# Normal families of meromorphic functions sharing one function

## Abstract

Suppose $p\left(z\right)$ is a holomorphic function, the multiplicity of its zeros is at most d, $P\left(z\right)$ is a nonconstant polynomial. Let be a family of meromorphic functions in a domain D, all of whose zeros and poles have multiplicity at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$. If for each pair of functions f and g in , $P\left(f\right){f}^{\left(k\right)}$ and $P\left(g\right){g}^{\left(k\right)}$ share a holomorphic function $p\left(z\right)$, then is normal in D. It generalizes and extends the results of Jiang, Gao and Wu, Xu.

MSC:30D35, 30D45.

## 1 Introduction and results

Let D be a domain in , let be a family of meromorphic functions in D. is said to be normal in D, in the sense of Montel, if for any sequence $\left\{{f}_{n}\right\}\in \mathcal{F}$ contains a subsequence $\left\{{f}_{nj}\right\}$ such that ${f}_{nj}$ converges spherically locally uniformly in D to a meromorphic function or ∞ [13].

Let $a\in \mathbb{C}\cup \left\{\mathrm{\infty }\right\}$, let f and g be two nonconstant meromorphic functions in D. If $f\left(z\right)-a$ and $g\left(z\right)-a$ have the same zeros (ignoring multiplicity), we say f and g share the value a in D.

In 1959, Hayman [1] proved that if f is a transcendental meromorphic function, then ${f}^{n}{f}^{\prime }$ assumes every finite nonzero complex number infinitely often for any positive integer $n\ge 3$. He [4] conjectured that this remains valid for $n=1$ and $n=2$. Further, the case of $n=2$ was confirmed by Mues [5] in 1979. The case $n=1$ was considered and settled by Clunie [6].

In 1994, Yang and Yang [7] proposed a conjecture: If f is an entire function and $k\ge 2$, then ${\left(f{f}^{\left(k\right)}\right)}^{n}-a\left(z\right)$ ($a\left(z\right)\not\equiv 0$) has infinitely many zeros.

Zhang and Song [8] proved the following theorem.

Theorem A Suppose that f is a transcendental meromorphic function, n, k are two positive integers, then when $n\ge 2$, ${\left(f{f}^{\left(k\right)}\right)}^{n}-a\left(z\right)$ has infinitely many zeros, where $a\left(z\right)\not\equiv 0$ is a small function of f.

In 2005, Wang [9] proved the following theorem.

Theorem B Let f be a transcendental meromorphic function, let $c\left(z\right)\not\equiv 0$ be a small function of f, and let n, k be two positive integers. If $n\ge 3$, then ${f}^{n}{f}^{\left(k\right)}-c\left(z\right)$ has infinitely many zeros.

In the case of $f{f}^{\left(k\right)}$, Yang and Yang [7] proposed a conjecture: If f is transcendental, then $f{f}^{\left(k\right)}$ assumes every finite nonzero complex number infinitely often. In 2006, Wang [10] proved that this conjecture holds when f has only zeros of multiplicity at least $k+1$ ($k\ge 2$).

In 2011, Meng and Hu [11] obtained the following theorem.

Theorem C Take a positive integer k and a nonzero complex number a. Let be a family of meromorphic functions in a domain $D\in \mathbb{C}$ such that each $f\in \mathcal{F}$ has only zeros of multiplicity at least $k+1$. For each pair $\left(f,g\right)\in \mathcal{F}$, if $f{f}^{\left(k\right)}$ and $g{g}^{\left(k\right)}$ share a, then is normal in D.

In 2011, Jiang and Gao [12] obtained the following theorem.

Theorem D Suppose that d (≥0) is an integer, $p\left(z\right)$ is an analytic function in D, and the multiplicity of its all zeros is at most d. Let be a family of meromorphic functions in D, let n be a positive integer. If $n\ge 2d+2$ and for each pair of functions f and g in , ${f}^{n}{f}^{\prime }$ and ${g}^{n}{g}^{\prime }$ share $p\left(z\right)$ in D, then is normal in D.

In 2012, Wu and Xu [13] got the following theorem.

Theorem E Let k be a positive integer, let $b\ne 0$ be a finite complex number, let P be a polynomial with either $degP\ge 3$ or $degP=2$ and P having only one distinct zero, and let be a family of meromorphic functions in D, all of whose zeros have multiplicity at least k. If for each pair of functions f and g in , $P\left(f\right){f}^{\left(k\right)}$ and $P\left(g\right){g}^{\left(k\right)}$ share b in D, then is normal in D.

It is natural to ask whether Theorem E can be improved by the idea of sharing a holomorphic function. In this paper, we study the problem and obtain the following theorems.

Theorem 1.1 Suppose that $d\ge 0$ is an integer, $p\left(z\right)\not\equiv 0$ is a holomorphic function in D, and the multiplicity of its all zeros is at most d. Let be a family of meromorphic functions in D, the multiplicity of all zeros and poles of $f\in \mathcal{F}$ is at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$. If for each pair of functions f and g in , $f{f}^{\left(k\right)}$ and $g{g}^{\left(k\right)}$ share $p\left(z\right)$ in D, then is normal in D.

Remark 1.1 Theorem 1.1 still holds when $p\left(z\right)$ is a nonzero finite constant.

Theorem 1.2 Suppose that $d\ge 0$ is an integer, $p\left(z\right)\not\equiv 0$ is a holomorphic function in D, and the multiplicity of its all zeros is at most d. Let P be a nonconstant polynomial, be a family of meromorphic functions in D, the multiplicity of all zeros and poles of $f\in \mathcal{F}$ is at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$. If for each pair of functions f and g in , $P\left(f\right){f}^{\left(k\right)}$ and $P\left(g\right){g}^{\left(k\right)}$ share $p\left(z\right)$ in D, then is normal in D.

