Applied Mathematics, 2011, 2, 718-724
doi:10.4236/am.2011.26095 Published Online June 2011 (http://www.SciRP.org/journal/am)
Mer omorphic Functions Sharing Thr ee Values*
Changjun Li, Limei Wang
School of Mathemat i cal Sci ences, Ocean University of China, Qingdao, Chin a
Received March 26, 2011; revised April 10, 2011; accepted April 13, 2011
Abstract
In this paper, we prove a result on the uniqueness of meromorphic functions sharing three values counting
multiplicity and improve a result obtained by Xiaomin Li and Hongxun Yi.
Keywords: Uniqueness, Meromorphic Functions, Sharing Three Values
1. Introductionand Main Results
Let
f
and
g
be two non-constant meromorphic func-
tions in the complex plane. It is assumed that the reader
is familiar with the standard notations of Nevanlinna’s
theory such as , 
,Trf


,mrf ,

,Nrf


,Nrf
and so on, which can be found in [1]. We use to
denote any set of positive real numbers of finite linear
measure, not necessarily the same at each occurrence.
The notation denotes any quantity satisfying
. A meromorphic
function
E

,Srf
 
,rf r

Srf

,r E 
 
,=T


b

is called a small function with respect
to
f
provided that Tr . A meromorphic
function
 
,=b Sr

,f


b

is called a exceptional function of
f
provided that

,= ,Nr Srf
fb



1
 .
Let be a complex number, we say that a
f
and
g
share the value a CM provided and
af a
g

have the same zeros counting multiplicities (see [2]). We
say that and
f
g
share CM provided that 1
f
and 1
g
share 0 CM.
Xiaomin Li and Hongxun Yi prove the following
theorem:
Theorem A ([3]). Let and f
g
be two distinct
nonconstant meromorphic functions sharing three values
and CM, if there exists a finite complex
number such that a is not a Picard value of
, and
0,1
f

0,1a


1
1
,,NrTrfSrf
fa




 ,,
then


1
12
,=, ,
k
NrTrf Srf
fa k






and one of the following cases will hold:
1)
 
11
e1e
=, =
e1e 1
kk
ss
fg




1



, with
 

11
11
11
=
1
ks ks
s
kk
asks
ak




 and 1k
a
s

;
2)
 
11
e1e 1
=, =
e1e
ss
kk
fg




1



,
with
 

1
1
1
1
1=
1
ks
s
ks
s
k
sk s
aa k





and 1
s
ak 

;
3)
 
11
e1e1
=, =
e1e
ss
ks ks
fg



 

1

, with




1
11
1
=
11
s
ks
s
kk
asks
ak





and 1
s
aks   ;
4)

e1e 1
=, =
e11e 1
kk
ss
fg







, with
and
0,1  
k
 
1=
ks ks
s
kk k
asks
ak 



;
5)

e1e 1
=, =
e11e 1
ss
kk
fg







, with
and
0,1
s
 
 
1=
ks
s
ks
ss
k
sks
aa k 


;
6)



e1e1
=, =
e11e
ss
ks ks
fg




 

1

, with
*The Project-sponsored by SRF for ROCS, SEM.

C. J. LI ET AL.719
s0,1
 and



=
1
s
ks
s
kk
asks
k
a



.
where

is a nonconstant entire function,
s
and
are positive integers such that

2k

s
and 1

k
are mutually prime and ks

1 in 1), 2), 3),
s
and
are mutually prime and k11

ks in 4), 5), 6).
Xinhou Hua and Mingliang Fang proved the following
theorem:
Theorem B ([4]). Let
f
and
g
be two non-
constant meromorphic functions sharing three values
and  CM, if 0,1
   
,,,TrfNrbz fSrf,
 
0,1,bz 
 is a small function of , then one of
the following holds:
f
1)
f
g;
2)
f
bg, and are exceptional functions of , 1b
f
;
3) , and are exceptional
functions of

f

11 1g 

b

, 0b
f
;
4)
 
