On the Uniqueness Results and Value Distribution of Meromorphic Mappings

Transcription

1 mathematics Article On the Uniqueness Results and Value Distribution of Meromorphic Mappings Rahman Ullah, Xiao-Min Li, Faiz Faizullah 2, *, Hong-Xun Yi 3 and Riaz Ahmad Khan 4 School of Mathematical Sciences, Ocean University of China, Qingdao 26600, China; rahman yzi@yahoo.com R.U.; xmli0267@gmail.com X.-M.L. 2 Department of BS & H, College of E & ME, National University of Sciences and Technology NUST, Islamabad 44000, Pakistan 3 Department of Mathematics, Shandong University, Jinan 25000, China; hxyi@sdu.edu.cn 4 SMME, National University of Sciences and Technology NUST, Islamabad 44000, Pakistan; riaz.ahmad@smme.nust.edu.pk * Correspondence: faiz math@ceme.nust.edu.pk; Tel.: Academic Editor: Lei Ni Received: 9 June 207; Accepted: 0 August 207; Published: 7 August 207 Abstract: This research concentrates on the analysis of meromorphic mappings. We derived several important results for value distribution of specific difference polynomials of meromorphic mappings, which generalize the work of Laine and Yang. In addition, we proved uniqueness theorems of meromorphic mappings. The difference polynomials of these functions have the same fixed points or share a nonzero value. This extends the research work of Qi, Yang and Liu, where they used the finite ordered meromorphic mappings. Keywords: meromorphic mappings; value distribution; difference polynomials; uniqueness results. Introduction Let Ω be the set of finite linear measure of positive real numbers, which may not be the same at every occurrence. Assume Tq, α denotes the Nevanlinna characteristic of a nonconstant meromorphic mapping α and Sq, α represents any quantity fulfilling Sq, α = o{tq, α}, as q and q Ω. Consider a point c in the extended plane. Indicate two nonconstant meromorphic mappings by α and β. The mappings α and β share the value c IM, if they have the same c-points ignoring multiplicities []. Also, c is called a small mapping of α, provided that c is a meromorphic mapping fulfilling Tq, c = Sq, α []. All through the current paper, we consider meromorphic mappings in the complex plane and represent the order of α by ρα. Consider the following result which was proved by Clunie [2] and Hayman [3]: Theorem. Suppose k be a positive integer. Let αy represents a transcendental entire mapping. Then there are infinitely many zeros of α k yα y. Reading Theorem, the following problem arises: Problem. Let λ = 0 be a complex number. What will be the conclusion of Theorem if α k yα y of Theorem is replaced with α k yαy + λ or α k y λ αy for a transcendental meromorphic mapping αy? In this direction, Laine and Yang [4] derived the following result to deal with Problem : Theorem 2. Let λ = 0 be a complex number and αy be a finite order transcendental entire mapping. Then αy k αy + λ assumes every finite nonzero value c infinitely often for k 2. Mathematics 207, 5, 42; doi:0.3390/math

2 Mathematics 207, 5, 42 2 of 8 We now give the following two examples, for details see [4,5]. Example. Let αy = + e y. Then αyαy + πi = e 2y has no zeros. This example shows that Theorem 2 does not remain valid if k =. Example 2. Let αy = e ey. Then αy 2 αy + λ 2 = and ρα =, where λ is a nonzero constant satisfying e λ = 2. Evidently, αy 2 αy + λ 2 have no zeros. This example shows that Theorem 2 does not remain valid if α is of infinite order. Recently Liu and Yang proved the following result [5]: Theorem 3. Let k 2 be an integer and λ = 0 be a complex number. Assume that αy be a finite order transcendental entire mapping. Let Py 0 be a polynomial. Then there are infinitely many zeros of α k yαy + λ Py. We recall the following two examples from [5]. Example 3. Let αy = e ey. Then αy k αy + λ Py = Py and ρα =, where η is a nonzero constant satisfying e λ = k, Py is a nonconstant polynomial, k is a positive integer. Evidently, αy k αy + λ Py has finitely many zeros. This example shows that the condition ρα < in Theorem 3 is necessary. In addition to Theorems 2 and 3 to deal with Problem we will prove the following theorem: Theorem 4. Let k 7 be an integer. Suppose that the order of a transcendental meromorphic mapping α is given by ρα = ρ <. Let λ be a nonconstant complex number and λ α 0. Assume P 0 be a polynomial. Then as q and q Ω k 6Tq, α N q, α k λ α P where Ω, + is a subset of finite logarithmic measure. Tq, α q ε + Olog q, The following definition is borrowed from [6] which will be used in the forthcoming work of this article. Definition. Let α be a nonconstant meromorphic function. We define difference operators as λ αy = αy + λ αy, n λαy = n λ λ αy, where λ is a nonzero complex number, n 2 is a positive integer. If λ =, we denote λ αy = αy. Moreover, n λ αy = n j=0 n n j αy + jλ. 2 j The proof of Theorem 4 yields the following interesting result, which will be proved in Section 3. Theorem 5. Let k 3 be an integer. Suppose that the order of a transcendental entire mapping α is given by ρα = ρ <. Let λ be a nonconstant complex number and λ α 0. Assume that P 0 be a polynomial. Then as q and q Ω, k 2Tq, α N q, α k λ α P where Ω, + is a subset of finite logarithmic measure. Tq, α q ε + Olog q, 3

