ON THE UNIQUENESS PROBLEM FOR MEROMORPHIC FUNCTIONS WITH HIGHER MULTIPILCITIES OF ZEROS AND POLES

In this paper, we show that the equation P(f1, ..., fs+1) = Q(g1, ..., gs+1), where P, Q are polynomials in a class of homogeneous polynomials of FermatWaring type, has entire solutions f1, · · · , fs+1; g1, · · · , gs+1. Some classes of unique range sets for linearly nondegenerate holomorphic curves are also obtained

ON THE UNIQUENESS PROBLEM FOR MEROMORPHIC FUNCTIONS WITH HIGHER MULTIPILCITIES OF ZEROS AND POLES VU HOAI AN, HA HUY KHOAI, AND NGUYEN XUAN LAI Abstract. Assume the polynomial P (z) = (z − a1 )...(z − aq ), associated to the set S = {a1 , ..., aq } ⊂ C, satisfies Fujimoto’s condition. We give some sufficient conditions for S to be a unique range set for meromorphic(entire) functions. With some additional conditions on multiplicities of zeros and poles, we reduce the cardinalities of unique range sets for meromorphic functions. 1. Introduction and main results In this paper, by a meromorphic function we mean a meromorphic function in the complex plane C. We assume that the reader is familiar with the notations in the Nevanlinna theory (see [2-4 ],[6-7],[11]). Let f be a non-constant meromorphic function on C. For every a ∈ C, define the function νfa : C → N by νfa (z) =  0 d if f (z) = a if f (z) = a with multiplicity d, and set νf∞ = ν 01 , and define the function ν af : C → N by ν af (z) = min {νfa (z), 1}, f 0 and set ν ∞ f = ν 1 . Let m be a positive integer. For every a ∈ C ∪ {∞}, define the f ≤m function ν(f,a) from C ∪ {∞} into N by ≤m ν(f,a) (z) =  0 if νfa (z) > m ν a (z) if ν a (z) ≤ m, f f ≤m ≤m and define the function ν ≤m (f,a) from C ∪ {∞} into N by ν (f,a) (z) = min {ν (f,a) (z), 1}. ≥m ≤m Similarly, we define ν(f,a) (r, f ), , ν ≥m (f,a) . Next, define the counting function N 1 N ≤m (r, f −a ), N N ≥m ≤m (r, f ), N ≤m 1 1 (r, f −a ), N ≥m (r, f ), N ≥m (r, f −a ), N ≥m (r, f ), 1 (r, f −a ). For f ∈ M(C) and S ⊂ C ∪ {∞}, we define {(z, νfa (z)) : z ∈ C}. Ef (S) = a∈S In case ν af (i.e., ignoring multiplicity) we denote E f (S) (this is the preimages of S). 2010 Mathematics Subject Classification. Primary 11S80, Secondary 30D35. 1 The work was supported by the National Foundation for Science and Technology Development (NAFOSTED) and the Vietnam Institute for Advanced study in Mathematics (VIASM). 1 Let m is a positive integer or ∞, we define Ef≤m (S) = ≤m {(z, ν(f,a) (z)) : z ∈ C}. a∈S Note that, if m is ∞, then Ef∞ (S) = Ef (S) and if m = 1, then Ef≤1 (S) = E f (S). Let F be a nonempty subset of M(C). Two functions f, g of F are said to share S, counting multiplicity, (share S CM), if Ef (S) = Eg (S) and to share S, ignoring multiplicity, (share S IM), if E f (S) = E g (S). Let a set S ⊂ C ∪ {∞} and f and g be two non-constant meromorphic (entire) functions. If Ef (S) = Eg (S) implies f = g for any