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Theory and Modern Applications

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

## 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

$$P(z,f)=\sum_{i=1}^{n}\lambda_{i}(z)f^{(k_{i})}(z+c_{i}),$$

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 $$f(z)$$:

$$\rho(f)=\limsup_{r\rightarrow\infty}\frac{\log T(r,f)}{\log r}.$$

### 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

\begin{aligned} \textstyle\begin{cases} f(z)F_{1}(z,f)-q(z)=p_{1}(z)e^{\alpha(z)}, \\ f(z)F_{2}(z,f)-q(z)=p_{2}(z)e^{\beta(z)}, \end{cases}\displaystyle \end{aligned}

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

([8, 18])

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

$$m \biggl(r,\frac{f^{(k)}}{f} \biggr)=S(r,f).$$

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

$$m \biggl(r,\frac{f(z+c)}{f(z)} \biggr)=O \biggl(\frac{\log r}{r}{T(r,f)} \biggr)=S(r,f)$$

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

$$m \biggl(r,\frac{P(z,f)}{f(z)} \biggr)=S(r,f)$$

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

$$\biggl\vert \frac{f^{(k)}(z)}{f^{(j)}(z)} \biggr\vert \leq \vert z \vert ^{(k-j)(\rho -1+\varepsilon)}.$$

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

\begin{aligned} &T\bigl(r,f(z+\eta)\bigr)=T(r,f)+O\bigl(r^{\rho-1+\varepsilon}\bigr)+O(\log r)=T(r,f)+S(r,f), \\ &N\bigl(r,f(z+\eta)\bigr)=N(r,f)+O\bigl(r^{\rho-1+\varepsilon}\bigr)+O(\log r)=N(r,f)+S(r,f). \end{aligned}

### Lemma 2.6

([18, Theorem 1.22])

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

$$T\bigl(r,f^{(n)}\bigr)\leq T(r,f)+n\overline{N}(r,f)+S(r,f).$$

### 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

$$P(z,f)=\sum_{i=1}^{n} \lambda_{i}(z)f^{(k_{i})}(z+c_{i}),$$

we have

$$T\bigl(r,P(z,f)\bigr)\leq \Biggl(n+\sum_{i=1}^{n}k_{i} \Biggr)T(r,f)+S(r,f).$$

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

$$T\bigl(r,P(z,f)\bigr)\leq T(r,f)+S(r,f).$$

### Lemma 2.7

([9, Lemma 2.4.2])

Let $$f(z)$$ be a transcendental meromorphic solution of

$$f^{n}A(z,f)=B(z,f),$$

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

$$a(z)f(z)F(z)-f'(z)F(z)-f(z)F'(z)=b(z)$$
(2.1)

gives

$$T(r,f)=N \biggl(r,\frac{1}{f} \biggr)+S(r,f)=N_{1} \biggl(r, \frac {1}{f} \biggr)+S(r,f),$$

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

\begin{aligned} 2m \biggl(r,\frac{1}{f} \biggr) & \leq m \biggl(r,\frac{F}{f} \biggr)+m \biggl(r,\frac{f'}{f} \frac{F}{f} \biggr) +m \biggl(r,\frac{F'}{f} \biggr)+S(r,f)=S(r,f) \end{aligned}
(2.2)

by Lemmas 2.12.3 and 2.5. By the Nevanlinna first main theorem we have

$$T(r,f)=N \biggl(r,\frac{1}{f} \biggr)+S(r,f).$$
(2.3)

From (2.1) it is easy to see that

$$N_{2} \biggl(r,\frac{1}{f} \biggr)\leq N \biggl(r,\frac{1}{b(z)} \biggr)=S(r,f),$$
(2.4)

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

$$T(r,f)=N \biggl(r,\frac{1}{f} \biggr)+S(r,f)=N_{1} \biggl(r, \frac {1}{f} \biggr)+S(r,f).$$

□

### 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

$$f(z)F_{1}(z,f)-q(z)=p_{1}(z)e^{\alpha(z)}$$
(3.1)

and

$$f(z)F_{2}(z,f)-q(z)=p_{2}(z)e^{\beta(z)},$$
(3.2)