3 Mathematics 207, 5, 42 3 of 8 Now consider the example given below, which indicates that the condition ρα < in Theorems 4 and 5 is necessary. Example 4. Let αy = e ey. Then ρα = and αy k λ αy = e k+ey has no zeros, where k Z and λ is a nonzero constant satisfying e λ = k. From Theorems 4 and 5 we can get the following results respectively. Corollary. Let k 7 be an integer. Suppose that the order of transcendental meromorphic mapping α is given by ρα <. Consider a nonconstant complex number such that λ α 0. Assume that P 0 be a polynomial. Then there are infinitely many zeros of α k λ α P. Corollary 2. Let k 3 be an integer. Suppose that the order of a transcendental entire mapping α is given by ρα <. Let λ be a nonconstant complex number such that λ α 0. Suppose that P 0 be a polynomial. Then there are infinitely many zeros of α k λ α P. Corresponding to Theorem 2, the following uniqueness theorem was derived by Qi-Yang-Liu [7]. Theorem 6. Suppose that k 6 be an integer and λ = 0 be a complex number. Let the distinct transcendental entire mappings α and β have finite orders. Assume that αy k αy + λ y and βy k βy + λ y share 0 CM. Then α = tβ, where t = is a constant fulfilling t k+ =. He further studied the following result [7]. Theorem 7. Let k 6 is an integer and λ = 0 is a complex number. Assume that the distinct transcendental entire mappings α and β have finite orders. Let αy k αy + λ and βy k βy + λ share CM. Then α = tβ, where t = is a constant fulfilling t k+ =. From Theorem 4 we will prove the following uniqueness results for meromorphic mappings associated to difference operators. Theorem 8. Suppose that k 2 be an integer and P 0 be a polynomial. Let the distinct transcendental meromorphic mappings α and β have finite orders. Assume that λ = 0 be a complex number such that λ α 0 and λ β 0. Suppose that α k λ α P and β k λ β P share 0 CM. Then i If k 2 and if α k λ α/p is a Möbius transformation of β k λ β/p, then [α k λ α][β k λ β] = P 2 or α k λ α = β k λ β. ii If k 6, then [α k λ α][β k λ β] = P 2 or α k λ α = β k λ β. Theorem 9. Let k 6 be an integer and λ = 0 be a complex number. Assume that the distinct nonconstant meromorphic mappings α and β have finite order. Suppose that α and β share 0, CM, α k λ α and β k λ β share CM. If lim sup q then one of the two cases given below holds: i f = c β, where c = is a constant fulfilling c k+ =. ii For all y C, αy = αy + λ and βy = βy + λ. N q, + N q, αy βy < 3 Tq, αy + Tq, βy k, 4 Proving Theorem 9 in Section 3, we can obtain the following interesting uniqueness results. In the complex plane, the difference polynomials of the following meromorphic mappings have the same fixed points.

4 Mathematics 207, 5, 42 4 of 8 Theorem 0. Suppose that k 6 be an integer and λ = 0 be a complex number. Let the distinct nonconstant meromorphic mappings α and β have finite orders. Suppose that α and β share 0, CM, α k y λ αy y and β k y λ βy y share 0 CM. If the inequality 4 holds, then one of the conclusions i and ii of Theorem 9 can occur. In view of Theorem 5 and Lemma 2.9, we will derive the following results for entire mappings. Theorem. Assume that k 5 be an integer, λ = 0 be a complex number and P 0 be a polynomial. Let the distinct transcendental meromorphic mappings α and β have finite orders. Suppose that λ α 0 and λ β 0. Let α n λ α P and β k λ β P share 0 CM. Then i If k 5 and if α k λ α/p is a Möbius transformation of β k λ β/p, then [α k λ α][β k λ β] = P 2 or α k λ α = β k λ β. ii If n 7, then [α k λ α][β k λ β] = P 2 or α k λ α = β k λ β. The above theorem gives us the following two uniqueness theorems of entire mappings. The difference polynomials of the mentioned mappings share a nonzero constant or have the same fixed points in the plane. Theorem 2. Suppose that k 7 be an integer, λ be a nonzero complex number. Let the distinct nonconstant entire mappings α and β have finite order. Assume that α and β share 0, CM, α k λ α and β k λ β share CM. Then one of the following arguments holds. i α = c β, where c = is a constant fulfilling c k+ =. ii For all y C, αy = αy + λ and βy = βy + λ. iii αy = e a y+a 0 and βy = e a y+b 0, where a, a 0 and b 0 are complex numbers such that a = 0 and e k+a 0+b 0 e a η e a η =. Theorem 3. Suppose that k 7 be an integer and λ be a nonzero complex number. Let the distinct nonconstant entire mappings α and β have finite orders. Assume that α and β share 0, CM, α k y λ αy y and β k y λ βy y share 0 CM. Then one of the conclusions i and ii of Theorem 2 holds. 2. Preliminaries Building on the previous ideas of meromorphic mapping and Nevanlinna theory, this section contains the fundamental definitions, notions and results required for the further study of the subject. For more details on the concepts briefly discussed, readers are suggested to consult the papers [8 4]. Let c C { }, p Z + and α be meromorphic mapping, which is not a constant. Then we give the following three definitions [5,6]. Definition 2. The counting mapping of those c-points of α whose multiplicities are not greater than p is denoted N p q, /α c. The corresponding reduced counting mapping ignoring multiplicities is indicated by N p q, /α c. N p q, /α c represents the counting mapping of those c-points of α counted with proper multiplicities whose multiplicities are not less than p. By N p q, /α c we present the corresponding reduced counting mapping ignoring multiplicities, where N p q, /α c, N p q, /α c, N p q, /α c and N p q, /α c mean N p q, α, N p q, α, N p q, α and N p q, α respectively, if c =. Definition 3. Assume that k is a nonnegative integer. Let α be a meromorphic mapping, which is not constant. Suppose that c be any value in the extended complex plane. Then we set N k q, = N q, α c + N α c 2 q, + + N α c k q,. α c

