# Applications Of Meromorphic Univalent Functions Associated With Differential Subordination

DOI : 10.17577/IJERTV2IS3595

Text Only Version

#### Applications Of Meromorphic Univalent Functions Associated With Differential Subordination

*Dr. S. M. Khairnar, **R. A. Sukne

*Professor and Head and Dean (R & D)

MITs Maharashtra Academy of Engineering, Alandi, Pune-412105

**Assistant Professor in Mathematics

Dilkap Research Institute of Engineering & Management Studies, Karjat, Dist. Raigad.

Abstract

(3) 14 < 2(), 1 < 1 .

14 2 14 2 2 2

In this paper authors introduced subclasses (, ) of meromorphic univalent functions in the punctured unit disk = : 0 < < 1 =D\ 0 . By using the method of differential subordinations, we derive some

certain properties of meromorphically univalent func- tions.

(4) Re > 12, 2 = 1, .

2

Several authors recently proved some interesting

properties of meromorphically univalent functions. In the present topic, we are going to prove some subordination properties for the class S.

When = = 1 + 0 , We define the

=

Key Words Analytic function, Subordination, Meromorphically univalent function, convolution.

1. Introduction

Hadamard product (convolution) of f z and g z by

=

=

= = 1 + 0 .

Where 0 = 0 , .

We define a linear operator by

, , , , ,

Let S denote the class of functions of the form:

1

=

=

0

0

= 1 +

+ (1+)

()

+ 1

(1) = 1 + =0 ,

Which are analytic and univalent in the punctured unit

disc .

= 1, .

, , ,

, , ,

Where

1, = 1 +

(1 + ) + 1

= : 0 < < 1 =\ 0 .

, , , , ,

=0

Let () and () be analytic in , then, we say that () is subordinate to () in .

> 0, 0 = 0 , , 0, 0,

& 0 < 1 , 0 < 1 , > 0, 0, > .

2 2

Where () (), if there exists an analytic function in , such that () and = , . If () is univalent in then the subordination

() ()() 0 = (0) and () ().

Let = 1 + 1 + 2 + be analytic in D,

For simplicity throughout the paper we are using

1

1

, , , , , = ()

and = + 0.

()

It is easy to verify that

1

1

(5) + , , , = ()

+ + ().

1 1 We note that

1

1

< such that

2 2

0 , , , , , = and

(2) () 1+2 ( ) If and only if

1 1 , 1,1,1, 1 , 1 = 2

1+2

1 2 2

= 2 + .

The above operators are analytic in D and satisfy the following condition

() > 0, .

Where

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

for , = 2 , 2 2

1,

1,

(6) , , , , , ; Definition 1.1.4 A function is said to be in the

1 1 ,1

class (; , , , , , ; ) if it satisfies the

= =0

()

1,

followin ubordination condition

= 1 + . g s

1

1

1,

1,

, , , , , ;

1+ () + +1 , , ,, ,

1+ , , ,, ,; + +1 , , ,, ,;

.

= 1 () +

1,

1,1

2

= 1 + .

1,

1,

, , , , , ;

1

, 0, 1, ; , , , , , ; and +1 , , , , , ; 0.

1,1

1,1

For simplicity we can write

0; , , , , , ; = , , , , , ; .

= 1 () +

2

= + .

1,

Where

1,

= 1 we have,

> 0, 0 = 0 , , 0, 0,

1,1

1,1

, , , , , ; = ().

We now introduce and investigate the following

0 <

1 , 0 <

1,

1,

2

1 , > 0, 0, > .

2

subclasses of the class of meromorphically univalent functions.

1. Definitions

Definition 1.1.5 A function is said to be in the class (; , , , , , ; ) if it satisfies the following subordination condition:

1+ () + +1 , , ,, ,

1 .

1+ , , , , ,; + +1 , , ,, ,;

1, 1,1

Definition 1.1.1 A function is said to be in the

class 1, (; , , , , , ; ) if it satisfies the

, 0, ; , , , , , ; and +1 , , , , ; 0.

1,

1,

1,

,

following subordination condition:

0; , , , , , ; = , , , , , ; .

1+ () + +1 , , ,, ,

1,

1,

1 .

