Abstract

The object of the this paper is to derive some interesting properties of certain subclasses of meromorphically -valent functions which are defined by using an integral operator.

1. Introduction

Let denote the class of functions of the formwhich are analytic and -valent in the punctured unit disc , where . For convenience, we write .

For functions given by (1) and defined bythe Hadamard product (or convolution) of and is given by

Let and be analytic in . The function is said to be subordinate to , or is superordinate to , written , if there exists a Schwarz function in with and , such that . If is univalent in , then the equivalence (cf., e.g., [1, 2])

For , and , Saleh et al. [3] introduced the -valent Rafid operator as follows:where is the Pochhammer symbol defined, in terms of the Gamma function , by

Note that (see [4]).

It follows from (5) that

By using the integral operator , we define a subclasses of as follows.

Definition 1. For fixed parameters , we say that a function is in the class if it satisfies the following condition:There are many papers about some subclasses of meromorphic functions associated with several families of linear operators (see, for example, [511]). In this paper, we obtain some properties of the class .

2. Preliminary Lemmas

To establish our main results, in this paper, we shall need the following lemmas.

Lemma 1 (see [12] and [2]). Suppose that the function is analytic and convex (univalent) in with and given byIfthenand is the best dominant of (10).

Let be the class of analytic in of the formwhich satisfies the following inequality:

Lemma 2 (see [13]). Let the function , given by (12), be in the class . Then,

Lemma 3 (see [14]). If , then

The result is the best possible.

Let , and be any real or complex numbers with , and consider the function given by

This function, called the Gauss hypergeometric function, is analytic and converges absolutely for (see [15]).

Lemma 4 (see [15]). Let , and any real or complex numbers with . Then,

3. Main Results

Unless otherwise mentioned, we shall assume throughout the sequel that

Theorem 1. If , thenwhere the function given byis the best dominant of (22). Furthermore,whereThe result is the best possible.

Proof. SetThen, the function is of form (9) and is analytic in . Differentiating (26) and with the aid of identity (7), we obtainNow, by using Lemma 1 for , we deduce thatwhere is the best dominant of (22) given byby change of variables followed by the use of identities (17) and (18) (with , , and ). This proves assertion (22) of Theorem 1. Next, in order to prove assertion (24) of Theorem 1, it suffices to show thatIndeed, for ,Settingwhich is a positive measure on [0, 1], we obtainso thatLetting in the above inequality, we obtain assertion (30). The result in (24) is best possible as the function is the best dominant of (22).
Putting in Theorem 1, we obtain the following corollary.

Corollary 1. If satisfiesthenwhere the function given byis the best dominant of (36). Furthermore,whereThe result is the best possible.

Remark 1. For and , the result (asserted by Corollary 1) was also obtained by Patel and Sahoo [16] and Lashin [17].
Applying Theorem 1 with , , and and making use of (19), we obtain the following corollary.

Corollary 2. If satisfies the following inequalitythenThe result is the best possible.

Remark 2. The result (asserted by Corollary 2) was also obtained by Srivastava and Patel [18].
Taking in Corollary 2, we have the following corollary.

Corollary 3. If satisfies the following inequalityby Replacing by , thenThe result is the best possible.

Remark 3. The result (asserted by Corollary 3) was also obtained by Pap [19].
Applying Theorem 1 with , , and and making use of (20), we obtain the following corollary.

Corollary 4. If satisfies the following inequalitythen

The result is the best possible.

Replacing by in (26) and applying the same method and technique as the proof of Theorem 1, we can prove the following result.

Theorem 2. If satisfiesthenwhere and are given as in Theorem 1. The result is the best possible.

Theorem 3. Let . If each of the functions satisfies the following subordination conditionthenwhereThe result is the best possible when .
Proof. If we letthen, by the hypothesis of Theorem 3, we haveUsing identity (7), (53) can be written asFrom (51) and (55), we obtainwhere

Since and it follows from Lemma 3 that

According to Lemma 2, we have

Now, by using (59) in (57) and then appealing to Lemma 4, we obtainwhich completes the proof of assertion (49).

When , we consider the functions defined by

Now, by using Lemma 4 and (57), we haveas , which ends the proof of Theorem 3.

Putting , and , in Theorem 3, we get the following result.

Corollary 5. If satisfiesthenwhere

Remark 4. For , the result (asserted by Corollary 5) was also obtained by Yang [20].

Data Availability

No data were used to support this study.

Conflicts of Interest

The author declares no conflicts of interest.