ON CERTAIN SUBCLASS OF BAZILEVI ˇC FUNCTIONS

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Subclass of Bazilevi ˇc Functions Dong Guo and Ming-Sheng Liu

vol. 8, iss. 1, art. 12, 2007

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ON CERTAIN SUBCLASS OF BAZILEVI ˇ C FUNCTIONS

DONG GUO AND MING-SHENG LIU

School of Mathematical Sciences, South China Normal University, Guangzhou 510631, Guangdong, China.

EMail:liumsh@scnu.edu.cn

Received: 22 September, 2006 Accepted: 26 February, 2007 Communicated by: H.M. Srivastava

2000 AMS Sub. Class.: Primary 30C45; Secondary 26A33, 33C05.

Key words: Starlike function, Bazileviˇc Function, Subordination relationships, Inclusion re- lationship, Coefficient estimates, Integral operator, Covering theorem, Fekete- Szegö inequalities.

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vol. 8, iss. 1, art. 12, 2007

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n=2anzⁿ, which are analytic in the unit disk U ={z :|z|<1}. In this paper, the authors introduce a subclassM(α, λ, ρ)ofHand study its some properties. The subordination relationships, inclusion relationships, coefficient estimates, the integral operator and covering theorem are proven here for each of the function classes. Furthermore, some interesting Fekete-Szegö inequalities are obtained. Some of the results, presented in this paper, gen- eralize the corresponding results of earlier authors.

Acknowledgements: This research is partly supported by the Doctoral Foundation of the Edu- cation committee of China (No. 20050574002).

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1. Introduction

LetHdenote the class of functionsf of the form

(1.1) f(z) = z+

∞

X

n=2

a_nzⁿ,

which are analytic in the open unit disk U = {z : |z| < 1}, and let S denote the class of all functions inHwhich are univalent in the diskU. Suppose also thatS^∗,K and α− Kdenote the familiar subclasses of H consisting of functions which are, respectively, starlike inU, convex inU andα−convex inU. Thus we have

S^∗ =

f :f ∈ HandR

zf⁰(z) f(z)

>0, z ∈ U

,

K=

f :f ∈ HandR

1 + zf⁰⁰(z) f⁰(z)

>0, z ∈ U

and α−K=

f :f ∈ HandR

α

1 + zf⁰⁰(z) f⁰(z)

+ (1−α)zf⁰(z) f(z)

>0, z ∈ U

. Letf(z)andF(z)be analytic inU. Then we say that the functionf(z)is subordi- nate toF(z)inU, if there exists an analytic functionω(z)inU such that|ω(z)| ≤ |z|

andf(z) = F(ω(z)), denotedf ≺ F orf(z) ≺ F(z). IfF(z)is univalent inU, then the subordination is equivalent tof(0) =F(0)andf(U)⊂F(U)(see [18]).

Assuming that α > 0, λ ≥ 0, ρ < 1, a function p(z) = 1 +p₁z+p₂z²+· · · is said to be in the class P_ρ if and only if p(z) is analytic in the unit disk U and

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Rp(z)> ρ, z ∈U. A functionf(z) ∈H is said to be in the classB(λ, α, ρ)if and only if it satisfies

(1.2) R(1−λ)

f(z) z

α

+λzf⁰(z) f(z)

f(z) z

α

> ρ, z ∈ U, where we choose the branch of the power

f(z) z

α

such that f(z)

z

α z=0

= 1.It is obvious that the subclassB(1, α,0)is the subclass of Bazileviˇc functions, which is the subclass of univalent functions S, we setB(α, ρ) ≡ B(1, α, ρ). The function classB(λ, α, ρ)was introduced and studied by Liu [10]. Some special cases of the function classB(λ, α, ρ)had been studied by Bazileviˇc [1], Chichra [2], Ding, Ling and Bao [3], Liu [9] and Singh [19], respectively.

