RESEARC H Open Access
Uniqueness of meromorphic functions
concerning differential polynomials share one
value
Chun Wu
1,2*
, Chunlai Mu
1
and Jiangtao Li
1
* Correspondence: xcw919@gmail.
com
1
College of Mathematics and
Statistics, Chongqing University,
Chongqing, 401331, People’s
Republic of China
Full list of author information is
available at the end of the article
Abstract
In this paper, we study the uniqueness of meromorphic functions whose differential
polynomial share a non-zero finite value. The results in this paper improve some
results given by Fang (Math. Appl. 44, 828-831, 2002), Banerjee (Int. J. Pure Appl.
Math. 48, 41-56, 2008) and Lahiri-Sahoo (Arch. Math. (Brno) 44, 201-210, 2008).
2010 Mathematics Subject Classification: 30D35
Keywords: Uniqueness, Meromorphic functions, Differential polynomials
1 Introduction and main results
In this paper, by meromorphic functions, we will always mean meromorphic functions
in the complex plane. We adopt the standard notations in the Nevanlinna theory of
meromorphic functions as explained in [1-3]. It will be convenient to let E denote any
set of positive real numbers of finite linear measure, not necessarily the same at each

the following three definitions:
Definition 1.1 Let f be a non-constant meromorphic function, and let p be a positive
integer and a Î C ∪ {∞}. Then by N
p)
(r,1/(f - a)), we denote the counting function of
those a-points of f (counted with proper multiplicities) whose multiplicities are not
Wu et al. Journal of Inequalities and Applications 2011, 2011:133
http://www.journalofinequalitiesandapplications.com/content/2011/1/133
© 2011 Wu et al; licensee Springer. This is an Open Access article distributed under the terms of the Creati ve Commons Attribution
License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distributi on, and reproduction in any medium,
provided the original work is properly ci ted.
greater than p,by
¯
N
p
)
(r,1/(f − a)
)
we denote the corresponding reduced counting
function (ignoring multiplicities). By N
(p
(r,1/(f - a)),wedenotethecountingfunction
of those a-poi nts of f (counted with proper multiplicities) whose multiplicities are not
less than p, by
¯
N
(
p
(r,1/(f − a)
)

(
p
(r,f
)
,
and
¯
N
(
p
(r, f
)
, respectively, if a = ∞.
Definition 1.2 Let f be a non-constant meromorphic function, and let a be any value
in the extended complex plane, and let k be an arbitrary nonnegative integer. We
define
δ
k
(a, f )=1− lim
n→∞
N
k

r,
1
f − a

T
(
r, f

N
(k

r,
1
f − a

.
(2)
Remark 1.1. From (1) and (2), we have 0 ≤ δ
k
(a, f) ≤ δ
k-1
(a, f) ≤ δ
1
(a, f) ≤ Θ(a, f) ≤ 1.
Definition 1.3 Let f be a non-constant meromorphic function, and let a be any value
in the extended complex plane, and let k be an arbitrary nonnegative integer.
We define

k)
(a, f )=1− lim
n→∞
¯
N
k)

r,
1
f − a

N
g>k

r,
1
f − 1

is defined analogously.
It is natural to ask the following question:
Question 1.1 What can be said about the relationship between two meromorphic
functions f,g when two differential polynomials, generated by f and g, respectively,
share certain values?
Regarding Question 1.1, we first recall the following result by Yang and Hua [4]:
Theorem A.Letf(z)andg(z) be two non-constant meromorphic functions, n ≥ 11
an integer and a Î C - {0}. If f
n
f’ and g
n
g’ share the value a CM, then either f = tg for
aconstantt with t
n+1
=1org(z)=c
1
e
cz
and f(z)=c
2
e
-cz
,wherec, c

for a constant t with t
n
=1orf(z)=c
1
e
cz
and g(z)=c
2
e
-cz
,wherec, c
1
andc
2
are con-
stants satisfying ( -1)
k
(c
1
c
2
)
n
(nc)
2k
=1.
Theorem C.Letf(z)andg(z) be two non-constant entire functions, and let n, k be
two positive integers with n ≥ 2k +8.If[f
n
(z)(f(z) - 1)]

