On Picard Value Problem of Some Difference Polynomials

TL;DR

This paper investigates the distribution of zeros in specific nonlinear difference polynomials derived from entire functions of finite order, contributing to the understanding of their value distribution properties.

Contribution

It introduces new results on the zero distribution of difference polynomials of entire functions of finite order, expanding existing value distribution theory.

Findings

Zeros of certain nonlinear difference polynomials are extensively characterized.

New theorems on the value distribution of these polynomials are established.

Results enhance understanding of the behavior of entire functions under difference operations.

Abstract

In this paper, we study the value distribution of zeros of certain nonlinear difference polynomials of entire functions of finite order.

Equations

$$f\left(z\right)f\left(z+\pi i\right)-1=\left(1+e^z\right)\left(1-e^z\right)-1=-e^{2z}.$$

$$\left\{\begin{array}[]{l}f^{n}\left(z\right)f\left(z+c_{1}\right)-q\left(z\right)=p_{1}\left(z\right)e^{\alpha\left(z\right)}\\ f^{n}\left(z\right)f\left(z+c_{2}\right)-q\left(z\right)=p_{2}\left(z\right)e^{\beta\left(z\right)}\end{array}\right.,$$

$$f^{n}P\left(z,f\right)=Q\left(z,f\right),$$

$$m\left(r,P\left(z,f\right)\right)=o\left(\frac{T\left(r+\left|c\right|,f\right)}{r^{\delta }}\right)+o\left(T\left(r,f\right)\right)+O\left(\stackrel{s}{\sum _{j=1}}m\left(r,a_j\right)\right).$$

$$T\left(r,f\left(z+c\right)\right)=T\left(r,f\left(z\right)\right)+S\left(r,f\right).$$

$$\left|\frac{f^{\left(k\right)}\left(z\right)}{f^{\left(j\right)}\left(z\right)}\right|\le {\left|z\right|}^{\left(k-j\right)\left(\rho -1+\epsilon \right)}.$$

$$m\left(r,\frac{f^{\left(k\right)}}{f}\right)=S\left(r,f\right),$$

$$m\left(r,\frac{f^{\left(k\right)}}{f}\right)=O\left(\log r\right).$$

$$m\left(r,\frac{f\left(z+{\eta }_1\right)}{f\left(z+{\eta }_2\right)}\right)=O\left(r^{\sigma -1+\epsilon }\right).$$

$$\left\{\begin{array}[]{l}f\left(z\right)f\left(z+c_{1}\right)-q\left(z\right)=p_{1}\left(z\right)e^{\alpha\left(z\right)},\\ f\left(z\right)f\left(z+c_{2}\right)-q\left(z\right)=p_{2}\left(z\right)e^{\beta\left(z\right)},\end{array}\right.$$

$$\mathrm{deg}\alpha =\mathrm{deg}\beta =\mathrm{deg}\left(\alpha +\beta \right)=\rho \left(f\right)>0.$$

$$f\left(z\right)f\left(z+c_1\right):=F=q\left(z\right)+p_1\left(z\right)e^{\alpha \left(z\right)}\in S\left(f\right).$$

$$f^{2}P\left(z,f\right)=p_1{p}_2{e}^{\alpha +\beta }+q^2,$$

$$P\left(z,f\right)=a\left(z\right)f^2-b\left(z\right)$$

$$a=\frac{f_{c_1}}{f}\frac{f_{c_2}}{f},b=q\left(\frac{f_{c_1}}{f}+\frac{f_{c_2}}{f}\right).$$

$$m\left(r,P\left(z,f\right)\right)=S\left(r,f\right)$$

$$2T\left(r,f\right)=m\left(r,\frac{b\left(z\right)+P\left(z,f\right)}{a\left(z\right)}\right)=S\left(r,f\right)$$

$$f\left(z\right)f\left(z+c_1\right)-q\left(z\right)=p_1\left(z\right)e^{\alpha \left(z\right)}$$