## 2 Some lemmas

Lemma 2.1 (see [14])

Let k be a positive integer, let be a family of meromorphic functions in D such that each function $f\in \mathcal{F}$ has only zeros with multiplicities at least k, and suppose that there exists $A\ge 1$ such that $|{f}^{\left(k\right)}\left(z\right)|\le A$ whenever $f\left(z\right)=0$, $f\in \mathcal{F}$. If is not normal at ${z}_{0}\in D$, then for each α, $0\le \alpha \le k$, there exists a sequence of complex numbers ${z}_{n}\in D$, ${z}_{n}\to {z}_{0}$, a sequence of positive numbers ${\rho }_{n}\to 0$, and a sequence of functions ${f}_{n}\in \mathcal{F}$ such that

${g}_{n}\left(\xi \right)=\frac{{f}_{n}\left({z}_{n}+{\rho }_{n}\xi \right)}{{\rho }_{n}^{\alpha }}\to g\left(\xi \right)$

locally uniformly with respect to the spherical metric, where g is a nonconstant meromorphic function on , all of whose zeros have multiplicity at least k, such that ${g}^{\mathrm{♯}}\left(\xi \right)\le {g}^{\mathrm{♯}}\left(0\right)=kA+1$. Moreover, $g\left(\xi \right)$ has order at most 2.

Lemma 2.2 (see [15])

Let $f\left(z\right)$ be a meromorphic function and k be a positive integer. If ${f}^{\left(k\right)}\not\equiv 0$, then

$N\left(r,\frac{1}{{f}^{\left(k\right)}}\right)\le N\left(r,\frac{1}{f}\right)+k\overline{N}\left(r,f\right)+S\left(r,f\right).$

Lemma 2.3 (see [1])

Let ${f}_{1}\left(z\right)$, ${f}_{2}\left(z\right)$ be two meromorphic functions defined in $D=\left\{z:|z|, then

$N\left(r,{f}_{1}{f}_{2}\right)-N\left(r,\frac{1}{{f}_{1}{f}_{2}}\right)=N\left(r,{f}_{1}\right)+N\left(r,{f}_{2}\right)-N\left(r,\frac{1}{{f}_{1}}\right)-N\left(r,\frac{1}{{f}_{2}}\right).$

Lemma 2.4 (see [16])

Let f be a transcendental meromorphic function, let ${P}_{f}\left(z\right)$, ${Q}_{f}\left(z\right)$ be two differential polynomials of f. If ${f}^{n}{P}_{f}={Q}_{f}$ holds and the degree of ${Q}_{f}$ is at most n, then $m\left(r,{P}_{f}\right)=S\left(r,f\right)$.

Lemma 2.5 Let d (≥0) be an integer, let k be a positive integer, and let $p\left(z\right)={a}_{d}{z}^{d}+{a}_{d-1}{z}^{d-1}+\cdots +{a}_{1}z+{a}_{0}$ be a polynomial, where ${a}_{d}\ne 0$, ${a}_{d-1},\dots ,{a}_{0}$ are constants. Suppose that f is a transcendental meromorphic function, all of whose zeros and poles have multiplicity at least 2, $p\left(z\right)$ is a small function of $f\left(z\right)$, then $f{f}^{\left(k\right)}\left(z\right)-p\left(z\right)$ has infinitely many zeros.

Proof Let

$\psi \left(z\right)=f{f}^{\left(k\right)}-p\left(z\right).$
(2.1)

Suppose $f{f}^{\left(k\right)}-p\left(z\right)$ has only finitely many zeros, then $N\left(r,\frac{1}{\psi \left(z\right)}\right)=S\left(r,f\right)$. By (2.1), then

${\left(\frac{\psi }{p}\right)}^{\prime }=\frac{{f}^{\prime }{f}^{\left(k\right)}}{p}+\frac{f{f}^{\left(k+1\right)}}{p}+{\left(\frac{1}{p}\right)}^{\prime }f{f}^{\left(k\right)}.$
(2.2)

Let

${\psi }_{1}=\frac{\psi }{p}.$

Since the multiplicity of zeros of $f\left(z\right)$ is at least 2, we can get from (2.2) that

$N\left(r,\frac{1}{f}\right)\le N\left(r,\frac{1}{{\left({\psi }_{1}\right)}^{\prime }}\right)+S\left(r,f\right).$
(2.3)

By Lemma 2.2, we know that

$N\left(r,\frac{1}{{\left({\psi }_{1}\right)}^{\prime }}\right)\le N\left(r,\frac{1}{{\psi }_{1}}\right)+\overline{N}\left(r,f\right)+S\left(r,f\right).$
(2.4)

We can get from (2.2) that

$\frac{f{f}^{\left(k\right)}}{p}\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}-\frac{{f}^{\prime }{f}^{\left(k\right)}}{p}-\frac{f{f}^{\left(k+1\right)}}{p}-{\left(\frac{1}{p}\right)}^{\prime }f{f}^{\left(k\right)}=\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}},$

i.e.,

$f\left(\frac{{f}^{\left(k\right)}}{p}\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}-\frac{{f}^{\prime }{f}^{\left(k\right)}}{fp}-\frac{{f}^{\left(k+1\right)}}{p}-{\left(\frac{1}{p}\right)}^{\prime }{f}^{\left(k\right)}\right)=\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}.$
(2.5)

Let

$fH=\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}},$
(2.6)

where $H=\frac{{f}^{\left(k\right)}}{p}\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}-\frac{{f}^{\prime }{f}^{\left(k\right)}}{fp}-\frac{{f}^{\left(k+1\right)}}{p}-{\left(\frac{1}{p}\right)}^{\prime }{f}^{\left(k\right)}$. By Lemma 2.4, we get $m\left(r,H\right)=S\left(r,f\right)$.

