ON APPLICATION OF DIFFERENTIAL SUBORDINATION FOR CERTAIN SUBCLASS OF MEROMORPHICALLY p-VALENT FUNCTIONS WITH POSITIVE COEFFICIENTS DEFINED BY LINEAR OPERATOR Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page WAGGAS GALIB ATSHAN S. R. KULKARNI Department of Mathematics College of Computer Science And Mathematics University of Al-Qadisiya, Diwaniya - Iraq EMail: waggashnd@yahoo.com Department of Mathematics Fergusson College, Pune - 411004, India EMail: kulkarni_ferg@yahoo.com Contents JJ II J I Page 1 of 24 Received: 06 January, 2008 Accepted: 02 May, 2009 Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 30C45. Key words: Meromorphic functions, Differential subordination, convolution (or Hadamard product), p-valent functions, Linear operator, δ-Neighborhood, Integral representation, Linear combination, Weighted mean and Arithmetic mean. Go Back Full Screen Close Abstract: This paper is mainly concerned with the application of differential subordinations for the class of meromorphic multivalent functions with positive coefficients defined by a linear operator satisfying the following: − z p+1 (Ln f (z))0 1 + Az ≺ (n ∈ N0 ; z ∈ U ). p 1 + Bz In the present paper, we study the coefficient bounds, δ-neighborhoods and integral representations. We also obtain linear combinations, weighted and arithmetic means and convolution properties. Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page Acknowledgement: The first author, Waggas Galib, is thankful of his wife (Hnd Hekmat Abdulah) for her support of him in his work. Contents JJ II J I Page 2 of 24 Go Back Full Screen Close Contents 1 Introduction 4 2 Coefficient Bounds 7 3 Neighbourhoods and Partial Sums 10 4 Integral Representation 15 5 Linear Combination 16 6 Weighted Mean and Arithmetic Mean 17 Title Page 7 Convolution Properties 19 Contents Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 JJ II J I Page 3 of 24 Go Back Full Screen Close 1. Introduction Let L(p, m) be a class of all meromorphic functions f (z) of the form: (1.1) f (z) = z −p + ∞ X ak z k for any m ≥ p, p ∈ N = {1, 2, . . . }, ak ≥ 0, k=m Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni which are p-valent in the punctured unit disk U ∗ = {z : z ∈ C, 0 < |z| < 1} = U/{0}. vol. 10, iss. 2, art. 53, 2009 Definition 1.1. Let f, g be analytic in U . Then g is said to be subordinate to f, written g ≺ f , if there exists a Schwarz function w(z), which is analytic in U with w(0) = 0 and |w(z)| < 1 (z ∈ U ) such that g(z) = f (w(z)) (z ∈ U ). Hence g(z) ≺ f (z) (z ∈ U ), then g(0) = f (0) and g(U ) ⊂ f (U ). In particular, if the function f (z) is univalent in U , we have the following (e.g. [6]; [7]): g(z) ≺ f (z)(z ∈ U ) if and only if g(0) = f (0) and g(U ) ⊂ f (U ). Definition 1.2. For functions f (z) ∈ L(p, m) given by (1.1) and g(z) ∈ L(p, m) defined by (1.2) g(z) = z −p + ∞ X bk z k , (bk ≥ 0, p ∈ N, m ≥ p), we define the convolution (or Hadamard product) of f (z) and g(z) by (1.3) (f ∗ g)(z) = z + ∞ X k=m Contents JJ II J I Page 4 of 24 Go Back Full Screen Close k=m −p Title Page ak b k z k , (p ∈ N, m ≥ p, z ∈ U ). Definition 1.3 ([9]). Let f (z) be a