IMPACT: International Journal of Research in
Engineering & Technology (IMPACT: IJRET)
ISSN(E): 2321-8843; ISSN(P): 2347-4599
Vol. 2, Issue 2, Feb 2014, 1-14
© Impact Journals
ON A SUBCLASS OF ANALYTIC AND UNIVALENT FUNCTION DEFINED BY
AL-OBOUDI OPERATOR
N. D. SANGLE1, A. N. METKARI2 & D. S. MANE3
1,2
Department of Mathematics, Annasaheb Dange College of Engineering, Ashta, Maharashtra, India
3
Department of Mathematics, Tatayasaheb Kore Institute of Engineering and Technology,
Warananagar, Maharashtra, India
ABSTRACT
In the present paper, a subclass of analytic and univalent function is defined by Al-Oboudi Operator and we have
obtained among other results like, Coefficient estimates, Growth and distortion theorem, external properties for the classes
Tn S pλ (α , β , ξ , γ , δ , A, B ) and TnV λ (α , β , ξ , γ , δ , A, B ) .
2000 AMS Subject Classification: Primary 30C45, Secondary 30C50
KEYWORDS: Al-Oboudi Operator, Starlike Functions, Analytic Function, Univalent Function
1. INTRODUCTION
Let S denote the class of functions of the form
∞
f ( z ) = z + ∑ an z n
(I )
n=2
(1)
that are analytic and univalent in the disc
z < 1 . For 0 ≤ α < 1 S*(α) and K(α) denote the subfamilies of S
consisting of functions starlike of order α and convex of order α respectively.
The subfamily T of S consists of functions of the form
∞
f ( z ) = z − ∑ an z n ,
n=2
an ≥ 0 , for n = 2,3,...;
( II )
(2)
Silverman [6] investigated function in the classes T
Let
z ∈U .
*
(α ) = T ∩ S * (α ) and C (α ) = T ∩ K (α ) .
n ∈ N and λ ≥ 0 . Denote by Dλn the Al-Oboudi operator [3] defined by, Dλn : A → A,
Dλ0 f ( z ) = f ( z )
Dλ1 f ( z ) = (1 − λ ) f ( z ) + λ z f ' ( z ) = Dλ f ( z )
Dλn f ( z ) = Dλ [ Dλn −1 f ( z )].
Note that for f(z) is given by (I),
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2
N. D. Sangle, A. N. Metkari & D. S. Mane
∞
Dλn f ( z ) = z + ∑ [1 + ( j − 1)λ ]β a j z j when λ = 1,
j =1
Dλn is the Salagean differential operator Dλn : A → A , n ∈ N defined as:
D0 f ( z) = f ( z)
D 1 f ( z ) = Df ( z ) = z f ' ( z )
D n f ( z ) = D [ D n −1 f ( z )].
Definition 1.1: [8] Let β , λ ∈ R , β ≥ 0 , λ ≥ 0 and
∞
f ( z ) = z + ∑ a j z j we denote by Dλβ the linear operator
j =2
β
defined by Dλ
∞
: A → A, Dλβ f ( z ) = z + ∑ [1 + ( j − 1) λ ]β a j z j .
j=2
Remark 1.1: If
f ( z) ∈ T ,
∞
f ( z ) = z − ∑ a j z j , a j ≥ 0, j = 2,3,..., z ∈U
j =2
Then Dλβ f ( z ) = z −
In this paper
∞
∑ [1 + ( j − 1) λ ]β a
j=2
using the
operator
j
z j.
Dλβ we introduce the classes Tn S λp (α , β , ξ , γ , δ , A, B) and
Tn V λ (α , β , ξ , γ , δ , A, B) and obtain coefficient estimates for these classes when the functions have negative
coefficients. We also obtain growth and distortion theorems, closure theorem for functions in these classes.
Definition 1.2: We say that a function f ( z ) ∈ T is in the class Tn S p (α , β , ξ , γ , δ , A, B) if and only if
λ
Dλβ +1 f ( z )
−1
Dλβ f ( z )
<δ
 Dλβ +1 f ( z )

 Dλβ +1 f ( z ) 
− α  + Aγ  β
− 1
( B − A)ξ  β
 Dλ f ( z )

 Dλ f ( z )

Where
z < 1,0 < δ ≤ 1,1/ 2 ≤ ξ ≤ 1, λ ≥ 0,0 ≤ α ≤ 1/ 2ξ ,1/ 2 < γ ≤ 1, β ≥ 0,0 < B ≤ 1, −1 ≤ A < B ≤ 1.
Definition 1.3: A function f ( z ) ∈ T is said to belong to the class Tn S p (α , β , ξ , γ , δ , A, B) if and only if
λ
Dλβ + 2 f ( z )
−1
Dλβ +1 f ( z )
< δ , where
 Dλβ + 2 f ( z )

