The Fekete Szegö problem for a subclass of quasi-convex functions with respect to symmetric points - UMS INSTITUTIONAL REPOSITORY

The Fekete Szeg¨ o Problem for a Subclass of Quasi-Convex Functions of Order β Type γ with

Respect to Symmetric Points

Chuah Puoi Choo School of Science and Technology

Universiti Malaysia Sabah

Jalan UMS, 88400 Kota Kinabalu, Sabah, Malaysia Aini Janteng

School of Science and Technology Universiti Malaysia Sabah

Jalan UMS, 88400 Kota Kinabalu, Sabah, Malaysia aini jg@ums.edu.my

Abstract

In [2], Janteng et al. introduced new subclass ofSdenoted byK_s^∗(α, β, γ) for 0≤α <1,0≤β <1,0 ≤γ <1. This paper obtained sharp upper bounds for |an|, n= 2,3 and the Fekete-szeg¨o inequalities for functions f ∈K_s^∗(α, β, γ).

Mathematics Subject Classification: Primary 30C45

Keywords: starlike w.r.t symmetric points, quasi-convex with respect to (w.r.t) symmetric points, Fekete-szeg¨o theorem

1 Introduction

Let S be the class of functionsf which are analytic and univalent in the open unit disc D={z:|z|<1}given by

f(z) =z+ ∞ n=2

a_nzⁿ (1)

where a_n is a complex number.

For 0 ≤ α <1,0≤ β < 1,0 ≤γ < 1, Janteng et al. in [2] were introduced a new class of functions denoted by K_s^∗(α, β, γ).

Definition 1.1 Let f be given by (1). Then for 0≤α < 1,0≤β <1,0≤ γ <1, f ∈K_s^∗(α, β, γ)if there exists a g ∈C_s(γ) such that for z ∈D,

Re

2α(z²f(z))

(g(z)−g(−z)) + 2(zf(z)) (g(z)−g(−z))

> β.

Note: The definition is also equivalent to the following:

f ∈K_s^∗(α, β, γ) if there exists a h=zg(z) ∈S_s^∗(γ) such that Re

2αz(z²f(z))

h(z)−h(−z) + 2z(zf(z)) h(z)−h(−z)

> β. (2)

We note that the class K_s^∗(α,0,0) =K_s^∗(α) in [1].

2 Preliminary Results

Here is a preliminary lemma required for proving our results.

Lemma 2.1 ([3]) Let k be analytic in D with Re{k(z)} > 0 and be given by k(z) = 1 +c₁z+c₂z²+. . . for z ∈D, then

|c_n| ≤2(n≥1) and

c₂− c²₁ 2

≤2− |c₁|² 2 .

3 Main Result

Theorem 3.1 Let f ∈K_s^∗(α, β, γ),0≤α <1,0≤β <1,0≤γ <1 and be given by by (1), then

2(α+ 1)|a₂| ≤1−β and

9(2α+ 1)|a₃| ≤3−2β−γ

Proof.

Since h ∈S_s^∗(γ), it follows that

2zh(z) = [γ + (1−γ)H(z)](h(z)−h(−z))

for z ∈D, with Re{H(z)}>0 where H(z) = 1 +p₁z+p₂z²+p₃z³+... Upon equating coefficients, we obtain

2b₂ = (1−γ)p₁,2b₃ = (1−γ)p₂ (3) It follows from (2) that

2αz(z²f(z))+ 2z(zf(z)) = [β+ (1−β)k(z)](h(z)−h(−z)) (4) whereRe{k(z)}>0. Writingk(z) = 1+c₁z+c₂z²+...and equating coefficients in (4) gives

4(α+ 1)a₂ = (1−β)c₁,9(2α+ 1)a₃ = (1−β)c₂+b₃ (5) The result now follows on using classical inequalities |p_n| ≤2,|c_n| ≤2, n ≥ 2 and the inequality |b₃| ≤1−γ which follow from (3).

Now we consider the functional |a₃−μa²₂| for a complex μ.

Theorem 3.2 For f ∈K_s^∗(α, β, γ),0≤α < 1,0≤β <1,0≤γ <1 and μ complex,

|a₃−μa²₂| ≤ 3−2β−γ 9(2α+ 1) max

1,

4(1−γ)(α+ 1)²+ (1−β)|8(α+ 1)²−9μ(2α+ 1)(1−β)| 4(3−2β−γ)(α+ 1)²

The result obtained is sharp.

Proof.

