International Journal of Scientific & Engineering Research, Volume 5, Issue 2, February-2014 8

ISSN 2229-5518

SANDWICH-TYPE THEOREMS FOR MEROMORPHIC MULTIVALENT FUNCTIONS ASSOCIATED WITH A LINEAR OPERATOR

H. E. Darwish, A. Y. Lashin, and B. H. Rizqan

Abstract— the purpose of this article is to obtain some subordination and superordination preserving properties of meromorphic

p

multivalent functions in the punctured open unit disk associated with the linear operator meromorphic multivalent functions are also considered.

Qα , β , γ

the sandwish- type results for these

2010 Mathematics Subject Classification: Primary 30C45.

Keywords— Meromorphic function, Hadamard product, Subordination, Superordination, Meromorphic multivalent function, Linear operators,

Sandwich result.

—————————— ——————————

open unit disk

U = {*z *∈ C :

z < 1}.

a dominant, if

p q

for all p satisfying (1.1). A domnant

ET *H *(U)

denote the class of analytic functions in the

solutions of the differential subordination, or more simply

For n ∈ N = {1, 2,...}

and

a ∈ C,

let

that satisfies for all dominants q of (1.1) is said

H [a, n] = { f ∈ H :

f ( z) = a + an

z n + a

n+1

z n+1 + ...}.

to be the best dominant.

Let f and F be members of

H . the function f is

Definition 2 [10]. Let ϕ

: C 2 → C

and let h be analytic

said to be subordinate to

F , or F is said to be superordi-

in U . If p and

ϕ ( *p*( *z*), *zp*′( *z*))

are univalent in U

nate to f , if there exists a function w analytic in U ,

and satisfy the differential superordination:

with w(0) = 0

and

w( z) < 1

( *z *∈ U),

such that

*h*( *z*) ϕ ( *p*( *z*), *zp*′( *z*))

( *z *∈ U),

(1.2)

f ( z) = F (w( z))

( *z *∈ U).

then p is called a solution of the differential superordination.

An analytic function q is called a subordinant of the

In such a case, we write

soltions of the differential superordination, or more simply

f F

( *z *∈ U) or

f ( z) F ( z)

( *z *∈ U).

a subordinant if

q p

for all p satisfying (1.2). A uni-

If the function F is univalent in U , then we have (cf. [11])

f ( z) F ( z)

( *z *∈ U) ⇔

f (0) = F (0)

vlent subordinant that for all subordinants q of

(1.2) is said to be the best subordinant.

and

*f *(U) ⊂ *F *(U).

**Definition 3 [10]. **Denote by Q the class of functions f

Definition 1 [9]. Let φ

: C 2 → C

and let h be univalent

that are analytic and injective on

U \ *E*( *f *),

where

in U . If p is analytic in U and satisfies the differential

subordination:

*E*( *f *) = {ζ ∈ ∂*U*

: lim *f *( *z*) = ∞},

z →ζ

φ ( *p*( *z*), *zp*′( *z*)) *h*( *z*)

( *z *∈ U),

(1.1)

and are such that

then p is called a solution of the differential subordination.

The univalent function q is called a dominant of the

*f *′(ζ ) ≠ 0(ζ ∈ ∂U \ *E*( *f *)).

For any integer

m > − p,

Let

Σ p , m

denote the class of

————————————————

Department of Mathematics, Faculty of Science,

all meromorphic functions f of the form

∞

• *H. E. Darwish, Mansoura University, Mansoura, Egypt*

• *Y. Lashin, Mansoura University, Mansoura, Egypt*

f ( z) = z − p + ∑ a z k

k =m

( *p *∈ N = {1, 2,...}),

(1.3)

E-mail: aylashin@mans.edu.eg

• *B. H. Rizqan, Mansoura University, Mansoura, Egypt E-mail :*

which are analytic and p -valent in the punctured unit disk

bakeralhaaiti@yahoo.com

U ∗ = {*z *∈ C :

0 < *z*

< 1} = U \ {0}.

For convenience, we

write

Σ p ,1− p = Σ p .