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

\begin{aligned} T(r,f) \leq& \overline{N}(r,f)+\overline{N} \biggl(r,\frac{1}{f(z)-\frac {q(z)}{F_{1}(z,f)}} \biggr)+ \overline{N} \biggl(r,\frac{1}{f(z)-\frac {q(z)}{F_{2}(z,f)}} \biggr)+S(r,f) \\ =&\overline{N} \biggl(r,\frac{1}{\frac{p_{1}(z)}{F_{1}(z,f)}} \biggr)+\overline{N} \biggl(r, \frac{1}{\frac{p_{2}(z)}{F_{2}(z,f)}} \biggr)+S(r,f)=S(r,f), \end{aligned}

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

$$T(r,F_{1})=m(r,F_{1})=S(r,f),$$

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

$$f(z)^{2}F_{1}(z,f)F_{2}(z,f)-q(z)f(z) \bigl[F_{1}(z,f)+F_{2}(z,f)\bigr]+q(z)^{2}=p_{1}(z)p_{2}(z)e^{\alpha +\beta}.$$

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

$$a_{1}fF_{1}-f'F_{1}-fF'_{1}=b_{1},$$
(3.3)

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

$$a_{2}fF_{2}-f'F_{2}-fF'_{2}=b_{2},$$
(3.4)

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

$$\bigl(f'F_{1}+b_{1}\bigr) (z_{0})=0$$
(3.5)

and

$$\bigl(f'F_{2}+b_{2}\bigr) (z_{0})=0.$$
(3.6)

Hence we have

$$(b_{2}F_{1}-b_{1}F_{2}) (z_{0})=0.$$
(3.7)

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

$$h=\frac{b_{2}F_{1}-b_{1}F_{2}}{f}.$$
(3.8)

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

\begin{aligned} N(r,h) =&N \biggl(r,\frac{b_{2}F_{1}-b_{1}F_{2}}{f} \biggr) \\ =&N_{1} \biggl(r,\frac{b_{2}F_{1}-b_{1}F_{2}}{f} \biggr)+S(r,f)=S(r,f). \end{aligned}
(3.9)

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

$$F_{1}=\frac{h}{b_{2}}f+\frac{b_{1}}{b_{2}}F_{2}.$$
(3.10)

By differentiating (3.10) we have

$$F'_{1}= \biggl(\frac{h}{b_{2}} \biggr)'f+\frac{h}{b_{2}}f'+ \biggl( \frac {b_{1}}{b_{2}} \biggr)'F_{2}+\frac{b_{1}}{b_{2}}F'_{2}.$$
(3.11)

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

\begin{aligned} \biggl[a_{1} \frac{h}{b_{2}}- \biggl(\frac{h}{b_{2}} \biggr)' \biggr]f^{2}-2\frac{h}{b_{2}}ff' + \biggl[a_{1}\frac{b_{1}}{b_{2}}- \biggl(\frac{b_{1}}{b_{2}} \biggr)' \biggr]fF_{2} &-\frac{b_{1}}{b_{2}}f'F_{2}- \frac{b_{1}}{b_{2}}fF'_{2}=b_{1}. \end{aligned}
(3.12)

In addition, equation (3.4) can be written as

$$a_{2}\frac{b_{1}}{b_{2}}fF_{2}- \frac{b_{1}}{b_{2}}f'F_{2}-\frac {b_{1}}{b_{2}}fF'_{2}=b_{1}.$$
(3.13)

Combining (3.12) and (3.13), we get

$$\biggl[a_{1}\frac{h}{b_{2}}- \biggl( \frac{h}{b_{2}} \biggr)' \biggr]f-2\frac {h}{b_{2}}f'+ \biggl[a_{1}\frac{b_{1}}{b_{2}}- \biggl(\frac{b_{1}}{b_{2}} \biggr)'-a_{2}\frac{b_{1}}{b_{2}} \biggr]F_{2}=0.$$
(3.14)