 Dλβ + 2 f ( z ) 
− α  + Aγ  β +1
− 1
( B − A ) ξ  β +1
 Dλ f ( z )

 Dλ f ( z )

z < 1, 0 < δ ≤ 1,1/ 2 ≤ ξ ≤ 1, λ ≥ 0, 0 ≤ α ≤ 1/ 2ξ ,1/ 2 < γ ≤ 1, β ≥ 0, 0 < B ≤ 1, − 1 ≤ A < B ≤ 1
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3
On a Subclass of Analytic and Univalent Function Defined by Al-Oboudi Operator
If we replace
β = 0 , λ = 1 we obtain the corresponding results of S.M. Khairnar and Meena More [4].
If we replace β = 0 , λ = 1 and
γ = 1 we obtain the results of Aghalary and Kulkarni [2] and Silverman and Silvia [7].
If we replace β = 0 , λ = 1 and
ξ = 1 , we obtain the corresponding results of [9].
2. MAIN RESULTS COEFFICIENT ESTIMATES
Theorem 2.1: A function f ( z ) = z −
∞
∑a
j=2
∞
∑ [1 + ( j − 1) λ ] β [ ( j − 1) λ {1 + Aγδ
j=2
j
λ
z j , ( a j ≥ 0 ) is in Tn S p (α , β , ξ , γ , δ , A, B) if and only if
+ ( B − A ) δξ } + ( B − A )δξ (1 − α ) ] a j ≤ ( B − A ) δξ (1 − α )
Proof: Suppose,
∞
∑[1 + ( j − 1)λ ]β [( j − 1)λ{1 + Aγδ + ( B − A)δξ } + ( B − A)δξ (1 − α )] a
j =2
j
≤ ( B − A)δξ (1 − α )
we have
Dλβ +1 f ( z ) − Dλβ f ( z ) − δ ( B − A)ξ  Dλβ +1 f ( z ) − α Dλβ f ( z )  + Aγ  Dλβ +1 f ( z ) − Dλβ f ( z )  < 0
with the provision,
∞
∞
j=2
j =2
z − ∑ (1 + ( j − 1)λ ) β +1 a j z j −z + ∑ (1 + ( j − 1)λ ) β a j z j −

∞
∞


j =2
j =2

δ ( B − A)ξ  z − ∑ (1 + ( j − 1)λ ) β +1 a j z j −α z + α ∑ (1 + ( j − 1)λ ) β a j z j 
∞
∞


+ Aγ  z − ∑ (1 + ( j − 1)λ ) β +1 a j z j −z + ∑ (1 + ( j − 1)λ ) β a j z j  < 0
j =2
j =2


For
z = r < 1 it is bounded above by
∞
∑[1 + ( j − 1)λ ]β [( j − 1)λ{1 + Aγ + ( B − A)δξ } + ( B − A)δξ (1 − α )] a r
j
j =2
Hence
f ( z ) ∈ Tn S pλ (α , β , ξ , γ , δ , A, B) .
Now we prove the converse result.
Dλβ +1 f ( z )
−1
Dλβ f ( z )
Let,
<δ
 Dλβ +1 f ( z )

 Dλβ +1 f ( z ) 
− α  + Aγ  β
− 1
( B − A) ξ  β
 Dλ f ( z )

 Dλ f ( z )

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j
≤( B − A)δξ (1 − α ).
4
N. D. Sangle, A. N. Metkari & D. S. Mane
∞
z − ∑ (1 + ( j − 1)λ ) β +1 a j z j
j =2
∞
z − ∑ (1 + ( j − 1)λ ) a j z
=
β
−1
j
j =2
∞


