Uniqueness of the power of a meromorphic functions with its differential polynomial sharing a set

This paper is devoted to the uniqueness problem of the power of a meromorphic function with its differential polynomial sharing a set. Our result will extend a number of results obtained in the theory of normal families. Some questions are posed for future research.


Introduction Definitions and Results
In this paper we assume that readers are familiar with the basic Nevanlinna Theory ( [5]). Let f and g be two non constant meromorphic functions in the complex plane C. If for some a ∈ C ∪ {∞}, f and g have same set of a-points with the same multiplicities, we say that f and g share the value a CM (counting multiplicities) and if we do not consider the multiplicities then f , g are said to share the value a IM (ignoring multiplicities).
When a = ∞ the zeros of f − a means the poles of f . The problem of meromorphic functions sharing values with their derivatives is a special subclass in the literature of uniqueness theory. The subject of sharing values between entire functions and their derivatives was first studied by Rubel and Yang ([10]). In 1977, they proved that if a non-constant entire function f and f ′ share two distinct finite numbers a, b CM, then f = f ′ . In 1979, analogous result for IM sharing was obtained by Mues and Steinmetz in the following manner.
Theorem A. ( [9]) Let f be a non-constant entire function. If f and f ′ share two distinct values a, b IM then f ′ ≡ f .
To proceed further we consider the following well known definition of set sharing. Let S be a set of complex numbers and E f (S) = a∈S {z : f (z) = a}, where each zero is counted according to its multiplicity. If we do not count the multiplicity, then the set If E f (S) = E g (S) we say that f and g share the set S CM. On the other hand, if E f (S) = E g (S), we say that f and g share the set S IM. Evidently, if S contains only one element, then it coincides with the usual definition of CM (respectively, IM) sharing of values.
In view of the above definition it will be interesting to study the relation between f and its derivative f ′ when they share a set. We see from the following example that results of In 2007 Chang, Fang and Zalcman( [2]) further extended the above result by considering an arbitrary set having three elements in the following manner.
Theorem C. ( [2]) Let f be a non-constant entire function and let S = {a, b, c}, where a, b and c are distinct complex numbers.
In the next year, Chang and Zalcman( [3]) replaced the entire function by meromorphic function with at most finitely many simple poles in Theorem B and C and obtained similar results as follows.
Theorem D. ( [3]) Let S = {0, a, b}, where a, b are two non-zero distinct complex numbers. If f is a meromorphic function with at most finitely many poles and Theorem E. ( [3]) Let f be a non-constant meromorphic function with at most finitely many simple poles; and let S = {0, a, b}, where a, b are distinct non zero complex numbers. If (1) f (z) = Ce z ; or So we observe from the above mentioned results that the researchers were mainly involved to find the uniqueness of an entire or meromorphic function with its first derivative sharing a set at the expanse of allowing several constraints. But all were practically tacit about the uniqueness of an entire or meromorphic function with its higher order derivatives. In 2007 Chang, Fang and Zalcman ([2]) consider the following example to show that in Theorem C, one can not relax the CM sharing to IM sharing of the set S. In other words, when multiplicity is disregarded, the uniqueness result ceases to hold. Thus it is natural to ask the following question :-Question 1.1. Does there exist any set which when shared by a meromorphic function together with its higher order derivative or even a power of a meromorphic function together with its differential polynomial, lead to wards the uniqueness ? To seek the possible answer of the above question is the motivation of the paper. We answer the above question even under relaxed sharing hypothesis. To this end, we resort to the notion of weighted sharing of sets appeared in the literature in 2001 ( [6]).
, we say that f, g share the value a with weight k.
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 any integer 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.
, then we say f , g share the set S with weight k.
Throughout the paper we use the following notation for L. Definition 1.3. Let k(≥ 1), l(≥ 1) be positive integers and a i ∈ C for i = 0, 2, ..., k − 1. For a non constant meromorphic function f , we define the differential polynomial in f as where n ≥ 3 is an integer and a and b are two nonzero complex numbers satisfying ab n−2 = 2.
We have from (1.1) We note that P ′ (0) = 0 and so from (1.1) P ′ (z) = 0 implies az n − 2(n − 1)z 2 + 2(n − 2)bz = 0. Now at each root of P ′ (z) = 0 we get implies that a zero of P ′ (z) is not a zero of P (z). In other words each zero of P (z) is simple.

Lemmmas
We define R(z) = az n n(n−1)(z−α1)(z−α2) , where α 1 and α 2 are the distinct roots of the equation Lemma 2.1. For any two non-constant meromorphic functions f 1 and f 2 , Combining the two cases we get the proof.

Proof. Let us define
As f m and L(f ) share (∞, 0), so if N (r, f ) = S(r, f ) then A = 1, i.e., F = G, which is not possible. So N (r, f ) = S(r, f ). Thus the lemma holds. Case-2 V ≡ 0 Let z 0 be a pole of f of order t, then it is a pole of L(f ) of order (t + k)l and that of F and G are tm(n − 2) and (t + k)l(n − 2) respectively. Clearly z 0 is a zero of ( F ′ F −1 − F ′ F ) order at least tm(n − 2) − 1 and zero of V of order atleast λ, where λ = min{m(n − 2) − 1, (1 + k)l(n − 2) − 1} Thus Proof. The proof is obvious if we are keeping the following in our mind : But simple zeros of f m − α i are not poles of H and multiple zeros of f m − α i are zeros of (f m ) ′ . Similar explanation for G is also hold.

Now by integration we have
which is a contradiction as n > 6. Thus γ + δ = 1 and γ = 0. So, From above we get N (r, 0; G + 1−γ γ ) = N (r, F ). If γ = 1, then using the Second Fundamental Theorem and (3.15), we get which is a contradiction as n > 6. Thus γ = 1 and F G ≡ 1 which gives As n > 6 from the above equation it is clear that f has no pole. Let z 0 be a α 1i point of f of order s, where (α 1i ) m = α 1 , then it can't be a pole of L(f ) as f has no pole, so z 0 is a zero of L(f ) of order q satisfying n ≤ nq = s.