CUBO A Mathematical Journal Vol.16, No¯ 01, (49–61). March 2014 On a result of Q. Han, S. Mori and K. Tohge concerning uniquesness of meromorphic functions. Indrajit Lahiri Department of Mathematics, University of Kalyani, Kalyani, Kalyani, West Bengal 741235, India. ilahiri@hotmail.com Nintu Mandal Department of Mathematics, A. B. N. Seal College, Cooch Behar, West Bengal 736101, India. nintu311209@gmail.com ABSTRACT In the paper we prove a result on the uniqueness of meromorphic functions that is related to a result of Q. Han, S. Mori and K. Tohge and is originated from a result of H.Ueda and two subsequent results of G. Brosch. RESUMEN En este art́ıculo probamos un resultado de unicidad de funciones meromórficas que se relaciona a un resultado de Q. Han, S. Mori y K. Tohge, y se origina de un resultado de H. Ueda y dos resultados derivados de G. Brosch. Keywords and Phrases: Meromorphic function, Uniqueness, Weighted sharing. 2010 AMS Mathematics Subject Classification: 30D35. 50 Indrajit Lahiri & Nintu Mandal CUBO 16, 1 (2014) 1 Introduction, Definitions and Results Let f and g be two non-constant meromorphic functions defined in the open complex plane C. For a ∈ C ∪ {∞} we say that f and g share the value a CM ( counting multiplicities ) if f, g have the same a-points with the same multiplicities. If we do not take the multiplicities into account then f, g are said to share the value a IM ( ignoring multiplicities ). For the standard notations and definitions of the value distribution theory we refer to [5] and [15] . However we require following notations. Definition 1. Let k be a positive integer or infinity. For a ∈ C ∪ {∞} we denote by Ek)(a; f) and Ek)(a; f) the collection of those a-points of f whose multiplicities does not exceed k, with counting multiplicities and with ignoring multiplicities respectively. Definition 2. Let k be a positive integer and a ∈ C ∪ {∞}. Then by N(r, a; f| ≤ k) we denote the counting function of those a-points of f (counted with proper multiplicities) whose multiplicities are not greater than k. By N(r, a; f| ≤ k) we denote the corresponding reduced counting funcion. In an analogous manner we define N(r, a; f| ≥ k) and N(r, a; f| ≥ k). Also by N(r, a; f| = k) and N(r, a; f| = k) we denote respectively the counting function and reduced counting function of those a-points of f whose multiplicities are exactly k. In 1980 H.Ueda[14]{see also p. 327 [15]}prove the following result. Theorem A. [14] Let f and g be nonconstant entire functions sharing 0, 1 CM, and a(6= 0, 1, ∞) be a complex number. If E ∞)(a; f) ⊂ E∞)(a; g), then f is a bilinear transformation of g. Improving Theorem A in 1989 G.Brosch[2] proved the following result. Theorem B. [2] Let f and g be two nonconstant meromorphic functions sharing 0, 1, ∞ CM, and a(6= 0, 1, ∞) be a complex number. If