Advances in Pure Mathematics, 2011, 1, 210-217
doi:10.4236/apm.2011.14037 Published Online July 2011 (http://www.SciRP.org/journal/apm)
Copyright © 2011 SciRes. APM
Normal Criteria and Shared Values by Differential
Polynomials*
Jihong Wang1, Qian Lu2, Qilong Liao3
1Department of Mathematics, Southwest University of Science and Technology, Mianyang, China
2Department of Mathematics, Southwest University of Science and Technology, Mianyang, China
3Department of Material Science and Engineer, Southwest University of Science and Technology,
Mianyang, China
E-mail: wangjihong@swust.edu.cn, luqiankuo1965@ho tmail.com, liaoql@swust.edu.cn
Received April 5, 2011; revised April 28, 2011; accepted May 8, 2011
Abstract
For a family
F
of meromorphic functions on a domain D, it is discussed whether
F
is normal on D if for
every pair functions

f
z,

g
zF, n
f
af
,b
and share value on D when , where a,
b are two complex numbers, . Finally, the following result is obtained:Let
n
agg
d2, 3n
0,a
F
be a family of
meromorphic functions in D, all of whose poles have multiplicity at least 4 , all of whose zeros have multi-
plicity at least 2. Suppose that there exist two functions
az not idendtically equal to zero, analytic
in D, such that for each pair of functions
dz
f
and
g
in
F
,
2
f
az f
and

2
azg
share the function
. If has only a multiple zeros and

dz
az
fz
whenever
az 0
, then
F
is normal in D.
Keywords: Normal Family, Meromorphic Function, Shared Value, Differential Polynomial
1. Introduction and the Main Result
In 1959,Hayman[4] proved
Theorem 1.1.
Let
f
be meromorphic functions in
C, n be a positive integer and a, b be two constant such
that , . If
5n0,aandb 
n
f
af b

then
f
is a constant.
Corresponding to Theorem 1.1 there is the following
theorems which confirmed a Hayman’s well-known con-
jecture about normal families in [5].
Theorem 1.2. Let F be a meromorphic function family
in D, be a positive integer and a, b be two constant
such that . If and for each
function
n
0,aandb 3n
f
Fn
,
f
af b
, then F is normal in D.
This result is due to S. Y. Li [8](), X. J. Li [9]
(), X. C. Pang [10](), H. H. Chen and M. L.
Fang [2]().
5n
5n4n
3n
In 2001, M. L. Fang and W. J. Yuan [3] obtained
Theorem 1.3. Let F be a meromorphic function family
in D, a, b be two constants such that .
If, for each function
0,aandb 
f
F
,2
f
af
b
and the poles
of
f
z are of multiplicity 3 at least, then F is nor-
mal in D.
Let D be a domain in C,

f
z be meromorphic on
D,and. aC


1:
f
Eaf aDZDfz a

Two functions f and g are said to share the value a if
f
Ea Ega. For a case in Theorem 1.2, Q.
C. Zhang [14] improved Theorem 1.2 by the idea of
shared values and obtained the following result.
4n
Theorem 1.4. Let F be a family of meromorphic func-
tions in D, n be a positive integer and a, b be two con-
stant such that ,
4n0,aandb

n
. If, for each
pair of functions f and g in F,
f
af
and n
g
ag
share the value b, then F is normal in D.
In this paper, we shall discuss a condition on which F
still is normal in D for the case and obtain the
following result.
2n3
Theorem 1.5. Let F be a family of meromorphic func-
tions in D, all of whose poles have multiplicity 2 at least,
and a, b be two constant such that .
0,aandb 
*Supported by China Industrial Technology Development Program
(B3120110001).
J. H. WANG ET AL.
211
If, for each pair of functions f and g in F, 3
f
af
and
3
g
ag
share the value b in D, then F is normal in D.
We denote
 

#
2
1
f
z
fz
f
z
for the spherical de-
rivatives of

f
z.The following example imply that the
restriction of poles in Theorem 1.5 is necessary.
Example 1. [14] Let
:1Dzz and
n
F
f,
where

1,,1,2,

(1
)
n
fz zDn
nz n

Then for each pair m, n, 3
mm
f
f
and 3
nn
f
f
z
share the value 0 in D. But F is not normal at 0
since

