International Journal of Analysis and Applications ISSN 2291-8639 Volume 2, Number 2 (2013), 124-136 http://www.etamaths.com VALUE DISTRIBUTION AND UNIQUENESS THEOREMS FOR DIFFERENCE OF ENTIRE AND MEROMORPHIC FUNCTIONS SUBHAS S. BHOOSNURMATH∗ AND SMITA R KABBUR Abstract. We investigate the value distribution and uniqueness problems of difference polynomials of entire and meromorphic functions. 1. Introduction and Main Results Let f(z) be a meromorphic functions of finite order. We define difference oper- ator as , ∆cf = f(z + c) −f(z), and ∆nc f = ∆ n−1 c (∆cf), n ≥ 2, where c is a non-zero constant. In particular, if c = 1 we use the usual difference notation ∆cf = ∆f. Certain estimates involving the derivative f 7→ f′ of a meromorphic function play key roles in the construction and applications of classical Nevanlinna theory. Recently, there has been an increasing interest in studying difference equations in the complex plane. Halburd and Korhonen [2] established a version of Nevanlinna theory based on difference operator. Bergweiler and Langley [ 12] considered the value distribution of zeros of difference operators that can be viewed as discrete analogues of the zeros of f′(z). Ishizaki and Yanagihara [13] developed a version of Wiman - Valiron theory for difference equations of entire functions of small growth. Growth estimates for the difference analogue of logarithmic derivative f(z+c) f(z) were given by Halburd and Korhonen [1] . This result has potentially large number of applications in the study of difference equation. Many ideas and methods from the theory of differential equations are utilized to obtain information about mero- morphic solutions of difference equations. The analogue of Clunie Lemma used to ensure that finite order meromorphic solutions of certain non-linear difference equations have large number of poles. All concepts of Nevanlinna theory related to ramification have natural difference analogue. Let f be a transcendental entire function and n be a positive integer. Hayman [14] and Clunie [22] proved that fnf′ assumes every non-zero value a ∈ C infinitely often. Let f be a transcendental entire function. As for the value distribution of dif- ferential polynomial fn(f −1)f′, Fang [15] showed that fn(f −1)f′ assumes every non-zero value a ∈ C infinitely often for n ≥ 4. We recall the following uniqueness theorem due to Lin and Yi [16, 17]. 2010 Mathematics Subject Classification. 