$$f\left(z\right)f\left(z+c_2\right)-q\left(z\right)=p_2\left(z\right)e^{\beta \left(z\right)}.$$

$$A_{1}f{f}_{c_1}-f^{\prime }{f}_{c_1}-f{f}_{c_1}^{\prime }=B_1,$$

$$\mathrm{deg}\alpha =\mathrm{deg}\beta =\mathrm{deg}\left(\alpha +\beta \right)=\rho \left(f\right)>0.$$

$$A_{2}f{f}_{c_2}-f^{\prime }{f}_{c_2}-f{f}_{c_2}^{\prime }=B_2,$$

$$2m\left(r,\frac{1}{f}\right)\le m\left(r,\frac{f_{c_i}}{f}\right)+m\left(r,\frac{f^{\prime }}{f}\frac{f_{c_i}}{f}\right)+m\left(r,\frac{f_{c_i}^{\prime }}{f_{c_i}}\frac{f_{c_i}}{f}\right)+O\left(\log r\right)$$

$$=O\left(r^{\rho -1+\epsilon }\right)+O\left(\log r\right)=S\left(r,f\right).$$

$$T\left(r,f\right)=N\left(r,\frac{1}{f}\right)+O\left(r^{\rho -1+\epsilon }\right)+O\left(\log r\right).$$

$$N_{(2}\left(r,\frac{1}{f}\right)\le N\left(r,\frac{1}{B_i}\right)=O\left(\log r\right),$$

$$T\left(r,f\right)=N_{1)}\left(r,\frac{1}{f}\right)+O\left(r^{\rho -1+\epsilon }\right)+O\left(\log r\right),$$

$$\left(f^{\prime }{f}_{c_1}+B_1\right)\left(z_0\right)=0$$

$$\left(f^{\prime }{f}_{c_2}+B_2\right)\left(z_0\right)=0.$$

$$\left(B_2{f}_{c_1}-B_1{f}_{c_2}\right)\left(z_0\right)=0$$

Full text

On Picard Value Problem of Some Difference Polynomials

Zinelâabidine **LATREUCH and Benharrat BELAÏDI111Corresponding author **

**Department of Mathematics **

**Laboratory of Pure and Applied Mathematics **

**University of Mostaganem (UMAB) **

B. P. 227 Mostaganem-(Algeria)

z.latreuch@gmail.com

benharrat.belaidi@univ-mosta.dz

**Abstract. **In this paper, we study the value distribution of zeros of certain nonlinear difference polynomials of entire functions of finite order.

2010 Mathematics Subject Classification:30D35, 39A05.

Key words: Entire functions, Non-linear difference polynomials, Nevanlinna theory, Small function.

1 Introduction and Results

Throughout this paper, we assume that the reader is familiar with the fundamental results and the standard notations of the Nevanlinna’s value distribution theory $($ $\left[10\right],$ $\left[13\right])$. In addition, we will use $\rho\left(f\right)$ to denote the order of growth of $f$, we say that a meromorphic function $a\left(z\right)$ is a small function of $f\left(z\right)$ if $T\left(r,a\right)=S\left(r,f\right),$ where $S\left(r,f\right)=o\left(T\left(r,f\right)\right),$ as $r\rightarrow+\infty$ outside of a possible exceptional set of finite logarithmic measure, we use $S\left(f\right)$ to denote the family of all small functions with respect to $f\left(z\right)$. For a meromorphic function $f\left(z\right),$ we define its shift by $f_{c}\left(z\right)=f\left(z+c\right).$

In 1959, Hayman proved in $\left[11\right]$ that if $f$ is a transcendental entire function, then $f^{n}f^{\prime}$ assume every nonzero complex number infinitely many times, provided that $n\geq 3.$ Later, Hayman $\left[12\right]$ conjectured that this result remains to be valid when $n=1$ and $n=2$. Then Mues $\left[18\right]$ confirmed the case when $n=2$ and Bergweiler-Eremenko $\left[2\right]$ and Chen-Fang $\left[3\right]$ confirmed the case when $n=1$, independently. Since then, there are many research publications $\left(\text{see }\left[17\right]\right)$ regarding this type of Picard-value problem. In 1997, Bergweiler obtained the following result.

