2_834_frasin.dvi EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 4, No. 1, 2011, 14-19 ISSN 1307-5543 – www.ejpam.com (α,β ,δ)−Neighborhood for Certain Analytic Functions with Neg- ative Coefficients B.A. Frasin Faculty of Science, Department of Mathematics, Al al-Bayt University, P.O. Box: 130095 Mafraq, Jordan Abstract. In this paper, we introduce (α,β ,δ)−neighborhoods of analytic functions with negative coefficients. Furthermore, we obtain some interesting results for functions belonging to this neighbor- hoods. 2000 Mathematics Subject Classifications: 30C45 Key Words and Phrases: Analytic functions, Neighborhood 1. Introduction and definitions Let T denote the class of functions of the form : f (z) = z − ∞ ∑ n=2 anzn, (an ≥ 0). (1) which are analytic in the open unit disk U = {z : |z| < 1}.For a function f (z) ∈ T , we define D0 f (z) = f (z), D1 f (z) = D f (z) = z f ′(z), and Dk f (z) = D(Dk−1 f (z)) = z − ∞ ∑ n=2 nkanzn (k ∈ N0 = N∪ {0}). The differential operator Dk was introduced by Sălăgean [12]. Email address: bafrasin�yahoo. om http://www.ejpam.com 14 c© 2010 EJPAM All rights reserved. B. Frasin / Eur. J. Pure Appl. Math, 4 (2011), 14-19 15 Following a recent investigation by Frasin and Darus [6] [see also 1], if f (z) ∈ T and µ ≥ 0, then we define the (k,µ)-neighborhood for the function f (z) by N k µ ( f ) = {g ∈ T : g(z) = z − ∞ ∑ n=2 bnzn, ∞ ∑ n=2 nk+1 � �an− bn � � ≤ µ}. (2) In particular, for the identity function e(z) = z, we immediately have N k µ (e) = {g ∈ T : g(z) = z − ∞ ∑ n=2 bnzn, ∞ ∑ n=2 nk+1 � �bn � � ≤ µ}, (3) We observe that N 0 µ ( f ) ≡ N µ ( f ) and N 1 µ ( f ) ≡ M µ ( f ), where N k µ ( f ) and Mµ( f ) denote, respectively, the µ-neighborhoods of f as defined by Ruscheweyh [11] and Silverman [13]. For further details about the neighborhood of analytic functions see (as examples) the papers in [2, 3, 4, 7, 5, 8, 9]. Very recently, Orhan et al. [10], introduced new definition of (α,δ)-neighborhood for analytic function f (z) in the form f (z) = z + ∞ ∑ n=2 anzn, (4) In this paper, we introduce the following new definition of (α,β ,δ)-neighborhood for a func- tion given by 1. Definition 1. A function f (z) ∈ T is said to be (α,β ,δ)-neighborhood for g(z) = z − ∞ ∑ n=2 bnzn ∈ T if it satisfies � �eiα(Dk f (z))′ − eiβ (Dk g(z))′ � � < δ (z ∈ U ) (5) for some −π≤ α,β ≤ π and δ > p 2(1− cos(α−β)). We denote this neighborhood by (α,β ,δ)−N (g). Now we show some results for functions belonging to (α,β ,δ)−N (g). 