#
n1fn.
But we also have the following examples which imply
that on the same as restriction of poles in Theorem 1.5 F
is not normal in D if for each pair of functions
f
and
g
in F, 2
f
af
and 2
g
ag
share the value b on
D.
Example 2. [3] Let


2
1
n
fznz zn
for
1, 2,n
, and
:1zz 
Clearly,


4
210
nn
fz fnzn
 
,
and
n
f
z

only a double pole and a simple zero. Since

#0
n
fn

, as from Marty’s criterion we n
have that n
f
z is not normal in In fact, in the-
present paper we also obtain two results as follows.
Theorem 1.6. Let F be a family of meromorphic func-
tions in D, all of whose poles have multiplicity 4 at least,
all of whose zeros have multiplicity 2 at least, and a, b be
two constant such that . If, for each
pair of functions f and g in F,
0,aandb 
2
f
af
and 2
g
ag
share the value b in D, then F is normal in D.
Theorem 1.7. Let F be a family of meromorphic func-
tions in D, all of whose poles have multiplicity at least 4 ,
all of whose zeros have multiplicity at least 2. Suppose
that there exist two functions not idendtically equal
to zero, analytic in D, such that for each pair of
functions f and g in F,

az


dz
2
f
az f
and
2
az

0az
g
share the function in D. If az has only a
multiple zeros and whenever
dz

fz

then
F is normal in D.
The following example shows that the condition

fzwhen in Theorem 1.7 is necessary.

0az
Example 3. [7] Let
:1Dzz and
n
F
f
where

4
1,,1,2,
n
fz zDn
nz

. We take
3
4az z
and
0dz
. Clearly, F fails to be nor-
mal at 0z
However, all poles of

n
f
zare of multi-
plicity 4, and for each pair m, n,
2
mm
f
az f
and
2
nn
f
az f
share analytic functions in

dz
.
2. Lemmas
To prove the above theorems, we need some lemma as
follows:
Lemma 2.1. ([1,2]) Let

f
zbe a meromorphic
function in C, n be a positive integer and b be a non-zero
constant. If n
f
fb
, then
f
is a constant. Moreover
if f is a transcendental meromorphic function, then
n
f
fz
assumes every finite non-zero value finitely
often.
Lemma 2.2. ([1]) Let
f
z be a transcendental me-
romorphic function with finite order in C. If
f
z has
only multiple zeros, then it’s first derivative
f
assumes
every finite value except possibly zero infinitely often.
Lemma 2.3. ([12]) Let

f
z

be a non-polynomials
rational function in C. If
f
z has only zeros of multi-
plicity 2 at least, then

2
cz d
faz b
where a, b, c, d
are four constants, 0, 0ac
.
Lemma 2.4. ([4]) If
f
zbe a transcendental mero-
morphic function in C, then either

f
z assumes every
finite value infinitely often or every derivative()l
f
as-
sumes every finite value except possibly zero infinitely
often. If
f
z is a non-constant rational function and
f
za
, a is a finite value, then ()l
f
assumes every
finite value except possibly zero at least once.
Lemma 2.5. ([11]) Let

f
z
k
be a transcendental
meromorphic function with finite order, all of whose ze-
roes are of multiplicity at least , and let
1
Pz be
a polynomial,
Pz is not idendtically equal to zero.
Then
()k
f
zPz has infinitely many zeros often.
Lemma 2.6. ([6]) Let
f
z

be a non-polynomial ra-
tional functions in C, all of whose zeroes are of multi-
plicity at least 4. Then r
f
z
z
has a zeros at least
often.
Lemma 2.7. ([13]) Let F be a family of meromorphic
functions on the unit disc , all of whose zeroes have
multiplicity p at least, all of whose poles have multiplic-
ity q at least. Let
be a real number satisfying
pq
. Then F is not normal at a point 0
z
if
and only if there exist
1) points n
z
, ;
0n
zz
2) functions n
f
F
; and
3) positive numbers 0
n
such that
Copyright © 2011 SciRes. APM
212 J. H. WANG ET AL.