30D35. Key words and phrases. Meromorphic functions ,difference ,uniqueness. c©2013 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 124 VALUE DISTRIBUTION AND UNIQUENESS THEOREMS 125 2. Preliminaries and lemmas. Theorem A. Let f and g be two non-constant (resp transcendental) entire func- tions and let n ≥ 7 be an integer. If fn(f − 1)f′ and gn(g − 1)g′ share 1 (resp z) CM then f = g. Recently , value distribution in difference analogue has become a subject of great interest. For analogue results in difference Laine and Yang [4] proved the following result. Theorem B. Let f be a transcendental entire function of finite order and c be a non-zero complex constant. Then for n ≥ 2, f(z)nf(z + c) assumes every non - zero value a ∈ C infinitely often. Following analogue results in difference are proved by J. Zhang [5]. Theorem C. Let f(z) be a transcendental entire function of finite order and α(z) be a small function with respect to f(z). Suppose that c is non-zero complex con- stant and n ≥ 2 is an integer then f(z)n(f(z)−1)f(z +c) has infinitely many zeros. Theorem D. Let f(z) and g(z) be a transcendental entire function of finite or- der and α(z) be a small function with respect to both f(z) and g(z). Suppose that c is non-zero complex constant and let n ≥ 7. If f(z)n(f(z) − 1)f(z + c) and g(z)n(g(z) − 1)g(z + c) share α(z) CM, then f(z) ≡ g(z). In this paper we consider the difference analogue of fn(fm − 1)f′ and prove Theorem 2 . For m = 1 our result reduces to Theorem 4. Also proved Theorem 3 where we investigate the uniqueness problem when the difference polynomial share α(z) with ignoring multiplicity. For a non-constant meromorphic function f and a set S ⊂ C∪{∞} , we define Ef (S) = ⋃ a∈S {z ∈ f(z) −a = 0, counting multipilicities} . We say that two non-constant meromorphic functions f and g share a CM , if Ef (S) = Eg(S) and S = {a} . In 1976, Gross [18] proved that there exists three finite sets Sj(j = 1, 2, 3) such that for any two non-constant entire functions f and g , Ef (Sj) = Eg(Sj)(j = 1, 2, 3) imply f = g. In the same paper , Gross posed the following question: Question 1 . Can one find two (or possibly even one ) finite sets Sj(j = 1, 2) such that any two entire functions f and g satisfying Ef (Sj) = Eg(Sj)(j = 1, 2) must be identical? Many authors have worked on it and got related results.We recall the following result by Li and Yang [19]. Theorem E. Let m ≥ 2, n > 2m+ 6 with n and n−m having no common factors. Let a and b be two non-zero constants such that the equation ωn + aωn−m + b = 0 has no multiple roots. Let S = {ω|ωn + aωn−m + b = 0} . Then for any two non-constant meromorphic functions f and g, the conditions Ef (S) = Eg(S) and Ef {∞} = Eg {∞} imply f = g. J Zhang [5] considered the difference analogue of this result and proved the follow- ing result. Theorem F. Let m ≥ 2, n ≥ 2m+ 4 with n and n−m having no common factors. Let a and b be two non-zero constants such that the equation ωn + aωn−m + b = 0 has no multiple roots. Let S = {ω|ωn + aωn−m + b = 0} . Suppose that f is a non-constant meromorphic function of finite order. Then Ef(z)(S) = Ef(z+c)(S) 126 BHOOSNURMATH AND KABBUR and Ef(z) {∞} = Ef(z+c) {∞} imply f(z) = f(z + c). In 1998 , Frank - Reinders [20] proved following