**Theorem A. $\left(\left[1\right]\right)$ **If $f$ is a transcendental meromorphic function of finite order and $q$ is a not identically zero polynomial, then $ff^{\prime}-$ $q$ has infinitely many zeros.

In 2007, Laine and Yang studied the difference analogue of Hayman’s theorem and proved the following result.

**Theorem B. $\left(\left[14\right]\right)$ **Let $f\left(z\right)$ be a transcendental entire function of finite order, and $c$ be a nonzero complex constant. Then for $n\geq 2,$ $f^{n}\left(z\right)f\left(z+c\right)$ assume every non-zero value $a\in\mathbb{C}$ infinitely often.

In the same paper, Laine and Yang showed that Theorem B does not remain valid for the case $n=1.$ Indeed, take $f\left(z\right)=e^{z}+1.$ Then

[TABLE]

After their, a stream of studies on the value distribution of nonlinear difference polynomials in $f$ has been launched and many related results have been obtained, see e.g. $\left[5,14,15,16\right].$ For example, Liu and Yang improved the previous result and obtained the following.

**Theorem C. $\left(\left[15\right]\right)$ **Let $f\left(z\right)$ be a transcendental entire function of finite order, and $c$ be a nonzero complex constant. Then for $n\geq 2,$ $f^{n}\left(z\right)f\left(z+c\right)-p\left(z\right)$ has infinitely many zeros, where $p\left(z\right)\not\equiv 0$ is a polynomial in $z.$

Hence, it is natural to ask: *What can be said about the value distribution of $f\left(z\right)f\left(z+c\right)-q\left(z\right),$ when $f$ is a transcendental meromorphic function and $q$ be a not identically zero small function of $f$?  *In this paper, as an attempt in resolving this question, we obtain the following results.

Theorem 1.1 Let $f$ be a transcendental entire function of finite order, let $c_{1},$ $c_{2}$ be two nonzero complex numbers such that $f\left(z+c_{1}\right)\not\equiv f\left(z+c_{2}\right)$ and $q$ be not identically zero polynomial. Then $f\left(z\right)f\left(z+c_{1}\right)-q\left(z\right)$ and $f\left(z\right)f\left(z+c_{2}\right)-q\left(z\right)$ at least one of them has infinitely many zeros.

The following corollary arises directly from Theorem 1.1 and Theorem C.

**Corollary 1.1 **Let $n\geq 1$ be an integer and let $c_{1},$ $c_{2}$ $\left(c_{1}c_{2}\neq 0\right)$ be two distinct complex numbers. Let $\alpha,\beta,$ $p_{1},$ $p_{2}$ and $q$ $\left(\not\equiv 0\right)$ be nonconstant polynomials. If $f$ is a finite order transcendental entire solution of

[TABLE]

then, $n=1$ and $f$ must be a periodic function of period $c_{1}-c_{2}.$

2 Some lemmas

The following lemma is an extension of the difference analogue of the Clunie lemma obtained by Halburd and Korhonen $\left[8\right]$.

Lemma 2.1 $\left[4\right]$ Let $f\left(z\right)$ be a non-constant, finite order meromorphic solution of

[TABLE]

where $P\left(z,f\right),$ $Q\left(z,f\right)$ are difference polynomials in $f\left(z\right)$ with meromorphic coefficients $a_{j}\left(z\right)$ $\left(j=1,\cdots,s\right),$ and let $\delta<1.$ If the degree of $Q\left(z,f\right)$ as a polynomial in $f\left(z\right)$ and its shifts is at most $n,$ then

[TABLE]

for all $r$ outside an exceptional set of finite logarithmic measure.