2. Main results In our first theorem, we introduce a sufficient condition to be in (α,β ,δ)−N (g). Theorem 1. If f (z) ∈ T satisfies ∞ ∑ n=2 nk+1 � �eiαan− eiβ bn � �≤ δ− p 2(1− cos(α− β)) (6) for some −π≤ α,β ≤ π and δ > p 2(1− cos(α−β)) then f (z) ∈ (α,β ,δ)−N (g). B. Frasin / Eur. J. Pure Appl. Math, 4 (2011), 14-19 16 Proof. We observe that � �eiα(Dk f (z))′ − eiβ(Dk g(z))′ � � = � � � � � eiα − eiβ − ∞ ∑ n=2 nk+1(eiαan − eiβ bn)z n−1 � � � � � ≤ � �eiα − eiβ � �+ ∞ ∑ n=2 nk+1 � �eiαan − eiβ bn � � |z|n−1 ≤ p 2(1− cos(α− β)) + ∞ ∑ n=2 nk+1 � �eiαan− eiβ bn � � . If ∞ ∑ n=2 nk+1 � �eiαan − eiβ bn � �≤ δ− p 2(1− cos(α−β)), then we have � �eiα(Dk f (z))′ − eiβ(Dk g(z))′ � �< δ (z ∈ U ). This shows that f (z) ∈ (α,β ,δ)−N (g). Corollary 1. Let f (z) ∈ T . Then for 0< µ ≤ δ, we have N k µ (g)⊆ (α,α,δ)−N k δ (g). Proof. Assuming that f (z) ∈ N k µ (g). We find from the definition (2) that ∞ ∑ n=2 nk+1 � �an − bn � � ≤ µ. Now ∞ ∑ n=2 nk+1 � �eiαan− eiαbn � � = ∞ ∑ n=2 nk+1 � �eiα � � � �an− bn � � = ∞ ∑ n=2 nk+1 � �an − bn � � ≤ δ. Thus by Theorem 1, we have f (z) ∈ (α,α,δ)−N (g). Corollary 2. If f (z) ∈ T satisfies ∞ ∑ n=2 nk+1 � � � �an � �− � �bn � � � � ≤ δ− p 2(1− cos(α− β)) (7) for some −π≤ α,β ≤ π, δ > p 2(1− cos(α− β)) and arg an− arg bn = β −α (n= 2,3,4, ...), then f (z) ∈ (α,β ,δ)−N (g). B. Frasin / Eur. J. Pure Appl. Math, 4 (2011), 14-19 17 Proof. Let arg an− arg bn = β −α and arg an = θ n. Then arg bn = θ n+α−β . Therefore, eiαan − eiβ bn = � �an � � ei(α+θ n) − � �bn � � ei(α+θ n), which implies � �eiαan− eiβ bn � �= � � � �an � �− � �bn � � � � . (8) From the hypotheses (7) and (8), we get (6). Thus by Theorem 1, it follows that f (z) ∈ (α,β ,δ)−N (g). Furthermore, from Theorem 1, we easily get Corollary 3. If f (z) ∈ T satisfies ∞ ∑ n=2 nk+1( � �an � �+ � �bn � �)≤ δ− p 2(1− cos(α− β)) (9) for some −π≤ α,β ≤ π and δ > p 2(1− cos(α−β) then f (z) ∈ (α,β ,δ)−N (g). Next, we prove Theorem 2. If f (z) ∈ (α,β ,δ)−N (g) and arg(eiαan − eiβ bn) = (n− 1)ϕ (n = 2,3,4, . . .), then ∞ ∑ n=2 nk+1 � �eiαan− eiβ bn � �> δ+ cosα− cosβ . (10) Proof. Let f (z) ∈ (α,β ,δ)−N (g) and arg z = −ϕ. Then for all z ∈ U , we have � �eiα(Dk f (z))′− eiβ (Dk g(z))′ � � = � � � � � (eiα− eiβ )− ∞ ∑ n=2 nk+1(eiαan − eiβ bn)z n−1 � � � � � = � � � � � (eiα− eiβ )− ∞ ∑ n=2 nk+1 � �eiαan− eiβ bn � � ei(n−1)ϕ |z|n−1 e−i(n−1)ϕ � � � � � = � � � � � (eiα− eiβ )− ∞ ∑ n=2 nk+1 � �eiαan− eiβ bn � � |z|n−1 � � � � � = [(cosα− cosβ)− ∞ ∑ n=2 nk+1 � �eiαan− eiβ bn � � |z|n−1]2+ (sinα− sinβ)2 �1/2 < δ for z ∈ U . This implies that (cosα− cosβ)− ∞ ∑ n=2 nk+1 � �eiαan − eiβ bn � � |z|n−1 < δ REFERENCES 18 for z ∈ U . Letting |z| → 1−, we have ∞ ∑ n=2 nk+1 � �eiαan− eiβ bn � �> δ+ cosα− cosβ . Finally, we prove Theorem 3. If f (z) ∈ T satisfies ∞ ∑ n=2 nk+1 � �eiαan− eiβ bn � � < µ− p 2(1− cos(α−β)) (11) for some −π≤ α,β ≤ π and µ > p 2(1− cos(α− β)), then Re � eiα(Dk f (z))′ eiβ (Dk g(z))′ � > 0 (12) where g(z) ∈ N k 1−µ(e). Proof. 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