5 Mathematics 207, 5, 42 5 of 8 Definition 4. Let k 2 be an integer. Assume that α is a meromorphic mapping, which is not constant. The difference operators are defined by λ αy = αy + λ αy, k λαy = k λ λ αy, where λ is a nonzero complex number. If λ =, we represent λ αy = αy. Also, k k k λ αy = k j αy + jλ. 5 j j=0 Now we state some important lemmas. These lemmas will be used in the proof of our forthcoming results. The following first lemma is borrowed from [3] while second and third lemmas can be found in [7]. Lemma. In the complex plane, consider a nonconstant meromorphic mapping α. Let c 0, c,, c k, c k be arbitrary constants and Pα = c k αy k + c k αy k + + c αy + c 0, 6 where c k = 0. Then mq, Pα = kmq, α + O. Lemma 2. Let λ C. Consider a meromorphic mapping α, which is not constant. If α is of finite order, then m q, αy + λ = O αy Tq, αy log q q for every q outside of a set Ω fulfilling lim sup q Ω [,q dt/t = 0, log q i.e., outside of a set Ω of zero logarithmic density. If ρ 2 α = ρ 2 < and ε > 0. Then for every q outside of a finite logarithmic measure αy + λ Tq, αy m q, = o αy q ρ, 2 ε where ε is a positive number. Lemma 3. Let s R +. Consider a continuous mapping T : [0, + [0, +, which is nondecreasing. If lim sup q log log Tq log q = ζ <, and δ 0, ζ, i.e., the hyper-order of T is strictly less than one. Then Tq Tq + s = Tq, where outside of a set of finite logarithmic measure, q runs to infinity. For the next four lemmas see [,8]. q δ

6 Mathematics 207, 5, 42 6 of 8 Lemma 4. Consider two meromorphic mappings F and G, which are nonconstant and G is a Möbius transformation of F. Assume that a subset I R + with its linear measure mesi = + exists and N q, + Nq, F + N q, + Nq, G < η Tq, F, F G as q I and q, where λ <. If a point y 0 C exists in such a way that Fy 0 = Gy 0 =, then F = G or FG =. Lemma 5. Consider two meromorphic mappings F and G, which are nonconstant. Let F and G share CM. Assume that a subset I R + with its linear measure mesi = exists and N 2 q, F + N 2 q, G + N 2 q, + N 2 q, < µ Tq, F G where µ <, Tq = max{tq, F, Tq, G}. Then F = G or FG =. Lemma 6. Consider two meromorphic mappings F and G, which are nonconstant. Let F and G share, CM. Assume that a subset I R + with its linear measure mesi = + exists and N 2 q, + N 2 q, + 2Nq, F < λtq + Sq, F G as q I and q, where λ <, Tq = max{tq, F, Tq, G} and Sq = o{tq}, as q I and q. Then F = G or FG =. Lemma 7. Consider the nonconstant meromorphic mappings α, α 2,, α n. Let α k+ 0 be a meromorphic mapping such that k+ f j =. If a subset I R + fulfilling mesi = exists and j= k+ N q, αi i= + k k+ i=, i =j as q and q I, where λ <, then α k+ =. The following lemma can be found in [9]. Nq, α i < {λ }Tq, α j, j =, 2,, k, Lemma 8. Consider two rational mapping α and β, which are nonconstant. Let they share 0,, CM. Then α = β. Now let Py = a + iby k + be a polynomial of degree k, where a and b are real numbers such that a + ib = 0, and let y = qe iθ. Then Re{a + ibe kθi } = a cos kθ b sin kθ =: δp, θ. The following results will be utilized to prove Theorem 9. For its proof see [20]. Lemma 9. Let Py be a polynomial of degree k, and let ε > 0 be a given constant. Then we have i If δp, θ > 0, then there exists an qθ > 0 such that for any q > qθ, we have e Pqeiθ exp εδp, θq k. ii If δp, θ < 0, then there exists an qθ < 0 such that for any q > qθ, we have e Pqeiθ exp εδp, θq k. The proof of the following lemma can be found on page 77 of [2].

