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Periodicals of Engineering and Natural Sciences ISSN 2303-4521
Vol.6, No.1, June 2018, pp. 241~250
Available online at: http://pen.ius.edu.ba
DOI: 10.21533/pen.v6i1.285 241
Coefficient Estimates and Fekete- Szegὅ Inequality for a Subclass of
Bi-Univalent Functions Defined by Symmetric Q-Derivative Operator
by Using Faber Polynomial Techniques
G. Saravanan1, Muthunagai. K2
1,2School of Advanced Sciences, VIT University, Chennai - 600 127, Tamil Nadu, India.
1Present Address: Department of Mathematics , Patrician College of Arts and Science, Adyar,
Chennai-600020, Tamil Nadu, India.
Article Info
ABSTRACT
Article history:
Received Jan 12th, 2018
Revised Apr 20th, 2018
Accepted May 26th, 2018
In this article we have defined a subclass of Bi-univalent functions using
symmetric q- derivative operator and estimated the bounds for the coefficients
using Faber polynomial techniques. We also have obtained the bounds for the
linear functional which is popularly known as Fekete- Szegὅ problem.
.
Keyword:
Bi-univalent
Faber Polynomials
Symmetric Q-Derivative
Fekete- Szegὅ problem
Correspondng Author:
G. Saravanan,
School of Advanced Sciences, VIT University,
Chennai - 600 127, Tamil Nadu, India.
Present Address: Department of Mathematics ,
Patrician College of Arts and Science, Adyar,
Chennai-600020, Tamil Nadu, India.
Email: gsaran825@yahoo.com
1. Introduction
Let A be the class of all normalized functions of the form
2
n
n
n
f z z a z
(1)
Which are analytic in the Unit disk U.
A function that is regular (holomorphic) in U is said to be univalent in U if it assumes no value more than
once in U. Denote by S, the subclass of A, of all univalent functions in U.
For
(z)f
and
(z)g
analytic in U, we say that
(z)f
is subordinate to
(z)g
, written,
(z) g(z)f
, if there
exists a Schwarz function
(z)w
with
(z) 0w
and
| w(z) | 1
in U such that
(z) g(w(z))f
. That is if the
range of one holomorphic function is contained in that of the second and these functions agree at a single
point, then a sharp comparison of these two functions can be made.
The problem of finding sharp bounds for the linear functional
2
32
|a |a
of any compact family of functions
is popularly known as the Fekete-Szegὅ problem. This coefficient functional on the normalized analytic
G. Saravanan et al. PEN Vol. 6, No. 1, 2018, pp. 241 – 520
242
functions in the unit disk represents various geometric quantities. For example, for
1
, the functional
represents Schwarzian derivative, which plays a significant role in the theory of univalent functions,
conformal mapping and hypergeometric functions.
A function
(z) Af
is said to be bi-univalent in U, if
(z)fS
and its inverse has an analytic continuation
to
| w | 1
. The class of all bi-univalent functions is denoted by
.
The concept of bi-univalent functions was introduced by Lewin [18] who proved that if
(z)f
is bi-univalent,
then
2
| | 1.51a
. Brannan and Clunie [10] improved Lewin's result to
2
| | 2a
. There is a rich literature on
the estimates of the initial coefficients of bi-univalent functions (see [11, 13, 16, 24, 25, 26]). However not
much is known about the estimates of higher coefficients. It is well known that every function
fS
has an
inverse
1
f
, satisfying
1( ( )) ,( )f f z z z U
and
100 1
( ( )) , ( ); ( ) 4
f f w w w r f r f
∣∣
, where
1 2 2 3 3 4
2 2 3 2 2 3 4
( ) (2 ) (5 5 )f w w a w a a w a a a a w
(2)
Let K be simply connected, compact set in the Complex plane. Let h be analytic on K. It is possible to
approximate h by polynomials uniformly on K called Faber polynomials, introduced by Faber [12]. These
polynomials play an important role in geometric function theory.
A detailed discussion about Faber polynomial expansion for functions
fS
of the form has been carried out
in [[1], [2], [3]].