Liu [11] introduced the following classB(λ, α, A, B, g(z))of analytic functions, and studied its some properties.

B(λ, α, A, B, g(z))

=

f ∈ H:

1−λzg⁰(z) g(z)

f(z) g(z)

α

+λzf⁰(z) f(z)

f(z) g(z)

α

≺ 1 +Az 1 +Bz

, whereα >0, λ≥0,−1≤B < A≤1, g(z)∈S^∗.

Fekete and Szegö [4] showed that forf ∈S given by (1.1),

|a₃−µa²₂| ≤











3−4u, ifµ≤0, 1 + 2e^{−2/(1−µ)}, if0≤µ <1, 4−3µ, ifµ≥1.

As a result, many authors studied similar problems for some subclasses ofH or S (see [6, 7,8, 13, 14, 15, 20]), which is popularly referred to as the Fekete-Szegö

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inequality or the Fekete-Szegö problem. Li and Liu [12] obtained the Fekete-Szegö inequality for the function classB(λ, α, ρ).

Recently, Patel [17] introduced the following subclassM_p(λ, µ, A, B)ofp−valent Bazileviˇc functions, and studied some of its properties.

An analytic functionf(z) = z^p+P∞

n=p+1a_nzⁿis said to be in the classM_p(λ, µ, A, B) if and only if there exists ap−valent starlike functiong(z) =z^p+P∞

n=p+1b_nzⁿsuch that

zf⁰(z) f(z)

f(z) g(z)

µ

+λ

1 + zf⁰⁰(z)

f⁰(z) − zf⁰(z) f(z) +µ

zf⁰(z)

f(z) −zg⁰(z) g(z)

≺p1 +Az 1 +Bz, whereµ≥0, λ >0,−1≤B < A≤1.

In the present paper, we introduce the following subclass of analytic functions, and obtain some interesting results.

Definition 1.1. Assume that α ≥ 0, λ ≥ 0, 0 ≤ ρ < 1, f ∈ H. We say that f(z)∈M(α, λ, ρ)if and only iff(z)satisfies the following inequality:

R

zf⁰(z) f(z)

f(z) z

α

+λ

1 + zf⁰⁰(z)

f⁰(z) − zf⁰(z) f(z) +α

zf⁰(z) f(z) −1

> ρ, z ∈ U. It is evident thatM(α,0, ρ) =B(α, ρ)(α ≥0)andM(0, α,0) =α− K(α ≥0).

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2. Preliminaries

To derive our main results, we shall require the following lemmas.

Lemma 2.1 ([16]). If−1≤B < A≤1, β >0and the complex numberγ satisfies R(λ)≥ ^−β(1−A)_1−B , then the differential equation

q(z) + zq⁰(z)

βq(z) +γ = 1 +Az

1 +Bz, z ∈ U, has a univalent solution inU given by

(2.1) q(z) =











z^β+γ(1 +Bz)^β(A−B)/B βRz

0 t^β+γ−1(1 +Bt)^β(A−B)/Bdt −^γ_β, B 6= 0, z^β+γexp^(βAz)

βRz

0 t^β+γ−1exp(βAt)dt − ^γ_β, B = 0.

Ifφ(z) = 1 +c₁z+c₂z²+· · · is analytic inU and satisfies

(2.2) φ(z) + zφ⁰(z)

βφ(z) +γ ≺ 1 +Az

1 +Bz, (z ∈ U), then

φ(z)≺q(z)≺ 1 +Az

1 +Bz, (z ∈ U), andq(z)is the best dominant of (2.2).

Lemma 2.2 ([11]). Suppose thatF(z)is analytic and convex inU, and0≤λ ≤1, f(z)∈ H, g(z)∈ H. Iff(z)≺F(z)andg(z)≺F(z). Then

λf(z) + (1−λ)g(z)≺F(z).

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Lemma 2.3 ([18]). Letp(z) = 1 +P∞

n=1p_nzⁿ ∈ P₀. Then

p₂−1 2p²₁

≤2−1 2|p²₁| and|p_n| ≤2for alln ∈N+.

