f z z − ∑∞a z z ∈U which are analytic in the unit disc U.

(1)

CERTAIN PROPERTIES FOR ANALYTIC FUNCTIONS DEFINED BY A GENERALISED DERIVATIVE OPERATOR

(Sifat Tertentu bagi Fungsi Analisis yang Ditakrif oleh Pengoperasi Terbitan Teritlak) AISHA AHMED AMER & MASLINA DARUS

ABSTRACT

In this paper, some important properties of analytic functions with negative coefficients defined by a generalised derivative operator are investigated. The properties include the necessary and sufficient conditions, radius of starlikeness, convexity and close-to-convexity.

Keywords: derivative operator; radius of starlikeness; convexity; close-to-convexity ABSTRAK

Dalam makalah ini dikaji beberapa sifat penting bagi fungsi analisis berpekali negatif yang ditakrif oleh pengoperasi terbitan teritlak. Sifat tersebut termasuklah syarat perlu dan cukup, jejari kebakbintangan, kecembungan dan dekat-dengan-kecembungan.

Kata kunci: pengoperasi terbitan; jejari kebakbintangan; kecembungan; dekat-dengan- kecembungan

1. Introduction

Let A denote the class of functions f in the open unit disc

= { z ∈ :| |<1}, z

U C

and let T denote the subclass of A consisting of analytic functions of the form

=2

( ) =

_k ^k

, ( ),

k

f z z − ∑∞a z z ∈U which are analytic in the unit disc U.

Definition 1.1 Let

f ∈ A

. Then f is said to be convex of order

µ (0 ≤ µ <1)

if and only if

1 ( ) > , .

( )

zf z z

f z ′′ µ

⎧ ⎫

ℜ + ⎨ ⎩ ′ ⎬ ⎭ ∈

U

Definition 1.2 Let f ∈A. Then f is said to be starlike of order

µ (0 ≤ µ <1)

if and only if

( ) > , .

( )

zf z z

f z ′ µ

⎧ ⎫

ℜ ⎨ ⎬ ∈

U

(2)

Definition 1.3 Let f ∈A, then the generalised derivative operator is given by

1 2

=2

( , , , ) ( ) = ,

m k

k k k

I λ λ l n f z z − ∑∞ζ a z (1) where

1 1

1 2

(1 ( 1) )

= ( , ),

(1 ) (1 ( 1))

m

k m m

k l c n k

l k

ζ λ

λ

−

+ − +

+ + − n m N , ∈

₀

={0,1,2,...},

2 1

0, l 0,

λ λ ≥ ≥ ≥

and ¹

1

( 1)

( , ) = .

(1)

_k ^k

c n k n

⁻

−

+

Definition 1.4 Let a function f be in T. Then f is said to be in the class of

1 2

( , , , , , ) l n τ α β λ λ

if,

ℜ I^m(λ₁,λ₂,l,n)f(z) z[I^m(λ₁,λ₂,l,n)f(z)]′

⎧⎨

⎪

⎩⎪

⎫⎬

⎪

⎭⎪>α I^m(λ₁,λ₂,l,n)f(z)

z[I^m(λ₁,λ₂,l,n)f(z)]′−1+β,

(2) where

n m N , ∈

₀

={0,1,2,...}, l ≥ 0, 0 ≤ α <1,

and

0 ≤ β <1.

The family

τ α β λ λ ( , , , , , )

₁ ₂

l n

is a special interest as it contains many well-known classes of analytic univalent functions. This family is studied by (Najafzadeh & Ebadian 2009), and also (Tehranchi & Kulkarni 2006a; 2006b).

2. Necessary and Sufficient Conditions

Theorem 2.1 Let

f T ∈ .

Then

f ∈ τ α β λ λ ( , , , , , )

₁ ₂

l n

if and only if,

=2

[(1 ) ( )] <1.

