Uniqueness of Meromorphic Functions and Differential Polynomials
Abstract
We study the uniqueness of meromorphic functions and differential polynomials sharing one value with weight and prove two main theorems which generalize and improve some results earlier given by M. L. Fang, S. S. Bhoosnurmath and R. S. Dyavanal, and so forth.
1. Introduction and Results
Let f be a nonconstant meromorphic function defined in the whole complex plane ℂ. It is assumed that the reader is familiar with the notations of the Nevanlinna theory such as T(r, f), m(r, f), N(r, f), and S(r, f), that can be found, for instance, in [1–3].
In 2002, C. Y. Fang and M. L. Fang [4] proved the following result.
Theorem A (see [4].)Let f(z) and g(z) be two nonconstant entire functions, and let n(≥8) be a positive integer. If [fn(z)(f(z) − 1)]f′(z) and [gn(z)(g(z) − 1)]g′(z) share 1 CM, then f(z) ≡ g(z).
Fang [5] proved the following result.
Theorem B (see [5].)Let f(z) and g(z) be two nonconstant entire functions, and let n, k be two positive integers with n > 2k + 8. If and share 1 CM, then f(z) ≡ g(z).
In [6], for some general differential polynomials such as , Liu proved the following result.
Theorem C (see [6].)Let f(z) and g(z) be two nonconstant entire functions, and let n, m, k be three positive integers such that n > 5k + 4m + 9. If and share 1 IM, then either f(z) ≡ g(z) or f and g satisfy the algebraic equation R(f, g) ≡ 0, where .
The following example shows that Theorem A is not valid when f and g are two meromorphic functions.
Example 1.1. Let f = (n + 2)(h − hn+2)/(n + 1)(1 − hn+2), g = (n + 2)(1 − hn+1)/(n + 1)(1 − hn+2), where h = ez. Then [fn(z)(f(z) − 1)]f′(z) and [gn(z)(g(z) − 1)]g′(z) share 1 CM, but f(z)≢g(z).
Lin and Yi [7] and Bhoosnurmath and Dyavanal [8] generalized the above results and obtained the following results.
Theorem D (see [7].)Let f(z) and g(z) be two nonconstant meromorphic functions with Θ(∞, f) > 2/(n + 1), and let n(≥12) be a positive integer. If [fn(z)(f(z) − 1)]f′(z) and [gn(z)(g(z) − 1)]g′(z) share 1 CM, then f(z) ≡ g(z).
Theorem E (see [8].)Let f(z) and g(z) be two nonconstant meromorphic functions satisfying Θ(∞, f) > 3/(n + 1), and let n, k be two positive integers with n > 3k + 13. If and share 1 CM, then f(z) ≡ g(z).
Naturally, one may ask the following question: is it really possible to relax in any way the nature of sharing 1 in the above results?
To state the next result, we require the following definition.
Definition 1.2 (see [9].)Let k be a nonnegative integer or infinity. For one denotes by Ek(a, f) the set of all a-points of f, where an a-point of multiplicity m is counted m times if m ≤ k and k + 1 times if m > k. If Ek(a, f) = Ek(a, g), one says that f, g share the value a with weight k.
We write that 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 with 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.
Recently, with the notion of weighted sharing of values, Xu et al. [10] improved the above results and proved the following theorem.
Theorem F (see [10].)Let f(z) and g(z) be two nonconstant meromorphic functions, and let n, k be two positive integers with n > 5k + 11. If Θ(∞, f) + Θ(∞, g) > 4/n, , and share 1 (1,2), then f(z) ≡ g(z).
Theorem G (see [10].)Let f(z) and g(z) be two nonconstant meromorphic functions, and let n, k be two positive integers with n > 7k + 23/2. If Θ(∞, f) + Θ(∞, g) > 4/n, , and share 1 (1,1), then f(z) ≡ g(z).
Remark 1.3. The proof of Theorem E contains some mistakes: for example, one cannot get formulas (6.9) and (6.10) in [8]. Therefore, the last inequality in page 1203 of [8] does not hold. So, Theorem E will not stand. Similarly, in [10], the proof of Case 1 in Theorem F is incorrect (see page 63 of paper [10]). Hence, the conclusion of Theorems F and G will not stand.
Now one may ask the following questions which are the motivations of the paper.
- (i)
What happens if the conclusions of Theorems E, F, and G can not stand?
- (ii)
Can the value n be further reduced in the above results?
In the paper, we investigate the solutions of the above questions. We improve and generalize the above related results by proving the following theorems.
Theorem 1.4. Let f(z) and g(z) be two nonconstant meromorphic functions, and let n, k be two positive integers with n > 3k + 11. If Θ(∞, f) > 2/n, , and share 1 (1,2), then f(z) ≡ g(z) or .