1
and c
2
are constants satisfying ( -1)
k
(c
1
c
2
)
n
(nc)
2k
= b
2
.
Recently, Lahiri and Sahoo [7] proved the following theorem.
Theorem E. Let f and g be two non-constant meromorphic functions, and
α(
≡ 0, ∞
)
be a small function of f and g. Let n and m(≥ 2) be two positive integers with n > max
{4, 4m +22-5Θ(∞, f)-5Θ(∞, g) -min[Θ(∞, f), Θ( ∞, g)]}. If f
n
(f
m
- a)f’ and g
n
(g
m

c
1
andc
2
are constants satisfying (-1)
k
( c
1
c
2
)
n
( nc )
2k
=1orf = tg for a constant t
with t
n
=1.
(ii) When m =1,n >9k + 18 and
(∞, f ) >
2
n
, then f ≡ g.
(iii) When m ≥ 2, n >4m +9k + 14, then f ≡ g or f and g satisfies the algebraic
equation R(x, y)=x
n
(x-1)
m
- y
n

(g -1)
m
]
(k)
share the value
1 IM. Then, one of the following holds:
Wu et al. Journal of Inequalities and Applications 2011, 2011:133
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Page 3 of 13
(i) When m =0andn >5k + 7, then either f(z)=c
1
e
cz
and g(z)=c
2
e
-cz
,wherec,
c
1
andc
2
are constants satisfying ( -1)
k
( c
1
c
2
)
n

f ≡ g or f ≡ -g.
The possibility f ≡ -g does not arise if n and m are both odd or if n is even and m is
odd or if n is odd and m is even.
Remark 1.4.Ifm = 0, then the cases becomes Theorem 1.3 (i).
2 Some lemmas
Lemma 2.1. (See [2,3].) Let f(z) be a non-constant meromorphic function, k apositive
integer and let c be a non-zero finite complex number. Then,
T(r, f ) ≤
¯
N( r, f )+N

r,
1
f

+ N

r,
1
f
(k)
− c

− N

r,
1
f
(k+1)


(4)
where
N
0

r,
1
f
(k+1)

is the counting function, which only counts those points such
that f
(k+1)
= 0 but f(f
(k)
-c) ≠ 0
Lemma 2.2. (See [8].) Let f(z) be a non-constant meromorphic function, and let k be
a positive integer.
Suppose that
f
(k)
≡
0
, then
N

r,
1
f
(k)

+ S(r, f )
.
Clearly,
¯
N

r,
1
f
(k)

= N
1

r,
1
f
(k)

.
Lemma 2.4. (See [10].) Let f, g share (1,0). Then
(i)
¯
N
f >1

r,
1
g − 1


≤
¯
N

r,
1
g

+
¯
N( r, g) − N
0

r,
1
g


+ S(r, g
)
.
Lemma 2.5.Letf(z)andg(z) be two non-constant meromorphic functions such that
f
(k)
and g
(k)
share 1 IM, where k be a positive integer. If
Wu et al. Journal of Inequalities and Applications 2011, 2011:133
http://www.journalofinequalitiesandapplications.com/content/2011/1/133
Page 4 of 13

)
+2δ
k+1
(
0, g
)
> 4k+1
1
then either f
(k)
g
(k)
≡ 1orf ≡ g.
Proof. Let
(z)=
f
(k+2)
f
(k+1)
− 2
f
(k+1)
f
(k)
− 1
−
g
(k+2)
g
(k+1)

Taylor series at z
0
into (5), we see that z
0
is a zero of F(z). Thus, we have
N
11

r,
1
f
(k)
− 1

= N
11

r,
1
g
(k)
− 1

≤
¯
N

r,
1


g

+ N
0

r,
1
f
(k+1)

+ N
0

r,
1
g
(k+1)