Lemma 2.4 ([1]). Letα≥0, f ∈ Hand for|z|< R ≤1, R

zf⁰(z) f(z)

f(z) z

α

>0, thenf(z)is univalent in|z|< R.

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3. Main Results and Their Proofs

Theorem 3.1. Letα ≥0andλ >0. Iff(z)∈M(α, λ, ρ). Then (3.1) zf⁰(z)

f(z)

f(z) z

α

≺q(z)≺ 1 + (1−2ρ)z

1−z , (z ∈ U), where

q(z) = λz^1/λ(1−z)^{−2(1−ρ)/λ} Rz

0 t^(1−λ)/λ(1−t)^{−2(1−ρ)/λ}dt, andq(z)is the best dominant of (3.1).

Proof. By applying the method of the proof of Theorem 3.1 in [17] mutatis mutandis, we can prove this theorem.

With the aid of Lemma2.4, from Theorem 3.1, we have the following inclusion relation.

Corollary 3.2. Letα ≥0,0≤ρ <1andλ≥0, then

M(α, λ, ρ)⊂M(α,0, ρ)⊂M(α,0,0)⊂ S.

Theorem 3.3. Letα ≥0andλ₂ > λ₁ ≥0,1> ρ₂ ≥ρ₁ ≥0, then M(α, λ₂, ρ₂)⊂M(α, λ₁, ρ₁).

Proof. Suppose thatf(z)∈M(α, λ₂, ρ₂). Then, by the definition ofM(α, λ₂, ρ₂), we have

(3.2) R

zf⁰(z) f(z)

f(z) z

α

+λ₂

1 + zf⁰⁰(z)

f⁰(z) − zf⁰(z) f(z) +α

zf⁰(z) f(z) −1

> ρ₂ (z ∈ U).

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Sinceα≥0andλ₂ > λ₁ ≥0, by Theorem3.1, we obtain

(3.3) R

zf⁰(z) f(z)

f(z) z

α

> ρ₂ (z ∈ U).

Settingλ= ^λ_λ¹

2, so that0≤λ <1, we find from (3.2) and (3.3) that R

zf⁰(z) f(z)

f(z) z

α

+λ₁

1 + zf⁰⁰(z)

f⁰(z) −zf⁰(z) f(z) +α

zf⁰(z) f(z) −1

=λR

zf⁰(z) f(z)

f(z) z

α

+λ₂[1 + zf⁰⁰(z)

f⁰(z) − zf⁰(z) f(z) +α

zf⁰(z) f(z) −1

+ (1−λ)Rzf⁰(z) f(z)

f(z) z

α

> ρ₂ ≥ρ₁ (z ∈ U), that is,f(z) ∈ M(α, λ₁, ρ₁). Hence, we haveM(α, λ₂, ρ₂) ⊂ M(α, λ₁, ρ₁), and the proof of Theorem3.3is complete.

Remark 1. Theorem3.3 obviously provides a refinement of Corollary3.2. Setting α= 0, ρ₂ =ρ₁ = 0in Theorem3.3, we get Theorem 9.4 of [5].

With the aid of Lemma2.2, by using the method of our proof of Theorem3.3, we can prove the following inclusion relation.

Theorem 3.4. Letµ≥0,−1≤B < A≤1andλ2 > λ1 ≥0, then M_p(λ₂, µ, A, B)⊂ M_p(λ₁, µ, A, B).

By applying the method of the proof of Theorem 3.13, Theorem 3.6 and Theorem 3.11 in [17] mutatis mutandis, we can prove the following three results.

