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# The zeros on complex differential-difference polynomials of certain types

*Advances in Difference Equations*
**volume 2018**, Article number: 262 (2018)

## Abstract

In this paper, we consider the zeros distribution of \(f(z)P(z,f) -q(z)\), where \(P(z,f)\) is a linear differential-difference polynomial of a finite-order transcendental entire function \(f(z)\), and \(q(z)\) is a nonzero polynomial. To a certain extent, Theorem 1.1 generalizes the recent results (Latreuch and Belaïdi in Arab. J. Math. 7(1):27–37, 2018; Lü et al. in Kodai Math. J. 39(3):500–509, 2016) related to Hayman conjecture (Hayamn in Ann. Math. 70:9–42, 1959).

## 1 Introduction

We assume that the readers are familiar with the basic symbols and fundamental results of Nevanlinna theory [8, 9, 18]. A function \(a(z)\not\equiv0, \infty\) is a small function with respect to \(f(z)\) if \(T(r,a)=S(r,f)\), where \(S(r,f)=o(T(r,f))\) as \(r\rightarrow \infty\) outside a possible exceptional set of finite logarithmic measure. We use \(S(f)\) to denote the family of all small functions with respect to \(f(z)\). In the paper, a linear differential-difference polynomial of a meromorphic function \(f(z)\) is defined by

where \(\lambda_{i}(z)\in S(f)\) and \(c_{i}\in\mathbb{C}\), \(k_{i}\) \((i=1,\ldots,n)\) are nonnegative integers.

In 1959, Hayman [7] considered the value distribution of the differential polynomial \(f^{n}f'\) and obtained the following result.

### Theorem A

*Let*
\(f(z)\)
*be a transcendental meromorphic function*, *and let*
\(n\geq3\)
*be an integer*. *Then*
\(f(z)^{n}f'(z)-d\)
*has infinitely many zeros*, *where*
*d*
*is a nonzero constant*.

Since then, there were many studies on the zeros distribution of differential polynomials, such as [1, 2, 17]. Recently, some researchers considered the difference analogues of Theorem A, and many related results have been obtained. Laine and Yang [10] investigated the value distribution of \(f(z)^{n}f(z+c)\) and proved the following result; for statement of this and others results, we recall the definition of the order of a meromorphic function \(f(z)\):

### Theorem B

*Let*
\(f(z)\)
*be a transcendental entire function of finite order*, *and let*
*c*
*be a nonzero complex constant*. *If*
\(n\geq2\), *then*
\(f(z)^{n}f(z+c)-a\)
*has infinitely many zeros*, *where*
*a*
*is a nonzero complex constant*.

Some researchers improved Theorem B in different ways; for example, the constant *a* was replaced by a nonzero polynomial in [13]. In addition, the papers [12, 16, 19] are devoted to the cases of meromorphic functions *f* or more general difference products. Liu, Liu, and Zhou [14] obtained results related to Theorem B in differential-difference polynomials, which can be stated as follows.

### Theorem C

*Let*
\(f(z)\)
*be a finite*-*order transcendental entire function*, *and let*
*k*
*be a positive integer*. *If*
\(n\geq2\), *then*
\(f(z)^{n}f^{(k)}(z+c)-a(z)\)
*has infinitely many zeros*, *where*
\(a(z)\)
*is an entire function with*
\(\rho(a)<\rho(f)\).

### Theorem D

*Let*
\(f(z)\)
*be a finite*-*order transcendental entire function with a Borel exceptional polynomial*
\(d(z)\), *and let*
*k*
*be a positive integer*. *If*
\(n\geq1\), *then*
\(f(z)^{n}f^{(k)}(z+c)-b\)
*has infinitely many zeros*, *where*
*b*
*is a nonzero constant*.

