A uniqueness polynomial for equi-polar meromorphic functions

Analysis 33, 13–23 (2013) / DOI 10.1524/anly.2013.1151c� Oldenbourg Wissenschaftsverlag, Munchen 2013

A uniqueness polynomial for equi-polarmeromorphic functions

Abhijit Banerjee�, Indrajit Lahiri

Received: November 9, 2011

Summary: In the paper we present a uniqueness polynomial for a class of meromorphic functionshaving the same set of poles.

1 Introduction, definitions and results

Let f be a non-constant meromorphic function in the open complex plane C and S bea set of distinct elements of C[ ¹1º. We put Ef .S/DS

a2S¹z W f .z/�a D 0º; wherezeros are counted with multiplicities.

F. Gross [6] exhibited the existence of three finite sets Sj .j D 1;2;3/ such that forany two non-constant entire functions f and g, Ef .Sj /D Eg.Sj / .j D 1;2;3/, impliesf � g. In 1982 F. Gross and C. C. Yang [7] found an infinite set S of complex numberssuch that for two non-constant entire functions f and g,Ef .S/DEg.S/ implies f � g.

Gross and Yang [7] called a set S a unique range set for entire functions (URSE inshort) if Ef .S/DEg.S/ implies f � g for any pair of entire functions. In an analogousmanner a unique range set for meromorphic functions (URSM in short) is defined.

It is seen that the finite URSE and URSM are zero sets of some polynomials. Thisobservation led P. Li and C. C. Yang [13] to introduce the idea of uniqueness polynomialfor entire and meromorphic functions. Let P.z/ be a polynomial in C. If P.f /� P.g/

implies f � g for any two non-constant meromorphic (entire) functions f and g, thenP.z/ is called a uniqueness polynomial for meromorphic (entire) functions. We say P isa UPM (UPE) in brief.

On the other hand, H. Fujimoto [5] introduced the idea of strong uniqueness poly-nomial for meromorphic (entire) functions (this terminology is used by T. T. H. An, J.T. Wang and P. Wang [1]) as a polynomial P.z/ in C such that P.f /� cP.g/ impliesf � g for any pair of non-constant meromorphic (entire) functions, where c is a suitablenon-zero constant. We say P is a SUPM (SUPE) in brief.

� Corresponding author: Abhijit BanerjeeAMS 2010 subject classification: 30D35Key words and phrases: Meromorphic function, uniqueness polynomial, unique range set

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14 Banerjee – Lahiri

The following polynomial introduced by G. Frank and M. Reinders [4] is one whichproduces a URSM with smallest available cardinality:

.n�1/.n�2/2

zn�n.n�2/zn�1C n.n�1/2

zn�2�b;where n� 11 and b 6D 0;1.

Carefully examining the anatomy of this polynomial, H. Fujimoto [5] discovereda unique property of a polynomial which characterises almost all available URSM pro-ducing polynomials. Fujimoto’s property (H) means that a polynomial is injective on theset of distinct zeros of its derivative. Since the set of distinct zeros of the derivative ofa polynomial is the set of its critical points, we may refer the property (H) as criticalinjective property (see [3]).

In 2011 T. T. H. An [2] first considered the possibility of a polynomial, not necessarilycritically injective, to be a UPM. In fact An [2] established a kind of equivalence betweenUPM and URSM for a kind of polynomial.

In 2004 T. T. H. An, J. T. Wang and P. Wong proved that P.z/ D .z�˛/nCa.z�˛/m C b is a UPM if and only if n � 5, n�m � 2, gcd .m;n/ D 1 and a 6D 0 (see [1,Corollary 1(iii)]). But the situation is completely different for UPE’s. Using the methodof proof of Theorem 2 [11] we see that zn Cazn�1 C b produces a URSE and so it isa UPE, where n� 7 and a, b are such that znCazn�1Cb has no multiple zero. This factsuggests that the poles have considerable influence on the characterisation of UPM’s.

In the paper we see that a hypothesis on the poles of a meromorphic function givesrise to a new class of UPM’s.