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Theorem 3.5. Letα ≥0, λ > 0andγ >0. Iff(z)∈ Hsatisfies γ

zf⁰(z) f(z)

f(z) z

α +λ

1 + zf⁰⁰(z)

f⁰(z) − zf⁰(z) f(z) +α

zf⁰(z) f(z) −1

6=it, (z ∈ U), wheretis a real number satisfying|t|>p

λ(λ+ 2γ), then R

zf⁰(z) f(z)

f(z) z

α

>0, (z ∈ U).

Theorem 3.6. Suppose thatα >0and0≤ρ <1. Iff(z)∈Hsatisfies Re

zf⁰(z) f(z)

f(z) z

α

> ρ, (z ∈ U),

thenf(z)∈M(α, λ, ρ)for|z|< R(λ, ρ), whereλ >0, and

R(λ, ρ) =







(1+λ−ρ)−√

(1+λ−ρ)²−(1−2ρ)

1−2ρ , ρ6= ¹₂,

1

1+2λ, ρ= ¹₂.

The boundR(λ, ρ)is the best possible.

For a functionf ∈ H, we define the integral operatorF_α,δas follows:

(3.4) F_α,δ(f) =F_α,δ(f)(z) =

α+δ z^δ

Z z 0

t^δ−1f(t)^αdt _α¹

(z ∈ U), whereαandδare real numbers withα >0, δ >−α.

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Theorem 3.7. Letαandδbe real numbers withα >0,0≤ρ <1,δ >max{−α,−αρ}

and letf(z)∈ H. If

arg

zf⁰(z) f(z)

f(z) z

α

−ρ

≤ π

2β (0≤ρ <1; 0< β≤1),

then

arg

zF_α,δ⁰ (f) Fα,δ(f)

F_α,δ(f) z

α

−ρ

≤ π 2β, whereF_α,δ(f)is the operator given by (3.4).

Now we derive the Fekete-Szegö inequality for the function classM(α, λ, ρ).

Theorem 3.8. Suppose thatf(z) =z+P∞

n=2a_nzⁿ ∈M(α, λ, ρ). Then

|a₂| ≤ 2(1−ρ) (1 +λ)(1 +α), and for eachµ∈C, the following bound is sharp

|a₃−µa²₂| ≤ 2(1−ρ) (1 + 2λ)(2 +α)

×max

1,

1 + (1−ρ)[2λ(3 +α)−(2 +α)(α−1 + 2µ+ 4µλ)]

(1 +λ)²(1 +α)²

. Proof. Sincef(z) ∈ M(α, λ, ρ), by Definition 1.1, there exists a function p(z) = 1 +P+∞

k=1p_kz^k ∈ P₀, such that zf⁰(z)

f(z)

f(z) z

α

+λ

1 + zf⁰⁰(z)

f⁰(z) − zf⁰(z) f(z) +α

zf⁰(z) f(z) −1

= (1−ρ)p(z) +ρ, z ∈ U.

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Equating coefficients, we obtain

a₂ = 1−ρ

(1 +λ)(1 +α)p₁, a3 = 1−ρ

(1 + 2λ)(2 +α)p2+

(1−ρ)²h

λ(3 +α)− ^{(α+2)(α−1)}₂ i (1 +λ)²(1 +α)²(1 + 2λ)(2 +α)p²₁. Thus, we have

a3−µa²₂ = 1−ρ (1 + 2λ)(2 +α)

p2 −1

2p²₁

+(1−ρ)²[2λ(3+α)−(2+α)(α−1)−2µ(1+2λ)(2+α)]+(1−ρ)(1+λ)²(1+α)² 2(1+λ)²(1+α)²(1+2λ)(2+α) p²₁. By Lemma2.3, we obtain that|a₂|= _(1+λ)(1+α)^1−ρ |p₁| ≤ _(1+λ)(1+α)^2(1−ρ) , and

|a₃ −µa²₂| ≤H(x) =A+ABx² 4 , wherex=|p₁| ≤2,

A= 2(1−ρ)

(1 + 2λ)(2 +α), B = |C| −(1 +λ)²(1 +α)² (1 +λ)²(1 +α)² , and

C = (1 +λ)²(1 +α)²+ (1−ρ)[2λ(3 +α)−(2 +α)(α−1 + 2µ+ 4µλ)].

