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Generalization of Uniqueness of Meromorphic Functions Sharing Fixed Point ()

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*Applied Mathematics*,

**7**, 939-952. doi: 10.4236/am.2016.79084.

Received 7 January 2016; accepted 24 May 2016; published 27 May 2016

1. Introduction and Main Results

Let be a non constant meromorphic function in the whole complex plane. We will use the following standard notations of value distribution theory: (see [2] [3] ). We de-

note by any function satisfying as possibly outside of a set with

finite linear measure.

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

counted. Set

Let be a non constant meromorphic function. We denote by the counting function for

a-points of both and about which has larger multiplicity than, where multiplicity

is not counted. Similarly, we have notation.

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

In 2004, Lin and Yi [4] obtained the following results.

Theorem A. Let f and g be two transcendental meromorphic functions, an integer. If and share z CM, then either or

where h is a non constant meromorphic function.

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

In 2013, Subhas S. Bhoosnurmath and Veena L. Pujari [1] extended the above theorems A and B with respect to differential polynomials sharing fixed points. They proved the following results.

Theorem C. Let f and g be two non constant meromorphic functions, a positive integer. If and share z CM, f and g share ¥ IM, then either or

where h is a non constant meromorphic function.

Theorem D. Let f and g be two non constant meromorphic functions, a positive integer. If and share z CM, f and g share ¥ IM, then.

Theorem E. Let f and g be two non constant entire functions, an integer. If and share z CM, then.

In this paper, we generalize theorems C, D, E and obtain the following results.

Theorem 1. Let f and g be two non constant meromorphic functions, an integer. If and share z CM, f and g share ¥ IM, then .

For, we get Theorem C.

For, we get Theorem D.

Theorem 2. Let f and g be two non constant entire functions, an integer. If and share z CM, then.

2. Some Lemmas

Lemma 2.1 (see [5] ). Let and be non constant meromorphic functions such that. If and are linearly independent, then

where and

Lemma 2.2 (see [2] ). Let and be two non constant meromorphic functions. If, where and are non-zero constants, then

Lemma 2.3 (see [2] ). Let f be a non constant meromorphic function and let k be a non-negative integer, then

Lemma 2.4 (see [6] ). Suppose that is a meromorphic function in the complex plane and

, where are small meromorphic functions of. Then

Lemma 2.5 (see [7] ). Let and be three meromorphic functions satisfying,

let and. If and are linearly independent then and are linearly independent.

Lemma 2.6 (see [8] ). Let, then

where which are distinct respectively.

The following lemmas play a cardinal role in proving our results.

Lemma 2.7 Let f and g be two non constant meromorphic functions. If and share z CM and, then

Proof. Applying Nevanlinna’s second fundamental theorem (see [3] ) to, we have

(1)

By first fundamental theorem (see [3] ) and (1), we have

(2)

We know that,

(3)

Therefore, using Lemma 2.3, (2) becomes

Using we get

(4)

since, we have

This completes the proof of Lemma 2.7.

Lemma 2.8 Let f and g be two non constant entire functions. If and share z CM and, then

(5)

Proof. Since f and g are entire functions, we have. Proceeding as in the proof of Lemma 2.7, we can easily prove Lemma 2.8.

3. Proof of Theorems

Proof of Theorem 1. By assumption, and share z CM, f and g share ¥ IM. Let

(6)

Then, H is a meromorphic function satisfying

By (3), we get

Therefore,

(7)

From (6), we easily see that the zeros and poles of H are multiple and satisy

(8)

Let

(9)

Then, and denote the maximum of

We have, (10)

(11)

Therefore,

and thus

(12)

Now, we discuss the following three cases.

Case 1. Suppose that neither nor is a constant. If and are linearly independent, then by Lemma 2.1 and 2.4, we have

(13)

Using (8), we note that

since, , We obtain that,

(14)

But, so we get

(15)

Using (14) and (15) in (13), we get

Since f and g share ¥ IM, we have

Using this with (8), we get

(16)

If is a zero of f with multiplicity p, then is a zero of with multiplicity , we have

(17)

Similarly,

(18)

Let

(19)

By Lemma 2.6, we have

Since, we have

By the first fundamental theorem, we have

(20)

we have

(21)

where are distinct roots of algebraic equation,

From (16)-(21), we get

Using Lemma 2.3, we get

(22)

Let

Then. By Lemma 2.5, and are linearly independent. In the same manner as above, we get expression for.

Note that. We have,

Simplifying, we get

(23)

or

(24)

Combining (23) and (24), we get

(25)

By and (12), we get a contradiction. Thus and are linearly dependent. Then, there exists three constants such that

(26)

If from (26), and

On integrating, we get

(27)

Since, we get a contradiction. Thus, and by (26), we have

Substituting this in, we get

that is,

From (9), we obtain

(28)

Applying Lemma 2.2, to the above equation, we get

(29)

Note that,

Using (29), we get

(30)

By, Lemmas 2.3, 2.4 and (30), we have

We obtain, which contradicts.

Case 2. Suppose that where c is constant If then, we have

(31)

Applying Lemma 2.2 to the above equation, we have

(32)

By Lemmas 2.3, 2.4 and (32), we have

Using Lemma 2.7, we get

(33)

Since, we get contradiction

Therefore, and by (6), (8), we have

(34)

On integrating, we get

(35)

We claim that. Suppose that, then

(36)

We have,

similarly,

Using Lemma 2.4, we have

(37)

Thus,

(38)

similarly,

Therefore, (36) becomes,

which contradicts. Thus we have

(39)

Let substituting in the above equation, we can easily get

(40)

If h is not a constant, then with simple calculations we get contradiction (refer [9] ). Therefore h is a constant. We have from (40) that, which imply. Hence.

Case 3. Suppose that where c is a constant. If, then

(41)

Applying Lemma 2.2 to above equation, we have

(42)

Using Lemmas 2.4, 2.3 and (42), we have

Using Lemma 2.7, we get

(43)

Since, we get contradiction.

Therefore

Hence,

(44)

Let be a zero of f of order p. From (44) we know that is a pole of g. Suppose is a pole of g of order q, from (44), we obtain

Hence,

(45)

Let be a zero of of order. From (44) we know that is a pole of g. (say order). From (44), we obtain

(46)

Let be a zero of of order, that is not zero of, then from (44), is a pole of g of order. From (44), we have

(47)

In the same manner as above, we have similar results for zeros of. From (44)-(47), we have

(48)

(49)

By Nevanlinna’s second fundamental theorem, we have from (45), (46) and (49) that,

(50)

Similarly,

(51)

From (50) and (51), we get

since, we get a contradiction.

This completes the proof of Theorem 1.

Proof of Theorem 2. By the assumption of the theorems, we know that either both f and g are two transcendental entire functions or both f and g are polynomials. If f and g are transcendental entire functions, using and similar arguments as in the proof of Theorem 1, we can easily obtain Theorem 2. If f and g are polynomials, and share z CM, we get

(52)

where k is a non-zero constant. Suppose that, (52) can be written as,

(53)

Apply Lemma 2.2 to above equation, we have

Since f is a polynomial, it does not have any poles. Thus, we have

Therefore,

(54)

Using Lemmas 2.4, 2.3 and (54), we have

Using Lemma 2.8, we get

(55)

since, we get a contradiction. Therefore,. So, (52) becomes

(56)

On Integrating, we get

(57)

We claim that. Suppose that, then

(58)

Proceeding as in Theorem 1,

we get.

Conflicts of Interest

The authors declare no conflicts of interest.

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