Normal Criteria for Family Meromorphic Functions Sharing Holomorphic Function

Abstract

In this paper, we study the value distribution of differential polynomial with the form $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)},$ where f is a transcendental meromorphic function. Namely, we prove that $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)$ has infinitely zeros, where P(z) is a nonconstant polynomial and $n\in {\mathbb {N}},$ $k, n_1, \dots , n_k, t_1, \dots , t_k$ are positive integer numbers satisfying $n+\sum _{v}^{k}n_v\ge \sum _{v=1}^{k}t_v+3.$ Using it, we establish some normality criterias for family of meromorphic functions under a condition where differential polynomials generated by the members of the family share a holomorphic function with zero points. Our results generalize some previous results on normal family of meromorphic functions.

1 Introduction

Let D be a domain in the complex plane $\mathbb {C}$ and ${\mathcal {F}}$ be a family of meromorphic functions in D. The family ${\mathcal {F}}$ is said to be normal in D,  in the sense of Montel, if for any sequence $\{f_v\}\subset {\mathcal {F}},$ there exists a subsequence $\{f_{v_i}\}$ such that $\{f_{v_i}\}$ converges spherically locally uniformly in D,  to a meromorphic function or $\infty .$

In 1989, Schwick [8] proved that

Theorem A

Let ${\mathcal {F}}$ be a family of meromorphic functions defined in a domain D and k, n are positive integer numbers satisfying $n\ge k+3.$ If $(f^n)^{(k)}\ne 1$ for every $f\in {\mathcal {F}},$ then ${\mathcal {F}}$ is normal.

In 2014, Dethloff et al. [4] came up with new normality criteria, which extended the result given by Schwick.

Theorem B

Let a be a nonzero complex value, let n be a non-negative integer and $n_1, n_2, \dots , n_k,$ $t_1, t_2, \dots , t_k$ be positive integers. Let ${\mathcal {F}}$ be a family of meromorphic functions in a complex domain D such that for every $f\in {\mathcal {F}},$ $f^n(f^{n_1})^{(t_1)}\cdots (f^{n_k})^{(t_k)}-a$ is nowhere vanishing on D. Assume that

• (a) $n_v\ge t_v \text { for all } 1\leqslant v\leqslant k,$

• (b) $n+\sum _{v=1}^kn_v\ge 3+\sum _{v=1}^kt_v.$

Then ${\mathcal {F}}$ is normal on D.

In 2009, Li and Gu [7] improved Theorem A in the following manner

Theorem C

Let ${\mathcal {F}}$ be a family of meromorphic functions in a domain D, $k, n(n\ge k+2)$ be positive integers and $a\in \mathbb {C}{\setminus } \{0\}$. If $(f^n)^{(k)}$ and $(g^n)^{(k)}$ share the value $a-IM$ in D for each pair of functions $f, g \in {\mathcal {F}}$, then ${\mathcal {F}}$ is normal.

In 2014, Datt and Kumar [3], by idea sharing value, they proved the result corresponding Theorem B.

Theorem D

Let $\alpha (z)$ be a holomorphic function defined in $D \subset C$ such that $\alpha (z)\ne 0.$ Let n be a non-negative integer and $n_1, n_2, \dots , n_k$, $t_1, t_2, \dots , t_k$ be positive integers such that

• (a) $n_v\ge t_v \text { for all } 1\leqslant v\leqslant k;$

• (b) $n+\sum _{v=1}^kn_v\ge 3+\sum _{v=1}^kt_v.$

Let ${\mathcal {F}}$ be a family of meromorphic functions in a domain D such that for every pair $f, g \in {\mathcal {F}}$, $f^n(z)(f^{n_1})^{(t_1)}(z)\cdots (f^{n_k})^{(t_k)}(z)$ and $g^n(z)(g^{n_1})^{(t_1)}(z)\cdots (g^{n_k})^{(t_k)}(z)$ share $\alpha (z)-IM$ on D. Then ${\mathcal {F}}$ is normal in D.

In 2012, Zeng and Lahiri [12] proved the result concerning Theorem C.

Theorem E

Let ${\mathcal {F}}$ be a family of meromorphic functions defined in a domain D, $a\in \mathbb {C}{\setminus } \{0\}$ and k, n be positive integers such that $n\ge 1$ if $k=1$ and $n\ge 2$ if $k\ge 2$. If $f^n(f^{k+1})^{(k)}$ and $g^n(g^{k+1})^{(k)}$ share the value $a-IM$ in D for each pair of functions $f, g \in {\mathcal {F}}$, then ${\mathcal {F}}$ is normal.

We see that the value $a \ne 0$ in Theorems C and E is a holomorphic function nowhere vanishing.

Question 1

Can we extend Theorems C, D and E by idea sharing a holomorphic function with zero point?

In 2012, Yunbo and Zongsheng [10] proved that

Theorem F

Let $n, k \ge 2$, $m \ge 0$ be three integers, and m be divisible by $n+1$. Suppose that $a(z )\not \equiv 0$ is a holomorphic function with zeros of multiplicity m in a domain D. Let ${\mathcal {F}}$ be a family of holomorphic functions in D, for each $f \in {\mathcal {F}},$ f has only zeros of multiplicity $k+m$ at least. For each pair $(f , g) \in F,$ $f (f^{(k)})^n$ and $g (g^{(k)})^n$ share $a(z ) - IM$, then ${\mathcal {F}}$ is normal in D.

For each meromorphic function f on D, we call that $N(f, f', \dots , f^{(t_1)}, \dots , f^{(t_k)})$ is a monomial differential polynomial of f and defined by

$$\begin{aligned} N(f, f', \dots , f^{(t_1)}, \dots , f^{(t_k)})=f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}, \end{aligned}$$

where $n \in \mathbb {N}, n_1, \dots , n_k, t_1, \dots , t_k$ are positive integer numbers and $k \in \mathbb {N}^{*}.$ We denote

$$\begin{aligned} \Gamma _{N}=n+n_1+\dots +n_k, \Upsilon _{N}=t_1+\dots +t_k. \end{aligned}$$

In this paper, we consider the differential polynomial with the form

$$\begin{aligned}&f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}+\sum _{I}a_{I}f^{n_{I}}(f^{n_{1I}})^{(t_{1I})}\dots (f^{n_{kI}})^{(t_{kI})}\nonumber \\&= N(f, f', \dots , f^{(t_1)}, \dots , f^{(t_k)})+\sum _{I}a_{I}N_{I}(f, f', \dots , f^{(t_{1I})}, \dots , f^{(t_{kI})}), \end{aligned}$$

(1.1)

where $a_{I}$ are holomorphic functions on D, and $n_{I}$, $n_{jI}$, $t_{jI}$, $j=1,\dots ,k$ are non-negative integer numbers, and $I \subset \mathbb {N}$ is the set index finitely.

Now, connection with result of Theorem F, we prove the results as following:

Theorem 1

Let $n, m \in \mathbb {N}$ and $n_v, t_v, k$ $(v=1,2,\dots ,k)$ be positive integer numbers such that m is divisible by $n+\sum _{v=1}^{k}n_v$ and

$$\begin{aligned} n_v\geqslant t_v, v=1,\dots ,k, n+\sum _{v=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+3, n+\sum _{v=1}^{k}n_v = \Gamma _{N_{I}}. \end{aligned}$$

Let ${\mathcal {F}}$ be a family of meromorphic functions in a complex domain D with all poles and zeros of multiplicity at least $\Big [\dfrac{2m+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \Big ]+1.$ Let $a(z)\not \equiv 0$ be a holomorphic functions with zeros of multiplicity m in a domain D. If

$$\begin{aligned}&f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}+\sum _{I}a_{I}f^{n_{I}}(f^{n_{1I}})^{(t_{1I})}\dots (f^{n_{kI}})^{(t_{kI})},\\&g^n(g^{n_1})^{(t_1)}\dots (g^{n_k})^{(t_k)}+\sum _{I}a_{I}g^{n_{I}}(g^{n_{1I}})^{(t_{1I})}\dots (g^{n_{kI}})^{(t_{kI})} \end{aligned}$$

share $a(z)-$ IM in D for each pair (f, g) in ${\mathcal {F}}$, where $t_{jI}$ satisfy

$$\begin{aligned} \sum _{v=1}^{k}t_v>\sum _{v=1}^{k}t_{vI}, \end{aligned}$$

then ${\mathcal {F}}$ is a normal family. Here, we denote [x] by integer part of the number x.

Remark 2

Theorem 1 is an extension of Theorems D and E for case sharing holomorphic function with zero point.

From Theorem 1, we get a corollary as following:

Corollary 3

Let $n, m \in \mathbb {N}$ and $n_v, t_v, k$ $(v=1,2,\dots ,k)$ be positive integer numbers such that m is divisible by $n+\sum _{v=1}^{k}n_v$ and

$$\begin{aligned} n_v\geqslant t_v, v=1,\dots ,k, n+\sum _{v=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+3, n+\sum _{v=1}^{k}n_v = \Gamma _{N_{I}}. \end{aligned}$$

Let ${\mathcal {F}}$ be a family of meromorphic functions in a complex domain D with all poles and zeros of multiplicity at least $\Big [\dfrac{2m+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \Big ]+1.$ Let $a(z)\not \equiv 0$ be a holomorphic functions with zeros of multiplicity m in a domain D. If

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}+\sum _{I}a_{I}f^{n_{I}}(f^{n_{1I}})^{(t_{1I})}\dots (f^{n_{kI}})^{(t_{kI})} \ne a(z) \end{aligned}$$

for every f in ${\mathcal {F}}$, where $t_{jI}$ satisfy

$$\begin{aligned} \sum _{v=1}^{k}t_v>\sum _{v=1}^{k}t_{vI}, \end{aligned}$$

then ${\mathcal {F}}$ is a normal family.

We see that Corollary 3 is an extension of Theorems A and B.

Theorem 4

Let $n, m \in \mathbb {N}$ and $n_v, t_v, k$ $(v=1,2,\dots ,k)$ be positive integer numbers such that m is divisible by $n+\sum _{v=1}^{k}n_v$ and

$$\begin{aligned} n_v\geqslant t_v, v=1,\dots ,k, n+\sum _{v=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+2, n+\sum _{v=1}^{k}n_v = \Gamma _{N_{I}}. \end{aligned}$$

Let ${\mathcal {F}}$ be a family of entire functions in a complex domain D with all zeros of multiplicity at least $\Big [\dfrac{2m+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \Big ]+1.$ Let $a(z)\not \equiv 0$ be a holomorphic functions with zeros of multiplicity m in a domain D. If

$$\begin{aligned}&f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}+\sum _{I}a_{I}f^{n_{I}}(f^{n_{1I}})^{(t_{1I})}\dots (f^{n_{kI}})^{(t_{kI})},\\&g^n(g^{n_1})^{(t_1)}\dots (g^{n_k})^{(t_k)}+\sum _{I}a_{I}g^{n_{I}}(g^{n_{1I}})^{(t_{1I})}\dots (g^{n_{kI}})^{(t_{kI})} \end{aligned}$$

share $a(z)-$ IM in D for each pair (f, g) in ${\mathcal {F}}$, where $t_{jI}$ satisfy

$$\begin{aligned} \sum _{v=1}^{k}t_v>\sum _{v=1}^{k}t_{vI}, \end{aligned}$$

then ${\mathcal {F}}$ is a normal family.