1

^{k k}

k

k a

α α β ζ β

∞

+ − +

∑ − (3) Proof: Let us assume that f ∈ τ α β λ λ ( , , , , , )1 2 l n . So by using the fact that

( ) > | ω α ω 1| β

ℜ − +

if, and only if

ℜ [ (1 ω + α e

ⁱ^θ

) − α

ⁱ^θ

] > β

and letting ¹ ²

1 2

( , , , ) ( )

= [ ( , , , ) ( )]

m m

I l n f z

z I l n f z ω λ λ

λ λ ′

in (2), we obtain

ℜ [ (1 ω + α e

ⁱ^θ

) − α

ⁱ^θ

] > . β

So

(3)

=2

1

=2

(1 ) > 0,

1

k k k

i i

k k k k k

z a z

e e

z k a z

θ θ

ζ α α β

ζ

∞

∞ −

⎡ ⎤

⎢ − ⎥

⎢ ⎥

ℜ + − −

⎢ ⎛ ⎜ − ⎞ ⎟ ⎥

⎢ ⎝ ⎠ ⎥

⎣ ⎦

∑

then

1 1

=2 =2

1

=2

1 (1 ) (1 )

> 0.

1

k i k

k k k k

k k

k k k k

k a z e k a z

k a z

β β ζ α

θ

ζ

∞ ∞

− −

∞ −

⎡ − − − − − ⎤

⎢ ⎥

ℜ ⎢ ⎢ ⎣ − ⎥ ⎥ ⎦

∑ ∑

∑

The above inequality must hold for all z in U

.

Letting

z re =

⁻ⁱ^θ

where

0 ≤ r < 1

, we obtain

1

=2

1

=2

1 [(1 ) (1 )]

> 0.

1

i k

k k k

k k k k

k e k a r

k a r

β β α

θ

ζ

∞ −

⎡ − − − + − ⎤

⎢ ⎥

ℜ ⎢ ⎢ ⎣ − ⎥ ⎥ ⎦

∑

By letting

r → 1

through half line

z re =

⁻ⁱ^θ

and by mean value theorem, we have

1

=2

1 [(1 ) (1 )]

_{k k} ^k

> 0,

k

k k a r

β

^∞

β α ζ

⁻

⎡ ⎤

ℜ − − ⎢ ⎣ ∑ − + − ⎥ ⎦

then we get

=2

[(1 ) ( )] <1.

1

^{k k}

k

k a

α α β ζ β

∞

+ − +

∑ −

Conversely, let (3) holds. We will show that (2) is satisfied and so

f ∈ τ α β λ λ ( , , , , , ).

₁ ₂

l n

By using the fact that

ℜ ( ) > ω β

if and only if

| ω − + (1 β ) |<| ω + − (1 β ) |,

it is enough to show that

| ω − + (1 α ω | − + 1| β ) |<| ω + − (1 α ω | − + 1| β ) |,

if

R =| ω + − (1 α ω | − + 1| β ) |

1 2 =2

= 1 2 [1 (1 ) ] .

| [ ( , , , ) ( )] |

k k k

m k

z z k a z

z I l n f z β β α α ζ

λ λ

− −

∞

+ − + −

′ ∑

This implies that

(4)

Similarly, if

L =| ω − − (1 α ω | − + 1| β ) |,

we get

1 2 =2

< | | [ (1 ) ( ) ] .

| [ ( , , , ) ( )] |

k k k

m k

L z k k k a z

z I l n f z β α α β ζ

λ λ

+

∞

− + + − +

′ ∑

It is easy to verify that

R L − > 0

and so the proof is complete.

Corollary 2.1 Let

f ∈ τ α β λ λ ( , , , , , ),

₁ ₂

l n

then

< 1 .

[(1 ) ( )]

k

a k

β

α α β ζ

−

+ − +

Proof: For

0 ≤ µ < 1

, we need to show that

( ) 1 <1 . ( )

zf z

f z ′ − − µ

Now, let us show that

1

=2

1

=2

( 1) ( ) ( ) =

( ) 1

k k k

k a z zf z f z

f z a z

∞ −

− −

′ −

−

∑

1

=2

1

=2

( 1) | |

<1

1 | |

k k k

k a z

a z µ

∞ −

−

≤ −

−

∑

1

=2

| | <1.

