On meromorphic solutions of certain differential-difference equations

1.   Introduction and main results

We assume that the reader is familiar with the basic notions of Nevanlinna theory (see ^[4,6,22]). Recently, a number of papers (including ^{[1,2,3,5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,23]}) have focused on solvability and existence of meromorphic solutions of difference equations or differential-difference equations in complex plane. In 2009, Liu ^[10] obtianed the Fermat type equation $l(z)^{2}+[l(z+c)-l(z)]^{2} = 1$ has a nonconstant entire solution of finite order. In 2012, Liu et al. ^[11] proved that $l(z)^{2}+l(z+c)^{2} = 1$ has a transcendental entire solution of finite order. In 2018, Zhang ^[23] obtained the difference equations $l(z)^{2}+[l(z+c)-l(z)]^{2} = R(z)$ has no finite order transcendental meromorphic solutions with finitely many poles. In 2020, Wang et al. ^[18] further discussed the existence and the forms of the solutions for some differential-difference equations, they obtained

Theorem A. Let $c$ be a nonzero constant, $R(z)$ be a nonzero rational function, and $\alpha, \beta\in\mathbb{C}$ satisfy $\alpha^{2}-\beta^{2}\neq1$. Then the following difference equation of Fermat-type

has no finite order transcendental meromorphic solutions with finitely many poles.

Theorem B. Let $c(\neq0)$, $\alpha(\neq0)$, $\beta\in\mathbb{C}$, and $P(z)$, $Q(z)$ be nonzero polynomials satisfying one of two following cases:

$(i)$ $deg_{z}P(z)\geq1$, $deg_{z}Q(z)\geq1$;

$(ii)$ $P(z)$, $Q(z)$ are two constants and $P^{2}(\alpha^{2}-\beta^{2}) \ne 1$. Then the following Fermat-type difference equation

has no transcendental entire solutions with finite order.

For further study, we continue to discuss the existence and the forms of solutions for certain differential-difference equations with more general forms than the previous forms by Liu et al. ^{[10,11,18,23]} and obtain the following results.

Theorem 1.1. Let $c_{j}(j = 1, 2, \cdots, m)$ be distinct constants, $a\in\mathbb{C}\backslash\{0\}$, $\varrho_{i}\in\mathbb{C}\; (i = 1, 2, \cdots, m)$, $R(z)$ be a nonzero rational function, and $\sum\limits_{i = 1}^m{{\varrho_i}} ({exp^{a{c_{{i}}}}}+{exp^{{{-}}a{c_{{i}}}}}){{\neq0}}$. Then the following difference equation

has no finite order transcendental meromorphic solutions with finitely many poles.

Theorem 1.2. Let $c_{j}(j = 1, 2, \cdots, m)$ be distinct constants, $a\in\mathbb{C}\backslash\{0\}$, $\varrho_{i}\in\mathbb{C}\; (i = 1, 2, \cdots, m)$, and $P(z)$, $Q(z)$ be nonzero polynomials satisfying one of two following cases:

$(i)$ $deg_{z}P(z)\geq1$;

$(ii)$ $P$ is a constant and ${P^2}[\sum\limits_{i{{ = }}1}^m{{\varrho _i}}{exp^{a{c_{{i}}}}}\sum\limits_{i{{ = }}1}^m{{\varrho_i}}{exp ^{{{-}}a{c_{{i}}}}}] \ne 1$. Then the following difference equation

has no transcendental entire solutions with finite order.

Theorem 1.3. Let $c_{j}(j = 1, 2, \cdots, m)$ be distinct constants, $a\in\mathbb{C}\backslash\{0\}$, $\varrho_{i}\in\mathbb{C}\; (i = 1, 2, \cdots, m)$. Let ${l(z)}$ be a transcendental finite order meromorphic solution of difference-differential equation

where $R(z)$ is a nonzero rational function. If $l(z)$ has finitely many poles, and $\sum\limits_{j = 1}^m {{c}_j}^2 {\varrho _j}{exp ^{a{c_j}}}\sum\limits_{j = 1}^m {{c}_j}^2{\varrho _j}{exp ^{{\rm{ - }}a{c_j}}} \neq0$, then $R(z)$ is a nonconstant polynomial with $deg_{z}R(z)\leqslant2$, and $\sum\limits_{j = 1}^m {{c_j}} {\varrho _j}{exp ^{a{c_j}}} \sum\limits_{j = 1}^m {{c_j}} {\varrho _j}{exp ^{ - a{c_j}}} = 1$. Furthermore,