2 Some Lemmas

To prove our results, we need the following lemmas.

Lemma 1

(Zalcman’s Lemma, [11]) Let ${\mathcal {F}}$ be a family of meromorphic functions defined in the unit disc $\bigtriangleup .$ Then if ${\mathcal {F}}$ is not normal at a point $z_0\in \bigtriangleup ,$ there exist, for each real number $\alpha$ satisfying $-1<\alpha <1,$

• 1. a real number $r,\;0<r<1,$

• 2. points $z_n,\;|z_n|<r,$ $z_n\rightarrow z_0,$

• 3. positive numbers $\rho _n\rightarrow 0^+,$

• 4. functions $f_n\in {\mathcal {F}}$

such that

$$\begin{aligned} g_n(\xi )=\frac{f_n(z_n+\rho _n\xi )}{\rho _n^\alpha }\rightarrow g(\xi ) \end{aligned}$$

spherically uniformly on compact subsets of $\mathbb {C},$ where $g(\xi )$ is a nonconstant meromorphic function and $g^{\#}(\xi )\leqslant g^{\#}(0)=1.$ Moreover, the order of g is not greater than 2. Here, as usual, $g^\#(z)=\frac{|g'(z)|}{1+|g(z)|^2}$ is the spherical derivative.

Lemma 2

[2] Let g be a entire function, and M is a positive constant. If $g^{\#}(\xi )\leqslant M$ for all $\xi \in \mathbb {C},$ then g has the order at most one.

Remark 5

In Lemma 1, if ${\mathcal {F}}$ is a family of holomorphic functions, then by Hurwitz’s Theorem, g is a holomorphic function. Therefore, by Lemma 2, the order of g is not greater than 1.

We consider a nonconstant meromorphic function g in the complex plane $\mathbb {C},$ and its first p derivatives. A differential polynomial P of g is defined by

$$\begin{aligned} P(z):=\sum _{i=1}^n\alpha _i(z)\prod _{j=0}^p(g^{(j)}(z))^{S_{ij}}, \end{aligned}$$

where $S_{ij} \;(0\leqslant i,j\leqslant n)$ are non-negative integers, and $\alpha _i \;(1\leqslant i\leqslant n)$ are small (with respect to g) meromorphic functions. Set

$$\begin{aligned} d(P):=\min _{1\leqslant i\leqslant n}\sum _{j=0}^pS_{ij}\;\text {and}\; \theta (P):=\max _{1\leqslant i\leqslant n}\sum _{j=0}^pjS_{ij}. \end{aligned}$$

In 2002, Hinchliffe [6] generalized theorems of Hayman [5] and Chuang [1] and obtained the following result.

Proposition 1

Let g be a transcendental meromorphic function, let P(z) be a nonconstant differential polynomial in g with $d(P)\ge 2.$ Then

$$\begin{aligned} T(r,g)\leqslant \frac{\theta (P)+1}{d(P)-1}\overline{N}\left( r,\frac{1}{g}\right) +\frac{1}{d(P)-1}\overline{N}\left( r,\frac{1}{P-1}\right) +o(T(r,g)), \end{aligned}$$

for all $r\in [1,+\infty )$ excluding a set of finite Lebesgues measure.

By argument as Proposition 1, we are easy to get the result as following for small function. However, for convenience of the reader, we prove it here.

Lemma 3

Let g be a nonconstant meromorphic function and P(z) be a nonconstant differential polynomial in g with $d(P)\ge 1.$ Let $a(z) \not \equiv 0$ be a small function of P(g). Then

$$\begin{aligned} T(r,g)\leqslant \frac{\theta (P)+1}{d(P)} \overline{N}\left( r,\frac{1}{g}\right) +\frac{1}{d(P)}\overline{N}(r,g)+\frac{1}{d(P)}\overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,g)), \end{aligned}$$

for all $r\in [1,+\infty )$ excluding a set of finite Lebesgues measure.

Moreover, in the case where g is a nonconstant entire function, we have

$$\begin{aligned} T(r,g)\leqslant \frac{\theta (P)+1}{d(P)}\overline{N}\left( r,\frac{1}{g}\right) +\frac{1}{d(P)}\overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,g)), \end{aligned}$$

for all $r\in [1,+\infty )$ excluding a set of finite Lebesgues measure.

Remark 6

Let g be a nonconstant meromorphic function and P(z) be a nonconstant differential polynomial in g with $d(P)\ge 2.$ Let $a(z) \not \equiv 0$ be a small function of P(g). Then

$$\begin{aligned} T(r,g)\leqslant \frac{\theta (P)+1}{d(P)-1}\overline{N}\left( r,\frac{1}{g}\right) +\frac{1}{d(P)-1}\overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,g)), \end{aligned}$$

for all $r\in [1,+\infty )$ excluding a set of finite Lebesgues measure.

Proof of Lemma 3 and Remark 6

For any z such that $|g(z)|\leqslant 1,$ since $\sum _{j=0}^pS_{ij}\ge d(P)\;(1\leqslant i\leqslant n),$ we have

$$\begin{aligned} \frac{1}{|g(z)|^{d(P)}}&=\frac{1}{|P(z)|}\cdot \frac{|P(z)|}{|g(z)|^{d(P)}}\\&\leqslant \frac{1}{|P(z)|}\cdot \sum _{i=1}^n\big (|\alpha _i(z)|\prod _{j=0}^p\big |\frac{g^{(j)}(z)}{g(z)}\big |^{S_{ij}}\big ). \end{aligned}$$

This implies that for all $z\in \mathbb {C},$

$$\begin{aligned} \log ^+\frac{1}{|g(z)|^{d(P)}}\leqslant \log ^+\big (\frac{1}{|P(z)|}\cdot \sum _{i=1}^n\big (|\alpha _i(z)|\prod _{j=0}^p\big |\frac{g^{(j)}(z)}{g(z)}\big |^{S_{ij}}\big )\big ). \end{aligned}$$

Therefore, by the Lemma on logarithmic derivative and by the first main theorem, we have

$$\begin{aligned} d(P) m\left( r,\frac{1}{g}\right)&\leqslant m\left( r,\frac{1}{P}\right) +o(T(r,g))\\&=T\left( r,\frac{1}{P}\right) -N(r,\frac{1}{P})+o(T(r,g))\\&=T(r,P)-N\left( r,\frac{1}{P}\right) +o(T(r,g)). \end{aligned}$$

On the other hand, by the second main theorem for small function [5, 9], we have

$$\begin{aligned} T(r,P)\leqslant \overline{N}(r,P)+\overline{N}\left( r,\frac{1}{P}\right) +\overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,g)). \end{aligned}$$

Hence,

$$\begin{aligned} d(P)m(r,\frac{1}{g})&\leqslant \big (\overline{N}(r,P)+\overline{N}\big (r,\frac{1}{P}\big )+\overline{N}\big (r,\frac{1}{P-a}\big )\big )\\&\quad -\,N\big (r,\frac{1}{P}\big )+o(T(r,g)). \end{aligned}$$

By First Main Theorem, we have

$$\begin{aligned} d(P)T(r, g)&=d(P)T\big (r,\frac{1}{g}\big )+O(1)\nonumber \\&=d(P)m\big (r,\frac{1}{g}\big )+d(P)N(r,\frac{1}{g})+O(1)\nonumber \\&\leqslant \big (\overline{N}(r,P)+\overline{N}\big (r,\frac{1}{P}\big )+ \overline{N}\big (r,\frac{1}{P-a}\big )\big )\nonumber \\&\quad +\,d(P)N\big (r,\frac{1}{g}\big )-N\big (r,\frac{1}{P}\big )+o(T(r,g)). \end{aligned}$$

(2.1)

We see

$$\begin{aligned} \frac{1}{g^{d(P)}}=\frac{1}{P(z)}\sum _{i=1}^n\big (\alpha _ig^{S_{ij}-d(P)}\prod _{j=0}^p\big (\frac{g^{(j)}}{g}\big )^{S_{ij}}\big ). \end{aligned}$$

Note that $\sum _{j=0}^pS_{ij}-d(P)\ge 0,$ and therefore we get

$$\begin{aligned} d(P)\nu _{\frac{1}{g}}&\leqslant \nu _{\frac{1}{P}}+\max _{{1\leqslant i\leqslant n}}\left\{ \nu _{\alpha _i}+\sum _{j=0}^pjS_{ij}\overline{\nu }_{\frac{1}{g}}\right\} \\&\leqslant \nu _{\frac{1}{P}}+\sum _{i=1}^n\nu _{\alpha _i}+\theta (P)\overline{\nu }_{\frac{1}{g}}, \end{aligned}$$

where $\nu _\phi$ is the pole divisor of the meromorphic $\phi$ and $\overline{\nu }_\phi :=\min \{\nu _\phi ,1\}.$

This implies

$$\begin{aligned} d(P)\nu _{\frac{1}{g}}-\nu _{\frac{1}{P}}+\overline{\nu }_{\frac{1}{P}}\leqslant (\theta (P)+1)\overline{\nu }_{\frac{1}{g}}+\sum _{i=1}^n\nu _{\alpha _i} \end{aligned}$$

(note that for any $z_0,$ if $\nu _{\frac{1}{g}}(z_0)=0$ then $d(P)\nu _{\frac{1}{g}}(z_0)-\nu _{\frac{1}{P}}(z_0)+\overline{\nu }_{\frac{1}{P}}(z_0)\leqslant 0).$ Then, we obtain

$$\begin{aligned} d(P)N\left( r,\frac{1}{g}\right) -N\left( r,\frac{1}{P}\right) +\overline{N}\left( r,\frac{1}{P}\right)&\leqslant (\theta (P)+1)\overline{N}(r, \frac{1}{g})+\sum _{i=1}^nN(r,\alpha _i)\\&=(\theta (P)+1)\overline{N}\left( r, \frac{1}{g}\right) +o(T(r,g)). \end{aligned}$$

Combining with (2.1), we have

$$\begin{aligned} d(P)T(r, g)\leqslant \big (\overline{N}(r,P)+\overline{N}\big (r,\frac{1}{P-a}\big )\big ) +(\theta (P)+1)\overline{N}\big (r, \frac{1}{g}\big )+o(T(r,g)). \end{aligned}$$

On the other hand, by the definition of the differential polynomial P,  Pole$(P)\subset \cup _{i=1}^n$ Pole$(\alpha _i)\cup$ Pole(g). Hence,

$$\begin{aligned} d(P)T(r, g)\leqslant&\big (\overline{N}(r,g)+\overline{N}\big (r,\frac{1}{P-a}\big )\big )\nonumber \\&+\,(\theta (P)+1)\overline{N}\big (r, \frac{1}{g}\big )+o(T(r,g)). \end{aligned}$$

(2.2)

This implies that

$$\begin{aligned} T(r, g)\leqslant \dfrac{(\theta (P)+1)}{d(P)}\overline{N}\big (r, \frac{1}{g}\big )+\dfrac{1}{d(P)}\overline{N}(r,g)+\dfrac{1}{d(P)}\overline{N}\big (r,\frac{1}{P-a}\big )\big )+o(T(r,g)). \end{aligned}$$

(2.3)

From (2.3), we conclude the statement of Lemma 3 for nonconstant meromorphic function.