7 Mathematics 207, 5, 42 7 of 8 Lemma 0. Assume that αy be an analytic mapping of y = q iθ, regular in the region D between two straight lines making an angle π/a at the origin and on the lines themselves. Let f y M on the lines, where M > 0 be some constant, and that, as y, αy = Oe qβ, where b < a, uniformly in the angle. Then actually the inequality αy M holds throughout the region D. 3. Proof of Results In this section, we provide the proof of theorems, stated in first section. Proof. Theorem 4: In view of Lemmas and 2 we obtain k + Tq, αy = Tq, α k+ y + O αy = T q, η αy αk y λ αy + O T q, λαy + T q, α k y αy λ αy + O αy + λ αy + λ = m q, + N q, αy αy N q, + N q, αy + λ + T αy + T q, α k y η αy q, α k y λ αy + O Tq, αy, q ε 7 as q Ω and q. Noting that lim sup q log log N q, αy log q ρ 2 α = 0. By virtue of Lemma 3 as q and q Ω Tq, αy N q, αy + λ N q + λ, αy = N q, αy, 8 where Ω, indicates a subset with logarithmic measure log mesω <. Similarly and q ε q ε Tq, βy N q, βy + λ N q, βy, 9 Tq, αy T q, αy + λ Tq, αy q ε Tq, αy N q, αy + λ Nq, αy, as q Ω and q. By virtue of 7 and 8 we get k + Tq, αy + O N q, αy Tq, αy q ε 2Tq, αy + T q ε + N q, αy + T q, α k y λ αy q, α k y λ αy Tq, αy q ε, 0 i.e., k Tq, αy T q, α k y λ αy Tq, αy q ε, 2

8 Mathematics 207, 5, 42 8 of 8 as q Ω and q. On the other hand, by 0, and Theorem.36 of [] we get T q, α k y λ αy N q, α k y λ αy + Olog q + N q, α k + N q, y λ αy α k y λ αy Py Nq, αy + Nq, αy + λ + N q, + N q, αy + N q, α k + Olog q y λ αy Py 2Nq, αy + N q, + N q, αy αy + λ αy Tq, αy + N q, α k y λ αy Py q ε + Olog q, 4Tq, αy + Tq, αy + λ + N q, α k y λ αy Py Tq, αy q ε + Olog q Tq, αy 5Tq, αy + N q, α k y λ αy Py αy + λ αy q ε + Olog q, 3 as q Ω and q. From 2 and 3 we can get the conclusion of Theorem.. Thus the proof stands completed. Proof. Theorem 8: To prove this theorem let us set F y = αyk λ αy Py, G y = βyk λ αy. 4 Py Applying similar arguments as used in the proof of Theorem 4 one can derive 8 2. From 2 and the left equality of 4 yields as q Ω and q. Similarly Tq, αy k Tq, αy T q, F y + Olog q, 5 q ε Tq, βy k Tq, gy T q, G y + Olog q, 6 From the condition k 6 and the condition that α, β are transcendental meromorphic functions, we can deduce from 5, 6 and Lemma 3 that F, G are transcendental meromorphic mappings. Suppose that y 0 C is a zero of F of multiplicity µ. Then, by the condition that Py 0 is a polynomial we can see that y 0 is a zero of αy n αy + λ Py of multiplicity µ + ν, where ν 0 is the multiplicity of y 0 as a zero of Py. Hence y 0 is a zero of gy k gy + λ Py of multiplicity µ + ν by the value sharing assumption. Now one sees that y 0 is a zero of G of multiplicity µ. This also works in the other direction. Therefore, F and G indeed share CM. As the order of α as well as β is finite, so 4 and Lemma 3 yields that the same is true for F and G as well. We now study the following two cases: q ε

9 Mathematics 207, 5, 42 9 of 8 Case. Consider a Möbius transformation F of G. By virtue of the Valiron-Mokhon ko lemma [22] and 3.8 we obtain Tq, F y = Tq, αy k λ αy + Olog q = Tq, βy k λ βy + Olog q = Tq, G y + O. 7 From Theorem 4 we get k 6Tq, α N q, α k λ α P Tq, α q ε + O. 8 The inequality 8 together with Lemma 3 and the condition that αy k λ αy Py and βy k λ βy Py share 0 CM gives k 6 Tq, αy N q, β k y λ βy Py Tq, αy Tq, G y q ε Tq, αy q ε + Olog q + O 9 as q and q Ω. In a similar way k 6 Tq, βy Tq, βy N q, α k y λ αy Py q ε + O Tq, βy Tq, F y + Olog q, q ε as q Ω and q. From Lemma 2 and the left equality of 4 we have mq, G m q, βy k λ βy + m q, Py βy + λ m q, βy k+ + O βy m q, βy k+ βy + λ + m q, + O βy βy + λ k + mq, βy + m q, + O βy Tq, βy k + mq, βy, q ε 20 2 as q Ω and q. By similar arguments as used in the proof of Theorem 4 we derive 9. From 9 and the left equality of 4 we obtain Nq, G N q, βy k λ βy + N q, Py knq, βy + Nq, βy + λ βy + Olog q Tq, βy k + 2Nq, βy + Olog q, q ε 22