1

1

f
bgb bb
f, and are ex-
ceptional functions of .
, b
As we all know, many results on constants are also
valid for small functions, although some times they are
more difficult. In this paper, we improve the above
theorems and obtain the following result.
Theorem 1.1. Let and f
g
be two distinct
nonconstant meromorphic functions sharing three values
and CM, if there exists a small function
of such that is a exceptional
function of , and
0,1

bz  1,f
 
0,
f 
bz


  
1,,, ,NrbzfTrf Srf

(1.1)
then

  
1
2
,,=, ,
k
NrbzfTrf Srf
k


(1.2)
and one of the following cases will hold:
1)

11
e1e
=, =
e1e 1
kk
ss
fg






1
, with
 

11
11
11
1
ks ks
s
kk
bsks
bk
 


 and 1k b
s


;
2)

11
e1e 1
=, =
e1e
ss
kk
fg





1
, with
 

1
1
1
1
1
ks
s
ks
s
k
sk s
bb k





 and 1
s
bk 
;
3) 1
1
=,
1
1
=)1()1( 





 



sk
s
sk
s
e
e
g
e
e
f, with
1
1
11)(
)1(
)(1
)(







k
sks
k
s
k
sks
b
b and sk
s
b 
1;
4) 1))(1(1/
1
=,
1)(1
1
=










s
k
s
k
ecb
e
g
ecb
e
f,
with and 0,1
k




 
 
2
1
112 1
121 ;
k
s
ks
kb b
s
bcbbcb cbb
cb bks bcb


 

 





5)
  
e1e 1
=, =
1e1 11e1
ss
kk
fg
cb cb






 
,
with and 0,1
s
 
 



 
2
1
112 1
121 ;
s
k
sk
sb b
kbcbbcb cbb
cb bskbcb


 





  
6)

 

e1e 1
=, =
e111e
ss
ks ks
fg
cb




 

1


, with
and 0,1
s
 




 

   
2
1
112 1
121 .
s
s
k
k
sb b
sk bcbbcbcbb
cb bkbcb







 

 
where

is a nonconstant entire function,
s
and


2k are positive integers such that
s
and 1k

are mutually prime and ks 

1 in 1), 2), 3),
s
and
are mutually prime and in 4), 5), 6),
and
k1k1 sc

are constants.
2. Some Lemmas
Lemma 2.1 ([4]). Let and f
g
be two nonconstant
meromorphic functions sharing three values and 0,1

CM. If gf

, then for any small function




0,1,bz


 we have









33
,, ,,=,NrbzfNrbzg Srf.
Lemma 2.2 ([3]). Let be a nonconstant mero-
rphic function, 1 and be three distinct small
functions of , if
faa 2
a3
f

11
11
,,=Nr NrSrf
fa fa





,,

C. J. LI ET AL. 720
then


1
3
1
,=, ,NrTrfSrf
fa
 


 .
Using the same method of [3] in Lemma 2.2, we get
the following result:
Lemma 2.3. Let and f
g
be two nonconstant
meromorphic functions sharing three values and
CM. If is a fractional linear transformation of
0,1
f
g
, for any small function


bz

0,1,
f
, then either
is a exceptional function of , or

bz



1
1
,=,,NrTrf Srf
fb
 


 .
Lemma 2.4. Let
s
and
t
are two integers, and

be a nonconstant meromorphic function and


bz is a
small function of

, if , then 1

s
b


0,1, =,
st
Nrb Sr,

 
where denotes the reduced coun-
ting function of the common zero of and

0,1,
st
Nr b  

1
s
 tb 

.
Proof. If is a zero of and , then we
have
0
z1
s
 t  b

0=1,
sz
 (2.1)
and



0
=
tzbz

0
.
,
(2.2)
From (2.1) and (2.2) we get , thus

0=1
s
bz


0,1, =,
st
Nrb Sr

  since . 1
s
b

Lemma 2.5. Let

=
nm
Pa   ,b

 (2.3)
where =e

 ,

is a nonconstant entire function,

a


and



b

are two small functions of

,
and m are positive integers such that . n>nm
1)

 
3
1
,=,.Nr Sr
P





 (2.4)
2) If
 
,
n
bab mab
amna amna
 
 

 
 

 


 

m
(2.5)
then

 
2
1
,=,Nr Sr
P.