two non-constant meromorphic (entire) functions f, g, then S is called a unique range set for meromorphic(entire) functions or, in brief, URSM (URSE). A set S ⊂ C ∪ {∞} is called a unique range set for meromorphic (entire) functions ignoring multiplicity (URSM-IM) (URSE-IM) if E f (S) = E g (S) implies f = g for any pair of non-constant meromorphic (entire) functions. A set S ⊂ C ∪ {∞} is called a URSM≤m (URSE≤m ) if for any two non-constant meromorphic (entire) functions f, g, Ef≤m (S) = Eg≤m (S), implies f = g. In 1926, R. Nevanlinna discovered his famous five-value uniqueness theorem which says that if two non-constant meromorphic functions f and g on the complex plane C share ve distinct values IM then f = g. A few years later, he showed that when multiplicities are considered, 4 points are sufficient to determine the functions and in this case either the functions coincide or one is the bilinear transformation of the other. In 1982 Gross and Yang [ 8 ] proved the following theorem: Theorem A [ 8 ] Let S = {z ∈ C: ez + z = 0}. If two entire functions f, g satisfy Ef (S) = Eg (S), then f = g. It is to be observed that since the range set S given in Theorem A is an innite set, Theorem A cannot be considered as an exact solution to the problem of Gross. After the introduction of the novel idea of unique range sets researchers were getting more involved to nd new unique range sets with cardinalities as small as possible. In 1994, Yi [13] exhibited a URSE with 15 elements and in 1995 Li and Yang [12] exhibited a URSM with 15 elements and a URSE with 7 elements. Till date the URSM with 11 elements are the smallest available URSM, obtained by Frank and Reinders [5]. The URSM discovered by Frank and Reinders is highlighted by a number of researchers. A polynomial P in C, is called a uniqueness polynomial for meromorphic (entire) functions, if for any two non-constant meromorphic (entire) functions f and g, P (f ) = P (g) implies f = g. We say P is a UPM (UPE) in brief. A polynomial P in C, is called a strong uniqueness poly- nomial for meromorphic (entire) functions, if P (f ) = cP (g) implies f = g for any two non-constant meromorphic (entire) functions f and g , and any nonzero constant c ∈ C. We sayP is a SUPM (SUPE) in brief. H. Fujimoto [6] proved the following theorem 2 Theorem B [6] Suppose that P (z) is a strong uniqueness polynomial of form (1.1) satisfying condition (1.2), and that k ≥ 3, or k = 2 and min{q1 , q2 } ≥ 2. Then, S is a URSM (resp. URSE) whenever q > 2k + 6 (resp. q > 2k + 2),while a URSM-IM (resp. URSE-IM) whenever q > 2k + 12 (resp.q > 2k + 5). In 2009 Bai, Han and Chen [2] proved the following truncated sharing version of Theorem B. Theorem C [2] In addition to the hypothesis of Theorem B we suppose that m is a positive integer or ∞. Let S be the set of zeros of P. If 1. m ≥ 3 or ∞ and q > 2k + 6(2k + 2); 2. m = 2 and q > 2k + 7(2k + 2); 3. m = 1 and q > 2k + 10(2k + 4), then S is a URSM≤m (URSE≤m ). We recall the URSM introduced by Frank and Reinders [5] which is the zero set of PF R (z) = (n − 1)(n − 2) n n(n − 1) n−2 z − n(n − 2)z n−1 + z − c(c = 0, 1). 