10 Mathematics 207, 5, 42 0 of 8 as q and q Ω. Equations 2 and 22 yield Tq, βy Tq, G y ktq, βy + Nq, βy q ε Tq, βy k + Tq, βy q ε + Olog r, + Olog q 23 as q and q Ω. Similarly Tq, αy Tq, F y k + Tq, αy + Olog q, 24 as q Ω and q. From 23, 24, the condition k 6, Definition 2 and Lemma..2 of [23] we obtain ρα ρg ρβ. 25 Similarly, from 20 and 25 we have q ε ρβ ρf ρα. 26 From 25 and 26 we get ρα = ρβ = ρf = ρg. 27 From 0 and 4 we derive Nq, F y + N q, F y N q, αy + N q, αy + λ + N + Olog q q, 3Tq, αy + 2Tq, αy + λ + Olog q Tq, αy 5Tq, αy + Olog q, q ε + N q, αy αy + λ αy 28 as q Ω and q. Similarly Nq, G y + N q, 5Tq, βy G y Tq, βy q ε + Olog q, 29 as q Ω and q. Applying similar arguments as utilized in the proof of Theorem. we can derive 3. From 5 and 6 we get and Tq, αy k Tq, F y + Tq,αy o q ε as q Ω and q. From 2, 27 3 we derive N q, F 30 Tq, βy k Tq, G y + Tq,βy o, 3 q ε + Nq, F + N q, 5 k {Tq, F + Tq, G } = 0 k Tq, F, + Nq, G Tq, αy Tq, βy G q ε q ε + Olog q 32

11 Mathematics 207, 5, 42 of 8 as q Ω and q. This together with 32, Lemma 4 and the condition k 2 gives F G = or F = G. This proves the conclusion i of Theorem 8 Case 2. Suppose that k 6. In the same manner as in the proof of Case we can get 30 and 3. From 0 and 4 we have and N 2 r, F y + N 2 q, F y 2N q, αy + N q, αy + λ + 2N + Olog q q, 5Tq, αy + 2Tq, αy + λ + Olog q Tq, αy 7Tq, αy + Olog q N 2 r, G y + N 2 q, q ε 7Tq, βy G y as q Ω and q. From 30, 3, 33 and 34 we have and N 2 q, F y + N 2 q, N 2 q, G y + N 2 q, + N q, αy 7 F y k Tq, F y 7 G y k Tq, G y as q Ω and q. From 27, 35 and 36 we have N 2 q, F y + N 2 q, Tq, βy q ε Tq, αy αy + λ αy q ε Tq, βy q ε 33 + Olog q, 34 + Olog r 35 + Olog q, 36 + N F y 2 q, G y + N 2 q, 4 G y k T q, 37 as q Ω and q, where T q = max{tq, F, Tq, G }. From 37, Lemma 5 and the condition k 6 we have F = G or F G =. This reveals the conclusion ii of Theorem 8. Thus the proof stands completed. Proof. Theorem 9: This theorem is proved by considering the below two cases: Case. Let one of α and β, say α, is a rational mapping. Then, β is a rational function. In fact, if β is a transcendental meromorphic mapping, then, in the same manner as in the proof of 20 we can get from the assumption of Theorem 9 that k 6 Tq, βy N q, α k y λ αy Tq, βy o q ε + Olog q, Tq, βy q ε + O as q and q Ω. From k 6 and 38 we can deduce Tq, βy = Olog q, as q Ω and q. This implies that β is a rational function, which is impossible. Therefore, by virtue of the condition that α and β share 0, CM we derive 38 α = c β, 39