function in the class L(p, m) given by (1.1). We define a linear operator Ln by L0 f (z) = f (z), L1 f (z) = z −p + ∞ X (p + k + 1)ak z k = k=m (z p+1 f (z))0 zp and in general (1.4) Ln f (z) = L(Ln−1 f (z)) ∞ X −p =z + (p + k + 1)n ak z k = (z p+1 k=m n−1 L zp vol. 10, iss. 2, art. 53, 2009 Title Page f (z))0 , (n ∈ N). It is easily verified from (1.4) that (1.5) Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni z(Ln f (z))0 = Ln+1 f (z) − (p + 1)Ln f (z), (f ∈ L(p, m), n ∈ N0 = N ∪ {0}). 1. Liu and Srivastava [4] introduced recently the linear operator when m = 0, investigating several inclusion relationships involving various subclasses of meromorphically p-valent functions, which they defined by means of the linear operator Ln (see [4]). 2. Uralegaddi and Somanatha [10] introduced the linear operator Ln when p = 1 and m = 0. 3. Aouf and Hossen [2] obtained several results involving the linear operator Ln when m = 0 and p ∈ N. Contents JJ II J I Page 5 of 24 Go Back Full Screen Close We introduce a subclass of the function class L(p, m) by making use of the principle of differential subordination as well as the linear operator Ln . Definition 1.4. Let A and B (−1 ≤ B < A ≤ 1) be fixed parameters. We say that a function f (z) ∈ L(p, m) is in the class L(p, m, n, A, B), if it satisfies the following subordination condition: (1.6) 1 + Az z p+1 (Ln f (z))0 ≺ p 1 + Bz (n ∈ N0 ; z ∈ U ). By the definition of differential subordination, (1.6) is equivalent to the following condition: p+1 n z (L f (z))0 + p Bz p+1 (Ln f (z))0 + pA < 1, (z ∈ U ). We can write 2β L p, m, n, 1 − , −1 = L(p, m, n, β), p where L(p, m, n, β) denotes the class of functions in L(p, m) satisfying the following: Re{−z p+1 (Ln f (z))0 } > β (0 ≤ β < p; z ∈ U ). Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page Contents JJ II J I Page 6 of 24 Go Back Full Screen Close 2. Coefficient Bounds Theorem 2.1. Let the function f (z) of the form (1.1), be in L(p, m). Then the function f (z) belongs to the class L(p, m, n, A, B) if and only if (2.1) ∞ X k(1 − B)(p + k + 1)n ak < (A − B)p, Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni k=m where −1 ≤ B < A ≤ 1, p ∈ N, n ∈ N0 , m ≥ p. The result is sharp for the function f (z) given by f (z) = z −p + (A − B)p zm, k(1 − B)(p + k + 1)n vol. 10, iss. 2, art. 53, 2009 m ≥ p. Proof. Assume that the condition (2.1) is true. We must show that f ∈ L(p, m, n, A, B), or equivalently prove that p+1 n z (L f (z))0 + p (2.2) Bz p+1 (Ln f (z))0 + Ap < 1. Title Page Contents JJ II J I Page 7 of 24 We have ∞ P p+1 −(p+1) n k−1 p+1 n z (−pz + k(p + k + 1) ak z ) + p 0 z (L f (z)) + p k=m ∞ Bz p+1 (Ln f (z))0 + Ap = Bz p+1 (−pz −(p+1) + P k(p + k + 1)n ak z k−1 ) + Ap k=m ∞ P k(p + k + 1)n ak z k+p k=m = ∞ P n a z k+p (A − B)p + B k(p + k + 1) k k=m Go Back Full Screen Close ≤ ∞ P n k(p + k + 1) ak k=m < 1. ∞ P n k(k + p + 1) ak (A − B)p + B k=m The last inequality by (2.1) is true. Conversely, suppose that f (z) ∈ L(p, m, n, A, B). We must show that the condition (2.1) holds true. We have p+1 n z (L f (z))0 + p Bz p+1 (Ln f (z))0 + Ap < 1, Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 