$(i)$ If $R(z)$ is a nonconstant polynomial with $deg_{z}R(z)\leqslant2$, and $\sum\limits_{i = 1}^m{{\varrho_i}}\ne0$, then we have

where $R(z) = (m_{1}+as_{1}(z))(m_{2}-as_{2}(z)), a\neq0, b\in\mathbb{C}$ and $a, b, c_{j}, \varrho_{i}$ satisfy $i({\varrho_1}{exp^{a{c_1}}}+\cdots+ {\varrho_m}{exp^{a{c_m}}}) = a$ and $i({\varrho_1}{exp^{-a{c_1}}}+\cdots+{\varrho_m}{exp^{-a{c_m}}}) = a$, where $s_{j}(z) = m_{j}z+n_{j}, m_{j}, n_{j}\in\mathbb{C}(j = 1, 2)$.

$(ii)$ If $R(z)$ is a nonzero constant, and $\sum\limits_{i = 1}^m{{\varrho_i}}\ne0$, then

where $R(z) = -a^{2}n_{1}n_{2}, a\neq0, b\in\mathbb{C}$.

Theorem 1.4. Let $c_{j}(j = 1, 2, \cdots, m)$ be distinct constants, $a\in\mathbb{C}\backslash\{0\}$, $\varrho_{i}\in\mathbb{C}\; (i = 1, 2, \cdots, m)$. Let ${l(z)}$ be a transcendental meromorphic solution of the following difference-differential equation

where $R(z)$ is a nonzero rational function.

$(i)$ If $\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{a{c_{{i}}}}}+\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{{{-}}a{c_{{i}}}}}\neq0$, then (1.4) has no finite order transcendental meromorphic solution with finitely many poles.

$(ii)$ If $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{a{c_j}}} \neq 2a$, $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{ - a{c_j}}} \neq 2a$, and (1.4) has a finite order transcendental meromorphic solution $l(z)$ with finitely many poles, then $R(z)$ is a constant. Furthermore if $\sum\limits_{i = 1}^m{{\varrho_i}}\neq 0$, then we have

where $a, b, t_{1}, t_{2}, \varrho_{i}, c_{j}$ satisfy $\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{a{c_{{i}}}}}+\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{{{-}}a{c_{{i}}}}} = 0$, $R(z) = a^{4}t_{1}t_{2}$, $b\in\mathbb{C}$.

2.   Preliminary lemmas

The following two lemmas play an important role in the proof of our results.

Lemma 2.1. (^[22]) Suppose that $f_{1}, f_{2}, \cdots, f_{n} (n\geq2)$ are meromorphic functions and $g_{1}, g_{2}, \cdots, g_{n}$ are entire functions satisfying the following conditions:

$(i)$ $\sum\limits_{j = 1}^n {{f_j}} {exp ^ {{g_j}}}\equiv0;$

$(ii)$ $g_{j}-g_{k}$ are not constants for $1\leq j < k \leq n;$

$(iii)$ For $1\leq j\leq n, 1\leq h < k \leq n$, $T(r, {f_j}) = o\{T(r, {exp ^{{g_h} - {g_k}}})\} (r \to \infty, r \notin E)$, where $E$ is a set of $r\in (0, \infty)$ with finite linear measure.

Then $f_{j}\equiv0 (j = 1, 2, \cdots, m)$.

Lemma 2.2. (^[22]) Let $l(z)$ be a meromorphic function of finite order $\rho(l)$. Write

near $z = 0$ and let $\{a_{1}, a_{2}, \cdots\}$ and $\{b_{1}, b_{2}, \cdots\}$ be the zeros and poles of $l$ in $\mathbb{C}\setminus \{0\}$, respectively. Then

where $P_{1}(z)$ and $P_{2}(z)$ are the canonical products of $l$ formed with the non-null zeros and poles of $l$, respectively, and $Q(z)$ is a polynomial of degree $\leqslant\rho(l)$.