result. Let the polynomial P be defined as, P(ω) = (n− 1)(n− 2) 2 ωn −n(n− 2)ωn−1 + n(n− 1) 2 ωn−2 − c, where n(≥ 3) is an integer and c(6= 0, 1) is a constant. Theorem G.Let S = {ω|P(ω) = 0} , where P(ω) is as defined above and n ≥ 11 be an integer. Then for any two non-constant meromorphic functions f and g the condition Ef(z)(S) = Eg(z)(S) implies f = g. In this section, we consider the difference analogue of this result and prove Theorem 4. The techniques used here greatly improves the con- dition on n from ′n ≥ 11′ to ′n ≥ 8′. For a non-constant meromorphic function f and a set S ⊂ C∪{∞} , we define Ef (S) = ⋃ a∈S {z ∈ f(z) −a = 0, counting multipilicities} . We say that two non-constant meromorphic functions f and g share a CM , if Ef (S) = Eg(S) and S = {a} . In 1976, Gross [18] proved that there exists three finite sets Sj(j = 1, 2, 3) such that for any two non-constant entire functions f and g , Ef (Sj) = Eg(Sj)(j = 1, 2, 3) imply f = g. In the same paper , Gross posed the following question: Question 1 . Can one find two (or possibly even one ) finite sets Sj(j = 1, 2) such that any two entire functions f and g satisfying Ef (Sj) = Eg(Sj)(j = 1, 2) must be identical? Many authors have worked on it and got related results.We recall the following result by Li and Yang [19]. Theorem H. Let m ≥ 2, n > 2m+ 6 with n and n−m having no common factors. Let a and b be two non-zero constants such that the equation ωn + aωn−m + b = 0 has no multiple roots. Let S = {ω|ωn + aωn−m + b = 0} . Then for any two non-constant meromorphic functions f and g, the conditions Ef (S) = Eg(S) and Ef {∞} = Eg {∞} imply f = g. J Zhang [5] considered the difference analogue of this result and proved the follow- ing result. Theorem I. Let m ≥ 2, n ≥ 2m + 4 with n and n−m having no common factors. Let a and b be two non-zero constants such that the equation ωn + aωn−m + b = 0 has no multiple roots. Let S = {ω|ωn + aωn−m + b = 0} . Suppose that f is a non-constant meromorphic function of finite order. Then Ef(z)(S) = Ef(z+c)(S) and Ef(z) {∞} = Ef(z+c) {∞} imply f(z) = f(z + c). In 1998 , Frank - Reinders [20] proved following result. Let the polynomial P be defined as, P(ω) = (n− 1)(n− 2) 2 ωn −n(n− 2)ωn−1 + n(n− 1) 2 ωn−2 − c, where n(≥ 3) is an integer and c(6= 0, 1) is a constant. Theorem J. Let S = {ω|P(ω) = 0} , where P(ω) is as defined above and n ≥ 11 be an integer. Then for any two non-constant meromorphic functions f and g the condition Ef(z)(S) = Eg(z)(S) implies f = g. In this section, we consider the difference analogue of this result and VALUE DISTRIBUTION AND UNIQUENESS THEOREMS 127 prove Theorem 4. The techniques used here greatly improves the con- dition on n from ′n ≥ 11′ to ′n ≥ 8′. 