Lemma 2.2 $\left[6\right]$ Let $f\left(z\right)$ be a non-constant, finite order meromorphic function and let $c\neq 0$ be an arbitrary complex number. Then

[TABLE]

Lemma 2.3* $\left[7\right]$* Let $f\left(z\right)$ be a transcendental meromorphic function of finite order $\rho,$ and let $\varepsilon>0$ be a given constant. Then, there exists a set $E_{0}\subset\left(1,+\infty\right)$ that has finite logarithmic measure, such that for all $z$ satisfying $\left|z\right|\notin E_{0}\cup\left[0,1\right],$ and for all $k,j,$ $0\leq j<k,$ we have

[TABLE]

The following lemma is the lemma of the logarithmic derivative.

Lemma 2.4 $\left[10\right]$ Let $f$ be a meromorphic function and let $k\in\mathbb{N}.$ Then

[TABLE]

where $S\left(r,f\right)=O\left(\log T\left(r,f\right)+\log r\right),$ possibly outside a set $E_{1}\subset\left[0,+\infty\right)$ of a finite linear measure. If $f$ is of finite order of growth, then

[TABLE]

The following lemma is a difference analogue of the lemma of the logarithmic derivative for finite order meromorphic functions.

Lemma 2.5 $\left[6,8,9\right]$ Let $\eta_{1},\eta_{2}$ be two arbitrary complex numbers such that $\eta_{1}\neq\eta_{2}$ and let $f\left(z\right)$ be a finite order meromorphic function. Let $\sigma$ be the order of $f\left(z\right)$. Then for each $\varepsilon>0,$ we have

[TABLE]

Lemma 2.6* Let $f\left(z\right)$ be a transcendental meromorphic solution of the system*

[TABLE]

where $\alpha,\beta$ are polynomials and $p_{1},$ $p_{2},$ $q$ are not identically zero rational functions. If $N\left(r,f\right)=S\left(r,f\right),$ then

[TABLE]

Proof**. First, **we prove that $\deg\alpha=\rho\left(f\right)$ and by the same we can deduce that $\deg\beta=\rho\left(f\right).$ It’s clear from $\left(2.1\right)$ that $\deg\alpha\leq\rho\left(f\right).$ Suppose that $\deg\alpha<\rho\left(f\right),$ this means that

[TABLE]

Applying Lemma 2.1 and Lemma 2.2 into $\left(2.2\right),$ we obtain  $T\left(r,f_{c}\right)=T\left(r,f\right)=S\left(r,f\right)$ which is a contradiction. Assume now that $\deg\left(\alpha+\beta\right)<\rho\left(f\right),$ this leads to $p_{1}p_{2}e^{\alpha+\beta}\in S\left(f\right).$ From this and $\left(2.1\right)$ we have

[TABLE]

where

[TABLE]

and

[TABLE]

It’s clear that $P\left(z,f\right)\not\equiv 0,$ and by using Lemma 2.1, we get

[TABLE]

which leads to

[TABLE]

which is a contradiction. Hence, $\deg\left(\alpha+\beta\right)=\deg\alpha=\deg\beta.$ Finally, by using Lemma 2.1, it’s easy to see that both of $\alpha$ and $\beta$ are nonconstant polynomials.

3 Proof of Theorem 1.1

We shall prove this theorem by contradiction. Suppose contrary to our assertion that both of $f\left(z\right)f\left(z+c_{1}\right)-q\left(z\right)$ and $f\left(z\right)f\left(z+c_{2}\right)-q\left(z\right)$ have finitely many zeros. Then, there exist four polynomials $\alpha$, $\beta,$ $p_{1}$ and $p_{2}$ such that

[TABLE]

and

[TABLE]