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ISSN 2229-5518

For functions

f ∈ Σ p , m

given by (1.3), and

g ∈ Σ p , m

If is not subordinate to , then there exist points

iθ

defined by

∞

z0 = r0 e

∈ U and ζ 0 ∈ ∂U \ *E*( *f *),

g ( z) = z − p + ∑ b z k

=

(*m *> − *p*; *p *∈ N),

(1.4)

for which

q(U

) ⊂ *p*(U),

q( z

) = *p*(ζ )

k m r0 0 0

then the Hadamard product (or convolution) of f and g

is

and

z0 q′( z0

) = *m*ζ 0

*p*′(ζ 0 )

(*m *≥ *n*).

∞ A function

L( z, t )

defined on

U × [0, ∞),

is the subordi-

( *f *∗ *g *)( *z*) = *z *− p + ∑ *a b z *k

= ( *g *∗ *f *)( *z*)

nation chain (or Löwner chain) if

L( z,.)

is analytic and uni-

k k

k =m

(1.5)

valent in U for all

t ∈ [0, ∞),

L( z,.)

is continuously

(*m *> − *p*; *p *∈ N).

Differentiable on

[0, ∞)

for all

z ∈ U

and

For

f ∈ Σ p , m

we now define the integral operator

L( z, s) L( z, t )

( *z *∈ U; 0 ≤ *s *< *t *).

p

α , β , γ

: Σ p , m

→ Σ p , m

which was introduced and studied

by El-Ashwah et al. [3] as follows:

*q *∈ H[*a*,1]

and ϕ

: C 2 → C.

Q p f ( z) =

Γ(α + β − γ + 1) 1 ⋅

Also set

α , β , γ

Γ(β )Γ(α − γ + 1) *z *β + p

ϕ (*q*( *z*), *zq*′( *z*)) ≡ *h*( *z*)

( *z *∈ U).

z α −γ

If *L*( *z*, *t *) = ϕ (*q*( *z*), *tzq*′( *z*))

is a subordination chain and

− *t * *t*

f t dt

*p *∈ H[*a*,1] ∩ *Q*,

1

0 *z *

β + *p *−1 ( )

then.

= 1 + Γ(α + β − γ + 1) ⋅

*z *p Γ(β )

∞ Γ(β + *p *+ *k *)

∑

a z k

(1.6)

implies that

*h*( *z*) ϕ ( *p*( *z*), *zp*′( *z*))

( *z *∈ U).

k =m

Γ(α + β + *p *+ *k *− γ + 1)

Furthermore, if

ϕ (*q*( *z*), *zp*′( *z*)) = *h*( *z*)

has a univalent

and

p

(β > 0; α > γ − 1; γ

> 0;

p ∈ N;

*z *∈ U∗ ),

∗

solution

q ∈ Q,

then q is the best subordinate.

2

Qα −1, β , γ f ( z) =

f ( z)

(β > 0; γ

> 0;

p ∈ N;

*z *∈ U ).

From (1.6), it is easy to verify that

p p

satisfies the following condition:

Re{*H *(*is*, *t *)} ≤ 0

*z*(*Q*α +1, β , γ *f *( *z*))′ = (α + β − γ + 1)*Q*α , β , γ *f *( *z*) −

(α + β + *p *− γ + 1)*Q *p *f *( *z*).

(1.7)

for all real s and

t ≤ −n(1 + s 2 ) / 2

(*n *∈ N).

If the function

p( z) = 1 + pn

z n + ....

is analytic in U and

Remark:

p p

p Re{*H *( *p*( *z*), *zp*′( *z*))} > 0

( *z *∈ U),

(i) For

γ = 1, *Q*α , β ,1 = *Q*α , β ,

where the operator

Qα , β

then

was introduced and studied by Aqlan et al. [2] (see also [1]);

Re{ *p*( *z*)} > 0

( *z *∈ U).

1 *p*

(ii) For

p = γ

= 1, *Q*α , β ,1 = *Q*α , β ,

where the operator

Qα , β ,

was introduced and studied by Lashin [6].

β , γ ∈ C

with

β ≠ 0

and

*h *∈ H(U)

with

h(0) = c.

If

Re{β*h*( *z*) + γ } > 0

( *z *∈ U),

The following lemmas will be required in our present

then, the solution of the differential equation

investigation.

p ∈ Q

with

p(0) = a

and let

q( z) +

zq′( z)

βq( z) + γ

= *h*( *z*)

( *z *∈ U;

q(0) = c),

q( z) = a + an

z n + .....

is analytic in U satisfies the inequality

be analytic in U with

Re{β*q*( *z*) + γ } > 0

( *z *∈ U).

q( z) ≠ a

and

n ≥ 1.