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

$$a_{1}\frac{h}{b_{2}}- \biggl(\frac{h}{b_{2}} \biggr)'\not\equiv0\quad \mbox{and} \quad a_{1} \frac{b_{1}}{b_{2}}- \biggl(\frac {b_{1}}{b_{2}} \biggr)'-a_{2} \frac{b_{1}}{b_{2}}\not\equiv0.$$
(3.15)

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

$$p_{1}e^{\alpha}=C_{2} \frac{h}{b_{2}},$$
(3.16)

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

$$\frac{p_{1}}{p_{2}}e^{\alpha-\beta}=C_{3}\frac{b_{1}}{b_{2}}:=\gamma,$$

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

$$f(F_{1}-\gamma F_{2})=q(1-\gamma).$$
(3.17)

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

$$m(r,F_{1}-\gamma F_{2})+S(r,f)=T(r,F_{1}-\gamma F_{2})=S(r,f).$$

From this and from (3.17) we have

$$T(r,f)=T \biggl(r,\frac{q(1-\gamma)}{F_{1}-\gamma F_{2}} \biggr)=S(r,f),$$

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

$$F_{2}=mf+nf',$$
(3.18)

where

$$m=\frac{(\frac{h}{b_{2}})'-a_{1}\frac{h}{b_{2}}}{(a_{1}-a_{2})\frac {b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'},\qquad n=\frac{\frac {2h}{b_{2}}}{(a_{1}-a_{2})\frac{b_{1}}{b_{2}}-(\frac{b_{1}}{b_{2}})'}.$$

Furthermore, (3.10) and (3.18) give

$$F_{1}=sf+tf',$$
(3.19)

where

$$s=\frac{b_{1}m+h}{b_{2}},\qquad t=\frac{b_{1}}{b_{2}}n.$$
(3.20)

Differentiating (3.18), we have

$$F'_{2}=m'f+ \bigl(m+n'\bigr)f'+nf''.$$
(3.21)

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

$$\bigl(a_{2}m-m'\bigr)f^{2}+ \bigl(a_{2}n-2m-n'\bigr)ff'-n \bigl(f'\bigr)^{2}-nff''=b_{2}.$$
(3.22)

Differentiating (3.22), we have

\begin{aligned}[b] & \bigl(a_{2}m-m' \bigr)'f^{2}+ \bigl[2\bigl(a_{2}m-m' \bigr)+\bigl(a_{2}n-2m-n'\bigr) \bigr]ff' \\ &\quad {}+\bigl(a_{2}n-2m-2n'\bigr) \bigl[\bigl(f' \bigr)^{2}+ff'' \bigr]-n\bigl(3f'f''+ff''' \bigr)=b_{2}'. \end{aligned}
(3.23)

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

$$\bigl(n\bigl(f'\bigr)^{2}+b_{2} \bigr) (z_{0})=0$$

and

$$\bigl[\bigl(a_{2}n-2m-2n'\bigr) \bigl(f' \bigr)^{2}-3nf'f''-b_{2}' \bigr](z_{0})=0,$$

which leads to

$$\bigl(\bigl[b_{2}\bigl(a_{2}n-2m-2n' \bigr)+b_{2}'n\bigr]f'-3b_{2}nf'' \bigr) (z_{0})=0.$$

Let

$$H=\frac{[b_{2}(a_{2}n-2m-2n')+b_{2}'n]f'-3b_{2}nf''}{f}.$$

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

$$f''=\frac{b_{2}(a_{2}n-2m-2n')+b_{2}'n}{3b_{2}n}f'- \frac{H}{3b_{2}n}f.$$
(3.24)

Substituting (3.24) into (3.22), we have

$$g_{1}f^{2}+g_{2}ff'+g_{3} \bigl(f'\bigr)^{2}=b_{2},$$
(3.25)

where

\begin{aligned} &g_{1}=a_{2}m-m'+\frac{H}{3b_{2}}, \\ &g_{2}=\frac{1}{3} \biggl(2a_{2}-\frac{b_{2}'}{b_{2}} \biggr)n-\frac {4}{3}m-\frac{1}{3}n',\qquad g_{3}=-n. \end{aligned}