### Remark 1

If \(n=1\), then Theorem B is not true. For example, if \(f(z)=e^{z}+1\) and \(e^{c}=-1\), then \(f(z)f(z+c)-1=-e^{2z}\) has no zeros. Chen, Huang, and Zheng [3] considered the case \(n=1\) in Theorem B with \(f(z)\) having a Borel exceptional value. In fact, the above function also shows that Theorem D happens: \(f(z)f^{(k)}(z+c)-b=-e^{2z}-e^{z}-b\) has infinitely many zeros, and the value 1 is the Borel exceptional value of \(e^{z}+1\).

However, it is still an open question whether Theorem D is true for a general transcendental entire function \(f(z)\), that is, whether the condition that \(f(z)\) has a Borel exceptional polynomial can be removed.

### Question 1

Let \(f(z)\) be a finite-order transcendental entire function, and let *c* be a nonzero constant and *k* be a positive integer. Have the differential-difference polynomials \(f(z)f^{(k)}(z+c)-a(z)\) infinitely many zeros or not?

More generally, we can raise the following Question 2.

### Question 2

What about the zeros distribution of \(f(z)P(z,f)-a(z)\)? Here \(P(z,f)\) is a linear differential-difference polynomial in \(f(z)\), which is a transcendental entire function of finite order, and \(a(z)\) is a small function with respect to \(f(z)\).

Two papers [11, 15] contribute greatly to this paper. In fact, assume that \(f(z)\) is a transcendental entire function of finite order. Lü et al. [15] obtained that one of \(f(z)f^{(k)}(z)-p(z)\) and \(f(z)f^{(l)}(z)-p(z)\) must have infinitely many zeros, provided that *k* and *l* are nonzero distinct constants. Latreuch and Belaïdi [11] showed that one of \(f(z)f(z+c_{1})-p(z)\) and \(f(z)f(z+c_{2})-p(z)\) must have infinitely many zeros, where \(f(z+c_{1})\not\equiv f(z+c_{2})\). We obtain the following result.

### Theorem 1.1

*Let*
\(f(z)\)
*be a transcendental entire function of finite order*, *let*
\(F_{1}(z,f)\)
*and*
\(F_{2}(z,f)\)
*be two linear differential*-*difference polynomials in*
\(f(z)\)
*with entire coefficients such that*
\(F_{1}(z,f)\not \equiv F_{2}(z,f)\), *and let*
\(q(z)\)
*be a nonzero polynomial*. *Then at least one of*
\(f(z)F_{1}(z,f)-q(z)\)
*and*
\(f(z)F_{2}(z,f)-q(z)\)
*has infinitely many zeros except the case where only one of*
\(F_{1}(z,f)\)
*or*
\(F_{2}(z,f)\)
*is a small function with respect to*
\(f(z)\).

### Remark 2

Considering the case that all \(c_{i}\) and \(k_{i}\) are zeros, that is, \(F_{1}(z,f)=\lambda_{1}(z)f(z)\) and \(F_{2}(z,f)=\lambda_{2}(z)f(z)\), where \(\lambda_{1}(z)\not\equiv\lambda_{2}(z)\) are small functions with respect to \(f(z)\). In this case, it is easy to get that one of \(f(z)F_{1}(z,f)-p(z)\) and \(f(z)F_{2}(z,f)-p(z)\) has infinitely many zeros by the second main theorem for three small functions [8, Theorem 2.5]. In addition, if \(F_{1}(z,f)=f^{(k)}(z)\) and \(F_{2}(z,f)=f^{(l)}(z)\) in Theorem 1.1, then it is the result given in [15]. If \(F_{1}(z,f)=f(z+c_{1})\) and \(F_{2}(z,f)=f(z+c_{2})\) in Theorem 1.1, it is the result considered in [11]. If \(F_{1}(z,f)=f^{(k)}(z+c_{1})\) and \(F_{2}(z,f)=f^{(k)}(z+c_{2})\), Theorem 1.1 partially answers Question 1.

### Remark 3

The condition that \(q(z)\not\equiv0\) and \(F_{1}(z,f)\not\equiv F_{2}(z,f)\) cannot be removed, which can be seen by taking \(f(z)=e^{z}\) and \(c_{1}=2\pi i\), \(c_{2}=4\pi i\). In this case, we have that \(f(z)f^{(k)}(z+c_{1})=f(z)f^{(k)}(z+c_{2})=e^{2z}\) has no zeros.

