extreme points in a class of polynomials having univalent sequential

TRANSACTIONS OF THE AMERICAN MATHEMATICAL Volume 163, January 1972

SOCIETY

EXTREME POINTS IN A CLASS OF POLYNOMIALS HAVING UNIVALENT SEQUENTIAL LIMITS BY

T. J. SUFFRIDGEC1) Abstract. This paper concerns a class ^ (defined below) of polynomials of degree less than or equal to n having the properties: each polynomial which is Univalent in the unit disk and of degree n or less is in 3^ and if {Pnk)k= l is a sequence of polynomials such that P„k e 0^k and limk_a, Pnk=f/(«+!)

were introduced and shown to be Univalent. These polynomials are related to the polynomials Qp(z; n) by the equation P(z; n,p) = ((n+ \)jn)Qp(z;n) —(\ln)zQ'p(z;n). If Pe0>n, let P*(z) = ((«+l)/«)P(z)-(l/«)zP'(z). We show below that if Pe^n and an= 1 then P* is Univalent in the disk. We require the following lemma.

Lemma 1. IfP(z) = Ji%1aizie0,nthen

p (z)- 2 —£— a,z' j =i

ei

Proof. Observe that

\kP*(z) = 5±1 AkP(z)-i AJzP'(z)] =

A,P(z)-i

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z(AkP)'(z).

228

T. J. SUFFRIDGE

[January

Since AfcF(z)= z Yl]=l 0 -z/z,), \zj\ ä 1, we have

Re \^P)'{z)-\

I AfcP(z)J

\l_Jl_ElzL_]

KeV k\~z\z\

< {+n^l

= 1+ 2

= n+

2

Hence AtP*(z)

A*F(z)

z

nz

n+\

z(\P)'(z)

|AfcF(z)| n+\

AfcF(z)

/ 0

when |z| < 1. Remark. It is also clear in the above proof that AkP*(z0) = 0 for some z0 on |z| = 1 if and only if AkP(z) has a double zero at z = z0. Theorem

3. If P(z) = £"=, c^z5'e ^n a/jfl" \a„\ = 1,

P* is Univalent in |z| < 1.

Proof. We may assume an=l. Using the coefficient relation a, +1 = an_j and proceeding as in [6, pp. 497-498] we find eieP*'(eie) = eiin+1)912 R(8) where R is real

valued and Re [e'eP*"(e'e)IP*'(eie)+ l] = (n+1)/2 when P*'(ei9)/0. That is, the tangent line to the curve P*(e'e), 0^8^2tt, turns at a constant rate in a counterclockwise direction as 8 increases except at the cusps where it reverses direction. We wish to show that for each 8, OsS80 such that S(d) assumes the values tk and —tk, n—l—p times. This means that all the zeros of

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232

T. J. SUFFRIDGE

AkP(z) + tAkR(z) and AkP(z)-tAkR(z) t = m'm1Skgn tk the proof is now complete.

lie

[January

on

|z| = l

when

t^tk.

Setting

Examples, n=1. ^ = {z}. n = 2. Theorem 2 implies that the extreme points of 2P2are rotations of z + z2. « = 3. Theorem 2 implies that the extreme points of ^3 are rotations of polynomials of the form />(z) = z4-a2z2 + z3 where a2 is real. Clearly we may assume a2^0. Theorem 6 implies that one of the polynomials A1P(z) = z + \/2a2z2 + z3 or A2P(z) = z —z3 has a double zero on |z| = 1. It follows that a2 = \/2 so all extreme points of ^3 are rotations of z+ \/2z2 + z3. n = 4. Theorem 2 implies that the extreme points of ^ are rotations of polynomials of the form P(z) = z + a2z2 + ä2z3 + zi. Theorem 6 implies that 1 +2o2 cos (7t/5)z + 2ö2 cos (7r/5)z2 + z3

and 1 + 2a2 cos (277/5)z —2ä2 cos (277/5)z2 —z3

each have a double zero on |z| = 1. Applying this to P*, each of the polynomials, 1 + f a2 cos (tt/5)z + a2 cos (tt/5)z2 + ^z3

(5)

1 + \a2 cos (2it/5)z —ö2 cos (27r/5)z

and

2_i73

has exactly one zero on |z| = l. By Cohn's rule [l] and [5, p. 149], iff(z) = c0 + c1z + ■■■+ckzk satisfies |c-0| > then /*(z)

= c0f(z)-ckzkf{\jz)

has the same zeros as / on |z| = l and the same number of zeros as /in |z| < l. Applying Cohn's rule twice to each of the polynomials (5) leads to the linear polynomials /12cos2t7/5 \

ä2

2ä2\

36 cos2 2nl5

,

a2j

and /l2 cos 77/5^2^2^ \

ä2

a2;

36 cos2 7t/5

^

\a2\2

(we have used the fact that cos 2tt/5 = (-\/5—l)/4 and cos 77/5= (\/5+1)/4

so

cos (2tt/5) cos 77/5= 1/4) each of which has a zero on |zj = l. This fact yields the equations

(6)

|a2|4-16

Re a| cos 27r/5 + 72|a2|2 cos2 2^/5-432

|a2|4+16

Rec7lcos

77/5+ 72|fl2|2cos2

77/5-432

cos4 2t7/5 = 0, cos47r/5 = 0.