 
nn nnn
fzg g
 
 
spherically uniformly on each compact subset of C,
where

g
is a non-constant meromorphic function
satisfying the zeros of

g
are of multiplicities p at
least and the poles of
g
are of multiplicities q at
least. Moreover, the order of
g
is not greater than 2.
3. Proofs of Theorem 1.5.-1.7.
3.1. Proof of Theorem 1.5.
Suppose that there exists one point 0 such that F
is not normal at point 0. Without loss of generality we
assume that . By Lemma 2.7, there exist points,
,0n, functions
zD
z
00z
z
n
z zn
f
F
and positive numbers
0
n
such that



1
1n
jjjjj
gfzg


(3.1)
spherically uniformly on each compact subset of C,
where

g
is a non-constant meromorphic function
with order , all of whose poles are of multiplicities k
at least.
2
From (3.1) we have


 
1
1
n
n
n
jjjj jjj
n
nn
n
jjj
fzaf zb
g
agb gag
 

 

 
(3.2)
By the same method as [14], from Lemma 2.1 it is not
difficult to find that n
g
ag
has just a unique zero
0
.
Set 1g
again, if then 3n
2nn
gaga n
 


 

thus 2n
an
 


has just a unique zero 0

.
Thus 0
is a multiple pole of
or else a zero of
.
2
na

If 0
is a multiple pole of
, since
2nn
a
 


has only one zero 0
, then . By Lemma
2.1 again,
20
na


is a constant which contradicts with g is
not any constant.
So we have that
has no multiple poles and
a


have only a unique zero. By Lemma 2.1, and
Lemma 2.4, we have
is not transcendental.
If
is non-constant polynomial, then

2
0
l
naA
 
 .
Since all zeros of
are of multiplicity 2, then . 3l
Denoting
for

1n1n
,
11
nn

, we
have

0
l
A
a


and

1
0
l
Al

 . Since
all zeros of
are of multiplicity, then

214n
0
0,

.
If
00

, then
0
0

which contradicts with
0a
0

 . So
is a constant.
Next we prove that there exists no rational functions
such as
. Noting that

11
nn

 and
has
no multiple pole, we may set
 




12
1
m
mm
s
n
n
t



 
1, 1
12
1
12
,
()
s



1
n
A

 
(3.3)
where A is a non-zero constant,
s
t,12
,,,
s
mm m

2,,)js
are s positive integers,j,(1 . For
a convenience of stating, we denote

21 ,
,
mn
12
s
mm mm
  (3.4)
then
21mns
 

.
From (3.3), we have





11
1
1
m
n
A

1
s
m
s
n
t
h
 
1
1
,
p
q


 



(3.5)
where
1hmtn

12st
st
a

20
st
 a
 


 
1
11
m
 

1
s
m
s
ph


1
 


,
n
t


a


 
11
n
q


0
 (3.6)
are three polynomials. Since has only a
unique zero
then there exists a non-zero constant B
such that
 


0
12
l
nn
t





,
n
B


a
 

 
(3.7)
so

1
02
11 n
nn
t
Bp


 1
12
,
l


 
 (3.8)
where
plnt

1tt
b
21t0
b
 
 


is a poly-
nomial. From (3.5) we also have


1
m
A
2
13
1
s
m
s
n
t
p
2
1
1
n
 





 
 (3.9)
where
3
p
is a polynomial also.
We denote
degp for the degree of a polynomial
p
, from (3.5) and (3.6) we may obtain

deg
deg
hst
pm


11
deg qn
1
1,
t t
 (3.10)
Copyright © 2011 SciRes. APM
J. H. WANG ET AL.
213
t
From (3.8), (3.9) and (3.10) we may obtain

2
deg ,p
(3.11)

3
deg2 22.pts
 (3.12)
Since has only a unique zero

a

0
and
21(1,2,,),
j
mjs
then 0j
(1 rom (3.8), (3.9) and (3.11)
it follows
,2,,j.