 (2.6)
3) If and are mutually prime, and n m
 
,
nm
bab mab
amnaamna
 
 

 
 




 





(2.7)
then

 
2
1
,=2, ,NrTr Sr
P.



 


 (2.8)
Proof. 1) Differentiating

P

two times and eli-
minating n

and m

from the three equations we
obtain




12
=1,PhP hP  
  (2.9)
with






,=, =1,2
i
Trh Sri  . Thus (2.4) holds.
2) Suppose

 
2
1
,NrSr
P

,


 , and let be
0
z



a zero of


P

with multiplicity , then from (2.3)
we have
2




0000
=0,
nm
zaz zbz   (2.10)
and


  

 
0
00
0
00
00
0
(
=0.
n
m
nz zaz
z
az zzbz
z






 





(2.11)
From (2.10) and (2.11) we get




 
   
0000
0
000
=,
mbz nbzzz
zaz mnazzz





 0
(2.12)
and




 
  
00 00
0
000
=.
nma zb zzz
zaz mnazzz





 0
(2.13)
Since =e

 ,

is a nonconstant entire function, we
have


,=,TrSr .



 (2.14)
From (2.7) (2.12) (2.13) and (2.14), we get (2.6) holds.
3) Let 0 be a zero of z

P

with multiplicity ,
using proceeding as in 2) we can get (2.12) and (2.13).
On the other hands, since and are mutually
prime, there exist one and only one pair of integers
2
n m
s
and
t
such that


=1 0<<,0<<ns mtsmtn (2.15)
From (2.12) (2.13) and (2.15) we can get is a root
of
0
z
 
=
=
ns mt
s
t
bnb ab
amna amna

 




  



 








C. J. LI ET AL.721
which implies (2.5) holds since

has two distinct
exceptional functions.
Lemma 2.6 ([5]) Let 1 and 2 be two non-
constant meromorphic functions satisfying
f f
 
1
,,=, =
ii
NrfNrSr i
f


 1,2.

Then either

012
,1; ,=NrffSr
or there exist two integers
s
,
t

>0st

such that
12 1ff
st

where 2
denotes the reduced counting
function of and related to the common 1-point
and

01
,1; ,Nr ff
1
f2 f


 


12
=,,,
= ,
Tr TrfTrf
SroTrrr E

 

only depending on and .
1 2
Lemma 2.7 ([6]) Let be a nonconstant mero-
f f
f
morphic function and
 

=Pf Rf Qf , where
 
=1 =1
= and =
pq
kj
k
kj
PfafQfbf

j
are two mutually prime polynomials in . If the co-
efficients
f


k
az,


j
bz are small functions of and
, , then
f
0
p
a
0
q
b

  
,=max,, ,TrRfpqTrfSrf.
3. Proof of Theorem 1.1
If is a fractional transformation of f
g
, by Lemma
2.3 we have that either is a exceptional function

bz
of , or f


1
1
,=,, NrTrf Srf
fb
 


 ,
which
contradicts with the assumption of Theorem 1.1. Thus
is not a fractional transformation of f
g
. By Theorem
B we have