E ∞)(a; f) ⊂ E∞)(a; g), then f is a bilinear transformation of g. Following example shows that in Theorem B the condition E ∞)(a; f) ⊂ E∞)(a; g) cannot be replaced by E ∞)(a; f) ⊂ E∞)(b; g) for b 6= a, 0, 1, ∞. Example 1. Let f = e2z + ez + 1, g = e−2z + e−z + 1, a = 3 4 and b = 3. Then f, g share 0, 1, ∞ CM and f − a = 1 4 (2ez + 1)2, g − b = e−2z(1 + 2ez)(1 − ez). So Ē ∞)(a; f) ⊂ Ē∞)(b; g) but f is not a bilinear transformation of g. Considering the possibility a 6= b, G.Brosch[2] proved the following theorem. Theorem C. [2] Let f and g be two nonconstant meromorphic functions sharing 0, 1, ∞ CM, and a, b be two complex numbers such that a, b 6∈ {0, 1, ∞} . If E ∞)(a; f) = E∞)(b; g), then f is a bilinear transformation of g. CUBO 16, 1 (2014) On a result of Q. Han, S. Mori and K. Tohge concerning . . . 51 In 2001 the idea of weighted sharing of values was introduced {cf.[6], [7]} which provides a scaling between IM sharing and CM sharing of values. We now explain this notion in the following definition. Definition 3. [11] Let k be a nonnegative integer or infinity. For a ∈ C ∪ {∞} we denote by Ek(a, f) the set of all a-points of f, where an a-point with multiplicity m is counted m times if m ≤ k and k + 1 times if m > k. If Ek(a, f) = Ek(a, g), we say that f, g share the value a with weight k. The definition means that z0 is a zero of f − a with multiplicity m(≤ k) if and only if z0 is a zero of g with multiplicity m(≤ k) and z0 is a zero of f − a with multiplicity m(> k) if and only if z0 is a zero of g 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 f, g share (a, 0) or (a, ∞) respectively. In 2004 using the idea of weighted value sharing T.C. Alzahari and H.X.Yi [1] improved Theorem C in the following manner . Theorem D. [1] Let f, g be two nonconstant meromorphic functions sharing (a1, 1), (a2, ∞), (a3, ∞), where {a1, a2, a3} = {0, 1, ∞}, and let a, b be two finite complex numbers such that a, b 6∈ {0, 1} . If Ē ∞)(a; f) = Ē∞)(b; g), then f is a bilinear transformation of g. Moreover f and g satisfy exactly one of the following relations: (i) f ≡ g; (ii) fg ≡ 1; (iii) bf ≡ ag; (iv) f + g ≡ 1; (v) f ≡ ag; (vi) f ≡ (1 − a)g + a; (vii) (1 − b)f ≡ (1 − a)g + (a − b); (viii) (1 − a + g)f ≡ ag; (ix) f{(b − a)g + (a − 1)b} ≡ a(b − 1)g; 52 Indrajit Lahiri & Nintu Mandal CUBO 16, 1 (2014) (x) f(g − 1) ≡ g; The cases (ii) and (v) may occur if ab = 1, cases (iv) and (viii) may occur if a + b = 1, cases (vi) and (x) may occur if ab = a + b. Improving Theorem D recently I.Lahiri and P.Sahoo [12] proved the following theorem. Theorem E. [12] Let f, g be two distinct nonconstant meromorphic functions sharing (a1, 1), (a2, m), (a3, k), where {a1, a2, a3} = {0, 1, ∞} and (m − 1)(mk − 1) > (1 + m) 2.If for two values a, b 6∈ {0, 1, ∞} the functions f − a and g − b share (0, 0) then f, g share (0, ∞), (1, ∞), (∞, ∞) and f − a, g − b share (0, ∞). Also there exists a non-constant entire function λ such that f and g are one of the following forms: (i) f = aeλ and g = be−λ, where ab = 1; (ii) f = 1 + aeλ and g = 1 + (1 − 1 b )e−λ, where ab = a + b; (iii) f = a a+eλ and g = e λ 1−b+eλ , where a + b = 1: (iv) f = e λ −a eλ−1 and g = be λ −1 eλ−1 , where ab = 1; (v) f = be λ −a beλ−b and g = be λ −a aeλ−a , where a 6= b; (vi) f = a 1−eλ and g = be λ eλ−1 , where ab = a + b; (vii) f = b−a (b−1)(1−eλ) and g = (b−a)e λ (a−1)(1−eλ) , where a 6= b; (viii) f = a + eλ and g = b(1 + 1−b eλ ), where a + b = 1; (ix) f = eλ − a(b−1) a−b and g = b(a−1) a−b {1 − a(b−1) (b−a)eλ }, where a 6= b; Q.Han, S.Mori and K.Tohge [4] further improved Theorem C, Theorem D, Theorem E and proved the following. Theorem F. [4] Let f and g be two distinct nonconstant meromorphic functions sharing (a1, k1), (a2, k2) and (a3, k3) for three distinct values a1, a2, a3 ∈ C∪{∞}, where k1k2k3 > k1 +k2 +k3 +2. Furthermore if Ek)(a4; f) = Ek)(a5; g) for values a4, a5 in C∪{∞}\{a1, a2, a3} and for some positive integer k(≥ 2), then f is a bilinear transformation of g. CUBO 16, 1 (2014) On a result of Q. Han, S. Mori and K. Tohge concerning . . . 53 Example 1 with a = b = 3 4 shows that the conclusion of Theorem F does not hold for k = 1. This suggests that some further investigation is necessary for the case k = 1. In the paper we take up this problem and prove the following result. Theorem 1.1. Let f, g be two distinct nonconstant meromorphic functions sharing (a1, k1), (a2, k2), (a3, k3) where a1, a2, a3 ∈ C ∪ {∞} are distinct and k1k2k3 > k1 + k2 + k3 + 2. Further let E1)(a; f) ⊂ E∞)(b; g) for two complex numbers a, b 6∈ {a1, a2, a3} and E1)(0; f ′) ⊂ E ∞)(0; g ′). Then f is a bilinear transformation of g. If, in particular, {a1, a2, a3} = {0, 1, ∞},then there exists a non-constant entire function λ such that f and g assume exactly one of the following forms: (i) f = aeλ and g = be−λ where ab = 1; (ii) f = 1 + aeλ and g = 1 + (1 − 1 b )e−λ where ab = a + b; (iii) f = a a+eλ and g = e λ 1−b+eλ where a + b = 1: (iv) f = e λ −a eλ−1 and g = e λ −a aeλ−a where E ∞)(a; f) = φ; (v) f = be λ −a beλ−b and g = be λ −a aeλ−a where a 6= b; (vi) f = a 1−eλ and g = ae λ (1−a)(1−eλ) where E ∞)(a; f) = φ; (vii) f = b−a (b−1)(1−eλ) and g = (b−a)e λ (a−1)(1−eλ) where a 6= b; (viii) f = a + eλ and g = (1 − a)(1 + a eλ ) where E(a; f) = φ; (ix) f = eλ − a(b−1) a−b and g = b(a−1) a−b {1 − a(b−1) (b−a)eλ } where a 6= b; Considering Example 1 we see that the condition E1)(0; f ′) ⊂ E ∞)(0, g ′) is essential for Theorem 1.1. 2 Lemmas In the section we present some necessary lemmas. Lemma 2.1. [3] Let f and g share (0, 0), (1, 0), (∞, 0).Then T(r, f) ≤ 3T(r, g)+S(r, f) and T(r, g) ≤ 3T(r, f) + S(r, f). From this we conclude that S(r, f) = S(r, g). Henceforth we denote either of them by S(r). 