From (2.2), we have

$$\begin{aligned} d(P)T(r, g)\leqslant \,&\big (T(r,g)+\overline{N}\big (r,\frac{1}{P-a}\big )\big )\nonumber \\&+\,(\theta (P)+1)\overline{N}\big (r, \frac{1}{g}\big )+o(T(r,g)). \end{aligned}$$

Therefore, if $d(P)\ge 2,$ we get

$$\begin{aligned} T(r,g)\leqslant \frac{\theta (P)+1}{d(P)-1}\overline{N}\left( r,\frac{1}{g}\right) +\frac{1}{d(P)-1}\overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,g)). \end{aligned}$$

(2.4)

From (2.4), we obtain Remark 6.

In the case where g is nonconstant holomorphic function, the inequality in (2.2) becomes

$$\begin{aligned} d(P)T(r, g)\leqslant \overline{N}\left( r,\frac{1}{P-a}\right) +(\theta (P)+1)\overline{N}\left( r, \frac{1}{g}\right) +o(T(r,g)). \end{aligned}$$

This implies that

$$\begin{aligned} T(r, g)\leqslant \frac{\theta (P)+1}{d(P)})\overline{N}\left( r, \frac{1}{g}\right) +\frac{1}{d(P)}\overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,g)). \end{aligned}$$

We have completed the proof of Lemma 3. $\square$

Lemma 4

Let f be a nonconstant rational function, $P(z)=a_dz^d+a_{d-1}z^{d-1}+\dots +a_0, d\in \mathbb {N}, a_d\ne 0, a_{d-1}, \dots , a_0$ be complex numbers and $n\in \mathbb {N}$, $k, n_v, t_v \in \mathbb {N}^{*}$, $v=1,\dots ,k.$

If $d=0,$ $P(z)=a_0 \ne 0,$

$$\begin{aligned} n_v\geqslant t_v, n+\sum _{v=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+2, v=1,\dots , k; \end{aligned}$$

and if $d\ge 1,$

$$\begin{aligned} n_v\geqslant t_v, n+\sum _{v=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+2, v=1,\dots , k, \end{aligned}$$

suppose that all zeros and poles of f having multiplicity at least

$$\begin{aligned} \Big [\dfrac{2d+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \Big ]+1, \end{aligned}$$

then the equation

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}=P(z) \end{aligned}$$

have at least two distinct zeros.

Proof

We consider two cases as following:

Case 1. f is a polynomial. Then $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)$ is a polynomial with degree at least $2d+2$ and when $P(z)=a_0\ne 0,$ $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)$ is polynomial with degree at least 2. Indeed, when $\deg P \ge 1,$ then all zeros of f have multiple at least

$$\begin{aligned} \Big [\dfrac{2d+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \Big ]+1>\dfrac{2d+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}. \end{aligned}$$

Hence $\deg f>\dfrac{2d+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}.$ This implies that $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)$ is polynomial with degree at least

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) \deg f-\sum _{j=1}^{k}t_v>2d+2+\sum _{v=1}^{k}t_v- \sum _{v=1}^{k}t_v=2d+2. \end{aligned}$$

We suppose that $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)$ has unique zero $z_0,$ then

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)=A(z-z_0)^{l}, l\ge 2d +2 \end{aligned}$$

(2.5)

and $A\ne 0$ is a constant. Take derivative both sides (2.5) to d and $d+1$ times, we have

$$\begin{aligned} \Big (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}\Big )^{(d)}-Al(l-1)\dots (l-d+1)(z-z_0)^{l-d}=P^{(d)}(z). \end{aligned}$$

(2.6)

and

$$\begin{aligned} \Big (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}\Big )^{(d+1)}=Al(l-1)\dots (l-d+1)(l-d)(z-z_0)^{l-d-1}. \end{aligned}$$

(2.7)

Since $l\ge 2d+2>d+1,$ from (2.7), we get that $z_0$ is uniqueness zero of $\Big (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}\Big )^{(d+1)}.$ We see that all zeros of f belong to zeros of $\Big (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}\Big )^{(d+1)}.$ Thus, f has uniqueness zero $z_0.$ From (2.7), we get

$$\begin{aligned} 0= \Big (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}\Big )^{(d)}(z_0)=P^{(d)}(z_0) \ne 0. \end{aligned}$$

This is a contradiction. Hence,

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z) \end{aligned}$$

have at least distinct two zeros in two cases $P(z)=a_0 \ne 0$ and $\deg P \ge 1.$

Case 2. f is not a polynomial. By hypothesis, we can express f as following

$$\begin{aligned} f=A\dfrac{(z-a_1)^{p_1}(z-a_2)^{p_2}\dots (z-a_s)^{p_s}}{(z-b_1)^{q_1}(z-b_2)^{q_2}\dots (z-b_t)^{q_t}}, \end{aligned}$$

(2.8)

where $p_i \ge 1, i=1, \dots , s$, $q_j\ge 1$, $j=1, \dots , t$ if $P(z)=a_0\ne 0$ and $p_i \ge \Big [\dfrac{2d+2+\sum _{j=1}^{k}t_j}{n+\sum _{j=1}^{k}n_j} \Big ]+1, i=1, \dots , s$, $q_j\ge \Big [\dfrac{2d+2+\sum _{j=1}^{k}t_j}{n+\sum _{j=1}^{k}n_j} \Big ]+1$, $j=1, \dots , t$ if $\deg P \ge 1.$

Take

$$\begin{aligned} p=\sum _{i=1}^{s}p_i\ge s\dfrac{2d+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}, q=\sum _{j=1}^{t}q_j\ge t\dfrac{2d+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \end{aligned}$$

(2.9)

if $\deg P\ge 1$, and $p\ge s$, $q\ge t$ if $P(z)=a_0\ne 0.$

From (2.8), we have

$$\begin{aligned} f^{n_v}=A^{n_v}\dfrac{(z-a_1)^{p_1n_v}(z-a_2)^{p_2n_v}\dots (z-a_s)^{p_sn_v}}{(z-b_1)^{q_1n_v}(z-b_2)^{q_2n_v} \dots (z-b_t)^{q_tn_v}}, \quad v=1, \dots , k. \end{aligned}$$

Then

$$\begin{aligned} (f^{n_v})^{(t_v)}=\,&A^{n_v}\dfrac{(z-a_1)^{p_1n_v-t_v}(z-a_2)^{p_2n_v-t_v}\dots (z-a_s)^{p_sn_v-t_v}}{(z-b_1)^{q_1n_v+t_v} (z-b_2)^{q_2n_v+t_v}\dots (z-b_t)^{q_tn_v+t_v}}g_v(z),\nonumber \\ v=\,&1, \dots , k, \end{aligned}$$

(2.10)

where

$$\begin{aligned} g_v(z)&=(n_vp-n_vq)(n_vp-n_vq-1)\dots (n_vp-n_vq-t_v+1)z^{t_v(s+t-1)}\\&\quad +\,b_{t_v(s+t-1)-1}z^{t_v(s+t-1)-1}+\dots +b_0 \end{aligned}$$

and $b_e$, $e=0, \dots , t_v(s+t-1)-1$ are complex numbers.

From (2.10), we see

$$\begin{aligned} f^n&(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}\nonumber \\&=A^{n+\sum _{v=1}^{k}n_v}\dfrac{\prod _{i=1}^{s}(z-a_i)^{(n+\sum _{v=1}^{k}n_v)p_i-\sum _{v=1}^{k}t_v}}{ \prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v}}g(z)\nonumber \\&=\dfrac{P_1(z)}{Q_1(z)}, \end{aligned}$$

(2.11)

where $g(z)=\prod _{v=1}^{k}g_v(z)$, $\deg g \le (\sum _{v=1}^{k}t_v)(s+t-1).$

Case 2.1. $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)$ has uniqueness a zero, we denote by $z_0.$ Thus, we can write

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)=\dfrac{B(z-z_0)^l}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v} }, \end{aligned}$$

(2.12)

where $l \in \mathbb {N}^{*}$ and $B\ne 0$ is a complex number. From (2.11), taking derivative both sides d times, we get

$$\begin{aligned} (f^n&(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})^{(d)}\nonumber \\&=A^{n+\sum _{v=1}^{k}n_v}\dfrac{\prod _{i=1}^{s}(z-a_i)^{(n+\sum _{v=1}^{k}n_v)p_i-\sum _{v=1}^{k}t_v-d}}{ \prod _{j=1}^{t} (z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+d}}G_1(z), \end{aligned}$$

(2.13)

where $\deg G_1 \le (\sum _{v=1}^{k}t_v+d)(s+t-1).$ Similar to (2.13), taking derivative both sides (2.11) $d+1$ times, we get

$$\begin{aligned} (f^n&(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})^{(d+1)}\nonumber \\&=A^{n+\sum _{v=1}^{k}n_v}\dfrac{\prod _{i=1}^{s}(z-a_i)^{(n+\sum _{v=1}^{k}n_v)p_i-\sum _{v=1}^{k}t_v-d-1}}{\prod _{j=1}^{t} (z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+d+1}}G_2(z), \end{aligned}$$

(2.14)

where $\deg G_2 \le (\sum _{v=1}^{k}t_v+d+1)(s+t-1).$

Note that in the case $P(z)=a_0\ne 0,$ we take the derivative both sides (2.11) with 0, 1 times, respectively, we obtain the (2.13) and (2.14), respectively.