We say that two meromorphic functions are equi-polar if they have the same set ofpoles (ignoring multiplicities). A unique range set and a uniqueness polynomial for theclass of equi-polar meromorphic functions are, in short, expressed by URSM(EP) andUPM(EP). We present a polynomial for which these two notions are equivalent.

In 2001 the notion of weighted sharing was introduced [9, 10]. Let k be a non-negativeinteger or infinity. For a 2 C[ ¹1º we denote by Ek.aIf / the set of all a-points of f ,where an a-point of multiplicitym is countedm times ifm� k and kC1 times ifm> k.

If for two meromorphic functions f and g we have Ek.aIf / D Ek.aIg/, then wesay that f and g share the value a with weight k. The IM and CM sharing respectivelycorrespond to weight 0 and 1.

For S � C[ ¹1º we define Ef .S;k/ as

Ef .S;k/D[a2S

Ek.aIf /;

where k is a nonnegative integer or infinity. Clearly Ef .S/DEf .S;1/.A set S � C[ ¹1º is called a unique range set for meromorphic (entire) functions

with weight k if for any two non-constant meromorphic (entire) functions f and g,Ef .S;k/D Eg.S;k/ implies f � g. We write S is URSMk (URSEk) in short. If S isa unique range set for equi-polar meromorphic functions with weight k, we write S isURSMk(EP).

For standard definitions and notations of the value distribution theory we refer thereader to [8] and [14].

We now state the result of the paper.Brought to you by | Penn State - The Pennsylvania State University

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A uniqueness polynomial for equi-polar meromorphic functions 15

Theorem 1.1 Let P.z/D aznCQ.z/, Q.z/DmPjD0

aj zj and .z/D nQ.z/�zQ0.z/,

where aa0am 6D 0, a1 D am�1 D 0 and m � 2. We suppose that P.z/ and .z/ haveno common zero, P.z/ and Q.z/ have only simple zeros and l.� m/ be the number ofdistinct zeros of .z/.

Let S D ¹z W P.z/D 0º and

A.l;m/ D 2mC6�2min¹1;m� lº if N.r;1If / 6D S.r;f /; (1.1)

D max¹2lC5;2mC2º if N.r;1If /D S.r;f /: (1.2)

If n� A.l;m/, then the following statements are equivalent :

(i) P is a SUPM (EP);

(ii) S is a URSM2 (EP);

(iii) S is a URSM (EP);

(iv) P is a UPM (EP).

Further P is a UPM(EP) if any one of the following holds: (i) gcd.m;n/ D 1; (ii)am�t ¤ 0 for some t 2 ¹2;3; : : : ;m�2º and gcd.m;t/D 1.

2 LemmasIn this section we present some necessary lemmas. Let b;b1;b2; : : : ;bm bemC1 distinct

complex numbers. We put zj D b�bj for j D 1;2;: : : ;m and define �0 D 1, �1 DmPjD1

zj ,

�2 DmPj<k

zj zk , �3 DmP

i<j<k

zizj zk , : : :, �m D z1z2 : : : zm. It is said that a constant C

satisfies the condition (I) if �j .C j �1/D 0 for j D 1;2; : : : ;m. It is said that a constantK satisfies the condition (II) if Kj�m�j D �j�m for j D 1;2; : : : ;m (see [14, p. 482]).

Lemma 2.1 ( [14, p. 482]) Let f and g be non-constant meromorphic functions satisfy-ing N.r;0If /CN.r;1If / D S.r;f / and N.r;0Ig/CN.r;1Ig/ D S.r;g/. Supposethat S0 D ¹b1; b2; : : : ; bmº be a set of finite nonzero complex numbers. If Ef .S0;1/DEg.S0;1/, then either f � Cg or fg �K , where C and K satisfy the conditions (I)and (II) respectively (for b D 0).