3. Lemmas. In order to prove our results, we need following lemmas. Lemma 1. [12]. Let f(z) be a meromorphic function of finite order ρ and let c be a fixed non-zero complex constant. Then for each � > 0, we have m ( r, f(z + c) f(z) ) + m ( r, f(z) f(z + c) ) = O(rρ−1+�). Lemma 2. [6]. Let f(z) be a meromorphic function of finite order ρ and let c be a fixed non-zero complex constant . Then for each � > 0, we have T(r, f(z + c)) = T(r, f) + O(rρ−1+�). It is evident that S(r, f(z + c)) = S(r, f). Lemma 3. [7]. Let f(z) be a meromorphic function of finite order ρ and let c be a fixed non-zero complex constant . Then N ( r, 1 f(z + c) ) ≤ N ( r, 1 f ) + S(r, f), N(r, f(z + c)) ≤ N(r, f) + S(r, f), N ( r, 1 f(z + c) ) ≤ N ( r, 1 f ) + S(r, f), N(r, f(z + c)) ≤ N(r, f) + S(r, f), outside of possible exceptional set with finite logarithmic measure. Lemma 4. Let f(z) be a transcendental entire function of finite order ρ. Let F = f(z)n(f(z)m − 1)f(z + c). Then (1) T(r, F) = (n + m + 1)T(r, f) + S(r, f) From this lemma it is clear that S(r, F) = S(r, f) and similarly S(r, G) = S(r, g) Proof. Since f is entire function of finite order , we deduce from Lemma 1 and the standard Valiron Mohon’ko theorem that, (n + m + 1)T(r, f(z)) = T(r, f(z)n+1(f(z)m − 1)) + S(r, f) = m(r, f(z)n+1(f(z)m − 1)) + S(r, f) or, (n + m + 1)T(r, f(z)) ≤ m ( r, f(z)n+1(f(z)m − 1) f(z)n(f(z)m − 1)f(z + c) ) + m(r, F) + S(r, f) ≤ m ( r, f(z) f(z + c) ) + m(r, F) + S(r, f) ≤ T(r, F) + S(r, f) Therefore, we have (2) (n + m + 1)T(r, f(z)) ≤ T(r, F) + S(r, f) 128 BHOOSNURMATH AND KABBUR On the other hand , by Lemma 2 and the fact that f is a transcendental entire function of finite order, we get T(r, F ) ≤ T(r, f(z)n(f(z)m − 1)) + T(r, f(z + c)) + S(r, f) = (n + m)T(r, f(z)) + T(r, f(z + c)) + S(r, f) ≤ (n + m + 1)T(r, f(z)) + S(r, f) (3) i.e, T(r, F) ≤ (n + m + 1)T(r, f(z)) + S(r, f) Thus (3.1) follows from (3.2) and (3.3). Lemma 5. Let f(z) and g(z) be a meromorphic function of finite order. If n ≥ m + 6, n, m are positive integers and (4) f(z)n(f(z)m − 1)f(z + c) = g(z)n(g(z)m − 1)g(z + c) then f = tg, where tm = 1. Proof. Let h(z) = f(z) g(z) . If h(z)n+mh(z + c) 6= 1, then from (3.4) we have g(z)nh(z)n(g(z)mh(z)m − 1)g(z + c)h(z + c) = g(z)n(g(z)m − 1)g(z + c) h(z)n(g(z)mh(z)m − 1)h(z + c) = g(z)m − 1 h(z)n+mh(z + c)g(z)m −h(z)nh(z + c) −g(z)m + 1 = 0 g(z)m(h(z)n+mh(z + c) − 1) = h(z)nh(z + c) − 1 (5) or, g(z)m = h(z)nh(z + c) − 1 h(z)n+mh(z + c) − 1 If 1 is a Picard exceptional value of h(z)n+mh(z + c), applying the Nevanlinna second main theorem with Lemma 2, we get T(r, h(z)n+mh(z + c)) ≤ N(r, h(z)n+mh(z + c)) + N ( r, 1 h(z)n+mh(z + c) ) +N ( r, 1 h(z)n+mh(z + c) − 1 ) + S(r, h) ≤ 2T(r, h(z)) + 2T(r, h(z + c)) + S(r, h) (6) or, T(r, h(z)n+mh(z + c)) ≤ 4T(r, h(z)) + S(r, h) On the other hand, combining the standard Valiron - Mohon’ko theorem with (3.6) and Lemma 2, we get (n + m)T(r, h(z)) = T(r, h(z)n+m) + S(r, h) ≤ T(r, h(z)n+mh(z + c)) + T(r, h(z + c)) + S(r, h) ≤ 