From (2.5) and Lemma 2.3, we obtain that

$\begin{array}{rcl}m\left(r,\frac{1}{f}\right)& \le & m\left(r,H\right)+m\left(r,\frac{{\psi }_{1}}{{\left({\psi }_{1}\right)}^{\prime }}\right)\\ \le & N\left(r,\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}\right)-N\left(r,\frac{{\psi }_{1}}{{\left({\psi }_{1}\right)}^{\prime }}\right)+m\left(r,\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}\right)+S\left(r,f\right)\\ \le & N\left(r,{\left({\psi }_{1}\right)}^{\prime }\right)+N\left(r,\frac{1}{{\psi }_{1}}\right)-N\left(r,\frac{1}{{\left({\psi }_{1}\right)}^{\prime }}\right)-N\left(r,{\psi }_{1}\right)+S\left(r,f\right)\\ \le & \overline{N}\left(r,f\right)+N\left(r,\frac{1}{\psi }\right)-N\left(r,\frac{1}{{\left({\psi }_{1}\right)}^{\prime }}\right)+S\left(r,f\right).\end{array}$
(2.7)

We can get from (2.6) that

$f{f}^{\left(k\right)}\left(\frac{{\left({\psi }_{1}\right)}^{\prime }}{p{\psi }_{1}}-\frac{{f}^{\prime }}{fp}-\frac{{f}^{\left(k+1\right)}}{p{f}^{\left(k\right)}}-{\left(\frac{1}{p}\right)}^{\prime }\right)=\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}.$
(2.8)

Let

$f{f}^{\left(k\right)}G=\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}},$
(2.9)

where $G=\frac{{\left({\psi }_{1}\right)}^{\prime }}{p{\psi }_{1}}-\frac{{f}^{\prime }}{fp}-\frac{{f}^{\left(k+1\right)}}{p{f}^{\left(k\right)}}-{\left(\frac{1}{p}\right)}^{\prime }$. By Lemma 2.4, then $m\left(r,G\right)=S\left(r,f\right)$.

By (2.9), we have that

$\begin{array}{rcl}m\left(r,{f}^{\left(k\right)}\right)& \le & m\left(r,\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}\right)+m\left(r,\frac{1}{f}\right)+m\left(r,\frac{1}{G}\right)\\ \le & m\left(r,\frac{1}{f}\right)+N\left(r,G\right)-N\left(r,\frac{1}{G}\right)+S\left(r,f\right).\end{array}$
(2.10)

Since $\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}$ has only simple poles, and by (2.9) we know that the poles of f are impossible G’s. Hence the poles of G are only possible from the zeros and poles of $p\left(z\right)$ or the zeros of ${\psi }_{1}$, f and ${f}^{\left(k\right)}$.

Hence by (2.8) and (2.9), we obtain that

$\begin{array}{rcl}N\left(r,G\right)& \le & \overline{N}\left(r,\frac{1}{{\psi }_{1}}\right)+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+N\left(r,\frac{1}{p}\right)\\ \le & \overline{N}\left(r,\frac{1}{\psi }\right)+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+S\left(r,f\right).\end{array}$
(2.11)

Since $\frac{{\left({\psi }_{1}\right)}^{\prime }}{{\psi }_{1}}$ has only simple poles, so by (2.9) we know that

$N\left(r,\frac{1}{G}\right)\ge N\left(r,f\right)+N\left(r,{f}^{\left(k\right)}\right)-\overline{N}\left(r,f\right).$
(2.12)

Combining (2.7) and (2.10)-(2.12), we have

$\begin{array}{rcl}m\left(r,{f}^{\left(k\right)}\right)& \le & \left\{\overline{N}\left(r,\frac{1}{\psi }\right)+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)\right\}-\left\{N\left(r,f\right)+N\left(r,{f}^{\left(k\right)}\right)-\overline{N}\left(r,f\right)\right\}\\ +\left\{\overline{N}\left(r,f\right)+N\left(r,\frac{1}{\psi }\right)-N\left(r,\frac{1}{{\left({\psi }_{1}\right)}^{\prime }}\right)\right\}+S\left(r,f\right).\end{array}$

Hence

$\begin{array}{rcl}T\left(r,{f}^{\left(k\right)}\right)& \le & \overline{N}\left(r,\frac{1}{\psi }\right)+\overline{N}\left(r,\frac{1}{f}\right)+\overline{N}\left(r,\frac{1}{{f}^{\left(k\right)}}\right)-N\left(r,f\right)+\overline{N}\left(r,f\right)\\ +\overline{N}\left(r,f\right)+N\left(r,\frac{1}{\psi }\right)-N\left(r,\frac{1}{{\left({\psi }_{1}\right)}^{\prime }}\right)+S\left(r,f\right).\end{array}$

Since the multiplicity of the zeros and poles of $f\left(z\right)$ is at least 2, by an elementary calculation and combing with Lemma 2.2, (2.3) and (2.4), the above inequality yields

$\begin{array}{rcl}T\left(r,{f}^{\left(k\right)}\right)& \le & N\left(r,\frac{1}{{f}^{\left(k\right)}}\right)+2N\left(r,\frac{1}{\psi }\right)+S\left(r,f\right)\\ \le & N\left(r,\frac{1}{f}\right)+k\overline{N}\left(r,f\right)+2N\left(r,\frac{1}{\psi }\right)+S\left(r,f\right)\\ \le & \left(k+1\right)\overline{N}\left(r,f\right)+3N\left(r,\frac{1}{\psi }\right)+S\left(r,f\right).\end{array}$
(2.13)

Since the multiplicity of the poles of $f\left(z\right)$ is at least 2, we can get from (2.13) that

$\begin{array}{rcl}T\left(r,{f}^{\left(k\right)}\right)& \le & \left(1-\frac{1}{k+2}\right)N\left(r,{f}^{\left(k\right)}\right)+3N\left(r,\frac{1}{\psi }\right)+S\left(r,f\right)\\ \le & \left(1-\frac{1}{k+2}\right)N\left(r,{f}^{\left(k\right)}\right)+S\left(r,f\right).\end{array}$

This implies $T\left(r,{f}^{\left(k\right)}\right)=S\left(r,f\right)$, then ${f}^{\left(k\right)}$ is a rational function, thus f is a rational function which contradicts with f is transcendental. Hence $f{f}^{\left(k\right)}\left(z\right)-p\left(z\right)$ has infinitely many zeros. □

Remark 2.1 When $p\left(z\right)$ is a nonzero finite constant or a small function of $f\left(z\right)$, similarly we can get the same conclusion.

Lemma 2.6 Let d (≥0) be an integer, let k be a positive integer, and let $p\left(z\right)={a}_{d}{z}^{d}+{a}_{d-1}{z}^{d-1}+\cdots +{a}_{1}z+{a}_{0}$ be a polynomial, where ${a}_{d}\ne 0$, ${a}_{d-1},\dots ,{a}_{0}$ are constants. If $f\left(z\right)$ is a nonconstant polynomial, all of whose zeros and poles have multiplicity at least $k+d$, then $f{f}^{\left(k\right)}\left(z\right)-p\left(z\right)$ has at least two distinct zeros, and $f{f}^{\left(k\right)}\left(z\right)-p\left(z\right)\not\equiv 0$.