j
β +1
 z − ∑ (1 + ( j − 1)λ ) a j z

j =2
 + Aγ
α
( B − A)ξ 
−
∞
 z − (1 + ( j − 1)λ ) β a z j

∑
j


j=2


∞


z
−
(1 + ( j − 1)λ ) β +1 a j z j 
∑

j =2

 −1
∞
 z − (1 + ( j − 1)λ ) β a z j 
∑
j


j=2


<δ
∞
∑ (1 + ( j − 1)λ ) β ( j − 1)λ a z
=
j =2
∞
j
j
( B − A)ξ z (1 − α ) − ∑ (1 + ( j − 1)λ ) [ ( B − A)ξ (1 − α ) + (( B − A)ξ + Aγ )( j − 1)λ ] a j z
β
<δ
j
j =2
As
Re f ( z ) ≤ z for all z , we have
∞
∑ (1 + ( j − 1)λ ) β (−( j − 1)λ )a z
Re
j =2
∞
j
j
( B − A)ξ z (1 − α ) − ∑ (1 + ( j − 1)λ ) [ ( B − A)ξ (1 − α ) + (( B − A)ξ + Aγ )( j − 1)λ ] a j z
β
<δ
j
j =2
we choose values of z on real axis such that
Dλβ +1
is real and clearing the denominator of above expression and
Dλβ
allowing z → 1 through real values, we obtain.
∞
∑[1 + ( j − 1)λ ]β [( j − 1)λ{1 + Aγδ + ( B − A)δξ } + ( B − A)δξ (1 − α )] a
j =2
j
≤( B − A)δξ (1 − α )
∞
⇒ ∑ (1 + ( j − 1)λ ) β {( j − 1)λ (1 + Aγδ + ( B − A)δξ ) + ( B − A)δξ (1 − α )}a j − ( B − A)δξ (1 − α ) ≤ 0
j =2
Remark 2.1: If
aj ≤
f ( z ) ∈ Tn S pλ (α , β , ξ , γ , δ , A, B) then
( B − A)δξ (1 − α )
, j = 2,3, 4,.....
(1 + ( j − 1)λ ) {( j − 1)λ (1 + Aγδ + ( B − A)δξ ) + ( B − A)δξ (1 − α )}
β
and equality holds for,
f ( z) = z −
( B − A)δξ (1 − α )
z j.
(1 + ( j − 1)λ ) {( j − 1)λ (1 + Aγδ + ( B − A)δξ ) + ( B − A)δξ (1 − α )}
Corollary 2.1: If
β
f ( z ) ∈ Tn S 1p (α , β , ξ , γ , δ , −1,1) (In particular if A = −1, B = 1 ) then we get,
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5
On a Subclass of Analytic and Univalent Function Defined by Al-Oboudi Operator
aj ≤
[1 + ( j − 1)λ ]
2ξδ (1 − α )
β
{( j − 1)λ ( 2ξδ + 1 − γδ ) + 2δξ (1 − α )}
, j = 2, 3, 4,.....
and equality holds for,
f ( z) = z −
[1 + ( j − 1)λ ]
2ξδ (1 − α )
β
{( j − 1)λ ( 2ξδ + 1 − γδ ) + 2δξ (1 − α )}
zj
This corollary is due to [11].
Corollary 2.2: If
f ( z ) ∈ Tn S 1p (α , 0, ξ , γ , δ , −1,1) (in particular β = 0, λ = 1, B = 1, A = −1 ) then we get,
2δξ (1 − α )
, j = 2, 3, 4,.....
( j − 1) − δ {2αξ − 2ξ j + γ j − γ }
aj ≤
and equality holds for,
f ( z) = z −
2ξδ (1 − α )
zj
( j − 1) − δ {2αξ − 2ξ j + γ j − γ }
This corollary is due to [4].
f ( z ) ∈ Tn S 1p (α , 0, ξ ,1, δ , −1,1) (In particular β = 0, λ = 1, γ = 1, A = −1and B = 1 )
Corollary 2.3: If
then we get, a j
≤
2ξδ (1 − α )
, j = 2,3, 4,.....
( j − 1) − δ {2ξα − 2ξ j + j − 1}
and equality holds for,
f ( z) = z −
2ξδ (1 − α )
z j.
( j − 1) − δ {2ξα − 2ξ j + j − 1}
This corollary is due to [2] and [7].
f (z) ∈Tn S1p (α ,0,1,1, δ , −1,1) (in particular β = 0, λ = 1, γ = 1, ξ = 1, A = −1 and B = 1 )
Corollary 2.4: If
then we get,
aj ≤
2δ (1 − α )
, j = 2, 3, 4,.....
( j − 1) − δ {2α − j − 1}
and equality holds for,
f ( z) = z −
2δ (1 − α )
z j.
( j − 1) − δ {2α − j − 1}
This corollary is due to [9].
Corollary 2.5: If
f (z) ∈TnS1p (α,0,1,1,1, −1,1) (in Particular β = 0, λ = 1, γ = 1, ξ = 1, δ = 1, A = −1, B = 1 )
∞
if and only if
∑( j −α ) a
j =2
j
≤ (1 − α ).
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6
N. D. Sangle, A. N. Metkari & D. S. Mane
Theorem 2.2: A function f ( z ) = z −
∞
∑a
j=2
∞
j
z j , ( a j ≥ 0 ) is in
Tn S λp (α , β , ξ , γ , δ , A, B) if and only if
∑[1 + ( j − 1)λ ]β [( j − 1)λ{1 + Aγ + ( B − A)δξ } + ( B − A)δξ (1 − α )] a
+1
j =2
j
≤( B − A)δξ (1 − α ).
Proof: The proof of this theorem is analogous to that of Theorem 1, because a function
f ( z ) ∈ TnV λ (α , β , ξ , γ , δ , A, B ) if and only if zf ' ( z ) ∈ Tn S pλ (α , β , ξ , γ , δ , A, B). So it is enough that replacing
β with β + 1 in Theorem 2.1.
Remark 2.2: If
aj ≤
f ( z ) ∈ TnV λ (α , β , ξ , γ , δ , A, B ) then
[1 + ( j − 1)λ ]
β +1
and equality holds for,
Corollary 2.6: If
aj ≤
( B − A)δξ (1 − α )
[ ( j − 1)λ{1 + Aγ + ( B − A)δξ } + ( B − A)δξ (1 − α )]
f ( z) = z −
( B − A)δξ (1 − α )
z j.
(1 + ( j − 1)λ ) {( j − 1)λ (1 + Aγδ + ( B − A)δξ ) + ( B − A)δξ (1 − α )}
β +1
f ( z ) ∈ TnV λ (α , β , ξ , γ , δ , −1,1) (in particular β = 0, λ = 1, A = −1, B = 1 ) then we get,
(1 + ( j − 1)λ )
2δξ (1 − α )
β +1
 ( j − 1)λ (2δξ + 1 − γδ ) + 2δξ (1 − α ) 
, j = 2,3, 4,.....
and equality holds for,
f ( z) = z −
(1 + ( j − 1)λ )
2δξ (1 − α )
β +1
 ( j − 1)λ (2δξ + 1 − γδ ) + 2δξ (1 − α ) 
zj
This corollary is due to [11].
Corollary 2.7: If
aj ≤
f ( z ) ∈ TnV 1 (α , 0, ξ , γ , δ , −1,1) (in particular β = 0, λ = 1, A = −1, B = 1 ) then we get,
2δξ (1 − α )
, j = 2,3, 4,.....
j [ ( j − 1) − δ {2αξ − 2ξ j + γ j − γ }]
and equality holds for,
f ( z) = z −
2δξ (1 − α )
z j.
j [ ( j − 1) − δ {2αξ − 2ξ j + γ j − γ }]
This corollary is due to [4].
Corollary 2.8: If
then we get,
aj ≤
f ( z ) ∈ TnV 1 (α , 0, ξ ,1, δ , −1,1) (in particular β = 0, λ = 1, γ = 1, A = −1 and B = 1 )
2δξ (1 − α )
, j = 2,3, 4,.....
j [ ( j − 1) − δ {2αξ − 2ξ j + j − 1}]
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7
On a Subclass of Analytic and Univalent Function Defined by Al-Oboudi Operator
and equality holds for,
f ( z) = z −
2δξ (1 − α )
z j.
j [ ( j − 1) − δ {2αξ − 2ξ j + j − 1}]
This corollary is due to [2] and [7].
f ( z ) ∈ TnV 1 (α , 0,1,1, δ , −1,1) (in particular β = 0, λ = 1, γ = 1, ξ = 1, A = −1 and B = 1 )
Corollary 2.9: If
then we get,
aj ≤
2δ (1−α)
, j = 2,3,4,.....
j [( j −1) −δ{2α − j −1}]
and equality holds for,
Corollary 2.10: If
f ( z) = z −
2δ (1 − α )
z j.
j [ ( j − 1) − δ {2α − j − 1}]
f ( z ) ∈ Tn v1 (α , 0,1,1,1, −1,1) (in particular β = 0, λ =1,γ =1,ξ =1, δ =1, A = −1 and B =1)
∞
if and only if
∑ j ( j −α ) a ≤ (1−α ) .
j
j =2
3. GROWTH AND DISTORTION THEOREM
Theorem 3.1: If
λ
f (z)∈TS
n p (α, β,ξ,γ ,δ, A, B) then