By differentiating $\left(3.1\right)$ and eliminating $e^{\alpha},$ we get

[TABLE]

where $A_{1}=\frac{p_{1}^{\prime}}{p_{1}}+\alpha^{\prime},$ $B_{1}=\left(\frac{p_{1}^{\prime}}{p_{1}}+\alpha^{\prime}\right)q-q^{\prime}.$ By Lemma 2.6 we have

[TABLE]

Now, we prove that $A_{1}\not\equiv 0.$ To show this, we suppose the contrary. Then, there exists a constant $A$ such that $A=p_{1}\left(z\right)e^{\alpha},$ which implies the contradiction $\deg\alpha=\rho\left(f\right)=0.$ By the same, we can prove that $B_{1}\not\equiv 0.$ By the same arguments as above, $\left(3.2\right)$ gives

[TABLE]

where $A_{2}=\frac{p_{2}^{\prime}}{p_{2}}+\beta^{\prime}$ and $B_{2}=\left(\frac{p_{2}^{\prime}}{p_{2}}+\beta^{\prime}\right)q-q^{\prime}.$ Obviously, $A_{2}\not\equiv 0$ and $B_{2}\not\equiv 0.$ Dividing both sides of $\left(3.3\right)$ and $\left(3.4\right)$ by $f^{2},$ we get for each $\varepsilon>0$

[TABLE]

[TABLE]

So, by the first fundamental theorem, we deduce that

[TABLE]

It’s clear from $\left(3.3\right)$ and $\left(3.4\right)$ that any multiple zero of $f$ is a zero of $B_{i}$ $\left(i=1,2\right).$ Hence

[TABLE]

where $N_{(2}\left(r,\frac{1}{f}\right)$ denotes the counting function of zeros of $f$ whose multiplicities are not less than 2. It follows by this and $\left(3.5\right)$ that

[TABLE]

where $N_{1)}\left(r,\frac{1}{f}\right)$ is the counting function of zeros, where only the simple zeros are considered. From $\left(3.3\right)$ and $\left(3.4\right),$ for every zero $z_{0}$ such that $f^{\prime}\left(z_{0}\right)\neq 0$ which is not zero or pole of $B_{1}$ and $B_{2},$ we have

[TABLE]

and

[TABLE]

By $\left(3.7\right)$ and $\left(3.8\right),$ we obtain

[TABLE]

which means that the function $\frac{B_{2}f_{c_{1}}-B_{1}f_{c_{2}}}{f}$ has at most a finite number of simple poles. We consider two cases:

Case 1. $B_{2}f_{c_{1}}-B_{1}f_{c_{2}}\not\equiv 0.$ Set

[TABLE]

Then, from the lemma of logarithmic differences, we have $m\left(r,h\right)=O\left(r^{\rho-1+\varepsilon}\right)+O\left(\log r\right).$ On the other hand

[TABLE]

[TABLE]

Thus, $T\left(r,h\right)=O\left(r^{\rho-1+\varepsilon}\right)+O\left(\log r\right)=S\left(r,f\right).$ From the equation $\left(3.10\right),$ we have

[TABLE]

By differentiating $\left(3.11\right),$ we get

[TABLE]

Substituting $\left(3.11\right)$ and $\left(3.12\right)$ into $\left(3.3\right)$

[TABLE]

[TABLE]

Equation $\left(3.4\right),$ can be rewritten as

[TABLE]

By adding this to $\left(3.13\right)$, we get

[TABLE]

Its clear that $-\frac{2h}{B_{2}}\not\equiv 0$. In order to complete the proof of our theorem, we need to prove

[TABLE]

Suppose contrary to our assertion that $\frac{A_{1}h}{B_{2}}-\left(\frac{h}{B_{2}}\right)^{\prime}\equiv 0.$ Then, by the definition of $A_{1}$ and by simple integration, we get