### Remark 4

The exceptional case in Theorem 1.1 may happen. For example, consider \(F_{1}(z,f)\equiv1\) and \(F_{2}(z,f)=f(z+i\pi)\) with \(f(z)=e^{z}+1\). Then \(f(z)F_{1}(z,f)-1=e^{z}\) and \(f(z)F_{2}(z,f)-1=-e^{2z}\) have no zeros.

The following corollary follows directly from Theorem 1.1.

### Corollary 1.2

*Let*
\(\alpha, \beta, p_{1}, p_{2}\)
*and*
\(q\not\equiv0\)
*be nonconstant polynomials*. *Then the system of equations*

*has no transcendental entire functions of finite order for every*
\(F_{1}(z,f)\neq F_{2}(z,f)\)
*such that*
\(F_{1}(z,f)\)
*and*
\(F_{2}(z,f)\)
*are not small functions with respect to*
\(f(z)\).

## 2 Some lemmas

The following lemma on the logarithmic derivative of meromorphic functions plays a crucial role in the paper.

### Lemma 2.1

*Let*
*f*
*be a finite*-*order meromorphic function*, *and let*
\(k\in\mathbb {N}\). *Then*

The difference analogue of the logarithmic derivative lemma, which is also very important in the proof of Theorem 1.1, was independently found by Halburd and Korhonen [6] and Chiang and Feng [4]. Let us state the result as follows.

### Lemma 2.2

*Let*
*f*
*be a transcendental meromorphic function of finite order*, *and let*
\(c\in\mathbb{C}\). *Then*

*for all*
*r*
*outside a set*
*E*
*of finite logarithmic measure*.

The following result is trivial by Lemma 2.1 and Lemma 2.2.

### Lemma 2.3

*Let*
*f*
*be a transcendental meromorphic function of finite order*, *and let*
\(P(z,f)\)
*be a linear differential*-*difference polynomial*. *Then*

*for all*
*r*
*outside a set*
*E*
*of finite logarithmic measure*.

### Lemma 2.4

([5])

*Let*
\(f(z)\)
*be a transcendental meromorphic function of finite order*
\(\rho(f)=\rho\), *and let*
\(\varepsilon>0\)
*be a given constant*. *Then there exists a set*
\(E_{0}\subset(0,+\infty)\)
*of finite logarithmic measure such that*, *for all*
*z*
*satisfying*
\(|z|\notin E_{0}\cup[0,1]\)
*and all*
\(k,j\)
*such that*
\(0\leq j\leq k\), *we have*

We also need the following lemma to estimate the counting function and the characteristic function for transcendental meromorphic functions of finite order.

### Lemma 2.5

([4, Theorems 2.1, 2.2])

*Let*
\(f(z)\)
*be a transcendental meromorphic function with finite order*
\(\rho(f)=\rho\), *and let*
*η*
*be a fixed nonzero complex number*. *Then*, *for each*
\(\varepsilon> 0\), *we have*

### Lemma 2.6

([18, Theorem 1.22])

*Let*
\(f(z)\)
*be a transcendental meromorphic function*. *Then*

### Remark 5

Let \(f(z)\) be a meromorphic function of finite order. Combining Lemma 2.5 and Lemma 2.6, for a linear differential-difference polynomial

we have

If \(f(z)\) is a transcendental entire function of finite order, this inequality reduces to

### Lemma 2.7

([9, Lemma 2.4.2])

*Let*
\(f(z)\)
*be a transcendental meromorphic solution of*

*where*
\(A(z,f)\)
*and*
\(B(z,f)\)
*are differential polynomials in*
*f*
*and its derivatives with small meromorphic coefficients*
\(a_{\lambda}\)
*in the sense of*
\(m(r,a_{\lambda})=S(r,f)\)
*for all*
\(\lambda\in I\) (*I*
*is a set of distinct complex numbers*). *If the total degree of*
\(B(z,f)\)
*as a polynomial in*
*f*
*and its derivatives is less than or equal to*
*n*, *then*
\(m(r,A(z,f))=S(r,f)\).