hence we get ∞ P k(p + k + 1)n ak z k+p k=m < 1. ∞ (A − B)p + B P k(p + k + 1)n ak z k+p Title Page Contents JJ II J I k=m Since Re(z) < |z|, so we have ∞ P n k+p k(p + k + 1) ak z k=m Re < 1. ∞ P n k+p (A − B)p + B k(p + k + 1) a z k Page 8 of 24 Go Back Full Screen k=m We choose the values of z on the real axis and letting z → 1− , then we obtain ∞ P n k(p + k + 1) ak k=m < 1, ∞ P n k(p + k + 1) ak (A − B)p + B k=m Close then ∞ X k(1 − B)(p + k + 1)n ak < (A − B)p k=m and the proof is complete. Corollary 2.2. Let f (z) ∈ L(p, m, n, A, B), then we have ak ≤ (A − B)p , k ≥ m. k(1 − B)(p + k + 1)n Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Corollary 2.3. Let 0 ≤ n2 < n1 , then L(p, m, n2 , A, B) ⊆ L(p, m, n1 , A, B). Title Page Contents JJ II J I Page 9 of 24 Go Back Full Screen Close 3. Neighbourhoods and Partial Sums Definition 3.1. Let −1 ≤ B < A ≤ 1, m ≥ p, n ∈ N0 , p ∈ N and δ ≥ 0. We define the δ - neighbourhood of a function f ∈ L(p, m) and denote Nδ (f ) such that ( ∞ X −p bk z k , and (3.1) Nδ (f ) = g ∈ L(p, m) : g(z) = z + k=m ∞ X k(1 − B)(p + k + 1)n |ak − bk | ≤ δ (A − B)p k=m ) . Goodman [3], Ruscheweyh [8] and Altintas and Owa [1] have investigated neighbourhoods for analytic univalent functions, we consider this concept for the class L(p, m, n, A, B). Theorem 3.2. Let the function f (z) defined by (1.1) be in L(p, m, n, A, B). For −p every complex number µ with |µ| < δ, δ ≥ 0, let f (z)+µz ∈ L(p, m, n, A, B), then 1+µ Nδ (f ) ⊂ L(p, m, n, A, B), δ ≥ 0. Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page Contents JJ II J I Page 10 of 24 Proof. Since f ∈ L(p, m, n, A, B), f satisfies (2.1) and we can write for γ ∈ C, |γ| = 1, that p+1 n z (L f (z))0 + p (3.2) 6= γ. Bz p+1 (Ln f (z))0 + pA Equivalently, we must have (3.3) (f ∗ Q)(z) 6= 0, z −p z ∈ U ∗, Go Back Full Screen Close where Q(z) = z −p + ∞ X ek z k , k=m γk(1−B)(p+k+1)n n such that ek = , satisfying |ek | ≤ k(1−B)(p+k+1) and k ≥ m, p ∈ N, (A−B)p (A−B)p n ∈ N0 . −p ∈ L(p, m, n, A, B), by (3.3), Since f (z)+µz 1+µ f (z) + µz −p 1 ∗ Q(z) 6= 0, z −p 1+µ Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 and then 1 (f ∗ Q)(z) + µz −p 1+µ 6= 0. z −p Now assume that (f ∗Q)(z) < δ. Then, by (3.4), we have z −p (3.4) 1 f ∗Q µ |µ| 1 (f ∗ Q)(z) |µ| − δ 1 + µ z −p + 1 + µ ≥ |1 + µ| − |1 + µ| > |1 + µ| ≥ 0. z −p This is a contradiction as |µ| < δ. Therefore (f ∗Q)(z) ≥ δ. z −p Letting ∞ X −p bk z k ∈ Nδ (f ), g(z) = z + k=m Title Page Contents JJ II J I Page 11 of 24 Go Back Full Screen Close then (g ∗ Q)(z) ((f − g) ∗ Q)(z) ≤ δ − z −p z −p ∞ X k ≤ (ak − bk )ek z ≤ k=m ∞ X Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni |ak − bk ||ek ||z|k k=m m < |z| ∞ X k(1 − B)(p + k + 1)n (A − B)p k=m vol. 10, iss. 2, art. 53, 2009 |ak − bk | Title Page ≤ δ, Contents therefore (g∗Q)(z) z −p 6= 0, and we get g(z) ∈ L(p, m, n, A, B), so Nδ (f ) ⊂ L(p, m, n, A, B). Theorem 3.3. Let f (z) be defined by (1.1) and the partial sums S1 (z) and Sq (z) be defined by S1 (z) = z −p and Sq (z) = z + X II J I Page 12 of 24 Go Back m+q−2 −p JJ k ak z , q > m, m ≥ p, p ∈ N. Full Screen k=m Also suppose that Close P∞ k=m Ck ak ≤ 