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

$$\frac{g_{2}}{g_{3}}=\frac{2}{3}\frac{h'}{h}+ \frac{1}{3}\frac {n'}{n}-\frac{1}{3}\frac{b_{2}'}{b_{2}}- \frac{2}{3}(a_{1}+a_{2})=0,$$
(3.26)

since, by the definition of $$a_{1}$$ and $$a_{2}$$,

$$\alpha'+\beta'+\frac{p_{1}'}{p_{1}}+\frac{p_{2}'}{p_{2}}= \frac {h'}{h}+\frac{1}{2} \biggl(\frac{n'}{n}-\frac{b_{2}'}{b_{2}} \biggr).$$

By integration we have

$$p_{1}p_{2}e^{\alpha+\beta}=C_{3}h\biggl( \frac{n}{b_{2}}\biggr)^{\frac{1}{2}},$$

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

$$g_{1}'f^{2}+ \bigl(2g_{1}+g_{2}'\bigr)ff'+ \bigl(g_{2}+g_{3}'\bigr) \bigl(f' \bigr)^{2}+g_{2}ff''+2g_{3}f'f''=b_{2}'.$$
(3.27)

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

$$f''=\frac{b_{2}'g_{3}-b_{2}(g_{2}+g_{3}')}{2b_{2}g_{3}}f'+ \frac {R}{2b_{2}g_{3}}f,$$
(3.28)

where

$$R=\frac{[b_{2}(g_{2}+g_{3}')-b_{2}'g_{3}]f'+2b_{2}g_{3}f''}{f} \quad \mbox{and}\quad T(r,R)=S(r,f).$$

Substituting (3.28) into (3.27), we get

\begin{aligned}[b] &\biggl(g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}} \biggr)f^{2}+ \biggl(2g_{1}+g_{2}'+\frac{1}{2} \frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2}\bigl(g_{2}+g_{3}' \bigr)\frac{g_{2}}{g_{3}} +\frac{R}{b_{2}} \biggr)f'f \\ &\quad {}+\frac{b_{2}'g_{3}}{b_{2}}\bigl(f'\bigr)^{2}=b_{2}'. \end{aligned}
(3.29)

Combining (3.29) and (3.25), we have

$$\biggl(g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}}- \frac{b_{2}}{b_{2}}'g_{1} \biggr)f+ \biggl(2g_{1}+g_{2}'- \frac{1}{2}\frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2} \bigl(g_{2}+g_{3}'\bigr)\frac{g_{2}}{g_{3}}+ \frac{R}{b_{2}} \biggr)f'=0.$$
(3.30)

Let

$$q_{1}=g_{1}'+\frac{g_{2}R}{2b_{2}g_{3}}- \frac{b_{2}}{b_{2}}'g_{1}$$

and

$$q_{2}=2g_{1}+g_{2}'- \frac{1}{2}\frac{b_{2}'}{b_{2}}g_{2}-\frac {1}{2} \bigl(g_{2}+g_{3}'\bigr)\frac{g_{2}}{g_{3}}+ \frac{R}{b_{2}}.$$

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

$$T(r,f)=N_{1} \biggl(r,\frac{1}{f} \biggr)+S(r,f)\leq N \biggl(r,\frac {1}{q_{2}} \biggr)+S(r,f)=S(r,f),$$

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

\begin{aligned}[b] &g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)\frac {b_{2}'}{b_{2}}+g_{2}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)+2g_{2}g_{2}'g_{3}-g_{2}^{2}g_{3}'-4g_{1}'g_{3}^{2} \\ &\quad =g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)\frac {b_{2}'}{b_{2}}+g_{2}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)-g_{3}\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)'+g_{3}'\bigl(4g_{1}g_{3}-g_{2}^{2} \bigr)=0. \end{aligned}
(3.31)

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

$$\frac{g_{2}}{g_{3}}=\frac {(4g_{1}g_{3}-g_{2}^{2})'}{4g_{1}g_{3}-g_{2}^{2}}-\frac {b_{2}'}{b_{2}}-\frac{g_{3}'}{g_{3}}.$$