So, we have

|a₃−µa²₂| ≤







H(0) =A, |c| ≤(1 +λ)²(1 +α)², H(2) = _(1+λ)^A|C|2(1+α)², |c| ≥(1 +λ)²(1 +α)².

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Here equality is attained for the function given by (3.5) zf⁰(z)

f(z)

f(z) z

α

=











λz^1/λ(1−z²)^(ρ−1)/λ Rz

0 t^(1−λ)/λ(1−t²)^(ρ−1)/λdt, λ >0, |c| ≤(1 +λ)²(1 +α)², 1 + (1−2ρ)z²

1−z² , λ = 0, |c| ≤(1 +λ)²(1 +α)² λz^1/λ(1−z)^2(ρ−1)/λ

Rz

0 t^(1−λ)/λ(1−t)^2(ρ−1)/λdt, λ >0, |c| ≥(1 +λ)²(1 +α)², 1 + (1−2ρ)z

1−z , λ = 0, |c| ≥(1 +λ)²(1 +α)².

Settingλ= 0in Theorem3.8, we have the following corollary.

Corollary 3.9. Iff(z)∈B(α, ρ)given by (1.1), then

|a₂| ≤ 2(1−ρ) 1 +α , and for eachµ∈C, the following bound is sharp

|a₃−µa²₂| ≤ 2(1−ρ) 2 +α max

1,

1 + (1−ρ)(2 +α)(1−2µ−α) (1 +α)²

. Notice thatM(0, α,0) ≡α− K, and from Theorem3.8, we have the following corollary.

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Corollary 3.10. Letα ≥0. Iff(z)∈α− Kgiven by (1.1). Then

|a₂| ≤ 2 1 +α, and for eachµ∈C, the following bound is sharp

|a₃−µa²₂| ≤ 1

1 + 2αmax

1,

1 + 6α+ 2−4µ−8µα (1 +α)²

.

Theorem 3.11 (Covering Theorem). Let α ≥ 0, λ ≥ 0and f(z) ∈ M(α, λ, ρ), then the unit diskU is mapped byf(z)on a domain that contains the disk|ω|< r₁, where

r₁ = (1 +α)(1 +λ)

2(1 +α)(1 +λ) + 2(1−ρ).

Proof. Letω₀ be any complex number such thatf(z) 6= ω₀(z ∈ U), thenω₀ 6= 0 and (by Corollary3.2) the function

ω₀f(z)

ω₀−f(z) =z+

a₂+ 1 ω₀

z²+· · · , is univalent inU, so that

a2+ 1 ω₀

≤2, Therefore, according to Theorem3.8, we obtain

|ω0| ≥ (1 +α)(1 +λ)

2(1 +α)(1 +λ) + 2(1−ρ) =r1. Thus we have completed the proof of Theorem3.11.

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Remark 2. Setting α = λ = ρ = 0 in Theorem3.11, we get the well-known ¹₄− covering theorem for the familiar classS^∗ of starlike functions.

If0≤µ≤µ₁ andµis a real number, Theorem3.8can be improved as follows.

Theorem 3.12. Suppose that f(z) = z +P∞

n=2a_nzⁿ ∈ M(λ, α, ρ) and µ ∈ R.