Proof We discuss the following two cases.

Case 1. If $f{f}^{\left(k\right)}-p\left(z\right)\ne 0$, then $f{f}^{\left(k\right)}-p\left(z\right)\equiv C$, where C is a nonzero constant. So $f{f}^{\left(k\right)}\equiv p\left(z\right)+C$. Since the multiplicity of all the zeros of f is at least $k+d$, thus $deg\left(f{f}^{\left(k\right)}\right)\ge k+2d$, which contradicts with $deg\left(p\left(z\right)\right)=d$.

Case 2. If $f{f}^{\left(k\right)}-p\left(z\right)$ has only one zero ${z}_{0}$, we assume $f{f}^{\left(k\right)}-p\left(z\right)\equiv A{\left(z-{z}_{0}\right)}^{l}$, where A is a nonzero constant, l is a positive integer.

We discuss the following two cases.

1. (i)

If $l\le d+1$, then $f{f}^{\left(k\right)}\equiv p\left(z\right)+A{\left(z-{z}_{0}\right)}^{l}$. Since $deg\left(f{f}^{\left(k\right)}\right)\ge k+2d$, the degree of the right of the equation is at most $d+1$, which is smaller than the degree of the left of the equation. We get a contradiction.

2. (ii)

If $l>d+1$, then $f{f}^{\left(k\right)}\equiv p\left(z\right)+A{\left(z-{z}_{0}\right)}^{l}$. So ${\left(f{f}^{\left(k\right)}\right)}^{\left(d\right)}\equiv {a}_{d}+Al\cdots \left(l-d+1\right){\left(z-{z}_{0}\right)}^{l-d}$. Since ${a}_{d}\ne 0$, so ${\left(f{f}^{\left(k\right)}\right)}^{\left(d\right)}$ has only simple zeros, which contradicts with the multiplicity of all the zeros of f is at least $k+d$.

By Case 1 and Case 2, $f{f}^{\left(k\right)}-p\left(z\right)$ has at least two distinct zeros.

If $f{f}^{\left(k\right)}-p\left(z\right)\equiv 0$, then similar to the proof of Case 1, we get a contradiction. Hence $f{f}^{\left(k\right)}-p\left(z\right)\not\equiv 0$. □

Lemma 2.7 Let d (≥0) be an integer, let k be a positive integer, and let $p\left(z\right)={a}_{d}{z}^{d}+{a}_{d-1}{z}^{d-1}+\cdots +{a}_{1}z+{a}_{0}$ be a polynomial, where ${a}_{d}\ne 0$, ${a}_{d-1},\dots ,{a}_{0}$ are constants. If $f\left(z\right)$ is a nonconstant rational function and not a polynomial, and the multiplicity of whose zeros and poles is at least $\frac{k}{2}+d+1$, then $f{f}^{\left(k\right)}\left(z\right)-p\left(z\right)$ has at least two distinct zeros, and $f{f}^{\left(k\right)}\left(z\right)-p\left(z\right)\not\equiv 0$.

Proof Since $f\left(z\right)$ is a nonconstant rational function and not a polynomial, then obviously $f{f}^{\left(k\right)}\left(z\right)-p\left(z\right)\not\equiv 0$. Let

$f\left(z\right)=B\frac{{\left(z-{\alpha }_{1}\right)}^{{m}_{1}}{\left(z-{\alpha }_{2}\right)}^{{m}_{2}}\cdots {\left(z-{\alpha }_{s}\right)}^{{m}_{s}}}{{\left(z-{\beta }_{1}\right)}^{{n}_{1}}{\left(z-{\beta }_{2}\right)}^{{n}_{2}}\cdots {\left(z-{\beta }_{t}\right)}^{{n}_{t}}},$
(2.14)

where B is a nonzero constant. Since the multiplicity of the zeros and poles of f is at least $\frac{k}{2}+d+1$, we have ${m}_{i}\ge \frac{k}{2}+d+1$ ($i=1,2,\dots ,s$), ${n}_{j}\ge \frac{k}{2}+d+1$ ($j=1,2,\dots ,t$). For simplicity, we denote

${m}_{1}+{m}_{2}+\cdots +{m}_{s}=m\ge \left(\frac{k}{2}+d+1\right)s,\phantom{\rule{2em}{0ex}}{n}_{1}+{n}_{2}+\cdots +{n}_{t}=n\ge \left(\frac{k}{2}+d+1\right)t.$

By (2.19), we get

${f}^{\left(k\right)}\left(z\right)=B\frac{{\left(z-{\alpha }_{1}\right)}^{{m}_{1}-k}{\left(z-{\alpha }_{2}\right)}^{{m}_{2}-k}\cdots {\left(z-{\alpha }_{s}\right)}^{{m}_{s}-k}g\left(z\right)}{{\left(z-{\beta }_{1}\right)}^{{n}_{1}+k}{\left(z-{\beta }_{2}\right)}^{{n}_{2}+k}\cdots {\left(z-{\beta }_{t}\right)}^{{n}_{t}+k}},$
(2.15)

where $g\left(z\right)=\left(m-n\right)\left(m-n-1\right)\cdots \left(m-n-k+1\right){z}^{k\left(s+t-1\right)}+\cdots +{c}_{1}z+{c}_{0}$ is a polynomial, ${c}_{i}$ ($i=0,1$) are constants and $deg\left(g\left(z\right)\right)\le k\left(s+t-1\right)$. Thus (2.14) together with (2.15) implies

$f{f}^{\left(k\right)}\left(z\right)=B\frac{{\left(z-{\alpha }_{1}\right)}^{2{m}_{1}-k}{\left(z-{\alpha }_{2}\right)}^{2{m}_{2}-k}\cdots {\left(z-{\alpha }_{s}\right)}^{2{m}_{s}-k}g\left(z\right)}{{\left(z-{\beta }_{1}\right)}^{2{n}_{1}+k}{\left(z-{\beta }_{2}\right)}^{2{n}_{2}+k}\cdots {\left(z-{\beta }_{t}\right)}^{2{n}_{t}+k}}.$
(2.16)

By (2.16), we obtain

${\left(f{f}^{\left(k\right)}\left(z\right)\right)}^{\left(d+1\right)}=B\frac{{\left(z-{\alpha }_{1}\right)}^{2{m}_{1}-k-d-1}{\left(z-{\alpha }_{2}\right)}^{2{m}_{2}-k-d-1}\cdots {\left(z-{\alpha }_{s}\right)}^{2{m}_{s}-k-d-1}{g}_{1}\left(z\right)}{{\left(z-{\beta }_{1}\right)}^{2{n}_{1}+k+d+1}{\left(z-{\beta }_{2}\right)}^{2{n}_{2}+k+d+1}\cdots {\left(z-{\beta }_{t}\right)}^{2{n}_{t}+k+d+1}},$
(2.17)

where $deg\left({g}_{1}\left(z\right)\right)\le \left(k+d+1\right)\left(s+t-1\right)$.