5T(r, h(z)) + S(r, h) (7) or, (n + m− 5)T(r, h(z)) ≤ S(r, h) which contradicts the hypothesis that n ≥ m + 6. Therefore 1 is not a picard exceptional value of h(z)n+mh(z+c). Thus there exists z0 such that h(z0) n+mh(z0 + c) = 1 then by (3.5) , we have h(z0) nh(z0 + c) = 1. Hence h(z0) m = 1, and (8) N ( r, 1 h(z)n+mh(z + c) − 1 ) ≤ N ( r, 1 h(z)m − 1 ) ≤ mT(r, h) + O(1) VALUE DISTRIBUTION AND UNIQUENESS THEOREMS 129 Denote, (9) H(z) = h(z)n+mh(z + c) We have T(r, H) ≤ (n + m + 1)T(r, h) + S(r, h). Applying the second main theorem to H and using Lemma 2 and (3.8), we get T(r, H) ≤ N(r, H(z)) + N ( r, 1 H(z) ) + N ( r, 1 H(z) − 1 ) + S(r, H) ≤ N(r, H(z)) + N ( r, 1 H(z) ) + mT(r, H(z)) + S(r, h) ≤ (m + 4)T(r, h) + S(r, h) Therefore, we have (10) T(r, H(z)) ≤ (m + 4)T(r, h) + S(r, h). On the other hand using (3.9) and (3.10) we have (n + m)T(r, h) = T(r, h(z)n+m) + S(r, h) ≤ T(r, H(z)) + T(r, h(z + c)) + S(r, h) ≤ (m + 5)T(r, h) + S(r, h) or, (n−5)T(r, h) ≤ S(r, h). which contradicts our hypothesis , n ≥ m+6. Therefore , h(z)n+mh(z + c) ≡ 1, and h(z)nh(z + c) ≡ 1. Thus h(z)m ≡ 1. Hence we get f(z) = tg(z), where tm = 1. Lemma 6. [21] . Let F and G be two non-constant meromorphic function. If F and G share 1 CM, then one of the following cases hold: i)max{T(r, F), T(r, G)} ≤ N2 ( r, 1 F ) +N2(r, F)+N2(r, G)+ ( r, 1 G ) +S(r, F )+S(r, G), ii) FG ≡ 1, iii) F ≡ G, where N2 ( r, 1 F ) denoted the counting function of zeros of F such that simple zeros are counted once and multiple zeros are counted twice. Lemma 7. [8] . Let F and G be two non-constant meromorphic function. Let F and G share 1 IM. Let, (11) H = F ′′ F ′ − 2 F ′ F − 1 − G′′ G′ + 2 G′ G− 1 If H 6= 0, then T(r, F) + T(r, G) ≤ 2 [ N2 ( r, 1 F ) + N2(r, F) + N2(r, G) + N2 ( r, 1 G )] +3 [ N(r, F) + N(r, G) + N ( r, 1 F ) + N ( r, 1 G )] (12) +S(r, F) + S(r, G) Lemma 8. [20].Let Q(ω) = (n− 1)2(ωn − 1)(ωn−2 − 1) −n(n− 2)(ωn−1 − 1)2 be a polynomial of degree 2n− 2 (n ≥ 3). Then Q(ω) = (ω − 1)4(ω −β1)(ω −β2), . . . , (ω −β2n−6) 130 BHOOSNURMATH AND KABBUR where βj ∈ C \{0, 1}. 4. Statement and Proof of Main Result. Theorem 1. Let f(z) be entire function of finite order and α(z) be a small func- tion with respect to f(z). Suppose that c is a non-zero complex constant and n is an integer. If n ≥ 2, then f(z)n(f(z)m−1)f(z+c)−α(z) has infinitely many zeros. Proof. Let F(z) = f(z)n(f(z)m − 1)f(z + c). Contrary to the assumption, suppose f(z)n(f(z)m − 1)f(z + c) −α(z) has finitely many zeros. By second fundamental theorem, Lemma 3 , we have T(r, F) ≤ N(r, F(z)) + N ( r, 1 F(z) ) + N ( r, 1 F(z) − 1 ) + S(r, F) ≤ N ( r, 1 f(z)n(f(z)m − 1)f(z + c) ) + N ( r, 1 F(z) −α(z) ) + S(r, f) ≤ (m + 2)T(r, f) + S(r, f) using Lemma 4 we get (n− 1)T(r, f) ≤ S(r, f), which contradicts our assumption. Theorem 2. Let f(z) and g(z) be two transcendental entire function of finite order and