JJ II

(1)

volume 5, issue 4, article 98, 2004.

Received 20 September, 2004;

accepted 09 October, 2004.

Communicated by:Th.M. Rassias

Abstract Contents

JJ II

J I

Home Page Go Back

Close Quit

Journal of Inequalities in Pure and Applied Mathematics

ON SOME CLASSES OF ANALYTIC FUNCTIONS

KHALIDA INAYAT NOOR AND M.A. SALIM

Department of Mathematics and Computer Science College of Science

United Arab Emirates University P.O. Box 17551, Al-Ain United Arab Emirates.

EMail:[email protected]

2000c Victoria University ISSN (electronic): 1443-5756 189-04

(2)

On Some Classes Of Analytic Functions

Khalida Inayat Noor and M.A.

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page2of13

J. Ineq. Pure and Appl. Math. 5(4) Art. 98, 2004

Abstract

We define some classes of analytic functions related with the class of functions with bounded boundary rotation and study these classes with reference to certain integral operators.

2000 Mathematics Subject Classification:30C45, 30C50.

Key words: Close-to-convex functions, Univalent functions, Bounded boundary rota- tion, Integral operator.

1. Introduction

Let A denote the class of functions f of the form f(z) = z +P∞

m=2a_mz^m which are analytic in the unit disk E = {z : |z| < 1}.LetC, S^?, K andS be the subclasses ofAwhich are respectively convex, starlike, close-to-convex and univalent inE.It is known that C ⊂S^? ⊂K ⊂S.In [1], Kaplan showed that f ∈K if, and only if, forz ∈E, 0≤θ₁ < θ₂ ≤2π, 0< r <1,

Z θ2

θ1

Re

1 + reîθf⁰⁰(reîθ) f⁰(reîθ)

dθ >−π, z =re^iθ.

Let V_k(k ≥ 2)be the class of locally univalent functions f ∈ A that mapE conformally onto a domain whose boundary rotation is at most kπ. It is well known thatV₂ ≡CandV_k⊂K for2≤k ≤4.

Definition 1.1. Letf ∈ Aandf⁰(z)6= 0.Thenf ∈Tk(λ), k ≥2, 0≤λ < 1 if there exists a functiong ∈V_k such that, forz ∈E

Re

f⁰(z) g⁰(z)

> λ.

The class T_k(0) = T_k was considered in [2, 3] and T₂(0) = K, the class of close-to-convex functions.

Definition 1.2. Let f ∈ A and ^f^(z)f_z⁰^(z) 6= 0, z ∈ E. Then f ∈ Tk(a, γ, λ), Rea≥0, 0≤γ ≤1if, and only if, there exists a functiong ∈T_k(λ)such that (1.1) zf⁰(z) +af(z) = (a+ 1)z(g⁰(z))^γ, z ∈E.

(4)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page4of13

We note that T_k(0,1, λ) = T_k(λ)andT₂(0,1, λ) = K(λ) ⊂ K,and it follows thatf ∈Tk(a, γ, λ)if, and only if, there existsF ∈Tk(∞, γ, λ)such that

f(z) = a+ 1 z^a

Z z 0

t^a−1F(t)dt.

(5)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page5of13

2. Preliminary Results

Lemma 2.1 ([2]). Let f ∈ A.Then, for0 ≤ θ₁ < θ₂ ≤ 2π, z = re^iθ, 0 <

r <1, f ∈T_k if and only if Z θ2

θ1

Re

zf⁰(z))⁰ f⁰(z)

dθ >−k 2π.

Lemma 2.2. Let q(z)be analytic inE and of the formq(z) = 1 +b₁z+· · · for|z|= r∈ (0,1).Then, fora, c₁, θ₁, θ₂ witha ≥1, Re(c₁) ≥0, 0 ≤θ₁ <

θ₂ ≤2π, Z θ2

θ1

Re

q(z) + azq⁰(z) c₁a+q(z)

dθ >−β₁π; (0< β₁ ≤1) implies

Z θ2

θ1

Req(z)dθ > −β₁π, z =re^iθ.

This result is a direct consequence of the one proved in [4] forβ₁ = 1.