Where

1+ , , ,, ,; + +1 , , ,, ,;

> 0,

= 0 , , 0, 0,

1,

0, ,

1,1

0

0 < 1 , 0 < 1 , > 0, 0, > .

; , , , , , ; , +1 , , , , , ; 0 . 2 2

1, 1,1

For simplicity we can write

0; , , , , , ; = , , , , , ; .

Definition 1.1.6 A function is said to be in the

1,

Where

1,

class 1, (; , , , , , ; ) if it satisfies following

1+ () + +1 , , ,, ,

= 0 , , 0, 0,

1 .

0 1+ , , , , ,; + +1 , , ,, ,;

1 1 .

1,

1,1

0 < , 0 < , > 0, 0, >

2 2

Definition 1.1.2 A function is said to be in the

, 0, ; , , , , , ; and +1 , , , , , ; 0.

1,

1,

1,

1,

0; , , , , , ; = , , , , , ; .

1

1

class (; , , , , , ; ) if it satisfies the

1,

Where

1,

following subordination condition:

> 0, 0 = 0 , , 0, 0,

1

1

1+ () + +1 , , ,, ,

1+ , , ,, ,; + +1 , , ,, ,;

0 <

1 , 0 <

2

1 , > 0, 0, > .

2

1 1

1

1

1

1

Where , 0 , ; , , , , , ; , and +1 , , , , , ; 0.

0; , , , , , ; = , , , , , ; .

2. Remarks

Remark 1.2.1 Putting = = = 1, = 1 , = 0,

1 1 2

= 2, and = 1+.

Definition 1.1.3 A function is said to be in the class (; , , , , , ; ) if it satisfies the following subordination condition:

1

1

1+ () + +1 , , ,, ,

1

In definition 1.1.1, we have the class

1+3 +

1+3 + > 0,

1 .

1+ 1 , , ,, ,; +1 +1 , , ,, ,;

1

1

1

1

, 0, ; , , , , , ; and +1 , , , , , ; 0.

where = 1 .

2

2

Remark 1.2.2 For = 0, = = 1, = 1 we

1 0; , , , , , ; = 1 , , , , , ; .

2

have the class

0; 1 , 1,1, , , ; = , , ; .

Lemma 1.3.3 Let , ; and () is convex and

1, 2

,

univalent in D with

Where the class (, , ; ) consisting of

functions ,

1,

this

satisfies the following

0 = 1 and + > 0.

If () is analytic in with 0 = 1, then the

subordination condition

, ,

1

1

.

following subordination:

Where ,

1,

1,

, , ;

1, (, ,;)

+ ()

+

() () () ( ).

=0 1

=0 1

= 1 1 , , , , , 0,

and

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

Lemma 1.3.4 Let , ; and () is convex and univalent in with 0 = 1 , + > 0. Also let () () ( ).

If is analytic in D with 0 = 1, then the following subordination

2 2

+ ()

() () () ( ).

Remark 1.2.3 For = 0, = = = 1, = 1 we

2

have the class

+

Lemma 1.3.5 Let ; , , , , , ; then

1

1,

1, 0; , 1,1,1/2,1,1; = 1, . Where the class

1+ , , ,, , + +1 , , ,, ,

2 1 1

() consisting of functions () which

1+ , , , , ,; + +1 , , ,, ,;

1,

satisfies the following subordination condition

1,

() (z D).

1,1

1 where

1 where

,

> 0, 0 = 0 , , 0, 0,

1,

1,

1

0 < 1 , 0 <

1 , > 0, 0, > .

; = 1 0, 2 2

1,

and

=0

1

Proof For 0, 1, 2,, 1 we have obtained

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

, , , ; =

1,

1,

1 1 , , , , , .

2 2

=0 1

Remark 1.2.4 Putting = 0, = 2, = = = 1,

Hence

(7) , , , , , ;

= 1

and = 1+.

1,

1

+

2 1

= 1 , , , , ,

In definition 1.1.3, we have the class

=0 1

= 1 1 (+ ) , , , , , +

1+3 +

> 0

=0 1

1+3 +

= , , , , , ;

where = 1 [ ( )] and

And

1,

2

1,

1,

(8) , , , , , ;

= 1 1 2 , , , , , .