L( z, t ) = a1 (t ) z + ...

with

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a1 (t ) ≠ 0

and

lim*a*1 (*t *) = ∞

t →∞

is a subordination chain if

then,

Re{*q*( *z*)} > 0

( *z *∈ U).

and only if

*z*∂*L*( *z*, *t *)

Taking the logarithmic differentiation on both sides of the se- cond equation in (3.5) and using the equation (1.7) we obtain

Re > 0

( *z *∈ U; 0 ≤ *t *< ∞),

*p*(α + β − γ )φ ( *z*) = *p*(α + β − γ )*G*( *z*) + ( *p *− η ) *zG*′( *z*).

__=__∂*L*( *z*, *t *)

(3.7)

and

∂*t *

Now, by differentiating both sides of (3.7), We obtain the rela- tionship:

L( z; t ) ≤ K 0 a1 (t ) ,

z < r0 < 1, t ≥ 0,

1 + *z*φ ′ ( *z*) = 1 + *zG*′ ( *z*) +

zq′( z)

φ ′( *z*)

G′( z)

*q*( *z*) + *p*(α + β − γ ) /( *p *− η )

For some positive constants

a subordination chain.

K 0 and

r0 , then

L( z, t ) is

= *q*( *z*) +

zq′( z)

*q*( *z*) + *p*(α + β − γ ) /( *p *− η )

≡ *h*( *z*).

(3.8)

We also note from (3.1) that

We begin with proving the following subordination theorem

p

Re*h*( *z*)

+ *p*(α + β − γ

p −η

)

> 0

( *z *∈ U),

involving the operator

Qα , β , γ f ( z)

defined by (1.6).

and so by Lemma 4, we conclude that the differential equation

(3.8) has a solution

*q *∈ H(U)

with

*f *, *g *∈ Σ p , m , α > γ ,

β > 0, γ

> 0,

p ∈ N,

0 ≤ η < *p*

q(0) = h(0) = 1.

and

*z *∈ U.

Suppose that

*z*φ ′

Re1 +

( *z*)

> −δ ,

(3.1)

Let us put

H (u, v) = u +

v + δ

*u *+ *p*(α + β − γ ) /( *p *− η )

(3.9)

setting

where

φ ′( *z*)

where δ is given by (3.2).

From (3.1), (3.8) and (3.9), we obtain

Now, we proceed to show that

( *p *− η ) 2 + *p *2 (α + β − γ ) 2

Re{

( , )} 0 ;

1 (1

2 ) .

δ = −

4 *p*( *p *− η )(α + β − γ )

H is t ≤

*s *∈ R

t ≤ −

2

+ *s *

(3.10)

( *p *− η ) 2 − *p *2 (α + β − γ ) 2

(0 < δ ≤ 1/ 2).

(3.2)

Indeed, From (3.9), we have

*t*

Re{*H *(*is*, *t *)} = Re*is *+

+ δ

4 *p*( *p *− η )(α + β − γ )

*is *+ *p*(α + β − γ ) /( *p *−η )

Then, the following subordination relation

= *tp*(α + β − γ ) / *p *− η + δ

*p*(α + β − γ ) /( *p *− η ) + *is *2

implies that

(3.3)

≤ − *E*δ (*s*) ,

2 *p*(α + β − γ ) /( *p *− η ) + *is *2

(3.11)

z p Q p f ( z) z

p p

α , β , γ

( *z *∈ U).

(3.4)

where

p p *p *α + β − γ

*p *α β γ

Moreover, the function

z Qα , β , γ g ( z)

is the best dominant.

E (s) := (

) − 2δ *s *2 − ( +

− ) ⋅

Proof. Let us define the functions F and G,

δ

respectively,

*p *− η

*p *− η

by

F ( z) := z

Qα , β , γ f ( z)

and

G( z) := z

Qα , β , γ g ( z). (3.5)

2δ

*p*(α + β − γ )

.

*p *− η

(3.12)

We first show. If the function is defined by

q( z) := 1 + zG ′ ( z)

( *z *∈ U),

(3.6)

For δ given by (3.2), we can prove easily that the expression

G ′( z)

Eδ (s)

given by (3.12) is greater than or equal to zero. Hence,

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from (3.9), we see that (3.10) holds true. Thus using Lemma 3, we conclude that.