Lemma 2.2 ( [14, p. 92]) Suppose that f1;f2; : : : ;fn (n� 3) are meromorphic functions

which are not constants except for fn. Furthermore, letnP

jD1fj � 1. If fn 6� 0 and

nXjD1

N.r;0Ifj /C .n�1/nXjD1

N.r;1Ifj / < .�Co.1// T .r;fk/ (2.1)

as r �! 1 except possibly for a set of finite linear measure, where k D 1;2; : : : ;n� 1and 0 < � < 1, then fn � 1.




Lemma 2.3 ( [14, p. 39]) Suppose f is a non-constant meromorphic function and k bea positive integer. Then

N.r;0If .k// �N.r;0If /CkN.r;1If /CS.r;f /:

Lemma 2.4 ([12]) Let f be a non-constant meromorphic function. Then

N.r;0If .1/f 6D 0/�N.r;0If /CN.r;1If /CS.r;f /;

where N.r;0If .1/f 6D 0/ is the counting function of those zeros of f .1/ which are notthe zeros of f .

Lemma 2.5 ( [14, p. 28 ]) Let f be a non-constant meromorphic function in the complexplane and P.z/, Q.z/ are polynomials with constant coefficients of respective degree pand q. Then

T

�r;P.f /

Q.f /

�D max¹p;qº T .r;f /CO.1/:

3 Proof of the theorem

Proof of Theorem 1.1: Suppose that (i) holds and for two non-constant equi-polarmeromorphic functions f and g, Ef .S;2/D Eg.S;2/. Let S D ¹�1; �2; : : : ; �nº D ¹z WP.z/D 0º and S� D ¹˛1;˛2; : : : ;˛mº D ¹z WQ.z/D 0º. Suppose thatR.z/D azn

Q.z/, then

R0.z/D azn�1 .z/

¹Q.z/º2 .

We put F D R.f /, G D R.g/ and H D . F00

F 0� 2F 0

F�1 /� . G00

G0� 2G0

G�1 /. Therefore

F 0 DR0.f /f 0 D af n�1 .f /f 0

¹Q.f /º2 and G0 D agn�1 .g/g 0

¹Q.g/º2 .

We suppose that H 6� 0. Then by lemma of logarithmic derivative we get m.r;H/DS.r;f /C S.r;g/.WD S.r//. Since Ef .S;2/ D Eg.S;2/, we see that F and G share.�1;2/. By Laurent expansion of H we can easily verify that each double zero of F C1

and GC1 is not a pole of H and each simple zero of F C1 and GC1 is a zero of H .Hence

N.r;�1IF j� 1/ D N.r;�1IG j� 1/ (3.1)

� N.r;0IH/� T .r;H/CO.1/

D N.r;1IH/CS.r/;

where N.r;�1IF j� 1/ denotes the counting function of simple zeros of F C1.Brought to you by | Penn State - The Pennsylvania State University



Since f and g are equi-polar functions and Ef .S;2/D Eg.S;2/, we obtain

N.r;1IH/ (3.2)

D N.r;1IH/� N.r;0If /CN.r;1If /CN.r;0Ig/CN �.r;�1IF;G/

ClX

jD1

®N.r; j If /CN.r; j Ig/¯CN 0.r;0If 0/CN 0.r;0Ig0/;

whereN �.r;�1IF;G/ denotes the reduced counting function of those common �1-pointsof F and G, where each such common �1-point of F and G has different multiplicitiesrelated toF andG. AlsoN0.r;0If 0/ (N 0.r;0If 0/) denotes the counting function (reduced

counting function) of those zeros of f 0 which are not the zeros of f .F C1/lQ

jD1.f � j /;

where j ’s are distinct zeros of .z/.By the second fundamental theorem we get from (3.1) and (3.2) by Lemma 2.3

.nC l/T .r;f / (3.3)