From (5), we write a₃−μa²₂ = 1−β

9(2α+ 1)

c₂ −c²₁ 2

+(1−β){8(α+ 1)²−9μ(2α+ 1)(1−β)}c²₁ 144(2α+ 1)(α+ 1)²

+ (1−γ)p₂

18(2α+ 1) (6)

It follows from (6) and Lemma 2.1 that

|a₃−μa²₂| ≤ 1−β 9(2α+ 1)

c₂−c²₁ 2

+(1−β)|8(α+ 1)²−9μ(2α+ 1)(1−β)||c₁|² 144(2α+ 1)(α+ 1)²

+(1−γ)|p₂|

18(2α+ 1) (7)

≤ 1−β 9(2α+ 1)

2− |c₁|² 2

+(1−β)|8(α+ 1)²−9μ(2α+ 1)(1−β)||c₁|² 144(2α+ 1)(α+ 1)²

+(1−γ)|p₂| 18(2α+ 1)

= 2(1−β)

9(2α+ 1) + (1−β){|8(α+ 1)²−9μ(2α+ 1)(1−β)| −8(α+ 1)²} |c₁|² 144(2α+ 1)(α+ 1)²

+(1−γ)|p₂|

18(2α+ 1) (8)

which, on using |c₁| ≤2 and|p₂| ≤2 gives a₃−μa²₂≤

_3−2β−γ

9(2α+1), if κ(α, β)≤8(α+ 1)²;

9(2α+1)1−γ + (1−β)|8(α+1)²−9μ(2α+1)(1−β)|

36(2α+1)(α+1)² , if κ(α, β)≥8(α+ 1)². where κ(α, β) = |8(α+ 1)²−9μ(2α+ 1)(1−β)|.

Lettingc₁ = 0, c₂ =p₂ = 2 andc₁ =c₂ =p₂ = 2 respectively in (6) shows that the result is sharp.

Next, we consider the real number μas follows.

Theorem 3.3 For f ∈K_s^∗(α, β, γ),0≤α < 1,0≤β <1,0≤γ <1 and μ real,

a₃ −μa²₂≤

⎧⎪

⎪⎨

⎪⎪

⎩

3−2β−γ

9(2α+1) − ^μ(1−β)_4(α+1)²2, if μ≤0;

3−2β−γ

9(2α+1), if 0≤μ≤ 9(2α+1)(1−β)^16(α+1)2 ;

μ(1−β)²

4(α+1)² −^1−2β+γ_9(2α+1), if μ≥ 9(2α+1)(1−β)^16(α+1)2 . The result obtained is sharp.

Proof.

We consider two cases. At first, we suppose that μ ≤ 9(2α+1)(1−β)^8(α+1)2 . From (8) and using the fact that |c₁| ≤2,|p₂| ≤2, we obtain

a₃−μa²₂≤

_3−2β−γ

9(2α+1) − ^μ(1−β)_4(α+1)²2, if μ≤0;

3−2β−γ

9(2α+1), if 0≤μ≤ 9(2α+1)(1−β)^8(α+1)2 .

Lettingc₁ =c₂ =p₂ = 2 andc₁ = 0, c₂=p₂ = 2 respectively in (6) shows that the result is sharp.

Next, we suppose that μ ≥ 9(2α+1)(1−β)^8(α+1)2 . In this case, it follows from (8) and Lemma 2.1 that

a₃−μa²₂≤

_3−2β−γ

9(2α+1), if 9(2α+1)(1−β)^8(α+1)2 ≤μ≤ 9(2α+1)(1−β)^16(α+1)2 ;

μ(1−β)²

4(α+1)² − ^1−2β+γ_9(2α+1), if μ≥ 9(2α+1)(1−β)^16(α+1)2 .

Letting c₁ = 0, c₂ = p₂ = 2 and c₁ = 2i, c₂ = −2, p₂ = 2 respectively in (6) shows that the result is sharp.

ACKNOWLEDGEMENTS. This work was supported by FRG0268-ST- 2/2010 Grant, Malaysia. The authors express their gratitude to the referee for his valuable comments.

References

[1] A. Janteng, S.A. Halim and M. Darus, Subclasses of functions quasi- convex with respect to other points, Proceedings of the 2nd IMT-GT Re- gional Conference on Mathematics, Statistics and Appl.,(2006)

[2] A. Janteng, S.A. Halim and M. Darus, New subclasses of univalent func- tions, International J. of Pure and Appl. Math.,44(5)(2008), 673-681.

[3] Ch. Pommerenke, Univalent functions, Vandenhoeck and Ruprecht, G¨ottingen, 1975.

Received: March, 2012