3.   Proof of Theorem 1.1

Suppose that (1.1) has a finite order transcendental meromorphic solution ${l}(z)$ with finitely many poles. Rewriting (1.1) as follows

Since ${l}(z)$ has finitely many poles, $R(z)$ is a nonzero rational function, then ${l}(z)+i(\varrho_{1}{l}(z+c_{1})+\varrho_{2}{l}(z+c_{2})+\cdots +\varrho_{m}{l}(z+c_{m}))$ and ${l}(z)-i(\varrho_{1}{l}(z+c_{1})+\varrho_{2}{l}(z+c_{2})+\cdots +\varrho_{m}{l}(z+c_{m}))$ both have finitely many poles and zeros. Together Lemma 2.2 with (3.1), we obtain that

and

where $R_{1}(z), R_{2}(z)$ are two nonzero rational functions such that $R_{1}(z)R_{2}(z) = R(z)$, and $p(z)$ is a nonconstant polynomial. (3.2) and (3.3) imply that

and

Substituting (3.4) into (3.5), we have

By Lemma 2.1 and (3.6), we have

and

Since $R_{1}(z), R_{2}(z)$ are two nonzero rational functions and that ${l(z)}$ is of finite order, we obtain that $p(z)$ is a polynomial of degree one. If $deg_{z}p(z)\geq2$, then we obtain that $deg_{z}[p(z+c_{j})-p(z+c_{i})]\geq1$. Hence, we have $T(r, i\varrho_{j}R_{j}(z+c_{j})) = S(r, ex{p^{ p(z + {c_i}) - p(z+c_j)}})$, Lemma 2.1 and (3.7) imply that $R_{1}(z)\equiv0$. This is impossible. By the similar method as above, we also have $R_{2}(z)\equiv0$, a contradiction. So we have $deg_{z}p(z) = 1$. Set $p(z) = az+b, a\neq0, b\in\mathbb{C}$. By (3.7) and (3.8), we have

and

Thus, it yields that $\sum\limits_{i = 1}^m{{\varrho_i}} ({exp^{a{c_{{i}}}}}+{exp^{{{-}}a{c_{{i}}}}}){\rm{ = 0}}$, this is a contradiction with the assumption of Theorem 1.1. Hence, Theorem 1.1 holds.

4.   Proof of Theorem 1.2

If ${l}(z)$ is a transcendental entire solution with finite order of (1.2), then by the similar method as the proof of Theorem 1.1, we have

and

where $p(z)$ is a nonconstant polynomial and $Q_{1}(z)Q_{2}(z) = Q(z)$, $Q_{1}(z)$, $Q_{2}(z)$ are nonzero polynomials. Together (4.1) with (4.2), we have

By Lemma 2.1 and $p(z)$ is a nonconstant polynomial, we have

and

If $deg_{z}p(z)\geq2$, then we have that $deg_{z}[p(z+c_{j})-p(z+c_{i})]\geq1$. Hence, we have $T(r, i{\varrho _j}P(z){Q_1}(z + {c_j})) = S(r, ex{p^{ p(z + {c_i}) - p(z+c_j)}})$, Lemma 2.1 and (4.4) imply that $Q_{1}(z)\equiv0$. A contradiction. By the similar method as above, we also obtain that $Q_{2}(z)\equiv0$, this is also impossible. Hence, $deg_{z}p(z) = 1$. Let $p(z) = az+b, a\neq0, b\in\mathbb{C}$. (4.4) and (4.5) imply that

and

By this, we have

Set $deg_{z}P(z) = p$ and $deg_{z}Q(z) = q$, then $p\geq0, q\geq0$ and $p, q\in\mathbb{N_{+}}$. Next we divided the following proof into four cases:

Case 1. $p\geq1$ and $\sum\limits_{i{{ = }}1}^m{{\varrho _i}}{exp^{a{c_{{i}}}}}\sum\limits_{i{{ = }}1}^m{{\varrho_i}}{exp ^{{{-}}a{c_{{i}}}}} = 0$. If $q\geq1$, by comparing the order both sides of (4.6), we have $2p+q-1\leqslant q$, that is, $p\leqslant\dfrac{1}{2}$, this is impossible. If $q = 0$, that is, $Q(z)$ is a constant. Hence, by (4.6), we have $Q(z) = 0$, a contradiction.

