On the Uniqueness of Meromorphic Functions on Annuli in terms of Deficiencies

Abstract

The purpose of this article is to study the uniqueness of meromorphic functions on annuli. Under a certain condition about deficiencies, we prove some new uniqueness theorems of meromorphic functions on the annulus , where .

1. Introduction and Main Results

In this article, we assume that the readers are familiar with the classical notations and definitions of Nevanlinna theory (refer to [1, 2]). The main purpose of this article is to study the uniqueness of meromorphic functions on annuli. For the necessary concepts and notations of the Nevanlinna theory of meromorphic functions on annuli, such as , refer to the excellent summarizations [3–11].

Let a be a value in the extended complex plane , and let be two meromorphic functions on the annulus , where . Then, we say that f and share a IM (ignoring multiplicities) when and have the same zeros, and furthermore, we say that f and share a CM (counting multiplicities) when and have the same zeros with the same multiplicities. As mentioned in [3, 4], the reduced counting function is defined bywhere and are the functions counting (counting only once) the zeros of in and , respectively. Similarly, we denote by the reduced counting function of common zeros (different zeros) of and on , where . It is obvious that f and share a IM on if . Following the definitions in [2], we say that f and share a “IM” if and we say that f and share a “CM” if , in which denotes the reduced counting function of common zeros with the same multiplicities of and .

For a nonconstant meromorphic function f on the annulus , it is named as a transcendental meromorphic function on provided thatrespectively. In fact, the transcendental meromorphic functions are also known as admissible meromorphic functions. If f is a transcendental meromorphic function on , then we have for all except for a set such that or a set such that , respectively.

There existed many famous results about the uniqueness theory of meromorphic functions sharing values. In 1926, Nevanlinna [12] proved the celebrated five-value theorem.

Theorem 1. Let f and g be two nonconstant meromorphic functions in , and let be five distinct values in If f and g share the values IM for in , then .

Since that time, a series of results emerged in large numbers, which discussed and generalized the five-value theorem (Theroem 1). For the main results about the generalizations of Theroem 1 in simply connected regions, we can refer to [7, 13–16]. For instance, Zheng [15, 16] and Fang [13] obtained the generalization of the five-value theorem in an angular domain and in the unit disc, respectively. For the generalizations on multiply-connected regions, we can refer to [5, 17, 18].

Recently, Cao et al. [17, 18] proved the following five-value theorem on annuli (the case of was derived from [5] by Kondratyuk and Laine).

Theorem 2. Let f and be two transcendental meromorphic functions on the annulus , where Let be five distinct values in If f and g share the values IM for on , then .

From the very point of sharing small functions, we studied above theorems in [19] and provided the following uniqueness theorem of meromorphic functions sharing four small functions on annuli.

Theorem 3. Let f and be two transcendental meromorphic functions on the annulus where . Let be four distinct small functions with respect to f and g on . If f and g share IM andthen , where is the reduced counting function which counts the multiple common zeros of and on .

In this article, we mainly investigate whether Theroem 2 holds if f and dissatisfy the condition of sharing values. From the very point of deficiencies, we deal with this question and propose the following uniqueness theorem without conditions of sharing values. This theorem generalizes Theroem 2.

Theorem 4. Let f and be two transcendental meromorphic functions on the annulus , where and let be five distinct complex numbers in Then, we have provided thatwhere the deficiencies are defined aswhen , orwhen respectively.

In special, if f and share “IM,” then it is obvious that which satisfies the conditionof Theroem 4. In view of the discussion above, we deduce a corollary as follows. This corollary partly improves Theroem 2 in the sense that IM is replaced with “IM.”

Corollary 1. Let f and be two transcendental meromorphic functions on the annulus where and let be five distinct complex numbers in If f and g share “IM,” then .

2. Some Lemmas

In this section, we will give some necessary lemmas, where the third lemma is motivated by the ideas of [20–22].

Lemma 1 (see [4], Theroem 1). Let f be a nonconstant meromorphic function on the annulus where and let . Then,(i)If then for except for a set such that .(ii)If , then for except for a set such that .

Lemma 2 (see [18], Theorem 2.3). Let f be a nonconstant meromorphic function on the annulus where Let be q distinct complex numbers in the extended complex plane . Then,

Inspired by the ideas of [20–22], we propose the following lemma and give the proof.

Lemma 3. Let f and be two transcendental meromorphic functions on where and let be five distinct complex numbers in . If , thenwhere is the reduced counting function of the common (different) zeros of and on .

Proof. Without loss of generality, we suppose that , , , , and , in which are two distinct complex numbers such that . Otherwise, a Möbius transformation aswill be done. Then, setwhereNoting that can be expressed byBy Lemma 1, we have . Similarly, we get and thus,holds.
Next, we will estimate the counting function A simple computation yieldsThen, it is easy to see that the poles of h only come from the zeros of and the poles of on . Now, let be a common zero of f and on with multiplicity p and q, respectively. Without loss of generality, assume that Then, it follows that is a zero of with multiplicity at least and that is a pole ofwith multiplicity 2. We consequently know that is not a pole of h, and hence the poles of h cannot occur at the common zeros of f and . By similar methods, we can conclude that the poles of h cannot occur at the common zeros of and , the common zeros of and and the common poles of f and , so the poles of h only come from the different zeros of and the different poles of on . In order to analyze these different zeros and different poles, we distinguish the following distinct cases. Case 1: let be a zero of which is not a zero of . Then, by using the equation (15), we find that is a pole of h with multiplicity at most 1 if otherwise, is a pole of h with multiplicity at most 2. Case 2: let be a zero of which is not a zero of It is clear that is a pole of h with multiplicity at most 1 if otherwise, is a pole of h with multiplicity at most 2. Case 3: let be a pole of which is not a pole of It is clear that is a pole of h with multiplicity at most 1 if otherwise, is a pole of h with multiplicity at most 2. Case 4: let be a zero of which is not a zero of It is clear that is a pole of h with multiplicity at most 1 if otherwise, is a pole of h with multiplicity at most 2. Case 5: let be a zero of which is not a zero of It is clear that is a pole of h with multiplicity at most 1 if otherwise, is a pole of h with multiplicity at most 2. Case 6: let be a zero of which is not a zero of It is clear that is a pole of h with multiplicity at most 1 if otherwise, is a pole of h with multiplicity at most 2. Case 7: let be a pole of which is not a pole of It is clear that is a pole of h with multiplicity at most 1 if otherwise, is a pole of h with multiplicity at most 2. Case 8: let be a zero of which is not a zero of It is clear that is a pole of h with multiplicity at most 1 if otherwise, is a pole of h with multiplicity at most 2.In view of these cases, we obtainwhich meansCombining (14) with (18), we getIf then This implies that which is impossible, so holds. Furthermore, it follows from (15) that the common zeros of and must be the zeros of . This implies thatwhich further implies thatcombined with (19). Similarly, we can derive other inequations asTherefore, we have proved Lemma 3.

3. The Proof of Theorem 4

On the contrary, we suppose that , and then it follows from Lemma 3 thatfor Thus, noting thatwe know

This yields thatfor which further yields that

On the other hand, by utilizing Lemma 2, we find

Therefore, it follows from (27) that which means

Consequently, from (30), we havewhich is a contradiction. Hence, the proof is completed.

Data Availability

The data used to support the findings of this study are included within the article.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This work was supported by the NSF of China (11271227, 11561033, and 61373174) and the Fundamental Research Funds for the Central Universities (JB180708).