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International Journal of Mathematics and Mathematical Sciences

Volume 2010 (2010), Article ID 507454, 11 pages

http://dx.doi.org/10.1155/2010/507454

## Unicity of Meromorphic Function Sharing One Small Function with Its Derivative

^{1}Department of Mathematics, Shandong University, Jinan, Shandong 250100, China^{2}Department of Mathematics, Xinxiang University, Xinxiang, Henan 453002, China

Received 26 October 2009; Revised 2 February 2010; Accepted 3 February 2010

Academic Editor: Stanisława R. Kanas

Copyright © 2010 Ang Chen et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We deal with the problem of uniqueness of a meromorphic function sharing one small function with its k's derivative and obtain some results.

#### 1. Introduction and Main Results

In this article, a meromorphic function means meromorphic in the open complex plane. We assume that the reader is familiar with the Nevanlinna theory of meromorphic functions and the standard notations such as and so on.

Let and be two nonconstant meromorphic functions; a meromorphic function is called a small functions with respect to provided that Note that the set of all small function of is a field. Let be a small function with respect to and We say that and share CM(IM) provided that and have same zeros counting multiplicities (ignoring multiplicities).

Moreover, we use the following notations.

Let be a positive integer. We denote by the counting function for the zeros of with multiplicity and by the corresponding one for which the multiplicity is not counted. Let be the counting function for the zeros of with multiplicity , and let be the corresponding one for which the multiplicity is not counted. Set And we define

Obviously, For more details, reader can see [1, 2].

Brück (see [3]) considered the uniqueness problems of an entire function sharing one value with its derivative and proved the following result.

Theorem 1 A. *Let be nonconstant entire function. If and share the value CM and if , then for some constant .*

Yang [4], Zhang [5], and Yu [6] extended Theorem A and obtained many excellent results.

Theorem 1 B (see[5]). *Let be a nonconstant meromorphic function and, let be a positive integer. Suppose that and share CM and
**
for where is a set of infinite linear measure and satisfies then for some constant *

Theorem 1 C (see[6]). * Let be a nonconstant, nonentire meromorphic function and be a small function with respect to If *(1)* and have no common poles,*(2)* and share the value CM,*(3)* then where is a positive integer.*

In the same paper, Yu [6] posed four open questions. Lahiri and Sarkar [7] and Zhang [8] studied the problem of a meromorphic or an entire function sharing one small function with its derivative with weighted shared method and obtained the following result, which answered the open questions posed by Yu [6].

Theorem 1 D (see[8]). *Let be a non-constant meromorphic function and, let be a positive integer. Also let be a meromorphic function such that Suppose that and share IM and
**
for and is a set of infinite linear measure. Then for some constant .*

In this article, we will pay our attention to the value sharing of and that share a small function and obtain the following results, which are the improvements and complements of the above theorems.

Theorem 1.1. * Let 11 be integers and let be a non-constant meromorphic function. Also let be a small function with respect to . If and share IM and
**
or and share CM and
**
for , , and is a set of infinite linear measure, then for some constant *

Theorem 1.2. * Let 11 be integers and be a non-constant meromorphic function. Also let be a small function with respect to . If and share IM and
**
or and share CM and
**
then .*

Clearly, Theorem 1.1 improves and extends Theorems B and D, while 1.2 improves and extends Theorem C.

#### 2. Some Lemmas

In this section, first of all, we give some definitions which will be used in the whole paper.

*Definition 2.1. *Let and be two meromorphic functions defined in assume, that and share IM; let be a zero of with multiplicity and a zero of with multiplicity We denote bythe counting function of the zeros of where and bythe counting function of zeros of where We denotes by the counting function of the zeros of where each point is counted according to its multiplicity, and denote its reduced form. In the same way, we can define*, * and so on.