[TABLE]

where $C_{1}$ is a nonzero constant. This implies that $\deg\alpha=\rho\left(f\right)-1,$ which is a contradiction. Hence, $\frac{A_{1}h}{B_{2}}-\left(\frac{h}{B_{2}}\right)^{\prime}\not\equiv 0.$ Next, we shall prove $\frac{A_{1}B_{1}}{B_{2}}-\left(\frac{B_{1}}{B_{2}}\right)^{\prime}-\frac{B_{1}A_{2}}{B_{2}}\not\equiv 0.$ Suppose that $\frac{A_{1}B_{1}}{B_{2}}-\left(\frac{B_{1}}{B_{2}}\right)^{\prime}-\frac{B_{1}A_{2}}{B_{2}}\equiv 0.$ Then we obtain

[TABLE]

where $C_{2}$ is a nonzero constant and $\gamma$ is a small function of $f.$ From $\left(3.1\right)$ and $\left(3.2\right)$ we get

[TABLE]

If $\gamma\not\equiv 1,$ then by applying Clunie’s lemma to $\left(3.15\right),$ we obtain

[TABLE]

By this and $\left(3.15\right),$ we have

[TABLE]

which is a contradiction. If $\gamma\equiv 1,$ then we obtain the contradiction $f_{c_{1}}\left(z\right)\equiv f_{c_{2}}\left(z\right).$ Thus, $\frac{A_{1}B_{1}}{B_{2}}-\left(\frac{B_{1}}{B_{2}}\right)^{\prime}-\frac{B_{1}A_{2}}{B_{2}}\not\equiv 0.$ From the above discussion and $\left(3.14\right),$ we have

[TABLE]

and

[TABLE]

where

[TABLE]

and

[TABLE]

Differentiation of $\left(3.16\right),$ gives

[TABLE]

Substituting $\left(3.16\right)$ and $\left(3.18\right)$ into $\left(3.4\right),$ we get

[TABLE]

Differentiating $\left(3.19\right),$ we get

[TABLE]

[TABLE]

Suppose $z_{0}$ is a simple zero of $f$ and not a zero or pole of $B_{2}.$ Then from $\left(3.19\right)$ and $\left(3.20\right),$ we have

[TABLE]

[TABLE]

It follows that $z_{0}$ is a zero of $\left[B_{2}\left(2N^{\prime}-A_{2}N+2M\right)-B_{2}^{\prime}N\right]f^{\prime}+3B_{2}Nf^{\prime\prime}.$ Therefore the function

[TABLE]

satisfies $T\left(r,H\right)=S\left(r,f\right)$ and

[TABLE]

Substituting $\left(3.21\right)$ into $\left(3.19\right),$ we get

[TABLE]

where

[TABLE]

[TABLE]

We prove first $q_{2}\not\equiv 0.$ Suppose the contrary. Then

[TABLE]

which leads to

[TABLE]

By simple integration of both sides of the above equation, we get

[TABLE]

where $c$ is a nonzero constant, this leads to the contradiction $\deg\left(\alpha+\beta\right)<\deg\alpha=\deg\beta.$ Hence, $q_{2}\not\equiv 0.$ Differentiating $\left(3.22\right),$ we obtain

[TABLE]

Let $z_{0}$ be a simple zero of $f$ which is not a zero or pole of $B_{2}.$ Then from $\left(3.22\right)$ and $\left(3.24\right)$ we have

[TABLE]

[TABLE]

Therefore $z_{0}$ is a zero of $\left(B_{2}\left(q_{2}+q_{3}^{\prime}\right)-B_{2}^{\prime}q_{3}\right)f^{\prime}+2B_{2}q_{3}f^{\prime\prime}.$ Hence the function

[TABLE]

satisfies $T\left(r,R\right)=S\left(r,f\right)$ and

[TABLE]

Substituting $\left(3.25\right)$ into $\left(3.24\right)$

[TABLE]

[TABLE]

Combining $\left(3.26\right)$ and $\left(3.22\right),$ we obtain

[TABLE]

From $\left(3.27\right),$ we deduce that

[TABLE]

and

[TABLE]

By eliminating $R$ from the above two equations, we obtain

[TABLE]

Thus, equation $\left(3.25\right)$ can be rewritten as

[TABLE]

Subcase 1. If $4q_{1}q_{3}-q_{2}^{2}\not\equiv 0,$ then from $\left(3.28\right)$ we have

[TABLE]

On the other hand

[TABLE]

Hence

[TABLE]

By the definition of $A_{i}$ $\left(i=1,2\right)$ and simple integration, we deduce that

[TABLE]

which is a contradiction.

