Abstract and Applied Analysis

Volume 2014, Article ID 256032, 7 pages

http://dx.doi.org/10.1155/2014/256032

## Further Results on Uniqueness of Meromorphic Functions concerning Fixed Points

College of Science, Civil Aviation University of China, Tianjin 300300, China

Received 30 April 2014; Accepted 29 July 2014; Published 16 October 2014

Academic Editor: Adrian Petrusel

Copyright © 2014 Xiao-Bin Zhang. 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 will study the uniqueness problems of meromorphic functions of differential polynomials sharing fixed points. Our results improve or generalize some previous results on meromorphic functions sharing fixed points.

#### 1. Introduction and Main Results

Let denote the complex plane and let be a nonconstant meromorphic function on . We assume the reader is familiar with the standard notion used in the Nevanlinna value distribution theory such as , , and (see, e.g., [1–3]), and denotes any quantity that satisfies the condition as outside of a possible exceptional set of finite linear measure.

Let and be two nonconstant meromorphic functions. Let ; we say that , share CM (counting multiplicities) if , have the same zeros with the same multiplicities and we say that , share IM (ignoring multiplicities) if we do not consider the multiplicities. We denote by the counting function of those -points of whose multiplicities are greater (less) than the multiplicities of the corresponding -points of , where each -point is counted only once. denotes the truncated counting function bounded by .

We say that a finite value is called a fixed point of if or is a zero of .

The following theorem in the value distribution theory is well-known [4, 5].

Theorem A. *Let be a transcendental meromorphic function and a positive integer. Then has infinitely many solutions.*

Related to Theorem A, Fang [6] proved that a meromorphic function has infinitely many fixed points when is transcendental and is a positive integer. Then Fang and Qiu [7] obtained the following uniqueness theorem.

Theorem B. *Let and be two nonconstant meromorphic (entire) functions and a positive integer. If and share CM, then either , , where , , and are three constants satisfying , or for a constant such that .*

For more related results, see [8, 9]. Recently, Cao and Zhang [10] replaced with and obtained the following.

Theorem C. *Let and be two transcendental meromorphic functions, whose zeros are of multiplicities at least , where is a positive integer; let be a positive integer. If and share z CM, and and share IM, then one of the following two conclusions holds:*(1)*;*(2)*, , where , , and are constants such that .*

*Regarding Theorem C, it is natural to ask the following.*

*Problem 1. *Does Theorem C still hold without the “transcendental” condition?

*Problem 2. *Does Theorem C still hold without the “multiplicity of zeros of and ” condition?

*Problem 3. *Can the lower bound of be reduced in Theorem C?

*We consider Problems 1–3 and give affirmative answers to them, and we get the following.*

*Theorem 1. Let and be two nonconstant meromorphic functions and , two positive integers with . If and share CM, and and share IM, then one of the following two conclusions holds:(1);(2), , where , , and are constants such that .*

*One may ask whether the condition “” can be further reduced. We have proved the following.*

*Theorem 2. Let and be two nonconstant meromorphic functions and , two positive integers with . If and share CM, and and share IM, then one of the following two conclusions holds:(1), possibly except for at most one exceptional case, namely,
where , , , are 8 distinct constants and , are two nonzero constants;(2), , where , , and are constants such that .*

*If and are two transcendental meromorphic functions, the lower bound of in Theorem 2 can be further reduced. We have the following.*

*Theorem 3. Let and be two transcendental meromorphic functions and , two positive integers with . If and share CM, and and share IM, then one of the following two conclusions holds:(1);(2), , where , , and are constants such that .*

*2. Preliminary Lemmas*

*2. Preliminary Lemmas**Let
where and are meromorphic functions.*

*Lemma 4 (see [11]). Let be a nonconstant meromorphic function and let be small functions with respect to . Then
*

*Lemma 5 (see [2]). Let be a nonconstant meromorphic function, and let be a positive integer. Suppose that ; then
*

*By using the similar method to Yang and Hua [12, Lemma 3], we can prove the following lemma.*

*Lemma 6. Let , , and be defined as in (2). If and share 1 CM and IM, and , then , and
the same inequality holding for .*

*Lemma 7 (see [13]). Let , , and be defined as in (3). If and share IM, and , then .*

*Lemma 8. Let , be two nonconstant meromorphic functions, defined as in (3), where , , and , , and three integers. If , and share 1 CM, and and share IM, then
*

*Proof. *Note that , and share IM, suppose that is a pole of with multiplicity , then is a pole of with multiplicity . Thus is a zero of with multiplicity , and a zero of with multiplicity . Hence is a zero of with multiplicity at least . Suppose that is a pole of with multiplicity , by the similar discussion as above, we get that is a zero of with multiplicity at least . So we have
By the logarithmic derivative lemma, we have . Note that and share 1 CM, so we have
Obviously,
From (8)–(10) we get (7). This proves Lemma 8.

*Lemma 9 (see [1, Theorem 3.10]). Suppose that is a nonconstant meromorphic function; is an integer. If
then , where , are constants.*

*Lemma 10. Let , be two nonconstant meromorphic functions and , two positive integers. If , and and share IM, then , , where , , and are constants such that .*

*Proof. *Since and share IM, from
we get that both and are entire functions.

The case has been proved by Fang and Qiu [7, Propostion 2]; here we only need to consider the case .

Let , . Then we have
We obtain from (13) that
Note that
We obtain from (15) that
Thus from (14) and (16) we have . Similarly we have . It follows from (12) that ; we get .