54 Indrajit Lahiri & Nintu Mandal CUBO 16, 1 (2014) Lemma 2.2. [16] Let f and g share (0, k1), (1, k2), (∞, k3) and f 6≡ g, where k1k2k3 > k1 + k2 + k3 + 2.Then N(r, 0; f |≥ 2) + N(r, 1; f |≥ 2) + N(r, ∞; f |≥ 2) = S(r). Following can be proved in the line of Theorem 3.2 of [11]. Lemma 2.3. Let f and g be two distinct nonconstant meromorphic functions sharing (0, k1), (1, k2), (∞, k3), where k1k2k3 > k1 + k2 + k3 + 2. If N0(r) + N1(r) ≥ λT(r, f) + S(r) for some λ > 1 2 , then f is a bilinear transformation of g and N0(r) + N1(r) = T(r, f) + S(r) = T(r, g) + S(r), where N0(r)(N1(r)) denotes the counting function of those simple(multiple) zeros of f − g which are not the zeros of f(f − 1) and 1 f . Lemma 2.4. [13] Let f and g be two distinct noncostant meromorphic functions sharing (0, 0), (1, 0), (∞, 0). Further suppose that f is a bilinear transformation of g and E1)(a; f) ⊂ E∞)(b; g),where a, b 6∈ {0, 1, ∞}. Then there exists a nonconstant entire function λ such that f and g assume exactly one of the forms given in Theorem1.1. Following can be proved in the line of Lemma 2.4 [13]. Lemma 2.5. Let f and g share (0, k1), (1, k2), (∞, k3) and f 6≡ g, where k1k2k3 > k1 +k2 +k3 +2. If f is not a bilinear transformation of g, then for a complex number a 6∈ {0, 1, ∞} each of the following holds: (i) N(r, a; f |≥ 3) + N(r, a; g |≥ 3) = S(r); (ii) T(r, f) = N(r, a; f ≤ 2) + S(r); (iii) T(r, g) = N(r, a; g ≤ 2) + S(r). In the line of Lemma 5 [9] we can prove the following. Lemma 2.6. Let f, g share (0, k1), (1, k2), (∞, k3) and f 6≡ g, where k1k2k3 > k1 + k2 + k3 + 2. If α = f−1 g−1 and β = g f , then N(r, a; α) = S(r) and N(r, a; β) = S(r) for a = 0, ∞. Following is an analogue of Lemma 2.6 [13]. Lemma 2.7. Let f and g be two distinct meromorphic functions sharing (0, k1), (1, k2), (∞, k3), where k1k2k3 > k1 + k2 + k3 + 2.Then T(r, α (p) α ) + T(r, β (p) β ) = S(r), where p is a positive integer and α, β are defined as in Lemma 2.6. CUBO 16, 1 (2014) On a result of Q. Han, S. Mori and K. Tohge concerning . . . 55 Using the techniques of [8] and [10] we can prove the following. Lemma 2.8. Let f, g share (0, k1), (1, k2), (∞, k3) and f 6≡ g, where k1k2k3 > k1 + k2 + k3 + 2. If f is not a bilinear transformation of g, then each of the following holds : (i) T(r, f) + T(r, g) = N(r, 0; f |≤ 1) + N(r, 1; f |≤ 1) + N(r, ∞; f |≤ 1) + N0(r) + S(r), (ii) T(r, f) = N(r, 0; g′ |≤ 1) + N0(r) + S(r), (iii) T(r, g) = N(r, 0; f′ |≤ 1) + N0(r) + S(r), (iv) N1(r) = S(r), (v) N0(r, 0; g ′ |≥ 2) = S(r), (vi) N0(r, 0; f ′ |≥ 2) = S(r), (vii) N(r, 0; g′ |≥ 2) = S(r), (viii) N(r, 0; f′ |≥ 2) = S(r), (ix) N(r, 0; f − g |≥ 2) = S(r), (x) N(r, 0; f − g | f = ∞) = S(r), where N0(r, 0; g ′ |≥ 2)(N0(r, 0; f ′ |≥ 2)) is the counting function of those multiple zeros of g′(f′) which are not the zeros