### Lemma 2.8

*Let*
\(f(z)\)
*be a transcendental meromorphic function of finite order*, *let*
\(F(z)\)
*be a linear differential*-*difference polynomial of*
\(f(z)\), *and let and*
\(a(z)\)
*and*
\(b(z)\)
*be nonzero small functions with respect to*
\(f(z)\). *The equation*

*gives*

*where*
\(N_{1}\)
*denotes the counting function of the simple zeros of*
*f*.

### Proof

Dividing both sides of (2.1) by \(f^{2}\), since \(a(z)\) and \(b(z)\) are small functions of *f*, we get that

by Lemmas 2.1–2.3 and 2.5. By the Nevanlinna first main theorem we have

From (2.1) it is easy to see that

where \(N_{2}\) denotes the counting function of zeros of *f* with multiplicities not less than 2. Inequality (2.4) implies that the zeros of *f* are mainly simple zeros. Thus, by (2.3) and (2.4) we deduce that

□

### Remark 6

If \(a(z)\) and \(b(z)\) are rational functions in (2.1), then there are at most finitely many multiple zeros of \(f(z)\).

## 3 Proof of Theorem 1.1

The proof utilizes the ideas of the papers [11, 15], although some details are different. Suppose contrary to the assertion that both \(f(z)F_{1}(z,f)-q(z)\) and \(f(z)F_{2}(z,f)-q(z)\) have finitely many zeros. Since \(f(z)\) is of finite order, by the Hadamard factorization theorem we can write

and

where \(\alpha(z)\), \(\beta(z)\), \(p_{1}(z)\), \(p_{2}(z)\) are polynomials. From Remark 5 we know that \(T(r,F_{1}(z,f))\leq T(r,f)+S(r,f)\) and \(T(r,F_{2}(z,f))\leq T(r,f)+S(r,f)\). First, if \(T(r,F_{1}(z,f))=S(r,f)\) and \(T(r,F_{2}(z,f))=S(r,f)\), then from the second main theorem for three small functions [8, Theorem 2.5] we know that

which is impossible.

Second, suppose that \(T(r,F_{1}(z,f))\neq S(r,f)\) and \(T(r,F_{2}(z,f))\neq S(r,f)\). We affirm that \(e^{\alpha}\), \(e^{\beta}\), and \(e^{\alpha+\beta}\) are not small functions with respect to \(f(z)\). Otherwise, if \(e^{\alpha}\) is a small function with respect to \(f(z)\), we have \(f(z)F_{1}(z,f)=t(z)\) from (3.1), where \(t(z)=q(z)+p_{1}(z)e^{\alpha(z)}\) is a small function with respect to \(f(z)\). Lemma 2.7 and Lemma 2.3 imply that

and we get \(T(r,f)=S(r,f)\), a contradiction. We can use a similar method to obtain that \(e^{\beta}\) is not a small function of \(f(z)\). From (3.1) and (3.2) we have

If \(e^{\alpha+\beta}\) is a small function, by Lemma 2.7 and Lemma 2.3 we have \(T(r, F_{1}F_{2})=S(r,f)\) and \(T(r, \frac {F_{1}F_{2}}{f})=S(r,f)\), and hence \(T(r,f)=S(r,f)\), which is impossible.

Since *α* is a polynomial, *α* and \(\alpha'\) are small functions of *f*. Differentiating (3.1) and eliminating \(e^{\alpha }\), we obtain

where \(a_{1}=\frac{p_{1}'}{p_{1}}+\alpha'\) and \(b_{1}=(\frac {p_{1}'}{p_{1}}+\alpha')q-q'\). Now, we will show that \(a_{1}\not\equiv 0\). Indeed, if \(a_{1}\equiv0\), then by a simple integration there exists a nonzero constant \(C_{1}\) such that \(C_{1}=p_{1}e^{\alpha}\), which implies a contradiction to the fact that \(e^{\alpha}\) is not a small function with respect to \(f(z)\). Similarly, we have \(b_{1}\not \equiv0\).