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

=0 1

Replacing + 1 in (A) we get

2 2 (9) +1 , , , , , ;

1,

= +1 , , , , , ; .

3. Preliminary Lemmas

1,

R

by + 1 in (B) we get

Lemma 1.3.1 Let be analytic and starlike univalent in

with 0 = 0. If g z is analytic in and

eplacing

1,

1,

(10) +1 , , , , , ;

=0 1

=0 1

= 1 1 2 +1 , , , , , .

,

()

From (7) and (10) we obtained

1+ , , , , , + +1 , , ,, ,

then 0 +

.

1,

1,

0

1+ , , , , ,; + +1 , , ,, ,;

Lemma 1.3.2 Let q z be analytic and other than

1,

1

1,

2 1+ 1 , , ,, ,

= 1

+1

constant in D with q 0 = 1. 0 < z0 < 1and

=0

1+ 1, , , ,, ,; +1,

, , ,, ,;

0 = 0 ,

2 +1 , , ,, ,

1( ) 2

+ 1

=

then 0 .

1+ , , ,, ,; + +1 , , ,, ,;

0 0 2 1 (0)

1,

1,

1 1+ , , , , ,

> 0, 0 = 0 , , 0, 0,

1 1 1 1 .

=0

1+ , , ,, ,; + +1 , , ,, ,;

0 <

, 0 <

, > 0, 0, > ,

1,

1, 2 2

1

1

+1 , , ,, ,

+ 1

, , ,, ,;

() .

1+ , , ,, ,; + +1 , , ,, ,;

, , ,, ,;

1,

It is clear that

1, 1

1+ , , ,, , + +1 , , ,, ,

Lemma 1.3.7: Let 1, ; , , , , , ; then

1

1

+1

1+ , , , , ,; + +1 , , , , ,;

1+ 1 , , ,, , + 1

, , , , ,

1,

1,

1+ , , , , ,; + +1 , , , , ,;

.

Noting that (z) is convex and univalent in D we conclude that Lemma 1.3.5 holds true. Using equations (5) and (6) we get

1,

1,

, , , , ,

+ 1 + , , , , , ;

1 1

() ( ).

Therefore, if with

1 2 + 1 + 1 () > 0.

Where

> 0, 0 = 0 , , 0, 0,

(+)

1,

0 < 1 , 0 < 1 , > 0, 0, > .

=

1 +1 , , , , , 2 2

(+)

=0 1

, , , , ,;

=

+1 , , , , , ; .

1

1

1

1

() .

(+)

1,

, , ,, ,;

1,

1,

Let ; , , , ; and suppose

, , ,, ,;

, , ,, ,;

, , ,, ,;

= 1, .

1,

1,

1,

Then is analytic in D and 0 = 1.

2. Main Results

Section 1

Hence 1 + 1

=

1 +1 , , ,, ,;

1, , , ,;

1,

1,

+1 , , , ; =

() + 1 + 1 , , , , , ;

Some properties of meromorphically univalent functions

.

1,

Theorem 2.1:

Let 0 < 1 and 0 < < 1. If () satisfies

From above relations we obtained

1+ , , ,, , + +1 , , ,, ,

() 0 and

1,

1,

= (11) 1

() + 1 < z D .

1+ , , , , ,; + +1 , , ,, ,;

()

()

1, 1,

1, 1,

1, 1,

1+ , , , , ,; + 1+1 , , ,, ,

1+ , , ,, ,; + 1+1 , , ,, ,

Where is the minimum positive root of the equation

(12) 2 + 1 = 0.

1,

1,

2 2 2 2

1+ + ()+ 1+1 Then

= 1

p>

1+ + 1+

+ 1+ + 1+1

(13) () < .

2 2

The bound b is the best possible for each 0 < 1

= 1 .

1+ + 1+

1 1

1 1

+

+2 +1

. . Since

Proof: Let

(14) = 2

1 2 2

+ 2 + 1 > 0

+ 1

.

and by Lemma 3 we get

2 2

It is cle that the quation

ar E

1, , , ,, ,;

2 + 1 = 0

=

, , ,, ,; () ( ). 2 2 2 2

1,

Lemma 1.3.6: Let ; , , , , , ; then

have two positive roots.