Case 2. $p\geq1$ and $\sum\limits_{i{{ = }}1}^m{{\varrho _i}}{exp^{a{c_{{i}}}}}\sum\limits_{i{{ = }}1}^m{{\varrho_i}}{exp ^{{{-}}a{c_{{i}}}}}\neq0$. If $q\geq1$, by comparing the order both sides of (4.6), we have $2p+q = q$, that is, $p = 0$, a contradiction. If $q = 0$, that is, $Q(z)$ is a constant. Hence, by (4.6), we have $P(z)$ is a constant, this is impossible.

Case 3. $p = 0$ and $\sum\limits_{i{\rm{ = }}1}^m{{\varrho _i}}{exp^{a{c_{{i}}}}}\sum\limits_{i{\rm{ = }}1}^m{{\varrho_i}}{exp ^{{\rm{-}}a{c_{{i}}}}} = 0$. That is, $P(z) = K(\neq0)$. If $q\geq1$, we have $q-1 = q$, this is impossible. If $q = 0$, we have $Q(z)\equiv0$. A contradiction.

Case 4. $p = 0$ and $\sum\limits_{i{\rm{ = }}1}^m{{\varrho _i}}{exp^{a{c_{{i}}}}}\sum\limits_{i{\rm{ = }}1}^m{{\varrho_i}}{exp ^{{\rm{-}}a{c_{{i}}}}}\neq0$. If $q\geq1$, set $P(z) = K(\neq0)$, $Q({{z}}) = {b_q}{z^q} + {b_{q - 1}}{z^{q - 1}} + \cdots + {b_0}, {b_q} \ne 0, {b_{q - 1}}, \cdots, {b_0}$ are constants. By comparing the coefficients of $z^{q}$ both sides of (4.6), we have

This is a contradiction with the condition of Theorem 1.2. If $q = 0$, then ${K^2}[\sum\limits_{i{\rm{ = }}1}^m{{\varrho _i}}{exp^{a{c_{{i}}}}}\sum\limits_{i{\rm{ = }}1}^m{{\varrho_i}}{exp ^{{\rm{-}}a{c_{{i}}}}}] = 1$, this is impossible.

Hecne, Theorem 1.2 holds.

5.   Proof of Theorem 1.3

Suppose that (1.3) has a finite order transcendental meromorphic solution ${l}(z)$ with finitely many poles. Rewriting (1.3) as follows

Since ${l}(z)$ has finitely many poles, and $R(z)$ is a nonzero rational function, then $l'(z) + i({\varrho _1}l(z + {c_1}) + {\varrho _2}l(z + {c_2}) + \cdots + {\varrho _m}l(z + {c_m}))$ and $l'(z) - i({\varrho _1}l(z + {c_1}) + {\varrho _2}l(z + {c_2}) + \cdots + {\varrho _m}l(z + {c_m}))$ both have finitely many poles and zeros. Hence, by Lemma 2.2, (5.1) can be written as

and

where $R_{1}(z), R_{2}(z)$ are two nonzero rational functions such that $R_{1}(z)R_{2}(z) = R(z)$, and $p(z)$ is a nonconstant polynomial. (5.2) and (5.3) imply that

and

(5.5) implies that

where $A_{1}(z) = R_{1}'+R_{1}(z)p'$ and $B_{1}(z) = R_{2}'-R_{2}(z)p'$. Substituting (5.4) into (5.6), we have

Together Lemma 2.1 with (5.7), we have

and

Since $R_{1}(z), R_{2}(z)$ are two nonzero rational functions and $l(z)$ is of finite order, by the similar method as the proof of Theorem 1.1, we have $deg_{z}p(z) = 1$. Let $p(z) = az+b, a\neq0, b\in\mathbb{C}$. Substituting $p(z)$, $A_{1}(z)$, $B_{1}(z)$ into (5.8) and (5.9), as $z\longrightarrow\infty$, we have

and

That is

According to (5.8), (5.9) and (5.10), we have

and

If $R_{1}(z), R_{2}(z)$ are two nonzero constants, then (5.11) and (5.12) hold and $R_{1}(z)R_{2}(z) = R(z)$ is a constant.

