Key words and phrases: Close-to-convex functions, Univalent functions, Bounded boundary rotation, Integral operator

http://jipam.vu.edu.au/

Volume 5, Issue 4, Article 98, 2004

ON SOME CLASSES OF ANALYTIC FUNCTIONS

KHALIDA INAYAT NOOR AND M.A. SALIM DEPARTMENT OFMATHEMATICS ANDCOMPUTERSCIENCE

COLLEGE OFSCIENCE, UNITEDARABEMIRATESUNIVERSITY

P.O. BOX17551, AL-AIN, UNITEDARABEMIRATES. KhalidaN@uaeu.ac.ae

Received 20 September, 2004; accepted 09 October, 2004 Communicated by Th.M. Rassias

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.

Key words and phrases: Close-to-convex functions, Univalent functions, Bounded boundary rotation, Integral operator.

2000 Mathematics Subject Classification. 30C45, 30C50.

1. INTRODUCTION

LetAdenote the class of functionsfof the formf(z) =z+P∞

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

r <1,

Z θ2

θ1

Re

1 + re^iθf⁰⁰(re^iθ) f⁰(re^iθ)

dθ > −π, z =re^iθ.

LetV_k(k ≥2)be the class of locally univalent functionsf ∈ Athat mapEconformally onto a domain whose boundary rotation is at mostkπ.It is well known thatV₂ ≡ C andV_k ⊂ K for 2≤k≤4.

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

Re

f⁰(z) g⁰(z)

> λ.

ISSN (electronic): 1443-5756

c 2004 Victoria University. All rights reserved.

189-04

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. Letf ∈ Aand ^f(z)f_z^0(z) 6= 0, z∈E.Thenf ∈T_k(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.

We note that T_k(0,1, λ) = T_k(λ) and T₂(0,1, λ) = K(λ) ⊂ K, and it follows that f ∈ T_k(a, γ, λ)if, and only if, there existsF ∈T_k(∞, γ, λ)such that

f(z) = a+ 1 z^a

Z z 0

t^a−1F(t)dt.

2. PRELIMINARYRESULTS

Lemma 2.1 ([2]). Letf ∈ A.Then, for 0≤ θ₁ < θ₂ ≤ 2π, z =re^iθ, 0< r < 1, f ∈T_kif and only if

Z θ2

θ1

Re

zf⁰(z))⁰ f⁰(z)

dθ > −k 2π.

Lemma 2.2. Letq(z)be analytic inEand 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 Lemma 2.1, we can easily have the following:

Lemma 2.3. A function f ∈ T_k(∞, γ, λ) if and only if, it may be represented as f(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),

whereG⁰(z) = (G⁰₁(z))^γ ∈V_kandp(z) = (p₁(z))^γ, Rep(z)> λ,since0≤γ ≤1.

The converse case follows along similar lines.

Using Lemma 2.1 and Lemma 2.3, we have:

Lemma 2.4.

(i) Letf ∈T_k(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^iθand0≤θ1 < θ2 ≤2π, Z θ2

θ1

Re

zf⁰(z) f(z)

dθ >−kγ 2 .

3. MAINRESULTS

Theorem 3.1. For0 < α < _1−λ+λβ¹ , 0< β < _1−λ^λ , 0≤ λ < ¹₂ andf, g ∈ Tk(∞, 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 inE andH(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 .

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

(1−λ)zH⁰ Rz

0 t

1 α−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

1α−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.

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 Lemma 2.2 witha = _{(1+β)(1−λ)}¹ ≥ 1, c1 = ₁

α −1

+ (1 +β)λ ≥ 0,we obtain

the required result.

Theorem 3.2. Let f, g ∈ Tk(∞, γ, λ), α, c, δ and ν be positively real, 0 < α ≤ _1−λ¹ , 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 ≤ ²_γ, 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+· · · 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)

α1

=z[L(z)]

1α

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

α1

which equals 1 when z = 0.Thus F 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 . 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, where we have used Lemma 2.4.

Now using Lemma 2.2 witha= _α(1−λ)¹ >1,forα < _1−λ¹ and c₁ =c−α+αλ=c−α(1−λ)≥0,

we obtain the required result.

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.