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In the paper, we take up a new method to prove a result of value distribution of meromorphic functions: let f be a meromorphic function in , and let , where P is a polynomial. Suppose that all zeros of f have multiplicity at least , except possibly finite many, and as . Then has infinitely many zeros.

The value distribution theory of meromorphic functions occupies one of the central places in Complex Analysis which now has been applied to complex dynanics, complex differential and functional equations, Diophantine equations and others.

In his excellent paper [

In this paper, we study the value distribution of transcendental meromorphic functions, all but finitely many of whose zeros have multiplicity at least, where is a positive integer.

In 2008, Liu et al. [

Theorem A Let be an integer, let be a meromorphic function of infinite order in, and let, where is a polynomial. Suppose that 1) all zeros of have multiplicity at least, except possibly finitely many, and 2) all poles of are multiple, except possibly finitely many.

Then has infinitely many zeros.

Theorem B Let be an integer, let be a meromorphic function of finite order in, and let, where is a polynomial. Suppose that 1) all zeros of have multiplicity at least, except possibly finitely many, and 2).

Then has infinitely many zeros.

In the present paper, we prove the following result, which is a significant improvement of Theorem 1.

Theorem 1 Let be an integer, let be a meromorphic function of order in, and let, where is a polynomial. Suppose that all zeros of have multiplicity at least, except possibly finitely many. Then has infinitely many zeros.

Theorem 1 and Theorem 2 taken together imply the following result.

Theorem 2 Let be an integer, let be a meromorphic function in, and let, where is a polynomial. Suppose that 1) all zeros of have multiplicity at least, except possibly finitely many, and 2) as.

Then has infinitely many zeros.

We use the following notation. Let be complex plane and be a domain in. For and, and. We write in to indicate that the sequence converges to in the spherical metric uniformly on compact subsets of and in if the convergence is in the Euclidean metric.

Let be a meromorphic function in. Set

The Ahlfors-Shimizu characteristic is defined by

Remark Let denote the usual Nevanlinna characteristic function. Since is bounded as a function of, we can replace with in the paper.

The order of the meromorphic function is defined as

Lemma 1 [

Then is quasinormal of order 1 in. If, moreover, no subsequence of is normal at, then

locally uniformly in and there exists such that for all.

Remark Since Lemma 1 is not stated explicitly in [

Lemma 2 [5, Lemma 2] Let be a family of functions meromorphic in, all of whose zeros have multiplicity at least, and suppose that there exists such that whenever. Then if is not normal at, there exist, for each1) points,;

2) functions; and 3) positive numbers

such that in, where is a nonconstant meromorphic function inall of whose zeros have multiplicity at least, such that.

Lemma 3 Let be a meromorphic function of order in, then there exist and such that

Proof We claim that there exist and such that

Otherwise there would exist and such that

for all. From this follows

and hence

Now we have which contradicts the hypothesis that.

Observing that hence there exists a sequence such that and as. Let. Obviously, and , and hence as.

Lemma 4 Let and. Let be a transcendental meromorphic function, all of whose zeros have multiplicity at least. Set. Suppose that. Then there exists a sequence

and such that

as.

Proof Since and, we have. By Lemma 3, there exist and such that

Set. Clearly,. Thus is not normal at 0. Obviously, all zeros of have multiplicity at least in, and hence all zeros of have multiplicity at least in for sufficiently large. Using Lemma 2 for, there exist points, and positive numbers and a subsequence of (that we continue to call) such that

in, where is a nonconstant meromorphic function in, all of whose zeros have multiplicity at least.

We claim that, where is a constant. Otherwise, , where and are constants. Then, either is a constant function, or all zeros of have multiplicity at most. A contradiction.

Let be not a zero or pole of, and let. Now we have

where. Since and is not a zero or pole of, we have, and as, where.

Set and, where. Clearly,

where satisfying as.

Now, we have and

Set. Obviously, and, and hence as.

Proof We assume that has at most finitely many zeros and derive a contradiction. Let as, where and.

Set. By Lemma 4, there exists a sequence and such that

and

Set. By (1.4),

Hence, no subsequence of is normal at.

Since has at most finitely many zeros, we have for sufficiently large,

Observing that

in. It follows from Lemma 1 (applied to in), and there exists such that for all

Set. Then

and hence

Using the simple inequality

for, we have

The second term on the right of (1.7) is

Putting (1.7), (1.8), and (1.9) together, we have for and sufficiently large,

It follows from (1.1), (1.6), and (1.10),

Thus,

which contradicts (1.3).

This work was supported by National Natural Science Foundation of China (No.11001081, No.11226095).

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