Subcase 2. If $4q_{1}q_{3}\equiv q_{2}^{2},$ then from $\left(3.29\right)$ and $\left(3.21\right)$ we have

[TABLE]

On the other hand

[TABLE]

Combining $\left(3.30\right)$ and $\left(3.31\right),$ we obtain

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Dividing both sides of the above equation by $\frac{\left(A_{1}+A_{2}\right)^{2}}{2}$ and since $\underset{z\rightarrow\infty}{\lim}\frac{R^{\prime}\left(z\right)}{R\left(z\right)}=0$ if $R$ is a nonzero rational function, we obtain

[TABLE]

On the other hand, since $\rho\left(h\right)\leq\rho\left(f\right)-1$ and by Lemma 2.3

[TABLE]

for all $z$ satisfying $\left|z\right|\notin E_{0}\cup\left[0,1\right],$ where $E_{0}\subset\left(1,\infty\right)$ is a set of finite logarithmic measure. By combining $\left(3.32\right)$ and $\left(3.33\right),$ we deduce

[TABLE]

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

[TABLE]

which implies that $\frac{a_{m}}{b_{m}}=2$ or $\frac{1}{2}.$ We consider first the case $\frac{a_{m}}{b_{m}}=\frac{1}{2},$ we get from $\left(3.1\right)$ and $\left(3.17\right)$

[TABLE]

and

[TABLE]

where $A=p_{1}e^{a_{m-1}z^{m-1}+\cdots+a_{0}}$ and $B=p_{2}e^{b_{m-1}z^{m-1}+\cdots+b_{0}}.$ From $\left(3.34\right)$ and $\left(3.35\right),$ we get

[TABLE]

Hence

[TABLE]

Therefore

[TABLE]

[TABLE]

[TABLE]

where $E_{1}=\left\{\theta:\left|f\left(re^{i\theta}\right)\right|\leq 1\right\}$ and $E_{2}=\left\{\theta:\left|f\left(re^{i\theta}\right)\right|>1\right\}.$ Now

[TABLE]

[TABLE]

[TABLE]

On the other hand

[TABLE]

[TABLE]

[TABLE]

Hence

[TABLE]

which is a contradiction. If $\frac{a_{m}}{b_{m}}=2,$ then by the same argument we have

[TABLE]

which implies the contradiction

[TABLE]

Case 2. $B_{2}f_{c_{1}}-B_{1}f_{c_{2}}\equiv 0,$ by using the same arguments as in the proof of $\left(3.14\right),$ we obtain that

[TABLE]

which leads to

[TABLE]

where $k$ is a nonzero complex constant. By this $\left(3.1\right)$ and $\left(3.2\right),$ we have

[TABLE]

If $k\neq 1,$ then by applying Clunie lemma to $\left(3.37\right),$ we deduce the contradiction $T\left(r,f_{c_{i}}\right)=S\left(r,f\right).$ Hence, $k=1$ and from the equation $\left(3.36\right),$ we conclude that $f_{c_{1}}\equiv f_{c_{2}}$ which exclude the hypothesis of our theorem. This shows that at least one of $f\left(z\right)f\left(z+c_{1}\right)-q\left(z\right)$ and $f\left(z\right)f\left(z+c_{2}\right)-q\left(z\right)$ has infinitely many zeros.

Acknowledgements. The authors are grateful to the anonymous referee for his/her valuable comments and suggestions which lead to the improvement of this paper.

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