Case 2.1.1. $d\ge l.$ Then from (2.12), we have

$$\begin{aligned} (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})^{(d+1)}=\dfrac{BR_{d+1}(z)}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+d+1}},\nonumber \\ \end{aligned}$$

(2.15)

where

$$\begin{aligned} R_{d+1}(z)&=\prod _{h=0}^{d}\left( l-\left( n+\sum _{v=1}^{k}n_v\right) q-\left( \sum _{v=1}^{k}t_v\right) t-h\right) z^{(d+1)t-(d-l+1)}\\&\quad +\,c_{(d+1)t-(d-l+1)-1}z^{(d+1)t-(d-l+1)-1}+\dots +c_0 \end{aligned}$$

and $c_h,$ $h=0, \dots , (d+1)t-(d-l+1)-1$ are complex numbers.

From (2.11) and (2.12), compare degree of the numerator after computing, we get

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) p-\left( \sum _{v=1}^{k}t_v\right) s+\deg g=\max \{l, \left( n+\sum _{v=1}^{k}n_v\right) q+\left( \sum _{v=1}^{k}t_v\right) t+d\}. \end{aligned}$$

(2.16)

From (2.16) and $\deg g \le (\sum _{v=1}^{k}t_v)(s+t-1),$ we obtain

$$\begin{aligned}&\left( n+\sum _{v=1}^{k}n_v\right) p-\left( \sum _{v=1}^{k}t_v\right) s+\left( \sum _{v=1}^{k}t_v\right) (s+t-1)\\&\quad \ge \left( n+\sum _{v=1}^{k}n_v\right) q+\left( \sum _{v=1}^{k}t_v\right) t+d. \end{aligned}$$

This implies

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) (p-q) \ge \sum _{v=1}^{k}t_v+d. \end{aligned}$$

Hence, $p\ge q+\dfrac{ \sum _{v=1}^{k}t_v+d}{n+\sum _{v=1}^{k}n_v}>q.$ From (2.14) and (2.15), we see

$$\begin{aligned} \deg \prod _{i=1}^{s}(z-a_i)^{(n+\sum _{v=1}^{k}n_v)p_i-\sum _{v=1}^{k}t_v-d-1} \le \deg R_{d+1}(z). \end{aligned}$$

Thus,

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) p-\left( \sum _{v=1}^{k}t_v +d+1\right) s \le (d+1)t-(d-l+1). \end{aligned}$$

This implies

$$\begin{aligned} l-d\ge \left( n+\sum _{v=1}^{k}n_v\right) p+1-\left( \sum _{v=1}^{k}t_v +d+1\right) s-(d+1)t. \end{aligned}$$

(2.17)

From (2.9) and $p>q$, we have

$$\begin{aligned} \left( \sum _{v=1}^{k}t_v+d+1\right) s+(d+1)t&\le \left( \sum _{v=1}^{k}t_v+d+1\right) p\dfrac{n+\sum _{v=1}^{k}n_v}{2d+2+\sum _{v=1}^{k}t_v}\nonumber \\&\quad +\,(d+1)q\dfrac{n+\sum _{v=1}^{k}n_v}{2d+2+\sum _{v=1}^{k}t_v}\nonumber \\&\quad <\left( n+\sum _{v=1}^{k}n_v\right) p. \end{aligned}$$

(2.18)

Combining (2.17) and (2.18), we have $l-d\ge 1.$ This contradicts with $d \ge l.$

Case 2.1.2. $d< l.$ If $d\ge 1$, we have

$$\begin{aligned} (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})^{(d)}-(P(z))^{(d)}=\dfrac{(z-z_0)^{l-d}U_d(z)}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+d}}, \end{aligned}$$

(2.19)

where $U_d(z)=\prod _{h=0}^{d-1}(l-(n+\sum _{v=1}^{k}n_v)q-(\sum _{v=1}^{k}t_v)t-h)z^{dt}+y_{dt-1}z^{dt-1}+\dots +y_0,$ $y_{j}$, $j=0, \dots , dt-1$ are complex numbers. We also have

$$\begin{aligned}&(f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})^{(d+1)}-(P(z))^{(d+1)}\nonumber \\&\quad =\dfrac{(z-z_0)^{l-d-1}U_{d+1}(z)}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+d}}, \end{aligned}$$

(2.20)

where $U_{d+1}(z)=\prod _{h=0}^{d}(l-(n+\sum _{v=1}^{k}n_v)q-(\sum _{v=1}^{k}t_v)t-h)z^{(d+1)t}+x_{(d+1)t-1}z^{(d+1)t-1}+\dots +x_0,$ $x_{j}$, $j=0, \dots , (d+1)t-1$ are complex numbers.

We distinguish two subcase:

Case 2.1.2.1. $l\ne (n+\sum _{v=1}^{k}n_v)q+(\sum _{v=1}^{k}t_v)t+d.$ From (2.11) and (2.12), we see $\deg P_1\ge \deg Q_1.$ This implies

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) p-\left( \sum _{v=1}^{k}t_v\right) s+\deg g \ge \left( n+\sum _{v=1}^{k}n_v\right) q+\left( \sum _{v=1}^{k}t_v\right) t. \end{aligned}$$

From $\deg g \le (\sum _{v=1}^{k}t_v)(s+t-1).$ Thus,

$$\begin{aligned} p\ge q+\dfrac{\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}>q. \end{aligned}$$

(2.21)

From (2.13) and (2.19), we see $z_0\ne a_i, i=1, \dots , s.$ Thus, from (2.14) and (2.20), we get

$$\begin{aligned} \deg \prod _{i=1}^{s}(z-a_i)^{(n+\sum _{v=1}^{k}n_v)p_i-\sum _{v=1}^{k}t_v-d-1} \le \deg U_{d+1}(z). \end{aligned}$$

Thus, we have

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) p- \left( \sum _{v=1}^{k}t_v\right) s-(d+1)s\le (d+1)t. \end{aligned}$$

From (2.9) and (2.21), we obtain

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) p&\le \left( \sum _{v=1}^{k}t_v+d+1\right) s+(d+1)t\\&\le \left( \sum _{v=1}^{k}t_v+d+1\right) p\dfrac{n+\sum _{v=1}^{k}n_v}{2d+2+\sum _{v=1}^{k}t_v}\\&\quad +\,(d+1)q\dfrac{n+\sum _{v=1}^{k}n_v}{2d+2+\sum _{v=1}^{k}t_v}\\&\quad < \left( n+\sum _{v=1}^{k}n_v\right) p. \end{aligned}$$

This is a contradiction.

Case 2.1.2.2. $l= (n+\sum _{v=1}^{k}n_v)q+(\sum _{v=1}^{k}t_v)t+d.$

If $p>q$, by argument as Case 2.1.2.1, we obtain the contradiction.

If $p\le q,$ from (2.14) and (2.20), we have

$$\begin{aligned} l-d-1\le \deg G_2\le \left( \sum _{v=1}^{k}t_v+d+1\right) (s+t-1). \end{aligned}$$

Therefore,

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) q=&l-\left( \sum _{v=1}^{k}t_v\right) t-d\\&\le \deg G_2-\left( \sum _{v=1}^{k}t_v\right) t+1\\&\le \left( \sum _{v=1}^{k}t_v+d+1\right) (s+t-1)-\left( \sum _{v=1}^{k}t_v\right) t+1\\&< \left( \sum _{v=1}^{k}t_v+d+1\right) s+(d+1)t\\&\le \left( \sum _{v=1}^{k}t_v+d+1\right) p\dfrac{n+\sum _{v=1}^{k}n_v}{2d+2+\sum _{v=1}^{k}t_v}\\&\quad +\,(d+1)q\dfrac{n+\sum _{v=1}^{k}n_v}{2d+2+\sum _{v=1}^{k}t_v}\\&\le \left( n+\sum _{v=1}^{k}n_v\right) q. \end{aligned}$$

This is an impossible.

If $d=0$, $P(z)=a_0\ne 0$, from (2.12), we have

$$\begin{aligned} (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})' =\dfrac{(z-z_0)^{l-1}H_1(z)}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_j)q_j+\sum _{v=1}^{k}t_v+1}} \end{aligned}$$

(2.22)

where $H_1(z)=B(l-(n+\sum \limits _{j=1}^{k}n_j)q-(\sum \limits _{j=1}^{k}t_j)t)z^{t}+w_{1}z^{t-1}+\dots +w_{t},$ $w_1, \dots , w_t$ are complex numbers and B is a nonzero constant. From (2.11), we see

$$\begin{aligned} (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})'=\dfrac{\prod _{i=1}^{s}(z-a_i)^{(n+\sum _{v=1}^{k}n_v)p_i+\sum _{v=1}^{k}t_v-1}H_2(z)}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+1}}, \end{aligned}$$

(2.23)

where $s+t-1 \le \deg H_2(z) \le (\sum _{v=1}^{k}t_{v}+1)(s+t-1).$ By argument as $d\ge 1,$ and remark that $p\ge s, q\ge t$, we get a contradiction.

Case 2.2. $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)$ has no zeros. Thus, we can write

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)=\dfrac{C}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v}}, \end{aligned}$$

(2.24)

where $C\ne 0$ is a complex number. Thus, (2.20) can be replaced by

$$\begin{aligned}&(f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})^{(d+1)}-(P(z))^{(d+1)}\nonumber \\&\quad =\dfrac{U_{d+1}^{*}(z)}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+d}}, \end{aligned}$$

(2.25)

where $U_{d+1}^{*}(z)=\prod _{h=0}^{d}(-(n+\sum _{v=1}^{k}n_v)q-(\sum _{v=1}^{k}t_v)t-h)z^{(d+1)(t-1)}+x_{dt}^{*}z^{dt}+\dots +x_0^{*},$ $x_{j}^{*},$ $j=0, \dots , (d+1)(t-1)-1$ are complex numbers.

From (2.24) and (2.11), we have $\deg P_1 \ge \deg Q_1.$ From (2.21), we see that $p\ge q+\dfrac{\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}>q.$ Thus, combine (2.14) and (2.25), we get

$$\begin{aligned} \deg \prod _{i=1}^{s}(z-a_i)^{(n+\sum _{v=1}^{k}n_v)p_i-\sum _{v=1}^{k}t_v -d-1}\le \deg U_{d+1}^{*}(z). \end{aligned}$$

Hence,

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) p-\left( \sum _{v=1}^{k}t_v\right) s -(d+1)s\le (d+1)(t-1). \end{aligned}$$

This implies

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) p&\le \left( \sum _{v=1}^{k}t_v+d+1\right) s +(d+1)(t-1)\\&<\left( \sum _{v=1}^{k}t_v+d+1\right) s +(d+1)t\nonumber \end{aligned}$$

(2.26)

From (2.26), and compute similarly to Case 2.1.2.1, we get a contradiction. $\square$

Lemma 5

Let f be a nonconstant rational function, $f \ne 0$, $P(z)=a_dz^d+a_{d-1}z^{d-1}+\dots +a_0, d\in \mathbb {N}^{*}, a_d\ne 0, a_{d-1}, \dots , a_0$ be complex numbers and $n\in \mathbb {N}$, $k, n_j, t_j \in \mathbb {N}^{*}$, $j=1,\dots ,k.$

If $d=0$, $P(z)=a_0 \ne 0,$

$$\begin{aligned} n_j\geqslant t_j, n+\sum _{j=1}^{k}n_j\geqslant \sum _{j=1}^{k}t_j+2, \quad j=1,\dots , k; \end{aligned}$$

and if $d\ge 1,$

$$\begin{aligned} n_j\geqslant t_j, n+\sum _{j=1}^{k}n_j\geqslant \sum _{j=1}^{k}t_j+2,\quad j=1,\dots , k, \end{aligned}$$

suppose that all poles of f having multiplicity at least

$$\begin{aligned} \Big [\dfrac{2d+2+\sum _{j=1}^{k}t_j}{n+\sum _{j=1}^{k}n_j} \Big ]+1. \end{aligned}$$

Then the equation

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}=P(z) \end{aligned}$$

has at least distinct two zeros.