1

k k k

a z k µ

µ

∞ −

⎛ − ⎞

⇒ ∑ ⎜ ⎝ − ⎟ ⎠

By Theorem 2.1, it is enough to consider

1

(1 )[(1 ) ( )]

| | < .

( )(1 )

k

z k

µ α α β ζ

µ β

−

− + − +

− −

� Theorem 2.2 Let

f ∈ τ α β λ λ ( , , , , , ).

₁ ₂

l n

Then f is convex of order

µ (0 ≤ µ <1)

in

2 1 2

| |< = ( , , , , , , ) z r r α β λ λ l n µ

where

1 1 2

( , , , , , , ) =

1 2

inf (1 )[(1 ( ) )(1 ( ) )] .

k k k

r l n k

k k

µ α α β

α β λ λ µ ζ

µ β

⎡ − + − + ⎤

−

⎢ − − ⎥

⎣ ⎦

Proof: For

0 ≤ µ < 1

( ) <1 . ( )

zf z

f z ′′ − µ

′

Now again let us show that

(5)

1

=2

1

=2

( 1) 1

k k k

k k a z ka z

∞ −

− −

−

∑

1

=2

1

=2

( 1) | |

<1

1 | |

k k k

k k a z

ka z µ

∞ −

−

≤ −

−

∑

1

=2

| | <1.

1

k k k

ka z k µ

µ

∞ −

⎛ − ⎞

⇒ ∑ ⎜ ⎝ − ⎟ ⎠

By Theorem 2.1, it is again enough to consider

1

(1 )[(1 ) ( )]

| | < .

( )(1 )

k

z k k

µ α α β ζ

µ β

−

− + − +

− −

f z ( ) ∈ τ α β λ λ ( , , , , , ).

₁ ₂

l n

Then

f z ( )

is close-to-convex of order

(0 <1)

µ ≤ µ

in

| |< = ( , , , , , , ) z r r

3

α β λ λ

1 2

l n µ

where

1 1 3

( , , , , , , ) =

1 2

inf (1 )[(1 (1 ) ) ( )] .

k k k

r l n k

k

µ α α β

α β λ λ µ ζ

β

⎡ − + − + ⎤

−

⎢ − ⎥

⎣ ⎦

Proof: For

0 ≤ µ < 1

, we must show that

f z ′ − ( ) 1 <1 − µ .

Similarly we show that

1

=2

( ) 1 =

_k ^k

k

f z ′ − ∑∞ka z − 1

=2

| |

^k

1

k k

ka z µ

∞ −

≤ ∑ ≤ −

1

=2

| | < 1.

1

k k k

k a z

µ

∞ −

⇒ ∑ −

By Theorem 2.1, the above inequality holds true if,

|z|^k−1< (1−µ)[(1+α)−k(α+β)]

k(1−β) ζ_k.

� 3. Radius of Starlikeness, Convexity and Close-to-convexity

In this section, we will calculate Radius of Starlikeness, Convexity and Close-to-convexity for

(6)

1 1 2

| |< = ( , , , , , , ) z r r α β λ λ l n µ

, where

1 1 1

( , , , , , , ) =

1 2

inf (1 )[(1 ( )(1 ) ( ) )] .

k k k

r l n k

k

µ α α β

α β λ λ µ ζ

µ β

⎡ − + − + ⎤

−

⎢ − − ⎥

⎣ ⎦

Proof: For

0 ≤ µ < 1

( ) 1 <1 . ( )

zf z

f z ′ − − µ

Now, we have to show that

1

=2

1

=2

( 1) ( ) ( ) =

( ) 1

k k k

k a z zf z f z

f z a z

∞ −

− −

′ −

−

∑

1

=2

1

=2

( 1) | |

<1

1 | |

k k k

k a z

a z µ

∞ −

−

≤ −

−

∑

1

=2

| | <1.

