`International Journal of AnalysisVolume 2013 (2013), Article ID 538027, 12 pageshttp://dx.doi.org/10.1155/2013/538027`
Research Article

## Uniqueness of Meromorphic Functions Sharing Fixed Point

Department of Mathematics, Karnatak University, Dharwad 580003, India

Received 28 September 2012; Revised 22 January 2013; Accepted 30 January 2013

Copyright © 2013 Subhas S. Bhoosnurmath and Veena L. Pujari. 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 study the uniqueness of meromorphic functions concerning differential polynomials sharing fixed point and obtain some significant results , which improve the results due to Lin and Yi (2004).

#### 1. Introduction and Main Results

Let be a nonconstant meromorphic function in the whole complex plane . We will use the following standard notations of value distribution theory: , (see [1, 2]). We denote by any function satisfying possibly outside of a set with finite measure.

Let be a finite complex number and a positive integer. We denote by the counting function for the zeros of in with multiplicity and by the corresponding one for which multiplicity is not counted. Let be the counting function for the zeros of in with multiplicity and the corresponding one for which multiplicity is not counted. Set

Let be a nonconstant meromorphic function. We denote by the counting function for -points of both and about which has larger multiplicity than , where multiplicity is not counted. Similarly, we have notation .

We say that and share CM (counting multiplicity) if and have same zeros with the same multiplicities. Similarly, we say that and share IM (ignoring multiplicity) if and have same zeros with ignoring multiplicities.

In 2004, Lin and Yi [3] obtained the following results.

Theorem A. Let and be two transcendental meromorphic functions, an integer. If and share CM, then either or where is a nonconstant meromorphic function.

Theorem B. Let and be two transcendental meromorphic functions, an integer. If and share CM, then .

In this paper, we study the uniqueness problems of entire or meromorphic functions concerning differential polynomials sharing fixed point, which improves Theorems A and B.

##### 1.1. Main Results

Theorem 1. Let and be two nonconstant meromorphic functions, a positive integer. If and share CM, and share IM, then either or where is a nonconstant meromorphic function.

Theorem 2. Let and be two nonconstant meromorphic functions, a positive integer. If and share CM, and share IM, then .

Theorem 3. Let and be two nonconstant entire functions, an integer. If and share CM, then .

#### 2. Some Lemmas

Lemma 4 (see [4]). Let , , and be nonconstant meromorphic functions such that . If , , and are linearly independent, then where and .

Lemma 5 (see [1]). Let and be two nonconstant meromorphic functions. If , where , , and are non-zero constants, then Lemmas 4 and 5 play a very important role in proving our theorems.

Lemma 6 (see [1]). Let be a nonconstant meromorphic function and let be a nonnegative integer, then The following lemmas play a cardinal role in proving our results.

Lemma 7. Let and be nonconstant meromorphic functions. If and share CM and , then

Proof. Applying Nevanlinna’s second fundamental theorem to , we have By the first fundamental theorem and (9), we have We know that Therefore using Lemma 6, (10) becomes since , we have This completes the proof of Lemma 7.

Lemma 8. Let and be nonconstant entire functions. If and share CM and , then

Proof. Applying Nevanlinna’s second fundamental theorem to , we have Since is an entire function, we have and the above equation becomes By the first fundamental theorem and (17), we have We know that Therefore using Lemma 6, (18) becomes or since , we have This completes the proof of Lemma 8.

Lemma 9 (see [5]). Suppose that is a meromorphic function in the complex plane and , where , are small meromorphic functions of . Then

Lemma 10 (see [6]). Let , and be three meromorphic functions satisfying , let , , and . If , and are linearly independent, then , and are linearly independent.

#### 3. Proof of Theorems

Proof of Theorem 1. By assumption, and share CM, and and share IM. Let Then, is a meromorphic function satisfying Therefore, From (24), we easily see that the zeros and poles of are multiple and satisfy Let Then, and denote the maximum of , .
We have Therefore, , and thus Now, we discuss the following three cases.
Case  1. Suppose that neither nor is a constant.
If , , and are linearly independent, then by Lemma 4 and (28), we have Using (27), we note that Since , we obtain that But , so we get Using (33) and (35) in (31), we get Since and share IM, we have .
Using this with (27), we get If is a zero of with multiplicity , then is a zero of with multiplicity ; we have Similarly, Let By Lemma 9, we have .
Since , we have By the first fundamental theorem, we have we have From (37)–(43), we get Using Lemma 6, we get Let Then . By Lemma 10, , and are linearly independent. In the same manner as above, we get Note that Adding (45) and (47) gives Using (48), we get or Combining (50) and (51), we get By and (30), we get a contradiction. Thus , , and are linearly dependent. Then, there exists three constants such that If , from (53) , , and On integrating, we get since , we get a contradiction.
Thus , and by (53) we have Substituting this in , we get that is, From (28), we obtain Applying Lemma 5 to the above equation, we get Note that Using (61), we get By Lemmas 9 and 6 and (63), we have we obtain , which contradicts .
Case  2. Suppose that , where is a constant.
If , then we have Applying Lemma 5 to the above equation, we have Note that Therefore, Using Lemmas 9 and 6 and (69), we have Using Lemma 7, we get since , we get a contradiction.
Therefore , and by (27) and (24) we have On integrating, we get We claim that . Suppose that , then We have Similarly, Using Lemma 9, we have Thus, Similarly, Therefore, (74) becomes which contradicts . Thus, we have Let . If , then we easily obtain that If , that is, .
Case  3. Suppose that , where is a constant.
If , then we have Applying Lemma 5 to the above equation, we have Note that Therefore using (84), we have Using Lemmas 9 and 6 and (86), we have Using Lemma 7, we get since , we get a contradiction. Thus, . Hence, Let be a zero of of order . From (89), we know that is a pole of . Suppose that is a pole of of order . From (89), we obtain which implies that and . Hence, Let be a zero of of order , then from (89) is a pole of (say order ). By (89), we get Let be a zero of of order that is not zero of , then from (89), is a pole of of order . Again by (89), we get In the same manner as above, we have similar results for the zeros of . From (89)–(93), we have that is, By Nevanlinna’s second fundamental theorem, we have from (91), (92), and (95) that Similarly, From (96) and (97), we get since , we get a contradiction.
This completes the proof of Theorem 1.

Using the same argument as in the proof of Theorem 1, we can prove Theorem 2.

Proof of Theorem 3. By the assumption of the theorem, we know that either both and are two transcendental entire functions or both and are polynomials.
If and are transcendental entire functions, putting , and using similar arguments as in the proof of Theorem 1, we easily obtain Theorem 3.
If and are polynomials, and share CM, we get where is a nonzero constant. Suppose that , (99) can be written as Applying Lemma 5 to the above equation, we have Since is a polynomial, so it does not have any poles. Thus, we have, Note that Therefore, Using Lemmas 9 and 6 and (104), we have Using Lemma 8, we get since , we get a contradiction.
Therefore, ; so (99) becomes On integrating, we get We claim that . Suppose that , then We have