Then

(3.6) |a₃−µa²₂|+µ|a₂|²

≤ 2(1−ρ) (1 + 2λ)(2 +α)

1 + (1−ρ)[2λ(3 +α)−(2 +α)(α−1)]

(1 +λ)²(1 +α)²

, 0≤µ≤µ₀,

(3.7) |a3−µa²₂|+ (µ₁−µ)|a2|² ≤ 2(1−ρ)

(1 + 2λ)(2 +α), µ₀ ≤µ≤µ₁, and these inequalities are sharp, where

µ₀ = 1

2+ 2λ−α(2 +α)

2(1 + 2λ)(2 +α) + (1 +λ)²(1 +α)² 2(1 + 2λ)(2 +α)(1−ρ), µ₁ = 1

2 + 2λ−α(2 +α)

2(1 + 2λ)(2 +α)+ (1 +λ)²(1 +α)² (1 + 2λ)(2 +α)(1−ρ). Proof. From Theorem3.8, we get

(3.8) |a3−µa²₂| ≤ 2(1−ρ)

(1 + 2λ)(2 +α) + 2(1−ρ) (1 + 2λ)(2 +α)

·

|(1−ρ)[2λ(3 +α)−(2 +α) [(α−1) + 2µ(1 + 2λ)]] + (1 +λ)²(1 +α)²|

4(1 +λ)²(1 +α)² −1

4

|p1|².

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Using (3.8) anda₂ = _(1+λ)(1+α)^1−ρ p₁, if0≤µ≤µ₀, we obtain

|a₃−µa²₂| ≤ 2(1−ρ)

(1 + 2λ)(2 +α) + 2(1−ρ) (1 + 2λ)(2 +α)

× (1−ρ)[2λ(3 +α)−(2 +α)(α−1)−2µ(1 + 2λ)(2 +α)]

4(1 +λ)²(1 +α)² |p1|²

= 2(1−ρ)

(1 + 2λ)(2 +α)+ 2(1−ρ)²[2λ(3 +α)−(2 +α)(α−1)]

4(1 + 2λ)(2 +α)(1 +λ)²(1 +α)² |p₁|²−µ|a₂|². Hence

|a₃−µa²₂|+µ|a₂|²

≤ 2(1−ρ)

(1 + 2λ)(2 +α) +2(1−ρ)²[2λ(3 +α)−(2 +α)(α−1)]

4(1 + 2λ)(2 +α)(1 +λ)²(1 +α)² |p₁|²

≤ 2(1−ρ) (1 + 2λ)(2 +α)

1 + (1−ρ)[2λ(3 +α)−(2 +α)(α−1)]

(1 +λ)²(1 +α)²

, 0≤µ≤µ₀. Ifµ₀ ≤µ≤µ₁, from (3.8), we obtain

|a₃−µa²₂|

≤ 2(1−ρ)

(1 + 2λ)(2 +α) + 2(1−ρ) (1 + 2λ)(2 +α)

×−2(1 +λ)²(1 +α)²−(1−ρ)[2λ(3 +α)−(2 +α)(α−1 + 2µ+ 4µλ)]

4(1 +λ)²(1 +α)² |p₁|²

= 2(1−ρ)

(1 + 2λ)(2 +α)−(µ₁−µ)|a2|².

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Therefore

|a₃−µa²₂|+ (µ₁−µ)|a₂|² ≤ 2(1−ρ)

(1 + 2λ)(2 +α), µ₀ ≤µ≤µ₁.

Here equality is attained for the function given by (3.5), and the proof of Theorem 3.12is complete.

Theorem 3.13. Letf(z)∈ H, α≥0, λ≥0and0< k≤1. If (3.9)

zf⁰(z) f(z)

f(z) z

α

+λ

1 + zf⁰⁰(z)

f⁰(z) − zf⁰(z) f(z) +α

zf⁰(z) f(z) −1

−1

< k, z ∈ U, then

|a₂| ≤ k

(1 +λ)(1 +α), and for eachµ∈C, the following bound is sharp

|a₃−µa²₂|

≤ k

(1 + 2λ)(2 +α)max





 1,

k(1 + 2λ)(2 +α)

1−2µ− _1+2λ^α + (1+2λ)(2+α)^2λ

2(1 +λ)²(1 +α)²





 . Proof. By (3.9), there exists a functionp(z)∈ P₀ such that for allz ∈ U

zf⁰(z) f(z)

f(z) z

α

+λ

1 + zf⁰⁰(z)

f⁰(z) −zf⁰(z) f(z) +α

zf⁰(z) f(z) −1

= 2k

1 +p(z) + 1−k.