We next consider the case that $R_{1}(z), R_{2}(z)$ are two nonzero rational functions. If $R_1(z)$ has a pole of multiplicity $v$ at $z_0$, by (5.11), we know that there exists at least on index $l_1\in \{1, 2, \cdots, m\}$ such that $z_0+c_{l_1}$ is a pole of $R_1(z)$ of multiplicity $v+1$, following the above step, we know $R_1(z)$ has a sequence of poles

Hence, we have $\lambda(\frac{1}{R_1(z)})\geq 1$, this is impossible. So $R_1(z)$ is a polynomial. Using the same method as above, we know that $R_2(z)$ is also a polynomial. If $R_i(z)$ is a nonconstant polynomial with $deg_{z}R_i(z)\geq2$. Let $R_i({z}) = {a_n}{z^n} + {a_{n - 1}}{z^{n - 1}} +\cdots + {a_0}$, then

where $i = 1, 2$. Substituting (5.13) and (5.14) into (5.11) and (5.12), comparing the coefficients of $z^{n - 1}$, $z^{n- 2}$, we have $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{a{c_j}}} = 1$, $\sum\limits_{j = 1}^m {c_j}^2 {\varrho _j}{exp ^{a{c_j}}} = 0$ and $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{ - a{c_j}}} = {\rm{ - }}1$, $\sum\limits_{j = 1}^m {c_j}^2 {\varrho _j}{exp ^{{\rm{ - }}a{c_j}}} = 0$, a contradiction with $\sum\limits_{j = 1}^m {{c}_j}^2 {\varrho _j}{exp ^{a{c_j}}}\sum\limits_{j = 1}^m {{c}_j}^2{\varrho _j}{exp ^{{\rm{ - }}a{c_j}}} \neq0$. Hence, $deg_{z}R_i(z)\leq 1$. So $deg_{z}R(z) = deg_{z}R_1(z)R_2(z)\leq 2$.

${(i)}$ If $R(z)$ is a nonconstant polynomial with $deg_{z}R(z)\leqslant2$, then by (5.4), we have

where $s_{j}(z) = m_{j}z+n_{j}, m_{j}, n_{j}\in\mathbb{C}, (j = 1, 2)$ and $\vartheta\in\mathbb{C}$;

Case 1. If $deg_{z}R(z) = 2$, then $m_{j}\neq0, j = 1, 2$. If $\sum\limits_{i = 1}^m{{\varrho_i}}\ne0$, substituting (5.15) into (5.5), we have $\vartheta\equiv0$, $R(z) = (m_{1}+as_{1}(z))(m_{2}-as_{2}(z))$. Hence, we have

$R(z) = (m_{1}+as_{1}(z))(m_{2}-as_{2}(z)), a\neq0, b\in\mathbb{C}$.

Case 2. If $deg_{z}R(z) = 1$, then one of $m_{1}, m_{2}$ is zero, we can assume that $m_{1} = 0$. Substituting (5.15) into (5.5), we have $R_{1}(z)$ is a constant and $R_{2}(z)$ is a polynomial of degree one. Using the same method as case 1, we have $\vartheta\equiv0$. Hence, we obtain that

$R(z) = (m_{1}+as_{1}(z))(m_{2}-as_{2}(z)), a\neq0, b\in\mathbb{C}$.

${(ii)}$ If $R(z)$ is a nonzero constant, by (5.4), we have

where $n_{1}, n_{2}\in\mathbb{C}$ and $d\in\mathbb{C}$. Substituting (5.16) into (5.5), we have $d = 0$, $R(z) = -a^{2}n_1 n_2$. Hence, Theorem 1.3 holds.

6.   Proof of Theorem 1.4

Suppose that (1.4) has a finite order transcendental meromorphic solution ${l}(z)$ with finitely many poles. Rewriting (1.4) as follows

Since ${l}(z)$ has finitely many poles, $R(z)$ is a nonzero rational function, then $l''(z) + i({\varrho _1}l(z + {c_1}) + {\varrho _2}l(z + {c_2}) + \cdots+ {\varrho _m}l(z + {c_m}))$ and $l''(z) - i({\varrho _1}l(z + {c_1}) + {\varrho _2}l(z + {c_2}) + \cdots + {\varrho _m}l(z + {c_m}))$ both have finitely many poles and zeros. Hence, we can rewrite (6.1) as follows

and

where $R_{1}(z), R_{2}(z)$ are two nonzero rational functions such that $R_{1}(z)R_{2}(z) = R(z)$, and $p(z)$ is a nonconstant polynomial. By (6.2) and (6.3), we obtain

and

(6.5) implies that

where $A_{2}(z) = A_{1}'+A_{1}(z)p'$ and $B_{2}(z) = B_{1}'-B_{1}(z)p'$. Together (6.4) with (6.6), we obtain that