( B − A)ξδ (1 − α )
r − r2 
 ≤ f ( z)
β
 (1 + λ ) {λ + ( B − A)ξδλ + Aγδλ + ( B − A)ξδ (1 − α )} 


( B − A)ξδ (1 − α )
≤ r + r2 

β
 (1 + λ ) {λ + ( B − A)ξδλ + Aγδλ + ( B − A)ξδ (1 − α )}
Equality holds for
f ( z) = z −
( B − A)δξ (1 − α )
z 2 at z ± r
{1 + 2( B − A)ξδ } + δ { Aγ − ( B − A)ξα }
Proof: By theorem 2.1, we have
f (z) ∈Tn Spλ (α, β ,ξ , γ ,δ , A, B) if and only if
∞
∑[1+ ( j −1)λ]β [( j −1)λ{1+ Aγδ + (B − A)δξ} + (B − A)δξ (1−α )] a
j =2
Let,
t = 1−
j
≤(B − A)δξ (1 − α )
( B − A)ξδ (1 − α )
λ + ( B − A)ξδλ + Aγδλ
∴ f ( z ) ∈ Tn S p (α , β , ξ , γ , δ , A, B) if and only if
λ
∞
∑ (1 + ( j − 1)λ )β ( j − t )a
j =2
when j=2
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j
≤ (1 − t )
(3)
(3)
8
N. D. Sangle, A. N. Metkari & D. S. Mane
∞
∞
j =2
j=2
(1 + λ ) β (2 − t )∑ a j ≤ ∑ (1 + ( j − 1) β a j ( j − t ) ≤ (1 − t )
∞
∞


