VOLUME 3, ISSUE 6, Jun.-2017
183 | P a g e
CERTAIN CLASSES OF UNIVALENT FUNCTIONS DEFINED BY
RUSCHEWEYH OPERATOR WITH SOME MISSING COEFFICIENT
RAHULKUMAR DIPAKKUMAR KATKADE
Dr. D.Y.Patil School of Engineering, Charholi (Bk), Lohegaon,Pune – 412105, India. Email: [email protected]
ABSTRACT:
In this paper, we introduce the class
O
n( , , , )
byusing the Ruscheweyh Operator. The aim of this paper is to study some properties of this class,like, coefficient inequality, Hadamard products, Extreme points, radius of starlikeness and convexity, closure theorem of a class of analytic and univalent function in the unit disc. The results obtained here are found to be sharp.
KEY WORDS: Analytic functions, Univalent functions, Hadamard product, Starlike functions, Rescheweyh operator.
1. INTRODUCTIONS:
Let
S
denote the class of function2 2 2
( )
kk k
f z
z
a z
which are analytic and univalentin the unit disc
{ :
1}.
U
z z
For 2 2 2g( )
k kk
z
z
b z
theconvolution or Hadamard product of
f z
( )
andg(z)
isdefined by 2
2 2 2
(
)(z)
k k k,
k
f
g
z
a b z
z
U
.Let
D f z
n( )
denote the thn
orderderivative introduced by Ruscheweyh [6].The Ruscheweyh derivative is defined as follows:
D
n:
S
S
such that1
( )
( ),
1
(1
)
n
n
z
D f z
f z
n
z
1
0
(
( ))
=
,
0,1, 2...
!
n n
z z
f z
n
N
n
2 2
2
( , )
kk k
z
a
n k z
, Where1
( , )
n k
n k
.
n
Notice that, 2
1
2
( , )
(1
)
k n
k
z
z
n k z
z
, where1
n
and 0( )
( ),
( )
( ).
D f z
f z D f z
zf z
We aim to study the class
( , , , )
which consists of functions
and satisfy
n
O
f
S
((
( ))
1)
((
( ))
1) { (
( ))
( )} (
)
,
n
n n n
z D f z
z
D f z
z D f z
D f z
z U
for
0
0
1, 0
1, 0
1, 0
1,
n
N
.
The investigation here is motivated by M.Darus [1].
Next we characterize the class
( , , , ) by
proving the Coefficient Inequality.
n
O
1. COEFFICIENT INEQUALITY: Theorem 1:
Let
f
S
.Then
f
O
n( , , , ) if and only if
22
2
1
,
(
)
k k
k
n k a
(1.1)
for 0
1, 0
1, 0
1, 0
1,
0
n
N
1
( , )
n k
n
k
n
The result (1.1) is sharp for the function
(
)
2( )
,
2
2
1
,
k
f z
z
z
k
k
n k
184 | P a g e
Proof:Assume inequality (1.1) is true then
(
( ))
1)
(
( ))
1)
{ (
( ))
( )}
(
)
n
n n n
z
D f z
z D f z
z D f z
D f z
2 2 2
2 2 2
2 2 2
2 2
2
2
( , )
2
( , )
2
( , )
( , )
(
)
k k k
k k k
k k k
k k
k
k
n k a z
k
n k a z
k
n k a z
a
n k z
2 2
[2 (1
)
] ( , )
k(
)
0
k
k
n k a
by maximum modulus principle,
( , , , ).
n
f
O
Conversely, Let
f
O
n( , , , ).Then
(
( ))
1)
(
( ))
1) { (
( ))
( )} (
)
,
n
n n n
z D f z
z
D f z
z D f z
D f z
z U
that is
2 2 2
2 2 2
2
( , )
(2
2
1) ( , )
(
)
k k k
k k k
k
n k a z
k
k
n k a z
(1.2)
Using the fact that Re( (z))
f
f z
( )
2 2 2
2 2 2
2
( , )
Re
(2
2
1) ( , )
(
)
k k k
k k k
k
n k a z
k
k
n k a z
Choosing on real axis and allowing
z
z
1
2 2
2 2
2
( , )
(2
2
1) ( , )
(
)
k k
k k
k
n k a
k
k
n k a
2 2
[2 (1
)
] ( , )
k(
)
0
k
k
n k a
Thus the proof is complete.
Corollary:
If
f
O
n( , , , ) then
2
(
)
,
2, 3,...
[2 (1
)
] ( , )
k
a
k
k
n k
with equality for
2
(
)
( )
,
2, 3,...
