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),
This article can be downloaded from www.impactjournals.us
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
Articles can be sent to editor@impactjournals.us
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 )
This article can be downloaded from www.impactjournals.us
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, Articles can be sent to editor@impactjournals.us
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 ).
This article can be downloaded from www.impactjournals.us
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}]
Articles can be sent to editor@impactjournals.us
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
This article can be downloaded from www.impactjournals.us
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, Articles can be sent to editor@impactjournals.us
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].
This article can be downloaded from www.impactjournals.us
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,
Articles can be sent to editor@impactjournals.us
(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 )} This article can be downloaded from www.impactjournals.us
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 ) Articles can be sent to editor@impactjournals.us
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]. This article can be downloaded from www.impactjournals.us
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.
REFERENCES 1.
M. Acu, Owa, “Note on a class of starlike functions”, Proceeding of the International short work on study on calculus operators in univalent function theory, Kyoto, Pp. 1-10, 2006.
2.
R. Aghalary and S. R. Kulkarni, “Some theorems on univalent functions”, J. Indian Acad. Math., Vol. 24, No. 1, Pp.81-93, 2002.
3.
F.M. Al-Oboudi, “on univalent functions defined by a generalized Salagean operator”, Ind. J. Math. Sci., No. 25-28, 1429-1436, 2004.
4.
S.M. Khairnar and Meena More, “Certain family of analytic and univalent functions”, Acta Mathematica Academiae Paedogical, Vol. 24, Pp. 333-344, 2008.
5.
S.R. Kulkarni, “Some problems connected with univalent functions”, Ph.D. Thesis, Shivaji University, Kolhapur, 1981.
6.
H. Silverman., Univalent functions with negative coefficients; Proc. Amer. Math. Soc.51, (1975), 109-116.
7.
H. Silverman and E. Silvia, “Subclasses of prestarlike functions”, Math., Japon, Vol.29, No. 6, Pp. 929-935, 1984.
8.
T. V. Sudharsan, R. Thirumalaisamy, K.G. Subramanian, Mugur Acu, “A class of analytic functions based on an extension of Al-Oboudi operator”, Acta Universitatis Apulensis, Vol. 21, Pp. 79-88, 2010.
9.
S. Owa and J. Nishiwaki, “Coefficient Estimate for certain classes of analytic functions”, JIPAM, J. Inequal. Pure Appl. Math, vol. 315, article 72, 5pp, (electronic), 2002.
10. N.D. Sangle, S.B. Joshi, “New classes of analytic and univalent functions”, Varahamihir Journal of Mathematical Sciences, 6(2), 2006, 537-550. 11. T. V. Sudharsan and S.P. Vijayalakshmi, “On certain classes of Analytic and univalent functions based on Al-Oboudi operator”, Bonfring international Journal of data Mining 2(2), 2012, 6-12.
Articles can be sent to editor@impactjournals.us