From (1.1) and Lemma2.1, we can easily have the following:

Lemma 2.3. A function f ∈ T_k(∞, γ, λ)if and only if, it may be represented asf(z) = p(z)·zG⁰(z),whereG∈V_kandRep(z)> λ, z ∈E.

Proof. Sincef ∈T_k(∞, γ, λ),we have

f(z) = z(g⁰(z))^γ, g ∈T_k(λ)

=z[G⁰₁(z)p₁(z)]^γ, G₁ ∈V_k, Rep₁(z)> λ

=zG⁰(z).p(z),

(6)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page6of13

where G⁰(z) = (G⁰₁(z))^γ ∈ V_k and p(z) = (p₁(z))^γ, Rep(z) > λ, since 0≤γ ≤1.

The converse case follows along similar lines.

Using Lemma2.1and Lemma2.3, we have:

Lemma 2.4.

(i) Letf ∈Tk(0, γ, λ).Then, withz =re^iθ, 0≤θ1 < θ−2≤2π, Z θ2

θ1

Re

(zf⁰(z))⁰ f⁰(z)

dθ > −kγ

2 , see also [3].

(ii) Letf ∈Tk(∞, γ, λ).Then, forz =re^iθ and0≤θ1 < θ2 ≤2π, Z θ2

θ1

Re

zf⁰(z) f(z)

dθ > −kγ 2 .

(7)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page7of13

3. Main Results

Theorem 3.1. For 0 < α < _1−λ+λβ¹ , 0 < β < _1−λ^λ , 0 ≤ λ < ¹₂ and f, g ∈T_k(∞, γ, λ), z ∈E,let

(3.1) F(z) =

β+ 1 α

z¹⁻

α1

Z z 0

t

α1−2

(f(t))^βg(t)dt

1 1+β

.

ThenF₁,withF = zF₁⁰ and0 < γ < 1, k ≤ ²_γ,is close-to-convex and hence univalent inE.

Proof. We can write (3.1) as (3.2) (F(z))^β+1 =

β+ 1

α

z¹⁻

α1

Z z 0

t

α1−2

(f(t))^βg(t)dt.

Let

(3.3) zF⁰(z)

F(z) = (zF₁⁰(z))⁰

F₁⁰(z) = (1−λ)H(z) +λ, whereH(z)is analytic inEandH(z) = 1 +c₁z+c₂z²+· · · .

We differentiate (3.2) logarithmically to obtain (β+ 1)zF⁰(z)

F(z) =

1− 1 α

+ z

α1−1

(f(z))^βg(z) Rz

0 t^α¹⁻²(f(z))^β(t)g(t)dt .

(8)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page8of13

Using (3.2) and differentiating again, we have after some simplifications,

(1−λ)zH⁰ Rz

0 t

α1−2

(f(t))^βg(t)dt z^α¹⁻¹(f(z))^βg(z)

+ (1−λ)H(z)

= β

1 +β · zf⁰(z) f(z) + 1

β+ 1 · zg⁰(z) g(z) −λ.

Now

z

α1−1

(f(z))^βg(z) Rz

0 t^α¹⁻²(f(t))^βg(t)dt

= 1

α −1

+ (1 +β)zF⁰(z) F(z) . Hence

−λ+ β

1 +β · zf⁰(z)

f(z) + 1

β+ 1 · zg⁰(z) g(z)

= (1−λ)H(z) + (1−λ)zH⁰(z)

(1−λ)(1 +β)H(z) + (_α¹ −1) +λ(1 +β) and we have

(3.4) 1

1−λ β

1 +β

zf⁰(z) f(z) −λ

+ 1

1 +β

zg⁰(z) g(z) −λ

=H(z) +

1

(1+β)(1−λ)zH⁰(z) H(z) +h ₍1

α−1)

(1+β)(1−λ) +_1−λ^λ i.

(9)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page9of13

Sincef, g∈T_k(∞, γ, λ),so with z =re^iθ, 0≤θ₁ < θ₂ ≤2π, β

1 +β Z θ2

θ1

Re 1

1−λ

zf⁰(z) f(z) −λ

dθ

+ 1

1 +β Z θ2

θ1

Re 1

1−λ

zg⁰(z) g(z) −λ

dθ > −kγ 2 π, and, therefore,

Z θ2

θ1

Re



H(z) +

1

(1+β)(1−λ)zH⁰(z) H(z) +

n ₍1 α−1)

(1+β)(1−λ)+ _1−λ^λ o



dθ > −kγ 2 π.

