UNIQUENESS OF ENTIRE OR MEROMORPHIC FUNCTIONS SHARING ONE VALUE OR A FUNCTION WITH FINITE WEIGHT

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1 Volume ), Issue 3, Article 88, 4 pp. UNIQUENESS OF ENTIRE OR MEROMORPHIC FUNCTIONS SHARING ONE VALUE OR A FUNCTION WITH FINITE WEIGHT HONG-YAN XU AND TING-BIN CAO DEPARTMENT OF INFORMATICS AND ENGINEERING JINGDEZHEN CERAMIC INSTITUTE XIANGHU XIAOQU) JINGDEZHEN, JIANGXI , CHINA xhyhhh@26.com DEPARTMENT OF MATHEMATICS NANCHANG UNIVERSITY NANCHANG, JIANGXI 33003, CHINA tbcao@ncu.edu.cn Received 3 December, 2008; accepted 27 August, 2009 Communicated by S.S. Dragomir ABSTRACT. The purpose of this paper is to deal with some uniqueness problems of entire functions or meromorphic functions concerning differential polynomials that share one value or fixedpoints with finite weight. We obtain a number of theorems which generalize some results due to M.L. Fang & X.H. Hua, X.Y. Zhang & W.C. Lin, X.Y. Zhang & J.F. Chen and W.C. Lin. Key words and phrases: Meromorphic function, Entire function, Weighted sharing, Uniqueness Mathematics Subject Classification. 30D30, 30D35.. INTRODUCTION AND MAIN RESULTS Let f be a non-constant meromorphic function in the whole complex plane. We shall use the following standard notations of value distribution theory: T r, f), mr, f), Nr, f), Nr, f),... see Hayman [6],Yang [3] and Yi and Yang [6]). We denote by Sr, f) any quantity satisfying Sr, f) = ot r, f)), as r +, possibly outside of a set with finite measure. A meromorphic function a is called a small function with respect to f if T r, a) = Sr, f). Let Sf) be the set of meromorphic functions in the complex plane C which are small functions with respect to f. For some a C, we define Nr, a; f) Θa, f) = lim r T r, f). The research was supported by the NNSF of China No ), the NSF of Jiangxi of China 2008GQS0075)

2 2 HONG-YAN XU AND TING-BIN CAO For a C and k a positive integer, we denote by Nr, a; f = ) the counting function of simple a-points of f, and denote by Nr, a; f k) Nr, a; f k)) the counting functions of those a-points of f whose multiplicities are not greater less) than k where each a-point is counted according to its multiplicity see [6]). Nr, a; f k)nr, a; f k)) are defined similarly, where in counting the a-points of f we ignore the multiplicities. Set N k r, a; f) = Nr, a; f) + Nr, a; f 2) + + Nr, a; f k). We define N k r, a; f) δ k a, f) = lim. r T r, f) Let f and g be two nonconstant meromorphic functions defined in the open complex plane C. If for some a Sf) Sg) the roots of f a and g a coincide in locations and multiplicities we say that f and g share the value a CM counting multiplicities) and if they coincide in locations only we say that f and g share a IM ignoring multiplicities). In 997, Yang and Hua [4] proved the following result. Theorem A [4]). Let f and g be two nonconstant entire functions, n 6 a positive integer. If f n f and g n g share the value CM, then either f = c e cz and g = c 2 e cz, where c, c, and c 2 are constants satisfying c c 2 ) n+ c 2 = or f = tg for a constant t such that t n+ =. Using the same argument as in [4], Fang [3] proved the following result. Theorem B [3]). Let f and g be two nonconstant entire functions and let n, k be two positive integers with n > 2k +4. If [f n ] k) and [g n ] k) share the value CM, then either f = c e cz, g = c 2 e cz, where c, c 2 and c are three constants satisfying ) k c c 2 ) n nc) 2k =, or f tg for a constant t such that t n =. Fang [5] obtained some unicity theorems corresponding to Theorem B. Theorem C [5]). Let f and g be two nonconstant entire functions, and let n, k be two positive integers with n > 2k + 8. If [f n f )] k) and [g n g )] k) share CM, then f g. Recently, Zhang and Lin [7], Zhang, Chen and Lin [8] extended Theorem C and obtained the following