*Definition 2.2. *In this paper denotes the counting function of the zeros of which are not the zeros of and and denotes its reduced form. In the same way, we can define and

Next we present some lemmas which will be needed in the sequel. Let be two non-constant meromorphic functions defined in We shall denote by the following function:

Lemma 2.3 (see[2]). *Let be two nonconstant meromorphic functions defined in If and are sharing IM, then
**
If and are sharing CM, then
*

Lemma 2.4 (see[1]). * Let be a meromorphic function and is a finite complex number. Then *(i)(ii)* for *(iii)*where are two meromorphic functions such that *

Lemma 2.5 (see[7]). * Let be a non-constant meromorphic function, and are two positive integers. Then
*

Lemma 2.6 (see[9]). * Let be a non-constant meromorphic function and let be a positive integer. where are meromorphic functions such that and Then
*

#### 3. Proof of Theorem 1.1

Let then

From the definitions of and recalling that and share value IM(CM), we get

We will distinguish two cases below.

*Case 1 (). *From (2.1) it is easy to see that *Subcase 1.1. *Suppose that and share IM. According to (3.1), and share IM except the zeros and poles of By (3.1), we have

Let be a simple zero of and but Through a simple calculation we know that is a zero of so

From (3.4)–(3.6) and Lemma 2.3, we have
It follows by the second fundamental theorem, (3.5), and (3.7) that
By Lemma 2.5, we have
which contradicts (1.4).*Subcase 1.2. *Suppose that and share CM.

Let be a simple zero of and , but By a simple calculation, we can still get Therefore

Noting that by (3.4) and Lemma 2.3, we can deduce

By the second fundamental theorem, (3.5), and (3.11), we have

Taking into account (3.1), we have

This contradicts (1.5).

*Case 2 (). *Integration yields
where are constants and It is easy to see that and share CM. Now we claim that

If then by (3.14) we get So our claim holds. Hence we can assume that
If then we can rewrite (3.14) as
So
If then by Lemma 2.4 and (3.17) we have
Hence
that is,
This is a contradiction with (1.4) and (1.5). If then from (3.14) we get We rewrite it as
So by Lemmas 2.4 and 2.6 and (3.15), we have
This implies that , since . This is impossible. Hence our claim is right. So . Theorem 1.1 is, thus, completely proved.

#### 4. Proof of Theorem 1.2

The proof is similar to the proof of Theorem 1.1. Let and be defined as in Theorem 1.1; hence, we have (3.1)–(3.5). We still distinguish two cases.

*Case 1. **Subcase 1.1. *Suppose that and share IM, then we can still get (3.6) and (3.7). Then by the second fundamental theorem, Lemma 2.3, and (3.5) we have
Applying Lemma 2.5 to the above inequality and noticing the definition of we get
This implies that
This contradicts (1.6).*Subcase 1.2. *Suppose that and share CM. Similarly as above, we can easily obtain by Lemma 2.3, we can deduce
So by the second fundamental theorem, (4.4), and using Lemma 2.5 again, we have
This implies that
This contradicts (1.7).

*Case 2 (). *Similarly, we can also get (3.14). Next we claim that . If then it follows that from (3.14). Hence, we may assume that (3.15) holds. If then
and so

Again by second fundamental theorem and (4.4) we have
that is,
Then we have and it follows that and from (3.15) we have then with (1.6) and (1.7) we may deduce It is impossible, and we can assume that thus, we can get
It shows that

If , by (4.11), then we have which with the above equality may lead to which is impossible. If then by second fundamental theorem, Lemma 2.5, (3.15), and (4.11) we have
which with (3.15) may deduce so which with and (1.6) may deduce which is impossible. Hence our claim holds.

Next we will prove that From (3.17) we have Then
If then we have
By Lemma 2.5, we get
It implies that
Combining (4.16) with (1.6) yields
that is, This is a contradiction.

Combining (4.16) with (1.7) yields
that is, , which is also a contradiction. Hence and Now Theorem 1.2 has been completely proved.

#### Acknowledgment

The authors would like to express their sincere thanks to the referee for helpful comments and suggestions.

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