1

k k k

a z k µ

µ

∞ −

⎛ − ⎞

⇒ ∑ ⎜ ⎝ − ⎟ ⎠

1

(1 )[(1 ) ( )]

| | < .

( )(1 )

k

z k

µ α α β ζ

µ β

−

− + − +

− −

f ∈ τ α β λ λ ( , , , , , ).

₁ ₂

l n

Then f is convex of order

µ (0 ≤ µ <1)

in

2 1 2

| |< = ( , , , , , , ) z r r α β λ λ l n µ

, where

1 1 2

( , , , , , , ) =

1 2

inf (1 )[(1 ( ) )(1 ( ) )] .

k k k

r l n k

k k

µ α α β

α β λ λ µ ζ

µ β

⎡ − + − + ⎤

−

⎢ − − ⎥

⎣ ⎦

Proof: For

0 ≤ µ < 1

( ) <1 . ( )

zf z

f z ′′ − µ

′

We have to show that

1

=2

1

=2

( 1) 1

k k k

k k a z ka z

∞ −

− −

−

∑

1

=2

1

=2

( 1) | |

<1

1 | |

k k k

k k a z

ka z µ

∞ −

−

≤ −

−

∑

(7)

1

=2

| | <1.

1

k k k

ka z k µ

µ

∞ −

⎛ − ⎞

⇒ ∑ ⎜ ⎝ − ⎟ ⎠

1

(1 )[(1 ) ( )]

| | < .

( )(1 )

k

z k k

µ α α β ζ

µ β

−

− + − +

− −

Theorem 3.3 Let

f ∈ τ α β λ λ ( , , , , , ).

₁ ₂

l n

Then f is close-to-convex of order

µ (0 ≤ µ <1)

in

| |< = ( , , , , , , ) z r r

₃

α β λ λ

₁ ₂

l n µ

, where

1 1 3

( , , , , , , ) =

1 2

inf (1 )[(1 (1 ) ) ( )] .

k k k

r l n k

k

µ α α β

α β λ λ µ ζ

β

⎡ − + − + ⎤

−

⎢ − ⎥

⎣ ⎦

Proof: For

0 ≤ µ < 1

, we must show that f′(z)−1 < 1−µ.

We have to show that

f′(z)−1 =

k=2

∑

∞ ^ka^k^z^k−1 ¹

=2

| |

^k

1

k k

ka z µ

∞ −

≤ ∑ ≤ −

1

=2

| | < 1.

1

k k k

k a z µ

∞ −

⇒ ∑ −

By Theorem 2.1, the above inequality holds true if

1

(1 )[(1 ) ( )]

| | < .

(1 )

k

z k

µ α α β ζ

β

−

− + − +

−

�

(8)

Acknowledgement

The work presented here was partially supported by UKM-ST-06-FRGS0244-2010.

References

Amer A. A. & Darus M. 2011. On some properties for new generalized derivative operator. Jordan Journal of Mathematics and Statistics (JJMS) 4(2): 91-101.

Amer A. A. & Darus M. 2012. On preserving the univalence integral operator. Applied Sciences 14: 15-25.

Najafzadeh Sh. & Ebadian A. 2009. Neighborhood and partial sum property for univalent holomorphic functions in terms of Komatu operator. Acta Universitatis Apulensis 19: 81-89.

Tehranchi A. & Kulkarni S.R. 2006a. Some integral operators defined on p-valent functions by using hypergeometric functions. Studia Univ. Babes, Bolyai Mathematica (Cluj) 1: 525-532.

Tehranchi. A & Kulkarni S.R. 2006b. Study of the class of univalent functions with negative coefficients defined by Ruscheweyh derivative. J. Rajasthan academy of Physical Science 5(1): 169-180.

School of Mathematical Sciences Faculty of Science and Technology Universiti Kebangsaan Malaysia 43600 UKM Bangi

Selangor DE, MALAYSIA

E-mail: eamer_80@yahoo.com, maslina@ukm.my^*

* Corresponding author