)s F
that then
, (3.13)
Since , then
w
3
deg 1pl


2
2degmsp t 

21
j
mn
2

21mns, so by (3.13)
e have
s
t.
If lnmt, fro (3.8), (3.9) and (3.12), we have
Then, . Combining with above inequality

3
11deg 222ntlpts
21ts
2
s
t, wout a contradiction.
nt, then from (3.5) and (3.7) w
e bring ab
If e have
that is . If , then
1
this is impossible. Thus,
l
 
deg degpq
11

degmsh nt 
2s
t . So

1mtn

degh
1
 


deg
deg
22
mtnsnth tn
st h
stst
 

 
1mtn
re,
and

deg 1hst.Therefo
11
. Then
m
2s
t
t
n
e have m21s
, this
contradicts to 2
. Again from
(3.8) and (3.9), wt
s
t.
This comple ptes theroof of Theorem 1.5.
.2. Proof of Theorem 1.6.
or any points , Without loss of generality, we
3
F0
zD
ose thatset 00z. Supp F is not normal at 00z
, then
by L 2.7, we have that there exist a uence
n
emma subseq
f
F, points sequence 0
zD
, and a positive num-
bers n
, 0
n
, such that

1,z

nnnnn
gf g

 (3.14)
spherically uniformly on each compact subset of C,
where
g
is a non-constant meromorphic function
with or, all of whose poles are ofmultiplicities at
least 2, all hose zeros are of multiplicities at least 4.
From (3.14) we have
der 2
of w
 


2
22
nn
n
1
g
a
gad
gg


  (3.15)
If , then

0ga


0
g
ac

, this contra-
dito which allfcts zeros o

g
have multiplicity at
least 4. If for any point C
,


0a
, then By
Lemma 2.2, we have that
g
g
is not transcendental in
C, so
g
is non-constantal function in C. By
Lemmwe also have that
ration
a 2.3

3
d
ab

c
g
a contradictions. Theref
ore,

2
gag




have a
zeros. We may claim that

2
gag
h
 as a uni-
que zero 0
. Otherwis*
e, suppose that 00
,
are two
distinguis of h zeros

2
gag


sitive nu
then there exists a pomber0
such that
*
00
,,NN

.On the other han by Hur-
n find two point sequences
d,
witzrem we c’s Theoa
0,
nN

,
**
0,
nN

Such that
**
00
,
nn


, and
nn

22
0
nn n
ggad
 


*
mm

2*2 0
mmm
ggad



then, we have

2
n n
af z
0
nnnnn d
 
n
fz
,

*2
m m
af z
*0
mmmmm d
 
m
fz

ev
.
Fr ypothesis thery pair functions om the hat for f,
g
in F,
2
fz af
and

2
g
zag
share comp
mber dave
lex
nu in D, we h
z af

2
n n
z
0
nnnnn d

m
f
,

*2
m n
af z
*0
n nnnn
fz d
 
m


.
Fix , the0m, let n n

2
00
mm
fafd.
Since
2
mm
f
zafzd
n w
0, m
has no accumulation points,
so for sufficiently large have
*
mm
zz
 
e
0
nnn

then
*
,
nn

nn
nn
zz

Thiradicts to s cont

*
00
,,NN
. Thus,
2
gag

has
zero0
a unique
. Further-
either 0
move thatre, we ha
is a me poles of ultipl
g
or 0
is a unzero of ique
g
a
. If
0
is a le poles of

gmultip
, then
0ga
,
C
for any
. By Lemma 2 Lemm
mediately deduce that
.2 anda 2.3, we im-
g
must be a constant in C,
Copyright © 2011 SciRes. APM
214 J. H. WANG ET AL.
which contradicts to

g
is a non-constant mero-
morphic functions in C. erefore,

g Th
has only a
simple poles and

g
a
has a un0
ique
. But
since
g
has otiple poles, so we that nly a mule hav
g
is entire in C and

g
a
has a unique
0
. Also by Lemma 2.2, hat we have t
g
is a
nstant polynomials, all of whose zeros mul-
tiplicity at least 4. Setting

1
non-coare of

12
2
s
m
,
2
s
m
m
gA

 m
s




12
11mm
we have

g
 
1
A h
12
 



 
12ss
a

02s
a
,A
Where

,
hm

2
0,
01
,,
s
aa t
,s are
a
1, 2,
are some complex constan s,
j
mj
s
positive integers, 4
j
m, and
1
js
j
j
. m mThus, we have

gaB

l
0


 ,
ga
where 3l. So we have that

0
Bl

1l

g




0
0g

.
If 0g, then

00
g

0
ctions.
.
Bu 0
t
0
efore,
F
ga
is norm
, a contr
a
adi
Ther l at 0z
.
.3.
or a
3 Proof of
ny
he
Theorem
az
.7
1.7.