L(ζ

0 , *t *) = *G*(ζ

) + ( *p *− η )(1 + *t *) ζ

0 *p*(α + β − γ )

0 G′(ζ 0 )

Re{*q*( *z*)} > 0

( *z *∈ U).

= + *p *− η

Moreover, we see that the condition:

G′(0) ≠ 0

F ( z0 )

*p*(α + β − γ )

z0 F ′( z0 )

is satisfied. Hence, the function defined by (3.5) is convex

( *p *− η ) p p

( ) η p p

( ) ( ),

in U.

= z0 Qα , β , γ f z0

p

+ z0 Qα , β , γ f p

z0 ∈ φ U

Next, we prove that the subordination condition (3.3) implies

by virtue of the subordination condition (3.3). This contra-

that

dicts the above observation

*L*(ζ 0 , *t *) ∉ φ (U). Therefore, the

F ( z) G( z)

( *z *∈ U)

(3.13)

subordination condition (3.3) must imply the subordination

for the functions F and G defined by (3.5). Without loss

given by (3.13). Considering

F ( z) = G( z), we see that the

of generality, we can assume that G is analytic and univa-

lent on U and

function G is the best dominant. This evidently completes the proof of Theorem 1.

*G*′(ζ ) ≠ 0

(ζ ∈ ∂U).

We next provide a dual problem of Theorem 1, in the

For this purpose, we consider the function

L( z, t )

given by

sense that the subordinations are replaced by superordina-

tions.

*L*( *z*, *t *) := *G*( *z*) + ( *p *− η )(1 + *t *) *zG*′( *z*)

*p*(α + β − γ )

(0 ≤ *t *< ∞; 0 ≤ η < *p*;

*z *∈ U).

*f *, *g *∈ Σ p , m , α > γ ,

β > 0, γ

> 0,

p ∈ N,

0 ≤ η < *p*

We note that

∂*L*( *z*, *t *)

= *G*′(0) ⋅

and

*z *∈ U.

Suppose that

Re1 +

zφ ′

( *z*)

> −δ ,

∂*z *z =0

φ ′( *z*)

*p *α + β − γ

+ *p *− η

+ *t *

setting

( ) (

)(1

) ≠ 0.

p − η η

(α + β − γ )

φ ( *z*) :=

z Qα

p

, β , γ g ( z) +

z Qα , β , γ g ( z),

p

This shows that the function

*L*( *z*, *t *) = α1 (*t *) *z *+ ...

−

where δ is given by (3.2), and

p − η η

satisfies the condition

α1 (*t *) ≠ 0

(0 ≤ *t *< ∞).

p p

p α −1

, β , γ f ( z) +

z p Q p f ( z),

p

p p

α β γ

∈ H ∩

Then,

Furthermore, we have

is univalent in U and

z Q , ,

f ( z)

[1,1] *Q*.

Re *z*∂*L*(

z, t) /

∂*z *

=

the following superordination relation

∂*L*( *z*, *t*) / ∂*t *

φ ( *z*)

p − η z p Q p

f ( z) + η z p Q p

*f *( *z*) ( *z *∈ U),

*p *α + β − γ

*zG*′ *z *

p α −1 , β , γ

p α , β , γ

( Re

) + (1 + *t *)1 +

( ) > 0.

(3.14)

*p *− η

*G*′( *z*)

implies that

p p p p

Therefore, by virtue of Lemma 5,

L( z, t )

is a subordination

z Qα , β , γ g ( z) z

p

Qα , β , γ f ( z)

p

( *z *∈ U).

chain. We observe from the definition of subordination chain

that

Moreover, the function

z Qα , β , γ g ( z)

is the best subordi-

*L*(ζ , *t *) ∉ *L*(U, 0) = φ (U)

(ζ ∈ ∂U; 0 ≤ *t *< ∞).

nant.

Proof. The first part of the proof is similar to that of Theorem 1

Now, suppose that F is not subordinate to G,

then by

and so we will use the same notation as in the proof of

Lemma 1, there exists points

that

z0 ∈ U

and

ζ 0 ∈ ∂U,

such

Therem1.

Now, let us define the functions F and G by (3.5). We

q is defined by (3.6), using

*F *( *z*0 ) = *G*(ζ 0 )

(0 ≤ *t *< ∞).

and

z0 F ( z0 ) = (1 + t )ζ 0 G′(ζ 0 )

first note that, if the function

(3.7), then we obtain

Hence, we have

φ ( *z*) = *G*( *z*) +

p − η zG′

*p*(α + β − γ )

:= ϕ (*G*( *z*), *zG*′( *z*)).