Since

1,

0 = 0 2 0 + 1

1+ , , , , , + +1 , , ,, , 2 2 2 2

1 1

1+ , , ,, ,; + +1 , , ,, ,;

= 1 > 0

1 1 2 2

() ( ).

Therefore,if P with

1 = 1 2 1 + 1 2 2 2 2

1

2 +

1 + 1 () > 0.

=

2

+

2 2 2

Wh e

= 2 < 0.

er 2 2

Hence we get (15) 0 <

2

Put

< 1

Theorem 2.2: If () satisfies , ( D)

and

(22) Re () () + 1 < .

(16) = + 1 .

()

2 2 (23) 0 < < 1

2log 2

Then from the assumption of the theorem, we see that

() is analytic in

with 0 = 1 and + 1 () 0 for all

then

(24) Re 1

()

> 1 2 2 ( )

2 2 The above inequality holds good.

. Taking the logarithmic differentiations on both sides of

= + 1 , 2 2

we obtained

(17) () + 1 = 2 ()

Proof: Let

(25) = ()

Then is analytic in with 0 = 1 and

0 for . In accordance with (17) and (20), we obtained

()

()

+ 2 ()

2 ()

1 ()

1+

. Let

2

2

(18) () + 1 = .

() + 2 ()

2() 1

Thus the inequality

(26) 1

2 . Now by Lemma 1, we obtain

1 () + 1 <

()

1

1

() ()

1 2 1 .

is equivalent to next equation as given in (19).

2 ()

()

Since the function 1 2 1 is convex

(19) + 2 () 2 .

By using Lemma 1, above inequality leads to

univalent in D and

Re 1 2 1 > 1 2 2 .

2 ()

From 1

2

we obtained the inequality

(20) 0 + 2 () 2 .

()

1

Or to

1 2

.

1

()

> 1 2 2.

+ 2 ()

To show that the bound

In view of above results it can be written as

1+

1

()

> 1 2 2 .

(21) 2 . Now by taking =

Cannot be increased, we consider

and =

1

in 1.2 , we have

2

= 1

[12 1 ]

.

2 2 We can verify that the function () satisfies the

2

2

2

= () < 2

2+

= . .

2

inequality

()

()

+ 1 < ( ).

Because of = 0. This proves statement. Next, we consider the function f(z) defined by

()

On the other hand we have

= 1

1

It is easy to see that

.

1 2 2 1.

Hence the Theorem holds good.

() () + = < .

()

Theorem 2.3: Let satisfies ,

(27) .If < 2

Since = 1+2 .

2 1

It follows from (3) that

2

2

, > 0 . Then

(28) () > 0 .

2

2

= 2

2+

= .

2

Proof: Let q z in D be defined as

() = then 0 = 1, 0,

Hence, we conclude that the bound b is the best possible for each (0,1].

And

(29) () () = () 1

Next, we derive the following.

()

()

(30) () > 0, < 0 and 0 = where is real and 0. Then by Lemma 1.2.2 we have

(1 2)

+1+1 , , ,, ,

(31) 0 0 2

+

1,

1

1+ , , ,, ,; + 1+ , , ,, ,

Thus it follows from above obtained results that

1,

1,

1

0 0

1+ + ()+ 1+

=

(32) = 0 .

1+ + 1+1

= + ( )

=

=

+ 1

0 0

In accordance with > 0 and from the statements

+ 1+ + 1+

1+ + 1+1

(2.2.1) and (2.2.2) we obtained

+(+2) 2

= +

1 1

1 1

+2 +1

() .

(33) 0

(34) 0

( + 2) ( > 0)

2

+(+2) 2 ( + 2) ( > 0)

2

Since 1

+ 2 1 + 1 > 0

But both The inequalities obtained above contradict the assumption given below.

And by Lemma 3 we get

, , ,, ,;

< 2 and

=

1,

1,

1,

, , ,, ,;

() .

> 0. Therefore we have, Re > 0 for all

( ). This shows that () > 0 .

By Lemma 1.2.2 we find that

.

; , , , , , ;

Theorem Holds true.

1,

, , , ; , , , , , ; .

1,

;

1,

Section 2

Subclasses of meromorphically univalent Fun- tions associated with Generalized multiplier transformations

1. Inclusion relationships

Corollary 2.1.1: Let P with

1 2 + 1 + 1 () > 0.