1, where Ck = k(1 − B)(p + k + 1)n . (A − B)p Then f ∈ L(p, m, n, A, B) f (z) 1 Re , >1− Sq (z) Cq (i) (3.5) (ii) (3.6) Re Sq (z) f (z) Cq > , 1 + Cq Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni z ∈ U, q > m. Proof. −p +µz −p (i) Since z 1+µ = z −p ∈ L(p, m, n, A, B), |µ| < 1, then by Theorem 3.2, we have N1 (z −p ) ⊂ L(p, m, n, A, B), p ∈ N(N1 (z −p ) denoting the 1-neighbourhood). Now since ∞ X Ck ak ≤ 1, k=m then f ∈ N1 (z −p ) and f ∈ L(p, m, n, A, B). (ii) Since {Ck } is an increasing sequence, we obtain m+q−2 (3.7) X k=m ∞ X ak + Cq ak ≤ k=q+m−1 vol. 10, iss. 2, art. 53, 2009 Title Page Contents JJ II J I Page 13 of 24 ∞ X Go Back Ck ak ≤ 1. Full Screen k=m Close Setting G1 (z) = Cq 1 f (z) − 1− Sq (z) Cq ∞ P Cq = ak z k+p k=q+m−1 1+ m+q−2 P k=m ak z k+p + 1, from (3.7) we get ∞ P k+p ak z Cq G1 (z) − 1 k=q+m−1 = G1 (z) + 1 m+q−2 ∞ P P k+p k+p 2 + 2 ak z ak z + Cq k=m k=q+m−1 Cq ∞ P ak k=q+m−1 ≤ 2−2 m+q−2 P k=m ak − Cq ∞ P ≤ 1. vol. 10, iss. 2, art. 53, 2009 ak k=q+m−1 n f (z) Sq (z) This proves (3.5). Therefore, Re(G1 (z)) > 0 and we obtain Re Now, in the same manner, we can prove the assertion (3.6), by setting Sq (z) Cq G2 (z) = (1 + Cq ) − . f (z) 1 + Cq This completes the proof. Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni o > 1− C1q . Title Page Contents JJ II J I Page 14 of 24 Go Back Full Screen Close 4. Integral Representation In the next theorem we obtain an integral representation for Ln f (z). Theorem 4.1. Let f ∈ L(p, m, n, A, B), then Z z p(Aψ(t) − 1) n L f (z) = dt, p+1 (1 − Bψ(t)) 0 t where |ψ(z)| < 1, z ∈ U ∗ . Proof. Let f (z) ∈ L(p, m, n, A, B). Letting − z y(z) ≺ p+1 (Ln f (z))0 p = y(z), we have 1 + Az 1 + Bz Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page y(z)−1 or we can write By(z)−A < 1, so that consequently we have y(z) − 1 = ψ(z), |ψ(z)| < 1, z ∈ U. By(z) − A We can write −z p+1 (Ln f (z))0 1 − Aψ(z) = , p 1 − Bψ(z) which gives p(Aψ(z) − 1) (L f (z)) = p+1 . z (1 − Bψ(z)) Hence Z z p(Aψ(t) − 1) n L f (z) = dt, p+1 (1 − Bψ(t)) 0 t and this gives the required result. n Contents JJ II J I Page 15 of 24 Go Back Full Screen 0 Close 5. Linear Combination In the theorem below, we prove a linear combination for the class L(p, m, n, A, B). Theorem 5.1. Let fi (z) = z −p + ∞ X ak,i z k , (ak,i ≥ 0, i = 1, 2, . . . , `, k ≥ m, m ≥ p) Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni k=m belong to L(p, m, n, A, B), then F (z) = ` X vol. 10, iss. 2, art. 53, 2009 ci fi (z) ∈ L(p, m, n, A, B), i=1 where P` i=1 ci Title Page = 1. Contents Proof. By Theorem 2.1, we can write for every i ∈ {1, 2, . . . , `} ∞ X k(1 − B)(p + k + 1)n ak,i < 1, (A − B)p k=m therefore F (z) = JJ II J I Page 16 of 24 ` X ci z −p + i=1 ∞ X ! ak,i z k = z −p + k=m ∞ X ` X k=m i=1 ! Go Back ci ak,i z k . Full Screen However, ∞ X k(1 − B)(p + k + 1)n (A − B)p k=m Close ` X i=1 ! ci ak,i " # ` ∞ X X k(1 − B)(p + k + 1)n = ak,i ci ≤ 1, (A − B)p i=1 k=m then F (z) ∈ L(p, m, n, A, B), so the proof is complete. 6. Weighted Mean and Arithmetic Mean Definition 6.1. Let f (z) and g(z) belong to L(p, m), then the weighted mean hj (z) of f (z) and g(z) is given by 1 hj (z) = [(1 − j)f (z) + (1 + j)g(z)]. 