Proof

Since f has not zeros, then we can write

$$\begin{aligned} f=\dfrac{A}{\prod _{j=1}^{t}(z-b_j)^{q_j}}, \end{aligned}$$

(2.27)

where $q_j \ge \dfrac{2d+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}, j=1, \dots , t$ if $\deg P \ge 1$ and $q_j \ge 1, j=1, \dots , t$ if $P(z)=a_0 \ne 0.$ Similar to (2.10), from (2.27), we have

$$\begin{aligned} (f^{n_v})^{(t_v)}=\dfrac{A^{n_v}}{\prod _{j=1}^{t}(z-b_j)^{q_jn_v+t_v}}g_v(z), \quad v=1, \dots , k, \end{aligned}$$

(2.28)

where

$$\begin{aligned} g_v(z)&=(-n_vq)(-n_vq-1)\dots (-n_vq-t_v+1)z^{t_v(t-1)}\\&\quad +\,b_{t_v(t-1)-1}^{*}z^{t_v(s+t-1)-1}+\dots +b_0^{*} \end{aligned}$$

and $b_e^{*}$, $e=0, \dots , t_v(t-1)-1$ are complex numbers.

We consider two cases:

Case 1.1. $f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)$ has uniqueness a zero, we denote by $z_0.$ Thus, we can write

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)=\dfrac{B(z-z_0)^l}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v} }, \end{aligned}$$

(2.29)

where $l \in \mathbb {N}^{*}$ and $B\ne 0$ is a complex number.

From (2.28), we have

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_v})^{(t_v)}&=\dfrac{A^{n+\sum _{v=1}^{k}n_v}g(z)}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v}}\nonumber \\&=\dfrac{P_2(z)}{Q_2(z)}, \end{aligned}$$

(2.30)

where $g(z)=\prod _{v=1}^{k}g_v(z)$, $\deg g \le (\sum _{v=1}^{k}t_v)(t-1).$ Similar to (2.14), we have

$$\begin{aligned} (f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})^{(d+1)}=\dfrac{A^{n+\sum _{v=1}^{k}n_v}G_2^{*}(z)}{\prod _{j=1}^{t} (z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+d+1}}, \end{aligned}$$

(2.31)

where $\deg G_2^{*}\le (\sum _{v=1}^{k}t_v+d+1)(t-1).$

Case 1.1.1. $d\ge l$.

From (2.29) and (2.30), we see

$$\begin{aligned} \deg g=\max \left\{ l, \left( n+\sum _{v=1}^{k}n_v\right) q+\left( \sum _{v=1}^{k}t_v\right) t+d\right\} . \end{aligned}$$

(2.32)

From $\deg g \le (\sum _{v=1}^{k}t_v)(t-1)$ and (2.32), we obtained

$$\begin{aligned} \left( \sum _{v=1}^{k}t_v\right) (t-1) \ge \left( n+\sum _{v=1}^{k}n_v\right) q+\left( \sum _{v=1}^{k}t_v\right) t+d. \end{aligned}$$

This is a contradiction.

Case 1.1.2. $d< l$.

If $l\ne (n+\sum _{v=1}^{k}n_v)q+(\sum _{v=1}^{k}t_v)t+d.$ From (2.29) and (2.30), we have $\deg P_2=\deg g \ge \deg Q_2.$ By argument Case 1.1.1, we get a contradiction. We have the expression as following

$$\begin{aligned}&(f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)})^{(d+1)}-(P(z))^{(d+1)}\nonumber \\&\quad =\dfrac{(z-z_0)^{l-d-1}U_{d+1}(z)}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v+d}}, \end{aligned}$$

(2.33)

where

$$\begin{aligned} U_{d+1}(z)= & {} \prod _{h=0}^{d}\left( l-\left( \left( n+\sum _{v=1}^{k}n_v\right) q\right) -\left( \sum _{v=1}^{k}t_v\right) t-h\right) z^{(d+1)t}\\&+\,x_{(d+1)t-1}z^{(d+1)t-1}+\dots +x_0, \end{aligned}$$

$x_{j}$, $j=0, \dots , (d+1)t-1$ are complex numbers.

If $l=(n+\sum _{v=1}^{k}n_v)q+(\sum _{v=1}^{k}t_v)t+d.$ From (2.31) and (2.33), we obtain

$$\begin{aligned} \deg G_2^{*}=l-d-1+\deg U_{d+1}\ge l-d-1. \end{aligned}$$

(2.34)

From (2.34) and $\deg G_2^{*} \le (\sum _{v=1}^{k}t_v+d+1)(t-1)$, we obtain

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) q&=l-\left( \sum _{v=1}^{k}t_v\right) t-d\\&\le \deg G_2^{*}-\left( \sum _{v=1}^{k}t_v\right) t+1\\&\le \left( \sum _{v=1}^{k}t_v+d+1\right) (t-1)-\left( \sum _{v=1}^{k}t_v\right) t+1\\&< (n+\sum ^k_{v=1} n_{v})q. \end{aligned}$$

This is a impossible.

Case 1.2. $f^n(f^{n_1})^{t_1}\dots (f^{n_k})^{t_k}-P(z)$ has no zeros. Thus, we can write

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}-P(z)=\dfrac{C}{\prod _{j=1}^{t}(z-b_j)^{(n+\sum _{v=1}^{k}n_v)q_j+\sum _{v=1}^{k}t_v} }, \end{aligned}$$

(2.35)

where $C\ne 0$ is a constant complex number. From (2.30) and (2.35), we have

$$\begin{aligned} \deg g=\left( n+\sum _{v=1}^{k}n_v\right) q+\left( \sum _{v=1}^{k}t_v\right) t+d. \end{aligned}$$

(2.36)

From (2.36) and $\deg g \le (\sum _{v=1}^{k}t_v)(t-1),$ we get a contradiction. $\square$

Lemma 6

Let f be a transcendental meromorphic function and $a(z)=a_dz^d+a_{d-1}z^{d-1}+\dots +a_0, d\in \mathbb {N}^{*}, a_d\ne 0, a_{d-1}, \dots , a_0$ be constant numbers complex. Let $n\in \mathbb {N}$, $k, n_v, t_v \in \mathbb {N}^{*}$, $v=1,\dots , k$ satistfy

$$\begin{aligned} n+\sum _{j=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+3. \end{aligned}$$

Then the equation

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}=a(z) \end{aligned}$$

has infinitely zeros. Furthermore, if f is a transcendental entire function, then the statement holds with

$$\begin{aligned} n+\sum _{j=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+2. \end{aligned}$$

Proof

We see $P(f)=f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}$ is a transcendental meromorphic function. By Remark 6, we have

$$\begin{aligned} T(r,f)\leqslant \frac{\theta (P)+1}{d(P)-1}\overline{N} \left( r,\frac{1}{f}\right) + \frac{1}{d(P)-1}\overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,f)). \end{aligned}$$

(2.37)

By easy computing, we have $d(P)=\sum _{v=1}^{k}n_v$, $\theta (P)=\sum _{v=1}^{k}t_v.$ From (2.37) we get

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v-\sum _{v=1}^{k}t_v-2\right) T(r,f)\leqslant \overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,f)). \end{aligned}$$

(2.38)

By $n+\sum _{v=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+3$ and (2.38), we obtain that the equation

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}=a(z) \end{aligned}$$

has infinitely zeros. In the case f is a transcendental entire function, by Lemma 3, we have

$$\begin{aligned} T(r,f)\leqslant \frac{\theta (P)+1}{d(P)}\overline{N}\left( r,\frac{1}{f}\right) +\frac{1}{d(P)}\overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,f)). \end{aligned}$$

Thus,

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v-\sum _{v=1}^{k}t_v-1\right) T(r,f)\leqslant \overline{N}\left( r,\frac{1}{P-a}\right) +o(T(r,f)). \end{aligned}$$

(2.39)

By $n+\sum _{v=1}^{k}n_v\geqslant \sum _{v=1}^{k}t_v+2$ and (2.39), we obtain that the equation

$$\begin{aligned} f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}=a(z) \end{aligned}$$

has infinitely zeros. We have completed the proof of Lemma 6. $\square$

3 Proof of Our Results

Proof of Theorem 1 and Theorem 4

First, we prove Theorem 1. Without loss of generality, we may assume that D is the unit disc and ${\mathcal {F}}$ is not normal at $z_0=0\in D.$ Then $a(0)=0$ or $a(0)\ne 0.$

Case 1. $a(0)=0$, then we may assume that $a(z) = a_mz^m + a_{m+1}z^{m+1}+\dots = z^mh(z)$, where h(z) is a holomorphic function on neighbourhood of 0,  $h(0)=a_m \ne 0$ and

$$\begin{aligned} m=\left( n+\sum _{v=1}^{k}n_v\right) s=\left( n_I+\sum _{v=1}^{k}n_{vI}\right) s. \end{aligned}$$

We consider the family $\mathcal G$ which defined as following

$$\begin{aligned} \mathcal G=\left\{ H_j: H_j(z)=\dfrac{f_j(z)}{z^s}\right\} . \end{aligned}$$

If $\mathcal G$ is not normal at $z=0,$ apply to Lemma 1, for $\alpha =\frac{\sum _{v=1}^kt_v}{n+\sum _{v=1}^kn_v}$ there exist

• (1) a real number $r, 0<r<1,$

• (2) points $z_j, |z_v|<r,$ $z_j\rightarrow 0,$

• (3) positive numbers $\rho _j\rightarrow 0^+,$

• (4) functions $H_j\in \mathcal G$

such that

$$\begin{aligned} g_j(\xi )=\frac{H_j(z_j+\rho _j\xi )}{\rho _j^\alpha }\rightarrow g(\xi ) \end{aligned}$$