of f(f − 1) and N(r, 0; f − g | f = ∞) is the counting function of those zeros of f − g which are poles of f. 3 Proof of Theorem 1.1 Proof. If necessary considering a bilinear transformation we may choose {a1, a2, a3} = {0, 1, ∞}. We now consider the following cases CASE 1. Let a = b. If possible, we suppose that f is not a blinear transformation of g. We put Φ = f′(f − a) f(f − 1) − g′(g − a) g(g − 1) . Let Φ 6≡ 0. Since Φ = aβ ′ β + (1 − a)α ′ α , by Lemma 2.7 we get T(r, Φ) = S(r). Since E1)(a; f) ⊂ E ∞)(a; g) and E1)(0; f ′) ⊂ E ∞)(0; g ′), it follows that N(r, a; f |≤ 2) ≤ 2N(r, 0; Φ) = S(r), which contradicts (ii) of Lemma 2.5. Therefore Φ ≡ 0 and so f′(f − a) f(f − 1) = g′(g − a) g(g − 1) (3.1) 56 Indrajit Lahiri & Nintu Mandal CUBO 16, 1 (2014) If z0 is a double zero of g − a, then from (3.1) we see that z0 is a common zero of f ′ and g′.Hence z0 is a zero of α ′ α = f ′ f−1 − g ′ g−1 . So by (i) of Lemma 2.5 and Lemma 2.7 we get N(r, a; g |≥ 2) = 2N(r, 0; α′ α ) + S(r) = S(r). (3.2) Again if z1 is a zero of g ′ which is not a zero of g(g−1)(g−a), then from (3.1) and the hypotheses of the theorem it follows that z1 is a zero of f ′ and so of α ′ α . Hence from Lemma 2.2, Lemma 2.7 and (3.2) we get N(r, 0; g′ |≤ 1) ≤ N(r, a; g |≥ 2) + N(r, 0; f |≥ 2) + N(r, 1; f |≥ 2) + N(r, 0; α′ α ) = S(r). (3.3) Now from (ii) and (iv) of Lemma 2.8 and (3.3) we obtain N0(r) + N1(r) = T(r, f) + S(r), which is impossible by Lemma 2.3. Therefore f is a bilinear transformation of g and so by Lemma 2.4 f and g take one of the forms (i)-(iv),(vi) and (viii). CASE 2. Let a 6= b. If f is a bilinear transformation of g, then by Lemma 2.4 f and g assume one of the forms (i) − (ix). So we suppose that f is not a bilinear transformation of g. Following two subcases come up for consideration. Subcase (i) Let N(r, a; f |≥ 2) 6= S(r). We put Ψ = f ′ (f−b) f(f−1) − g ′ (g−b) g(g−1) . Since a double zero of f − a is a zero of f′ and so a zero of g′, if Ψ 6≡ 0, then we get by Lemma 2.5(i) and Lemma 2.7, N(r, a; f |≥ 2) ≤ 2N(r, 0; Ψ) + S(r) = S(r) which is a contradiction. Hence Ψ ≡ 0 and so f′(f − b) f(f − 1) = g′(g − b) g(g − 1) . This shows that f − a has no simple zero because E1)(a; f) ⊆ E∞)(b; g). Since α ′ α = f ′ f−1 − g ′ g−1 . and E1)(0; f ′) ⊆ E ∞)(0; g ′), it follows that a double zero of f − a is a zero of α ′ α . So by Lemma 2.7 we get N(r, a; f |= 2) ≤ 2N(r, 0; α ′ α ) = S(r), which contradicts (ii) of Lemma 2.5. Subcase (ii) Let N(r, a; f |≥ 2) = S(r). Since f is not a bilinear transformation of g, we see that α, β and αβ are non-constant. Also we note that f = 1 − α 1 − αβ and g = (1 − α)β 1 − αβ . CUBO 16, 1 (2014) On a result of Q. Han, S. Mori and K. Tohge concerning . . . 