α(z) be a small function with respect to f(z) and g(z). Suppose that c is a non-zero complex constant and let n ≥ m + 6. If f(z)n(f(z)m −1)f(z + c) and g(z)n(g(z)m − 1)g(z + c) share α(z) CM, then f(z) ≡ tg(z), where tm = 1. Proof. Let (13) F(z) = f(z)n(f(z)m − 1)f(z + c) α(z) and G(z) = g(z)n(g(z)m − 1)g(z + c) α(z) Then F(z) and G(z) share 1 CM, except the zeros and poles of α(z). By Lemma 4 (14) T(r, F(z)) = (n + m + 1)T(r, f) + S(r, f) (15) and T(r, G(z)) = (n + m + 1)T(r, g) + S(r, g) Since f and g are transcendental entire functions, N2(r, F) = S(r, f), and N2(r, G) = S(r, g) By Lemma 3, we have N2 ( r, 1 F ) ≤ 2N ( r, 1 fn ) + N ( r, 1 fm − 1 ) + N ( r, 1 f(z + c) ) + S(r, f) ≤ 2N ( r, 1 f ) + mT(r, f) + T(r, f(z + c) + S(r, f) ≤ (m + 3)T(r, f) + S(r, f) Therefore (16) N2(r, F) + N2 ( r, 1 F ) ≤ (m + 3)T(r, f) + S(r, f) VALUE DISTRIBUTION AND UNIQUENESS THEOREMS 131 Similarly, (17) N2(r, G) + N2 ( r, 1 G ) ≤ (m + 3)T(r, g) + S(r, g) By condition of the theorem, Suppose case (i) of Lemma 6 holds. Substituting (4.4) and (4.5) in (i) of Lemma 6, we have T(r, F) + T(r, G) ≤ (2m + 6)(T(r, f) + T(r, g)) + S(r, f) + S(r, g) Using (4.2) and (4.3) we get (n + m + 1){T(r, f) + T(r, g)} ≤ (2m + 6)(T(r, f) + T(r, g)) + S(r, f) + S(r, g) or, (n−m− 5){T(r, f) + T(r, g)} ≤ S(r, f) + S(r, g) which contradicts our assumption, n ≥ m + 6. Hence by Lemma 6 F(z)G(z) ≡ α(z)2 or F(z) ≡ G(z). Suppose, F(z)G(z) ≡ α(z)2. That is f(z)n (f(z)m − 1) f(z + c) g(z)n (g(z)m − 1) g(z + c) = α(z)2 or, f(z)n (f(z)−1) (f(z)m−1 −···−1) f(z+c) g(z)n (g(z)−1) (g(z)m−1 −···−1) g(z+c) = α(z)2 Then, N ( r, 1 f ) = S(r, f) and N ( r, 1 f − 1 ) = S(r, f) From this we have, δ(0, f) + δ(1, f) + δ(∞, f) = 3, which is impossible. From this we conclude that, F(z) ≡ G(z). i, e. f(z)n(f(z)m − 1)f(z + c) ≡ g(z)n(g(z)m − 1)g(z + c) Using Lemma 5, we conclude that f(z) = tg(z), where tm = 1. Theorem 3. Let f(z) and g(z) be two transcendental entire function of finite order, and α(z) be a small function with respect to f(z) and g(z). Suppose that c is a non-zero complex constant and let n ≥ 4m + 12. If f(z)n(f(z)m − 1)f(z + c) and g(z)n(g(z)m − 1)g(z + c) share α(z) IM, then f(z) = tg(z), where tm = 1. Proof. Let (18) F(z) = f(z)n(f(z)m − 1)f(z + c) α(z) and G(z) = g(z)n(g(z)m − 1)g(z + c) α(z) Then F(z) and G(z) share 1 IM. Let H be as defined in Lemma 7. Using Lemma 3, we have N ( r, 1 F ) ≤ N ( r, 1 fn ) + N ( r, 1 fm − 1 ) + N ( r, 1 f(z + c) ) + S(r, f) (19) ≤ (m + 2)T(r, f) + S(r, f) 132 BHOOSNURMATH AND KABBUR Using (4.4), (4.5) and (4.7) in (4.10) of Lemma 7, we get T(r, F) + T(r, G) ≤ 2 [ N2 ( r, 1 f ) + N2 ( r, 1 G )] + 3 [ N(r, 1 F ) + N(r, 1 G ) ] +S(r, f) + S(r, g) ≤ 2(m + 3) [T(r, f) + T(r, g)] + 3(m + 2) [T(r, f) + T(r, g)] +S(r, f) + S(r, g) ≤ (5m + 12) [T(r, f) + T(r, g)] + S(r, f) + S(r, g) Using (4.2) and (4.3) , we have (20) (n + 4m− 