# Coefficient bounds for a subclass of bi univalent functions using new differential operator

## Full text

(1)

### Presidency College, Chennai

ARTICLE INFO ABSTRACT

In this present paper, we introduce new subclasses

defined in the open disk

third coefficients for functions in these new subclasses using differential operator.

use, distribution, and reproduction in any medium, provided the original work is properly cited.

### INTRODUCTION

Let A denote the class of functions f z

of the form

2

, n n n

f z z a z

 

### 

which are analytic in the open unit disk U

 

Further, by

### S

we shall denote the class of functions which are univalent in

### U

. Since univalent functions are one to-one, they are invertible and the inverse functions need not be defined on the entire unit disk

### U

. However, the famous Koebe one-quarter theorem ensures that the image of the unit disk

under every function f A contains a disk of radius Thus every univalent function f has an inverse

1

( )

ff zz

, (zU) and

1

### 

0 0

1 ( ) , ( ), ( )

4

f fw ww r f r f

 

 

*Corresponding author: Elumalai, M.

Presidency College, Chennai-600 005, Tamilnadu, India.

### ISSN: 0975-833X

Article History:

Received 08th November, 2016 Received in revised form 28th December, 2016 Accepted 18th January, 2017 Published online 28th February,2017

Key words:

Bi-univalent functions, Coefficient estimates, Starlike functions, Convex function, Differential operator.

Citation: Elumalai, M. and Selvaraj, C.2017. “Coefficient bounds for a subclass of bi study of tunisie telecom operators”, International Journal of Current Research

### Presidency College, Chennai-600 005, Tamilnadu, India

ABSTRACT

this present paper, we introduce new subclasses S( , )  and

defined in the open diskU

z :z 1 .

### 

Furthermore, we find upper bounds for the second and third coefficients for functions in these new subclasses using differential operator.

is an open access article distributed under the Creative Commons Attribution License, which use, distribution, and reproduction in any medium, provided the original work is properly cited.

f z

of the form

(1.1)

### 

: 1 .

z z

  

we shall denote the class of functions fA Since univalent functions are one-one, they are invertible and the inverse functions need not be

. However, the famous Koebe quarter theorem ensures that the image of the unit disk

### 1 / 4

.

has an inverse

1

f

satisfying

where

1 2 2 3

2 2 3

3 4

2 2 3 4

( ) (2 )

(5 5 ) .

f w w a w a a w

a a a a w

  

 

  

A function f A is said to be bi

f z

and

1 ( )

fz

are univalent in

We let

### 

to denote the class of bi

given by (1.1). If f z

###  

is

bi-the boundary of bi-the domain and such that it can be continued

across the boundary of the domain so that

and analytic throughout w 1.

class

are

### 

, 1

1

z

log z

z  

and so on.

The coefficient estimate problem for the class Bieberbach conjecture, is settled by de

proved that for a function

### 

f z z a z

n

an

, for n2,3,, with equality only for the rotations of the Koebe function

International Journal of Current Research

Vol. 9, Issue, 02, pp.46091-46095, February, 2017

INTERNATIONAL

Coefficient bounds for a subclass of bi-univalent functions using new differential operator

International Journal of Current Research, 09, (02), 46091-46095.

### UNIVALENT FUNCTIONS USING NEW

and C( , )  of bi-univalent functions

Furthermore, we find upper bounds for the second and

third coefficients for functions in these new subclasses using differential operator.

1 2 2 3

2 2 3

3 4

2 2 3 4

( ) (2 )

(5 5 ) .

f w w a w a a w

a a a a w

  

(1.2)

is said to be bi-univalent in

### U

if both are univalent in

### U

.

to denote the class of bi-univalent functions in

### U

-univalent, it must be analytic in the boundary of the domain and such that it can be continued

across the boundary of the domain so that

1 ( )

fz is defined

1. 

Examples of functions in the

log z

and so on.

The coefficient estimate problem for the class

### S

known as the Bieberbach conjecture, is settled by de-Branges , who

2

, n n n

f z z a z

 

### 

in the class S, , with equality only for the rotations of

INTERNATIONAL JOURNAL OF CURRENT RESEARCH

(2)

0( ) 2.

(1 )

z K z

z

 

In 1967, Lewin  introduced the class

### 

of bi-univalent

functions and showed that a2 1.51 for the functions

belonging to

### 

. It was earlier believed that for f  , the

bound was an 1 for every

### n

and the extremal function in

the class was 1

z z

. E.Netanyahu  in 1969, ruined this

conjecture by proving that in the set

### 

, 2

4

ax / 3

m

f a

. In

1969, Suffridge  gave an example of f   for which

2 4 / 3

a

and conjectured thata2 4 / 3. In 1981, Styer and

Wright  disproved the conjecture that a2 4 / 3. Brannan

and Clunie  conjectured that a2  2. Kedzierawski  in

1985 proved this conjecture for a special case when the

function f and

1

f

are starlike functions. Brannan and Clunie

 conjectured that a2  2.Tan  in proved that 2 1.485

a

which is the best known estimate for functions in the class of bi-univalent functions.

Brannan and Taha  introduced certain subclasses of the bi-univalent function class

### 

similar to the familiar subclasses

S 

and C

###  

 of the univalent function class

### 

. Recently, Ali et al.  extended the results of Brannan and Taha  by generalising their classes using subordination.

An analytic function f is subordinate to an analytic function

g

,written f z( )g z( ), provided there is a Schwarz function

defined on

### U

with w(0)0 and w z( ) 1 satisfying

( )

### 

f zg w z

. Ma and Minda , unified various subclasses of starlike and convex functions for which either of

the quantity

( ) ( )

zf z f z

or

( ) 1

( )

zf z f z

 

is subordinate to a more general superordinate function. For this purpose, they considered an analytic function  with positive real part in the

unit disk

### U

,(0)1, (0)0 and  maps

### U

onto a region starlike with respect to 1 and symmetric with respect to the real axis. Such a function has a series expansion of the form

2 3

1 2 3 1

( )z 1 B z B z B z , (B 0).

### 

     

(1.3)

In this paper, for f z( )A. Let a new differential operator be defined  on a class of analytic functions of the form (1.1) as follows:

0 ,

F f zf z

1 :

F f zzfzFf z

and in general

1

### 

0 0 .

n n

F f zF Ff z n  

We easily find that

0

2

.

k n

nk n n

F f z z C a z n

 

### 

(1.4)

where

! | ( ) | !

n k

n C

n k

 

Definition: 1.1. Let  be a non-zero complex number. A

function f z

###  

given by (1.1) is said to be in the class S

 ,

### 

if the following conditions are satisfied:

f  and

' 1

1 1 ,

( )

j

j

z F f z

z z

F f z

 

 

  

 

 

 U

(1.5) and

###  

1

1 1 , ,

( )

j

j

w F g w

w w

F g w

 

 

  

 

 

 U

(1.6)

where the function g is given by (1.2).

Definition: 1.2. Let  be a non-zero complex number. A

function f z

###  

given by (1.1) is said to be in the class C

 ,

### 

if the following conditions are satisfied:

f  and

1

1 ,

( )

j

j

z F f z

z z F f z

 



 

 

 

 

 

 U

(1.7) and

###  

1

1 , ,

( )

j

j

w F g w

w w F g w

 



 

 

 

 

 

 U

(1.8)

where the function g is given by (1.2).

2. Coefficient estimates

Lemma: 2.1.  If pP , then ck 2 for each k, where P is the family of functions p analytic in

for which

Rep z 0 ,

###  

2

1 2

1

p z  c zc z 

for zU .

Theorem: 2.2. Let the function f z

###  

A be given by (1.1). If

,

f S   , then

1 1 2

2 2 2

3 2 1 1 2 2

2 j j j

B B a

C C B B B C

  

and

1 2 1

3 2

3 2

.

2 j j

B B B

a

C C

### 

 

(3)

Proof: Sincef S

 ,

### 

, there exists two analytic functions

, :

r s UU, withr(0)0s(0),such that

### 

1

1 1 ( )

( )

j

j

z F f z

r z

F f z

             and

### 

1

1 1 ( ) .

( )

j

j

w F g w

s z

F g w

          

  (2.2)

Define the functions p and q by

2

1 2

1 ( )

1

1 ( )

r z

p z p z p z

r z

    

 

and

###  

2

1 2

1 ( )

1 .

1 ( )

s z

q z q z q z

s z         (2.3) Or equivalently,

###  

2 2 1 1 2 2 3

1 1 1 2

3 2

( ) 1 ( ) 1

2 1

2

2 2 2

p z r z

p z

p

p z p z

p p p p

p p z

                                    (2.4) and

###  

2 2 1 1 2 2 3

1 1 1 2

3 2

( ) 1 ( ) 1

2 1

. 2

2 2 2

q z s z

q z

q

q z q z

q q q q

q q z

                                   (2.5)

It is clear that p and q are analytic in U andp(0) 1 q(0). Also p and q have positive real part in U and hence pi 2

and qi 2.

In the view of (2.3), (2.4) and (2.5), clearly,

### 

1 ( ) 1

1 1

( ) 1 ( )

j

j

z F f z p z

p z

F f z

                and

### 

1 ( ) 1

1 1 .

( ) 1 ( )

j

j

w F g w q w

q w

F g w

             

  (2.6)

Using (2.5) and (2.6) together with (1.3), one can easily verify that

2

2 2

1 1 1 1

2 2 1

( ) 1 1

1

( ) 1 2 2 2 4

B p B p

p z

z p B p z

p z                   (2.7) and 2 2 2

1 1 1 1 2 1

2

( ) 1

1 .

( ) 1 2 2 2 4

B q B q B q

q w

w q w

q w                   (2.8)

Since f  has the Maclaurin series given by (1.1),

computation shows that its inverse

1

gf has the expansion

given by (1.2). It follows from (2.6), (2.7) and (2.8) that

2 2 1 1

1 , 2 j

C aB p

(2.9)

2 2 2 2

3 3 2 2 1 2 1 2 1

1 1 1

2

2 2 4

j j

C aC a  Bpp B p

  (2.10)

and

2 2 1 1

1 , 2 j

C a B q

 

(2.11)

2

### 

2 2 2

3 2 2 3 3 1 2 1 2 1

1 1 1

4 2 .