(3.1)

spherically uniformly on compact subsets of $\mathbb {C},$ where $g(\xi )$ is a nonconstant meromorphic function and $g^{\#}(\xi )\leqslant g^{\#}(0)=1.$

We consider two subcases:

Case 1.1. There exists the subsequence of $\dfrac{z_j}{\rho _j}$, we also still denote by $\dfrac{z_j}{\rho _j}$ such that $\dfrac{z_j}{\rho _j} \rightarrow c \in \mathbb {C}.$ Then

$$\begin{aligned} F_j(\xi )=\dfrac{f_j(\rho _j\xi )}{\rho _j^{s+\dfrac{\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}}}&=\dfrac{\xi ^sH_j\left( z_j+\rho _j\left( \xi -\dfrac{z_j}{\rho _j}\right) \right) }{\rho _j^{\alpha }}\\&\rightarrow \xi ^sg(\xi -c):=H(\xi ).\nonumber \end{aligned}$$

(3.2)

On the other hand, we see

$$\begin{aligned} \big (F_j^{n_v}(\xi )\big )^{(t_v)}=\big (\big (\frac{f_j(\rho _j\xi )}{\rho _j^\beta }\big )^{n_v}\big )^{(t_v)} =\frac{1}{\rho _j^{n_v\beta -t_v}}\big (f_j^{n_v}\big )^{(t_v)}(\rho _j\xi ), \end{aligned}$$

where $\beta =s+\dfrac{\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}.$ From $\sum _{v=1}^{k}t_v>\sum _{v=1}^{k}t_{vI},$ we have

$$\begin{aligned}&F_j^n(F_j^{n_1})^{(t_1)}\dots (F_j^{n_k})^{(t_k)}\\&\qquad +\sum _{I}\rho _j^{\dfrac{(n_I+\sum _{v=1}^{k}n_{vI})(\sum _{v=1}^{k}t_v)-(n+\sum _{v=1}^{k}n_v)(\sum _{v=1}^{k}t_{vI})}{n+\sum _{v=1}^{k}n_v}}a_{I}(\rho _j \xi )\\&\qquad \times \,F_j^{n_I}(F_j^{n_{1I}})^{(t_{1I})}\dots (F_j^{n_{kI}})^{(t_{kI})}-\dfrac{a(\rho _j\xi )}{\rho _j^{m}}\\&\qquad =\dfrac{f_j^{n}(\rho _j\xi )(f_j^{n_1})^{(t_1)}(\rho _j\xi )\dots (f_j^{n_k})^{(t_k)}(\rho _j\xi )}{\rho _j^{m}}\\&\qquad \quad +\sum _{I}\rho _j^{\dfrac{(n_I+\sum _{v=1}^{k}n_{vI})(\sum _{v=1}^{k}t_v)-(n+\sum _{v=1}^{k}n_v)(\sum _{v=1}^{k}t_{vI})}{n+\sum _{v=1}^{k}n_v}}a_I(\rho _j\xi ) \\&\qquad \times \dfrac{\rho _j^{\sum _{v=1}^{k}t_{vI}}}{\rho _j^{(n_{I}+\sum _{v=1}^{k}n_{vI})\left( s+\dfrac{\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}\right) }}f_j^{n_I}(\rho _j\xi )(f_j^{n_{1I}})^{(t_{1I})}(\rho _j\xi ) \dots (f_j^{n_{kI}})^{(t_{kI})}(\rho _j\xi )\\&\qquad -\dfrac{a(\rho _j\xi )}{\rho _j^{m}}. \end{aligned}$$

Thus,

$$\begin{aligned}&F_j^n(F_j^{n_1})^{(t_1)}\dots (F_j^{n_k})^{(t_k)}\\&+\sum _{I}\rho _j^{\dfrac{(n_I+\sum _{v=1}^{k}n_{vI})(\sum _{v=1}^{k}t_v)-(n+\sum _{v=1}^{k}n_v)(\sum _{v=1}^{k}t_{vI})}{n+\sum _{v=1}^{k}n_v}}a_{I}(\rho _j \xi )\\&\times F_j^{n_I}(F_j^{n_{1I}})^{(t_{1I})}\dots (F_j^{n_{kI}})^{(t_{kI})}-\dfrac{a(\rho _j\xi )}{\rho _j^{m}} \end{aligned}$$

$$\begin{aligned}&=\dfrac{f_j^{n}(\rho _j\xi )(f_j^{n_1})^{(t_1)}(\rho _j\xi )\dots (f_j^{n_k})^{(t_k)}(\rho _j\xi )}{\rho _j^{m}}\nonumber \\&\quad +\sum _{I}\rho _j^{\dfrac{(n_I+\sum _{v=1}^{k}n_{vI})(\sum _{v=1}^{k}t_v)-(n+\sum _{v=1}^{k}n_v)(\sum _{v=1}^{k}t_{vI})}{n+\sum _{v=1}^{k}n_v}}a_I(\rho _j\xi ) \nonumber \\&\quad \times \, \rho _j^{\dfrac{(\sum _{v=1}^{k}t_{vI})(n+\sum _{v=1}^{k}n_v)-(n_{I}+\sum _{v=1}^{k}n_{vI})(\sum _{v=1}^{k}t_v)}{n+\sum _{v=1}^{k}n_v}}\nonumber \\&\quad \times \dfrac{f_j^{n_I}(\rho _j\xi )(f_j^{n_{1I}})^{(t_{1I})}(\rho _j\xi ) \dots (f_j^{n_{kI}})^{(t_{kI})}(\rho _j\xi )}{\rho _j^m}-\dfrac{a(\rho _j\xi )}{\rho _j^{m}}\nonumber \\&=\dfrac{f_j^{n}(\rho _j\xi )(f_j^{n_1})^{(t_1)}(\rho _j\xi )\dots (f_j^{n_k})^{(t_k)}(\rho _j\xi )}{\rho _j^{m}}\nonumber \\&\quad +\dfrac{\sum _{I}a_I(\rho _j\xi )f_j^{n_I}(\rho _j\xi )(f_j^{n_{1I}})^{(t_{1I})}(\rho _j\xi ) \dots (f_j^{n_{kI}})^{(t_{kI})}(\rho _j\xi )}{\rho _j^{m}}-\dfrac{a(\rho _j(\xi ))}{\rho _j^{m}}\nonumber \\&\rightarrow H^n(\xi )(H^{n_1})^{(t_1)}(\xi ) \dots (H^{n_k})^{(t_k)}(\xi )-a_m\xi ^m \end{aligned}$$

(3.3)

spherically uniformly on compact subsets of $\mathbb {C}{\setminus } \{\text {poles of}\quad H\},$ all zeros and poles of H are multiple at least $\Big [\dfrac{2m+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \Big ]+1.$ We see that

$$\begin{aligned} H^n(\xi )(H^{n_1})^{(t_1)}(\xi ) \dots (H^{n_k})^{(t_k)}(\xi )\not \equiv a_m\xi ^m. \end{aligned}$$

Indeed, if

$$\begin{aligned} H^n(\xi )(H^{n_1})^{(t_1)}(\xi ) \dots (H^{n_k})^{(t_k)}(\xi ) \equiv a_m\xi ^m, \end{aligned}$$

(3.4)

then H has not poles on $\mathbb {C}$ and H has a unique zero $z=0.$ Thus, from (3.4), we have

$$\begin{aligned} T(r,H^{n+\sum _{v=1}^{k}n_v}(\xi ))&=T\left( r, \dfrac{1}{a_m\xi ^m}\dfrac{(H^{n_1})^{(t_1)}(\xi )}{H^{n_1}}\dots \dfrac{(H^{n_k})^{(t_k)}(\xi )}{H^{n_k}} \right) +O(1)\\&=m\left( r, \dfrac{(H^{n_1})^{(t_1)}(\xi )}{H^{n_1}}\dots \dfrac{(H^{n_k})^{(t_k)}(\xi )}{H^{n_k}}\right) \\&\quad +\,N\left( r, \dfrac{(H^{n_1})^{(t_1)}(\xi )}{H^{n_1}}\dots \dfrac{(H^{n_k})^{(t_k)}(\xi )}{H^{n_k}}\right) +O(\log r)\\&\le O(\log r). \end{aligned}$$

Thus, H is polynomial with the form $H(z)=az^p, a\ne 0,$ where

$$\begin{aligned} p\ge & {} \Big [\dfrac{2m+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \Big ]+1>\dfrac{2m+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}\\= & {} \dfrac{s(2m+2+\sum _{v=1}^{k}t_v)}{m}>2s. \end{aligned}$$

From (3.4), we see

$$\begin{aligned} \left( n+\sum _{v=1}^{k}n_v\right) p-\sum _{v=1}^{k}t_v=m=\left( n+\sum _{v=1}^{k}n_v\right) s. \end{aligned}$$

Thus $\sum _{v=1}^{k}t_v$ divisible by $n+\sum _{v=1}^{k}n_v$, this contradicts with

$$\begin{aligned} n+\sum _{v=1}^{k}n_v>\sum _{v=1}^{k}t_v. \end{aligned}$$

Then, by Lemma 4 to Lemma 6, we see

$$\begin{aligned} H^n(\xi )(H^{n_1})^{(t_1)}(\xi ) \dots (H^{n_k})^{(t_k)}(\xi )-a_m\xi ^m \end{aligned}$$

having at least two distinct zeros, we denote by $\xi _1, \xi _2.$ Thus, there exists $\delta >0$ such that $D(\xi _1, \delta ) \cap D(\xi _2,\delta )=\varnothing .$ From (3.3), by Hurwitz’s Theorem there exist two sequences $\xi _j\rightarrow \xi _1$ and $\xi _j^{*} \rightarrow \xi _2$ satisfying

$$\begin{aligned} f_j^{n}(\rho _j\xi _j)(f_j^{n_1})^{(t_1)}(\rho _j\xi _j)&\dots (f_j^{n_k})^{(t_k)}(\rho _j\xi _j)\\&+\sum _{I}a_I(\rho _j\xi _j)f_j^{n_I}(\rho _j\xi _j)(f_j^{n_{1I}})^{(t_{1I})}(\rho _j\xi _j) \dots \\&\quad (f_j^{n_{kI}})^{(t_{kI})}(\rho _j\xi _j)=a(\rho _j\xi _j),\\ f_j^{n}(\rho _j\xi _j^{*})(f_j^{n_1})^{(t_1)}(\rho _j\xi _j^{*})&\dots (f_j^{n_k})^{(t_k)}(\rho _j\xi _j^{*})\\&+\sum _{I}a_I(\rho _j\xi _j^{*})f_j^{n_I}(\rho _j\xi _j^{*})(f_j^{n_{1I}})^{(t_{1I})}(\rho _j\xi _j^{*}) \dots \\&\quad (f_j^{n_{kI}})^{(t_{kI})}(\rho _j\xi _j^{*})=a(\rho _j\xi _j^{*}). \end{aligned}$$