� N.r;1If /CN.r;0If /ClX

jD1N.r; j If /C

nXkD1

N.r;�k If /

�N0.r;0If 0/CS.r;f /

� N.r;1If /CN.r;0If /ClX

jD1N.r; j If /CN.r;�1IF /

�N0.r;0If 0/CS.r;f /

D 2N.r;1If /C2N.r;0If /CN.r;0Ig/ClX

jD1

®2N.r; j If /

CN.r; j Ig/¯CN.r;�1IG j� 2/CN.r;�1IG j� 3/

CN 0.r;0Ig0/CS.r/

� 2N.r;1If /C2N.r;0If /CN.r;0Ig/ClX

jD1

®2N.r; j If /

CN.r; j Ig/¯CN.r;0Ig0/CS.r/

� 3N.r;1If /C2N.r;0If /C2N.r;0Ig/C2lT .r;f /C lT .r;g/CS.r/

� .2lC2/T .r;f /C .lC2/T .r;g/C3N.r;1If /CS.r/:

We first suppose thatN.r;1If /D S.r;f /. Then from (3.3) we get .nC l/T .r;f /�.2lC2/T .r;f /C .lC2/T .r;g/CS.r/.

We denote by T .r/ the maximum of T .r;f / and T .r;g/. Then from above we obtain.nC l/T .r;f /� .3lC4/T .r/CS.r/.

Similarly we get .nC l/T .r;g/ � .3l C 4/T .r/C S.r/. Therefore .nC l/T .r/ �.3lC4/T .r/CS.r/, which is a contradiction as n� A.l;m/.




Next we suppose that N.r;1If / 6D S.r;f /. We put V D F 0

F.FC1/ � G0

G.GC1/ : SinceH 6� 0, we have F 6�G and so V 6� 0.

Suppose that z0 is a pole of f and g with multiplicities p and q respectively. Thenz0 is a pole of F and G with multiplicities .n�m/p and .n�m/q respectively. So z0 isa zero of V with multiplicity at least .n�m/�1. Since f and g are equi-polar, we getfrom Lemma 2.4

.n�m�1/N.r;1If /� N.r;0IV /� N.r;1IV /CS.r/

� N.r;0IF /CN.r;0IG/CN �.r;�1IF;G/CS.r/

� N.r;0If /CN.r;0Ig/C 1

2¹N.r;�1IF /�N.r;�1IF /º CS.r/


2

nXjD1

®N.r;�j If /�N.r;�j If /¯


2N.r;0If .1/ j f 6D 0/CS.r/

� 3

2N.r;0If /C 1

2N.r;1If /CN.r;0Ig/CS.r/:

Therefore �n�m� 3

2

�N.r;1If /� 3

2N.r;0If /CN.r;0Ig/CS.r/: (3.4)

Now from (3.3) and (3.4) we get

.nC l/T .r;f / � .2lC2/T .r;f /C .lC2/T .r;g/

C 3

n�m� 32

²3

2N.r;0If /CN.r;0Ig/

³CS.r/

��2lC2C 9

2n�2m�3�T .r;f /

C�lC2C 6

2n�2m�3�T .r;g/CS.r/

��3lC4C 15

2n�2m�3�T .r/CS.r/:

Similarly we have

.nC l/T .r;g/ ��3lC4C 15

2n�2m�3�T .r/CS.r/:

Therefore

.nC l/T .r/��3lC4C 15

2n�2m�3�T .r/CS.r/;




which is a contradiction as n� A.l;m/.Hence H � 0. By integrating two sides of H � 0 we get

F �QAGC QBQCGC QD ; (3.5)

where QA, QB , QC , QD are constants such that QA QD� QB QC 6D 0. Now by Lemma 2.5 we have

T .r;f /D T .r;g/CO.1/: (3.6)

We note that F is non-constant. For, if F is a constant, then af n D QK.f �˛1/.f �˛2/ � � �.f �˛m/, where QK is a constant. This shows that f omits the values 0;˛1;˛2; : : : ;˛m, which is impossible as m� 2.