By the same arguments as before, (3.2) gives

where \(a_{2}=\frac{p_{2}'}{p_{2}}+\beta'\) and \(b_{2}=(\frac {p_{2}'}{p_{2}}+\beta')q-q'\). We also obtain \(a_{2}\not\equiv0\) and \(b_{2}\not\equiv0\).

In view of Lemma 2.8 and Remark 6, we assume that \(z_{0}\) is a simple zero of *f* such that \(z_{0}\) is not a zero or pole of \(b_{i}\)
\((i=1,2)\). Equations (3.3) and (3.4) imply that

and

Hence we have

We will consider two cases depending on whether \(b_{2}F_{1}-b_{1}F_{2}\equiv0\) or not.

*Case* 1. \(b_{2}F_{1}-b_{1}F_{2}\not\equiv0\). Set

From Lemma 2.3 we have \(m(r,h)=S(r,f)\). On the other hand, from (3.7) and Lemma 2.8 we have

Thus \(T(r,h)=S(r,f)\). Rewrite (3.8) in the form

By differentiating (3.10) we have

Substituting (3.10) and (3.11) into (3.3), we get

In addition, equation (3.4) can be written as

Combining (3.12) and (3.13), we get

Note that \(-2\frac{h}{b_{2}}\not\equiv0\). We proceed to prove that

Otherwise, if \(a_{1}\frac{h}{b_{2}}-(\frac{h}{b_{2}})'\equiv0\), then by the definition of \(a_{1}\) and a simple integration we have

where \(C_{2}\) is a nonzero constant. Since \(T (r,\frac {h}{b_{2}} )=S(r,f)\), \(p_{1}e^{\alpha}\) is a small function of *f*, a contradiction. If \(a_{1}\frac{b_{1}}{b_{2}}-(\frac {b_{1}}{b_{2}})'-a_{2}\frac{b_{1}}{b_{2}}\equiv0\), then we have \(a_{1}-a_{2}\equiv\frac{b_{1}'}{b_{1}}-\frac{b_{2}'}{b_{2}}\). A simple integration yields that

where \(C_{3}\) is a nonzero constant, and *γ* is a small function of *f*. From (3.1) and (3.2) we have

If \(\gamma\not\equiv1\), then by Lemma 2.7 we get

From this and from (3.17) we have

a contradiction. If \(\gamma\equiv1\), then we have \(F_{1}\equiv F_{2}\), which contradicts the hypothesis of Theorem 1.1. Thus \(a_{1}\frac {b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'-a_{2}\frac{b_{1}}{b_{2}}\not\equiv0\).

From these discussions we can rewrite (3.14) as

where

Furthermore, (3.10) and (3.18) give

where

Differentiating (3.18), we have

Substituting (3.18) and (3.21) into (3.4), we obtain

Differentiating (3.22), we have

Since we have supposed that \(f(z_{0})=0\), \(f'(z_{0})\neq0\), and \(b_{2}(z_{0})\neq0 \), from (3.22) and (3.23), respectively, we obtain

and

which leads to

Let

It is obvious that *H* is a small function of *f* and

Substituting (3.24) into (3.22), we have

where

It is easy to see that \(g_{3}\not\equiv0\). We proceed to prove \(g_{2}\not\equiv0\). Otherwise, we get

since, by the definition of \(a_{1}\) and \(a_{2}\),

By integration we have

where \(C_{3}\) is a nonzero constant. Since \(T (r,C_{3}h(\frac {n}{b_{2}})^{\frac{1}{2}} )=S(r,f)\), we can deduce that \(e^{\alpha +\beta}\) is a small function of *f*, a contradiction. Thus \(g_{2}\not \equiv0\). Differentiating (3.25), we have

By the same method used to deal with (3.22) and (3.23) we have

where

Substituting (3.28) into (3.27), we get

Combining (3.29) and (3.25), we have

Let

and

Now we claim that \(q_{1}\) and \(q_{2}\) vanish identically. If \(q_{2}\not \equiv0\), then by the previous analysis we can get that \(q_{2}\) is a small function of *f*. From (3.30) we have \(q_{2}(z_{0})=0\). Thus

a contradiction. Again by (3.30) we get \(q_{1}\equiv0\). Eliminating *R* from \(q_{1}\equiv0\) and \(q_{2}\equiv0\), we have

We continue by discussing two subcases depending on whether \(4g_{1}g_{3}-g_{2}^{2}\) vanishes identically or not.