Where

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

2 2

Then

; , , , , , ; , , , , , ; .

1 1

Theorem 2.1.1: Let with

1 2 + 1 + 1 () > 0.

Wh e

Corollary 2.1.2: Let with

1 2 + 1 + 1 () > 0.

er

> 0, 0 = 0 , , 0, 0,

Where

0 < 1 , 0 < 1 , > 0, 0, > .

2 2

Then

; , , , , , ; , , , , , ; .

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

2 2

Then

1,

1,

; , , , , , ; , , , , , ; .

1 1

1,

1,

Proof: Let ; , , , , , ; and

, , ,, ,

Theorem 2.1.2: Let with

= 1

, , ,, ,;

.

2 +

1 + 1 () > 0.

Then

1

is analytic in and 0 = 1 hence

(+)

Where

1, , , , , , ;

1

1

= 1 1 , , , , ,

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

1

1

+ 1 + 1 , , , , , .

2 2

Then

; , , , , , ; , , , , , ; .

Differentiating both sides we get

1,

1,

+ 1

, , ,, ,;

+ 1 + 1,

, , ,, ,;

Proof Let ; , , , , , ; and suppose

= 1

that = 1

that = 1

1,

+1 , , ,, , ()

+1 , , ,, , ()

1

, , ,, ,;

, , ,, ,;

, , ,, ,;

1,

1,

.

1,

' Thus is analytic in and 0 = 1.

1,

1,

1+ , , ,, , + +1 , , , , ,

1 1

1, , , , , , ;

1+ , , ,, ,; + +1 , , ,, ,;

= 1 +1 , , , , ,

1,

1,

1

=

=

1+ 1, , , ,, ,;

1+ , , ,, ,; + 1+1 , , ,, ,

1

1

+ 1

+ 1 , , , , , .

1,

1,

Differentiating both sides we get

1 , , ,, ,;

Corollary 2.1.8: Let P with

+

+ 1 +

1,

, , ,, ,;

1 2 + 1 + 1 () > 0. Where

1,

1 +1 , , , , , ()

> 0, 0 = 0 , , 0, 0,

=

1

1,

1,

, , ,, ,;

0 <

1 , 0 <

2

1 , > 0, 0, > .

2

=

=

1, , , , , ,;

, , ,, ,;

.

Then

+1 , , , , , ; , , , , , ; .

1, 1 1

() .

1,

1,

This implies that ; , , , , , ; .

Corollary 2.1.3: Let P with

1 2 + 1 + 1 () > 0.

Corollary 2.1.9: Let with

1 2 + 1 + 1 () > 0.

Where

, = 0 , , 0, 0,

Where

> 0 0

0 < 1 , 0 < 1 , > 0, 0, > .

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

2 2

Then

+1 , , , , , ; , , , , , ; .

2 2 1 1

Then

; , , , , , ; , , , , , ; .

Corollary 2.1.10: Let P with

1 1

() 2 + 1 + 1 () > 0.

Corollary 2.1.4: Let with

1 2 + 1 + 1 () > 0.

(+)

Where

,

= 0 , , 0, 0,

> 0 0

Where

0 < 1 , 0 < 1 , > 0, 0, > .

> 0, 0 = 0 , , 0, 0,

2

Then m+1 , , , , ,

2

m , , , , , ; .

0 < 1 , 0 < 1 , > 0, 0, > .

2 2

Then

1 ; 1

2. Integral Representation

; , , , , , ; , , , , , ; .

1 1

In this section we are going to prove integral

Corollary 2.1.5: Let P with

1 2 + 1 + 1 () > 0.

representations associated with the function classes

, , , , , ; , , , , , , ;

1,

1,

, , , , , ; .

Where

and

1,

> 0, 0 = 0 , , 0, 0,

.

Theorem 2.2.1: Let , , , , , ; then

0 < 1 , 0 < 1 , > 0, 0, >

1,

2 2

Then

1, , , , , , ; =

+1 , , , , , ; , , , , , ; .

1 1

1

1

.

1,

1,

=0 0

Where , , , , , ;

Corollary 2.1.6: Let P with

1,

1

1 1 =

1 , , , , ,

2 + + 1 () > 0.