2 2 Clearly PF R (z) has two distinct zeros that is here k = 2 and has a zero at 0 of order n − 3. Second type of URSM is demonstrated by Yi in [14] which is the zero set of PY (z) = z n + az n−r + b, (a, b = 0), where n, r are two positive integers having no common factors, n ≥ 2r + 9, r ≥ 2 and a = 0 and b = 0 are so chosen so that P has n distinct zeros. Clearly PY (z) has a zero at 0 of order n − r − 1. In 2014, Banerjee[4 ] introduced the following polynomial p PB (z) = i=0 p i where (−1)i z n+p+1−i + a = QB (z) + a, n+p+1−i p QB (z) = i=0 p i (1.3) (−1)i z n+p+1−i , a = 0, n+p+1−i is a constant. Clearly PB (z) = z n (z − 1)p , and has a zero at 0 of order n. From this polynomial, Banerjee[4 ] proved the following theorem. Theorem D [4] Let n, p ≥ 3(p ≥ 2) be two positive integers. Now when a = 1, n ≥ p + 3(n ≥ p + 2), or when a = 1, n ≥ 3(n ≥ 2), then PB given by (1.3), is a SUPM (SUPE). To reduce the cardinalities of the range sets further in the application part of [4] the following theorem was proved. 3 Theorem E [4] In addition to the hypothesis of Theorem C we suppose that m is a positive integer or ∞. Let S be the set of zeros of P. If 1. m ≥ 3 or ∞ and min{Θ(∞; f ), Θ(∞; g)} > 6+2k−q ; 4 14+4k−2q ; 9 10+2k−q and min{Θ(∞; f ), Θ(∞; g)} > , 6 ≤m 2. m = 2 and min{Θ(∞; f ), Θ(∞; g)} > 3. m = 1 and then S is a URSM . In this paper we consider polynomials: P (z) l of the form (1.1) satisfying the condition (1.2) with P (z) = q(z − d1 )m1 (z − d2 )m2 ...(z − dk )mk , where d1 = 0; or p p P (z) = (n + p + 1) i i=0 (−1)i z n+p+1−i bi + a = Q(z) + a, n+p+1−i (1.4) where p Q(z) = (n + p + 1) i=0 p i (−1)i z n+p+1−i bi , a, b = 0, are constants, n+p+1−i QB (1)bn+p+1 = −2. (1.5) Clearly P (z) = (n + p + 1)z n (z − b)p , and has a zero at 0 of order n. By unsing these polynomials, we obtained the following results, which improve Theorems B, C, D, E. Theorem 1.1. Let f and g be two non-constant meromorphic(entire) functions and m be a positive integer, or ∞. Let P (z) be a strong uniqueness polynomial of the form (1.1) satisfying the condition (1.2) with P (z) = q(z − d1 )m1 (z − d2 )m2 ...(z − dk )mk , where d1 = 0. Suppose that q ≥ 5, k ≥ 3, or k = 2 and min{m1 , m2 } ≥ 2, and Ef≤m (S) = Eg≤m (S), and all of the zeros and poles of f and g are of multiplicities at least s, l, respectively. Assume that one of the following conditions is satisfied: 1. q > 2k − 2 + 4s + 4l (q > 2k − 2 + 4s ) when m ≥ 3 or ∞; 2. q > 2k − 32 + 4s + 2l9 (q > 2k − 32 + 4s ) when m = 2; 3. q > 2k + 4s + 6l (q > 2k + 4s ) when m = 1. Then f = g. Theorem 1.2. Let f and g be two non-constant meromorphic (entire) functions and P (z) be a polynomial of the form (1.4) satisfying the condition (1.5). Suppose that P (f ) = cP (g), c = 0, and all of the zeros and poles of f and g are of multiplicities at least s, l, respectively, and p > 1 + 1l , np > p + n. Assume that one of the following conditions is satisfied: 1. n > 1s + 1l (n > 1s ) when a = 1; 2. n > p + 1s + 1l (n > p + 1s ) when a = 1. Then f = g. 4 Theorem 1.3. Let f and g be two non-constant meromorphic (entire) functions and P (z) be a polynomial of the form (1.4) satisfying the condition (1.5) and m be a positive integer, or ∞. Suppose that Ef≤m (S) = Eg≤m (S), and all of the zeros and poles of f and g are of multiplicities at least s, l, respectively, and p > 1 + 1l , np > p + n. Assume that for the following two systems of conditions, one of (I) and one of (II) are satisfied. (I) 1. n + p > 2k − 3 + 4s + 4l (q > 2k − 3 + 4s ) when m ≥ 3 or ∞; 2. n + p > 2k − 25 + 4s + 2l9 (q > 2k − 52 + 4s ) when m = 2; 3. n + p > 2k − 1 + 4s + 6l (q > 2k − 1 + 4s ) when m = 1. (II) 4. n > 1s + 1l (n > 1s ) when a = 1; 5. n > p + 1s + 1l (n > m + 1s ) when a = 1. Then f = g. Remark 1.4. Take s = l = 1 in Theorem 1.1 (Theorem 1.2), we obtain Theorem C (Theorem D). Remark 1.5. Take some values of s, l in Theorem 1.1, we get the unique range sets with 11, 9 and 7 elements. (i) If s = l = 1(s = 1), then q > 10(q > 6); (ii) If either s = 2 and l = 1 (s = 2) or s = 1 and l = 2, then q > 8 (q > 4); (iii) If s = 2 and l = 2, then q > 6.... Remark 1.6. Take in Theorem 1.2 s = 2, l = 2 and p > 32 , np > p + n we have: 1. n > 1 (n > 21 ) when a = 1; 2. n > p + 1 (n > p + 12 ) when a = 1. Remark 1.7. Take in Theorem 1.3 m ≥ 3 or ∞ and s = 2, l = 2 and p > np > p + n we have: 1. n + p > 5 and n > 1 when a = 1; 2. n + p > 5 and n > p + 1 when a = 1. 3 , 2 2. Preliminaries H. Fujimoto [7] proved the following: Lemma 2.1. Let P (z) be a polynomial of the form (1.1) satisfying the condition (1.2). Then P (z) is a uniqueness polynomial if and only if k ql qm > ql . i=1 1≤l 2k − 2 + 4s + 4l (q > 2k − 2 + 4s ) when m ≥ 3 or ∞. ≤1 ≤1 1 1 By the hypothesis we have N (r, P (f ) = N (r, P (g) ). From this and Lemma 2.4 ) we have 1 1 1 1 1 ≥m+1 ≥m+1 ≤1 ) + N (r, )+N )+N ) − N (r, )= N (r, (r, (r, P (f ) P (g) P (f ) P (g) P (f ) 1 1 1 ≤1 1 1 1 ≤1 1 ≥m+1 ≥m+1 N (r, )+N (r, ) − N (r, )+N (r, ) − N (r, ) P (f ) P (f ) 2 P (f ) P (g) 2 P (f ) 1 1 1 q ≤ (N (r, ) + N (r, )) ≤ T (r) + S(r). (2.11) 2 P (f ) P (g) 2 From (2.10)- (2.11) we have q 4 4 1 1 (q+k−1)T (r) ≤ 2( + +k−1)T (r)+ T (r)+S(r); (q+2k+2− − )T (r) ≤ S(r). l s 2 s l 4 This is a contradiction to the assumption that q > 2k − 2 + s + 4l . Case2. q > 2k − 32 + 4s + 2l9 (q > 2k − 23 + 4s ) when m = 2. Applying Lemma 2.2 we have N ≥3 (r, 1 )= P (f ) k N i=1 ≥3 (r, 1 1 1 1 1 1 1 ≥2 ) ≤ N (r, ) ≤ N (r, ) ≤ N (r, )+ N (r, f ) f − ai f 2 f 2 f 2 1 1 1 1 +S(r, f ) ≤ T (r, f ) + N (r, f ) + S(r, f ) ≤ (1 + )T (r, f ) + S(r, f ). (2.12) 2 2 2 l Similarly, 1 1 1 ≥3 N (r, ) ≤ (1 + )T (r, g) + S(r, g). (2.13) P (g) 2 l From (2.12), (2.13) and Lemma 2.4 we obtain 1 1 1 1 1 ≥3 ≥3 ≤1 N (r, ) + N (r, ) + N (r, ) + N (r, ) − N (r, )= P (f ) P (g) P (f ) P (g) P (f ) 11 1 ≥3 1 ≤1 1 ≥3 1 1 1 1 ) + N (r, ) − N (r, ) + N (r, ) P (f ) 2 P (f ) 2 P (f ) 2 P (f ) 1 ≥3 1 ≤1 1 ≥3 1 1 1 1 +N (r, ) + N (r, ) − N (r, ) + N (r, ) P (g) 2 P (g) 2 P (g) 2 P (g) 1 1 1 1 1 1 1 1 ≤ N (r, ) + T (r, f ) + N (r, f ) + N (r, ) + T (r, g) + N (r, g) + S(r) 2 P (f ) 4 4 2 P (g) 4 4 q 1 1 q 1 1 ≤ T (r) + T (r) + T (r) + S(r) = ( + + )T (r) + S(r). (2.14) 2 4 4l 2 4 4l By (2.10), (2.14) we obtain q 1 1 1 1 (q + k − 1)T (r) ≤ 2( + + k − 1)T (r) + ( + + )T (r) + S(r); l s 2 4 4l 3 4 9 (q − 2k + − − )T (r) ≤ S(r). 2 s 2l This is a contradiction to the assumption that q > 2k − 32 + 4s + 2l9 . Case3. q > 2k + 4s + 6l (q > 2k + 4s ) when m = 1. Applying Lemma 2.2 we have N (r, N ≥2 (r, 1 )= P (f ) k N ≥2 i=1 (r, 1 1 1 ) ≤ N (r, ) ≤ N (r, ) + N (r, f ) + S(r, f ) f − ai f f 1 ≤ T (r, f ) + N (r, f ) + S(r, f ) ≤ (1 + )T (r, f ) + S(r, f ). l (2.15) Similarly, 1 1 1 ) ≤ (1 + )T (r, g) + S(r, g). (2.16) P (g) 2 l From (2.15), (2.16) and Lemma 2.4 we obtain 1 1 1 1 1 ≥2 ≥2 ≤1 N (r, ) + N (r, ) + N (r, ) + N (r, ) − N (r, )= P (f ) P (g) P (f ) P (g) P (f ) 1 ≤1 1 ≤1 1 1 1 1 1 ≥2 N (r, ) − N (r, ) + N (r, ) + N (r, ) − N (r, ) P (f ) 2 P (f ) P (f ) P (g) 2 P (g) 1 1 1 1 1 1 1 ≥2 +N (r, ) ≤ N (r, ) + (1 + )T (r, f ) + N (r, ) + (1 + )T (r, g) P (g) 2 P (f ) l 2 P (g) l q 1 +S(r) ≤ T (r) + (1 + )T (r) + S(r). (2.17) 2 l By (2.10), (2.17) we obtain q 1 1 1 (q + k − 1)T (r) ≤ 2( + + k − 1)T (r) + ( + 1 + )T (r) + S(r); l s 2 l 4 6 (q − 2k − − )T (r) ≤ S(r). s l This is a contradiction to the assumption that q > 2k + 4s + 6l . So L ≡ 0. Therefore, 1 = Pc(f0 ) + c1 for any two constants c0 (= 0) and c1 . By Lemma 2.5 we have c1 = 0. P (f ) Thus, there is constant c(= 0) such that P (f ) = cP (g). N ≥2 (r, 12 In particular, in the case where f and g are entire, we applying the result of the case where f and g are meromorphic and take l = ∞. Banerjee[4] proved the following: Lemma 2.7. QB (1) is not an integer. In particular, when a = 1, P (1) = 1, wheren ≥ 3, m ≥ 3 are integers. 3. Proof of Theorems Now we use the above Lemmas to prove the main result of the paper. Proof of Theorem 1.1. Since P (z) isa strong uniqueness polynomial ,we obtain f = g. Proof of Theorem 1.2. We have P (0) = Q(0) = a, P (b) = Q(b)+a = QB (1)bn+m+1 + a. By QB (1) = 0, b = 0, we have P (b) = P (0). Set F = P (f ), G = P (g). From P (f ) = cP (g), c = 0, it implies F = cG, T (r, f ) = T (r, g) + S(r, g); S(r, f ) = S(r, g). (3.1) Now we consider the following cases: Case1. c = 1. Subcase1.1. c = P (b). Then P (b) = c = 1 and F − P (b) = P (b)(G − 1). We have P (0) = a. Suppose a = 1. Then P = 0 has zeros at 0, b and P (0) − 1 = 0, P (b) − 1 = 0. It follows that all the zeros of P (z) − 1 are simple, and let them be bi , i = 1, 2, ..., n + p + 1. Clearly ,P (z) − P (b) has a zero at b of order p + 1. Let the remaining distinct zeros of P (z) − P (b) be ci , i = 1, 2, ..., n. Then N (r, n+p+1 n 1 1 1 )= ). )+ N (r, N (r, f −b f − ci f − bi i=1 i=1 (3.2) Applying the second main theorem to the functions f and the values b1 , b2 , ..., bn+p+1 , ∞, and by (3.1), (3.2) we get n+p+1 (n + p)T (r, g) ≤ N (r, g) + N (r, i=1 1 ) + S(r, g) g − bi n 1 1 1 ≤ T (r, g) + N (r, )+ N (r, ) l f −b f − ci i=1 1 1 ≤ T (r, g) + (n + 1)T (r, f ) + S(r, f ) ≤ (n + 1 + )T (r, g) + S(r, g) l l 1 (p − 1 − )T (r, g) ≤ S(r, g). l This is a contradiction to the assumption that p > 1 + 1l . Next suppose a = P (0) = 1. Then from (3.1) we have 1 F − 1 = P (b)(G − ). P (b) 13 (3.3) We consider P (z)− P 1(b) . By P (0) = a = 1 and P (b) = c = 1 we obtain P (0)− P 1(b) = 0. Moreover P (b) = QB (1)bn+p+1 +a = QB (1)bn+p+1 +1 = −1. Therefore, P (z)− P 1(b) has only simple zeros. Let they be given by bi , i = 1, 2, ..., n + p + 1. We also let ci , i = 1, 2, ..., p, be distinct simple zeros of P (z) − P (0) = P (z) − a = P (z) − 1. Applying the second main theorem to the functions f and the values b1 , b2 , ..., bn+p+1 , ∞, and by (3.1), (3.3) we get n+p+1 (n + p)T (r, g) ≤ N (r, g) + N (r, i=1 1 ) + S(r, g) g − bi n 1 1 1 ≤ T (r, g) + N (r, ) + N (r, ) l f f − ci i=1 1 1 1 1 ≤ T (r, g) + T (r, f ) + pT (r, f ) + S(r, f ) ≤ (p + + )T (r, g) + S(r, g) l s l s 1 1 (n − − )T (r, g) ≤ S(r, g). s l This is a contradiction to the assumption that n > 1s + 1l . Subcase1.2. c = P (b). Suppose a = 1. Then from (3.1) we have F − ac = c(G − a). (3.4) Consider P (z) − ac. By c = 1 we have ac = a. So P (0) − ac = a − ac = 0. Suppose ac = P (b). Then P (z) − ac has only simple zeros. Let they be given by ei , i = 1, 2, ..., n + p + 1. Clearly, P (z) − P (0) has a zero at 0 of order n + 1. Let the remaining distinct zeros of P (z) − P (0) be ei , i = 1, 2, ..., p. Applying the second main theorem to the functions f and the values ei , i = 1, 2, ..., n + p + 1, ∞, and by (3.1), (3.4) we get n+p+1 (n + p)T (r, f ) ≤ N (r, f ) + N (r, i=1 1 ) + S(r, f ) f − ei p 1 1 1 ≤ T (r, f ) + N (r, ) + N (r, ) l g g − ei i=1 1 1 1 1 ≤ T (r, g) + T (r, f ) + pT (r, f ) + S(r, f ); (n − − )T (r, g) ≤ S(r, g). l s s l 1 1 This is a contradiction to the assumption that n > s + l . Suppose ac = P (b). Then P (z) − P (b) has a zero at b of order p + 1. Let the remaining distinct zeros of P (z) − P (b) be ci , i = 1, 2, ..., n. Applying the second main theorem to the functions f and the values b, ci , i = 1, 2, ..., n. we have n 1 1 nT (r, f ) ≤ N (r, f ) + N (r, )+ N (r, ) + S(r, f ) f −b f − ci i=1 p 1 1 1 ≤ T (r, f ) + N (r, ) + N (r, ) l g g − ei