12 Mathematics 207, 5, 42 2 of 8 where c is some nonzero complex number. Thus α k λ α = c k+ β k λ β. 40 Suppose that λ α = 0. Then, if αy has a zero at some point y 0, then αy has a zero at y 0 + λ by λ α = 0. Continuing, αy 0 + 2λ = 0, αy 0 + 3λ =, and so on. Therefore, αy would have infinitely many zeros, which is impossible. Similarly, one can obtain a contradiction, if αy has a pole at some point y C. Therefore, λ α 0, and so λ β 0 by 40. Combining this with 40 and the assumption that α and β share 0, CM, we find that α k λ α and β k λ β share 0, CM. This together with Lemma 8 and the assumption that α k λ α and β k λ β share CM gives α k λ α = β k λ β. 4 From 40 and 4 we have c k+ =. Together with 39, this proves the conclusion i of Theorem 9. Case 2. Consider two transcendental meromorphic mappings α and β. Then, from Theorem 8 we have [α k λ α][β k λ β] = or α k λ α = β k λ β. The present case is divided in the below two subcases: Subcase 2.. Let [α k λ α][β k λ β] =. 42 One can derive 8 and 9 in a similar manner as in the proof of Theorem 4. Combining this with Definition 4, Lemma 2 and the assumption that α and β share 0, CM we deduce 2nN q, = kn q, αy + kn q, αy βy Nq, αy + λ + Nq, βy + λ = 2Nq, αy + λ Tq, αy 2Nq, αy q ε Tq, αy 2Tq, αy q ε and i.e., 2kN q, αy = kn q, αy + kn q, βy N q, αy + λ αy + N q, βy + λ βy T q, αy + λ αy + T q, βy + λ βy αy + λ m q, αy + Nq, αy + λ + Nq, αy αy βy + λ + m q, βy + Nq, βy + λ + Nq, βy βy mq, αy + 2Nq, αy + mq, βy + 2Nq, βy Tq, αy Tq, βy q ε q ε Tq, αy Tq, βy 2Tq, αy + 2Tq, βy N q, αy q ε Tq, αy k Tq, αy q ε N q, αy k Tq, αy + k Tq, βy Tq, αy q ε q ε Tq, βy q ε 43, 44

13 Mathematics 207, 5, 42 3 of 8 as q Ω and q. By 43, 44 and the second fundamental theorem we obtain Tq, αy Nq, αy + N q, αy + N q, + Olog q αy 2 k Tq, αy + q, k Tq, βy + N + Olog q αy Tq, αy Tq, βy + Olog q, as q Ω and q. Similarly q ε q ε Tq, βy 2 k Tq, βy + q, k Tq, αy + N βy Tq, αy Tq, βy + Olog q, q ε as q Ω and q. From 45 and 46 we have Tq, αy + Tq, βy 3 k Tq, βy + 3 q, k Tq, αy + N αy Tq, αy Tq, βy + N q, βy + Olog q, q ε q ε q ε as q Ω and q. Applying similar arguments as utilized in the proof of 27 we determine ρ f = ρβ <. This together with 47 gives lim sup r which contradicts the assumption 4. N q, + N q, αy βy 3 Tq, αy + Tq, βy k, 48 Subcase 2.2. Suppose that α k λ α = β k λ β. 49 In view of the hypothesis that α and β share 0, CM we get α β = ep, 50 where α is an entire function. Noting that ρα = ρβ := ρ <, we can get from 50 that ρe P ρ, and so P is a polynomial with degree ρ. Suppose that P is some constant, then e P is some nonzero constant, say e P = c 2. Thus from 49 and 50 we get c k+ 2 β k λ β = 0. 5 If λ β = 0, then we can get the conclusion ii from 49. Next we suppose that λ β 0, and so we have from 5 that c2 k+ =, which together with 50 and e P = c 2 reveals the conclusion ii of Theorem 9. Suppose that P is a nonzero polynomial. Then P y = a + ib y k + be a polynomial of degree k, where a and b are real numbers such that a + ib = 0. By 49 and 50 we have βy + λ βy = ek+p y e kp y+p y+η. 52

14 Mathematics 207, 5, 42 4 of 8 Given a positive number ε, we set T ε = k j=0 {y : arg y θ j < ε}, 53 where such that θ j = 2j + π θ k, 0 j k k 2k k, 54 α cos k θ j β sin k θ j = 0, 0 j k. 55 By and Lemma 9 we can find that βy + λ lim = y, y T ε βy Hence from 55, Lemma 0 and Liouville s Theorem we can find that βy + λ/βy is a constant. Therefore βy + λ = c βy 3, 57 where c 3 is a nonzero constant. Similarly αy + λ αy = c 4, 58 where c 4 is a nonzero constant. If one of c 3 and c 4 is equal to, then we can get the conclusion ii from 57, 58 and 49. Next we suppose that c 3 = and c 4 =. By substituting 57 and 58 into 49 we have α k+ = c 3 c 4 βk+, and so α = c 5 β, 59 where c 5 is a constant satisfying c5 k+ = c 3 c 4. Again from 59 and 49 we deduce c 5 = and c5 k+ =, which reveals the conclusion i of Theorem 9. The proof stands completed. Proof. Theorem 0: To prove the current theorem, we set F 2 y = αyk λ αy y, G 2 y = βyn λ βy. 60 y Then, applying similar arguments as utilized in the proof of Theorem 9 we can find that F 2 and G 2 share CM. We consider the following two cases: Case. Let one of α and β, say α is a nonconstant rational mapping. Then β is also a nonconstant rational mapping. In fact, if β is a transcendental meromorphic mapping, then we can derive in the same manner as in the proof of 20 that k 6 Tq, βy N q, α k y λ αy y Tq, βy Tq, F 2 y q ε Tq, βy = o + Olog q, q ε Tq, βy q ε + log q + O + O 6