[2 (1
)
] ( , )
k
f z
z
z
k
k
n k
3. GROWTH AND DISTORTION THEOREM: Theorem 2:
If the function ( )
f z
O
( , , , ) then
n
4
4
(
)
[4(1
)
] ( , 2)
(
)
( )
[4(1
)
] ( , 2)
z
z
n
f z
z
z
n
The result is sharp for
4
(
)
( )
[4(1
)
] ( , 2)
f z
z
z
n
Proof: We have
2 2 2
( )
k kk
f z
z
a z
2 2 2
( )
k kk
f z
z
a
z
(2.1)
4
(
)
( )
[4(1
)
] ( , 2)
f z
z
z
n
Similarly
2 2 2
( )
k kk
f z
z
a
z
4
(
)
[4(1
)
] ( , 2)
z
z
n
Combining (2.1) and (2.2) we get the result.
Theorem 3:
If the function ( )
f z
O
( , , , ) then
n
3 3
4 (
)
1
( )
[4(1
)
] ( , 2)
4 (
)
1
[4(1
)
] ( , 2)
z
f z
n
z
n
VOLUME 3, ISSUE 6, Jun.-2017
185 | P a g e
The result is sharp for
4
(
)
( )
[4(1
)
] ( , 2)
f z
z
z
n
4. RADIUS OF STARLIKENESS AND CONVEXITY: Theorem 4: Let
f z
( )
O
n( , , , )
,then( )
f z
is starlike inz
R
where1 2 1
[2 (1
)
] ( , )
inf
(2
2) (
)
,
2,3,...
k k
k
n k
R
k
s
k
The estimate is sharp for the function
2
(
)
( )
[2 (1
)
] ( , )
k
f z
z
z
k
n k
for some k. Proof:
is starlike of order , 0
f
s
s
1 if
( )
Re
>
( )
f z
z
s
f z
(3.1)that is if
( )
1
1
( )
f z
z
s
f z
(3.2) This simplifies to
2 1 2 2
(2
2)
1
1
k k k
k
s
a
z
s
by (1.1) we have
2
(
)
,
2
[2 (1
)
] ( , )
k
a
k
k
n k
(3.4)Using (3.3) and (3.4) we get
2 1
[2 (1
)
] ( , )
(2
2) (
)
k
k
n k
z
k
s
thus,
1 2 1 2
[2 (1
)
] ( , )
,
(2
2) (
)
inf
2,3,...
k k k
k
n k
a
k
s
z
R
k
Theorem 5:
If
( , , , ) then ( ) is convex of order ,
0
1 in
where
n
O
f z
c
c
z
R
1 2 1
(1
)[2 (1
)
] ( , )
inf
,
2 (2
2
) (
)
2,3...
k k
c
k
n k
R
k
k
c
k
estimate is sharp for
2
(
)
( )
, for some .
[2 (1
)
] ( , )
k
f z
z
z
k
k
n k
Proof:
( , , , ) is convex of order c, 0
1 if
n
f
O
c
( )
Re 1
> c
( )
zf
z
f z
which is equivalent to
( )
< 1
( )
zf
z
c
f z
That is
2 1 2 2
2 1 2 2
2 (2
1)
1
2
k k k
k k k
k
k
a z
k a z
1
c
Using the arguments similar to theorem 4,We get
1 2 1
(1
)[2 (1
)
] ( , )
<
inf
2 (2
2
) (
)
k k
c
k
n k
z
R
k
k
c
5. EXTREME POINTS:Theorem 6: Let
f z
1( )
z
,2
(
)
,
2, 3...
[2 (1
)
] ( , )
k k
f
z
z
k
k
n k
Then
f
O
n( , , , )
if and only if it can be expressed inthe form
1
( )
k k( )
k
f z
f z
, where1
0, and
1
k k
k
. Proof: Suppose1
( )
k k( )
k
f z
f z
186 | P a g e
2 1
(
)
( )
[2 (1
)
] ( , )
k k
k
f z
z
z
k
n k
( )
f z
21
(
)
[2 (1
)
] ( , )
k k
k
z
z
k
n k
(5.1)Now,
f z
( )
O
n( , , , )
since2
[2 (1
)
] ( , )
.
(
)
(
)
[2 (1
)
] ( , )
k
k
k
n k
k
n k
1 2
1
1.
k k
Conversely, suppose that
f
O
n( , , , )
then by(1.1)
2
(
)
,
2, 3,...
[2 (1
)
] ( , )
k
a
k
k
n k
Setting
2
[2 (1
)
] ( , )
,
2, 3,...