1
,=,,. (3.1) NrTrf Srf
fb
 



From (1.1) and (3.1) we obtain


2
1
,Nr Srf
fb





,.
(3.2)
By Lemma 2.1 we have


3
1
,=,NrSrf
fb




Combining (3.2) and (3.3) we get


2
1
, Nr Srf
fb





, . (3.4)
Noting that and f
g
share 0,1 and CM, we
have

1
=e, =e.
1
ff
gg



 (3.5)
where

and

are two entire functions. From (3.5)
we get
e1e 1
=, =
e1e 1
fg



,



 (3.6)
and
ee1
=
e1
bb
fb



. (3.7)
Assume that



,=,Tre Srf

, Noting 0 and

are Picard values of , from (3.6) we have

e1
e1


and


are exceptional functions of , by Lemma 2.2 we
get
f


1
1
,=,,NrTrf Srf
fb
 


 ,
which contradicts with the assumption of Theorem 1.1.
Thus




,,TreSrf

.
Similarly, we have



,Tre Srf

,

and




,,Tre Srf
 
z.fb Let 0 be a multiple zero of , but not a zero of
,,




and





. From (3.7) we obtain






00
00
1=0.
zz
ebzebz

 (3.8)
and
 


 


00
00 000
ee
zz
zbzbz zbz


 
 =0.
(3.9)
From (3.8) and (3.9) we have







 

2
000 00
0
00000
0000
0
00000
e=
e= .
z
z
bzbzzbzz
bz bzz bzz
bzz bzz
bz bzz bzz










 



,

(3.10)
Set
12
2
=e, =
bb bbb b
ff
bb bb be,


 
 

 
 

 
 (3.11)
and



 
 


12
=, ,,
= ,
Tr TrfTrf
SroTrrr E

 


.
(3.12)
. (3.3)

C. J. LI ET AL. 722
.
From (3.5) (3.11) and (3.12) we get

,=SrfSr (3.13)
From (3.11) (3.12) and (3.13) we get
 
1
,,=, =1,2
ii
NrfNrSr i
f


 .
(3.14)
From (3.10) and (3.11) we have
. Thus

10
=1,fz

20
=1fz


012
2
1
,,1;,NrNrff Srf
fb






,.

(3.15)

01
,1; ,Nr ff
2
denotes the reduced counting function
of the common 1-points of 1
f
and 2
f
. From (3.4)
(3.13) and (3.15), we obtain

012
,1; ,.Nr ffSr

(3.16)
From (3.16) and Lemma 2.5, we know there exist two
integers and
p

>0qp q

such that
12 1.
pq
ff

(3.17)
Noting , and
, from (3.11) and (3.17), we have
, and , and


,e ,Tr Srf



,Srf
0pq


,e ,TrSrf




,eTr
 
0pq
2
e=
pq
pq bbbbb
bb bbb b

 
 

 
  
 
  

 .
(3.18)
Let

=b Qz bbb








.
, then from (3.18) we get

e=11
p
q
pq QbQ   (3.19)
Noting that is a small function of , we
obtain that

bz f

,Qz c (3.20)
where is a constant. From (3.19) and (3.20) we
obtain
c
 
e=11.
p
q
pq cbc
  (3.21)
Without loss of generality, From (3.21) we may
assume that and are mutually prime and .
p q>0q
Let

=1
p
q
c   and


q=, where

is an entire
function. Then from (3.6) and (3.21) we obtain



e1
=,
1e1
e1
=.
11 e1
q
p
q
p
fcb
gcb












(3.22)
Noting that , and , We discuss
the following three cases.
0p0qqp 
Case 1. Suppose that , we discuss the
following two subcases.
0>>pq 
Subcase 1.1. If


1
q
cb  1

. Setting 1=kq

and p
s

=, let


=1cb   and e=e
s
s



. From
(3.22) and (3.7) we get
 
11δ
δ
e1e
=, =
e1e 1
kk
ss
fg
 
δ
1
.



(3.23)
And
 1δδ
δ
ee
=.
e1
ks
s
bb
fb 1


 (3.24)
Since in this subcase b is a constant, let


=,
(3.23) assume the form (1) in Theorem 1.1. From the
proof of Theorem A we know (1.2) holds with
 

11
11
11
1
ks ks
s
kk
bsks
bk






and
1k
b
s



Subcase 1.2. If


1
q
cb  1


. Setting and qk =
=
s
p

. From (3.22) we get



e1
=,
1e1
e1
=.
11e 1
k
s
k
s
fcb
gcb












(3.25)
which assume the form (iv) in Theorem 1.1.
We have from (3.25) and (3.7)


e1e
=1e1
ks
s
bcb b
fb cb




1




1
(3.26)
Since
 
1
k
cb 


, from (3.25) (3.26) Lemma 2.4
and Lemma 2.7 we get





,= ,,Trf kTreSrf

. (3.27)
 



0,0;e1e 1,1e
=,
ks
Nrbcbbcb
Srf

 1
s 


(3.28)
where




0,0;e1e 1,1e
ks
Nrbcbbcb

 1
s 


denotes the reduced counting function of common zeros
of


e1e
ks
bcb b

 1

 1 and .