+
¯
N
L

r,
1
f
(k)
− 1

+

g

+
¯
N

r,
1
g
(k)
− 1

− N
0

r,
1
g
(k+1)

+ S(r, g)
.
(8)
Since
¯
N

r,
1
g


r,
1
f
(k)
− 1

.
(9)
Thus, we deduce from (6)-(9) that
T(r, g) ≤ 2
¯
N(r, g)+
¯
N(r, f )+N
k+1

r,
1
g

+
¯
N
(k+2

r,
1
f


N
L

r,
1
f
(k)
− 1

+
¯
N
L

r,
1
g
(k)
− 1

+
¯
N
g
(k)
>1

r,
1
f

f
(k)
− 1

+ N
(2

r,
1
f
(k)

−
¯
N
(2

r,
1
f
(k)

≤ N

r,
1
f
(k+1)

.

r,
1
f
(k)

+
¯
N
(2

r,
1
f
(k)

≤ N

r,
1
f
(k)

− N
(2

r,
1
f
(k)


Substituting (11) in (10), we get
T(r, g) ≤ 2
¯
N(r, g)+
¯
N(r, f )+N
k+1

r,
1
g

+
¯
N
(k+2

r,
1
f

+
¯
N
(k+2

r,
1
g


1
g
(k)
− 1

+
¯
N
g
(k)
>1

r,
1
f
(k)
− 1

+ S(r, f )+S(r, g)
≤ 2
¯
N(r, g)+2
¯
N(r, f )+N
k+2

r,
1
g


¯
N
L

r,
1
g
(k)
− 1

+
¯
N
g
(k)
>1

r,
1
f
(k)
− 1

+ S(r, f )+S(r, g).
(12)
According to Lemma 2.3,
¯
N

r,

1
f
(k)
− 1

≤ N

r,
1
f
(k)
− 1

−
¯
N

r,
1
f
(k)
− 1

≤ N

r,
f
(k)
f
(k+1)

+(k +1)
¯
N( r, f )+S(r, f ).
similarly,
¯
N
L

r,
1
g
(k)
− 1

≤ N
k+1

r,
1
g

+(k +1)
¯
N( r, g)+S(r, g)
.
Combining the above inequality, Lemma 2.4 and (12), we obtain
T(r, g) ≤ (2k +4)
¯
N( r, g)+(2k +3)
¯


r,
1
g
(k+1)

+ S(r, f )+S(r, g
)
≤ (2k +4)
¯
N( r, g)+(2k +3)
¯
N( r, f )+N
k+2

r,
1
g

+ N
k+2

r,
1
f

+ N
k+1

r,

(
r, g
)
+ S
(
r, g
)
.
for Î I and 0 <ε < Δ -(4k +11)
Therefore, we can get T(r, g) ≤ S(r, g),r Î I, by the condition, a contradiction.
Hence, we get F(z) ≡ 0. Then, by (5), we have
f
(k+2)
f
(k+1)
−
2f
(k+1)
f
(k)
− 1
≡
g
(k+2)
g
(k+1)
−
2g
(k+1)
g

=
(1 + b)g
(k)
− 1
b
g
(k)
.
(15)
From (15), we get
¯
N

r,
1
g
(k)
− 1/
(
1+b
)

=
¯
N

r,
1
f
(k)


+ S(r, g
)
≤
¯
N(r, g)+N
k+1

r,
1
g

+ k
¯
N(r, f )+N
k+1

r,
1
f

+ S(r, f )+S(r, g).
(17)
From (17), we get

(
∞, g
)
+ k
(

r,
1
g
(k)
−
(
a +1
)