2 In the theorem below we will show the weighted mean for this class. Theorem 6.2. If f (z) and g(z) are in the class L(p, m, n, A, B), then the weighted mean of f (z) and g(z) is also in L(p, m, n, A, B). Proof. We have for hj (z) by Definition 6.1, " ! !# ∞ ∞ X X 1 −p k −p k ak z + (1 + j) z + bk z hj (z) = (1 − j) z + 2 k=m k=m =z −p ∞ X 1 + ((1 − j)ak + (1 + j)bk )z k . 2 k=m Since f (z) and g(z) are in the class L(p, m, n, A, B) so by Theorem 2.1 we must prove that ∞ X 1 n 1 k(1 − B)(p + k + 1) (1 − j)ak + (1 + j)bk 2 2 k=m ∞ ∞ X X 1 1 n = (1 − j) k(1 − B)(p + k + 1) ak + (1 + j) k(1 − B)(p + k + 1)n bk 2 2 k=m k=m 1 1 ≤ (1 − j)(A − B)p + (1 + j)(A − B)p. 2 2 Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page Contents JJ II J I Page 17 of 24 Go Back Full Screen Close The proof is complete. Theorem 6.3. Let f1 (z), f2 (z), . . . , f` (z) defined by (6.1) fi (z) = z −p + ∞ X ak,i z k , (ak,i ≥ 0, i = 1, 2, . . . , `, k ≥ m, m ≥ p) k=m be in the class L(p, m, n, A, B), then the arithmetic mean of fi (z) (i = 1, 2, . . . , `) defined by ` h(z) = (6.2) 1X fi (z) ` i=1 Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page is also in the class L(p, m, n, A, B). Proof. By (6.1), (6.2) we can write h(z) = ` ∞ 1 X −p X z + ak,i z k ` i=1 k=m Contents ! = z −p + ∞ X k=m ` ! 1X ak,i z k . ` i=1 Since fi (z) ∈ L(p, m, n, A, B) for every i = 1, 2, . . . , `, so by using Theorem 2.1, we prove that ! ` ∞ X 1X n ak,i k(1 − B)(p + k + 1) ` i=1 k=m ! ` ∞ ` 1X X 1X n = k(1 − B)(p + k + 1) ak,i ≤ (A − B)p. ` i=1 k=m ` i=1 The proof is complete. JJ II J I Page 18 of 24 Go Back Full Screen Close 7. Convolution Properties Theorem 7.1. If f (z) and g(z) belong to L(p, m, n, A, B) such that (7.1) f (z) = z −p + ∞ X k ak z , g(z) = z −p + k=m ∞ X bk z k , k=m then T (z) = z −p + ∞ X Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni (a2k + b2k )z k vol. 10, iss. 2, art. 53, 2009 k=m is in the class L(p, m, n, A1 , B1 ) such that A1 ≥ (1 − B1 )µ2 + B1 , where √ 2(A − B) . µ= p m(m + 2)n (1 − B) Proof. Since f, g ∈ L(p, m, n, A, B), Theorem 2.1 yields 2 ∞ X k(1 − B)(p + k + 1)n ak ≤ 1 (A − B)p k=m and ∞ X k(1 − B)(p + k + 1)n k=m (A − B)p Contents JJ II J I Page 19 of 24 Go Back Full Screen 2 bk Title Page ≤ 1. We obtain from the last two inequalities 2 ∞ X 1 k(1 − B)(p + k + 1)n (7.2) (a2k + b2k ) ≤ 1. 2 (A − B)p k=m Close However, T (z) ∈ L(p, m, n, A1 , B1 ) if and only if ∞ X k(1 − B1 )(p + k + 1)n (7.3) (a2k + b2k ) ≤ 1, (A 1 − B1 )p k=m where −1 ≤ B1 < A1 ≤ 1, but (7.2) implies (7.3) if 2 k(1 − B1 )(p + k + 1)n 1 k(1 − B)(p + k + 1)n < . (A1 − B1 )p 2 (A − B)p Hence, if 1 − B1 k(p + k + 1)n 2 1−B < α , where α = . A1 − B1 2p A−B In other words, k(k + 2)n 2 1 − B1 < α . A1 − B1 2 This is equivalent to A1 − B1 2 > . 1 − B1 k(k + 2)n α2 So we can write A1 − B1 2(A − B)2 (7.4) > = µ2 . 