Geometric Properties of Certain Analytic Functions Associated with the Dziok-Srivastava Operator

The objective of the present paper is to derive certain geometric properties of analytic functions associated with the Dziok–Srivastava operator.


Introduction
Throughout this paper, we assume that: n, p ∈ N, −1 ≤ B < A ≤ 1, α > 0 and β < 1. (1) Let A n (p) denote the class of functions of the form: which are analytic in the open unit disk U = {z : |z| < 1}.If f (z) = z p + ∑ ∞ k=n+p a k z k ∈ A n (p) and g(z) = z p + ∑ ∞ k=n+p b k z k ∈ A n (p), then the Hadamard product (or convolution) of f and g is defined by: For: where (x) k is the Pochhammer symbol given by (x is given by ( 2), then we have: For convenience, we write: It is noteworthy to mention that the Dziok-Srivastava operator is a generalization of certain linear operators considered in earlier investigations.
Next, we consider the function h(A, B; z) = (1 + Az)/(1 + Bz) for z ∈ U.It is known that the function h(A, B; z) is the conformal map of U onto a disk, symmetrical with respect to the real axis, which is centered at the point (1 − AB)/(1 − B 2 ) (B = ±1) and with its radius equal to and: For two functions f and g analytic in U, f is said to be subordinate to g, written by f (z) ≺ g(z) (z ∈ U), if there exists a Schwarz function w in U such that: Furthermore, if the function g is univalent in U, then: Many properties of analytic functions have been investigated by several authors(see [1][2][3][4][5][6][7][8][9][10][11]).In this paper, we derive certain geometric properties of analytic functions associated with the well-known Dziok-Srivastava operator.

Main Results
Theorem 1.Let f belong to the class A n (p).Furthermore, let: Then: where: The result is sharp.
Case (I).u n ≥ 1+Ar n 1+Br n , that is M n (A, B, α, r) ≤ 0. In view of ψ n (u n ) = 0 and (15), the function ψ n (u) is increasing on the segment 1−Ar n 1−Br n , 1+Ar n 1+Br n .Thus, we deduce from (12) that, if M n (A, B, α, r) ≤ 0, then: This gives (4).Next, we consider the function f defined by: which satisfies the condition (3).It is easy to check that: which implies that the inequality (4) is sharp.
Case (II).u n ≤ 1+Ar n 1+Br n , that is M n (A, B, α, r) ≥ 0. In this case, we easily have: In view of (6), ψ n (u) in (12) can be written as: Therefore, if M n (A, B, α, r) ≥ 0, then it follows from ( 16), (18), and (19) that: To show the sharpness, we take: where c n ∈ R is determined by: , and so, f satisfies the condition (3).Since: from the above argument, we find that: The proof of the theorem is now completed.
The result is sharp.

Proof. By considering
instead of H l m (α 1 ) f (z)/z p , we only need to prove the corollary for β = 0. Putting n = A = 1 and B = −1 in (6), we get: Consequently, an application of (4) in Theorem 2.1 yields: The sharpness follows immediately from that of Theorem 1.
, and satisfy: Then: The result is sharp for each k ≥ n + p.
Proof.It is well known that if: From (20), we have: In view of the function z 1+Bz being convex univalent in U, it follows from (22) that: which gives (21).Next, we consider the function f k−p (z) defined by: Since: and: for each k ≥ n + p, the proof of Theorem 2 is completed.
Furthermore, the boundary circle of this disk intersects the real axis at the points (1 − A)/(1 − B) and (1 + A)/(1 + B) with B = ±1.Let P[A, B] denote the class of functions of the form p(z) = 1 + p 1 z + • • • , which are analytic in U and satisfy the subordination p(z) ≺ h(A, B; z).It is clear that p ∈ P[A, B] if and only if: