We start with the basic facts of Nevanlinna's value distribution theory. For a meromorphic function f, let n(r,f) N(r,f),m(r,f) and T(r,f) denote the un-integrated counting function, integrated counting function, proximity function and characteristic function respectively. We will use the first main theorem and second main theorem of Nevanlinna for a meromorphic function f, see [6, 7, 20]. Recall the elementary definitions of order of growth ρ(f), hyper-order of growth ρ2(f) and exponent of convergence of zeros for a meromorphic function f:

and

A meromorphic function g(z) is a small function of f(z) if T(r,g)=S(r,f), where S(r,f) denotes all functions that grow with order o(T(r,f)) as r→∞, outside of a possible exceptional set of finite logarithmic measure.

In 1959, Hayman [5] gave a significant result regarding the zero distribution of complex differential polynomials. It states that `If n≥2 is a natural number, and f(z) is a transcendental entire function, then fnf'-a has infinitely many zeros, where a is a non-zero constant.' Later, Clunie [2] proved that Hayman's result is also true for the case when n=1.

In [5], Hayman also posed a conjecture: If n≥1 is a natural number and f is a transcendental meromorphic function, then fnf'-a has infinitely many zeros, where a is a non-zero constant. Note that the Hayman conjecture has been proved completely, see [1, 5, 15, 21]. After Hayman, Laine and Yang [8] studied the zero distribution of complex difference polynomials and proved the following result.

Theorem 1. Suppose that f(z) is a transcendental entire function of finite order and c is a non-zero constant, then fn(z)f(z+c)-a has infinitely many zeros, where n≥2 and a is a non-zero constant.

Later on, many authors made improvements to Theorem 1; the constant a can be replaced by any non-zero polynomial, f(z+c) and a can be replaced by f(z+c)-f(z) and any non-zero polynomial respectively, fn(z) and a have been replaced by polynomials of degree n and non-zero function of growth o(T(r,f)) respectively, see [11, 14].

In 2011, Liu et al. [12] also proved Theorem 1 for the case when f(z) is a transcendental meromorphic function with ρ2(f)<1 and n≥6. They also gave counter examples for n≤3. After Liu et al. [12], the same result was proved for n≥4 by Wang and Ye [17]. In 2014, Liu et al.[13] studied delay-differential version of Hayman conjecture. Then many improvements have been made by other authors, see [17, 9, 10].

Recently Gao and Liu [3] considered the Hayman conjecture of paired differential polynomials and paired delay-differential polynomials. In particular, they considered the zeros distribution of fn(z)L(g)-a(z) and gn(z)L(f)-a(z), where n∈N, a(z) is a non-zero small function of f(z) and g(z). Also, L(h) holds one of the following conditions:

• L(h)=h(k)(z),k≥1

• L(h)=h(z+c), c∈C∖{0}

• L(h)=h(z+c)-h(z), c∈C∖{0}

• L(h)=h(k)(z+c),k≥1 and c∈C∖{0}.

Let M denotes the function class of all transcendental meromorphic functions, M' denotes function class of transcendental meromorphic functions of hyper-order less than 1, E denotes function class of all transcendental entire functions and E' denotes function class of transcendental entire functions of hyper-order less than 1. Then Gao and Liu [3] proved the following result:

Theorem 2. If L(h) satisfies any one of the following conditions:

• L(h)=h(k)(z),n≥k+4 and h∈M or n≥3 and h∈E;

• L(h)=h(z+c),n≥4 and h∈M' or n≥3 and h∈E';

• L(h)=h(z+c)-h(z),n≥5 and h∈M' or n≥3 and h∈E';

• L(h)=h(k)(z+c),n≥k+4 and h∈M' or n≥3 and h∈E'.

Then at least one of fn(z)L(g)-a(z) and gn(z)L(f)-a(z) have infinitely many zeros.

In the same paper they raised a question namely `Question 1': Can we reduce n≥3 to n≥2 for entire functions f and g in E or E'? And what is the sharp value of n for meromorphic functions f and g in M or M'? We give partial answer to this question.

Let M* denote the class of transcendental meromorphic functions f for which N(r,f)+N(r,1/f)=S(r,f) and M** denote the class of transcendental meromorphic functions f such that N(r,f)+N(r,1/f)=S(r,f) with ρ2(f)<1. Then the following results hold:

Theorem 3. Suppose that f and g are functions from the class M*. At least one of f2(z)g(k)(z)-a(z) and g2(z)f(k)(z)-a(z) have infinitely many zeros, where k≥1, a(z) is a small function of f(z) and g(z).

Example 4. If f(z)=ez, g(z)=e-z and a(z) is any non-zero polynomial, then f2(z)g'(z)-a(z) and g2(z)f'(z)-a(z) both have infinitely many zeros.

Theorem 5. Suppose that f and g are functions from the class M**, a(z) is a small function of f(z) and g(z), c is a non-zero complex number, and L(h) satisfies any one of the following conditions:

• L(h)=h(z+c);

• L(h)=h(z+c)-h(z);

• L(h)=h(k)(z+c),k≥1.

Then at least one of f2(z)L(g)-a(z) and g2(z)L(f)-a(z) have infinitely many zeros.

Example 6. Let f(z)=ez, g(z)=e-z, a(z) be any non-zero polynomial and c be a non-zero number, then

• f2(z)g(z+c)-a(z) and g2(z)f(z+c)-a(z) both have infinitely many zeros.

• f2(z)(g(z+c)-g(z))-a(z) and g2(z)(f(z+c)-f(z))-a(z) both have infinitely many zeros.

• f2(z)g(k)(z+c)-a(z) and g2(z)f(k)(z+c)-a(z) both have infinitely many zeros.

It was observed that Theorem 2 is not true for n=1, see [3, Remark 1.2]. As we can see that if f(z)=ze4z and g(z)=z2e-4z, then f(z)g'(z)-a(z) and g(z)f'(z)-a(z) both have finitely many zeros provided a(z) is any polynomial in z except 2z2-4z3 and z2+4z3. Next we are going to study part (1) of Theorem 2 for the case when n=1 and k=1. The following result shows that under the certain conditions on f and g, we get the same conclusion as in Theorem 2 for n=1. Before stating the result, let F*={f=α(z)eP(z)-k:α(z) and P(z) be non-zero polynomial and non-constant polynomial respectively, and k∈C∖{0}}.

Theorem 7. Let f and g be the functions from F* class, then at least one of fg'-a(z) and gf'-a(z) have infinitely many zeros, where a(z) is a non-zero polynomial.

Example 8. Let f(z)=e2z+3, g(z)=e-2z+4 and a(z) be any non-zero polynomial then at least one of fg'-a(z) and gf'-a(z) have infinitely many zeros.

In this section we present some known basic facts and results which we will use in the next section. For a set I⊂(0,∞), the linear measure is defined by m(I)=∫Idt and for a set J⊂(1,∞), the logarithmic measure is defined by ml(J)=∫J1tdt.

Lemma 1. [4] Suppose that T:[0,∞)→[0,∞) is a non-decreasing continuous function having ρ2(T)<1 and c is a non-zero real number. If δ∈(0,1-ρ2(T)), then

The following lemma gives an estimate of a counting function corresponding to the zeros of derivative of non-constant meromorphic function f, see [19, Theorem 1.24].

Lemma 2. Suppose that f is a non-constant meromorphic function, then

where k≥1.

The following two lemmas can be easily obtained by applying basic facts of Nevanlinna theory.

Lemma 3. If f is a function from the class M*, then

N(r,f(k))=S(r,f) and Nr,1f(k)=S(r,f).

Proof. Given that f is a transcendental meromorphic function satisfying

We know that if f has a pole of order m at z0, then f(k) has a pole of order m+k≤(k+1)m at z0. Thus

Using equation (2.1), we have N(r,f(k))=S(r,f).

Using equation (2.1) with Lemma 10, we get Nr,1f(k)=S(r,f).

Lemma 4. If f is a function from the class M**, and c is a non-zero complex number and k≥1, then

• N(r,f(z+c))=S(r,f) and Nr,1f(z+c)=S(r,f).

• N(r,f(z+c)-f(z))=S(r,f) and Nr,1f(z+c)-f(z)=S(r,f).

• N(r,f(k)(z+c))=S(r,f) and Nr,1f(k)(z+c)=S(r,f).

Proof.

• By the simple observation and using Lemma 1, we have

• N(r,f(z+c)-f(z))=S(r,f) is obvious by (1) part and

• As we know that N(r,f(k)(z+c))≤(k+1)N(r,f(z+c)), then applying (1) part, we get N(r,f(k)(z+c))=S(r,f). Also applying (1) part into Lemma 10, we get Nr,1f(k)(z+c)=S(r,f).

Lemma 5. [4] Suppose that f is a transcendental meromorphic function such that ρ2(f)<1 and k≥0, then

outside of a possible exceptional set of finite logarithmic measure.

The following lemma is obtain by [7, Corollary 2.3.4].

Lemma 6. If f is a function from the class M*, then Tr,1f(k)≤T(r,f)+S(r,f).

Proof. Given that f is a transcendental meromorphic function satisfying

Next

Using first main theorem of Nevanlinna, we have

Applying Lemma 11, we have

Lemma 7. If f is a function from the class M**, then

• Tr,1f(z+c)≤T(r,f)+S(r,f).

• Tr,1f(z+c)-f(z)≤T(r,f)+S(r,f).

• Tr,1f(k)(z+c)≤T(r,f)+S(r,f).

Proof.

• This is easily obtained by applying first fundamental theorem of Nevanlinna, Lemma 13 and (1) part of Lemma 12.

• Applying first fundamental theorem of Nevanlinna, we have

• Similarly as we did for previous part, we can prove this one also.

Next lemma gives estimate of the characteristic function of an exponential polynomial f and this can also be seen in [18].

Lemma 8. [16] Suppose f is an entire function given by

where Ai(z) 0≤i≤m denote either exponential polynomial of degree <s or polynomial in z, wi 1≤i≤m denote the constants and s denotes a natural number. Then

here C(Co(W0)) is the perimeter of the convex hull of the set W0={0,w¯1,w¯2,...,w¯m}. Moreover,

• if A0(z)≡0, then

• if A0(z)≡0, then

Proof. Proof of Theorem 3 Let F(z)=f2(z)g(k)(z)-a(z), then

In a simple way, we have

Using second fundamental theorem of Nevanlinna for three small functions, we have

Applying Lemma 11, we have

Next using Lemma 14 on g(z), we have

From equations (3.1), (3.2) and (3.3), we have

Similarly, let G(z)=g2(z)f(k)(z)-a(z). Then proceeding to same technique as we have done above, we get

From equation (3.4) and (3.5), we have

This implies that at least one of F(z)=f2(z)g(k)(z)-a(z) and G(z)=g2(z)f(k)(z)-a(z) have infinitely many zeros.

Proof. Proof of Theorem 5 Let F(z)=f2(z)L(g)-a(z), then we have

Using second fundamental theorem of Nevanlinna for three small functions, we have

This gives

• Let L(g)=g(z+c), then applying (1) part of Lemma 12 to equation (3.7), we have

  (3.8)T(r,F+a)≤Nr,1F+S(r,f)+S(r,g).
  

  Applying (1) part of Lemma 7 to equation (3.6) and together with equation (3.8), we have

  (3.9)2T(r,f)−T(r,g)≤Nr,1F+S(r,f)+S(r,g).
  

  Similarly, let G(z)=g2(z)f(z+c)−a(z). Then proceeding on similar lines as we did for F(z) function, we get

  (3.10)2T(r,g)−T(r,f)≤Nr,1G+S(r,f)+S(r,g).
  

  From equations (3.9) and (3.10), we have

  T(r,f)+T(r,g)≤Nr,1F+Nr,1G+S(r,f)+S(r,g).
  

  This implies that at least one of F(z)=f2(z)g(z+c)−a(z) and G(z)=g2(z)f(z+c)−a(z) have infinitely many zeros.

• Let L(g)=g(z+c)-g(z), then applying (2) part of Lemma 12 to equation (3.7), we get equation (3.8). Also applying (2) part of Lemma 15 to equation (3.6) and together with equation (3.8), we get equation (3.9). Moreover, let G(z)=g2(z)(f(z+c)-f(z))-a(z), then with the same idea as for F(z), we get equation (3.10). Hence we obtain the required conclusion.

• Let L(g)=g(k)(z+c), then applying (3) part of Lemma 12 and 15 into the equations (3.6), (3.7) and (3.8), we get equation (3.9). Next, let G(z)=g2(z)f(k)(z+c)-a(z), then with the same idea as for F(z), we get equation (3.10). Hence we obtain the required conclusion.

Proof. Proof of Theorem 7 Suppose that

where α1, α2 are non-zero polynomials, k1,k2∈C∖{0} and P1,P2 are non-constant polynomials say P1(z)=amzm+am-1zm-1+...+a0 and P2(z)=bnzn+bn-1zn-1+...+b0.

Differentiating equation (3.11) gives

From equation (3.11) and (3.12), we have

and

Now we discuss the following three cases:

• If deg(P1)>deg(P2), then equations (3.13) and (3.14) can be written as

  (3.15)fg′−a(z)=H1(z)eamzm+H0(z)
  

  and

  (3.16)gf′−a(z)=J1(z)eamzm−a(z),
  

  where

  H1(z)=(α1α2′+α1α2P2′)eP2+am−1zm−1+...+a0,H0(z)=−(k1α2′+k1α2P2′)eP2−a(z),J1(z)=[(α2α1′+α1α2P1′)eP2−(k2α1′+k2α1P1′)]eam−1zm−1+...+a0.
  

  Applying Lemma 8 and first main theorem of Nevanlinna into equations (3.15) and 3.16 give

  Nr,1fg′−a(z)−a(z)=T(r,fg′−a(z))+O(1)
  

  and

  Nr,1gf′−a(z)−a(z)=T(r,gf′−a(z))+O(1).
  

  Hence λ(fg′−a(z))=ρ(fg′−a(z)) and λ(gf′−a(z))=ρ(gf′−a(z)). This implies that both fg′−a(z) and gf′−a(z) have infinite number of zeros.

• If deg(P2)>deg(P1), then equations (3.13) and (3.14) can be written as

  (3.17)fg′−a(z)=H3(z)ebnzn−a(z)
  

  and

  (3.18)gf′−a(z)=J3(z)ebnzn+J0(z),
  

  where

  H3(z)=[(α1α2′+α1α2P2′)eP1−(k1α2′+k1α2P2′)]ebn−1zn−1+...+b0,J3(z)=(α2α1′+α1α2P1′)eP1+bn−1zn−1+...+b0,J0(z)=−(k2α1′+k2α1P1′)eP1−a(z).
  

  Proceeding on similar manner as we deal with (1) part, we obtain that both fg′−a(z) and gf′−a(z) have infinite number of zeros from the equations (3.17) and 3.18.

• If deg(P1)=deg(P2), then equations (3.13) and (3.14) become

  (3.19)fg′−a(z)=H4(z)e(an+bn)zn+H5(z)ebnzn−a(z)
  

  and

  (3.20)gf′−a(z)=J4(z)e(an+bn)zn+J5(z)eanzn−a(z),
  

  where

  H4(z)=(α1α2′+α1α2P2′)e(an−1+bn−1)zn−1+...+(a0+b0),H5(z)=−(k1α2′+k1α2P2′)ebn−1zn−1+...+b0,J4(z)=(α2α1′+α1α2P1′)e(an−1+bn−1)zn−1+...+(a0+b0),J5(z)=−(k2α1′+k2α1P1′)ean−1zn−1+...+a0.
  

  Next we study the following subcases:

  • (a)

    If an≠±bn, then applying same reason to equations (3.21) and (3.22) as we done for (1) part, we obtain that both fg'-a(z) and gf'-a(z) have infinite number of zeros.

  • (b)

    If an=bn, then proceeding the same logic as we applied in (1) part, we get same conclusion from equations (3.21) and (3.22).

  • (c)

    If an=-bn, then equations (3.21) and (3.22) become

    (3.21)fg′−a(z)=H5(z)ebnzn+H4(z)−a(z)
    

    and

    (3.22)gf′−a(z)=J5(z)e−bnzn+J4(z)−a(z).
    

  • If H4(z)≡a(z) and J4(z)≡a(z), then applying same reason to equations (3.21) and (3.22) as we done for (1) part, we obtain that both fg'-a(z) and gf'-a(z) have infinite number of zeros.

  • If H4(z)≡a(z), then J4(z)≡a(z). Otherwise H4(z)≡J4(z), this gives α2(z)α1(z)=eP1-P2+C, which is not possible. Here C is an arbitrary constant. By the simple observation, we get fg'-a(z) has finitely many zeros from equation (3.21). Also proceeding with the same logic as we applied in (1) part, we obtain gf'-a(z) has infinitely many zeros from the equation (3.22).

  • If J4(z)≡a(z), then proceeding with the same logic as we applied in (II) part, we get fg'-a(z) has infinitely many zeros and gf'-a(z) has finitely many zeros.