Combining this equation with (3.26), we get that

$$2(a_{1}+a_{2})=2\frac{h'}{h}+\frac{n'}{n}+2 \frac{b_{2}'}{b_{2}}-3\frac {(4g_{1}g_{3}-g_{2}^{2})'}{4g_{1}g_{3}-g_{2}^{2}}.$$

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

$$f''= \biggl(\frac{b_{2}'g_{1}}{b_{2}g_{2}}-\frac{g_{1}'}{g_{2}} \biggr)f +\frac{1}{2} \biggl(\frac{b_{2}'}{b_{2}}-\frac{g_{2}}{g_{3}}- \frac {g_{3}'}{g_{3}} \biggr)f',$$

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

$$\frac{g_{1}'}{g_{2}}-\frac{b_{2}'}{b_{2}}\frac{g_{1}}{g_{2}}=\frac{H}{3b_{2}n}.$$

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

$$\frac{ g_{1}}{g_{3}}=\frac{m'}{n}-a_{2}\frac{m}{n}- \frac{H}{3b_{2}n}.$$

Therefore we get

$$\frac{g_{1}'}{g_{2}}-\frac{b_{2}'}{b_{2}}\frac{g_{1}}{g_{2}}+ \frac{ g_{1}}{g_{3}}-\frac{m'}{n}+a_{2}\frac{m}{n}=0.$$
(3.32)

For brevity, we denote

$$D=(a_{1}-a_{2})\frac{b_{1}}{b_{2}}-\biggl( \frac{b_{1}}{b_{2}}\biggr)'.$$

By calculation we have

$$\frac{ g_{1}}{g_{3}}=\frac{1}{4} \biggl(\frac{g_{2}}{g_{3}} \biggr)^{2},\qquad \frac{g_{1}}{g_{2}}=\frac{1}{4} \frac{g_{2}}{g_{3}},\qquad \frac{ g_{1}'}{g_{2}}=\frac{1}{2} \biggl( \frac{g_{2}}{g_{3}} \biggr)'+\frac{1}{4}\frac{g_{2}}{g_{3}} \frac{g_{3}'}{g_{3}},$$

and

$$\frac{m'}{n}= \biggl(\frac{m}{n} \biggr)'+ \frac{m}{n}\frac{n'}{n},\qquad \frac{m}{n}=\frac{1}{2} \biggl(\frac{h'}{h}-\frac {b_{2}'}{b_{2}}-a_{1} \biggr),\qquad \frac{n'}{n}=\frac{h'}{h}-\frac {b_{2}'}{b_{2}}-\frac{D'}{D}.$$

From (3.26) we have

$$\frac{g_{2}}{g_{3}}=\frac{h'}{h}-\frac{2}{3}\frac{b_{2}'}{b_{2}}- \frac {1}{3}\frac{D'}{D}-\frac{2}{3}(a_{1}+a_{2}).$$

Substituting these identities into (3.32), we have

\begin{aligned} &\frac{1}{4} \biggl(\frac{h'}{h}- \frac{2}{3}\frac{b_{2}'}{b_{2}}-\frac {1}{3}\frac{D'}{D}- \frac{2}{3}(a_{1}+a_{2}) \biggr) \biggl( \frac{8}{3}\frac{b_{2}'}{b_{2}}-2\frac{h'}{h}+\frac{4}{3} \frac {D'}{D}+\frac{2}{3}(a_{1}+a_{2}) \biggr) \\ &\quad {}+\frac{1}{2} \biggl(\biggl(\frac{h'}{h}\biggr)'- \biggl(\frac{b_{2}'}{b_{2}}\biggr)'-a_{1}' \biggr)+ \frac{1}{2} \biggl(\frac{h'}{h}-\frac{b_{2}'}{b_{2}}- \frac {D'}{D}-a_{2} \biggr) \biggl(\frac{h'}{h}- \frac{b_{2}'}{b_{2}}-a_{1} \biggr) \\ &\quad {}-\frac{1}{2} \biggl(\biggl(\frac{h'}{h}\biggr)'- \frac{2}{3}\biggl(\frac {b_{2}'}{b_{2}}\biggr)'- \frac{1}{3} \biggl(\frac{D'}{D} \biggr)'- \frac {2}{3}(a_{1}+a_{2})' \biggr)=0, \end{aligned}