1− t
∴ f ( z ) ≤ r + ∑ an r n ≤ r + r 2 ∑ an ≤ r + r 2 

β
j =2
j =2
 (1 + λ ) (2 − t ) 
similarly,
∞
∞


1− t
∴ f ( z ) ≥ r − ∑ an r n ≥ r − r 2 ∑ an ≥ r − r 2 

β
j =2
j =2
 (1 + λ ) (2 − t ) 
so,




1− t
1− t
r − r2 
≤ f ( z) ≤ r + r 2 

.
β
β
 (1 + λ ) (2 − t ) 
 (1 + λ ) (2 − t ) 
Hence the result.


( B − A)δξ (1 − α )
r − r2 
 ≤ f ( z)
β
 (1 + λ ) {λ + ( B − A)ξδλ + Aγδλ + ( B − A)ξδ (1 − α )} 
i.e.


( B − A)δξ (1 − α )
≤ r + r2 

β
 (1 + λ ) {λ + ( B − A)ξδλ + Aγδλ + ( B − A)ξδ (1 − α )} 
Corollary 3.1: If
f ( z ) ∈ Tn S 1p (α , 0, ξ , γ , δ , A, B) (i.e. replacing β = 0, λ = 1 ) then we get,




(B − A)δξ (1 − α )
( B − A)δξ (1 − α )
r − r2 
≤ f ( z) ≤ r + r 2 


1 + 2(B − A)ξδ + δ {Aγ − ( B − A)ξα}
1 + 2( B − A)ξδ + δ {Aγ − (B − A)ξα}
and equality holds for
f ( z) = z −
( B − A)δξ (1 − α )
z 2 at z = ± r
1 + 2( B − A)ξδ + δ { Aγ − ( B − A)ξα }
This corollary is due to [4].
Corollary 3.2: If
f ( z ) ∈ Tn S 1p (α , 0, ξ ,1, δ , A, B) (i.e. replacing β = 0, λ = 1 and γ = 1 ) then we get,




( B − A)δξ (1 − α )
( B − A)δξ (1 − α )
r − r2 
≤ f ( z) ≤ r + r 2 


1 + 2( B − A)ξδ + δ { A − ( B − A)ξα } 
1 + 2( B − A)ξδ + δ { A − ( B − A)ξα } 
and Equality holds for,
f ( z) = z −
( B − A)δξ (1 − α )
z j at z = ± r
1 + 2( B − A)ξδ + δ { A − ( B − A)ξα }
This corollary is due to [2] and [7].
Corollary 3.3: If
f ( z ) ∈ Tn S 1p (α , 0,1,1, δ , A, B) (i.e. replacing β = 0, λ = 1, γ = 1 and ξ = 1 ) then we get,
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9
On a Subclass of Analytic and Univalent Function Defined by Al-Oboudi Operator




( B − A)δ (1 − α )
( B − A)δ (1 − α )
r − r2 
≤ f ( z) ≤ r + r 2 


1 + 2( B − A)δ + δ { A − ( B − A)α } 
1 + 2( B − A)δ + δ { A − ( B − A)α } 
and Equality holds for,
f ( z) = z −
( B − A)δ (1 − α )
zj
1 + 2( B − A)δ + δ { A − ( B − A)α }
at
z = ±r
This corollary is due to [9].
Theorem 3.2: If
λ
f (z) ∈TV
(α, β,ξ,γ ,δ , A, B) then
n


( B − A)ξδ (1 − α )
r − r2 
≤ f ( z)

β +1
 (1 + λ ) {λ + ( B − A)ξδλ + Aγδλ + ( B − A)ξδ (1 − α )} 


( B − A)ξδ (1 − α )
≤ r + r2 

β +1
 (1 + λ ) {λ + ( B − A)ξδλ + Aγδλ + ( B − A)ξδ (1 − α )} 
Proof: The proof of this theorem is analogous to that of theorem 3.1, because a function
f ( z ) ∈ TnV λ (α , β , ξ , γ , δ , A, B ) if and only if zf ' ( z ) ∈ Tn S pλ (α , β , ξ , γ , δ , A, B) . So it is enough that replacing
β with β + 1 in Theorem: 2.1.
Corollary 3.4: If
f ( z ) ∈ TnV 1 (α , 0, ξ , γ , δ , A, B) (i.e. replacing β = 0, λ = 1 ) then we get,