results. Theorem D [7]). Let f and g be two nonconstant entire functions, n, m and k be three positive integers with n > 2k + m + 4, and λ, µ be constants such that λ + µ 0. If [f n µf m + λ)] k) and [g n µg m + λ)] k) share CM, then i) when λµ 0, f g; ii) when λµ = 0, either f tg, where t is a constant satisfying t n+m =, or f = c e cz, g = c 2 e cz, where c, c 2 and c are three constants satisfying ) k λ 2 c c 2 ) n+m [n + m)c] 2k = or ) k µ 2 c c 2 ) n+m [n + m)c] 2k =. Theorem E [8]). Let f and g be two nonconstant entire functions, and let n, m and k be three positive integers with n 3m + 2k + 5, and let P z) = a m z m + a m z m + + a z + a 0 or P z) c 0, where a 0 0, a,..., a m, a m 0, c 0 0 are complex constants. If [f n P f)] k) and [g n P g)] k) share CM, then i) when P z) = a m z m +a m z m + +a z+a 0, either f tg for a constant t such that t d =, where d = n + m,..., n + m i,..., n), a m i 0 for some i = 0,,..., m, or f and g satisfy the algebraic equation Rf, g) 0, where Rω, ω 2 ) = ω n a m ω m + a m ω m + + a ω + a 0 ) ω n 2 a m ω m 2 + a m ω m a ω 2 + a 0 );

3 UNIQUENESS OF ENTIRE OR MEROMORPHIC FUNCTIONS 3 ii) when P z) c 0, either f = c / n c 0 e cz, g = c 2 / n c 0 e cz, where c, c 2 and c are three constants satisfying ) k c c 2 ) n nc) 2k =, or f tg for a constant t such that t n =. Regarding Theorems D and E, it is natural to ask the following question. Problem.. In Theorems D and E, can the nature of sharing CM be further relaxed? For meromorphic functions, Yang and Hua [4] proved the following result corresponding to Theorem A. Theorem F [4]). Let f and g be two nonconstant meromorphic functions, n an integer, and a C {0}. If f n f and g n g share the value a CM, then either f = dg for some n + )th root of unity d or g = c e cz and f = c 2 e cz, where c, c, and c 2 are constants satisfying c c 2 ) n+ c 2 = a 2. Lin and Yi [7] obtained some unicity theorems corresponding to Theorem F. Theorem G [7]). Let f and g be two nonconstant meromorphic functions satisfying Θ, f) > 2 n+, n 2. If [f n f )]f and [g n g )]g share CM, then f g. Lin and Yi [8] extended Theorem G by replacing the value with the function z and obtained the following result. Theorem H [8]). Let f and g be two transcendental meromorphic functions, n 2 an integer. If f n f )f and g n g )g share z CM, then either f g or g = n+2) hn+ ) n+) h n+2 ) and f = n+2)h hn+ ), where h is a nonconstant meromorphic function. n+) h n+2 ) Recently, Zhang, Chen and Lin [8] extended Theorems F and G and obtained the following result. Theorem I [8]). Let f and g be two nonconstant meromorphic functions, and let n and m be two positive integers with n > max{m + 0, 3m + 3}, and let P z) = a m z m + a m z m + + a z + a 0, where a 0 0, a,..., a m, a m 0 are complex constants. If f n P f)f and g n P g)g share CM, then either f tg for a constant t such that t d =, where d = n + m +,..., n + m + i,..., n + ), a m i 0 for some i = 0,,..., m, or f and g satisfy the algebraic equation Rf, g) 0, where Rω, ω 2 ) = ω n am ω m n + m + + a m ω m n + m + + a ) 0 n + ω2 n am ω2 m n + m + + a m ω2 m n + m + + a ) 0. n + Regarding Theorem I, it is natural to ask the following questions. Problem.2. Is it possible that the value can be replaced by a function z in Theorem I? Problem.3. Is it possible to relax the nature of sharing z in Theorem I and if possible, how far? In 200, Lahiri [9, 0] first employed the idea of weighted sharing of values which measures how close a shared value is to being shared IM or to being shared CM. Recently, many mathematicians such as H. X. Yi, I. Lahiri, M. L. Fang, A. Banerjee, W. C. Lin, X. Yan) have been interested in investigating meromorphic functions sharing values with finite weight in the field of complex analysis. We first introduced the notion of weighted sharing of values as follows.