0,
by the
FzD
ore
, ifplete we may give the com
proof of Tm 1 same argument as Theorem
1.6, we emit the detail. In the sequel, we shall prove that
F is normal at which

0az
. Set

r
az zbz,
where

bz is analytic 0,
01b
positive integer, 2r.
at

z

, r is a

1
1
:,
r
F
FF
tion
zzf


fz F



z
uncFor every f
F
z in 1
F
, from the hypothesis
in Theorem 1.7, we ca that all zeros of n see
F
z are
of order at least 4, all poles of

F
z are of micity
at least 2.
Suppose
ultipl
that 1
F
ex
is not normal at z
n
0, then by
Lemma 2.7, there ists a subsequence 1
F
F, a point
sequence
,1
nn
zz r,and a positive nequence
n
umber s
, n
1z
0

, such th

nn
at

n
n
r
nn
11
n
zf

n
nn
z
n
gF
g


on com

act subsets of

rmly

p
rically uni
(3.16)
C, wsphe fohere

g
is a non-constant meromorphic function on C,
whose zeros are of multiplicity at least 4, and all of
whose poles are multiple. Moreover,

g
all of
has an or-
der at most 2.
Now we distinguish two cases:
Case 1. nn
z
 . Without loss of a generalization,
w exe assume that thereists a point zsuch that
,1
n
zzzr

, we have


 


1
1
()
11
rr
n
nn
r
g
22
2
2
nnn
n
nn n
nn nn
nn
r
n
nn n
fz
g
zg
zg
gz
r
g
z



note 1
S






 





(3.17)
For the sake of convenience, we defor
of
the set
all zeros of
g
,2
S for the set of all zeros of
g
, and 3
S for t sof all poles of

gheet
.
Since


22
nn
gg
lim ngg
 ,

 
11
lim
ng
gn
, and
 uni-
formly on compact subsets of 1
\CS

lim 0
nnn
r
z


ly on coms of C,
thus
uniformpact subset
lim nn n
fz

n
, uniformlympact sub- on co
sets of
123
\CS SS. Thus, it is not difficult to see
that
 





2
2
2
1
nnn nnn
nnnnn
nn
nnnnn
zazfz
azfzdz
dz
azfzd
g
bz
nn
nn
nn
z
f

 

 


 
 
(3.18)
uniformly on compact subsets of. If
\CS

123
S S

10
g
bz
 , then

g
bz
 , for any
123
\CS SS
 . Thus,
 
g
bz


for any
C
. By Lemma 2.5, we can see that
g
is not
ndental in C, but is a rational functioo from
Lemma 2.3, we deduce that

g
transce n. Als
is constant, which
contradicts to the fact that g

is non-constant. On
the other hand, it is easy to see that

g
is not iden-
tically equal to
bz
. Hence,

g
b
e as the
z

has one
zeros at least in Crguments
in Theorem 1.5 and Theorem 1.6, we deduce that
. In fact, by the sam a
g
has a unique zero 0
. By Lemma 2.5, we ca
that
n see
g
is not trdental in C, so

g
anscen
is non-
consttional function in C. For a non-cnt poly-ant raonsta
Copyright © 2011 SciRes. APM
J. H. WANG ET AL.
215
nomials

g
, and noting that

g
has only a zero
with multiplicity at least 4, we hav


e
0
l,3
g
bz Bl


 
Thus,


1
0
l
gBl

 
. Hence,
g
has a zero
0
at most. If 0
is a zero of

g
, then
 
g

0
00

0gg

 . But

00gbz

 ,
In the sequel,
a contradiction.
we denote r the degree of a
po

p fo
deg
lynomial

p
. If
g
i
s n
on polynomials ra-
tional functions, then we set
 


11
m
 
2
12
12
12
,
s
t
m
s
n
nn
t
 

(3.19)
Wher t.
t (3.20)
Then,
m


 

2
j
n,
st
jk
qn

gA
e j
m

4,1,2,,js; 1, 2,,j
11
4, 2
jk
mms

 







