Generalization of Uniqueness of Meromorphic Functions Sharing Fixed Point ()
Received 7 January 2016; accepted 24 May 2016; published 27 May 2016
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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
.