4 4 HONG-YAN XU AND TING-BIN CAO Definition. [9, 0]). Let k be a nonnegative integer or infinity. For a C { }, we denote by E k a; f) the set of all a-points where an a-point of multiplicity m is counted m times if m k and k + times if m > k. If E k a; f) = E k a; g), we say that f, g share the value a with weight k. We denote by E m) a; f) the set of all a-points of f with multiplicities not exceeding m, where an a-point is counted according to its multiplicity. If for some a C { }, E ) a; f) = E ) a; g), then we say that f, g share the value a CM. The definition implies that if f, g share a value a with weight k then z 0 is a zero of f a with multiplicity m k) if and only if it is a zero of g a with multiplicity m k); and z 0 is a zero of f a with multiplicity m> k) if and only if it is a zero of g a with multiplicity n> k), where m is not necessarily equal to n. We write f, g share a, k) to mean that f, g share the value a with weight k, clearly if f, g share a, k), then f, g share a, p) for all integers p 0 p k).also, we note that f, g share a value a IM or CM if and only if they share a, 0) or a, ), respectively. With the notion of weighted sharing of values, we investigate the solution of the above question and obtain the following results. Theorem.. Let f and g be two nonconstant entire functions, and let n, m and k be three positive integers with n 5m + 5k + 8. If [f n P f)] k) and [g n P g)] k) share, 0), then the conclusion of Theorem E still holds. Theorem.2. Let f and g be two nonconstant entire functions, and let n, m and k be three positive integers with n > 9 2 m + 4k If [f n P f)] k) and [g n P g)] k) share, ), then the conclusion of Theorem E still holds. Theorem.3. Let f and g be two nonconstant entire functions, and let n, m and k be three positive integers with n 3m + 3k + 5. If [f n P f)] k) and [g n P g)] k) share, 2), then the conclusion of Theorem E still holds. Remark. From Theorems..3, we obtain a positive answer to Question.. Theorem.4. Let f and g be two transcendental meromorphic functions, and let n and m be two positive integers with n > m + 0, and let P z) = a m z m + a m z m + + a z + a 0, where a 0 0, a,..., a m, a m 0 are complex constants. If f n P f)f and g n P g)g share z CM, then either f tg for a constant t such that t d =, where d = n + m +,..., n + m + i,..., n + ), a m i 0 for some i = 0,,..., m, or f and g satisfy the algebraic equation Rf, g) 0, where Rω, ω 2 ) = ω n am ω m n + m + + a m ω m n + m + + a ) 0 n + ω2 n am ω2 m n + m + + a m ω2 m n + m + + a ) 0. n + Theorem.5. Let f and g be two transcendental meromorphic functions, and let n and m be two positive integers with n > 4m + 22, and let P z) = a m z m + a m z m + + a z + a 0, where a 0 0, a,..., a m, a m 0 are complex constants. If f n P f)f and g n P g)g share z IM, then the conclusion of Theorem.4 still holds. Theorem.6. Let f and g be two transcendental meromorphic functions, let n, l and m be three positive integers, and let P z) = a m z m + a m z m + + a z + a 0, where a 0 0, a,..., a m, a m 0 are complex constants. If E l) z, f n P f)f ) = E l) z, g n P g)g ), i) If l = and n > 3m + 8, then the conclusion of Theorem.4 still holds.

5 UNIQUENESS OF ENTIRE OR MEROMORPHIC FUNCTIONS 5 ii) If l = 2 and n > 3 m + 2, then the conclusion of Theorem.4 still holds. 2 Remark 2. Theorem.4 is an improvement of Theorem H. Theorem.5 and.6 are complements to Theorem H. Though the standard definitions and notations of value distribution theory are available in [6, 3], we explain the ones which are used in the paper. Definition.2 [, 6]). When f and g share IM, We denote by N L r, ; f) the counting function of the -points of f whose multiplicities are greater than -points of g, where each zero is counted only once; Similarly, we have N L r, ; g). Let z 0 be a zero of f of multiplicity p and a zero of g of multiplicity q, we also denote by N r, ; f) the counting function of those -points of f where p = q = ; N 2 E r, ; f) denotes the counting function of those -points of f where p = q 2, each point in these counting functions is counted only once. In the same way, one can define N r, ; g), N 2 E r, ; g). Definition.3 [9, 0]). Let f, g share a value IM. We denote by N r, ; f, g) the reduced counting function of those -points of f whose multiplicities differ from the multiplicities of the corresponding -points of g. Clearly N r, ; f, g) N r, ; g, f) and N r, ; f, g) = N L r, ; f) + N L r, ; g). 2. SOME LEMMAS For the proof of our results we need the following lemmas. Lemma 2. [5, p. 27, Theorem.2]). Let f be a nonconstant meromorphic function and P f) = a 0 + a f + a 2 f a n f n, where a 0, a, a 2,..., a n are constants and a n 0. Then T r, P f)) = nt r, f) + Sr, f). Lemma 2.2 [8]). Let f be a transcendental entire function, let n, k, m be positive integers with n k +2, and P z) = a 0 +a z +a 2 z 2 + +a m z m, where a 0, a, a 2,..., a m are complex constants. Then [f n P f)] k) = has infinitely many solutions. Lemma 2.3 [8]). Let f and g be two nonconstant entire functions, and let n, k be two positive integers with n > k, and let P z) = a m z m + a m z m + + a z + a 0 be a nonzero polynomial, where a 0, a,..., a m, a m are complex constants. If [f n P f)] k) [g n P g)] k), then P z) is reduced to a nonzero monomial, that is, P z) = a i z i 0 for some i = 0,,..., m; further,f = c / n+i ai e cz, g = c 2 / n+i ai e cz, where c, c 2 and c are three constants satisfying ) k c c 2 ) n [n + )c)] 2k =. Let f be an entire function; we have Θ, f) =. Using the same argument as [2, Lemma 2.2], we can easily obtain the following lemma. Lemma 2.4. Let f and g be two entire functions, and let k be a positive integer.if f k) and g k) share, l) l = 0,, 2). Then i) If l = 0, 2.) Θ0, f) + δ k 0, f) + δ k+ 0, f) + δ k+ 0, g) + δ k+2 0, f) + δ k+2 0, g) > 5, then either f k) g k) or f g; ii) If l =, 2.2) 2 Θ0, f) + δ k0, f) + δ k+2 0, f)) + δ k+ 0, f) + δ k+ 0, g) + Θ0, g) + δ k 0, g) > 9 2, then either f k) g k) or f g;

6 6 HONG-YAN XU AND TING-BIN CAO iii) If l = 2, 2.3) Θ0, f) + δ k 0, f) + δ k+ 0, f)} + δ k+2 0, g) > 3, then either f k) g k) or f g. Lemma 2.5. Let f and g be two transcendental meromorphic functions, let n and m be three positive integers with n > 7, and let P z) = a m z m + a m z m + + a z + a 0, where a 0 0, a,..., a m, a m 0 are complex constants. If f n P f)f and g n P g)g share z IM, then Sr, f) = Sr, g). Proof. Using the same arguments as in [8] and [8], we easily obtain Lemma 2.5. Lemma 2.6. Let f and g be two transcendental meromorphic functions, and let n and m be three positive integers with n m + 3, F = f n P f)f and G z = gn P g)g, where n 4) is z a positive integer. If F G, then either f tg for a constant t such that t d =, where d = n + m +,..., n + m + i,..., n + ), a m i 0 for some i = 0,,..., m, or f and g satisfy the algebraic equation Rf, g) 0, where Rω, ω 2 ) is as stated in Theorem.4. Proof. Using the same arguments as those in [] and [8], we can easily get Lemma 2.6. Lemma 2.7. Let f and g be two transcendental meromorphic functions. Then where n m + 4 is a positive integer. f n P f)f g n P g)g z 2, Proof. Using the same argument as in [] and [8], we easily obtain Lemma 2.7. Lemma 2.8 [3]). Let f and g be two meromorphic functions. If f and g share CM, one of the following three cases holds: i) T r, f) N 2 r,, f) + N 2 r,, g) + N 2 r, 0, f) + N 2 r, 0, g) + Sr, f) + Sr, g), the same inequality holding for T r, g); ii) f g; iii) f g. Lemma 2.9 [4]). Let f and g be two meromorphic functions, and let l be a positive integer. If E l), f) = E l), g), then one of the following cases must occur: i): T r, f) + T r, g) N 2 r, ; f) + N 2 r, 0; f) + N 2 r, ; g) + N 2 r, 0; g) + Nr, ; f) + Nr, ; g) N r, ; f) + Nr, ; f l + ) + Nr, ; g l + ) + Sr, f) + Sr, g); ii): f = b+)g+a b ), where a 0), b are two constants. bg+a b) Lemma 2.0 [4]). Let f and g be two meromorphic functions. If f and g share IM, then one of the following cases must occur: i): T r, f) + T r, g) 2[N 2 r, ; f) + N 2 r, 0; f) + N 2 r, ; g) + N 2 r, 0; g)] ii): f = b+)g+a b ), where a 0), b are two constants. bg+a b) + 3N L r, ; f) + 3N L r, ; g) + Sr, f) + Sr, g);

7 UNIQUENESS OF ENTIRE OR MEROMORPHIC FUNCTIONS 7 Lemma 2.. Let f and g be two transcendental meromorphic functions, n > m + 6 be a positive integer, and let F = f n P f)f and G z = gn P g)g. If z 2.4) F = b + )G + a b ), bg + a b) where a 0), b are two constants, then the conclusion of Theorem.4 still holds. Proof. By Lemma 2. we know that T r, F ) = T r, f ) n P f)f 2.5) z T r, f n P f)) + T r, f ) + log r 2.6) So n + m)t r, f) = T r, f n P f)) + Sr, f) n + m)t r, f) + 2T r, f) + log r + Sr, f) = n + m + 2)T r, f) + log r + Sr, f), = Nr, ; f n P f)) + mr, f n P f)) + Sr, f) N r, ; f ) n P f)f Nr, ; f ) z + m r, f ) n P f)f + m r, f ) + log r + Sr, f) z T r, f ) n P f)f + T r, f ) Nr, ; f ) Nr, 0; f ) + log r + Sr, f) z T r, F ) + T r, f) Nr, ; f) Nr, 0; f ) + log r + Sr, f). 2.7) T r, F ) n + m )T r, f) + Nr, ; f) + Nr, 0; f ) + log r + Sr, f). Thus, by 2.5), 2.7) and n > m + 6, we get Sr, F ) = Sr, f). Similarly, 2.8) T r, G ) n + m )T r, g) + Nr, ; g) + Nr, 0; g ) + log r + Sr, g). Without loss of generality, we suppose that T r, f) T r, g), r I, where I is a set with infinite measure. Next, we consider three cases. Case. b 0,, If a b 0, then by 2.4) we know N r, a b ) b + ; G = Nr, 0; F ). Since 2.9) Nr, 0; g ) Nr, ; g) + Nr, 0; g) + Sr, g) 2T r, g) + Sr, g). By Nevanlinna s second fundamental theorem and 2.9) we have T r, G ) Nr, ; G ) + Nr, 0; G ) + N r, a b ) b + ; G + Sr, G ) Nr, ; g) + Nr, 0; g) + mt r, g) + Nr, 0; g ) + Nr, 0; f) + mt r, f) + Nr, 0; f) + Nr, ; f) + 2 log r + Sr, g) 2m + 4)T r, g) + Nr, ; g) + Nr, 0; g ) + 2 log r + Sr, g).

8 8 HONG-YAN XU AND TING-BIN CAO Hence, by n > m + 6 and 2.8), we know T r, g) Sr, g), r I, this is impossible. If a b = 0, then by 2.4) we know F = b + )G )/bg + ). Obviously, N r, ) b ; G = Nr, ; F ). By the Nevanlinna second fundamental theorem and 2.9) we have T r, G ) Nr, ; G ) + Nr, 0; G ) + N r, ) b ; G + Sr, G ) Nr, ; g) + Nr, 0; g) + mt r, g) + Nr, 0; g ) + Nr, ; f) + 2 log r + Sr, g) m + 2)T r, g) + Nr, ; g) + Nr, 0; g ) + 2 log r + Sr, g). Then by n > m + 6 and 2.8), we know T r, g) Sr, g), r I, a contradiction. Case 2. b =. Then 2.4) becomes F = a/a + G ). If a + 0, then Nr, a + ; G ) = Nr, ; F ). Applying a similar argument to that for Case, we can again deduce a contradiction. If a + = 0, then F G, that is, f n P f)f g n P g)g z 2. Since n m + 6, by Lemma 2.7 we get a contradiction. Case 3. b = 0. Then 2.4) becomes F = G + a )/a. If a 0, then Nr, a; G ) = Nr, 0; F ). Applying a similar argument to that for Case, we can again deduce a contradiction. If a = 0, then F G, that is f n P f)f g n P g)g. By Lemma 2.6, we obtain the conclusions of Lemma 2.. Thus we complete the proof of Lemma Proof of Theorem.. 3. THE PROOFS OF THEOREMS..3 Proof. i) P z) = a m z m + a m z m + + a z + a 0. By the assumptions of Theorem. and Lemma 2.2, we know that either both f and g are transcendental entire functions or both f and g are polynomials. First, we consider the case when f and g are transcendental entire functions. Let F = f n P f) and G = g n P g), from the condition of Theorem., we know that F, G share, 0). By Lemma 2. we can easily get Θ0, F ) = lim sup r = lim sup r = lim sup r Nr, 0; F ) T r, F ) Nr, 0; f n P f)) n + m)t r, f) Nr, 0; f) + Nr, 0; P f)), n + m)t r, f)

9 UNIQUENESS OF ENTIRE OR MEROMORPHIC FUNCTIONS 9 i.e. 3.) Θ0, F ) m + n + m = n n + m. Similarly, we have 3.2) Θ0, G) n n + m. Next, by the definition of N k r, a; f) we have Therefore δ k+ 0, f) = lim sup r N k+ r, 0; f) T r, f) δ k+ 0, F ) = lim sup r 3.3) δ k+ 0, F ) lim sup r Similarly we get lim sup r N k+ r, 0; f n P f)). T r, F ) m + k + )T r, f) n + m)t r, f) k + )Nr, 0; f), T r, f) = n k n + m. 3.4) δ k+ 0, G) n k n + m and 3.5) δ k+2 0, F ) n k 2 n + m, δ k+20, G) n k 2 n + m. From 3.) 3.5) and F, G share, 0), we can get Θ0, f) + δ k 0, f) + δ k+ 0, f) + δ k+ 0, g) + δ k+2 0, f) + δ k+2 0, g) n n + m + n k n + m + 2n k n + m + 2n k 2 n + m By n > 5m + 5k + 7, we have = 6n 5k 7 n + m. Θ0, f) + δ k 0, f) + δ k+ 0, f) + δ k+ 0, g) + δ k+2 0, f) + δ k+2 0, g) > 5. Therefore, by Lemma 2.4, we deduce either F k) G k) or F G. If F k) G k), that is 3.6) [f n a m f m + a m f m + + a 0 )] k) [g n a m g m + a m g m + + a 0 )] k), then by the assumptions of Theorem. and Lemma 2.3 we can get a contradiction. Hence, we deduce that F G, that is 3.7) f n a m f m + a m f m + + a 0 ) = g n a m g m + a m g m + + a 0 ). Let h = f/g. If h is a constant, then substituting f = gh into 3.7) we deduce a m g n+m h n+m ) + a m g n+m h n+m ) + + a 0 g n h n ) = 0, which implies h d =, where d = n + m,..., n + m i,..., n),a m 0 for some i = 0,,..., m. Thus f tg for a constant t such that t d =, where d = n + m,..., n + m i,..., n),a m i 0 for some i = 0,,..., m. If h is not a constant, then we know by 3.7) that f and g satisfy the algebraic equation Rf, g) = 0, where Rω, ω 2 ) = ω n a m ω m + a m ω m + + a ω + a 0 ) ω2 n a m ω2 m + a m ω2 m + + a ω 2 + a 0 ). This proves i) of Theorem..

10 0 HONG-YAN XU AND TING-BIN CAO Now we consider the case when f and g are two polynomials. From F, G share,0), we have 3.8) [f n a m f m + a m f m + + a 0 )] k) = c{[g n a m g m + a m g m + + a 0 )] k) }, where c is a nonzero constant. Let deg f = l. Then by 3.8) we know that deg g = l. Differentiating the two sides of 3.8), we can get 3.9) f n k q = g n k q 2, where q, q 2 are two polynomials with deg q = deg q 2 = m + k + )l k + ). By n 4m + 4k + 8, we have deg g n k = n k )l > deg q 2. Therefore by 3.9) we know that there exists z 0 such that fz 0 ) = gz 0 ) = 0. Hence, by 3.8) and fz 0 ) = gz 0 ) = 0, we deduce that c =, i.e., 3.0) [f n a m f m + a m f m + + a 0 )] k) [g n a m g m + a m g m + + a 0 )] k). Then we have 3.) f n a m f m + a m f m + + a 0 ) g n a m g m + a m g m + + a 0 ) = pz), where pz) is a polynomial of degree at most k. Next, we prove pz) = 0 by rewriting 3.0) as 3.2) f n k p = g n k p 2, where p, p 2 are two polynomials with deg p = deg p 2 = m + k)l k and deg f = l. Therefore, the total number of the common zeros of f n k and g n k is at least k. Then by 3.) we deduce that pz) 0, i.e., f n a m f m + a m f m + + a f + a 0 ) g n a m g m + a m g m + + a g + a 0 ). Then using the same argument of 3.7), we can also get the case i) of Theorem.. ii) P z) c 0. From Theorem B, we can easily see that the case ii) of Theorem. holds. Thus, we complete the proof of Theorem Proof of Theorem.2. Proof. From the condition of Theorem.2 and Lemma 2.4ii), using the same argument of Theorem., Theorem.2 can be easily proved Proof of Theorem.3. Proof. From the condition of Theorem.3 and Lemma 2.4iii), using the same argument of Theorem., Theorem.3 can be easily proved. and 4. THE PROOFS OF THEOREMS.4.6 Let F and G be defined as in Lemma 2. and F = a mf n+m+ n + m + + a m f n+m n + m + + a 0f n+ n + G = a mg n+m+ n + m + + a m g n+m n + m + + a 0g n+ n +.

11 UNIQUENESS OF ENTIRE OR MEROMORPHIC FUNCTIONS 4.. Proof of Theorem.4. Proof. From the condition of Theorem.4, then F and G share z CM. By Lemma 2., we have 4.) T r, F ) = n+m+)t r, f)+sr, f), T r, G ) = n+m+)t r, g)+sr, g). Since F ) = F z, we deduce ) ) m r, m r, F zf and by Nevanlinna s first fundamental theorem ) + Sr, f) m r, F + log r + Sr, f), 4.2) T r, F ) T r, F ) + Nr, 0; F ) Nr, 0; F ) + log r + Sr, f) T r, F ) + Nr, 0; f) + Nr, b ; f) + + Nr, b m ; f) Nr, c ; f) Nr, c m ; f) Nr, 0; f ) + log r + Sr, f), where b, b 2,..., b m are roots of the algebraic equation a m z m n + m + + a m z m n + m + + a 0 n + = 0, and c, c 2,..., c m are roots of the algebraic equation By the definition of F, G, we have a m z m + a m z m + + a z + a 0 = ) N 2 r, 0; F ) + N 2 r, ; F ) 2Nr, ; f) + 2Nr, 0; f) + Nr, c ; f) Similarly, we obtain + + Nr, c m ; f) + Nr, 0; f ) + 2 log r. 4.4) N 2 r, 0; G ) + N 2 r, ; G ) 2Nr, ; g) + 2Nr, 0; g) + Nr, c ; g) If Lemma 2.8i) holds, from 4.2) 4.4) we have + + Nr, c m ; g) + Nr, 0; g ) + 2 log r. 4.5) T r, F ) m + 5)T r, f) + m + 6)T r, g) + 4 log r + Sr, f) + Sr, g). Similarly, we obtain 4.6) T r, G ) m + 5)T r, g) + m + 6)T r, f) + 4 log r + Sr, f) + Sr, g). By 4.), 4.5), 4.6) and n > m + 0, we can obtain a contradiction. If Lemma 2.8ii) holds, then F G. By Lemma 2.6, we can get the conclusion of Theorem.4. If Lemma 2.8iii) holds, then F G. By Lemma 2.7 and n > m + 0, we can get a contradiction. Therefore, we complete the proof of Theorem.4.

12 2 HONG-YAN XU AND TING-BIN CAO 4.2. Proof of Theorem.5. Proof. Suppose that i) in Lemma 2.0 holds. Since N L r, ; F ) N r, ; F ) 4.7) F = N r, ; F ) + Sr, f) F Nr, ; F ) + Nr, 0; F ) + Sr, f) similarly, we have m + 4)T r, f) + 2 log r + Sr, f), 4.8) N L r, ; G ) m + 4)T r, g) + 2 log r + Sr, g). By 4.2) 4.4), 4.7) 4.8) and Lemma 2.0i), we have 4.9) n 4m 22)[T r, f) + T r, g)] 20 log r + Sr, f) + Sr, g). By n > 4m + 22, we get a contradiction. Hence F and G satisfy ii) in Lemma 2.0. By Lemma 2., we can get the conclusion of Theorem.5. Thus, we complete the proof of Theorem Proof of Theorem.6. Proof. i) If l =. Since Nr, ; F )+Nr, ; G ) N r, ; F ) 2 Nr, ; F ) + 2 Nr, ; G ) Suppose that i) in Lemma 2.9 holds, then we have 2 T r, F ) + 2 T r, G ) + Sr, f) + Sr, g). 4.0) T r, F ) + T r, G ) 2[N 2 r, 0; F ) + N 2 r, ; F ) + N 2 r, 0; G ) Since 4.) similarly, we have + N 2 r, ; G ) + Nr, ; F 2) + Nr, ; G 2)] + Sr, f) + Sr, g). N r, ; F 2) N r, ; F ) F = N r, ; F ) + Sr, f) F Nr, ; F ) + Nr, 0; F ) + Sr, f) m + 4)T r, f) + 2 log r + Sr, f), 4.2) Nr, ; G 2) m + 4)T r, g) + 2 log r + Sr, g). By 4.2) 4.4), 4.0) 4.2) and Lemma 2.9i), we have 4.3) n 3m 8)[T r, f) + T r, g)] 6 log r + Sr, f) + Sr, g). By n > 3m + 8, we get a contradiction. Hence F and G satisfy ii) in Lemma 2.9. By Lemma 2., we can get the conclusion of Theorem.6i).

13 UNIQUENESS OF ENTIRE OR MEROMORPHIC FUNCTIONS 3 ii) If l = 2. Since Nr, ; F ) + Nr, ; G ) N r, ; F ) + 2 N r, ; F 3) + 2 N r, ; G 3) 2 Nr, ; F ) + 2 Nr, ; G ) 2 T r, F ) + 2 T r, G ) + Sr, f) + Sr, g). Suppose that i) in Lemma 2.9 holds, then we have 4.4) T r, F ) + T r, G ) 2[N 2 r, 0; F ) + N 2 r, ; F ) + N 2 r, 0; G ) Since 4.5) similarly, we have + N 2 r, ; G ) + Nr, ; F 3) + Nr, ; G 3)] + Sr, f) + Sr, g). N r, ; F 3) 2 N r, ; F ) F = 2 N r, ; F ) F + Sr, f) m + 4)T r, f) + log r + Sr, f), 2 4.6) Nr, ; G 3) m + 4)T r, g) + log r + Sr, g). 2 By 4.2) 4.4), 4.4) 4.6) and Lemma 2.9i), we have 4.7) n 32 ) m 2 [T r, f) + T r, g)] 6 log r + Sr, f) + Sr, g). By n > 3m + 2, we get a contradiction. Hence F 2 and G satisfy ii) in Lemma 2.9. By Lemma 2., we can get the conclusion of Theorem.6ii). 5. REMARKS It follows from the proof of Theorem.4.5) that if the condition f n P f)f and g n P g)g share z CM IM) are replaced by the condition f n P f)f and g n P g)g share az) CM IM), where az) is a meromorphic function such that az) 0, and T r, f) = o{t r, f), T r, g)}, the conclusion of Theorem.4.5) still holds. Similarly, if the condition E l) z, f n P f)f ) = E l) z, g n P g)g )l =, 2) is replaced by the condition E l) az), f n P f)f ) = E l) az), g n P g)g )l =, 2) respectively, then the conclusion of Theorem.6 still holds. REFERENCES [] A. BANERJEE, Weighted sharing of a small function by a meromorphic function and its derivative, Comput. Math. Appl., ), [2] S.S. BHOOSNURMATH AND R.S. DYAVANAL, Uniqueness and value-sharing of meromorphic functions, Comput. Math. Appl., ), [3] M.-L. FANG AND X.-H. HUA, Entire functions that share one value, J. Nanjing Univ. Math., Biquarterly 3 996), [4] C.-Y. FANG AND M.-L. FANG, Uniqueness of meromorphic functions and differential polynomials, Comput. Math. Appl., )

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