Next, we discuss the following two cases.

Case 1. If $f{f}^{\left(k\right)}-p\left(z\right)$ has only one zero ${z}_{0}$, then let

$f{f}^{\left(k\right)}\left(z\right)-p\left(z\right)=C\frac{{\left(z-{z}_{0}\right)}^{l}}{{\left(z-{\beta }_{1}\right)}^{2{n}_{1}+k}{\left(z-{\beta }_{2}\right)}^{2{n}_{2}+k}\cdots {\left(z-{\beta }_{t}\right)}^{2{n}_{t}+k}}.$
(2.18)

Subcase 1.1. When $d\ge l$.

Combining (2.16) and (2.18), we get $d+2n+kt=deg\left(g\left(z\right)\right)+2m-ks\le k\left(s+t-1\right)+2m-ks$. That is, $2\left(m-n\right)\ge k+d$, and then $m>n$.

Differentiating both sides of (2.18), we have

$\begin{array}{r}{\left(f{f}^{\left(k\right)}\left(z\right)\right)}^{\left(d+1\right)}-{p}^{\left(d+1\right)}\left(z\right)\\ \phantom{\rule{1em}{0ex}}=C\frac{{g}_{2}\left(z\right)}{{\left(z-{\beta }_{1}\right)}^{2{n}_{1}+k+d+1}{\left(z-{\beta }_{2}\right)}^{2{n}_{2}+k+d+1}\cdots {\left(z-{\beta }_{t}\right)}^{2{n}_{t}+k+d+1}},\end{array}$
(2.19)

where $deg\left({g}_{2}\left(z\right)\right)\le t\left(d+1\right)+l-d-1$.

By (2.17) and (2.19), we know $2m-\left(k+d+1\right)s\le deg\left({g}_{2}\left(z\right)\right)\le t\left(d+1\right)+l-d-1$. Thus $2m-\left(k+d+1\right)s-t\left(d+1\right)\le l-d-1$. Since $2m-\left(k+d+1\right)s-t\left(d+1\right)\ge$ $2m-\left(k+d+1\right)\frac{m}{k/2+d+1}-\left(d+1\right)\frac{n}{k/2+d+1}>0$, then $0, which contradicts with $d\ge l$.

Subcase 1.2. When $d.

Differentiating both sides of (2.18), we have

$\begin{array}{r}{\left(f{f}^{\left(k\right)}\left(z\right)\right)}^{\left(d+1\right)}-{p}^{\left(d+1\right)}\left(z\right)\\ \phantom{\rule{1em}{0ex}}=D\frac{{\left(z-{z}_{0}\right)}^{l-d-1}{g}_{3}\left(z\right)}{{\left(z-{\beta }_{1}\right)}^{2{n}_{1}+k+d+1}{\left(z-{\beta }_{2}\right)}^{2{n}_{2}+k+d+1}\cdots {\left(z-{\beta }_{t}\right)}^{2{n}_{t}+k+d+1}},\end{array}$
(2.20)

where ${g}_{3}\left(z\right)=\left(l-2n-kt\right)\cdots \left(l-2n-kt-d\right){z}^{t\left(d+1\right)}+\cdots +{d}_{1}z+{d}_{0}$, where ${d}_{i}$ ($i=0,1$) are constants and $deg\left({g}_{3}\left(z\right)\right)\le t\left(d+1\right)$.

Differentiating both sides of (2.18) step by step for d times, we have ${z}_{0}$ is a zero of ${\left(f{f}^{\left(k\right)}\left(z\right)\right)}^{\left(d\right)}-{p}^{\left(d\right)}\left(z\right)$, as ${p}^{\left(d\right)}\left(z\right)={a}_{d}\ne 0$ and the multiplicity of all the zeros of $f\left(z\right)$ is at least $\frac{k}{2}+d+1$, thus ${\alpha }_{i}\ne {z}_{0}$ ($i=1,2,\dots ,s$). When $p\left(z\right)$ is a constant, from (2.18) we can also get ${\alpha }_{i}\ne {z}_{0}$ ($i=1,2,\dots ,s$).

Here, we discuss three subcases as follows.

Subcase 1.2.1. When $l<2n+kt+d$.

Combining (2.16) and (2.18), we get $d+2n+kt=deg\left(g\left(z\right)\right)+2m-ks\le k\left(s+t-1\right)+2m-ks$. That is, $2\left(m-n\right)\ge k+d$, and then $m>n$.

Since ${\alpha }_{i}\ne {z}_{0}$ ($i=1,2,\dots ,s$), by (2.17) and (2.20), we have $t\left(d+1\right)\ge deg\left({g}_{3}\left(z\right)\right)\ge 2m-\left(k+d+1\right)s$. Thus $2m\le \left(k+d+1\right)s+t\left(d+1\right)\le \left(k+d+1\right)\frac{m}{k/2+d+1}+\left(d+1\right)\frac{n}{k/2+d+1}<2m$, which is impossible.

Subcase 1.2.2. When $l=2n+kt+d$.

