Meromorphic Functions Sharing a Small Function

We will study meromorphic functions that share a small function, and prove the following result: let f (z) and g(z) be two transcendental meromorphic functions in the complex plane and let n ≥ 11 be a positive integer. Assume that a(z)( ≡ 0) is a common small function with respect to f (z) and g(z). If f n f ′ and gng ′ share a(z) CM, then either f n(z) f ′ (z)gn(z)g ′ (z) ≡ a2(z), or f (z) ≡ tg(z) for a constant satisfying tn+1 = 1. As applications, we give several examples.


Introduction and main result
In this paper, a meromorphic function always means a function which is meromorphic in the whole complex plane.Let f (z) be a nonconstant meromorphic function.We use the following standard notations of value distribution theory: (see [1][2][3]).We use S(r, f ) to denote any function satisfying as r → ∞, possibly outside of a set E with finite measure.
Let a(z) be a meromorphic function in the complex plane.If T(r,a) = S(r, f ), then a(z) is called a small function related to f (z).
Let b be a finite complex number.We denote by N 2) (r,1/( f − b)) the counting function for zeros of f (z) − b (or poles of 1/( f (z) − b)) with multiplicity at most 2, and by N 2) (r,1/( f − b)) the corresponding one for which multiplicity is not counted.Let N (2 (r,1/( f − b)) be the counting function for zeros of f (z) − b with multiplicity at least 2 and N (2 (r,1/( f − b)) the corresponding one for which multiplicity is not counted.Set Suppose that f (z) and g(z) are two meromorphic functions, and a(z) is a small function related to both of them.We say that f (z) and g(z) share the small function a(z) CM, if f (z) − a(z) and g(z) − a(z) assume the same zeros with the same multiplicities.We say that f (z) and g(z) share the value a CM if a(z) ≡ a(∈ C) is constant.
In the 1920's, Nevanlinna [2] proved his famous four-valued theorem, which is an important result about uniqueness of meromorphic functions.Then many results about meromorphic functions that share more than or equal to two values have been obtained (see [4]).In 1997, Yang and Hua [5] studied meromorphic functions sharing one value.
Fang and Qiu [6] investigated meromorphic functions sharing fixed point later.
Theorem 1.2 (see [6]).Let f (z) and g(z) be two nonconstant meromorphic(entire) functions and let n ≥ 11 (n ≥ 6) be a positive integer.If 2 , where c 1 , c 2 , and c are three constants satisfying Recently, Banerjee [7] also studied meromorphic functions sharing one value, generating Theorem 1.1.In this paper, we extend the results above as follows.
Theorem 1.3.Let f (z) and g(z) be two transcendental meromorphic functions, and let a(z)( ≡ 0) be a common small function with respect to them, and let n ≥ 11 be a positive integer.If f n (z) f (z) and g n (z)g (z) share a(z) CM, then either f n (z) f (z)g n (z)g (z) ≡ a 2 (z), or f (z) ≡ tg(z) for a constant such that t n+1 = 1.

Lemmas
In order to prove Theorem 1.3, we need the following lemmas.
S. Wang and Z. Gao 3 Lemma 2.1 (see [4]).Suppose that f (z) is a nonconstant meromorphic function in the complex plane, and k is a positive integer.Then N r, Lemma 2.2 (see [3,4]).Suppose that f (z) is a nonconstant meromorphic function in the complex plane, and a(∈ C ∪ ∞) is any complex number.Then, Lemma 2.3 (see [3,4,8]).Let f (z) and g(z) be two meromorphic functions in the complex plane.If f and g share 1 CM, then one of the following cases must occur:

Proof of Theorem 1.3
Let Then we know that F(z) and G(z) share 1 CM.From Lemma 2.1, we have By T(r,a) = S(r, f ) and (3.1), we obtain Thus we have, by elementary Nevanlinna theory, On the other hand, since we have S(r, f ) = S(r,F), and therefore, T(r,F) + S(r,F). (3.5) Similarly, we have Combining with Lemma 2.3, suppose first that Since both F and G are transcendental meromorphic functions and n ≥ 11, then we deduce a contradiction from (3.8).Therefore, we deduce that either where c is a constant.

S. Wang and Z. Gao 5
Suppose that c = 0, then which contradicts Lemma 2.2.Thus, f (z) ≡ tg(z) for a constant such that t n+1 = 1.The proof is complete.
Remark 3.1.At this time, it is not easy to obtain the representation of f (z) and g(z) like in Theorems 1.1 and 1.2.Suppose that either a(z) is an entire function or all of its poles are simple.Set we have In particular, .10) then we get f (z) = c 1 e z z 0 m(z)dz and g(z) = c 2 e z z 0 n(z)dz , where c 1 and c 2 are two nonzero constants, and the integral path [z 0 ,z](z 0 = z) does not pass the poles of either m(z) or n(z).Combining with the equality in Theorem 1.3

Example 3 . 2 .
where c is a constant such that (c 1 c 2 ) n+1 c 2 = −1.Hence (3.13) is one of the representations of f (z) and g(z) which can be obtained from (3.11) under the condition.If a(z) = e z , then by Theorem 1.3 and the Remark 3.1, we can obtain two representations of f (z) and g(z): f (z) ≡ tg(z) for a constant such that t n+1 = 1; f (z) = c 1 e ce z ,g(z) = c 2 e −ce z , where c 1 , c 2 , and c are three constants satisfying (c 1 c 2 ) n+1 c 2 = −1.3.1, we havef (z) = c 1 e c z z 0 a(z)dz = c 1 z − 1 z c e cz 2 , g(z) = c 2 e −c z z 0 a(z)dz = c 2 z z − 1 c e −cz 2 , (3.15)where c 1 , c 2 and c are three constants satisfying (c 1 c 2 ) n+1 c 2 = −1.This is one of the relations of f (z) and g(z).

Example 3 . 4 .
If a(z) has a pole of order m(> 1), then we cannot get(3.13).Suppose thata(z) = 1 + z 2 z 2 , (3.16)then f (z) = c 1 e −c/z e cz and g(z) = c 2 e c/z e −cz are not meromorphic functions in the complex plane.