2 2 4

j j j

CC aC a  B q  q B q

  (2.12)

From (2.9) and (2.11), it follows that

1 1.

p  q

(2.13)

Now (2.10), (2.12) and (2.13) gives

## 

3

1 2 2

2

2 2 2 2

3 2 1 2 1 2

.

4 2 j j j

B p q a

C C B C B B

       (2.14)

Using the fact that p2 2 and q2 2 gives the desired

estimate on a2 ,

1 1 2

2 2 2

3 2 1 1 2 2

.

2 j j j

B B a

C C B B B C

### 

  

From (2.10)-(2.12), gives

## 

### 

2 2 2

1

3 2 2 2 2 3 1 2 1

3 2 2

3 3 2

4

2 .

4 2

j j j j

j j j

B

C C p C q C p B B

a

C C C

   

Using the inequalities p1 2, p2 2 and q2 2 for

functions with positive real part yields the desired estimation of

3

a

.

For a choice of

###  

1 1 Az z Bz   

,  1 BA1, we have the following

Corollary: 2.3. Let 1 BA1. If

1 , 1 Az f Bz          S , then

2 2 2

3 2 2

2 j j 1 j

A B

a

C C A B B C

 

(4)

and

3 2

3 2

1 1

.

2 j j

A B B

a

C C

  

If we let

1 2 2

1 2 2

1

z

z z z

z

    

### 

  , 01, in the above theorem, we get the following:

Corollary: 2.4. Let 01. If f S

 ,

, then

2

2 2

3 2 2

2

2 2 j j 1 j

a

C C C

 

  

  

and

3 2

3 2

1 1 2

.

2 j j

a

C C

###  

 

Theorem: 2.5. Let the function f z

###  

A be given by (1.1). If

,

fC  

, then

1 1 2

2 2 2

3 2 1 1 2 2

2 3 j 2 j 2 j

B B a

C C B B B C

  

and

1 2 1

3 2

3 2

.

2(3 j 2 j)

B B B

a

C C

### 

 

(2.15)

Proof: Since fC

 ,

### 

, there exists two analytic functions

, :

r s UU, withr(0)0s(0)

, such that

1

1 ( )

( )

j

j

z F f z

r z F f z

 



 

 

 

 

 

and

### 

1

1 ( ) .

( )

j

j

w F g w

s z F g w

 



 

 

 

 

  (2.16)

Using (2.3), (2.4), (2.7) and (2.8), one can easily verified that

2 2 1 1

1

2 ,

2 j

C aB p

(2.17)

2 2 2 2

3 3 2 2 1 2 1 2 1

1 1 1

6 4

2 2 4

j j

C aC a  Bpp B p

  (2.18)

and

2 2 1 1

1

2 ,

2 j

C a B q

 

(2.19)

2

### 

2 2 2

3 2 2 3 3 1 2 1 2 1

1 1 1

12 4 6 .

2 2 4

j j j

CC aC a  B q  q B q

  (2.20)

From (2.17) and (2.19), it follows that

1 1.

p  q (2.21)

Now (2.18), (2.20) and (2.21) gives

## 

3

1 2 2

2

2 2 2 2

3 2 1 1 2 2

.

8 3 j 2 j 2 j

B p q a

C C B B B C

  

  

(2.22)

Using the fact that p2 2 and q2 2 gives the desired

estimate on a2 ,

1 1 2

2 2 2

3 2 1 1 2 2

.

2 3 j 2 j 2 j

B B a

C C B B B C

### 

  

From (2.18)-(2.20), gives

## 

### 

2 2 2 2

1

3 2 2 2 2 2 1 1 3

3 2

3 3 2

12 4 4 3

2 .

24 3 2

j j j j

j j j

B

C C p C q B B p C

a

C C C

   

Using the inequalities p1 2, p2 2 and q2 2 for

functions with positive real part yields

1 2 1

3 2

3 2

.

2 3 j 2 j

B B B

a

C C

 

For a choice of

###  

1 1

Az z

Bz

  

,  1 BA1, we have the following corollary.

Corollary: 2.6. Let 1 BA1. If

1 ,

1

Az f

Bz

 

 

 

S

, then

2

2 2

3 2 2

2 3 j 2 j 2 1 j

A B a

C C A B B C

 

   

and

3 2

3 2

1 1

.

2 3 j 2 j

A B B

a

C C

  

If we let

1 2 2

1 2 2

1

z

z z z

z

    

### 

  , 01, in the above theorem, we get the following:

Corollary: 2.7. Let 01. If fS

 ,

, then

2

2 2

3 2 2

3 j 2 j 1 j

a

C C C

 

 

(5)

and

### 

3 2

3 2

1 1

.

3 j 2 j

a

C C

  

 

Remark: 2.1. If we let  1,j0, Theorem: 2.2 and Theorem: 2.5 reduce to the result of R.M.Ali et.al , corollary 2.1 and corollary 2.2.

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### 1

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