Applied Mathematics, 2011, 2, 230-235
doi:10.4236/am.2011.22025 Published Online February 2011 (http://www.SciRP.org/journal/am)
Uniqueness of Meromorphic Functions Concerning
Differential Monomials*
Hui Huang, Bin Huang
College of Mathematics, Changsha University of Science and Technology, Changsha, China
Received October 21, 2010; revised December 6, 2010; accepted December 11, 2010
Abstract
Considering the uniqueness of meromorphic functions concerning differential monomials ,we obtain that, if
two non-constant meromorphic functions


f
z and


g
z satisfy

 
1, 1,
nn
kk
E
ffE gg


, where k
and n are tow positive integers satisfying 3k and 11n, then either
 
12
,
cz cz
zz
f
ce gce

 where
1
c, 2 c, c are three constants, satisfying

12
12 1
n
cc c


 or
f
tg

for a constant t such that 11
n
t

.
Keywords: Meromorphic Function, Sharing Value, Uniqueness
1. Introduction and Main Results
In this paper we use the standard notations and terms in
the value distribution theory [1].
Let

f
z be a nonconstant merom orop hi c funct i on on
the complex plane C. Define
  
,0Eafzf za,
where a zero point with multiplicity m is counted m
times in the set. If these zero points are only counted
once, then we deno te the set by

,Eaf . Let k a pos-
itive integer. Define
 



()
,0,,1,.0
i
k
Eafzfz aiikstfz 
,
where a zero point whit multiplicity m is counted m
times in the set.
Let

f
z and

g
z be two nonconstant meromor-
phic functions. If

,,Eaf Eag, then we say that

f
z and

g
zshare the value CM; if

,,Eaf Eag, then we say that

f
z and


g
z
share the value IM.
Additional, we denote by

,
k
Nrf the counting
function for poles of

f
z with multiplicity k

, and
by

k
N the corresponding one for which multiplicity is
not counted. Let


,
k
Nrf be the counting function for
poles of

f
z with multiplicityk, and by


,
k
Nrf the corresponding one for which multiplicity
is not counted. Set
   
(2
,, ,
k
k
N rfNrfNNrf, Similary, we
have the notation:

1
,
k
Nr
f






,

1
,
k
Nr
f






,

1
,,
k
Nr
f




1
,,
k
Nr
f



1
,
k
Nr
f



. If
 
1, 1,Ef Eg, we de-
note by 11 1 ,1 Nr
f




the counting function for com-
mon simple 1-points of both

f
z and


g
z where
multiplicity is no t counted.
In 1998, Wang and Fang [2] (cf. [3]) proved the fol-
lowing therem.
Theorem A Let


f
z be a transcendental mero-
morphic function, and n, k be tow positive integers
with 1nk
. Then


1
k
n
f has infinitely many
zeros.
It is interesting to establish the unicity theorem cor-
responding to the above result. In 2002, Fang [4] ob-
tained the following result.
Theorem B Let ,
f
g be tow nonconstant entire
function, and n, k be tow positive integers with


24nk. If


k
n
f and


k
n
g share 1 CM, then
either

1cz z
f
ce,

2cz z
g
ce

 where 1 c,2 c,c are
three constants, satisfying
 
2
12
11
kn
k
cc c

, or ftg

for a constant t such that 1
n
t.
Recently, Bhoosnurmath and Dyavanal [5] extended
Theorem B to the meromorphic case, as follows.
Theorem C Let ,fg be tow nonconstant meromor-
*This work is supported by the National Natural Science Foundation o
f
China (Grant No. 11071064) and Hunan Provincial Department of Edu-
cation, P.R. of China (No. 05C268).