Lemma 2.1 and (6.7) imply that

and

Since $R_{1}(z), R_{2}(z)$ are two nonzero rational functions and ${l(z)}$ is of finite order, using the similar method as the proof of Theorem 1.1, we know that $p(z)$ is a polynomial of degree one. Let $p(z) = az+b, a\neq0, b\in\mathbb{C}$. Substituting $p(z), A_{2}(z), B_{2}(z)$ into (6.8) and (6.9), and as $z\longrightarrow\infty$, we have

and

that is

So, we have $\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{a{c_{{i}}}}}+\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{{\rm{-}}a{c_{{i}}}}} = 0$.

$(i)$ If $\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{a{c_{{i}}}}}+\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{{\rm{-}}a{c_{{i}}}}}\neq 0$, this is a contradiction with $\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{a{c_{{i}}}}}+\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{{\rm{-}}a{c_{{i}}}}} = 0$. Hence, Theorem 1.4 (i) holds.

$(ii)$ If $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{a{c_j}}} \neq 2a$ and $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{ - a{c_j}}} \neq 2a$. By (6.8)–(6.10), we have

and

If $R_{1}(z), R_{2}(z)$ are two nonzero rational functions, using the similar method as the proof of Theorem 1.3, we know that $R_i(z)$ is a polynomial. If $\deg_z R_i(z)\geq 2$. Let $R_i({z}) = {a_n}{z^n} + {a_{n - 1}}{z^{n - 1}} + \cdots + {a_0}$, then

where $i = 1, 2$. Substituting (6.13) into (6.11) and (6.12), comparing the coefficients of $z^{n-1}$, $z^{n- 2}$, we have $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{a{c_j}}} = 2a$, $\sum\limits_{j = 1}^m {c_j}^2 {\varrho _j}{exp ^{a{c_j}}} = 2$ and $\sum\limits_{j = 1}^m{i{c_j}}{\varrho_j}{exp^{-a{c_j}}} = 2a$, $\sum\limits_{j = 1}^m{c_j}^2{\varrho_j}exp^{-ac_{j}} = -2$, a contradiction. Hence, $\deg_z R_i(z) \leq 1$.

If $\deg_z R_i(z) = 1$, then (6.11) and (6.12) imply that $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{a{c_j}}} = 2a$ and $\sum\limits_{j = 1}^m {i{c_j}} {\varrho _j}{exp ^{ - a{c_j}}} = 2a$, a contradiction. Hence, $R_{1}(z)$, $R_{2}(z)$ are two nonzero constants, $R(z) = R_{1}(z)R_{2}(z)$ is a constant. By (6.5), we have

where $a\neq0$, $b\in\mathbb{C}$, $t_{1}$, $t_{2}\in\mathbb{C}\setminus \{0\}$ and $P(z)$ is a polynomial of degree one. Since $\sum\limits_{i = 1}^m{{\varrho_i}}\neq 0$, then by (6.5), we have $P(z)\equiv0$. So, we have

where $\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{a{c_{{i}}}}}+\sum\limits_{i = 1}^m{{\varrho_i}}{exp^{{\rm{-}}a{c_{{i}}}}} = 0$, $b\in\mathbb{C}$, $R(z) = a^{4}t_{1}t_{2}$. Hence, Theorem 1.4 holds.

Acknowledgments

We thank the referee(s) for reading the manuscript very carefully and making a number of valuable and kind comments which improved the presentation. The work was supported by the NNSF of China (No.10771121, 11401387), the NSF of Zhejiang Province, China (No. LQ14A010007), the NSFC Tianyuan Mathematics Youth Fund (No. 11226094), the NSF of Shandong Province, China (No. ZR2012AQ020 and No. ZR2010AM030) and the Fund of Doctoral Program Research of Shaoxing College of Art and Science (20135018).

Conflict of interest

The authors declare that none of the authors have any competing interests in the manuscript.