which leads to

\begin{aligned} &\frac{7}{18}\frac{b_{2}'}{b_{2}}\frac{D'}{D}+\frac{1}{6} \biggl(\frac {D'}{D} \biggr)' -\frac{5}{18}(a_{1}+a_{2}) \frac{D'}{D}-\frac{1}{6}(a_{1}+a_{2}) \frac {h'}{h}+\frac{1}{18} \biggl(\frac{b_{2}'}{b_{2}} \biggr)^{2} \\ &\qquad {}-\frac{1}{18}(a_{1}+a_{2})\frac{b_{2}'}{b_{2}}+ \frac{1}{2}a_{1}\frac{D'}{D} +\frac{1}{2}a_{1}a_{2}- \frac{1}{6}\frac{b_{2}'}{b_{2}}-\frac {1}{2}a_{1}'+ \frac{1}{3}(a_{1}+a_{2})' \\ &\quad =\frac{1}{9}(a_{1}+a_{2})^{2}. \end{aligned}

Since

$$\lim_{z\rightarrow\infty}\frac{R'(z)}{R(z)}=0$$

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

$$\lim_{z\rightarrow\infty}\frac{a_{1}a_{2}}{(a_{1}+a_{2})^{2}}=\frac {2}{9}+\lim _{z\rightarrow\infty}\frac{1}{3}\frac{h'}{h}\frac{1}{(a_{1}+a_{2})}.$$

It follows by Lemma 2.4 that

$$\lim_{z\rightarrow\infty}\frac{a_{1}a_{2}}{(a_{1}+a_{2})^{2}}=\lim_{z\rightarrow\infty} \frac{\alpha'\beta'}{(\alpha'+\beta')^{2}}=\frac{2}{9}.$$

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

$$\lim_{z\rightarrow\infty}\frac{\alpha'\beta'}{(\alpha'+\beta ')^{2}}=\frac{a_{m}b_{m}}{(a_{m}+b_{m})^{2}}= \frac{2}{9},$$

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

$$sf^{2}+tff'-q=A(z)e^{2b_{m}z^{m}}$$

and

$$mf^{2}+nff'-q=B(z)e^{b_{m}z^{m}},$$

where

$$A(z)=p_{1}e^{a_{m-1}z^{m-1}+\cdots+a_{0}} \quad \mbox{and} \quad B(z)=p_{2}e^{b_{m-1}z^{m-1}+\cdots+b_{0}}.$$

Then

$$mf^{2}+nff'=q+B \biggl(\frac{sf^{2}+tff'-q}{A} \biggr)^{\frac{1}{2}}.$$

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

\begin{aligned}T(r,F_{2})&=T\bigl(r, mf+nf' \bigr) \\ &=m \biggl(r,\frac{q}{f}+B \biggl(\frac{sf^{2}+tff'-q}{Af^{2}} \biggr)^{\frac{1}{2}} \biggr)+S(r,f) \\ &\leq m\biggl(r,\frac{q}{f}\biggr)+m(r,B)+\frac{1}{2}m \biggl(r, \frac {sf^{2}+tff'-q}{Af^{2}} \biggr)+S(r,f) \\ &\leq m\biggl(r,\frac{s}{A}\biggr)+m\biggl(r,\frac{t}{A}\biggr)+m \biggl(r,\frac{f'}{f}\biggr)+m\biggl(r,\frac {q}{Af^{2}}\biggr)+S(r,f) \\ &=S(r,f). \end{aligned}

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

$$a_{1}\frac{b_{1}}{b_{2}}-\biggl(\frac{b_{1}}{b_{2}} \biggr)'-a_{2}\frac {b_{1}}{b_{2}}\equiv0,$$

which leads to

$$\frac{p_{1}}{p_{2}}e^{\alpha-\beta}=C_{3}\frac{b_{1}}{b_{2}},$$

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

$$fF_{1}-\frac{C_{3}b_{1}}{b_{2}}fF_{2}=q-q\frac{C_{3}b_{1}}{b_{2}}$$

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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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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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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