i=1 14 1 1 1 1 ≤ T (r, f ) + T (r, g) + pT (r, f ) + S(r, f ); (n − p − − )T (r, f ) ≤ S(r, f ). l s s l This is a contradiction to the assumption that n > p + 1s + 1l . Suppose a = 1. Then from (3.1) we have F − c = c(G − 1). (3.5) We consider P (z) − c. By P (0) = a = 1 and c = 1 we have P (0) − c = 1 − c = 0. Moreover c = P (b). So P (z) − c has only simple zeros. Now we consider P (z) − 1. By a = 1 we see that P (0) = 1, P (z) − P (0) has a zero at 0 of order n + 1. Since n > 1s + 1l , by a argument as above when ac = c = P (b), we can get a contradiction. Case2. c = 1. Then P (f ) = P (g) (3.6) Applying Lemma 2.1 for (3.6) we obtain f = g. Proof of Theorem 1.3. Let f, g be two non-constant meromorphic (entire) functions. 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Hai Duong Pedagogical College, Hai Duong, Viet Nam E-mail address: vuhoaianmai@yahoo.com Thang Long University, Hanoi, Vietnam E-mail address: hhkhoai@math.ac.vn Hai Duong College, Hai Duong Province, Viet Nam E-mail address: nguyenxuanlai @yahoo.com 16 [...]... get a contradiction Case2 c = 1 Then P (f ) = P (g) (3.6) Applying Lemma 2.1 for (3.6) we obtain f = g Proof of Theorem 1.3 Let f, g be two non-constant meromorphic (entire) functions If one of the conditions of (I) is satisfied, then by Theorem 1.1 we have P (f ) = cP (g), c = 0 Then, if one of the conditions of (II) is satisfied, then by Theorem 1.2 we have f = g References [1] Vu Hoai An and Tran... 1747- 1759 [10] Ha Huy Khoai and C C Yang, On the functional equation P (f ) = Q(g), Value Distribution Theory and Related Topics, Advanced Complex Analysis and Application, Vol 3, Kluwer Academic, Boston, MA, 2004, 201-208 [11] Hayman,W.K ,Meromorphic Functions, Clarendon, Oxford(1964) [12] Li, P., and C.C Yang, Some further results on the unique range sets of meromorphic functions Kodai Math J 18, 1995,... 1175-1203 [7] Fujimoto, H., On uniqueness polynomials for meromorphic functions, Nagoya Math J 170, 2003, 33-46 [8] Gross, F., and C.C Yang, On preimage and range sets of meromorphic functions, Proc Japan Acad 58, 1982, 17-20 [9] Ha Huy Khoai and Vu Hoai An and Le Quang Ninh Uniqueness Theorems for Holomorphic Curves with Hypersurfaces of Fermat-Waring Type, Complex Analysis and Operator Theory, Volume 8, Issue... This is a contradiction to the assumption that q > 2k + 4s + 6l So L ≡ 0 Therefore, 1 = Pc(f0 ) + c1 for any two constants c0 (= 0) and c1 By Lemma 2.5 we have c1 = 0 P (f ) Thus, there is constant c(= 0) such that P (f ) = cP (g) N ≥2 (r, 12 In particular, in the case where f and g are entire, we applying the result of the case where f and g are meromorphic and take l = ∞ Banerjee[4] proved the following:... 