Where

=0 1

= 1 + ,

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

2 2

Then

+1 , , , , , ; , , , , , ; .

() is analytic in with 0 = 0

and () < 1. .

1,

1,

Proof: Suppose that , , , , , ; .

1 = () .

1 = () .

1 1 , , ,, ,

, , ,, ,;

Corollary 2.1.7: Let P with

1 2 + 1 + 1 () > 0.

Where

> 0, 0 = 0 , , 0, 0,

0 < 1 , 0 < 1 , > 0, 0, > .

1,

Where () is analytic in with 0 = 0

and () < 1 .

Replacing by ( = 0,1,2,) above equation holds true.

That is

2 2

Then

+1 , , , , , ; , , , , , ; .

1 1

, , ,, ,

= 1. 1 1 .

1 = ( ) .

0

1,

1,

, , , , ,;

From above equations we get

We note that

, , ,;

, , , , , = 1, 2 ()

1, , , , , , ; = 1, , , , , , ; 1

Letting = 0,1,2, , successively and summing = 2 () 1 1 .

2 0

the resultant equations we obtained

Where () is analytic in with 0 = 0 and

1

1

, , ,, ,;

=

1 1 ( )

.

() < 1 ( , = 1,2). Integrating both

1,

1,

, , , , ,;

And

1

1

, , ,, ,; 1

1,

1,

, , ,, ,; +

=0

=

=

1 1

=0

( ) 1

.

sides of above integral, we readily approach to the assertion of above theorem.

1

1

1

1

Corollary 2.2.1: Let , , , , , ; then

Upon integration which yields,

1, , , , , , ;

, , , , ; =

1. 1 + 2

1 1

0

= 1 .

Where 1 , , , , , ;

=0 0

= 1 , , , , ,

1

1

Taking exponential theorem holds true. 2 1

Theorem 2.2.2: Let , , , , , ; then

+ , , , , ,

= 1 + .

1

1

, , , , , =

1,

Where () is analytic in with 0 = 0

and () < 1 ( )

2

1

,

Corollary 2.2.2: Let , , , , , ; then

0 Ã— 1 1

1

, , , , , =

=0

0

1

2

Where () is analytic in with 0 = 0

and () < 1 .

0 ()

Ã— 1 + 2 .

0

1,

1,

Proof: Suppose that , , , , , ; then

Where () is analytic in with 0 = 0

and () < 1 ( )

(, , , , , ) = 1, , , ,; ()

1

= 2 ()

1 1

1

.

Corollary 2.2.3: Let , , , , , ; then

1

1

, , , , , ;

=0 0 1

Integrating above equation, theorem holds true.

= 1. 1 2

Theorem 2.2.3: Let , , , , , ; then

Where

0

1

1

1

, , , , , ()

1,

, , , , , ; =

, , , , ,

2

1

1

1

1

=

=

1 + , , , , ,

= 1 + ( )

2 2 ()

Where () is analytic in with 0 = 0

0

1

1

1 1 ,

and () < 1. ( )

=0 0

Corollary 2.2.4: Let 1 , , , , , ; then

Where () is analytic in with 0 = 0 , , , , , =

and () < 1 , = 1,2 .

1

0

0

2

1 2

0

() .

Proof: Suppose that , , , , , ; then

Where () is analytic in with 0 = 0

1,

1,

, , ,, ,;

1,

1,

, , , , ,;

1,

= 1 () . ( )

and () < 1. ( )

3. Convolution Properties

Where 1 () is analytic in with 1 0 = 0

and w1 (z) < 1. (z D)

Thus by applying method of the proof of theorem

2.2.3 we find that

In this section we are going to derive several convolution properties for the function classes

, , , , , ; , , , , , , ;

, , , , , ;

1,

1,

1,

and 1, , , , , , ; .

1

1

Theorem 2.3.1: Let , , , , , ; then

= 2

=

+ + 1 1 .

0 =0

()

Ã— 1 1 1

Thus from above result, we can easily prove the

=0

0

Theorem.

()

1.

=0

+ +(+)

Theorem 2.3.3 Let and . Then

Where () is analytic in with 0 = 0

, , , , , ; if and only if

and () < 1 .