If $m>n$, by a similar discussion to Subcase 1.2.1, we can get a contradiction. Thus $m\le n$. Since ${\alpha }_{i}\ne {z}_{0}$ ($i=1,2,\dots ,s$), by (2.17) and (2.20), we have $l-d-1\le deg\left({g}_{1}\left(z\right)\right)\le \left(k+d+1\right)\left(s+t-1\right)$, since $l=2n+kt+d$, thus $2n+kt+d-d-1\le \left(k+d+1\right)\left(s+t-1\right)$. Then $2n\le \left(k+d+1\right)s+\left(d+1\right)t-\left(k+d\right)<\left(k+d+1\right)\frac{m}{k/2+d+1}+\left(d+1\right)\frac{n}{k/2+d+1}\le 2n$, which is impossible.

Subcase 1.2.3. When $l>2n+kt+d$.

By (2.16) and (2.18), we get $l=deg\left(g\left(z\right)\right)+2m-ks\le k\left(s+t-1\right)+2m-ks=2m+kt-k$. If $m>n$, by a similar discussion to Subcase 1.2.1, we get a contradiction. Thus $m\le n$.

Case 2. If $f{f}^{\left(k\right)}-p\left(z\right)$ has no zero. Then $l=0$ in (2.18), by a similar discussion to Subcase 1.1, we get a contradiction.

By Case 1 and Case 2, we get $f{f}^{\left(k\right)}-p\left(z\right)$ has at least two distinct zeros. □

## 3 Proof of Theorem 1.1

For any point ${z}_{0}$ in D, either $p\left({z}_{0}\right)=0$ or $p\left({z}_{0}\right)\ne 0$.

Case 1. When $p\left({z}_{0}\right)=0$. We may assume ${z}_{0}=0$. Then $p\left(z\right)={a}_{d}{z}^{d}+{a}_{d+1}{z}^{d+1}+\cdots ={z}^{d}h\left(z\right)$, where ${a}_{d}\left(\ne 0\right),{a}_{d+1},\dots$ are constants, $d\ge 1$, $h\left({z}_{0}\right)\ne 0$, without loss of generality, let $h\left({z}_{0}\right)={a}_{d}$, where $h\left(z\right)$ is a holomorphic function.

Let ${\mathcal{F}}_{1}=\left\{{F}_{j}|{F}_{j}=\frac{{f}_{j}}{{z}^{d/2}},{f}_{j}\in \mathcal{F}\right\}$. If ${\mathcal{F}}_{1}$ is not normal at 0, then by Lemma 2.1, there exists a sequence of complex numbers ${z}_{j}\to 0$, a sequence of positive numbers ${\rho }_{j}\to 0$ and a sequence of functions ${F}_{j}\in {\mathcal{F}}_{1}$ such that ${G}_{j}\left(\xi \right)={\rho }_{j}^{-\frac{k}{2}}{F}_{j}\left({z}_{j}+{\rho }_{j}\xi \right)\to G\left(\xi \right)$ spherically locally uniformly in , where $G\left(\xi \right)$ is a nonconstant meromorphic function in , and the multiplicity of the zeros and poles of $G\left(\xi \right)$ is at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$. Here, we discuss two cases as follows.

Case 1.1. There exists a subsequence of $\frac{{z}_{j}}{{\rho }_{j}}$, we may denote it as $\frac{{z}_{j}}{{\rho }_{j}}$ such that $\frac{{z}_{j}}{{\rho }_{j}}\to c$, c is a finite complex number. Then

${\varphi }_{j}\left(\xi \right)=\frac{{f}_{j}\left({\rho }_{j}\xi \right)}{{\rho }_{j}^{\frac{d+k}{2}}}=\frac{{\left({\rho }_{j}\xi \right)}^{\frac{d}{2}}{F}_{j}\left({z}_{j}+{\rho }_{j}\left(\xi -\frac{{z}_{j}}{{\rho }_{j}}\right)\right)}{{\rho }_{j}^{\frac{d}{2}}{\rho }_{j}^{\frac{k}{2}}}\to {\xi }^{\frac{d}{2}}G\left(\xi -c\right)=H\left(\xi \right)$

spherically locally uniformly in , so

${\varphi }_{j}\left(\xi \right){\varphi }_{j}^{\left(k\right)}\left(\xi \right)-\frac{p\left({\rho }_{j}\xi \right)}{{\rho }_{j}^{d}}=\frac{{f}_{j}\left({\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({\rho }_{j}\xi \right)-p\left({\rho }_{j}\xi \right)}{{\rho }_{j}^{d}}\to H\left(\xi \right){H}^{\left(k\right)}\left(\xi \right)-{a}_{d}{\xi }^{d}$

spherically locally uniformly in .

Since $\mathrm{\forall }f\in \mathcal{F}$, the multiplicity of whose zeros and poles is at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$, then the multiplicity of all zeros and poles of H is at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$, by Lemmas 2.5-2.7, we get $H\left(\xi \right){H}^{\left(k\right)}\left(\xi \right)-{a}_{d}{\xi }^{d}\not\equiv 0$, and $H\left(\xi \right){H}^{\left(k\right)}\left(\xi \right)-{a}_{d}{\xi }^{d}$ has at least two distinct zeros.

Suppose ${\xi }_{0}$, ${\xi }_{0}^{\ast }$ are two distinct zeros of $H\left(\xi \right){H}^{\left(k\right)}\left(\xi \right)-{a}_{d}{\xi }^{d}$. We may choose a proper $\sigma >0$ such that $D\left({\xi }_{0},\sigma \right)\cap D\left({\xi }_{0}^{\ast },\sigma \right)=\mathrm{\varnothing }$, where $D\left({\xi }_{0},\sigma \right)=\left\{\xi ||\xi -{\xi }_{0}|<\sigma \right\}$, $D\left({\xi }_{0}^{\ast },\sigma \right)=\left\{\xi ||\xi -{\xi }_{0}^{\ast }|<\sigma \right\}$.

By Hurwitz’s theorem, there exists a subsequence of ${f}_{j}\left({\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({\rho }_{j}\xi \right)-p\left({\rho }_{j}\xi \right)$, we may still denote it as ${f}_{j}\left({\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({\rho }_{j}\xi \right)-p\left({\rho }_{j}\xi \right)$, then there exist points ${\xi }_{j}\in D\left({\xi }_{0},\sigma \right)$ and points ${\xi }_{j}^{\ast }\in D\left({\xi }_{0}^{\ast },\sigma \right)$ such that for sufficiently large j, ${f}_{j}\left({\rho }_{j}{\xi }_{j}\right){f}_{j}^{\left(k\right)}\left({\rho }_{j}{\xi }_{j}\right)-p\left({\rho }_{j}{\xi }_{j}\right)=0$, ${f}_{j}\left({\rho }_{j}{\xi }_{j}^{\ast }\right){f}_{j}^{\left(k\right)}\left({\rho }_{j}{\xi }_{j}^{\ast }\right)-p\left({\rho }_{j}{\xi }_{j}^{\ast }\right)=0$.