Let QA QC 6D 0. If QB 6D 0, from (3.5) we get N�r;� QB

QA IG�

D N.r;0IF / and so by the

second fundamental theorem we get in view of (3.6)

nT .r;g/ D T .r;G/CO.1/

� N.r;0IG/CN.r;1IG/CN

r;�

QBQA IG

!CS.r;G/

� N.r;0Ig/CN.r;1Ig/CmXjD1

N.r; j Ig/CN.r;0If /CS.r;g/

� .mC3/T .r;g/CS.r;g/;

a contradiction as n� A.l;m/.Therefore QB D 0 and from (3.5) we obtain

F �QAG

QCGC QD: (3.7)

Since F is non-constant, we see that QD 6D 0. From (3.7) we have N.r;� QDQC IG/ D

N.r;1IF / and so by the second fundamental theorem and (3.6)

nT .r;g/

� N.r;0IG/CN.r;1IG/CN

r;�

QDQC IG

!CS.r;G/

� N.r;0Ig/CN.r;1Ig/CN.r;1If /CmXjD1

¹N.r; j If /CN.r; j Ig/º CS.r;g/

� .A.l;m/�1/T .r;g/CS.r;g/;

a contradiction as n� A.l;m/.Therefore QA QC D 0. Since QA QD� QB QC 6D 0, we have j QA j C j QC j6D 0 Let QA D 0 and

QC 6D 0. Then QB 6D 0 and from (3.5) we get F � QBQCGC QD . If QD 6D 0, by the second

fundamental theorem we get as above a contradiction because n� A.l;m/.Brought to you by | Penn State - The Pennsylvania State University



So QD D 0 and FG � QBQC . Since QA D 0 and QA QD� QB QC 6D 0, we have QB 6D 0. Let z0

be a pole of f and g. Then z0 is a pole of F and G, which is impossible as FG � QBQC .

So f has no pole. Now from FG � QBQC we see that a zero of f � j is a zero of gn for

j 2 ¹1;2; : : : ;mº. Therefore N.r; j If / � 1nN.r; j If / for j D 1;2; : : : ;m. Hence by

the second fundamental theorem we get

.m�1/T .r;f / �mXjD1

N.r; j If /CS.r;f /� m

nT .r;f /CS.r;f /;

a contradiction.Finally we suppose that QA 6D 0 and QC D 0. Since QA QD� QB QC 6D 0, we have QD 6D 0. If

QB 6D 0 then by the second fundamental theorem and (3.6) we get

nT .r;f / � N.r;0IF /CN.r;1IF /CN

r;

QBQD IF

!CS.r;F /

� N.r;0If /CN.r;1If /CN.r;0Ig/CmXjD1

N.r; j If /CS.r;f /

� .mC3/T .r;f /CS.r;f /;

a contradiction as n � A.l;m/. So QB D 0 and F � QAQDG. Now F C 1 D P.f /

Q.f /D

a.f��1/.f��2/:::.f��n/Q.f /

. Since f is non-constant andQ.�j / 6D 0 for j D 1;2;: : : ;n, we seethat F assumes the value �1 and soG assumes the value �1 at the same point. Therefore

QAQD D 1 and F �G. This implies

f nQ.g/� gnQ.f /: (3.8)

Since f and g are equi-polar, from (3.8) we see that f and g share .0;1/, .1;1/.We put h D f

g. Then h is an entire function which has no zero. We suppose that h is

non-constant. From (3.8) we get

mXkD0

8<:akf k

n�kXjD1

hn�k�j9=;� 0; (3.9)

where a1 D am�1 D 0.Now by (3.6) we get T .r;h/� T .r;f /CT .r;g/CO.1/D 2T .r;f /CO.1/. Let z0

be a pole of f with multiplicity p.� 1/. Since h is entire, from (3.9) we see that z0 isa zero of

n�mXjD1

hn�m�j D .h�u1/.h�u2/ : : : .h�un�m�1/;

