Let be a function transcendental and meromorphic in the plane. In what follow, we denote the convergence exponent of the zeros of by, the growth order of by, and the lower order of by.

Following Whittaker [1], define the forward differences to be k times iteration of the difference operator, that is,

Recently, a number of papers research on complex difference equations and differences analogues of Nevanlinna’s theory [2-6]. Bergweiler and Langley [7] firstly investigated the existence of zeros of, and obtained a result as follow.

Theorem 1.1. Let f be a function transcendental and meromorphic of lower order in the plane. Let be such that at most finitely many poles of satisfy. Then has infinitely many zeros.

In 2008, Z. X. Chen and K. H. Shon [8].

Theorem 1.2. Let and f be a function transcendental and meromorphic of lower order in the plane. Let and a set consist of all poles of, such that

at most except finitely many exceptions. Then has infinitely many zeros.

In 2009, Z. X. Chen and K. H. Shon [9] continue to investigate the existence of the zeros of the difference polynomials defined as follows

and obtained two results.

Theorem 1.3. Let f be a function transcendental and meromorphic of growth order, and be two complex numbers, such that, and. If has at most finitely many poles satisfying , then has infinitely many zeros, and.

In particular, if has at most finitely many zeros satisfying, then has also infinitely many zeros, and.

Theorem 1.4. Let satisfy the conditions in Theorem 1.3, If has at most finitely many poles satisfying

then has infinitely many zeros, and .

In particular, if has at most finitely many zeros such that, then has also infinitely many zeros, and

.

It is not difficult to understand that defined by (1.2) is more general difference polynomials than or and Theorem 1.3 extends Theorem 1.1. Therefore, we pose naturally one question whether more general difference polynomials than defined by (1.3) has also infinitely many zeros. In this paper, we focus on research a certain homogeneous difference polynomials and affirm to answer this problem.

Theorem 1.5. Suppose that k is a positive integer,. Let be a function transcendental and meromorphic of growth order, and there exists k complex numbers such that

. If has at most finitely many poles b_j

satisfying

Then has infinitely many zeros, and.

In particular, if has at most finitely many zeros satisfying, then has also infinitely many zeros, and.

Lemma 2.1. (see [7]) Let f be a function transcendental and meromorphic in the plane of growth order less than 1, and. Then there exists an E such that

as, uniformly in for.

Lemma 2.2. (see [7]) Let be a function transcendental and meromorphic in the plane of lower order. Then there exists arbitrarily large R with the following properties. First,

Second, there exists a set of linear measure, such that for

,

on.

Lemma 2.3. Let be a function transcendental and meromorphic in the plane with growth order

. Supposed that. If the homogeneous difference polynomials

or quotient of difference polynomials

is rational functions, then has at most finite many poles.

Proof. Without loss of generality, we assume that c₁ = 1. Because that the homogeneous difference polynomials is rational, there exists a rational functions such that

Set

and

.

So there exists no poles of in the domain

and

.

Now we complete the proof of the conclusion that has at most finite.

Now we complete the proof of the conclusion that has at most finite many poles. Suppose not, there exists one domain D_j, for example D₁, in which has infinitely many poles. We assume that the set consists of all poles of in D₁ and and divide it into two cases:

Case 1. There exists, such that for an arbitrary, there does not exist such that

, that is, for an arbitrary

, we have. In fact, since and, this case appears whenever for every. Thereforewe know and that there exists a unbounded subsequence set in which every

is the poles of. Hence we know that there are at least one in these signs, which takes every positive integer, for instance, m₁ takes every positive integer.

Thus, , which contradicts the hypothesis of Lemma 2.3.

Case 2. There exists, such that for every, there exists, such that

. From and

, we have that. As the set A is infinite and B has only a finite elementary, there exists, satisfying

By putting in order again, we have the following express

and

where

.