(B − A)δξ (1−α)
(B − A)δξ (1−α)
r − r2 
≤ f (z) ≤ r + r2 


 2[1+ 2(B − A)ξδ + δ{Aγ − (B − A)ξα)}]
 2[1+ 2(B − A)ξδ + δ{Aγ − (B − A)ξα)}]
And equality holds for

 j
( B − A)δξ (1 − α )
f ( z) = z − 
z
 2[1 + 2( B − A)ξδ + δ { Aγ − ( B − A)ξα )}] 
at z = ± r
This corollary is due to [4].
Corollary 3.5: If
f ( z ) ∈ TnV 1 (α , 0, ξ ,1, δ , A, B) (i.e. replacing β = 0, λ = 1 and γ = 1 ) then we get,




( B − A)δξ (1 − α )
( B − A)δξ (1 − α )
r − r2 
≤ f ( z) ≤ r + r 2 


 2[1 + 2( B − A)ξδ + δ { A − ( B − A)ξα )}] 
 2[1 + 2( B − A)ξδ + δ { A − ( B − A)ξα )}] 
And Equality holds for,

 j
( B − A)δξ (1 − α )
f ( z) = z − 
z
 2[1 + 2( B − A)ξδ + δ { A − ( B − A)ξα )}] 
at z = ± r
This corollary is due to [2] and [7].
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10
N. D. Sangle, A. N. Metkari & D. S. Mane
Corollary 3.6: If
f ( z ) ∈ TnV 1 (α , 0,1,1, δ , A, B) (i.e. replacing β = 0, λ = 1, γ = 1 and ξ = 1 ) then we get,




( B − A)δ (1 − α )
( B − A)δ (1 − α )
r − r2 
≤ f ( z) ≤ r + r 2 


 2[1 + 2( B − A)δ + δ { A − ( B − A)α )}] 
 2[1 + 2( B − A)δ + δ { A − ( B − A)α )}] 
and Equality holds for,
f ( z) = z −
( B − A)δ (1 − α )
zj
2[1 + 2( B − A)δ + δ { A − ( B − A)α )}]
at z = ± r
This corollary is due to [9].
Theorem 3.3: If
f ( z ) ∈ Tn S pλ (α , β , ξ , γ , δ , A, B) then


( B − A) 2 ξδ (1 − α )
1− r 
 ≤ f ( z)
β
(1
+
)
{
(1
+
A
)
+
(
B
−
A
)
(
+
1
−
)}
λ
λ
γδ
ξδ
λ
α


2


( B − A) ξδ (1 − α )
≤ 1+ r2 

β
 (1 + λ ) {λ (1 + Aγδ ) + ( B − A)ξδ (λ + 1 − α )} 
Proof: Since
f ( z ) ∈ Tn S pλ (α , β , ξ , γ , δ , A, B) we have by Theorem.3.1,
∞
∑ (1 + ( j − 1)λ )β ( j − t )a
j =2
j
≤ (1 − t )
In view of Theorem 3.1 we have
∞
∞
∑ ja = ∑ ( j − 1)a
j =2
j
j =2
∞
j
∞
f ' ( z ) ≤ 1 + ∑ nan z
+ t∑ a j ≤
j =2
n −1
n= 2
( B − A)(1 − t )
(1 + λ ) β (2 − t )
(4)
∞
 ( B − A)(1 − t ) 
≤1 + r ∑ nan ≤1 + r 

β
n=2
 (1 + λ ) (2 − t ) 
Similarly,
∞
f ' ( z ) ≥ 1 − ∑ nan z
n=2
So,
n −1
∞
 ( B − A)(1 − t ) 
≥1 − r ∑ nan ≥1 − r 

β
n=2
 (1 + λ ) (2 − t ) 
 ( B − A)(1 − t ) 
 ( B − A)(1 − t ) 
1− r 
≤ f ' ( z) ≤ 1 + r 


β
β
 (1 + λ ) (2 − t ) 
 (1 + λ ) (2 − t ) 
Substituting the value of t in above inequality,
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(4)
On a Subclass of Analytic and Univalent Function Defined by Al-Oboudi Operator
11


( B − A) 2 ξδ (1 − α )
'
1− r 
 ≤ f ( z)
β
 (1 + λ ) {λ (1 + Aγδ ) + ( B − A)ξδ (λ + 1 − α )} 


( B − A) 2 ξδ (1 − α )
≤ 1+ r 

β
 (1 + λ ) {λ (1 + Aγδ ) + ( B − A)ξδ (λ + 1 − α )} 
Corollary 3.7: If
f ( z ) ∈ Tn S 1p (α , 0, ξ , γ , δ , A, B) (i.e. replacing β = 0, λ = 1 ) then we get,