H. HUANG ET AL.
231
phic function, and n, k be tow positive integers with

38nk. If


k
n
f and


k
n
g share 1 CM, then
either
 
12
,
cz cz
zz
fcegce

 where 1 c,2 c,care three
constants, satisfying

2
12
11
kn
k
cc c, or ftg

for
a constant t such that 1
n
t.
Let 1k,

1
1
1n
f
nF

 and

1
1
1n
g
nG

 in
Theorem C. Then 1nn
f
FF

 

 and 1nn
Gg GG




 .
We see that the following result, which is proved by
Yang and Hua [6], is a direct consequence of Theorem
C.
Theorem D Let

f
z and

g
z be two noncons-
tant meromorphic functions, and 11n an integer. If
n
f
fand n
g
gshare 1 CM, then either

1cz
z
f
ce,

2cz z
g
ce

 where 1 c,2 c,c are three constants, satis-
fying

12
12 1
n
cc c
 or ftg

for a constant t such
that 11
n
t.
In this paper, we will extend the above result as fol-
lows.
Theorem 1 Let

f
z and

g
z be two noncons-
tant meromorphic functions,

3k,

11n be tow
positive integers. If

1, 1,
nn
kk
EffEgg

, then ei-
ther
 
12
,
cz cz
zz
fcegce

 where 1 c,2 c,c are three
constants, satisfying

12
12 1
n
cc c
 or ftg for a
constant t such that 11
n
t.
Theorem 2 Let

f
z and

g
z be two noncons-
tant meromorphic functions,

13n be a positive in-
teger. If




22
1, 1,
nn
EffEgg


, then the conclusion
of Theorem 1 holds.
Theorem 3 Let


f
z and

g
z be two noncons-
tant meromorphic functions,

19n be a positive in-
teger. If




11
1, 1,
nn
Eff Egg

 , then the conclusion
of Theorem 1 holds.
2. Some Lemmas
For the proof of our results, we need the following lem-
mas.
Lemma 1 [7]. Let f be a nonconstant meromorphic
function,and let 01
,,,
n
aa a be finite complex numbers
such that 0
n
a

. Then



11
110
,
,,
nn
nn
Trafa fafa
nT r fSr f




Lemma 2 [6]. Let


f
z and


g
z be two non-
constant meromorphic functions,

6n be a positive
integer, if 1
nn
ffgg



then either

1cz z
f
ce,

2cz z
g
ce

 where 1 c,2 c,c are three constants, satis-
fying

12
12 1
n
cc c


.
Lemma 3 [8]. Let
f
be a nonconstant meromorphic
function, k a positive integer, then


11
,,,,
k
NrNrkNrf Srf
f
f







 .
Lemma 4 [9]. Let
f
and
g
be two nonconstant
meromorphic functions,and let k be a positive integer.
If




1, 1,
kk
EfEg, then one of the following cases
must occur:

 

2222 22
11 11
1111
,, ,,, ,,,
11
111
,(,),,,
111
kk
TrfTrgNrfNrNrg NrNrNr
fgfg
NrNrNrSrf Srg
fff

  
 
  


  
 
 
 

 
(1)
 
 
11
,where 0, are tow constants.
bgab
fab
bga b


 (2)
Lemma 5. Let
f
and
g
be two nonconstant me-
romorphic functions,

6n be a positive integer, set
n
F
ff

, n Ggg

, if




11bGab
FbGa b

 (2.1)
where


0ab are two constants, then either

1cz z
f
ce

,

2cz z
g
ce


where 1 c,2 c,c are three constants, satis-
fying


12
12 1
n
cc c


 or
f
tg for a constant t such
that 11
n
t

.
Proof. By Lemma 1, we get




 
 
''
,,, ,,2,,
2,,
nn
TrFTrffTrfTrfnTrfTrfSrf
nTrfSrf

  (2.2)

H. HUANG ET AL.
232













 
 
,,, ,,,
1
,,,,,
1
,,,,,
1
,,, ,,
nnn
nn
n
nTrfTrf SrfNrf mrf Srf
Nrf fNrfmrffmrSrf
f
TrffTrf Nrf NrSrf
f
TrFTrfNrfNrSrf
f


 





 





 



(2.3)
So,
 
1
,1,,, ,TrFnTrf Nrf NrSrf
f

 