By hypothesis

$$\begin{aligned}&f^n(f^{n_1})^{(t_1)}\dots (f^{n_k})^{(t_k)}+\sum _{I}a_{I}f^{n_{I}}(f^{n_{1I}})^{(t_{1I})}\dots (f^{n_{kI}})^{(t_{kI})},\\&g^n(g^{n_1})^{(t_1)}\dots (g^{n_k})^{(t_k)}+\sum _{I}a_{I}g^{n_{I}}(g^{n_{1I}})^{(t_{1I})}\dots (g^{n_{kI}})^{(t_{kI})} \end{aligned}$$

share $a(z)-$ IM in D for each pair (f, g) in ${\mathcal {F}},$ then for any $r\in \mathbb Z^{+},$ we have

$$\begin{aligned} f_r^{n}(\rho _j\xi _j)(f_r^{n_1})^{(t_1)}(\rho _j\xi _j)&\dots (f_r^{n_k})^{(t_k)}(\rho _j\xi _j)\\&+\sum _{I}a_I(\rho _j\xi _j)f_r^{n_I}(\rho _j\xi _j)(f_r^{n_{1I}})^{(t_{1I})}(\rho _j\xi _j) \dots \\&\quad (f_r^{n_{kI}})^{(t_{kI})}(\rho _j\xi _j)=a(\rho _j\xi _j),\\ f_r^{n}(\rho _j\xi _j^{*})(f_r^{n_1})^{(t_1)}(\rho _j\xi _j^{*})&\dots (f_r^{n_k})^{(t_k)}(\rho _j\xi _j^{*})\\&+\sum _{I}a_I(\rho _j\xi _j^{*})f_r^{n_I}(\rho _j\xi _j^{*})(f_r^{n_{1I}})^{(t_{1I})}(\rho _j\xi _j^{*}) \dots \\&\quad (f_r^{n_{kI}})^{(t_{kI})}(\rho _j\xi _j^{*})=a(\rho _j\xi _j^{*}). \end{aligned}$$

Fix r,  taking $j \rightarrow \infty ,$ we get

$$\begin{aligned}&f_r^{n}(0)(f_r^{n_1})^{(t_1)}(0)\dots (f_r^{n_k})^{(t_k)}(0)\\&\quad +\sum _{I}a_I(0)f_r^{n_I}(0)(f_r^{n_{1I}})^{(t_{1I})}(0) \dots (f_r^{n_{kI}})^{(t_{kI})}(0)=a(0). \end{aligned}$$

Since the zeros of

$$\begin{aligned}&f_r^{n}(z)(f_r^{n_1})^{(t_1)}(z)\dots (f_r^{n_k})^{(t_k)}(z)\\&\quad +\sum _{I}a_I(z)f_r^{n_I}(z)(f_r^{n_{1I}})^{(t_{1I})}(z) \dots (f_r^{n_{kI}})^{(t_{kI})}(z)=a(z). \end{aligned}$$

have no accumulation points, in fact we have

$$\begin{aligned} \rho _{j}\xi _{j}=\rho _j\xi _j^{*}=0 \end{aligned}$$

or equivalently

$$\begin{aligned} \xi _j=\xi _j^{*}=0. \end{aligned}$$

This contradicts with $D(\xi _1,\delta )\cap D(\xi _2,\delta )=\varnothing .$ Hence, $\mathcal G$ is a normal family at 0.

Case 1.2. There exists the subsequence of $\dfrac{z_j}{\rho _j}$, we also still denote by $\dfrac{z_j}{\rho _j}$ such that $\dfrac{z_j}{\rho _j} \rightarrow \infty .$ By definition of H,  we have

$$\begin{aligned} f_j(z_j+\rho _j\xi )=(z_j+\rho _j\xi )^{s}H_j(z_j+\rho _j\xi ). \end{aligned}$$

This implies

$$\begin{aligned} \left( f_j^{n_v}(z_j+\rho _j\xi )\right) ^{(t_v)}= \sum _{l_v=0}^{t_v}C_{t_v}^{l_v}\left[ (z_j+\rho _j\xi )^{n_vs}\right] ^{(l_v)}\left[ H_j^{n_v}(z_j+\rho _j\xi )\right] ^{(t_v-l_v)}, \end{aligned}$$

for all $v=1, \dots , k.$ Hence,

$$\begin{aligned} (f_j^{n_v})^{(t_v)}(z_j+\rho _j\xi )&=(z_j+\rho _j\xi )^{n_vs}(H_{j}^{n_v})^{(t_v)}(z_j+\rho _j\xi )\\&\quad +\sum _{l_v=1}^{t_v}b_{l_v}(z_j+\rho _j\xi )^{n_vs-l_v}(H_{j}^{n_v})^{(t_v-l_v)}(z_j+\rho _j\xi ),\nonumber \end{aligned}$$

(3.5)

for all $v=1, \dots , k$, where $b_{l_v}$, $l_v=1, \dots , t_v$ are nonzero complex numbers. We also have

$$\begin{aligned} \rho _j^{\alpha }g_j(\xi ) =H_j(z_j+\rho _j\xi ). \end{aligned}$$

Thus,

$$\begin{aligned} (H_j^{n_v})^{(l_v)}(z_j+\rho _j\xi )=\rho _j^{n_v\alpha -l_v}(g_j^{n_v})^{(l_v)}(\xi ), \end{aligned}$$

(3.6)

for all $l_v=1, \dots , t_v$, $v=1, \dots , k.$ From (3.5) and (3.6), we get

$$\begin{aligned} (f_j^{n_v})^{(t_v)}(z_j+\rho _j\xi )&=\rho _j^{n_v\alpha -t_v}(z_j+\rho _j\xi )^{n_vs}(g_j^{n_v})^{(t_v)}(\xi )\\&\quad +\sum _{l_v=1}^{t_v}b_{l_v}\rho _j^{n_v\alpha }(z_j+\rho _j\xi )^{n_vs-l_v}\rho _j^{-(t_v-l_v)}(g_{j}^{n_v})^{(t_v-l_v)}(z_j+\rho _j\xi ),\nonumber \end{aligned}$$

(3.7)

for all $v=1, \dots , k.$ Thus,

$$\begin{aligned} f_j^n&(z_j+\rho _j\xi )(f_j^{n_1})^{(t_1)}(z_j+\rho _j\xi )\dots (f_j^{n_k})^{(t_k)}(z_j+\rho _j\xi )\\&=(z_j+\rho _j\xi )^m g_j^{n}(\xi )(g_j^{n_1})^{(t_1)}(\xi )\dots (g_j^{n_k})^{(t_k)}(\xi )\nonumber \\&\quad +\sum _{0\le l_1 \le t_1, \dots , 0\le l_k\le t_k,\sum _{v=1}^{k}l_v\ge 1}C_{l_1, \dots , l_k}(z_j+\rho _j\xi )^{m}\nonumber \\&\quad \times \prod _{v=1}^{k}\dfrac{1}{\left( \dfrac{z_j}{\rho _j}+\xi \right) ^{l_v}}g_j^{n}(\xi )(g_j^{n_1})^{(t_1)}(\xi )\dots (g_j^{n_k})^{(t_k)}(\xi ).\nonumber \end{aligned}$$

(3.8)

Similar to (3.8), we also have

$$\begin{aligned} f_j^{n_{I}}&(z_j+\rho _j\xi )(f_j^{n_{1I}})^{(t_{1I})}(z_j+\rho _j\xi )\dots (f_j^{n_{kI}})^{(t_{kI})}(z_j+\rho _j\xi )\nonumber \\&=\rho _j^{(n_I+\sum _{v=1}^{k}n_{vI})\alpha -\sum _{v=1}^{k}t_{vI}}(z_j+\rho _j\xi )^m g_j^{n_I}(\xi )(g_j^{n_{1I}})^{(t_{1I})}(\xi )\dots (g_j^{n_{kI}})^{(t_{kI})}(\xi )\nonumber \\&\quad +\,\rho _j^{(n_I+\sum _{v=1}^{k}n_{vI})\alpha -\sum _{v=1}^{k}t_{vI}}\sum _{0\le l_{1I} \le t_{1I}, \dots , 0\le l_{kI}\le t_{kI},\sum _{v=1}^{k}l_{vI}\ge 1}C_{l_{1I}, \dots , l_{kI}}(z_j+\rho _j\xi )^{m}\nonumber \\&\quad \times \prod _{v=1}^{k}\dfrac{1}{\left( \dfrac{z_j}{\rho _j}+\xi \right) ^{l_{vI}}}g_j^{n_I}(\xi )(g_j^{n_{1I}})^{(t_{1I})}(\xi )\dots (g_j^{n_{kI}})^{(t_{kI})}(\xi ). \end{aligned}$$

(3.9)

From (3.8) and (3.9) and condition

$$\begin{aligned} \sum _{v=1}^{k}t_v>\sum _{v=1}^{k}t_{vI}, \end{aligned}$$

we conclude that

$$\begin{aligned}&\dfrac{a_m f_j^n(z_j+\rho _j\xi )(f_j^{n_1})^{(t_1)}(z_j+\rho _j\xi )\dots (f_j^{n_k})^{(t_k)}(z_j+\rho _j\xi )}{a(z_j+\rho _j\xi )}\nonumber \\&\qquad +\dfrac{a_m\sum _{I}f_j^{n_I}(z_j+\rho _j\xi )(f_j^{n_{1I}})^{(t_{1I})}(z_j+\rho _j\xi )\dots (f_j^{n_{kI}})^{(t_{kI})}(z_j+\rho _j\xi )}{a(z_j+\rho _j\xi )}-a_m\nonumber \\&\quad =\dfrac{a_m}{h(z_j+\rho _j\xi )}g_j^{n}(\xi )(g_j^{n_1})^{(t_1)}(\xi )\dots (g_j^{n_k})^{(t_k)}(\xi )\nonumber \\&\qquad +\dfrac{a_m}{h(z_j+\rho _j\xi )}\sum _{0\le l_1 \le t_1, \dots , 0\le l_k\le t_k,\sum _{v=1}^{k}l_v\ge 1}\\&\quad \quad C_{l_1, \dots , l_k}\prod _{v=1}^{k}\dfrac{1}{\left( \dfrac{z_j}{\rho _j}+\xi \right) ^{l_v}}g_j^{n}(\xi )(g_j^{n_1})^{(t_1)}(\xi )\dots (g_j^{n_k})^{(t_k)}(\xi )\nonumber \\&\quad +\,\rho _j^{(n_I+\sum _{v=1}^{k}n_{vI})\alpha -\sum _{v=1}^{k}t_{vI}}\dfrac{a_m}{h(z_j+\rho _j\xi )} g_j^{n_I}(\xi )(g_j^{n_{1I}})^{(t_{1I})}(\xi )\dots (g_j^{n_{kI}})^{(t_{kI})}(\xi )\nonumber \\&\quad +\,\rho _j^{(n_I+\sum _{v=1}^{k}n_{vI})\alpha -\sum _{v=1}^{k}t_{vI}}\sum _{0\le l_{1I} \le t_{1I}, \dots , 0\le l_{kI}\le t_{kI},\sum _{v=1}^{k}l_{vI}\ge 1}\dfrac{a_m}{h(z_j+\rho _j\xi )}C_{l_{1I}, \dots , l_{kI}}\nonumber \\&\quad \times \prod _{v=1}^{k}\dfrac{1}{\left( \dfrac{z_j}{\rho _j}+\xi \right) ^{l_{vI}}}g_j^{n_I}(\xi )(g_j^{n_{1I}})^{(t_{1I})}(\xi )\dots (g_j^{n_{kI}})^{(t_{kI})}(\xi )\nonumber \\&\quad \rightarrow g^{n}(\xi )(g^{n_1})^{(t_1)}(\xi )\dots (g^{n_k})^{(t_k)}(\xi )-a_m \end{aligned}$$

spherically uniformly on compact subsets of $\mathbb {C}{\setminus } \{\text {poles of }g\}.$