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(3.15) Then using the same method as in the proof of Theorem 1.

p p

α , β , γ 2

are the best subordinate and the best

We can prove that

Re{*q*( *z*)} > 0

( *z *∈ U),

dominant, respectively.

The assumption of Theorem 3, that the functions

that is, G defined by (3.5) is convex (univalent) in U . Next, we prove that the subordination condition (3.14) implies that

need to be univalent in U , may be replaced by another

G( z) F ( z)

(3.16)

( *z *∈ U)

condition in the following result.

for the functions F and G defined by (3.5). Now, consider

Corollary 1. Let

the function

L( z, t )

defined by

f , g k ∈ Σ p , m

(*k *= 1, 2), α > γ ,

β > 0, γ

> 0,

p ∈ N,

L( z, t ) := G( z) +

( *p *− η )*t*

zG′( z)(0 ≤ t < ∞;

*z *∈ U).

0 ≤ η < *p*

*p*(α + β − γ )

and

*z *∈ U.

Suppose that condition (3.17) is satisfied and

As G is convex and

*p*(α + β − γ ) /( *p *− η ) > 0,

we can

Re1 +

zψ ′

( *z*)

> −δ ,

(3.18)

prove easily that

L( z, t )

is a subordination chain as in the

ψ ′( *z*)

proof of Theorem 1. Therefore, according to Lemma 2, we conclude that the superordination condition (3.14) must

setting

ψ ( *z*) :=

p − η z p Q p

f ( z) + η z p Q p

f ( z),

imply the superordination given by (3.16). Furthermore, as

the differential equation (3.15) has the univalent solution

p α −1 , β , γ

p α , β , γ

G , it is the best subordinate of the given differential su-

where δ is given by (3.2). Then, the following relation

perordination. Therefore, we complete the proof of Theo- rem 2. If we combine Theorems 1 and 2, then we obtain the

φ1 ( *z*)

p − η

p

p p

α −1

, β , γ f ( z) +

η z p

p

p

α , β , γ

following sandwich -type theorem.

φ2 ( *z*)

implies that

( *z *∈ U),

f , g k ∈ Σ p , m

(*k *= 1, 2), α > γ ,

β > 0, γ

> 0,

p ∈ N,

0 ≤ η < *p*

Moreover, the function

z Qα , β , γ g1 ( z)

and

and

*z *∈ U.

Suppose that

′′

p p

α , β , γ

g 2 ( z)

are the best subordinate and the best

zφ ( z)

Re1 + k > −δ ,

(3.17)

dominant, respectively.

φ ′ ( *z*)

Proof. To prove Corollary 1, we have to show that condi-

ψ and

setting

tion (3.18) implies univalent of

p p

( *z*)

F ( z) := z

Qα , β , γ g1 ( z).

As 0 < δ ≤ 1 / 2

from Theorem 1, condition (3.18) means

where δ is given by (3.2), and

that ψ is a close- to-convex function in U (see [4]) and

p − η

p

p p

α −1

, β , γ f ( z) +

η z p

p

Q p f ( z),

hence ψ is univalent in U . Furthermore, using the same techniques as in the proof of Theorem 1, we can prove the

is univalent in U and the following relation

z Qα , β , γ f ( z) ∈ H[1,1] ∩ Q.

Then,

convexity (univalent) of F and so the details may be omit- ted here. Therefore, by applying Theorem 3, we obtain

Corollary1.

φ1 ( *z*)

p − η

p

p p

α −1

, β , γ f ( z) +

η z p

p

p

α , β , γ

φ2 ( *z*)

( *z *∈ U)

f , g k ∈ Σ p , m

(*k *= 1, 2), α > ( *p *− η ) / *p*,

0 ≤ η < *p*

Implies that

and

*z *∈ U .

Suppose that

*z*φ ′′ ( *z*)

p p Re1 + ′

> −δ ,

(3.19)

Moreover, the function

z Qα , β , γ g1 ( z)

and

φk ( *z*)

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setting

z p Q p F ( z) ∈ H[1,1] ∩ Q.