Now using Lemma2.2witha= _{(1+β)(1−λ)}¹ ≥1, c₁ = ₁

α −1

+ (1 +β)λ ≥ 0,we obtain the required result.

Theorem 3.2. Letf, g∈T_k(∞, γ, λ), α, c, δandνbe positively real, 0< α≤

1

1−λ, c > α(1−λ), (ν+δ) = α.Then the functionF defined by (3.5) [F(z)]^α =cz^α−c

Z z 0

t^{(c−δ−ν)−1}(f(t))^δ(g(t))^νdt belongs toT_k(∞, γ, λ)fork≤ _γ², 0< γ <1.

Proof. First we show that there exists an analytic functionF satisfying (3.5).

Let

G(z) =z^−(ν+δ)(f(z))^δ(g(z))^ν

= 1 +d₁z+d₂z²+· · ·

(10)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page10of13

and choose the branches which equal1whenz = 0.For K(z) =z^{(c−ν−δ)−1}(f(z))^δ(g(z))^ν =z^c−1G(z), we have

L(z) = c z^c

Z z 0

K(t)dt = 1 + c

1 +cd₁z+· · · . HenceLis well-defined and analytic inE.

Now let

F(z) =

z^αL(z)

1

α =z[L(z)]

1 α , where we choose the branch of[L(z)]

α1

which equals 1 whenz = 0.ThusF is analytic inE and satisfies (3.5).

Set

(3.6) zF⁰(z)

F(z) = (1−λ)h(z) +λ, and let

zf⁰(z)

f(z) = (1−λ)h₁(z) +λ zg⁰(z)

g(z) = (1−λ)h₂(z) +λ.

Now, from (3.5), we have (3.7) z^(c−α)[F(z)]^α

(c−α) +αzF⁰(z) F(z)

=ch

z^{(c−δ−ν)−1}(f(z))^δ(g(z))^νi .

(11)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page11of13

We differentiate (3.7) logarithmically and use (3.6) to obtain α(1−λ)

h(z) + zh⁰(z)

(c−α) +α{λ+ (1−λ)h(z)}

+ (δ+ν−α)

=δzf⁰(z)

f(z) +νzg⁰(z) g(z) −αλ

=δ

zf⁰(z) f(z) −λ

+ν

zg⁰(z) g(z) −λ

. This gives us

h(z) + zh⁰(z)

(c−α) +α{λ+ (1−λ)h(z)}

= δ

α(1−λ)

zf⁰(z) f(z) −λ

+ ν

α(1−λ)

zg⁰(z) g(z) −λ

.

Sincef, g∈T_k(∞, γ, λ),we have, for0≤θ₁ < θ₂ ≤2π, z =re^iθ, Z θ2

θ1

Re

h(z) + zh⁰(z)

(c−α) +α{λ+ (1−λ)h(z)}

dθ

= δ

α Z θ2

θ1

Reh₁(z)dθ+ ν α

Z θ2

θ1

Reh₂(z)dθ

> δ α

−γk 2 π

+ ν

α

−γk 2 π

= δ+ν α

−γk 2 π

=−γk 2 π,

(12)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page12of13

where we have used Lemma2.4.

Now using Lemma2.2witha= _α(1−λ)¹ >1,forα < _1−λ¹ and c₁ =c−α+αλ=c−α(1−λ)≥0, we obtain the required result.

(13)

Salim

Title Page Contents

JJ II

J I

Go Back Close

Quit Page13of13

References

[1] W. KAPLAN, Close-to-convex Schlicht functions, Michigan Math. J., (1952), 169–185.

[2] K.I. NOOR, On a generalization of close-to-convexity, Intern. J. Math. &

Math. Sci., 6 (1983), 327–334.

[3] K.I. NOOR, Higher order close-to-convex functions, Math. Japonica, 37 (1192), 1–8.

[4] R. PARVATHANANDS. RADHA, On certain classes of analytic functions, Ann. Polon. Math., 49 (1988), 31–34.