Geometric function theory provides a platform to have a multiple dimensional view on the different subclasses
of analytic functions with help of q- calculus which is an effective tool of investigation. For example, the
theory of q- calculus is used to describe the extension of the theory of univalent functions. For basic
definitions, applications, terminologies, geometric properties and approximation one can refer [[5], [8], [9],
[14], [17], [19], [20], [21]]. Let us suppose
01q
throughout this paper.
Definition. 1
The q-derivative of a function f is defined on a subset of is given by
( ) ( )
( )( ) , 0,
(1 )
qf z f qz
D f z if z
qz
and
( )(0) (0)
q
D f f
provided
(0)f
exists.
Note that
11
( ) ( )
lim( )( ) lim ( )
(1 )
q
qq
f z f qz
D f z f z
qz
if
f
is differentiable. From (1), we have
1
2
1 [n] .
n
q q n
n
D f z a z
Where the symbol
[]
q
n
denotes the number
1
[] 1
n
qq
nq
Definition. 2
The symmetric q-derivative
q
Df
of a function
f
given by (1) is defined as follows:
G. Saravanan et al. PEN Vol. 6, No. 1, 2018, pp. 241 – 520
243
1
1
( ) ( )
( )( ) , 0,
()
qf qz f q z
D f z if z
q q z
(3)
and
( )(0) (0)
q
D f f
provided
(0)f
exists.
From (3), we have the deduction
1
2
( )( ) 1 [ ] ,
n
q q n
n
D f z n a z
(4)
Where the symbol
[]
q
n
denotes the number
1
[] nn
qqq
nqq
From (2) and (4), we also have
1
1
2 2 3 3
2 2 3 2 2 3 4
( ) ( )
( )( ) ()
1 [2] [3] (2 ) [4] (5 5 ) .
q
q q q
g qw g q w
D g w q q w
a w a a w a a a a w
(5)
Lemma. 1 [6, 22]
If the function
pP
is defined by
23
1 2 3
( ) 1 .p z p z p z p z
then
| | 2( {1,2,3, }),
n
pn
and
22
11
2||
| | 2 .
22
pp
p
Let
be an analytic function with positive real part in U, with
(0) 1
and
(0) 0
. Also, let
()U
be
starlike with respect to 1 and symmetric with respect to the real axis. Then,
has the Taylor series expansion
23
1 2 3 1
( ) 1 ( 0).z B z B z B z B
(6)
2. Main Results
Definition. 3
Let
fA
. Then
( , , ), {0}f b q b
R
if
f
,
1
1 ( )( ) 1 ( )
q
Re D f z u
b
(7)
and
1
1 ( )( ) 1 ( )
q
Re D g w v
b
(8)
Where
1
gf
.
2.1. COEFFICIENT BOUNDS FOR FUNCTIONS BELONGING TO THE CLASS
( , , )bq
R
Theorem 1
Let
( , , )f b q
R
and
1( , , )g f b q
R
. If
0
k
a
for
21kn
then
2; 3.
[]
nq
b
an
n
∣∣
∣∣
G. Saravanan et al. PEN Vol. 6, No. 1, 2018, pp. 241 – 520
244
Proof
Let
( , , )f b q
R
and
P
, then there exists two Schwarz functions
2
12
()u z c z c z
and
2
12
()v w d w d w
such that
1
1 ( )( ) 1 ( ( ))
q
D f z u z
b
(9)
1
1 ( )( ) 1 ( ( ))
q
D g w v w
b
(10)
Where
12
11
( ( )) 1 ( , , )
nkn
k n n
nk
u z D c c c z
(11)
and
12
11
( ( )) 1 ( , , )
nkn
k n n
nk
v w D d d d w
(12)
From (9) and (11) we have
1
12
1
[] ( , , ), 2.
n
qnk
k n n
k
na D c c c n
b
(13)
From (10) and (12), we have
1
12
1
[] ( , , ), 2.
n
qnk
k n n
k
nb D d d d n
b
(14)
For
0
k
a
for
21kn
, (13) and (14) respectively yield
11
[]
qnn
na c
b
and
11
[ ] [ ]
qq
nn
n
n b n a d
bb
By definition of
p
n
K
we have
nn
ba
.