(
)
k k
k
n k
a
k
(5.2)And 1
2
1
kk
We notice that,
2
( )
k k( )
k
f z
f z
.\ Hence the result.6. Hadamard Product
Theorem 7: Let
f g
,
O
n( , , , )
then
2 2 2 2
*
k k k n( , , , )
k
f
g
z
a b z
O
for 2
2 2
( )
k kk
f z
z
a z
, 22 2
( )
k kk
g z
z
b z
Where
2 2 2
2
(
)
[2 (1
)
]
( , )
(
)(2
2
1)
k
k
n k
k
k
Proof:
f g
,
O
n( , , , )
and so2 2
[2 (1
)
] ( , )
1
(
)
kk
k
n k
a
(6.1)2 2
[2 (1
)
] ( , )
1
(
)
kk
k
n k
b
(6.2)
We need to find a smallest number
such that2 2 2
[2 (1
)
] ( , )
1
(
)
k kk
k
n k
a b
(6.3) By Cauchy Schwarz inequality, we have2 2 2
[2 (1
)
] ( , )
1
(
)
k kk
k
n k
a b
(6.4) Thus, it is enough to show that2 2
[2 (1
)
] ( , )
(
)
k kk
n k
a b
2 2
[2 (1
)
] ( , )
(
)
k kk
n k
a b
That is
2k 2k
a b
[2 (1
)
]
[2 (1
)
]
k
k
(6.5)From (6.4)
2k 2k
a b
(
)
[2 (1
k
)
] ( , )
n k
(6.6)Therefore in view of (6.5) and (6.6) it is enough to show that
(
)
[2 (1
k
)
] ( , )
n k
[2 (1
)
]
[2 (1
)
]
k
k
This simplifies to
2 2 2
2
(
)
[2 (1
)
]
( , )
(
)(2
2
1)
k
k
n k
k
k
VOLUME 3, ISSUE 6, Jun.-2017
187 | P a g e
7. Closure Theorem
Theorem 8: Let
f
j
O
n( , , , ),
j
1, 2,...
then 2
2 , 1 2
( )
l
k
j k j
j k
g z
c
z
a
z
where
1
1
l j j
c
and 22 , 2
( )
kj k j
k
f z
z
a
z
.Proof: We have
2 2 , 1 2
( )
l
k
j k j
j k
g z
c
z
a
z
2 2 , 1 1 2
( )
l l
k
j j k j
j j k
g z
z
c
c a
z
2 2 , 2 1l
k k j j
k j
z
a
c z
(7.1)
2 2
k k k
z
e z
(7.2)
Where, 2 ,
1
l
k k j j
j
e
a
c
Since
f
j
O
n( , , , )
by (1.1)2 , 2
[2 (1
)
] ( , )
1
(
)
k jk
k
n k
a
(7.3) In view of (7.2),
g z
( )
O
n( , , , )
if2
[2 (1
)
] ( , )
1
(
)
kk
k
n k
e
Now,2
[2 (1
)
] ( , )
(
)
kk
k
n k
e
2 ,
2 1
[2 (1
)
] ( , )
(
)
l
k j j
k j
k
n k
a
c
2 , 1 2
[2 (1
)
] ( , )
(
)
l
j k j
j k
k
n k
c
a
1
l j j
c
, using (7.3)
1
Thus,
g z
( )
O
n( , , , ).
Theorem 9: Let
f g
,
O
n( , , , )
then
2 2
22 2 2
( )
k k kis in
n( , , , )
k
h z
z
a
b
z
O
Where
2
Proof :
f g
,
O
n( , , , )
and hence2 2 2
[2 (1
)
] ( , )
(
)
kk
k
n k
a
2 2 2 2
[2 (1
)
] ( , )
1
(
)
kk
k
n k
a
(8.1) 22 2
[2 (1
)
] ( , )
(
)
kk
k
n k
b
2 2 2 2
[2 (1
)
] ( , )
1
(
)
kk
k
n k
b
(8.2)Adding (8.1) and (8.2) we get
2
2 2 2 2 2
1 [2 (1
)
] ( , )
(
) 1
2
(
)
k kk
k
n k
a
b
(8.3) We must show that
h
O
n( , , , )
that is
2
2 2 2 2 2
[2 (1
)
] ( , )
1
(
)
k kk
k
n k
a
b
(8.4) In view of (8.3) and (8.4) it is enough to show that
[2 (1
)
] ( , )
1 [2 (1
)
] ( , )
(
)
2
(
)
k
n k
k
n k
This simplifies to
2
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coefficients, Kyungpook Math. J., 39 (1999), 133-139.
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XX (2003), XXX-XXX-1-6
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