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Let z0 be a zero of h� uk with multiplicity r . If r < mp, then z0 is a pole of

f mn�mPjD1

hn�m�j with multiplicity mp� r . From (3.9) we see that this pole must be

nutralised by the pole of some other terms. Suppose that it is nutralised by the pole

of the term f m�t n�mCtPjD1

hn�mCt�j . If h�uk is a factor ofn�mCtPjD1

hn�mCt�j , then we

have mp� r D .m� t/p� r and so p D 0, a contradiction. So h�uk is not a factor

ofn�mCtPjD1

hn�mCt�j and we have mp� r D .m� t/p. Hence r D tp � 2p � 2 because

am�1 D 0. Therefore r � min.mp;tp/� 2p � 2 and so we get by Lemma 2.3

N.r;1If / �n�m�1XkD1

N.r;uk Ih j� 2/

� 2N.r;0Ih0/� 2N.r;0Ih/C2N.r;1Ih/CS.r;h/

D S.r;f /;

where N.r;uk Ih j� 2/ denotes the counting function of multiple zeros of h�uk .Now putting f1 D 1

fin (3.9) we get

mXkD0

8<:akf m�k

1

n�kXjD1

hn�k�j9=;� 0: (3.10)

Since a zero of f is a pole of f1 and a1 D 0, proceeding as above we get from (3.10)

N.r;0If /DN.r;1If1/D S.r;f /:

Since f and g share .0;1/, .1;1/, we have by (3.6) that N.r;0Ig/D S.r;g/ andN.r;1Ig/D S.r;g/.

Also from (3.8) we see that Ef .S�;1/D Eg.S�;1/. Therefore by Lemma 2.1 we

get f � Cg or fg �K , where C and K are two constants satisfying the conditions (I)and (II) respectively. Also we note that �1 D 0, �2 D am�2

am, : : :, �m�1 D 0 and �m D a0

am.

First we suppose that fg �K . Then f and g do not have any zero and pole. From(3.8) we get

�mXjD1

aj

a0gj C

mXjD1

aj

a0

g2n�j

Kn�j C 1

Kng2n � 1; (3.11)

where a1 D am�1 D 0.Since the left hand side of (3.11) contains at least three terms, by Lemma 2.2 we get

g2n � Kn, a contradiction as g is non-constant. Therefore h is constant and f � Cg.From (3.8) we get

1C af n

Q.f /� 1C agn

Q.g/Brought to you by | Penn State - The Pennsylvania State University



i.e.P.f /

Q.f /� P.g/

Q.g/:

This together with (3.8) implies

f nP.g/� gnP.f / (3.12)

i.e.,C nP.g/� P.f /:

So by (i) f � g. Hence (i) implies (ii).Next we suppose that (ii) holds. If Ef .S;1/D Eg.S;1/ then clearly Ef .S;2/D

Eg.S;2/ and so by (ii) we have f � g. Therefore (ii) implies (iii).Now we suppose that (iii) holds. IfP.f /DP.g/, then clearlyEf .S;1/DEg.S;1/.

So by (iii) f � g. Hence (iii) implies (iv).Since every UPM(EP) is clearly a SUPM(EP), (iv) implies (i).We first suppose that gcd.m;n/D 1. Sincef , g are equi-polar, (3.12) implies that f , g

share .0;1/ and .1;1/ and so from (3.12) we haveEf .S;1/DEg.S;1/. Now usingLemma 2.1 and reasoning as above we get f �Dg, whereD is a constant satisfying thecondition (I), where �n D a0

a. Since f �Cg, we haveC DD andCm�1DDn�1D 0.

Now there exist two integers u and v such that umCvnD 1 and so C D C umCvn D 1.Hence f � g and S is a URSM2(EP). This implies P is a UPM(EP).

Next we suppose that am�t 6D 0 for some t 2 ¹2;3; : : : ;m� 2º with gcd.m; t/ D 1.Then �m D a0

am6D 0 and �t D am�t

am6D 0. So by the condition (I) we get Cm�1D 0 and

C t �1D 0. Therefore as above C D 1 and P is a UPM(EP). This proves the theorem. �

Acknowledgements. The authors are thankful to the referee for valuable suggestions.

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Abhijit BanerjeeDepartment of MathematicsWest Bengal State UniversityKolkata-700126India

Present address:Department of MathematicsUniversity of KalyaniWest Bengal 741235India

[email protected],[email protected]

Indrajit LahiriDepartment of MathematicsUniversity of KalyaniWest Bengal [email protected]



A uniqueness polynomial for equi-polar meromorphic functions

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