=
¯
N( r, f ).
(19)
From (19) and Lemma 2.1 and in the same manner as in the proof of (17), we get
T(r, g) ≤
¯
N( r, g)+N
k+1

r,
1
g

+
¯
N

r,
1
g

=
−a
b
2
[g
(k)
+
a − b
b
]
.
(20)
From (20), we get
¯
N
⎡
⎢
⎢
⎣
r,
1
g
(k)
+

a − b
b

⎤
⎥

g
(k)
+1−
1
a
,
(22)
f =
1
a
g + p(z)
.
(23)
where p(z) is a polynomial with its degree ≤ k.If
p
(
z
)
≡
0
, then by second funda-
mental theorem for small functions, we have
T(r, g) ≤
¯
N( r, g)+
¯
N

r,
1

+ S(r, g).
(24)
Using the argument as in Case 1, we get a contradiction. Therefore, p(z) ≡ 0. So
from (22) and (23), we obtain a = 1 and so f ≡ g. This proves the lemma.
Lemma 2.6.Letf(z)andg(z) be two non-constant entire functions such that f
(k)
and
g
(k)
share 1 IM, where k be a positive integer. If
 = δ
k+2
(
0, g
)
+ δ
k+2
(
0,
f )
+ δ
k+1
(
0,
f )
+2δ
k+1
(
0, g
)

Lemma 2.8. (See [12].) Let f(z) be a non-constant meromorphic function. Let k be a
positive integer, and let c be a non-zero finite complex number. Then,
T
(
r, a
n
f
n
+ a
n−1
f
n−1
+ ···+ a
0
)
= nT
(
r, f
)
+ S
(
r, f
).
3 Proof of theorems
3.1 Proof of Theorem 1.1
Let F = f
n
(f -1)
m
and G = g

T
(
r, f
)
≥
n + m − 1
m + n
,
δ
k+1
(0, F)=1− lim
n→∞
N
k+1

r,
1
F

T(r, F)
=1−
lim
n→∞
N
k+1

r,
1
f
n

, δ
k+2
(0, F) ≥
n − k − 2
m
+
n
, δ
k+2
(0, G) ≥
n − k − 2
m
+
n
.
Therefore,
 =(2k +4)(∞, G)+(2k +3)(∞, F)+δ
k+2
(0, G)+δ
k+2
(0, F)+δ
k+1
(0, F)+2δ
k+1
(0, G)
≥ (2k +4)·
m + n − 1
m
+
n

≡ 1orF ≡ G.
Case 1. F
(k)
G
(k)
≡ 1, that is,
(
f
n
(
f − 1
)
m
)
(k)
(
g
n
(
g − 1
)
m
)
(k)
≡ 1
.
(25)
Case 1.1 when m = 0, that is,
(
f

Therefore, we conclude that f ≠∞oo and g ≠∞.
From (26), we get
(
f
n
)
(k)
=0 and
(
g
n
)
(k)
=0
.
(27)
From (26)-(27) and Lemma 2.7, we get that f(z)=c
1
e
cz
and g(z)=c
2
e
-cz
,wherec, c
1
and c
2
are three constants satisfying ( -1)
k

+ k,i.e.,n(p
1
-
q
1
)=mq
1
+2k, which implies that p
1
≥ q
1
+ 1 and mq
1
+2k ≥ n.Fromn >4m +9k
+ 14, we can deduce p
1
≥ 6.
Let f - 1 has a zero z
2
of order p
2
,thenz
2
is a zero of [f
n
(f -1)
m
]
(k)
of order mp

(25), p
3
-(k-1) = (n + m)q
3
+ k, i.e., p
3
≥ n + m +2k -1.
Moreover , in the same manner as above, we have similar resul ts for the zeros of [g
n
(g-1)
m
]
(k)
.
On the other hand, Suppose z
4
is a pole of f, from (25), we get z
4
is a zero of [g
n
(g -
1)
m
]
(k)
.
Thus,
¯
N( r, f ) ≤
¯