 (2.4)
Thus, by (2.2 ) and (2.3) , we ge t



,,SrF Srf.
Similarly, we get
 



,1,,
1
,,
TrGnTrg Nrg
Nr Srg
g
 





(2.5)

,,SrG Srg
It is clear that the inequality



,,Trf Trg or
 
,,Trg Trf holds for a set of infinite measure of
r.
Without loss of generality, we may suppose that




,,TrfTrg, holds for rI, where
I
is a set
with infinite measure. Next we consider five cases.
Case 1. ,0,1abb

,
If 10ab

 , then by the 2.1 we known:

11
,, 1
1
NrNr ab
FGb




 
 

 
By the Nevalinna second fundamental theorem and
lemma 3, we have
  
 
 
  
'
11
,,, ,,
1
1
11
,, ,,
11 11
,, ,,, ,
11 1
,, ,,2,,
TrGNrG NrNrSrG
ab
GGb
NrG NrNrSrg
GF
Nrg NrNrNrNrSrg
ggf f
Nrg NrNrNrfNrSrg
gg f



 
 
 




  


  
 

 

 
 
 
 

 
 
 
1
,,,3,,
1
4(,),,,
TrgNrgNrTrfSrg
g
TrgNrgN rSrg
g

 








By 6n and (2.5), we get

,,Trg Srg, for
rI, a contradiction.
If 10ab, by (2.1 ) w e can obtain:

1
1
bG
FbG



We see that:

1
,,
1
NrF NrGb







Combining the Nevalinna second fundamental theo-
rem and lemma 3, we have

H. HUANG ET AL.
233
  
 
 
  
 
11
,,, ,,
1
1
,,,,
111
,, ,,,
1
,,,,,
1
2(,),,,
TrGNrGNrNrSrG
GGb
NrG NrNrF Srg
G
Nrg NrNrNrSrg
gg f
Trg NrgNrTrfSrg
g
TrgNrgN rSrg
g



 







 


 
 
 

 

 








By 6n and (2.5), we get
 
,,Trg Srg, rI

,
a contradiction.
Case 2. ,1abb, So

1
a
FaG 
We can get
 
1
,,
1
NrF NrGa





, similarly as
Case 1, it is impossible.
Case 3. ,0abb, So

1Ga
Fa

.
If 10a , then
F
G, so nn
f
fgg

.
It follows that:
11
11
11
nn
nn fg
ffggFC GC
nn



 , where C is a
constant.
We state that C is zero. If not, one we can get that
from the Nevalinna second fundamental theorem and
lemma 1.



 
 
 

1
111
111
1, ,,
11
,, ,,
1
11
,, ,,
11
,, ,,
3, ,
nTrgT rGSrg
NrG NrNrSrg
GG
NrG NrNrSrg
GF
Nrg NrNrSrg
gf
Trg Srg

 
 
 

 

 



 



Because 6n, we can get
 
,,Trg Srg, for
rI

, which is impossible. So C is zore.
Then 11
F
G

, it gives that 11nn
f
g

, so ,
f
tg

where t is constant satisfying 11
n
t.
Case 4. ,0,1abb

, from (2.1) we can get


11bG
FbG




1
,(,)NrF NrG , similarly as Case
1, it is impossible.
Since 0a

, now we consider the following case.
Case 6. 1ab


It yields 1FG

, that is: 1
nn
ffgg

. By the Lemma
2, we can get
 
12
,
cz cz
zz
f
ce gce

 where 1 c,2 c,c
are three constants, satisfying

12
12 1
n
cc c
 .
Now the proof of Lemma 5 is completed.
3. Proof of Theorems
Proof of theorem 1:
Noticing that 3k, we have
 
11
(1 (1
11 1
,, ,
11 1
11
,,
11
1111
,,
2121
11
,,1
22
kk
NrNrNr
FG F
Nr Nr
FG
Nr Nr
FG
TrFTrGO


















By lemma 4, we can get
  
  
2222
2222
11
,,2,,,,,,
11
2,,,,, ,
T rFT rGNrNrFNrNrGSrFS rG
FG
NrNrF NrNrGSrfSrg
FG

 
 