( B − A) 2 ξδ (1 − α )
( B − A) 2 ξδ (1 − α )
'
1− r 
≤ f ( z) ≤ 1 + r 

 for z = r
 (1 + Aγδ ) + ( B − A)ξδ (2 − α )} 
 (1 + Aγδ ) + ( B − A)ξδ (2 − α )} 
This corollary is due to [4].
f ( z ) ∈ Tn S 1p (α , 0, ξ ,1, δ , A, B) (i.e. replacing β = 0, λ = 1 and γ = 1 ) then we get,
Corollary 3.8: If




( B − A) 2 ξδ (1 − α )
( B − A) 2 ξδ (1 − α )
'
1− r 
≤
f
(
z
)
≤
1
+
r


 for z = r
 (1 + Aδ ) + ( B − A)ξδ (2 − α )} 
 (1 + Aδ ) + ( B − A)ξδ (2 − α )} 
This corollary is due to [2] and [7].
f ( z ) ∈ Tn S 1p (α , 0,1,1, δ , A, B) (i.e. replacing β = 0, λ = 1, γ = 1 and ξ = 1 ) then we get,
Corollary 3.9: If




( B − A) 2 δ (1 − α )
( B − A) 2 δ (1 − α )
'
1− r 
≤
f
(
z
)
≤
1
+
r


 for z = r
 (1 + Aδ ) + ( B − A)δ (2 − α )} 
 (1 + Aδ ) + ( B − A)δ (2 − α )} 
This corollary is due to [9].
Theorem 3.4: If
λ
f (z) ∈TV
(α, β,ξ,γ ,δ , A, B) then
n


( B − A) 2 ξδ (1 − α )
1− r 
 ≤ f ( z)
β +1
 (1 + λ ) {λ (1 + Aγδ ) + ( B − A)ξδ (λ + 1 − α )} 


( B − A) 2 ξδ (1 − α )
≤ 1+ r2 

β +1
 (1 + λ ) {λ (1 + Aγδ ) + ( B − A)ξδ (λ + 1 − α )} 
Proof: The proof of this theorem is analogous to that of theorem 3.3, because a function
f ( z ) ∈ TnV λ (α , β , ξ , γ , δ , A, B) if and only if zf ' ( z ) ∈ Tn S pλ (α , β , ξ , γ , δ , A, B) . So it is enough that replacing
β with β + 1 in Theorem: 3.3.
Corollary 3.10: If
f ( z ) ∈ TnV 1 (α , 0, ξ , γ , δ , A, B) (i.e. replacing β = 0, λ = 1 ) then we get,



( B − A) 2 ξδ (1 − α )
( B − A) 2 ξδ (1 − α )
2 
1− r 
≤
f
(
z
)
≤
1
+
r


 for z = r
 (1 + Aγδ ) + ( B − A)ξδ (2 − α )} 
 (1 + Aγδ ) + ( B − A)ξδ (2 − α )} 
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12
N. D. Sangle, A. N. Metkari & D. S. Mane
This corollary is due to [4].
Corollary 3.11: If
f ( z ) ∈ TnV 1 (α , 0, ξ ,1, δ , A, B) (i.e. replacing β = 0, λ = 1 and γ = 1 ) then we get,



( B − A) 2 ξδ (1 − α )
( B − A)2 ξδ (1 − α )
2 
1− r 
≤
f
(
z
)
≤
1
+
r


 for z = r
 (1 + Aδ ) + ( B − A)ξδ (2 − α )} 
 (1 + Aδ ) + ( B − A)ξδ (2 − α )} 
This corollary is due to [2] and [7].
Corollary 3.12: If
f ( z) ∈ TnV 1 (α ,0,1,1, δ , A, B) (i.e. replacing β = 0, λ = 1, γ = 1 and ξ = 1 ) then we get,



( B − A)2 δ (1 − α )
( B − A) 2 δ (1 − α )
2 
1− r 
≤
f
(
z
)
≤
1
+
r


 for z = r
 (1 + Aδ ) + ( B − A)δ (2 − α )} 
 (1 + Aδ ) + ( B − A)δ (2 − α )} 
This corollary is due to [9].
4. CLOSURE THEOREM
Theorem 4.1:Let
f j ( z) =
Then
f1 ( z ) = z and
( B − A)δξ (1 − α )
z j for j = 2,3, 4,.....
[1 + ( j − 1)λ ] [( j − 1)λ{1 + Aγδ + ( B − A)δξ } + ( B − A)δξ (1 − α )]
β
f ( z ) ∈ TnV λ (α , β , ξ , γ , δ , A, B ) if and only if f(z) can be expressed in the forms
∞
∞
f ( z ) = ∑ λ j f j ( z ) , where λ j ≥ 0 and
∑λ
j =1
j =1
j
=1
∞
Proof: Let
f ( z ) = ∑ λ j f j ( z ), λ j ≥ 0 , j = 1, 2,3,........... with
j =1
We have,
∞
∞
j =1
j =2
∞
∑λ
j =1
j
=1
f ( z ) = ∑ λ j f j ( z ) = λ1 f1 ( z ) + ∑ λ j f j ( z )
∞