1,

Proof: Since we know that linear operator given by

, , , , ,

1 + + 1 1 1 + Ã—

=1

=1

1 + + 1 1 1 1

1

= 1 + 0 (1 + ) + 1

=1

0

=0 1 1

=

= ()

= ()

1,

, , ,, ,

Where 1, = 1 + 0 (1 + ) + 1

; 0 < 2 .

, , ,, ,

=

Proof: Suppose that 1, , , , , , ; . since

and , , , , , =

1 2

0

the following subordination condition:

, , ,, , ()

1

1

Ã— 1 1

1

1

1

, , , , ,;

. .

=0

0

, , ,, , ()

1

1

It is equivalent to ( )

, , ,, ,;

2 () 1 1 1

; 0 < 2 .

1,

0

=0

0

=0

() 1 .

+ +(+)

It is easy to verify that the above condition can be written as follows:

(35) , , , , ,

1,

1,

Thus from above equation, we can easily get 1

0

0

= 2

1

1

+ , , , , , ; ( ) 0.

On the other hand we find from (10) that

Ã—

1

(36) , , , , , =

=0 0

1

1 +

+ 1 ( 1)1 ()

()

1.

=1

Moreover,

from the definition of

=0

1,

1,

+ +(+)

, , , , , ; ,

Theorem 2.3.2: Let , , , , , ; then

we obtained

1

2

(37) , , , , , ;

= 0 2

1,

0

0

1 1

Ã— 1

1

=0

1 1

= , , , , ,

1

1

=

1

0

1

1

1 1

()

1.

=

+ =1 + 1

1

=0

=0

=0

+ +(+)

. Substituting (36) and (37) in (35) we can easily

Where () is analytic in with 0 = 0

and () < 1 , = 1,2 .

Proof: Since we know that linear operator given by,

, , , , ,

arrive at the convolution property asserted by given theorem. In view of Corollaries 2.2.2 and 2.2.4 and by applying the method similar to method of Theorem

2.2.1 we can easily obtain the following result for the

function classes , , , and , , , , , .

1 1 1

= 1 + 0 (1 + ) + 1

=

Corollary 2.3.1: Let , , , , , , , then

= 1, .

1,

, , ,, ,

Where 1, =

1 +

(1 + ) + 1

0 2 2

, , ,, ,

=0

=

1 + 2

1

1

2 0

2 0

and , , , , , ()

Ã—

=0

=0

2 2

(+)

1

=

1

1

1 1

.

+ +(+)

0 Ã—

=0 0

1

Where w(z) is analytic in D with w 0 = 0

We obtained

0

0

1

1

2 2

Ã— 1 1

and () < 1. .

1,

1,

Corollary 2.3.2: Let , , , , , , , Then

=0

0

0

0

=

=

2 2

2 0

2 0

1 + 2

with the generalized hypergeometric function, Comput. Math Appl. 57, 571586 (2009).

1. Z. G. Wang, Y. Sun, Z. H. Zhang, Certain classes of meromorphically multivalent functions, Comput Math

()

1 .

Appl. 58, 14081417 (2009).

=0

+ +(+)

2. T. J. Suffridge, Some remarks on convex maps of the

Where () is analytic in with 0 = 0

and () < 1. .

Corollary 2.3.3: Let and . Then

1,

1,

, , , , , ;

if and only if

unit disk, Duke Math J. 37, 775777 (1970).

3. S. S. Miller, P. T. Mocanu, Second order differential inequalities in the complex plane, J Math Anal Appl. 65, 289305 (1978).

4. et al Xu, Some properties of meromorphically multi- valent functions, Journal of Inequalities and Applications 2012 2012:86.

n=1

n=1

f z1 +

ei I(z)

+ 1 m n 1 zn1 +

5. M. K. Aouf and H. M. Hossen, New criteria for meromorphic p-valent starlike functions, Tsukuba J. Math. 17 (1993) 481486.

2

+ ei

I f ( z ) 0 z D; 0 < 2 .

6. T. Bulboaca, Differential Subordinations and Super- ordinations, Recent Results (House of Scientific Book

2

Wher I(z) is given by

Publ., Cluj-Napoca, 2005).

e

1

1

7. N. E. Cho, O. S. Kwon and H. M Srivstava, Inclusion

= + =1 + 1 .