*Subcase* 1. If \(4g_{1}g_{3}-g_{2}^{2}\not\equiv0\), then from (3.31) we have

Combining this equation with (3.26), we get that

Similarly as in the proof of \(g_{2}\not\equiv0\), we can deduce that \(e^{\alpha+\beta}\) is a small function of *f*, a contradiction.

*Subcase* 2. If \(4g_{1}g_{3}-g_{2}^{2}\equiv0\), then, on the one hand, by using \(q_{1}\equiv0\) we can rewrite equation (3.28) as

and then from this equation and from (3.24) we have

On the other hand, by the definitions of \(g_{1}\) and \(g_{3}\) we have

Therefore we get

For brevity, we denote

By calculation we have

and

From (3.26) we have

Substituting these identities into (3.32), we have

which leads to

Since

if \(R(z)\) is a nonzero rational function, dividing both sides of the above equation by \(\frac{(a_{1}+a_{2})^{2}}{2}\) and taking the limit, we have

It follows by Lemma 2.4 that

By setting \(\alpha(z)=a_{m}z^{m}+\cdots+a_{0}\) and \(\beta (z)=b_{m}z^{m}+\cdots+b_{0}\) we have

which implies that \(\frac{a_{m}}{b_{m}}=2\) or \(\frac{a_{m}}{b_{m}}=\frac {1}{2}\). We first consider the case \(\frac{a_{m}}{b_{m}}=2\). From (3.1), (3.2), (3.18), and (3.19) we have

and

where

Then

Combining (2.2) with the expressions of *A* and *B*, we obtain

Hence we get \(T(r,F_{2})=S(r,f)\), a contradiction. For the case \(\frac{a_{m}}{b_{m}}=\frac{1}{2}\), by the same argument we can also deduce \(T(r,F_{1})=S(r,f)\), a contradiction.

*Case* 2. \(b_{2}F_{1}-b_{1}F_{2}\equiv0\).

Using the same arguments as in the proof of (3.14), we have

which leads to

where \(C_{3}\) is a nonzero constant. Then by (3.1) and (3.2), if \(\frac{C_{3}b_{1}}{b_{2}}\equiv1\), then \(F_{1}\equiv F_{2}\), which contradicts the hypothesis of Theorem 1.1. If \(\frac{C_{3}b_{1}}{b_{2}}\not\equiv1\), then from

we have \(T(r,f)=S(r,f)\), which is impossible.

Thus we obtain that at least one of \(f(z)F_{1}(z,f)-q(z)\) and \(f(z)F_{2}(z,f)-q(z)\) has infinitely many zeros.

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## Acknowledgements

The authors would like to thank the referee for his/her helpful suggestions and comments.

## Funding

This work was partially supported by the NSFC (No. 11661052), the NSF of Jiangxi (No. 20161BAB211005), the outstanding youth scientist foundation plan of Jiangxi (No. 20171BCB23003). This work was also partially supported by the Innovation Found for the Postgraduate of Nanchang University (No. cx2016148).

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### Cite this article

Song, C., Liu, K. & Ma, L. The zeros on complex differential-difference polynomials of certain types.
*Adv Differ Equ* **2018**, 262 (2018). https://doi.org/10.1186/s13662-018-1712-x

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DOI: https://doi.org/10.1186/s13662-018-1712-x

### MSC

- 30D35
- 39A45

### Keywords

- Differential-difference polynomials
- Zeros
- Finite order