Simple numerical calculations show that
<t>(a)<
1
I +a-az)
ffor3-*-642 ^ ^ 63
From (3) we conclude that |(£, - a)/(aÇx -l)|<l/(l+a-a
Lemma 1. D
contradicting
We now turn to the proofs of the lemmas.
Proof of Lemma 1. Assume by way of contradiction that \(ÇX- a)/(aÇx - 1)| <
1/(1 + a - a2). Hence we get (Ç, - a)/(a£x - 1) = reie, where 0 < r <
1/(1 + a - a2). This gives (, = (a - re'e)/(l - are'6) and so
\l-arel
which contradicts 1(a) > 1.
1 - ar
D
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
942
J. E. BROWN
Proof of Lemma 2. Suppose that p(z) is extremal and
n-l
(4)
p(z) = (z-a)Y[(z-zk),
\zk\<l
(l<k<n-l),
k=\
n-l
(5)
p'(z) = nH(z-CJ),
\Cj\<l
(l<j<n-l),
7=1
and 0 < a < I . Let z = T(w) = (w - a)/(aw - 1) and note that
p(T(w)) = p0(w)(aw - l)~" ,
where
p0(w) = Aw(w"~
+bn_xw"~
The zeros of p0(w) are 0, wx,w2,
n - 1. Hence we see that
+-\-bx).
... , wn_x, where wk - T(zk),
1< k <
n-l
*,=(-ir'n
w.
(6)
k=l
and
n-l
K-1--E
(7)
w,
k=l
Differentiating p(T(w))
gives
dp(T(w))
dp(T(w)) dz
-^dz-Ji-d^=A^w)[-a{aW-l)
dw
-n-l,
l
where
(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
Thus we get
(9)
p (T(w)) = Ax/ap0(w)[-a(aw
- 1)
(see Marden [7, p. 44]).
-«-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
;io)
Vj = aCj - 1 '
I <j <n-
I.
Next, we see after a simple check that
n-l
Ai/aPo(w) = B
w"-x + .-.+
n + ab n-l
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
7= 1
THE SENDOVCONJECTURE
943
and so
n-l
EN« n + ab
(11)
7= 1
n-l
By Lemma A there is a zero on each subarc of \z\ = 1 of length n . Hence
without loss of generality
\zn_x=ei6\
O<0o<7t/2
60 + n<dx<
Zn-2 = e
(If Im zn_x < 0, simply consider p(z).)
2n.
It is easy to check that
ae'a« - 1
ae^ - 1
2a-(1 -a2)cos0o
~ (I+a2)-2acosd0
2a + (1 + a2)cos0o
(I + a2)+ 2acosd0
_ 4fl(l+q2)(l-cos26i0)
4a
" (l+a2)2-4a2cos20o
" I + a2'
Using (6), (7), (11), and the above inequality, we obtain
n-l
n
7= 1
tj-°
i
<
i
«j
n-aTrkl\T(zk)\
<
1
n-aElZl^nz,)
_1_
M-4a2/(l+a2)-a(«-3)'
The proof of the lemma is complete.
D
Note that (11) immediately gives rj£ji IYj\^ l/(n-(n-l)a).
estimate is not good enough for our purposes.
However this
Proof of Lemma 3. Let p(z) = (z - a) n^=i(z _ zk) ^e extremal.
(i) 0 < a < §. Clearly 7(0) < 1 by the classical Gauss-Lucas Theorem.
Next, from Lemma 2 with n = 6 we see that
Í,-«
d2)
n «i
7=1
1
1
6-3a-4a2/(l+a2)'
Assuming |(C, - a)/(aÇx - 1)| < |(Ç, - a)/(a^j - 1)| for ; = 1, 2, ... , 5 we
have from (12) and an easy calculation that
C,-«
flC,-l
1
(6-3a-4a7(l+ait))
2^1/5
<
Thus, by Lemma 1, we get 1(a) < 1
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
1 + a - a1
J. E. BROWN
944
(ii) \ < a < §| and 1(a) > 1. For fixed a we apply Lemma B with
Q(z) =p(z + a), n = 6, R = 1, and X = 1 - (1 - a)x/6 to conclude that
(13)
|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\ < 1 and the degree of p(z) is six, we see that by Alexander's Theorem
(See Marden [7, p. 110, exercise 2]) that p(z) is univalent in \z - a\ < j.
