/ / Article

Research Article | Open Access

Volume 2015 |Article ID 825903 | 4 pages | https://doi.org/10.1155/2015/825903

# Some Applications of Ordinary Differential Operator to Certain Multivalent Functions

Accepted11 Aug 2015
Published18 Aug 2015

#### Abstract

The aim of this paper is to apply the well-known ordinary differential operator to certain multivalent functions which are analytic in the certain domains of the complex plane and then to determine some criteria concerning analytic and geometric properties of the related complex functions.

#### 1. Introduction, Notations, and Definitions

Let denote the class of functions of the following form:which are analytic and multivalent in the domain where is the set of complex numbers. As is known, the domains and are known as unit open disk and punctured open unit disk, respectively. Also let and when .

By differentiating both sides of the function in the form (1), -times with respect to complex variable , one can easily derive the following (ordinary) differential operator: where , , and

In this investigation, by applying the differential operator, defined by (3), to certain analytic functions which are multivalent in or meromorphic multivalent in , several criteria, which also include both analytic and geometric properties of univalent functions (see [1, 2]), for functions in the classes and , are then determined. In the literature, by using certain operators, several researchers obtained some results concerning functions belonging to the general class . In this paper, we also determined many results which include starlikeness, convexity, close-to-convexity, and close-to-starlikeness of analytic functions in the second section of this paper. One may refer to some results determined by ordinary differential operator in [3, 4], some properties of certain linear operators in , and also certain results appertaining to multivalent functions and some of their geometric and analytic properties in [8, 9] in the references.

For the proofs of the main results, we then need to recall the well-known method which was obtained by Jack  (see also ) and given by the following lemma.

Lemma 1. Let the function given by be analytic in with
If then where c is real number and .

#### 2. The Main Results and Their Applications

Theorem 2. Let and , and also let the following inequality:be true. Then

Proof. Let By applying the differential operator, defined in (3), to the function , one easily get that where and are defined by respectively. It is clear that the defined function has the form in Lemma 1; that is, it is analytic in with . Upon differentiating of the identity (9), one easily obtains that Now suppose that there exists a point such that by applying Lemma 1; we then have Thus, in view of the above equality, it can be calculated that which contradicts (7). Hence, we conclude that for all , and Definition (9) yields the inequality which is equivalent to (8). Therefore, this completes the desired proof.

Theorem 3. Let , , , and, also, the following inequality be satisfied. If then

Proof. Let the functions and be in the form respectively. Then, from related differential operator and definitions of the functions and , determine where and are defined by (10) and (11), respectively. Define by Clearly, it is easily seen that satisfies the conditions of Lemma 1. The definition in (21) clearly gives us Assume now that there exists a point such that Then, applying Lemma 1, it follows from (16) and (22) that which is contradicting to the assertion (17). Hence, for all in the disk . Thus, the modulus of the identity (21), that is, requires inequality (18). This completes the proof of Theorem 3.

Theorem 4. Let , , , , , and
If then

Proof. Let the functions and be given by (19). Then, in view of Definition (20), define again an analytic function by It is obvious that is analytic in with . By differentiating in (28) logarithmically, we find that and also suppose that there exists a point such that Then, applying Lemma 1, it follows from (29) that which contradicts the assumption given by (26). Thus, for all , holds. So, (28) immediately yields inequality (27). Therefore, the desired proof is completed.

Theorem 5. Let , , and , and also let the following inequality: hold. Then where the value of the above complex power is taken to be as its principal value and is given by (10).

Proof. Let . In view of (9), we again define a function by we can easily see that and also is analytic in . By differentiating logarithmically (34), we receive Now assuming that there exists a point such that , by using Lemma 1, we then have and also take . Thus, with the help of (35), it can be determined that which contradict the cases of (32), respectively. Hence, we conclude that for all , and Definition (34) immediately yields inequality (33). This completes the related proof.

This paper includes several useful results which will be important and/or interesting for analytic and geometric function theory [1, 2]. We want to point only two special results of the main results out. The others are here omitted.

Remark 6. If we take in Theorem 5, then Theorem 5 requires Theorem 2.

If we take in Theorem 5, then we receive the following corollary.

Corollary 7. Let , , and . Then or, equivalently, where the value of the above complex power is considered to be its principal value.

#### Conflict of Interests

The authors declare that there is no conflict of interests.

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