( B − A)δξ (1 − α )
∴ f ( z) = z − ∑ λ j 
zj
β
j =2
 [1 + ( j − 1)λ ] [( j − 1)λ{1 + Aγδ + ( B − A)δξ } + ( B − A)δξ (1 − α )] 
Then,
( B − A)δξ (1 − α )


 [1 + ( j − 1)λ ]β [( j − 1)λ{1 + Aγδ + ( B − A)δξ } + ( B − A)δξ (1 − α )] × ∞
 = ∑ λ = 1− λ ≤ 1
λj 
∑
 [1 + ( j − 1)λ ]β [( j − 1)λ{1 + Aγδ + ( B − A)δξ } + ( B − A)δξ (1 − α )]  j = 2 j
j =2


( B − A)δξ (1 − α )


∞
Hence,
f ( z ) ∈ TnV λ (α , β , ξ , γ , δ , A, B )
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13
On a Subclass of Analytic and Univalent Function Defined by Al-Oboudi Operator
f ( z ) ∈ TnV λ (α , β , ξ , γ , δ , A, B ) then remark of theorem 2.1 gives us
Conversely, suppose
aj ≤
( B − A)δξ (1 − α )
, j = 2,3, 4,.....
(1 + ( j − 1)λ ) {( j − 1)λ (1 + Aγδ + ( B − A)δξ ) + ( B − A)δξ (1 − α )}
β
We take
λj =
(1 + ( j − 1)λ ) β {( j − 1)λ (1 + Aγδ + ( B − A)δξ ) + ( B − A)δξ (1 − α )}
aj,
( B − A)δξ (1 − α )
j = 2,3, 4,.....
∞
And
λ1 = 1 − ∑ λ j .
j =1
∞
Then,
f ( z ) = ∑ λ j f j ( z ).
j =1
Corollary 4.1: If f1(z )= z and
f j ( z) = z −
Then
( B − A)δξ (1 − α )
z j for j = 2, 3, 4,.....
( j − 1) − δ {( B − A)ξα − ( B − A)ξ j − Aγ ( j − 1)}
f ( z ) ∈ TnV 1 (α , 0, ξ , γ , δ , A, B) if and only if f(z) can be expressed in the form
∞
f ( z ) = ∑ λ j f j ( z ) , where λ j ≥ 0 and
j =1
∞
∑λ
j =1
j
= 1, for j = 1, 2,3, 4,.....
This corollary is due to [4].
Corollary 4.2: If f1(z )= z and
f j ( z) = z −
Then
( B − A)δξ (1 − α )
z j for j = 2,3, 4,.....
( j − 1) − δ {( B − A)ξα − ( B − A)ξ j − A( j − 1)}
f ( z ) ∈ TnV 1 (α , 0, ξ ,1, δ , A, B) if and only if f(z) can be expressed in the form
∞
f ( z ) = ∑ λ j f j ( z ) , where
j =1
∞
∑λ
j =1
j
= 1 and λ j ≥ 0 , for j = 1, 2,3, 4,.....
This corollary is due to [2] and [7].
Corollary 4.3: If f1(z )= z and
Then
f j ( z) = z −
( B − A)δ (1 − α )
z j for j = 2,3, 4,.....
( j − 1) − δ {( B − A)α − Bj + A}
f ( z ) ∈ TnV 1 (α , 0,1,1, δ , A, B) if and only if f(z) can be expressed in the form
∞
f ( z ) = ∑ λ j f j ( z ), where
j =1
∞
∑λ
j =1
j
= 1 and λ j ≥ 0, for j = 1, 2,3, 4,.....
This corollary is due to [9].
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14
N. D. Sangle, A. N. Metkari & D. S. Mane
Corollary 4.4: If f1(z )= z and
Then
f j ( z) = z −
( B − A)
z j for j = 2,3, 4,.....
j ( B + 1) − ( A + 1)
f ( z ) ∈ TnV 1 (0, 0,1,1,1, A, B ) if and only if f(z) can be expressed in the form
∞
f ( z ) = ∑ λ j f j ( z ), where
j =1
∞
∑λ
j =1
j
= 1 and λ j ≥ 0, for j = 1, 2,3, 4,.....
5. CONCLUSIONS
In this paper making use of Al-Oboudi operator two new sub classes of analytic and univalent functions are
introduced for the functions with negative coefficients. Many subclasses which are already studied by various researchers
are obtained as special cases of our two new sub classes. We have obtained varies properties such as coefficient estimates,
growth distortion theorems, Further new subclasses may be possible from the two classes introduced in this paper.
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