g

+
m
m + n +2k
N

r,
1
g − 1

+
1
n + m +2k − 1
N

r,
1
g


.
We get
¯
N( r, f ) ≤

1
6
+
m

6
+
m
m + n +2k
+
1
n + m +2k − 1

T(r, g)+

1
6
+
m
m + n +2k

T(r, f)+S(r, f )+S(r, g)
.
Similarly, we have
T(r, g) ≤

1
6
+
m
m + n +2k
+
1
n + m +2k − 1


2
31
+
1
30

[T(r, f )+T(r, g )] + S(r, f )+S(r, g)
.
i.e., 0.57[T(r, f)+T(r, g)] ≤ S(r, f)+S(r, g),
which is contradiction.
Case 2. F ≡ G, i.e.,
f
n
(
f − 1
)
m
≡ g
n
(
g − 1
)
m
.
(28)
Wu et al. Journal of Inequalities and Applications 2011, 2011:133
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Page 10 of 13
Now we consider following three cases.
Case 2.1 when m = 0, then from (28), we get f = tg for a constant t such that t

=
1 − h
n
1 −
h
n+1
=constan
t
, a contradiction. So we suppose h is not a con-
stant. Since
f
≡
g
, we have
h

≡
1
.
From (29), we obtain
g
=
1 − h
n
1 −
h
n+1
and
f =
h(1 − h

(∞, f )=1−
¯
N( r, f )
T
(
r, f
)
≤
2
n
,
which contradicts the assumption
(∞, f ) >
2
n
, thus f ≡ g.
Case 2.3 when m ≥ 2, then from (28), we obtain
f
n
[f
m
+···+(−1)
i
C
m−i
m
f
m−i
+···+(−1)
m

g
m+n−i
(h
n+m−i
− 1) + ···+(−1)
m
g
n
(h
n
− 1) = 0
,
which implies h =1.Thus,f(z) ≡ g(z). If h is not a constant, then we know by (30)
that f and g satisfies the algebraic equation R(f, g) ≡ 0, where
R(w
1
, w
2
)=w
n
1
(w
1
− 1)
m
− w
n
2
(w
2

(f
m
− a))
(m + n)T(r, f )
≥ 1 −
lim
n→∞
T(r, f )
(
m + n
)
T
(
r, f
)
≥
m + n − 1
m + n
,
Wu et al. Journal of Inequalities and Applications 2011, 2011:133
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Page 11 of 13
and
δ
k+1
(0, F)=1− lim
n→∞
N
k+1


)
≥
n − k − 1
m + n
.
Similarly,
(∞, G) ≥
m + n − 1
m + n
, δ
k+1
(0, G) ≥
n − k − 1
m + n
, δ
k+2
(0, F) ≥
n − k − 2
m + n
, δ
k+2
(0, G)
≥
n − k − 2
m
+
n
.
Therefore,
 =(2k +4)(∞, G)+(2k +3)(∞, F)+δ

m
+
n
+2·
n − k − 1
m
+
n
Since n >4m +9k + 14, we get Δ >4k + 11, then by Lemma 2.5, we obtain either F
(k)
G
(k)
≡ 1orF ≡ G.
Let F
(k)
G
(k)
≡ 1, i.e.,
[f
n
(
f
m
− a
)
]
(k)
[g
n
(

(
g − a
m
)
]
(k)
≡ 1
,
(32)
where a
1
, a
2
, , a
m
are roots of w
m
- a =0.
By the similar argument for (32) of case 1.2 of Theorem 1.1, the case F
(k)
G
(k)
≡ 1
does not arise.
Let F ≡ G, i.e.,
f
n
(
f
m


≡−
1
. So from (33), we get
g
m
=
a(1 − h
n
)
1 −
h
n+m
.
Since g is non-constant, we see that h i s not a constant. Again since g
m
has no sim-
ple pole, h-h
k
has no simple zero, where
h
k
= exp

2π ki
n + m

and k = 1, 2, , n + m -1.
Hence,
(h

The authors declare that they have no competing interests.
Received: 24 July 2011 Accepted: 6 December 2011 Published: 6 December 2011
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doi:10.1186/1029-242X-2011-133
Cite this article as: Wu et al.: Uniqueness of meromorphic functions concerning differential polynomials share
one value. Journal of Inequalities and Applications 2011 2011:133.
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