Since ${f}_{j}{f}_{j}^{\left(k\right)}$ and ${g}_{j}{g}_{j}^{\left(k\right)}$ share $p\left(z\right)$ in D, it follows that for any positive integer m, ${f}_{m}\left({\rho }_{j}{\xi }_{j}\right){f}_{m}^{\left(k\right)}\left({\rho }_{j}{\xi }_{j}\right)-p\left({\rho }_{j}{\xi }_{j}\right)=0$, ${f}_{m}\left({\rho }_{j}{\xi }_{j}^{\ast }\right){f}_{m}^{\left(k\right)}\left({\rho }_{j}{\xi }_{j}^{\ast }\right)-p\left({\rho }_{j}{\xi }_{j}^{\ast }\right)=0$.

Fix m, let $j\to \mathrm{\infty }$ and note ${\rho }_{j}{\xi }_{j}\to 0$, ${\rho }_{j}{\xi }_{j}^{\ast }\to 0$, we obtain ${f}_{m}\left(0\right){f}_{m}^{\left(k\right)}\left(0\right)-p\left(0\right)=0$.

Since the zeros of ${f}_{m}\left(0\right){f}_{m}^{\left(k\right)}\left(0\right)-p\left(0\right)$ have no accumulation points, in fact when j is large enough, we have ${\rho }_{j}{\xi }_{j}={\rho }_{j}{\xi }_{j}^{\ast }=0$. Thus, when j is large enough, ${\xi }_{0}={\xi }_{0}^{\ast }=0$, which contradicts with $D\left({\xi }_{0},\sigma \right)\cap D\left({\xi }_{0}^{\ast },\sigma \right)=\mathrm{\varnothing }$. Thus, ${\mathcal{F}}_{1}$ is normal at 0.

Case 1.2. There exists a subsequence of $\frac{{z}_{j}}{{\rho }_{j}}$, we may denote it as $\frac{{z}_{j}}{{\rho }_{j}}$ such that $\frac{{z}_{j}}{{\rho }_{j}}\to \mathrm{\infty }$. Then

$\begin{array}{rcl}{f}_{j}\left({z}_{j}+{\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({z}_{j}+{\rho }_{j}\xi \right)& =& {\left({z}_{j}+{\rho }_{j}\xi \right)}^{\frac{d}{2}}{F}_{j}\left({z}_{j}+{\rho }_{j}\xi \right)\left[{\left({z}_{j}+{\rho }_{j}\xi \right)}^{\frac{d}{2}}{\left({F}_{j}\left({z}_{j}+{\rho }_{j}\xi \right)\right)}^{\left(k\right)}\\ +\sum _{i=1}^{k}{c}_{i}{\left({z}_{j}+{\rho }_{j}\xi \right)}^{\frac{d}{2}-i}{\left({F}_{j}\left({z}_{j}+{\rho }_{j}\xi \right)\right)}^{\left(k-i\right)}\right]\\ =& {\left({z}_{j}+{\rho }_{j}\xi \right)}^{d}{G}_{j}\left(\xi \right){G}_{j}^{\left(k\right)}\left(\xi \right)+\sum _{i=1}^{k}{c}_{i}{\left({z}_{j}+{\rho }_{j}\xi \right)}^{d-i}{\rho }_{j}^{i}{G}_{j}\left(\xi \right){G}_{j}^{\left(k-i\right)}\left(\xi \right),\end{array}$

where ${c}_{i}=\frac{d}{2}\left(\frac{d}{2}-1\right)\cdots \left(\frac{d}{2}-i+1\right){C}_{d/2}^{i}$ when $\frac{d}{2}\ge i$, and ${c}_{i}=0$ when $\frac{d}{2}.

Thus, we have

$\begin{array}{r}\frac{{a}_{d}{f}_{j}\left({z}_{j}+{\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({z}_{j}+{\rho }_{j}\xi \right)}{p\left({z}_{j}+{\rho }_{j}\xi \right)}-{a}_{d}\\ \phantom{\rule{1em}{0ex}}=\left({G}_{j}\left(\xi \right){G}_{j}^{\left(k\right)}\left(\xi \right)+\sum _{i=1}^{k}{c}_{i}\frac{{G}_{j}\left(\xi \right){G}_{j}^{\left(k-i\right)}\left(\xi \right)}{{\left(\frac{{z}_{j}}{{\rho }_{j}}+\xi \right)}^{i}}\right)\frac{{a}_{d}}{h\left({z}_{j}+{\rho }_{j}\xi \right)}-{a}_{d}\\ \phantom{\rule{1em}{0ex}}\to G\left(\xi \right){G}^{\left(k\right)}\left(\xi \right)-{a}_{d},\end{array}$

spherically locally uniformly in $\mathbb{C}-\left\{\xi |G\left(\xi \right)=\mathrm{\infty }\right\}$.

Since the multiplicity of all zeros and poles of G is at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$ and by Lemmas 2.5-2.7, we have $G\left(\xi \right){G}^{\left(k\right)}\left(\xi \right)-{a}_{d}\not\equiv 0$, and $G\left(\xi \right){G}^{\left(k\right)}\left(\xi \right)-{a}_{d}$ has at least two distinct zeros.

Suppose ${\xi }_{1}$, ${\xi }_{1}^{\ast }$ are two distinct zeros of $G\left(\xi \right){G}^{\left(k\right)}\left(\xi \right)-{a}_{d}$. We may choose a proper $\delta >0$ such that $D\left({\xi }_{1},\delta \right)\cap D\left({\xi }_{1}^{\ast },\delta \right)=\mathrm{\varnothing }$, where $D\left({\xi }_{1},\delta \right)=\left\{\xi ||\xi -{\xi }_{1}|<\sigma \right\}$, $D\left({\xi }_{1}^{\ast },\delta \right)=\left\{\xi ||\xi -{\xi }_{1}^{\ast }|<\delta \right\}$.