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Equating the coefficients, we obtain

a₂ =− k

2(1 +λ)(1 +α)p₁,

(1 + 2λ)(2 +α)a₃ =−k 2

p₂− 1

2p²₁

+ k²h

λ(3 +α)− ^{(2+α)(α−1)}₂ i 4(1 +λ)²(1 +α)² p²₁. Thus, we have

a₃−µa²₂ =− k

2(1 + 2λ)(2 +α)

p₂− 1 2p²₁

+ k²

h

λ(3 +α)−^{(2+α)(α−1)}₂ −µ(1 + 2λ)(2 +α) i

4(1 +λ)²(1 +α)²(1 + 2λ)(2 +α) p²₁, so that, by Lemma2.3, we get that|a₂|= 2(1+λ)(1+α)^k |p₁| ≤ _(1+λ)(1+α)^k , and

|a₃−µa²₂| ≤H(x) =A+ Bx² 4 , wherex=|p₁| ≤2,

A= k

(1 + 2λ)(2 +α), B = k²|C|

[(1 +λ)²(1 +α)²]− k

[(1 + 2λ)(2 +α)]

and

C = 1−2µ

2 − α

2(1 + 2λ)+ λ

(1 + 2λ)(2 +α).

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Therefore

|a3−µa²₂| ≤







H(0) =A, |c| ≤ ^(1+λ)k(1+2λ)(2+α)²^(1+α)², H(2) = Ak(1+2λ)(2+α)|C|

(1+λ)²(1+α)² , |c| ≥ ^(1+λ)k(1+2λ)(2+α)²^(1+α)². Here equality is attained for the function given by

zf⁰(z) f(z)

f(z) z

α

=











λz^1/λexp(−kz2)/(2λ)

Rz

0 t^(1−λ)/λexp(−kt2)/(2λ)dt, λ >0,|c| ≤ k(1+2λ)(2+α)^(1+λ)²^(1+α)², 1−kz², λ= 0,|c| ≤ k(1+2λ)(2+α)^(1+λ)²^(1+α)²,

λz^1/λexp^−kz/λ Rz

0 t^(1−λ)/λexp^−kt/λdt, λ >0,|c| ≥ k(1+2λ)(2+α)^(1+λ)²^(1+α)², 1−kz, λ= 0,|c| ≥ k(1+2λ)(2+α)^(1+λ)²^(1+α)². This completes the proof of Theorem3.13.

Settingλ= 0, we get the following corollary.

Corollary 3.14. Letf(z)∈ H, α≥0and0< k ≤1. If

zf⁰(z) f(z)

f(z) z

α

−1

< k, z ∈ U, then

|a₂| ≤ k (1 +α), and for eachµ∈C, the following bound is sharp

|a3−µa²₂| ≤ k

2 +αmax

1, k(2 +α)

2(1 +α)²|1−2µ−α|

.

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Corollary 3.15. Letf(z)∈ H, α≥0and0< k≤1. If

(1−α)zf⁰(z) f(z) +α

1 + zf⁰⁰(z) f⁰(z)

−1

< k, z ∈ U, then

|a₂| ≤ k 1 +α, and for eachµ∈Cthe following bound is sharp

|a₃ −µa²₂| ≤ k

2(1 + 2α)max (

1,k(1 + 2α)

1−2µ+ _1+2α^α (1 +α)²

) . Settingα= 1in Corollary3.15, we have the following corollary.

Corollary 3.16. Letf(z)∈ Hand0< k ≤1. If

zf⁰⁰(z) f⁰(z)

< k, z∈ U, then

|a₂| ≤ k 2, and for eachµ∈Cthe following bound is sharp

|a₃−µa²₂| ≤ k 6max

1,k|4−6µ|

4

.

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