Theorem 1.5. Let f(z) and g(z) be two nonconstant meromorphic functions, and let n, k be two positive integers with n > 5k + 14. If Θ(∞, f) > 2/n, and share 1 (1,1), then f(z) ≡ g(z) or .
2. Some Lemmas
For the proof of our result, we need the following lemmas.
Lemma 2.1 (see [1].)Let f be nonconstant meromorphic function, and let a0, a1, …, an be finite complex numbers such that an ≠ 0. Then
Lemma 2.2 (see [1].)Let f(z) be a nonconstant meromorphic function, k a positive integer, and c a nonzero finite complex number. Then
Lemma 2.3 (see [11].)Let f be a nonconstant meromorphic function, and let k, p be two positive integers. Then
Lemma 2.4 (see [1].)Let f(z) be a transcendental meromorphic function, and let a1(z), a2(z) be two meromorphic functions such that T(r, ai) = S(r, f), i = 1,2. Then
Lemma 2.5. Let f and g be two nonconstant meromorphic functions, and let k(≥1), l(≥1) be two positive integers. Suppose that f(k) and g(k) share (1, l).
- (i)
If l = 2 and
(2.5)then either f(k)g(k) ≡ 1 or f ≡ g. - (ii)
If l = 1 and
(2.6)then either f(k)g(k) ≡ 1 or f ≡ g.
Proof. Let
If z0 is a common simple 1-point of f(k) and g(k), substituting their Taylor series at z0 into (2.7), we see that z0 is a zero of h(z). Thus, we have
By our assumptions, h(z) have poles only at zeros of f(k+1) and g(k+1) and poles of f and g, and those 1-points of f(k) and g(k) whose multiplicities are distinct from the multiplicities of correspond to 1-points of g(k) and f(k), respectively. Thus, we deduce from (2.7) that
By Lemma 2.2, we have
- (i)
If l ≥ 2, it is easy to see that
(2.13)
Therefore, we have h ≡ 0, that is,
Case 1 (b ≠ 0, −1). For more details see the following subcases.
Subcase 1.1 (a − b − 1 ≠ 0). Then, by (2.19), we have .
By Lemma 2.2 and Lemma 2.3, we get
Subcase 1.2 (a − b − 1 = 0). Then, by (2.19), we have g(k) = (b + 1)f(k)/(bf(k) + 1).
Therefore .
By Lemma 2.2 and Lemma 2.3, we get
Case 2 (b = −1). For more details see the following subcases.
Subcase 2.1 (a + 1 ≠ 0). Then by (2.19), we get a/((a + 1) − f(k)) = g(k). So, we have . We can deduce a contradiction as in Case 1.
Subcase 2.2 (a + 1 = 0). Then by (2.19), we get f(k)g(k) ≡ 1.
Case 3 (b = 0). For more details see the following subcases.
Subcase 3.1 (a − 1 ≠ 0). Then by (2.19), we get (f(k) + a − 1)/a = g(k). So, we have . We can deduce a contradiction as in Case 1.
Subcase 3.2 (a − 1 = 0). Then by (2.19), we get f(k) ≡ g(k). From this, we obtain f = g + P(z), where P(z) is a polynomial, and so T(r, f) = T(r, g) + S(r, f). If P(z)≢0, then, by Lemma 2.4, we get
(ii) If l = 1, it is easy to see that
Without loss of generality, we suppose that there exists a set I with infinite measure such that T(r, g) ≤ T(r, f) for r ∈ I:
Therefore, we have h ≡ 0, that is,
The proof of Lemma 2.5 is completed.
3. Proof of Theorem 1.4
Proof. Let F(z) = fn(f − 1) and G(z) = gn(g − 1). We have
Next, we consider the following two cases.
Case 1. F(k)G(k) ≡ 1, that is, .
Case 2. F ≡ G, that is,
- (i)
Let h = f/g be a constant. Then from (3.7) it follows that h ≠ 1, hn ≠ 1, hn+1 ≠ 1 and g = (1 − hn)/(1 − hn+1) = constant, which leads to contradiction.
- (ii)
Let h = f/g be not a constant. Since f≢g, we have h≢1, and hence we deduce that g = (1 − hn)/(1 − hn+1) and g = ((1 − hn)/(1 − hn+1))h = (1 + h + h2 + ⋯+hn−1)h/(1 + h + h2 + ⋯+hn), where h is a nonconstant meromorphic function. It follows that
(3.8)
So we have
4. Proof of Theorem 1.5
Acknowledgments
This research was supported by the NSF of Key Lab of Geomathematics of Sichuan Province of China. The author wishes to thank the referee for his/her valuable comments and suggestions.