527-539 [4] Banerjee, A., A new class of strong uniqueness polynomial satisfying Fujimotos conditions, Annales Academi Scientiarum Fennic Mathematica Volumen 40, 2015, 465-474 [5] Frank, G., and M Reinders, A unique range set for meromorphic functions with 11 elements, Complex Var Theory Appl 37:1, 1998, 185-193 [6] Fujimoto, H., On uniqueness of meromorphic functions sharing nite sets, Amer J Math 122,... Yi, H.X. ,On a problem of Gross Sci China Ser A 24, 1994, 11371144 15 [14] Yi, H.X.,Unicity theorems for meromorphic or entire functions III Bull Austral Math Soc 53, 1996, 7182 [15] Yi, H.X. ,The reduced unique range sets for entire or meromorphic functions Complex Var Theory Appl 32, 1997, 191198 Hai Duong Pedagogical College, Hai Duong, Viet Nam E-mail address: vuhoaianmai@yahoo.com Thang Long University,... Now we consider the following cases: Case1 c = 1 Subcase1.1 c = P (b) Then P (b) = c = 1 and F − P (b) = P (b)(G − 1) We have P (0) = a Suppose a = 1 Then P = 0 has zeros at 0, b and P (0) − 1 = 0, P (b) − 1 = 0 It follows that all the zeros of P (z) − 1 are simple, and let them be bi , i = 1, 2, , n + p + 1 Clearly ,P (z) − P (b) has a zero at b of order p + 1 Let the remaining distinct zeros of P (z)... following: Lemma 2.7 QB (1) is not an integer In particular, when a = 1, P (1) = 1, wheren ≥ 3, m ≥ 3 are integers 3 Proof of Theorems Now we use the above Lemmas to prove the main result of the paper Proof of Theorem 1.1 Since P (z) isa strong uniqueness polynomial ,we obtain f = g Proof of Theorem 1.2 We have P (0) = Q(0) = a, P (b) = Q(b)+a = QB (1)bn+m+1 + a By QB (1) = 0, b = 0, we have P (b) = P (0)... a contradiction to the assumption that n > 1s + 1l Subcase1.2 c = P (b) Suppose a = 1 Then from (3.1) we have F − ac = c(G − a) (3.4) Consider P (z) − ac By c = 1 we have ac = a So P (0) − ac = a − ac = 0 Suppose ac = P (b) Then P (z) − ac has only simple zeros Let they be given by ei , i = 1, 2, , n + p + 1 Clearly, P (z) − P (0) has a zero at 0 of order n + 1 Let the remaining distinct zeros of. .. Dinh Duc, Uniqueness theorems and uniqueness polynomials for holomorphic curves, Complex Variables and Elliptic Equations, Vol 56, Nos 1-4, January-April 2011, pp.253-262 [2] Bai, X., Q Han, and A Chen, On a result of H Fujimoto, J Math Kyoto Univ 49:3, 2009, 631-643 [3] Banerjee, A., and I Lahir, A uniqueness polynomial generating a unique range set and vise versa, Comput Methods Funct Theory 12:2, ... to determine the functions and in this case either the functions coincide or one is the bilinear transformation of the other In 1982 Gross and Yang [ ] proved the following theorem: Theorem A [... (S), and all of the zeros and poles of f and g are of multiplicities at least s, l, respectively, and p > + 1l , np > p + n Assume that for the following two systems of conditions, one of (I) and. .. + 1s ) when a = Then f = g Theorem 1.3 Let f and g be two non-constant meromorphic (entire) functions and P (z) be a polynomial of the form (1.4) satisfying the condition (1.5) and m be a positive