Eliminating Re a2 from the equations (6) we obtain |a2|4 +18[gt2]2 —54 = 0 so |a2|2 = 3\/l5 —9. Substitution into either equation in (6) then yields

(7)

cos (3 arg a2) = /6V(9 + Sy/lS).

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1972]

EXTREME POINTS IN A CLASS OF UNIVALENT POLYNOMIALS

233

If we choose a value of arg a2 to satisfy (7) and choose \a2\ so the equations (6) are satisfied then the extreme points in SP^are rotations of the polynomials z + a2z2 + ä2z3 + zi and z + ä2z2 + a2z3 + z4.

n = 5. The extreme points of ^ are rotations of polynomials of the form P(z) = z + (a + bi)z2 + cz3 + (a —bi)zi + z5 where a, b and c are real. By considering —iP(iz), —P( —z) and (P(z))~ we see that we may assume a, b and c are nonnegative. Theorem 6 implies that among the roots of the equations

l+v'3(a (8)

+ bi)z + 2cz2 + V3(a-bi)z3 + zi = 0

l+ia + bjz 1

-(a-bjz3 -cz2

-z4 = 0 +z4 = 0

there must be three double roots on |z| = l (only half of the double roots of the third equation are to be counted). Observe that (1 +eiaz)2{ \ +e'ez)2= 1 + 2(e{a + eiß)z + (e2m + e2is + 4ei(a:+/!))z2+ 2(e«2a + a) + ei(2«+ a;))z3+ ei20, g(0o)0 where cos2 0O= f. Thus we conclude the system (10) has two solutions. Assume the values of corresponding to these solutions are 1 and 02 where 77> 0X> 0O> 02

>3w/4. From (11), we find a2 + b2 = 3 when cos 20=1, j while g(0)5. It then follows by the same argument as used in the proof of Theorem 6 that for t sufficiently small, Q(z) = Qi(z;ri) + it[Qi(z;ri)—Q2(.z;n)]E0'n and all zeros of AkQ are simple zeros,

l^kSn.

For fixed j, choosep odd so that

sin jpirl'n + V) < sin>/(«+!) sin pwl(n+\) sin tt/(«+ 1)' Applying the same argument, it follows that, for sufficiently small s,

F(z) = QAz; n)-s(Qp(z;

n)-Qx(z; n)) + //(ß4(z; «)- ß2(z; «))

is in &*n.Then \a}\ ä Re ^>(sin j7r/(n+ l))/(sin 1)). Actually the conclusion of the theorem holds for n>3 except for the case « = 5,7 = 3.

Theorem 9. Fuery function f in the class S of functions Univalent in \z\ < 1 and normalized by setting f(0) = 0,f'(0)=\, is the limit of polynomials of the form F(z)

= z + 2"= 2

6

which satisfy an = 1.

Proof. Let /eS. One can obtain a sequence of Univalent polynomials by taking appropriate partial sums of f^/lv) where {rk}k= 1 is a strictly increasing sequence of real numbers such that lim^^oo rk=\. Let {Qnic}be such a sequence where Qnk

is of degree nk. Define 2nk

PnS?) = Qnk(z)+ Z2n« +1Qnk(llz) - 2 ^ J=l

By an argument similar to that used to prove Theorem 1, Pnk e £?2nkand a2nk= 1. Also lim^oo Pnk(z)=f(z) so {FnJ"=i is the required sequence.

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236

T. J. SUFFRIDGE

[January

Theorem 10. The Univalent polynomials of the form z + Tj =2 ajz' which satisfy (j+ l)aj + 1= (n—j)an-j [and thus an = \jri] are dense in the class S. Proof. The polynomials {F*JiT=i have the same limit as {PnJ™=i above. We observe that in Theorem 9, if f e S has real coefficients then the sequence {P*k}k=i has real coefficients and we have a new proof of the Bieberbach conjecture for functions having real coefficients. Let 3>nbe the class of polynomials of degree n which are Univalent in |z| < 1 and

of the form z+2?=2«^> (j+^)aJ+i = (n~J)än-j- Applying Theorems 2 and 4 above and using the definition of P(z; n,j) in [6] we can represent any P in

the form

P(z) = P(z;n, 1)+ 2 a)[P{z;n,j)-P{z;n, (12)

1)]

i0M

+ i 2 ßmz\n,j)-P{z;n,2)l j even

Since the tangent line to the curve P(eie) (0^9^2-rr) turns at a constant rate (Re [eiBP"(ew)IP'(ew)+ \] = (n+ l)/2 when P'(ew)^0) as 9 varies the tangent line to the curve is horizontal when 9 is an odd multiple of 7r/(n+ 1) and vertical when 9 is an even multiple of 7r/(n+ 1). Recall that among all polynomials in 3sn having real coefficients, P(z; n, 1) maximizes every coefficient. Also AfcP(]; n, 1) = 0 when k is odd, k> \ (i.e. P{eikIlKn+ 1>\n, l) = P(