Then, the following relation

φk ( *z*) :=

p − η

p +1 p

α

, β , γ

g k ( z) +

η z p +1

p

α , β , γ

g k ( z),

φ ( *z*) *z*

p Q p

f ( z) φ

( *z*)

( *z *∈ U)

p −1 p

where

1

Implies that

p p

α , β , γ 2

p p

( *p *− η ) 2 + ( *p*(α + β − γ − 1) + η ) 2

z Qα , β , γ Fµ ( g1 )( z) z

Qα , β , γ Fµ ( f )( z)

δ = −

4( *p *− η )( *p*(α + β − γ − 1) + η )

z Q p F ( g )( z)

( *z *∈ U).

( *p *− η ) 2 − ( *p*(α + β − γ − 1) + η ) 2

Moreover, the functions

z p Q p F ( g )( z)

and

,

4( *p *− η )( *p*(α + β − γ − 1) + η )

z p Q p F ( g )( z)

are the best subordinate and the best

and

p − η

p +1 p

α

, β , γ f ( z) +

η z p +1

Q p f ( z),

dominant, respectively.**Proof**. Let us define the functions F and Gk

(*k *= 1, 2) by

p −1

p

p +1 p

F ( z) := z p Q p

Fµ ( f )( z)

Is univalent in U and z

Qα , β , γ f ( z) ∈ H[0,1] ∩ Q.

and

G ( z) := z p Q p

Fµ ( g k

)( *z*),

Then, the following relation

respectively. Without loss of generality, as in the proof of

φ1 ( *z*)

p − η

p

z p +1

p

α −1

, β , γ f ( z) +

η z p +1

p

p

α , β , γ

Theorem 1, we can assume that Gk

univalent on U , and

is analytic and

φ2 ( *z*)

( *z *∈ U),

Implies that

*G*′ (ζ ) ≠ 0

(ζ ∈ ∂U).

z p +1

Q p g ( z) z

p +1

p

α , β , γ

From the definition of the integral operator

Fµ defined by

p +1

p

α , β , γ 2

( *z *∈ U).

(3.20), we obtain

z(Q p

F ( f )( z))′ = µQ p

f ( z) −

Moreover, the functions z

p +1

p

α , β , γ 1

and dominant,

α , β , γ µ

p

α −1 , β , γ

respectively.

Next, we consider the integral operator by (ef. [5],[12])

z

µ + *p *−1

*F*µ (µ > 0)

defined

(µ + *p*)*Q*α , β , γ *F*µ ( *f *)( *z*).

Then, from (3.21) and (3.23), we have

µφ ( *z*) = µ*G *( *z*) + *zG*′ ( *z*).

Setting

(3.23)

(3.24)

Fµ ( f )( z) := µ + p ∫ t

0

f ( z)dt

( *f *∈ Σ p , m ;

µ > 0)

(3.20)

zG′′ ( z)

q ( z) = 1 + k

k G′ ( z)

(*k *= 1, 2;

*z *∈ U),

(3.25)

Now, we obtain the following result involving the integral

operator defined by (3.20).

k

And differentating both sides of (3.24), we obtain

f , g k ∈ Σ p , m

(*k *= 1, 2).

Suppose also

zφ ′′

1 + k

′

( *z*) = *q*

( *z*) +

zq′ ( z)

k .

q ( z) + µ

that

φk ( *z*) k

Re1

zφ ′′

+ k

( *z*)

> −δ

The remaining part of the proof is similar to that of

Theorem 1 and so is omitted the proof involved.

Setting

(3.21)

p p

φ ′ ( *z*)

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where

φk ( *z*) := *z*

Qα , β , γ g k ( z); k = 1, 2; z ∈ U,

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[2] E. Aqlan, J. M. Jahangiri and S. R. Kulkarni, "Certain integral operators ap- plied to meromorphic p-valent functions", J. Nat. Geom., 24 (2003), 111-120.

1 + µ 2 − 1 − µ 2

δ =

4µ

p p

(µ > 0),

(3.22)

[3] R. M. El-Ashwah, M. K. Aouf and Ahmed M. Abd-Eltawab, "Applications of differential subordination on certain subclasses of p-valent meromorphic functions", An. Univ. Oradea Fasc. Mat., 20 (2013), no. 1, 125-136.

[4] W. Kaplan, "Close-to-convex schlicht functions", Michigan Math. J., 2 (1952),

and

z Qα , β , γ f ( z)

is univalent in U and

169-185.

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International Journal of Scientific & Engineering Research, Volume 5, Issue 2, February-2014 14

ISSN 2229-5518

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