Upon simplification, we obtain
11
[]
nn
q
b
ac
n
(15)
11
[]
nn
q
b
ad
n
(16)
Taking the absolute values of (15) and (16) and using the facts that
1
| | 2
,
1
| | 1
n
c
and
1
| | 1
n
d
, we
obtain
2
[]
nq
b
an
∣∣
∣∣
Remark 1
If
1q
then the above theorem reduces to the results of Hamidi and Jahangiri [15].
Remark 2
If
1b∣∣
then above theorem reduces to the results of Altinkaya and Yalcin [7] (Theorem 7 for p=1).
G. Saravanan et al. PEN Vol. 6, No. 1, 2018, pp. 241 – 520
245
Theorem 2
Let
( , , )f b q
R
and
1( , , )g f b q
R
. Then
42
2 4 2
2
42
42
42
2 2 1
,
11
(i)
2 2 1
,.
1
1
q b q q
b
q q q
a
q b q q
bqq
qq
∣∣ ∣∣
∣∣
∣∣ ∣∣
2 2 2 2 2
2 2 4 2 4 2
32 2 2
4 2 4 2
4 2 ( 1)
,
( 1) 1 2( 1)
(ii) 4 ( 1)
,.
1 2( 1)
q b q b q
b
q q q q q
aq b q
b
q q q q
∣ ∣ ∣ ∣ ∣∣
∣∣
∣∣ ∣∣
2
2
32
42
23
(iii) , 1, ,2.
12
qb
aa
qq
∣∣
∣∣
Proof
Letting
2n
and
3n
in [13] and [14] respectively, we get
211
[2]qac
b
(17)
32
1 2 2 1
[3]qacc
b
(18)
and
211
[2]qbd
b
(19)
32
1 2 2 1
[3]qbdd
b
(20)
Comparing (5) with (19) and (20)
211
[2]qad
b
(21)
2
23 2
1 2 2 1
[3] (2 )
qaa dd
b
(22)
Using
11
cd
in either of (17) and (21), we deduce
22
2| | 2 | |
|| 1
[2]q
b q b
aq
From (18) and (22), we get
2
222
1 2 2 2 1 1
2[3] ( ) ( )
qac d c d
b
and thus
G. Saravanan et al. PEN Vol. 6, No. 1, 2018, pp. 241 – 520
246
242
2 | |
4| |
|| [3] 1
q
qb
b
aqq
Now the bounds for
2
||a
are justified since
242
2 | |
4| |
|| [3] 1
q
qb
b
aqq
for
42
42
21
| | .
1
qq
bqq
From (18), we get
22
1 2 2 1
342
| |(| |) 4| | 4 | |
|| 1
[3] [3]
qq
b c c b q b
aqq
(23)
On the other hand subtracting (22) from (18).
2
3 2 1 2 2
2[3] ( ) ( ).
qa a c d
b
(24)
Solving the above equation for
3
a
and taking absolute value
2 2 2
32 2 4 2
4 | | 2 | |
||
( 1) 1
q b q b
aq q q
(25)
Now, Theorem 2.2 (ii) follows from (23) and (25) upon noticing that
2 2 2 2 2 2
2 2 4 2 4 2 4 2
4 | | 2 | | 4| | ( 1)
if | |
( 1) 1 1 2( 1)
q b q b b q q
b
q q q q q q q
For the third part of the theorem, we rewrite (24) as
2
3 2 1 2 2
( ( ))
2[3]q
b
a a c d
(26)
Taking absolute values, we get
2
21 2 2
32 42
| || ( ) | 2| |
|| 1
2[3]q
b c d bq
aa qq
We rewrite (22) as
22
2 3 1 2 2 1
2 ( )
[3]q
b
a a d d
(27)
Taking absolute values, we get
22
22
3 2 1 2 2 1
4 2 4 2
| | 4| |
| 2 | | |
11
b q b q
a a d d
q q q q
Adding (26) and (27) and taking absolute value,
2
2
22
42
3 3| |
| | .
21
bq
aa
qq
2.2. FEKETE- SZEG
O
INEQUALITY FOR FUNCTIONS BELONGING TO THE CLASS
( , , )bq
R
Theorem 3
Let
f
given by (1) be in the class
( , , )bq
R
and
. Then
G. Saravanan et al. PEN Vol. 6, No. 1, 2018, pp. 241 – 520
247
1
2
32
1
1
0 ( )
4[3] 4[3]
1
4 ( ) ( ) .