Suppose that has a zero , say multiplicity ; then is a zero of with multiplicity . In view of (12), we get and . Moreover, has no zero. Therefore,
where , are nonconstant entire functions. We deduce that either both and are transcendental functions or both and are polynomials. From (17) we have
Moreover, we have
Thus we get
If , suppose that is a transcendental entire function. We deduce from Lemma 9 and (17) and (18) that is a polynomial, which is a contradiction. Thus is a polynomial and so is .

So from (12) we get
where , , and are differential polynomials in and of degree at most , respectively.

Since , , by (21) we immediately get a contradiction.

Thus has no zero; similarly, we get that has no zero. So we have
where , are nonconstant entire functions.

With similar discussion as above, we get that , where is a constant and and are both polynomials.

We deduce from (22) that
where and are differential polynomials in and of degree at most , respectively. Thus from (12) we obtain

If , since is not a constant, , by (24) we immediately get a contradiction.

This proves Lemma 10.

*3. Proof of Theorems 1–3*

*3. Proof of Theorems 1–3*

*Since the proof of Theorems 1 and 3 is quite similar to the proof of Theorem 2, here we only need to prove Theorem 2.*

*Let , , , and . Then and share 1 CM and IM.*

*Suppose that ; then , and .*

*By Lemma 6 we have
So
Obviously,
We have
It follows from (26)–(28) that
Similarly we have
Combining (29) and (30) gives
*

*Case **1.* Either or is transcendental; from (32) we get a contradiction since .

*Thus . Similar to the proof of [12, Lemma 3], we obtain(i), or(ii).*

*By Lemma 10, we get conclusion (2) from .*

*Case **2.* Both and are rational functions.

*If is a polynomial, so is . We get from (31) that
which implies , , and .*

*Set
where , , , are constants with . By our assumption, we have
where is a nonzero constant. By computation we have
which implies and ; thus for a constant . By the second fundamental theorem we get and for a constant such that .*

*If and are nonpolynomial rational functions, set
where , , , and are polynomials. Now we discuss three cases as follows.*

*Case **2.1.* Consider .

*Case **2.1.1.* If , from (31) we obtain
which implies that both and have only simple poles; thus
From (31) we get
which is a contradiction since , , , and .

*Case **2.1.2.* If , from (31) we obtain
a contradiction.

*Case **2.1.3*. Consider , and , . It follows from (31) that
If , then from (42) we get that both and have only simple poles; thus
From (31) we get
which implies that , , , and . Set
where , , , , , and are constants with . Therefore, we have
where , are nonzero constants. So has zeros and has zeros, which is a contradiction.

*If , then it follows from (42) that either or has only a pole of order at most 2, and both and can not have a pole of order 2. If has a pole of order 2, then has only simple poles. Thus from (42) we get
Obviously, . If , then . Set
where , , , , , , and are constants with . Therefore, we have
where is a polynomial with ; is a nonzero constant. So has zeros and has zeros, which is a contradiction.*

*If , then . So we have , a contradiction.*

*If has a pole of order 2, then has only simple poles. With similar discussion as above, we get a contradiction.*

*Therefore, both and have only simple poles and we also get (47). Thus . If , then ; we have , which is a contradiction. If and , we also get a contradiction. If and , then we get (45), which leads to a contradiction.*

*Case 2.1 has been ruled out.*

*Case **2.2.* Consider .

*With similar discussion as in Case 2.1, it is easy to get . Thus . Then from (31) we get
which implies that both and only have simple poles. Moreover, we have , and . Again from (31) we obtain
which is a contradiction.*

*Case 2.2 has been ruled out.*

*Case **2.3.* Thus we have . Similarly, we have .

*It follows from (31) that
Now we prove that and share CM. We discuss two cases below.*

*Case **2.3.1.* If , (52) implies that neither nor has a pole of order greater than 2; both and can not have poles of order 2. Thus only one of and may have a pole of order 2.

*Suppose that
We deduce from (53) that
where , are polynomials with , . We get that has zeros while has zeros, which is a contradiction because and share CM. Thus has only simple poles. Similarly, has only simple poles; thus and share CM.*

*Case **2.3.2.* If , (31) implies that
If and do not share CM, and if only has simple poles, (55) leads to
If has poles of order , then from (56) we get , , and . Set
With the similar discussion in Case we get a contradiction.

*If has poles of order 2, then from (56) we get .*

*If , then and . Set
With the similar discussion in Case we get a contradiction.*

*If , we get (53); with a similar discussion in Case we get a contradiction.*

*So must have multiple poles. Similarly, must have multiple poles.*

*From (52) we get that both and have one and only one multiple pole and their order is 2. Since the multiple poles are distinct, then from (55) we get
provided that , where , , , are 8 distinct constants and , are two nonzero constants. By our assumption, this case has been ruled out. So and share CM.*

*We have
where , , and are polynomials with , . Since and share CM, we have
where is a nonzero constant; then we get
which implies and . Thus .*

*This completes the proof of Theorem 2.*

*4. Discussion*

*4. Discussion*

*Remark 11. *The author can not assert whether (1) really exists because the calculation is rather complicated. The possibility of the existance of (1) is small because the 10 constants must satisfy at least 34 equations. Unfortunately, the author can not prove it. If there exists exceptional case (1), it has its own meaning. It will show that the condition “” of Theorem B can not be reduced to .

*Remark 12. *One can not get for a constant from . For example, let , , where and are polynomials with . Then for a constant but we still have .

*Problem 4. *Can guarantee for a constant when and are transcendental meromorphic functions?

*Conflict of Interests*

*Conflict of Interests*

*The author declares that there is no conflict of interests regarding the publication of this paper.*

*Acknowledgments*

*Acknowledgments*

*The author would like to thank the referee for valuable suggestions toward this paper. This work is supported by the NSFC (no. 11171184) and the Fundamental Research Funds for the Central Universities (Grant no. 3122013k008).*

*References*

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