1
1
1
1
A
1
1
1
1
s
t
m
m
s
n
n
t
h
g1
1
p
q

 

(3.21)
where
2
 

12
0
st
 
st
s
t
q a



 ,

deg hs
hm
pA
a

1
t
11
2
 


12 1
11
s
m
mm
sh


 
 
 

1
11
t
n
t
q
 


12
11
2
nn


Since

g
bz

has a uno ique zer0
, so we
set


1
0
1
l

 
nt. Then

1
1
() t
n
n
t
B
z
 


(3.22)
where Bnonzero constafrom (3.22), we
gb
is a

have


1
1
02
n
t
Bp


1tt
b


2
2
1
t
l
n


 (3.23)
where 1
bis a poly-
hat
g


2
deg

 
20
t
plqt


nomial,

pt.
follow tFrom (3.21), it




11 2
m
A
1
13
2
2
1
n
p
g
2
s
t
m
s
n
t




3
222 223
02
1
st st
23
s
t
pmqmq
cc

 

 
 
is also a polynomial,
3
deg22 2pst.
cases to derivative a cWe distinguish fiveontradiction:
Subcase 1.1. mq
. Then from (3.21), we have
lqt
. So,
222
deg, 1pti it
,
00
deg1, 11hst hhst
 
and
333
deg22 2,122 2pstitst
 
From (3.23) and (3.24), we have. S also
fro
23
1ii
e
o
m (3.23) and (3.24), we also hav
3
deg .
Thus, we have 3
22122lsti st
1
i
lp
2
 
Since lq
.
t
t and2qt, then we have 2
2si
.
On the ot, from and (3.24), we her hand (3.23) also have
2
2degms p . Since 4ms, we have 2
2
s
ti
.
.
Subcase 1.2. m
This is impossible
1q
. Then lqt,
22
deg pti
, 2
it1
,

deg 1hst
and
333
deg222, 1222psti tst

Similarly to Subcase (1.1), from (3.23) and (3.24), we
also have that 23
1ii
.
Also from (3(3
3
1deglp ,
.23) and .24), we have
then, we have 2
21ts i
 On the ot-
larly to the argucase (1.1), from (3.23) and
(3.24), we also have
her hand, simi
ment of Sub
22
2degmsp ti

, then
2
21
s
ti
 . This also is
3. 2mq
impossible.
Subcase 1.
. Then we still have

222 1lp tst
3 ,deg,1,degqttii th
 
and
,
3
deg22 2pst
. Therefor, 222lst, so
2s
2t
. Similarly, we 2
2sti, t
2
2
have 2mshen
s
ti
. . This is a contradiction
e 1.4. 1mq
Subcas
. Thenlqt
,
deg 1hst
,
32, deg22tps and
22
deg pt2
,i i0t

.
2.ti
From (3.23) and (3.24), we
have 2ms
 Thus,2
2
s
tiand 2
21 .ts i
This i
Subcase 1.5. m
s impossible.
. Then ,
2qlqt

  (3.24)
where
deg 1hs
t
,
3
deg 2pst22
d
2
gpt, ande
.
From (3.23) and (3.24), we have
3
1deg22 1lpst

and
2
2degp t
.
ms
21ts

So, we have that
2 and
s
t. This is a con-
tradiction.
Case 2. Suppose that there exists a complex number
C
and a subsequence of sequence

1
nn
z
, still
noting it 1
znn
, such that 1
znn
. Wea con- have
Copyright © 2011 SciRes. APM
216 J. H. WANG ET AL.
verges



11
ˆ
nnnnnnnnnn
F Fzz
gg
H
 
 



(3.25)
spherically uniform on compact subsets of C. Clearly, all
zeros of

ˆ
g
are of multiplicity at least 4, all poles of

ˆ
g
arultiplicity at least 2. For each 00e of m
, it
to see that there exists a neighborhood ),
is easy(0
N
of 0
, such that
 
ˆ
rr
n
Hg

, the con
bei spherically
,
vergence
ng uniform on
0
N
. For 00
,
since 0
is the pole of

g
, then there exists 0
,
such tt

ha ˆ
1g
is an on alytic
2:D
2
,
1Hn
aytic on re anal
:2D
2
for suffi
ciently la
-
rge n. Since