 
 


 


 
 

(3.1)

H. HUANG ET AL.
234
Because:



2222
11 11
,,, ,2,,2,
n
n
NrN rFNrN rffNrNrNrf
Fff
ff
  
 

  
 

  
 (3.2)
and
 
22
111
,,2,,2,NrNrGNrNrNrg
Ggg
 

 
 
 
   (3.3)
By (3.1)-(3.3) and lemma 3, we can get:
   
  

1111
,,22,2,,2,2, ,,,
1111
4,4, 2,, 4,4,2,,
11
5, 5,,
TrF TrGNrNrfNrNrNrgNrSrfSrg
ffgg
NrNrfNrSrfNrNrgNrSrg
ffgg
NrNrf NrS
ff

 
 
  

 

 
 
 

 
 

 

 
  
   
11
,5,5,,,
11
9,,,, 9,,,,
rfNrNrgNrSrg
gg
T rfNrfNrSrfT rgN rgNrSrg
fg
 

 

 
 
 
 

 
(3.4)
By 9nand (2.4), (2.5) we obtain



,,,,T rfT rgSrfSrg, which is impossi-
ble.
Therefor e, by lemma 4

 

11bgab
fbga b

, where

0a, b are tow constants, it follows by lemma 5
then either
 
12
,
cz cz
zz
f
ce gce

 where 1 c,2 c,c are
three constants, satisfying

12
12 1
n
cc c
 or
f
tg

for a constant t such that 11
n
t.
The proof of Theorem 1 is complete.
Proof of theorem 2:
We can see clearly:
 
(3 (3
11
11 11111
,,,,,
11 12121
11111 1
,,,,,,
21212 2
NrNrNrNrN r
FG FFG
NrNrTrFTrGSrfSrg
FG

 


  






By lemma 4, we can get:
 

2222
(3 (3
11
,,2,, ,,
11
,,,,
11
TrFTrGNrN rFNrN rG
FG
Nr NrSrfSrg
FG

 
 
 

 






(3.5)
Considering
 
 
'
(3 '
'
1111 11
,,,,,,,
1222 2
11 1
,,,,2,,
2
FF
N rNrNrSrfNrNrF Srf
FFF
F
NrNrNrfSrfTrf Srf
ff

 


 

 


 

 



(3.6)
Similarly, we can get
 (3 1
,2,,
1
Nr TrgSrg
G




 (3.7)
By from (3.4)-(3.7), we can get
  
11
,,11,,,,11 ,,,,TrFTrGTrfNrfNrSrfTrg NrgNrSrg
fg
 
 
 



H. HUANG ET AL.
235
Since 13n and (2.4), (2.5), we can get



,,,,T rfT rgSrfSrg impossible The proof
of Theorem 2 is complete.
Proof of theorem 3:
Since:
 
11
11 111
,, ,,
11 121
111 1
,,,,,
212 2
NrNrN rNr
FG FF
N
rTrFTrGSrfSrg
G

 


 






We can see clearly from lemma 4 that:
  

 

(2 (2
2222
(2 (2
2222
11 11
,,2,, ,,,,
11
,,
11 11
2,, ,,,,
11
,,
T rFTrGNrNrFNrNrGNrNr
FG FG
SrF SrG
Nr NrFNr NrGNrNr
FG FG
Srf Srg

 
  

 

 



 
 

 

 


(3.8)
Considering
  
 
(2 11
,,,,,,,
1
11
,,,,4,,
FF
N rNrNrSrfNrFNrSrf
FFF F
NrNrNrf SrfTrf Srf
ff

 

 


 
 
 
 

 
(3.9)
Similarly, we can get

(2 1
,4,,
1
Nr TrgSrg
G




 (3.10)
By from (3.8)-(3.10), we can get
   
11
,,17.,,,17 .,,,TrFTrGTrfNrfNrSrfTrgNrgNrSrg
fg
 
 
 


Since 19n and (2.4), (2.5), we can get



,,,,T rfT rgSrfSrg , impossible The
proof of Theorem 3 is complete.
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