15 Mathematics 207, 5, 42 5 of 8 as q Ω and q. From 6 and the the assumption k 6 we can deduce that β is a nonconstant rational mapping, this is impossible. Therefore, from 60 we can see that F 2 and G 2 are rational mapping. Next we prove that F 2 and G 2 are nonconstant rational mappings. In fact, if one of F 2 and G 2 is a constant, say F 2 = c 6, where c 6 is a finite complex number, then we can get from the first equality of 60 that αy k λ αy = c 6 y. 62 Set αy = P 2 P 3, 63 where P 2 and P 3 are nonzero relatively prime polynomials. Noting that at least one of P 2 and P 3 is not a constant. Applying similar arguments as used in the proof of Theorem 9, we determine η αy 0. Therefore, we can get from 62, 63 and the standard Valiron-Mokhonko lemma [22] that k max{degp 2 y, degp 3 y} log q + O = ktq, αy T q, P 2y P 3 y P 2y + λ + log r + O P 3 y + η T + T + O q, P 2y P 3 y q, P 2y + η P 3 y + λ = 2 max{degp 2 y, degp 3 y} log q + O, which implies that k 2, this contradicts the assumption k 6. Therefore, F 2 and G 2 are nonconstant rational functions. Combining this with 60 and the assumption that α and β share 0, CM and the assumption that F 2 and G 2 share CM, we have 39 and 40. Thus the conclusion i of Theorem 9 is proved. Case 2. Let α and β be transcendental meromorphic mappings. Then, by virtue of Theorems 8 and 0 we determine [α k λ α][β k λ β] = or α k λ α = β k λ β. If α k λ α = β k λ β. Then by using similar arguments as utilized in Subcase 2.2 of the proof of Theorem 9 we can get the conclusions i and ii. Next we suppose that [α k λ α][β k λ β] =. Noting that Tq, y = o{tq, α} and Tq, y = o{tq, β}, we can get in the same manner as in the proofs of 43 and 44 of Theorem 9 that and N q, αy Tq, αy k Tq, αy q ε + Olog q 64 N q, αy k Tq, αy + Tq, αy k Tq, βy q ε Tq, βy q ε + Olog q, 65 as q Ω and q, and so we can get 45 and 46. From 45 and 46 we have 47, and so we have 48, which contradicts 4. Thus the proof stands completed. Proof. Theorem 2: Let one of α and β are nonconstant polynomial. Then, by using similar arguments as utilized in Case of the proof of Theorem 9 we have the conclusion i of Theorem 9. Now assume that α and β are transcendental entire functions. Then, by Theorem and the assumptions of Theorem 2 we have [α k λ α][β k λ β] = or α k λ α = β k λ β. Suppose that α k λ α = β k λ β. Then, by using similar arguments as utilized in Case and Subcase 2.2 of the proof of Theorem 9 we can get the conclusions i and ii of Theorem 2. Suppose that [α k λ α][β k λ β] =. 66

16 Mathematics 207, 5, 42 6 of 8 Combining 66 with the assumption that α and β are entire functions sharing 0 CM, we have α = e f, α = e g, 67 where f and g are nonconstant polynomials. By substituting 67 into 66 we have e k f y+kgy [e f y+λ e f y ][e gy+λ e gy ] = 68 for all y C. By 68 we have e f y+λ e f y = e γy 69 for all y C, where γ is a polynomial. By 69 and Lemma 7 we can find that e f y+η f y = is a constant. Similarly e gy+λ gy = is also a constant. Set f y = c m y m + c m y m + + c y + a 0, 70 where c m = 0, c m,, c, c 0 are complex numbers. Suppose that m 2. Then f y + λ f y = mb m y m + b m 2 y m 2 + b y + b 0, 7 where b m 2, b m 3,, b, b 0 are complex numbers. Noting that c m = 0, we can find from 7 that f y + λ f y is not a constant, which contradicts the fact that e f y+λ f cy = is a constant. Therefore m =. Combining this with 68 and 70, we can deduce the conclusion iii of Theorem 2. This proves Theorem 2. Proof. Theorem 3: We set the equalities given in 60. Suppose that one of α and β are nonconstant polynomial. Then, by using similar arguments as utilized in Case of the proof of Theorem 0 we have the conclusion i of Theorem 2 from 60. Next assume that α and β are transcendental entire mappings. Then, by 60, Theorem and the assumptions of Theorem 3 we have [α n λ α][β n λ β] = z 2 or α k λ α = β k λ β. Suppose that α k λ α = β k λ β. Then, in the same manner as in Case and Subcase 2.2 of Theorem 9 we can get the conclusions i and ii of Theorem 2. Suppose that [α k λ α][β k λ β] = y Combining 7 with k 7 and the assumption that α and β are entire mappings sharing 0 CM, we have 67. By substituting 67 into 72 we have e k f y+kgy [e f y+λ e f y ][e gy+λ e gy ] = y 2 73 for all y C. From 73 we can find that at least one of e f y+λ f y and e gy+λ gy, say e f y+λ f y has a zero of y = 0, and so e f y+λ f y is transcendental entire mapping, which implies that e f y+λ f y has infinitely many zeros in the complex plane. But, from 73 we can find that e f y+λ f y at most has one zero of y = 0, this is a contradiction. Thus the proof stands completed. 4. Conclusions In the present article, we have proved several important results for value distribution of meromorphic mappings. It has been shown that the difference polynomials of the mentioned mappings have the same fixed points or share nonzero values. We have provided examples that the previous work of Laine and Yang need generalization. The results have been derived in more general domains. Several uniqueness results of meromorphic mapping have been explored. The research work of Qi, Yang and Liu has been generalized. The current work opens several new research directions. For instance, from Corollary, Corollary 2 and Example 3 we give the following problem:

