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

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

ISSN 2229-5518

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

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.

1 INTRODUCTION

——————————  ——————————
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

E-mail: Darwish333@yahoo.com

• 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 .

International Journal of Scientific & Engineering Research, Volume 5, Issue 2, February-2014 9

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

) = mζ 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 ).

α , β , γ

: Σ p , m

→ Σ p , m

which was introduced and studied
by El-Ashwah et al. [3] as follows:

Lemma 2 [10]. Let

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

(β > 0; α > γ − 1; γ

> 0;

p ∈ N;

z ∈ U∗ ),

∗

solution

q ∈ Q,

then q is the best subordinate.

Qα −1, β , γ f ( z) =

f ( z)

(β > 0; γ

> 0;

p ∈ N;

z ∈ U ).

Lemma 3 [7]. Suppose that the function H : C → C

From (1.6), it is easy to verify that

p p

satisfies the following condition:

Re{H (is, t )} ≤ 0

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

where the operator

Qα , β ,

was introduced and studied by Lashin [6].

Lemma 4 [8]. Let

β , γ ∈ C

with

β ≠ 0

and

h ∈ H(U)

2. A SET OF LEMMS

with

h(0) = c.

Re{βh( z) + γ } > 0

( z ∈ U),

The following lemmas will be required in our present
then, the solution of the differential equation
investigation.

Lemma 1 [9]. Let

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{βq( z) + γ } > 0

( z ∈ U).

q( z) ≠ a

and

n ≥ 1.

Lemma 5 [11]. The function

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

with

International Journal of Scientific & Engineering Research, Volume 5, Issue 2, February-2014 10

ISSN 2229-5518

a1 (t ) ≠ 0

and

lima1 (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 + zφ ′ ( 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)

3. MAIN RESULTS

We also note from (3.1) that
We begin with proving the following subordination theorem



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

Theorem 1. Let

(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

 zφ ′

Re1 +

( 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 ≤ −

+ 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)

≤ − Eδ (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,

respectively,

 p − η

 p − η

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

+ 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 )

Theorem 2. Let

(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



Re1 +

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

, β , γ g ( z) +

z Qα , β , γ g ( z),

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

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

Qα , β , γ f ( z)

( 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 ( z0 ) = 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)).

International Journal of Scientific & Engineering Research, Volume 5, Issue 2, February-2014 12

ISSN 2229-5518

(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



Re1 +

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

α −1

, β , γ f ( z) +

η z p

α , β , γ

following sandwich -type theorem.

Theorem 3. Let

 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)

Re1 + 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

α −1

, β , γ f ( z) +

η z 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

α −1

, β , γ f ( z) +

η z p

α , β , γ

Theorem 4. Let

 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

 zφ ′′ ( z) 

p p Re1 + ′

 > −δ ,

(3.19)
Moreover, the function

z Qα , β , γ g1 ( z)

and

 φk ( z) 

International Journal of Scientific & Engineering Research, Volume 5, Issue 2, February-2014 13

ISSN 2229-5518

setting

z p Q p F ( z) ∈ H[1,1] ∩ Q.

Then, the following relation

φk ( z) :=

p − η

p +1 p

, β , γ

g k ( z) +

η z p +1

α , β , γ

g k ( z),

φ ( z)  z

p Q p

f ( z)  φ

( z)

( z ∈ U)

p −1 p

where

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

, 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 +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

z p +1

α −1

, β , γ f ( z) +

η z p +1

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

α , β , γ

From the definition of the integral operator

Fµ defined by

p +1



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

α , β , γ 1

and dominant,

α , β , γ µ

α −1 , β , γ

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

µ + p −1

Fµ (µ > 0)

defined

(µ + p)Qα , β , γ Fµ ( f )( z).

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

µφ ( z) = µG ( z) + zG′ ( z).

Setting

(3.23)

(3.24)

Fµ ( f )( z) := µ + p ∫ t

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).

And differentating both sides of (3.24), we obtain

Theorem 5. Let

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



Re1

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) 

REFERENCES

[1] M. K. Aouf, A. Shamandy, A. O. Mostafa and F. Z. El-Emam, "On certain subclasses of meromorphically p-valent functions associated with integral op-

where

φk ( z) := z

Qα , β , γ g k ( z); k = 1, 2; z ∈ U,

erators", European J. Pure Appl. Math., 4 (2011), no. 4, 435-447.

[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.

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

ISSN 2229-5518

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[8] S. S. Miller and P. T. Mocanu, "Univalent solutions of Briot-Bouquet differen- tial equations", J. Different Eq., 57 (1985), 297-309.

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[10] S. S. Miller and P. T. Mocanu, "Subordinates of differential superordinations", Complex Var Theory Appl, 48 (2003), 815-826.

[11] C. Pommerenke, "Univalent functions", Vanderhoeck and Ruprech, Göttin- gen, 1975.

[12] H. M. Srivastava and S. Owa, "Current Topics in Analytic Function Theory", World Scientific Publishing Company, Singapore, New Jersey, London, Hong

Kong, 1992.