57 We put F = (f−a)(1−αβ) = aαβ−α+1−a and w = F′ F . Also we note that F = (f−a) g − f f(g − 1) . Since by Lemma 2.6 N(r, ∞; F) = S(r) and w has only simple poles (if there is any), we get T(r, w) = m(r, w) + N(r, w) = N(r, 0; F) + S(r). (3.4) Now by Lemma 2.2 and (ix), (x) of Lemma 2.8 we obtain N(r, 0; F |≥ 2) ≤ N(r, a; f |≥ 2) + N(r, 0; f − g |≥ 2) + N(r, ∞; f |≥ 2) +N(r, 0; f − g | f = ∞) = S(r). (3.5) Hence from (3.4) and (3.5) we get T(r, w) = N(r, 0; F |≤ 1) + S(r) = N(r, a; f |≤ 1) + N0(r) + N2(r) + S(r), (3.6) where N2(r) is the counting function of those simple poles of f which are non-zero regular points of f − g. From the definitions of α and β we get { g − α′β (αβ)′ }( α′ α + β′ β ) ≡ f′(g − f) f(f − 1) . (3.7) From (3.7) we see that a simple pole of f which is a non-zero regular point of f − g is a regular point of { g − α′β (αβ)′ }( α′ α + β′ β ) . Hence it is either a pole of α′β (αβ)′ or a zero of α′ α + β′ β . Therefore by Lemma 2.7 and the first fundamental theorem we get N2(r) ≤ T ( r, α′ α + β′ β ) + T ( r, α′β (αβ)′ ) ≤ T ( r, α′ α + β′ β ) + T ( r, 1 1 + αβ′ α′β ) ≤ 2T ( r, α′ α ) + 2T ( r, β′ β ) + O(1) = S(r). So from (3.6) we get T(r, w) = N(r, a; f |≤ 1) + N0(r) + S(r). (3.8) By (ii) of Lemma 2.5 we get from (3.8) T(r, w) = T(r, f) + N0(r) + S(r). (3.9) 58 Indrajit Lahiri & Nintu Mandal CUBO 16, 1 (2014) Let τ1 = a − 1 b − 1 (ξ − bδ), τ2 = 1 2 · a − 1 b − 1 {ξ′ + ξ2 − b(δ′ + δ2)} and τ3 = 1 6 · a − 1 b − 1 {ξ′′ + 3ξξ′ + ξ3 − b(δ′′ + 3δδ′ + δ3)}, where ξ = α′ α and δ = α′ α + β′ β . By Lemma 2.7 we see that T(r, ξ) = S(r) and T(r, δ) = S(r). If τ1 ≡ 0, from (3.7) we get (g − b)δ ≡ f′(g − f) f(f − 1) . (3.10) Since E1)(a; f) ⊂ E(b; g), it follows from (3.10) that a simple zero of f − a, which is neither a zero nor a pole of δ, is a zero of g − b and so is a zero of f′. Hence N(r, a; f |≤ 1) = S(r), which contradicts (ii) of Lemma 2.5. Therefore τ1 6≡ 0. Let z0 be a simple zero of f − a and τ1(z0) 6= 0. Then g(z0) = b and so α(z0) = a − 1 b − 1 and β(z0) = b a . Expanding F around z0 in Taylor’s series we get −F(z) = τ1(z0)(z − z0) + τ2(z0)(z − z0) 2 + τ3(z0)(z − z0) 3 + O((z − z0) 4). Hence in some neighbourhood of z0 we obtain w(z) = 1 z − z0 + B(z0) 2 + C(z0)(z − z0) + O((z − z0) 2), where B = 2τ2 τ1 and C = 2τ3 τ1 − ( τ2 τ1 )2 . We put H = w′ + w2 − Bw − A, (3.11) where A = 3C − B2 4 − B′. Clearly T(r, A) + T(r, B) + T(r, C) = S(r) and since w = F′ F and F = (f − a) g − f f(g − 1) , we get by Lemma 2.1 and (3.9) that S(r, w) = S(r). Let H 6≡ 0. Then it is easy to see that z0 is a zero of H. So N(r, a; f |≤ 1) ≤ N(r, 0; H) + S(r) ≤ T(r, H) + S(r) = N(r, H) + S(r). (3.12) CUBO 16, 1 (2014) On a result of Q. Han, S. Mori and K. Tohge concerning . . . 