Claim

$g^n(\xi )(g^{n_1}(\xi ))^{(t_1)}\dots (g^{n_k}(\xi ))^{(t_k)}$ is nonconstant.

Since g is nonconstant and $n_j\ge t_j \;(j=1,\dots ,k),$ it is easy to see that $(g^{n_j}(\xi ))^{(t_j)}\not \equiv 0,$ for all $j\in \{1,\dots ,k\}.$ Hence, $g^n(\xi )(g^{n_1}(\xi ))^{(t_1)}\dots (g^{n_k}(\xi ))^{(t_k)}\not \equiv 0.$

Suppose that $g^n(\xi )(g^{n_1}(\xi ))^{(t_1)}\dots (g^{n_k}(\xi ))^{(t_k)}\equiv a,$ $a\in \mathbb {C}{\setminus } \{0\}.$ From conditions of Theorem 1, we have that in the case $n=0,$ there exists $i\in \{1,\dots ,k\}$ such that $n_i>t_i.$ Therefore, since $a\ne 0,$ it is easy to see that g is entire having no zero. So, by Lemma 2, $g(\xi )=e^{c\xi +d},\; c\ne 0.$ Then

$$\begin{aligned} g^n(\xi )(g^{n_1}(\xi ))^{(t_1)}\cdots (g^{n_k}(\xi ))^{(t_k)}&=e^{nc\xi +nd}(e^{n_1c\xi +n_1d})^{(t_1)}\cdots (e^{n_kc\xi +n_kd})^{(t_k)}\\&=(n_1c)^{t_1}\cdots (n_kc)^{t_k}e^{(n+\sum _{j=1}^kn_j)c\xi +(n+\sum _{j=1}^kn_j)d}. \end{aligned}$$

Then $(n_1c)^{t_1}\cdots (n_kc)^{t_k}e^{(n+\sum _{j=1}^kn_j)c\xi +(n+\sum _{j=1}^kn_j)d}\equiv a,$ which is impossible. So,

$$\begin{aligned} g^n(\xi )(g^{n_1}(\xi ))^{(t_1)}\dots (g^{n_k}(\xi ))^{(t_k)} \end{aligned}$$

is nonconstant.

By Lemma 4 to Lemma 6, $g^{n}(\xi )(g^{n_1})^{(t_1)}(\xi )\dots (g^{n_k})^{(t_k)}(\xi )-a_m$ has at least two distinct zeros. Similar to Case 1.1, we have $\mathcal G$ is normal at $z=0.$

Hence, there exists $\Delta _{\rho }=\{z: |z|<\rho \}$ and a subsequence $\{H_{j_k}\}$ of $\{H_j\}$ such that $H_{j_k}$ converges spherically locally uniformly to a meromorphic function U(z) or $\infty$ $(k \rightarrow \infty )$ in $\Delta _\rho$. Now, we consider two case as following:

Case (i). When k is sufficiently large, $f_{j_k}(0) \ne 0$. So $U (0) = \infty$. Then, for arbitrary constant $R > 0,$ there exists $\sigma \in (0, \rho ),$ when $z \in \Delta _{\sigma }$, we have $|U (z )| > R$. Hence, for sufficiently large k, $|H_{j_k}(z)|>\dfrac{R}{2}.$ So $\dfrac{1}{f_{j_k}}$ is holomorphic in $\Delta _{\sigma }$ and when $\sigma =\dfrac{R}{2},$ we have

$$\begin{aligned} \Big |\dfrac{1}{f_{j_k}(z)}\Big | =\Big |\dfrac{1}{z^sH_{j_k}(z)}\Big |\le \dfrac{2^{s+1}}{R\sigma ^s}. \end{aligned}$$

By maximum principle and Montel’s theorem, ${\mathcal {F}}$ is normal at $z=0.$

Case (ii). There exists a subsequence of $f_{j_k}$, still denoted as $f_{j_k}$ such that $f_{j_k}(0)=0$. Since the multiplicity of every zero of $f_{j_k}$ is at least

$$\begin{aligned} \Big [\dfrac{2m+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v} \Big ]+1>\dfrac{2m+2+\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v}=\dfrac{s(2m+2+\sum _{v=1}^{k}t_v)}{m}>2s, \end{aligned}$$

and $H_{j_k}=\dfrac{f_{j_k}}{z^s}$, then $H_{j_k}(0)=0.$ Thus, there exists $0< \omega < \rho$ such that $H_{j_k}$ is holomorphic in $\Delta _{\omega } = \{z : |z | < \omega \}$. Then $H_{j_k}$ converges spherically locally uniformly to a holomorphic function U(z) in $\Delta _{\omega }=\{z : |z | < \omega \}.$ Since $H_{j_k}(0)=0$, then $U (0) = 0$. Hence, there exists $0< r < \rho$ such that U(z) is holomorphic in $\Delta _r = \{z : |z | < r\}$ and has a unique zero $z = 0$ in $\Delta _r.$ Thus, $H_{j_k}$ converges spherically locally uniformly to a holomorphic function U(z) in $\Delta _r$, then $f_{j_k}$ converges spherically locally uniformly to a holomorphic function $z^sU (z)$ in $\Delta _r$. Hence, ${\mathcal {F}}$ is normal at $z = 0.$ From Case (i) and Case (ii), we see ${\mathcal {F}}$ is normal at 0.

Case 2. $a(0)\ne 0.$

Apply to Lemma 1 with $\alpha =\dfrac{\sum _{v=1}^{k}t_v}{n+\sum _{v=1}^{k}n_v},$ we have

$$\begin{aligned} h_j(\xi )=\frac{f_j(z_j+\rho _j\xi )}{\rho _j^{\alpha }}\rightarrow h(\xi ) \end{aligned}$$

spherically uniformly on compact subsets of $\mathbb {C},$ where $h(\xi )$ is a nonconstant meromorphic function. We have

$$\begin{aligned}&h_j^{n}(\xi )(h_j^{n_1})^{(t_{1})}(\xi )\dots (h_j^{n_{k}})^{(t_{k})}(\xi )\\&\quad +\sum _{I}\rho _{j}^{(n_{I}+\sum _{v=1}^{k}n_{vI})\alpha -\sum _{v=1}^{k}t_{vI}}a_{I}(z_{j}+\rho _{j}\xi )h_{j}^{n_{I}}(\xi )(h_{j}^{n_{1I}})^{(t_{1I})}(\xi )\dots \\&\quad (h_{j}^{n_{kI}})^{(t_{kI})}(\xi )-a(z_j+\rho _j\xi )\\&\quad =f_{j}^{n}({{z}_{j}}+{{\rho }_{j}}\xi ){{(f_{j}^{{{n}_{1}}})}^{({{t}_{1}})}}({{z}_{j}}+{{\rho }_{j}}\xi )\dots {{(f_{j}^{{{n}_{k}}})}^{({{t}_{k}})}}({{z}_{j}}+{{\rho }_{j}}\xi )\\&\quad +\sum _{I}a_{I}(z_{j}+\rho _{j}\xi )f_{j}^{n_{I}}(z_{j}+\rho _{j}\xi )(f_{j}^{n_{1I}})^{(t_{1I})}(z_{j}+\rho _{j}\xi )\dots \\&\quad (f_{j}^{n_{kI}})^{(t_{kI})}(z_{j}+\rho _{j}\xi )-a(z_j+\rho _j\xi ).\\ \end{aligned}$$

By the condition

$$\begin{aligned} \sum _{v=1}^{k}t_{vI}<\sum _{v=1}^{k}t_{v}, \end{aligned}$$

we get

$$\begin{aligned} h^{n}(\xi )({h^{n_{1}}(\xi )})^{(t_{1})}\dots ({h^{n_{k}}(\xi )})^{(t_{k})}-a(0) \end{aligned}$$

is the uniform limit (with metric spherical) of

$$\begin{aligned}&f_{j}^{n}({{z}_{j}}+{{\rho }_{j}}\xi ){{(f_{j}^{{{n}_{1}}})}^{({{t}_{1}})}}({{z}_{j}}+{{\rho }_{j}}\xi )\dots {{(f_{j}^{{{n}_{k}}})}^{({{t}_{k}})}}({{z}_{j}}+{{\rho }_{j}}\xi )\\&\quad +\sum _{I}a_{I}(z_{j}+\rho _{j}\xi )f_{j}^{n_{I}}(z_{j}+\rho _{j}\xi )(f_{j}^{n_{1I}})^{(t_{1I})}(z_{j}+\rho _{j}\xi )\dots (f_{j}^{n_{kI}})^{(t_{kI})}(z_{j}+\rho _{j}\xi )\\&\quad -\,a(z_j+\rho _j\xi ) \end{aligned}$$

on each compact subset of $\mathbb {C}{\setminus } \{ \text {pole of } \quad h\}.$ By Lemma 4 to Lemma 6 and Lemma 2, the equation

$$\begin{aligned} h^{n}(\xi )({h^{n_{1}}(\xi )})^{(t_{1})}\dots ({h^{n_{k}}(\xi )})^{(t_{k})}=a(0). \end{aligned}$$

has at least two distinct zeros $\xi _1 \ne \xi _2.$ By argument as case 1.1, we get a contradiction.

By an argument of Theorem 1, we are easy to prove Theorem 4. $\square$