1H
r
nnnn
f
 
then 00
is a zero of
ˆ
1g
has order at least r,
we cace that n dedu

1r
n
H
converges uniform
to


ly
ˆ
1rg
on

2:2D


Hence, we have


 
1
1ˆ
rr
nn
r
nnn
GH
fg



(3.26)
spherically uniform on compact subsets of C. It follows
that

00G from

f
 whenever

0a
for
D
all of ze

, hence ros of G
have orst 4,
oles of
der at lea
all of p
G
have o least 2. Noting that

rder at

22rr
Gb Gd



 
22
2
nn
nnn
r
nnn nnnn
r
fa
f d
GG
 


  






(3.27)
If , then
 
20
r
GG





r
G


0, so

r
G
,

1
0
1
r
GC
r
 
for any C
. Since

00G
, then. Also 00C
since
G
ha
0, thi
s the zeroiplicity at least 4, then

G
s is a contradiction. Therefore,
 
2r
GG
s of mult



is not identically equal to zero.
If
 
2r
GG



for any C
, then
G
has no multiple poles and. Note that

0G


r
G
les, so
G
has only multiple po
at G
i on C
e th
s entire.
Also by Lemma 2.5, we hav

is not tran-
scendental in C, and then

G
isynomial. Thus, a pol
0
r
GC

, where 00C. We ha
ve
1r
Gr

  ,
then from
00G
and a multiplicities of every zeros
of
G
it follows that

0G
for any C
, this
Hence, is impossible.

2
GG
r


has some
zeros. In fact, by the sagument as the C
may dethat
me arase 1, we
duce

2r
GG

ha

 s a unique
zero 0
. Thus, we ha0
ve that either
is multi-
ple poles of
G
or 0
is a unique zero of
Gr
.
ilarlySim, if 0
is multiple poles of
G
, from
that
2r
GG
 

has a unique zero 0
it
atfollows th
r
G
for any C
. By Lemma
2.5, we have that
G
is not transcendentain by al. Ag
Lemm

G
a 2.6, we have that
is a constant, is which
a contradiction. Hence,
G
has nle pole and o multip
r
G
has a e zero 0
uniqu
. Thus,
G
is
entire on C and
r
G
has a unique zero 0
.
By Lemma 2.5, we have that
G
must be a polyno-
mial. Setting
 

2
2,
 
1
1
s
m
m
s
gA
 
 )
where, 12
,,,
m
(3.28
s
mmm
 are s posntegers, 4m itive ij
1, 2,,js
 ,
1
s
j
j
mm


G

0
l
rB
 
l is a positiv

)
 , .29 (3)
ve where e integer, l
Bl
3, we ha
1l
1l

r
Gr

0
, (3.30)
1
 

2
(3 1
0
1l
r
GrrBl
 
 . (3.31)
For
00G
, we have 00
and 0
j
. From
(3.29) it follows that 0j

,
d (3.31)
1, 2,,js
.
or 1,
From
ha
(3.29), (3.30)a, fn2,js,
, we
ve
l
0
r
jj
B


jj
rBl


1Bll
1
jj
(3.32)
(3.33)

0
1
1l
r

r

2
0
l

2
rr

, we have
 (3.34)
From (3.32) and (3.33)
,
j
rl 0,1r j

2,,s
 (3.35)
If lr
, then 00
, this is im
lr
possible. Therefore,
we have
, and so
Copyright © 2011 SciRes. APM
J. H. WANG ET AL.
Copyright © 2011 SciRes. APM
217
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f
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r
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From (3.33) and (3.34), we also have ,
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1
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r
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
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
00
1rr
 . Thus, we have 00
, a contra-
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
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k
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
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
0
n
F, we de-
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
theri
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al Families of Mero-
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
0
k
n
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k
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fz all

0,zN

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
k
n
f
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0, . 25
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There k
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
fore, for all e

112
,2
k
k
nrr
n
z z
M
zF
r
fz

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
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By Montel’s Theorem, is normal at ,
0
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Theorem 1.7 is given.
4.
ulo the referee for a numb
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Singularities of
Finite Order,”
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