1e
s
cb

 
If

C. J. LI ET AL.723





 
2
1
112 1
12 1
k
s
ks
kb b
s
bcbbcb cbb
cb bks bcb









 

by Lemma 2.5 (2), we get a contradiction with (3.4).
Thus From (3.27) (3.28) and Lemma 2.5 (3) we obtain
(1.2) holds with
 





2
1
112 1
121 .
k
s
ks
kb b
s
bcbbcb cbb
cb bks bcb


 

 




Case 2. Suppose that , we discuss the
following two subcases.
>>0pq
Subcase 2.1. If . Setting
and


11
q
cb  
=1=kp
s
q=1  , let and

cb

k
e=e
k


 . From (3.22) and (3.7) we get
 
δδ
1δ1δ
e1e 1
=, =
e1e
ss
kk
fg



1
(3.29)
and
δδ
δ
ee
=e1
sk
k
bb
fb  
1
(3.30)
Since in this subcase is a constant, let b=δ

,
(3.29) assume the form of 2) in Theorem 1.1. By the
proof of Theorem A we know (1.2) holds with
 

1
1
1
1
1
1
ks
s
ks
s
k
sk s
bb k





and
.
1
s
bk 


Subcase 2.2 If . Setting


1
q
cb  1p=k

and
=
s
q, from(3.22) we get

 
e1
=,
1e1
e1
=.
11e1
s
k
s
k
fcb
gcb












(3.31)
Which assume the form 5) in Theorem 1.1. We have
from (3.31) and (3.7)


e1e
=.
1e1
sk
k
bcbb
fb cb




 

1
(3.32)
In the same manner as Subcase 1.2 we know (1.2)
holds with





 
2
1
112 1
121 .
s
k
sk
sb b
s
bcbbcb cbb
cb bskbcb


 







 
Case 3. Suppose that , we discuss the follow-
ing two subcases.
>0p
Subcase 3.1. If


1
q
cb  1

. Setting
1=kpq

 and =
s
q 

, let and

=1cb


δ
e=
ks

e
ks


 . From (3.22) and (3.7) we get
 
δδ
1δ1δ
e1e 1
=, =
e1e
ss
ks ks
fg

 
.
1


 (3.33)
and


1δ
δ
1δ
ee 1
=.
e1
ks
s
ks
bb
fb 



 (3.34)
Since in this subcase b is a constant, let =δ

,
(3.33) assume the form (3) in Theorem 1.1. By the proof
of Theorem A we know (1.2) holds with




1
11
1
11
s
ks
s
kk
bsks
bk





and
1
s
bks 

Subcase 3.2. If
 
1
q
cb  1


. Setting
=kpq

and =
s
q, we have from (3.22)


 

e1
=,
1e1
e1
=.
11 e1
s
ks
s
ks
fcb
gcb














(3.35)
which assume the form (6) in Theorem 1.1. From (3.35)
and (3.7) we get




e1e
=.
1e1
ks
s
ks
bcbb
fb cb







1




(3.36)
In the same manner as Subcase 1.2 we get (1.2) holds
with





  
2
1
112 1
121 .
s
k
k
sb b
s
bcbbcb cbb
cb bkbcb










Theorem 1.1 is thus completely proved.

C. J. LI ET AL. 724
4. Acknowledgements
The authors want to express theirs thanks to the anony-
mous referee for his valuable suggestions and professor
Qizhi Fang for her support.
5. References
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