1 − B1 m(m + 2)n (1 − B)2 Hence we get A1 ≥ (1 − B1 )µ2 + B1 . Theorem 7.2. Let f (z) and g(z) of the form (7.1) belong to L(p, m, n, A, B). Then the convolution (or Hadamard product) of two functions f and g belong to the class, that is, (f ∗ g)(z) ∈ L(p, m, n, A1 , B1 ), where A1 ≥ (1 − B1 )v + B1 and (A − B)2 v= . m(1 − B)2 (m + 2)n Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page Contents JJ II J I Page 20 of 24 Go Back Full Screen Close Proof. Since f, g ∈ L(p, m, n, A, B), by using the Cauchy-Schwarz inequality and Theorem 2.1, we obtain ∞ X k(1 − B)(p + k + 1)n p (7.5) ak b k (A − B)p k=m ! 12 ! 12 ∞ ∞ X X k(1 − B)(p + k + 1)n k(1 − B)(p + k + 1)n ak bk ≤1. ≤ (A − B)p (A − B)p k=m k=m We must find the values of A1 , B1 so that ∞ X k(1 − B1 )(p + k + 1)n ak bk < 1. (7.6) (A1 − B1 )p k=m Therefore, by (7.5), (7.6) holds true if p (1 − B)(A1 − B1 ) (7.7) ak b k ≤ , k ≥ m, m ≥ p, ak 6= 0, bk 6= 0. (1 − B1 )(A − B) √ (A−B)p By (7.5), we have ak bk < k(1−B)(p+k+1) n , therefore (7.7) holds true if 2 k(1 − B)(p + k + 1)n k(1 − B1 )(p + k + 1)n ≤ , (A1 − B1 )p (A − B)p Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page Contents JJ II J I Page 21 of 24 Go Back Full Screen which is equivalent to (1 − B1 ) k(1 − B)2 (p + k + 1)n < . (A1 − B1 ) (A − B)2 p Alternatively, we can write (1 − B1 ) k(1 − B)2 (k + 2)n < , (A1 − B1 ) (A − B)2 Close to obtain A1 − B1 (A − B)2 > = v. 1 − B1 m(1 − B)2 (m + 2)n Hence we get A1 > v(1 − B1 ) + B1 . Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page Contents JJ II J I Page 22 of 24 Go Back Full Screen Close References [1] O. ALTINTAS AND S. OWA, Neighborhoods of certain analytic functions with negative coefficients, IJMMS, 19 (1996), 797–800. [2] M.K. AOUF AND H.M. HOSSEN, New criteria for meromorphic p-valent starlike functions, Tsukuba J. Math., 17 (1993), 481–486. [3] A.W. GOODMAN, Univalent functions and non-analytic curves, Proc. Amer. Math. Soc., 8 (1957), 598–601. Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 [4] J.-L. LIU AND H.M. SRIVASTAVA, Classes of meromorphically multivalent functions associated with the generalized hypergeometric functions, Math. Comput. Modelling, 39 (2004), 21–34. [5] J.-L. LIU AND H.M. SRIVASTAVA, Subclasses of meromorphically multivalent functions associated with a certain linear operator, Math. Comput. Modelling, 39 (2004), 35–44. [6] S.S. MILLER AND P.T. MOCANU, Differential subordinations and univalent functions, Michigan Math. J., 28 (1981), 157–171. [7] S.S. MILLER AND P.T. MOCANU, Differential Subordinations : Theory and Applications, Series on Monographs and Textbooks in Pure and Applied Mathematics, Vol. 225, Marcel Dekker, New York and Basel, 2000. Title Page Contents JJ II J I Page 23 of 24 Go Back Full Screen Close [8] St. RUSCHEWEYH, Neighborhoods of univalent functions, Proc. Amer. Math. Soc., 81 (1981), 521–527. [9] H.M. SRIVASTAVA AND J. PATEL, Applications of differential subordination to certain subclasses of meromorphically multivalent functions, J. Ineq. Pure and Appl. Math., 6(3) (2005), Art. 88. [ONLINE: http://jipam.vu. edu.au/article.php?sid=561] [10] B.A. URALEGADDI AND C. SOMANATHA, New criteria for meromorphic starlike univalent functions, Bull. Austral. Math. Soc., 43 (1991), 137–140. Application Of Differential Subordination Waggas Galib Atshan and S. R. Kulkarni vol. 10, iss. 2, art. 53, 2009 Title Page Contents JJ II J I Page 24 of 24 Go Back Full Screen Close