Thus from (13) 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 GraceHeawood Theorem (see [1, Lemma 1] 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\ = 1 with v0 > 0. It follows that p'(z) has a zero
if?
C0 = a + p0e ° with p0 > 1 and cos#0 > ^0(^0, v0) - a. (See Figure 1.) It
suffices to prove that p0(x0, y0) > 1/25 .
Figure 1
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
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) > 1/25 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-
(14)
*<*•'>--2(2^T^-'
ß2)
where /? = A - a . Observe that since a - X < x < x* we get
2
2
2
y = X - (x - a) < 2X(x - a + X) and 2ax + ß < -ß.
From (14) we conclude that
x(X2 + 3a2) + 2aß - \¡2X(x -a + X)(8ax + 4ßPaix, y) >-~~^2ß-~ßv
ß2)
It is enough to show that px > 1/25 .
Now px > 1/25 if and only if
(15)
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 + 0.08)2 - 2A/?(4- ß)(X - a),
1
1
(ß = X -a).
An easy check shows that cx > 0 and so (15) 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 + 0.16)(A2+ 3a2)}]
+ (X+ a)[l6a(X + a)(2a + 0.08)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 < -0.009.) Thus (15) holds and hence p0 >
1/25. 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.
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
J. E. BROWN
946
References
1. A. Aziz, On the zeros of a polynomial and its derivative, Bull. Austral. Math. Soc. 31 (1985),
245-255.
2. B. Bojanov, Q. I. Rahman, and J. Szynal, On a conjecture of Sendov about the critical points
of a polynomial,Math. Z. 190 (1985), 281-285.
3. D. A. Brannan, On a conjecture ofllieff, Proc. Cambridge Philos. Soc. 64 (1968), 83-85.
4. J. E. Brown, On the Ilieff-Sendovconjecture,Pacific J. Math. 135 (1988), 223-232.
5. A. W. Goodman, Q. I. Rahman, and J. Ratti, On the zeros of a polynomial and its derivative,
Proc. Amer. Math. Soc. 21 (1969), 273-274.
6. W. K. Hayman, Research problems in function theory, Athlone Press, London, 1967, 56 pp.
7. M. Marden, Geometry of polynomials, Amer. Math. Soc. Surveys, no. 3, 1966.
8. _,
Conjectures on the critical points of a polynomial, Amer. Math. Monthly 90 (1980),
267-276.
9. A. Meir and A. Sharma, On Ilyeff's conjecture,Pacific J. Math. 31 (1969), 459-467.
10. D. Phelps and R. Rodriguez, Some properties of extremal polynomials for the Ilieff Conjec-
ture, Kodai Math. Sem. Report 24 (1972), 172-175.
U.Z. Rubinstein, On a problem ofllyeff Pacific J. Math. 26 (1968), 159-161.
12. E. B. SafF'and J. Twomey, A note on the location of critical points of polynomials,
Proc.
Amer. Math. Soc. 27 (1971), 303-308.
13. G. Schmeisser, Bermerkungen zu einer Vermutung von Ilieff, Math. Z. 111 (1969), 121-125.
14. _,
Zur Lage der knitischen Punkte eines polynômes, Rend. Sem. Mat. Univ. Padova 46
(1971), 405-415.
15. _,
On Ilieff's conjecture,Math. Z. 156 (1977), 165-173.
Department
of Mathematics,
Purdue
University,
West Lafayette,
License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use
Indiana
47907