By Hurwitz’s theorem, there exists a subsequence of ${a}_{d}{f}_{j}\left({z}_{j}+{\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({z}_{j}+{\rho }_{j}\xi \right)-{a}_{d}p\left({z}_{j}+{\rho }_{j}\xi \right)$, we may still denote it as ${a}_{d}{f}_{j}\left({z}_{j}+{\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({z}_{j}+{\rho }_{j}\xi \right)-{a}_{d}p\left({z}_{j}+{\rho }_{j}\xi \right)$, then there exist points ${\xi }_{j}\in D\left({\xi }_{1},\delta \right)$ and points ${\xi }_{j}^{\ast }\in D\left({\xi }_{1}^{\ast },\delta \right)$ such that for sufficiently large j, ${a}_{d}{f}_{j}\left({z}_{j}+{\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({z}_{j}+{\rho }_{j}\xi \right)-{a}_{d}p\left({z}_{j}+{\rho }_{j}\xi \right)=0$, ${a}_{d}{f}_{j}\left({z}_{j}+{\rho }_{j}\xi \right){f}_{j}^{\left(k\right)}\left({z}_{j}+{\rho }_{j}\xi \right)-{a}_{d}p\left({z}_{j}+{\rho }_{j}\xi \right)=0$.

Similar to the proof of Case 1.1, we get a contradiction. Then ${\mathcal{F}}_{1}$ is normal at 0.

By Case 1.1 and Case 1.2, we know ${\mathcal{F}}_{1}$ is normal at 0. Hence there exists ${\mathrm{△}}_{\rho }=\left\{z:|z|<\rho \right\}$ and a subsequence of ${F}_{jk}$ of ${F}_{j}$ such that ${F}_{jk}$ converges spherically locally uniformly to a meromorphic function $F\left(z\right)$ or ∞ ($k\to \mathrm{\infty }$) in ${\mathrm{△}}_{\rho }$.

Here, we discuss the following two cases.

Case i. When k is large enough, ${f}_{jk}\ne 0$. Then $F\left(0\right)=\mathrm{\infty }$. Thus, for constant $R>0$, $\mathrm{\exists }\sigma \in \left(0,\rho \right)$, we have $|F\left(z\right)|>R$ when $z\in {\mathrm{△}}_{\rho }$. Thus, for sufficiently large k, $|{F}_{jk}\left(z\right)|>\frac{R}{2}$, $\frac{1}{{f}_{jk}}$ is a holomorphic function in ${\mathrm{△}}_{\rho }$. Hence when $|z|=\frac{\sigma }{2}$,

$|\frac{1}{{f}_{jk}}|=|\frac{1}{{F}_{jk}{z}^{d/2}}|\le \frac{{2}^{d/2+1}}{R{\sigma }^{d/2}}.$

By the maximum principle and Montel’s theorem, is normal at $z=0$.

Case ii. There exists a subsequence of ${f}_{jk}$, we may still denote it as ${f}_{jk}$, such that ${f}_{jk}\left(0\right)=0$. Since $\mathrm{\forall }f\in \mathcal{F}$, the multiplicity of whose zeros is at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$, then $F\left(0\right)=0$. Thus, there exists $0 such that $F\left(z\right)$ is holomorphic in ${\mathrm{△}}_{r}=\left\{z:|z| and has a unique zero $z=0$ in ${\mathrm{△}}_{r}$. Then ${F}_{jk}$ converges spherically locally uniformly to a holomorphic function $F\left(z\right)$ in ${\mathrm{△}}_{r}$, ${f}_{jk}$ converges spherically locally uniformly to a holomorphic function ${z}^{\frac{d}{2}}F\left(z\right)$ in ${\mathrm{△}}_{r}$. Hence is normal at $z=0$.

By Case i and Case ii, we obtain is normal at $z=0$.

Case 2. When $p\left({z}_{0}\right)\ne 0$.

Suppose that is not normal at ${z}_{0}$. Then by Lemma 2.1, there exists a sequence of complex numbers ${z}_{t}\to {z}_{0}$, a sequence of positive numbers ${\rho }_{t}\to 0$ and a sequence of functions ${f}_{t}\in \mathcal{F}$ such that ${g}_{t}\left(\xi \right)={\rho }_{t}^{-\frac{k}{2}}{f}_{t}\left({z}_{t}+{\rho }_{t}\xi \right)\to g\left(\xi \right)$ spherically locally uniformly in , where $g\left(\xi \right)$ is a nonconstant meromorphic function in , and the multiplicity of the zeros and poles of $g\left(\xi \right)$ is at least $max\left\{\frac{k}{2}+d+1,k+d\right\}$.

Hence by Lemmas 2.5-2.7, we have $g\left(\xi \right){g}^{\left(k\right)}\left(\xi \right)-p\left({z}_{0}\right)\not\equiv 0$, and $g\left(\xi \right){g}^{\left(k\right)}\left(\xi \right)-p\left({z}_{0}\right)$ has at least two distinct zeros. Similar to the proof of Case 1.1, we get a contradiction. Thus, is normal at ${z}_{0}$.

Hence, is normal in D as ${z}_{0}$ is arbitrary. The proof is complete.

## 4 Proof of Theorem 1.2

Because $P\left(z\right)$ has at least one zero, we may assume, with no loss of generality, that $P\left(z\right)={z}^{n}+{a}_{n-1}{z}^{n-1}+\cdots +{a}_{q}{z}^{q}$, where $q\ge 1$ is a positive integer and ${a}_{q}\ne 0$. Suppose that is not normal in D. Then similar to the proof of Theorem 1.1, we can get a contradiction. Hence is normal in D. The proof is complete.

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## Acknowledgements

The authors wish to thank the referees and editors for their very helpful comments and useful suggestions.

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Correspondence to FeiFei Hu.

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The authors declare that they have no competing interests.

### Authors’ contributions

LQ and FH performed and drafted manuscript. All authors read and approved the final manuscript.

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Qiu, L., Hu, F. Normal families of meromorphic functions sharing one function. J Inequal Appl 2013, 288 (2013). https://doi.org/10.1186/1029-242X-2013-288