4[3]
qq
q
Bb for h
aa
B h for h
∣∣ ∣∣
∣∣
∣ ∣ ∣ ∣
where
2
12
2
1 1 2
(1 )
() 4[ [3] [2] ( )]
qq
B
hb B B B
Proof
Let
( , , )f b q
R
and
g
be the analytic extension of
1
f
to U then there exists two functions
u
and
v
,
analytic in U with
(0) (0) 0uv
,
| ( )| 1,| ( )| 1u z v w
and
,z w U
such that
1
1 ( )( ) 1 ( ( ))
q
D f z u z
b
(28)
1
1 ( )( ) 1 ( ( ))
q
D g w v w
b
(29)
where
1
gf
.
Next, define the functions
,p q P
by
2
12
1 ( )
( ) 1
1 ( )
uz
p z p z p z
uz
(30)
2
12
1 ( )
( ) 1
1 ( )
vw
q z q w q w
vw
(31)
From the above definitions, one can derive
22
1 2 1
( ) 1 1 1 1
() ( ) 1 2 2 2
pz
u z p z p p z
pz
(32)
22
1 2 1
( ) 1 1 1 1
() ( ) 1 2 2 2
qw
v w q w q q w
qw
(33)
Combining (6), (28), (29), (32) and (33)
2 2 2
1 1 2 1 1 2 1
1 1 1 1 1
1 ( )( ) 1 1 2 4 2 2
q
D f z B p z B p B p p z
b
(34)
2 2 2
1 1 2 1 1 2 1
1 1 1 1 1
1 ( )( ) 1 1 2 4 2 2
q
D g w B q w B q B q q w
b
(35)
From (34) and (35), we deduce
211
[2] 1
2
qaBp
b
(36)
322
2 1 1 2 1
[3] 1 1 1
4 2 2
qaB p B p p
b
(37)
and
211
[2] 1
2
qaBq
b
(38)
G. Saravanan et al. PEN Vol. 6, No. 1, 2018, pp. 241 – 520
248
2
23 22
2 1 1 2 1
[3] (2 ) 1 1 1
4 2 2
qaa B q B q q
b
(39)
From (36) and (38), we get
11
pq
(40)
Subtracting (37) from (39) and applying (40)
21
3 2 2 2
()
4[3]q
bB
a a p q
(41)
By adding (37) to (39), we get
22
2 1 2 2 1 2 1
2[3] 11
( ) ( ).
22
qa B p q p B B
b
Using (36) and (38)
3
21 2 2
22
2
1 1 2
()
4 [3] [2] ( )
qq
bB p q
ab B B B
(42)
From (41) and (42), we get
2
3 2 1 2 2
11
( ) ( )
4[3] 4[3]
qq
a a bB h p h q
Where
2
12
2
1 1 2
(1 )
() 4[ [3] [2] ( )]
q
q
B
hb B B B
Then, by Lemma 1 and (6)
1
2
32
1
1
0 ( )
4[3] 4[3]
1
4 ( ) ( ) .
4[3]
qq
q
Bb for h
aa B h for h
∣∣ ∣∣
∣∣
∣ ∣ ∣ ∣
Corollary 1 If
( , , )f b q
R
then taking
1
, we get
2
21
32 42
||
| | .
4[ 1]
q b B
aa qq
(43)
Corollary 2 Let
22
1
( ) 1 2 2 ,(0 1).
1z
z z z
z
then from (43), we have
2
2
32 42
||
| | .
2[ 1]
qb
aa qq
Corollary 3 Let
2
1 (1 2 )
( ) 1 2(1 ) 2(1 ) ,(0 1).
1z
z z z
z
then the inequality (43) reduces to
G. Saravanan et al. PEN Vol. 6, No. 1, 2018, pp. 241 – 520
249
2
2
32 42
| |(1 )
| | .
2[ 1]
qb
aa qq
3. Conclusion
We have estimated the bounds for the coefficients and also the linear functional which is popularly known as
Fekete- Szeg
o
problem, for functions belonging to the class defined in this article. We also have seen our
results reducing to the results discussed in various other articles.
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