59 From (ii) of Lemma 2.5 and (3.12) we get T(r, f) ≤ N(r, H) + S(r). (3.13) Let z1 be a pole of F. Then z1 is a simple pole of w. So if z1 is not a pole of A and B, then z1 is at most a double pole of H. Hence by Lemma 2.6 we get N(r, ∞; H | F = ∞) ≤ 2N(r, ∞; F) + S(r) = S(r), (3.14) where N(r, ∞; H | F = ∞) denotes the counting function of those poles of H which are also poles of F. Let z2 be a multiple zero of F. Then z2 is a simple pole of w. So if z2 is not a pole of A and B, then z2 is a pole of H of multiplicity at most two. Hence by (3.5) we get N(r, ∞; H | F = 0, ≥ 2) ≤ 2N(r, 0; F |≥ 2) + S(r) = S(r), (3.15) where N(r, ∞; H | F = 0, ≥ 2) denotes the counting function of those poles of H which are multiple zeros of F. Let z3 be a simple zero of F which is not a pole of A and B. Then in some neighbourhood of z3 we get F(z) = (z−z3)h(z), where h is analytic at z3 and h(z3) 6= 0. Hence in some neightbourhood of z3 we obtain H(z) = ( 2h′ h − B ) 1 z − z3 + h1, where h1 = ( h′ h ) ′ + ( h′ h )2 − Bh′ h − A. This shows that z3 is at most a simple pole of H. Since a simple zero of f − a is a zero of H and N(r, 0; F | f = t) ≤ N(r, 0; f − g |≥ 2) for t = 0, 1 and F = (f − a) g − f f(g − 1) , we get from (3.14) and (3.15) in view of (ix) of Lemma 2.8 N(r, H) = N(r, ∞; H | F = ∞) + N(r, ∞; H | F = 0) + S(r) ≤ N(r, 0; F |≤ 1) − N(r, a; f |≤ 1) + S(r) = N0(r) + N2(r) + S(r) = N0(r) + S(r), (3.16) where N(r, 0; F | f = t) denotes the counting function of those zeros of F which are zeros of f − t and N(r, ∞; H | F = 0) denotes the counting function of those poles of H which are zeros of F From (3.13) and (3.16) we obtain T(r, f) ≤ N0(r) + S(r), which by (iv) of Lemma 2.8 and 60 Indrajit Lahiri & Nintu Mandal CUBO 16, 1 (2014) Lemma 2.3 implies a contradiction. Therefore H ≡ 0 and so w′ + w2 − Bw − A ≡ 0 i.e., w′ w ≡ A w − w + B i.e., F′′ ≡ AF + BF′. Since F′ = a(αβ)′ − α′ and F′′ = a(αβ)′′ − α′′, we get from above Kαβ + Lα ≡ A(f − a)(1 − αβ), (3.17) where K = a{ (αβ)′′ αβ − B (αβ)′ αβ } and L = B α′ α − α′′ α . By Lemma 2.7 we see that T(r, K) = S(r) and T(r, L) = S(r). Since αβ = g(f − 1) f(g − 1) and α = f − 1 g − 1 , we get from (3.17) Kg + Lf ≡ A(f − a)(g − f) (f − 1) (3.18) Let z0 be a simple zero of f−a which is not a pole of A. Since E1)(a; f) ⊂ E∞)(b; g), it follows from 3.18 that z0 is a zero of bK + aL. Hence N(r, a; f |≤ 1) ≤ N(r, 0; bK + aL) + N(r, ∞; A) ≡ S(r), which contradicts (ii) of Lemma 2.5.This proves the theorem. Received: April 2012. Accepted: September 2012. References [1] T. C. Alzahary and H.X.Yi, Weighted sharing three values and uniqueness of meromorphic functions, J. Math.Anal. Appl., Vol. 295 (2004), pp.247-257. [2] G. Brosch, Eindeutigkeissätze für Meromorphe Funktionen (Thesis), Technical University of Aachen, (1989). [3] G.G.Gundersen, Meromorphic functions that share three or four values, J.London Math. Soc.,Vol.20, No.2 (1979),pp.457-466. [4] Q.Han, S. Mori and K. Tohge, On results of H. Ueda and G. Brosch concerning the unicity of meromorphic functions, J. 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