Corollary 2.3.4: Let and . Then

1

1

, , , , , ; if and only if

=1

=1

1 + + 1 1 1

and argument properties for certain subclasses of meromorphic functions associated with a family of multiplier transformations, J. Math. Anal. Appl. 300 (2004) 505520.

8. N. E. Cho, O. S. Known and H. M. Srivastava, Inclusion relationships for certain subclasses of meromorphic functions associated with a family of

+ ()

2

multiplier transformations, Integral Transforms Spec. Funct. 16 (2005) 647659.

( ) ( ) 0 ; 0 < 2 .

9. P. Eenigenburg, S. S. Miller, P. T. Mocanu and M. O.

2

where is given by above equation.

Remark Specializing the parameters

, , , , , , , and in our results, we obtain corresponding results due to various researchers.

References

1. R. M. Ali, V. Ravichandran, N. Seenivasagan, Subordin-ation and ordination of the Liu-Srivastava linear operator on meromorphically functions, Bull Malays Math Sci Soc. 31, 193207 (2008).

2. M. K. Aouf, Certain subclasses of meromprphically multivalent functions associated with generalized hyper- geometric function, Comput. Math. Appl. 55, 494509 (2008).

3. N. E. Cho, O. S. Kwon, H. M. Srivastava, A class of integral operators preserving subordination and super- ordination for meromorphic functions, Appl Math Comput. 193, 463474 (2007).

4. J. L. Liu, H.M. Srivastava, A linear operator and associated families of meromorphically multivalent functions, J. Math Anal Appl. 259, 566581 (2001).

5. J. L. Liu, H. M. Srivastava, Classes of meromorphically multivalent functions associated with the generalized hyper- geometric function, math. Comput. Model. 39, 2134 (2004).

6. Z. G. Wang, Y. P. Jiang, H. M. Srivastava, Some sub- classes of meromorphic-ally multivalent functions associated

Reade, on a Briot- Bouquet differential subordination, in General Mathematics 3, International Series of Numerical Mathematics, Vol. 64 (BirkhÂ¨auser-Verlag, Basel, 1983), pp. 339348; see also Rev. Roumaine Math. Pures Appl. 29 (1984) 567573.

1. R. M. El-Ashwah, A note on certain meromorphic p- valent functions, to appear in Appl. Math. Lett.

2. S. S. Miller and P. T. Mocanu, Differential subordinations and univalent functions, Michigan Math. J. 28 (1981) 157171.

3. S. S. Miller and P. T. Mocanu, on some classes of first order differential subordinations, Michigan Math. J. 32 (1985) 185195.

4. S. S. Miller and P. T. Mocanu, Differential Subordinations, Theory and Applications, Series on Monographs and Textbooks in Pure and Applied Math, Vol. 225 (Marcel Dekker, New York, 2000).

5. K. S. Padmanabhan and R. Parvatham, Some applications of differential subordination, Bull. Austral. Math. Soc. 32 (1985) 321330.

6. H. M. Srivastava, D. G. Yang and N. E. Xu, Some subclasses of meromorphically multivalent functions associated with a linear operator, Appl. Math. Comput.195 (2008) 1123.

7. B. A. Uralegaddi and C. Somanatha, New criteria for meromorphic starlike functions, Bull. Austral. Math. Soc. 43 (1991) 137140.

8. B. A. Uralegaddi and C. Somanatha, Certain differential operators for meromorphic functions, Houston J. Math. 17 (1991) 279284.

9. Z. Z. Zou and Z. R. Wu, on meromorphically starlike functions and functions meromorphically starlike with

respect to symmetric conjugate points, J. Math. Anal. Appl.261 (2001) 1727.

10. Z.-Z. Zou and Z. R. Wu, on functions meromorphically starlike with respect to symmetric points,

J. Math. Res. Exposition 23 (2003) 7176.

*Dr. S. M. Khairnar,

*Professor and Head and Dean (R & D)

MITs Maharashtra Academy of Engineering, Alandi, Pune-412105

**R. A. Sukne

**Assistant Professor in Mathematics Dilkap Research Institute of Engineering & Management Studies,