ON A SUBCLASS OF STARLIKE FUNCTIONS WITH BOUNDED TURNING

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on the occasion of her 80th birthday

ON A SUBCLASS OF STARLIKE FUNCTIONS WITH BOUNDED TURNING

PETRU T. MOCANU

LetAnbe the class of analytic functions on the unit discU of the form f(z) =z+an+1zⁿ⁺¹+. . . , z∈U,

withA1=A. LetS^∗be the subclass ofAconsisting of starlike functions and let Rbe the class of functions inAwhich are of bounding turning (rotation), i.e.,

R={f∈ A; Ref⁰(z)>0, z∈U}.

It is well known thatS^∗6⊂ R and R 6⊂S^∗. The main result of this paper is to obtain a subclass ofS^∗which is contained inR.

AMS 2010 Subject Classification: 30C45.

Key words: starlike functions, bounded turning functions, differential subordination.

1. INTRODUCTION LetA_n denote the set of functions

f(z) =z+a_n+1zⁿ⁺¹+. . . , n≥1, that are analytic in the unit disc U ={z; |z|<1},A=A₁.

LetS^∗ be the usual class of starlike (univalent) functions inU, i.e., S^∗={f ∈ A; Re[zf⁰(z)/f(z)]>0, z ∈U}

and let

R={f ∈ A; Ref⁰(z)>0, z ∈U} be the class of functions with bounded turning.

In [3] we found some subsetsE of the right half-plane, such thatf ∈S^∗, wheneverf⁰(z)∈Efor allz∈U. On the other hand it is obvious thatS^∗ 6⊂ R.

The function f(z) =z/(1−z) shows that even the convex functions are not necessary of bounded turning. A natural problem is to find certain subsetsEof the right half-plane, such thatf ∈ Rwhenever zf⁰(z)/f(z)∈E, for allz∈U.

REV. ROUMAINE MATH. PURES APPL.,55(2010),5, 375–379

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In this paper we obtain some conditions in terms of zf⁰(z)/f(z) such that f ∈ R.

2. PRELIMINARIES

Since our main result is obtained by using the method of differential subordinations, we review here the definition of subordination. Let f and F be analytic functions on U. The function f is said to be subordinate to F, written f ≺F, if there exists a function w analytic in U with w(0) = 0 and

|w(z)|<1 such thatf =F ◦w.

LetH[a, n] denote the set of functions of the form f(z) =a+a_nzⁿ+. . . , n≥1, z∈U.

We shall use the following lemmas to prove our results.

Lemma A. Let h be starlike in U with h(0) = 0, and let a 6= 0. If p∈ H[a, n]satisfies zp⁰/p≺h, then p≺q, where

q(z) =a·exp 1

n Z z

0

h(t) t dt

and q is the best (a, n)-dominant.

Lemma B.LetE be a set in the complex plane Cand letqbe an analytic and univalent function onU. Suppose that the functionH :C×U →Csatisfies

H[q(ζ), mζq⁰(ζ);z]6∈E

whenever m≥n, |ζ|= 1 and z∈U. If p is analytic on U of the form p(z) = q(0) +pnzⁿ+. . . andpsatisfiesH[p(z), zp⁰(z);z]∈E, forz∈U, thenp≺q.

Lemma A is in [2, Corollary 3.1d.1] and a more general form of Lemma B can be found in [1].

3. MAIN RESULTS

Theorem 1. Let α, M > 0 be such that |1−α|< M < α. If f ∈ A_n satisfies the condition

(1)

zf⁰(z) f(z) −α

< M, z∈U,

where M =M(α, n) is given by the equation (2) M²−(1−α)²

n|1−α| arctan |1−α|

pM²−(1−α)² = arctan

√

α²−M²

M ifα 6= 1,

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and

cosM

n =M for α= 1, then f ∈ R, i.e., Ref⁰(z)>0, z∈U.

Proof. The inequality (1) can be written as

(3) zf⁰(z)

f(z) ≺ 1 +Az 1 +Bz, where

A= α(1−α)

M +M, B = 1−α M . If we let

(4) P(z) = f(z)

z , then P ∈ H[1, n] and by using (3) we have

zf⁰(z)

f(z) = 1 +zP⁰(z)

P(z) ≺ 1 +Az 1 +Bz. Hence

zP⁰(z)

P(z) ≺ (A−B)z 1 +Bz and by Lemma A we deduce

(5) P ≺Q,

where Qsatisfies

nQ⁰(z)

Q(z) = A−B 1 +Bz. Hence

Q(z) = (1 +Bz)^A−B^nB . Since

argQ(z) = A−B

nB arg(1 +Bz) we obtain

(6) |argQ(z)| ≤

A−B nB

arctan |B|

√

1−B² =ϕ and from (5) we deduce

(7) |ImP| ≤tanϕ·ReP,

where ϕis given by (6).

If we letp=f⁰, then (1) becomes (8)

p(z) P(z)−α

< M.

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In order to show that (8) implies Rep(z)>0 inU, according to Lemma B, it is sufficient to check the inequality

(9)

is P(z)−α

≥M²,

for all real sand all z∈U. The inequality (9) can be rewritten as s²−2αsImP + (α²−M²)|P|² ≥0

and this inequality holds for all real sif

α²|ImP|² ≤(α²−M²)|P|², which becomes

(10) |ImP(z)| ≤

√α²−M²

M ReP(z).

On the other hand from (5), (6) and (7) we deduce tanϕ≤

√α²−M²

M .

Since

A−B

B = M²−(1−α)² 1−α , from (6) we have

ϕ= M²−(1−α)²

n|1−α| arctan |1−α|

pM²−(1−α)²

and we deduce that the inequality (10) holds if M=M(α, n) satisfies (2).

We remark that forα= 1 this result was obtained by my former student X. Xanthopoulos [4].

For α = 2 we obtain M = √

2 and from Theorem 1 we deduce the following simple result

Corollary 1. If f ∈ Aand

zf⁰(z) f(z) −2

<

√

2 for z∈U, then Ref⁰(z)>0, z∈U.

In the extremal caseα=M from (2) we have α=M = ¹₂.

If the set En is the reunion of all discs with center α > ¹₂ and radius M = M(α, n), then by Theorem 1 an interesting problem is to prove, or disprove the following

Conjecture. If f ∈ A_n and zf⁰(z)/f(z) ∈ E_n for all z ∈ U, then Ref⁰(z)>0,z∈U.

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In Figures 1 and 2 we describe the setE1, for ¹₂ < α≤3 and ¹₂ < α≤20, respectively.

Fig. 1. The setE1, for ¹₂ < α≤3. Fig. 2. The setE1, for ¹₂ < α≤20.

REFERENCES

[1] S.S. Miller and P.T. Mocanu,Differential subordinations and inequalities in the complex plane. J. Differential Equations67(1987),2, 199–211.

[2] S.S. Miller and P.T. Mocanu,Differential Subordinations. Theory and Applications, Mar- cel Dekker Inc., New York, Basel, 1999.

[3] P.T. Mocanu,Some starlikeness conditions for analytic functions. Rev. Roumaine Math.

Pure Appl.33(1988),1-2, 117–124.

[4] X. Xanthopoulos,Subclasses of starlike functions withRef⁰(z)>0. Studia Univ. Babe¸s- Bolyai, MathematicaXXXVIII(1993),1, 39–47.

Received 16 May 2010 “Babe¸s-Bolyai” University

Faculty of Mathematics and Computer Science 1 M. Kog˘alniceanu Str.

400084 Cluj-Napoca, Romania pmocanu@math.ubbcluj.ro