17 Mathematics 207, 5, 42 7 of 8 Problem. What can be said about the conclusion of Corollary, if 2 n 6? From Theorems 2 and 3 we pose the following problem. Problem 2. What can be said about the conclusions of Theorems 2 and 3, if 2 n 6? We hope the techniques used in the present paper will play a key role to provide a framework for the concepts briefly discussed. Acknowledgments: The financial support of TWAS-UNESCO Associateship-Ref at Centro de Investigación en Matemáticas, A.C. CIMAT Jalisco S/N Valenciana A.P Guanajuato, GTO Mexico, is deeply appreciated and acknowledged. We are very grateful to NUST research directorate for providing publication charges and awards. Author Contributions: All of the authors provided equal contributions to the paper. Conflicts of Interest: this paper. References The authors declare that there is no conflict of interest regarding the publication of. Yang, C.C.; Yi, H.X. Uniqueness Theory of Meromorphic Functions; Kluwer Academic Publishers: Dordrecht, The Netherlands, Clunie, J. On a result of Hayman. J. Lond. Math. Soc. 967, 42, Hayman, W.K. Picard values of meromorphic functions and their derivatives. Ann. Math. 959, 70, Laine, I.; Yang, C.C. Value distribution of difference polynomials. Proc. Jap. Acad. Ser. A 2007, 83, Liu, K.; Yang, L.Z. Value distribution of the difference operator. Arch. Math. 2009, 92, Xu, J.F.; Yi, H.X. Uniqueness of entire functions and differential polynomials. Bull. Korean Math. Soc. 2007, 44, Qi, X.G.; Yang, L.Z.; Liu, K. Uniqueness and periodicity of meromorphic functions concerning the difference operator. Comput. Math. Appl. 200, 60, Chiang, Y.M.; Feng, S.J. On the Nevanlinna characteristic of f z + η and difference equations in the complex plane. Ramanujan J. 2008, 6, Halburd, R.G.; Korhonen, R.J. Nevanlinna theory for the difference operator. Ann. Acad. Sci. Fenn. Math. 2006, 3, Hayman, W.K. Meromorphic Functions; Clarendon Press: Oxford, UK, Heittokangas, J.; Korhonen, R.; Laine, I.; Rieppo, J. Uniqueness of meromorphic functions sharing values with their shifts. Complex Var. Elliptic Equ. 20, 56, Rahman, U.; Li, X.M.; Faizullah, F. On periodicity and uniqueness of meromorphic functions. J. Comput. Theor. Nanosci. 207, 4, Yang, C.C. On deficiencies of differential polynomials. Math. Z. 970, 6, Zhang, J.L. Value distribution and shared sets of differences of meromorphic functions. J. Math. Anal. Appl. 200, 367, Lahiri, I. Weighted sharing of three values and uniqueness of meromorphic functions. Kodai. Math. J. 200, 24, Whittaker, J.M. Interpolatory Function Theory. Cambridge University Press: Cambridge, UK, Halburd, R.; Korhonen, R.; Tohge, K. Holomorphic curves with shift-invariant hyperplane preimages. Trans. Am. Math. Soc. 204, 366, Yang, C.C.; Hua, X.H. Uniqueness and value sharing of meromorphic functions. Ann. Acad. Sci. Fenn. Math. 997, 22, Li, X.M.; Wen, Z.T. Uniqueness theorems of meromorphic functions sharing three values. Complex Var. Elliptic Equ. 20, 56, Bank, S.B.; Langley, J.K. On the oscillation of certain linear differential equation in the complex domain. Proc. Edinb. Math. Soc. 987, 30, Titchmarsh, E.C. The Theory of Functions; Oxford University Press: Oxford, UK, 986.

18 Mathematics 207, 5, 42 8 of Mokhon ko, A.Z. On the Nevanlinna characteristics of some meromorphic functions: Theory of Functions, Functional Analysis and Their Applications. Izd-vo Khar kovsk Un-ta Kharkov 97, 4, Laine, I. Nevanlinna Theory and Complex Differential Equations; Walter de Gruyter: Berlin, Germany; New York, NY, USA, 993. c 207 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution CC BY license