Now set

where

Since are between and, are between and, we know that, that is,. From

, and (2.4), we know that one of and is the pole of. If is the pole of, then from the some argument above we have one of is also the pole of. If is the pole of, then one of is also the pole of. On the analogy of this, it is not difficult to find there exists at least one of, for instance, we assume that is j, such that j takes all value of . From to, to, and to, repeating above proceeding, we have

where

Therefore, we can see that there exist infinite many poles of whose expressions are as follows

where

in which we can find that one of takes every positive integer. Thus, , which still contradict the hypothesis on the growth order of in Lemma 2.3.

By the similar method to above, it is easy to prove that has at most finite many poles whenever quotient of difference polynomials

is rational functions.                            

Lemma 2.4. Let be a function transcendental and meromorphic in the plane with growth order

. Supposed that, then the homogeneous difference polynomials

and

also are transcendental.

Proof. Suppose first that there exists a rational function, such that

By Lemma 2.3, has at most finite many poles. Again from Lemma 2.1, there exists such that as, we have

It follows that from (2.6) and (2.7)

We write for a polynomial formed by the pole of, and. So is an entire function, and. With the standard result in the Wiman-Valiron Theory, we know that there exists a subset with finite logarithmic measure, in which for an sufficiently large

the following equality holds

.

Thus,

where, and, as. Set. Since is, we have that F₁ also is of finite logarithmic measure. Therefore, for all z, , and

we immediately deduce that from (2.8) and (2.9)

Since and is transcendental, there exists a sequence, such that for arbitrary, we have that

Then, we induce that from (2.4) and (2.11)

Therefore, from (2.12) and (2.13) we have

By (2.10), (2.12), and (2.14), we deduce easily that, which contradicts the assumption on, that is, transcendental.

Lemma 2.5. Let be a function transcendental and meromorphic in the plane, whose growth orde . Supposed that, and. Then

Proof. For of growth order, from Hadamard’s factorization theorem we have

where and are respectively the canonical product of zeros and poles of , satisfying

.

From (2.15), we have

Therefore, if, we deduce that . For, the following equations hold

We have that from and (2.16)

If is a poles of with multiplicity m, then must be a poles of with multiplicity, so that we denote by, that is,

where is a canonical product of distinct poles of. By (2.16), we obtain that

From (2.15) and (2.18), we deduce that

Thus, if z₀ is the pole of (that is,), then, , but. Hence, we have that is not the zero of

.

So that

and

This completes the proof of Lemma 2.5.          

From Lemma 2.2 we see that there exists a sufficiently large, a positive number such that

and there exists a set with linear measure, such that for any, we have the following equation

where satisfies the following express,

here.

On the other hand, under the condition of Theorem 1.5 and from Lemma 2.4 we know transcendental.

Suppose that E concludes all of zeros and poles of, and. Setting

Since the property of and, we have that is with finite logarithmic measure, and has linear measure for sufficiently large.

We assume that is a set, such that

Noting that there exists many points

at most from (2.22), at which is not continuous, and also for any

holds for some whenever. Therefore, has linear measure

From (2.23)-(2.25), we know that there exists such that, , , , and have no zeros and poles on the circle. Therefore,

Applying Rouché’s Theorem to and, we obtain the following equation

Without loss of generality, we may assume that

for all poles of. From the assumption in Theorem 1.3, we know that there exists positive number, which does not depend on R and r, such that if is a pole of with multiplicity,

then by the expression of and,

we see that z₀, are respectively the pole of with multiplicity. Therefore, we deduce that

Since the pole z₀ of has multiplicity, we have the following equality

And obviously,

Substituting (2.28), (2.30) into (2.27), we obtain

If, then. Thus, we have that by (2.31)

then.

If, we have that from (2.31)

By Lemma 2.5 and (2.33), we deduce that

.

In particular, if is the zero of

then z₀ is, also the zero of. On the other hand, if z₁ is the zero of, but not the zero of, then z₁ must be the zero of, that is, for some j. From the assumption in Theorem 1.5 that has at most finitely many zeros satisfying, we have

Therefore,.