11)(T(r, f) + T(r, g)) ≤ S(r, f) + S(r, g) which contradicts the hypothesis that n ≥ 4m + 12. Thus we get H ≡ 0. Integrating H twice , we obtain (21) F = (b + 1)G + (a− b− 1) bG + (a− b) and G = (a− b− 1) − (a− b)F Fb− (b + 1) In the following , we will prove that FG ≡ 1 or F ≡ G. Case 1. b 6= 0, −1. If a− b− 1 6= 0, then by (4.9) we have (22) N ( r, 1 F ) = N ( r, 1 G− a−b−1 b+1 ) By the Nevanlinna second main theorem and Lemma 4, we have T(r, G) ≤ N(r, G) + N ( r, 1 G ) + N ( r, 1 G− a−b−1 b+1 ) + S(r, G) ≤ N ( r, 1 G ) + N ( r, 1 F ) + S(r, g) ≤ N ( r, 1 g(z)n(g(z)m − 1)g(z + c) ) + N ( r, 1 f(z)n(f(z)m − 1)f(z + c) ) + S(r, g) ≤ (m + 2)T(r, g) + (m + 2)T(r, f) + S(r, g) Therefore we have, (23) (n + m + 1)T(r, g) ≤ (m + 2)T(r, f) + (m + 2)T(r, g) + S(r, g) (24) (n + m + 1)T(r, f) ≤ (m + 2)T(r, g) + (m + 2)T(r, f) + S(r, f) From (4.11) and (4.12) , we get (n + m + 1)(T(r, f) + T(r, g)) ≤ (m + 2)(T(r, f) + T(r, g)) + S(r, f) + S(r, g) or, (n− 1)(T(r, f) + T(r, g)) ≤ S(r, f) + S(r, g) This contradicts the assumption that n ≥ 4m + 12. Thus a− b− 1 = 0, then by (4.9), we have (25) F = (b + 1)G bG + 1 Since F is an entire finction, from (4.13) we have N ( r, 1 G + 1 b ) = 0. VALUE DISTRIBUTION AND UNIQUENESS THEOREMS 133 Proceeding as above we deduce a contradiction. Case 2: b = −1, a 6= −1. Then by (4.9) we get F = a (a + 1) −G , and N ( r, 1 G− (a + 1) ) = N(r, F) = 0 Similarly as in Case 1, we get contradiction. Hence a = −1, Thus we get FG ≡ 1. Proceeding as in theorem 2 , we get contradiction. Case 3: If b = 0, a 6= 1. From (4.9) we have F = G + (a− 1) a and N ( r, 1 F ) = N ( r, 1 G− (a + 1) ) Proceeding as in Case 1 , we get a contradiction. Thus a = 1, implies F = G. Proceeding as in theorem 2 we get f = tg, where tm = 1. Theorem 4. Let n ≥ 8 be an integer and c(6= 0, 1) is a constant such that the equation P(ω) = 0 has no multiple roots. Where (26) P(ω) = (n− 1)(n− 2) 2 ωn −n(n− 2)ωn−1 + n(n− 1) 2 ωn−2 − c Let S = {ω|P(ω) = 0} , Suppose that f is a non-constant meromorphic function of finite order then Ef(z)(S) = Ef(z+c)(S) and Ef(z)({∞}) = Ef(z+c)({∞}) implies f(z) = f(z + c). Proof. From the condition of the theorem, we have Ef(z)(S) = Ef(z+c)(S) Then there exists a polynomial Q(z), such that (27) P(z + c) P(z) = eQ(z) and T(r, eQ(z)) = m(r, eQ(z)) = S(r, f) Rewriting (4.15), we have P(z + c) = eQ(z)P(z) or, (n− 1)(n− 2) 2 f(z + c)n −n(n− 2)f(z + c)n−1 + n(n− 1) 2 f(z + c)n−2 − c = eQ(z) [ (n− 1)(n− 2) 2 f(z)n −n(n− 2)f(z)n−1 + n(n− 1) 2 f(z)n−2 − c ] or, f(z + c)n−2 [ (n− 1)(n− 2) 2 f(z + c)2 −n(n− 2)f(z + c) + n(n− 1) 2 ] (28) = eQ(z) [ (n− 1)(n− 2) 2 f(z)n −n(n− 2)f(z)n−1 + n(n− 1) 2 f(z)n−2 − c + ce−Q(z) ] Let F(z) = (n− 1)(n− 2) 2 f(z)n −n(n− 2)f(z)n−1 + n(n− 1) 2 f(z)n−2, 134 BHOOSNURMATH AND KABBUR or, F(z) = f(z)n−2 [ (n− 1)(n− 2) 2 f(z)2 −n(n− 2)f(z) + n(n− 1) 2 ] , or, (29) F(z) = f(z)n−2(f −α1)(f −α2) where α1, and α2 are roots of the equation, (n− 1)(n− 2) 2 f(z)2 −n(n− 2)f(z) + n(n− 1) 2 = 0. Therefore (4.16) can be rewritten as, f(z + c)n−2 [ (n− 1)(n− 2) 2 f(z + c)2 −n(n− 2)f(z + c) + n(n− 1) 2 ] (30) = eQ(z) [ F(z) − (c− ce−Q(z)) ] Using standard Valiron Moho’nko theorem, we get (31) T(r, F) = nT(r, f) + S(r, f) Applying second main theorem to F and using (4.17) and (4.18), we have T(r, F) ≤ N(r, F) + N ( r, 1 F ) + N ( r, 1 F − (c− ce−Q(z) ) + S(r, F) ≤ N(r, F) + N ( r, 1 F ) + N ( r, 1 f(z + c)n−2 ) +N ( r, 1 (n−1)(n−2) 2 f(z + c)2 −n(n− 2)f(z + c) + n(n−1) 2 ) + S(r, f) or, T(r,F) ≤ N(r, f) + N ( r, 1 f(z)n−2 ) + N ( r, 1 f −α1 ) +N ( r, 1 f −α2 ) + N ( r, 1 f(z + c)n−2 ) +N ( r, 1 (n−1)(n−2) 2 f(z + c)2 −n(n− 2)f(z + c) + n(n−1) 2 ) + S(r, f) ≤ 4T(r, f) + 3T(r, f(z + c)) + S(r, f) By (4.19), we have n T(r, f) ≤ 4T(r, f) + 3T(r, f(z + c)) + S(r, f) (32) or, (n− 4)T(r, f(z)) ≤ 3T(r, f(z + c)) + S(r, f) (33) Similarly (n− 4)T(r, f(z + c)) ≤ 3T(r, f(z)) + S(r, f) Using (4.20) and (4.21), we have (n− 4) (T(r, f(z)) + T(r, f(z + c))) ≤ 3 (T(r, f(z)) + T(r, f(z + c))) + S(r, f) or, (n− 7) (T(r, f(z)) + T(r, f(z + c))) ≤ S(r, f), VALUE DISTRIBUTION AND UNIQUENESS THEOREMS 135 which contradicts the assumption n ≥ 8. Therefore eQ(z) = 1. Hence from (4.15), we get P(z + c) = P(z). or, (n− 1)(n− 2) 2 f(z + c)n −n(n− 2)f(z + c)n−1 + n(n− 1) 2 f(z + c)n−2 − c = (n− 1)(n− 2) 2 f(z)n −n(n− 2)f(z)n−1 + n(n− 1) 2 f(z)n−2 − c or, (n− 1)(n− 2) 2 (f(z)n −f(z + c)n) −n(n− 2) ( f(z)n−1 −f(z + c)n−1 ) (34) + n(n− 1) 2 ( f(z)n−2 −f(z + c)n−2 ) ≡ 0. Taking h(z) = f(z + c) f(z) , we get (35) (n− 1)(n− 2) 2 (hn − 1) f(z)2−n(n−2) ( hn−1 − 1 ) f(z)+ n(n− 1) 2 ( hn−2 − 1 ) = 0 Suppose h is not a constant. From (4.23), we have (36) { (n− 1)(n− 2)(hn − 1)f(z) −n(n− 2)(hn−1 − 1) }2 = −n(n− 2)Q(h) where Q(h) is defined as in Lemma 8. Using (4.24) and Lemma 8, we get{ (n− 1)(n− 2)(hn − 1)f(z) −n(n− 2)(hn−1 − 1) }2 (37) = −n(n− 2)(h− 1)4(h−β1)(h−β2), . . . , (h−β2n−6) From (4.24), all zeros of (h−βj) have order at least 2. Applying second fundamental theorem to h, we get (2n− 8)T(r, h) ≤ 2n−6∑ j = 1 N ( r, 1 h−βj ) + S(r, h) ≤ 1 2 2n−6∑ j = 1 N ( r, 1 h−βj ) + S(r, h) ≤ (n− 3)T(r, h) + S(r, h) which contradicts the assumption , n ≥ 8. Thus h is a constant, From (5.3.23 ) we obtain hn − 1 = 0. Therefore h = 1. Hence we conclude that f(z) ≡ f(z + c). Theorem 5. Let n ≥ 7 be an integer and c(6= 0, 1) is a constant such that the equation P(ω) = 0 has no multiple roots, where P(ω) is as defined in theorem 4 . Let S = {ω|P(ω) = 0} , Suppose that f is a non-constant entire function of finite order then Ef(z)(S) = Ef(z+c)(S) implies f(z) = f(z + c). Proof. f and g are entire functions. Taking N(r,f) = N(r,g) = 0, in the above theorem, we obtain conclusion of this theorem. 6. 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