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\title{Power of a meromorphic function that share a set with its derivative}
\author{Bikash Chakraborty}
\address{Ramakrishna Mission Vivekananda Centenary College, \\ Department of Mathematics\\
Kolkata-700 118,\\
India}
\email{bikashchakraborty.math@yahoo.com, bikashchakrabortyy@gmail.com}
%
\subjclass{30D35}
\keywords{Meromorphic function, weighted set sharing, uniqueness.}
\begin{abstract} In this article, we deal with the problem of the uniqueness of the power of a meromorphic function with its derivative counterpart sharing a set and thus improve our recent result under some constraints.
\end{abstract}
\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction and Definitions}
In this article, we assume that readers are familiar with basic Nevanlinna theory(\cite{3}). By $\mathbb{C}$ and $\mathbb{N}$, we mean the set of complex numbers and the set of natural numbers respectively.\par
Let $f$ and $g$ be two non-constant meromorphic functions and let $a$ be a finite complex number. If $f-a$ and $g-a$ have the same zeros with the same multiplicities, then we say that $f$ and $g$ share the value $a$ in counting multiplicities (in short, CM). Similarly, we say that $f$ and $g$ share the value $a$ in ignoring multiplicities (in short, IM), provided that $f-a$ and $g-a$ have the same set of zeros, where the multiplicities are not taken into account.\par
Also, we say that $f$ and $g$ share $\infty$ CM (resp. IM), if $1/f$ and $1/g$ share $0$ CM (resp. IM).\par
Next we shortly recall the notion of weighted sharing which appeared in the literature in 2001 (\cite{4}) as  scaling between IM sharing to CM sharing.
\begin{defi} (\cite{4}) Let $k$ be a non-negative integer or infinity. For $a\in\mathbb{C}\cup\{\infty\}$, we denote by $E_{k}(a;f)$, the set of all $a$-points of $f$, where an $a$-point of multiplicity $m$ is counted $m$ times if $m\leq k$ and $k+1$ times if $m>k$.\\ If $E_{k}(a;f)=E_{k}(a;g)$, we say that $f$ and $g$ share the value $a$ with weight $k$.
\end{defi}
We write $f$, $g$ share $(a,k)$ to mean that $f$, $g$ share the value $a$ with weight $k$. Clearly $f$ and $g$ share a value $a$ IM (resp. CM) if and only if $f$ and $g$ share $(a,0)$ (resp. $(a,\infty)$).
\begin{defi}(\cite{4}) Let $S\subset \mathbb{C}\cup\{\infty\}$ and $k$ be a non-negative integer or $\infty$. We denote by $E_{f}(S,k)$, the set $\cup_{a\in S}E_{k}(a;f)$.\par If $E_{f}(S,k)=E_{g}(S,k)$, then we say $f$, $g$ share the set $S$ with weight $k$. \end{defi}
\begin{defi} A set $S\subset \mathbb{C}\cup\{\infty\}$ is called a unique range set for meromorphic functions with weight $k$ (in short, $URSM_{k}$), if for any two non-constant meromorphic functions $f$ and $g$, $E_{f}(S,k)=E_{g}(S,k)$ implies $f\equiv g$.\par
Similarly, one can define unique range set for entire functions with weight $k$ (in brief, $URSE_{k}$).
\end{defi}
Next we recall following two definitions:
\begin{defi} (\cite{2}) Let $z_{0}$ be a zero of $f-a$ of multiplicity $p$ and a zero of $g-a$ of multiplicity $q$.
\begin{enumerate}
\item [i)] We denote by $\ol N_{L}(r,a;f)$, the counting function of those $a$-points of $f$ and $g$ where $p>q\geq 1$,
\item [ii)] by $N^{1)}_{E}(r,a;f)$, the counting function of those $a$-points of $f$ and $g$ where $p=q=1$ and
\item [iii)] by $\ol N^{(2}_{E}(r,a;f)$, the counting function of those $a$-points of $f$ and $g$ where $p=q\geq 2$, each point in these counting functions is counted only once.
\end{enumerate}
In the same way, we can define $\ol N_{L}(r,a;g),\; N^{1)}_{E}(r,a;g),\; \ol N^{(2}_{E}(r,a;g).$
\end{defi}
\begin{defi} (\cite{2}) Let $f$ and $g$ share a value $a$ IM. We denote by $\ol N_{*}(r,a;f,g)$, the reduced counting function of those $a$-points of $f$ whose multiplicities differ from the multiplicities of the corresponding $a$-points of $g$.\par
Clearly $\ol N_{*}(r,a;f,g)\equiv\ol N_{*}(r,a;g,f)$ and $\ol N_{*}(r,a;f,g)=\ol N_{L}(r,a;f)+\ol N_{L}(r,a;g)$.
\end{defi}
The subject of sharing values between entire functions and their derivatives was first studied by Rubel and Yang (\cite{7}).
In 1977, they proved that if non-constant entire functions $f$ and $f^{(1)}$ share two distinct finite numbers $a$, $b$ CM, then $f \equiv f^{(1)}$.\par
Later, in 1979, analogous result for IM sharing was obtained by Mues and Steinmetz (\cite{6}) in the following manner:
\begin{theoA}(\cite{6}) Let $f$ be a non-constant entire function. If $f$ and $f^{(1)}$ share two distinct values $a$, $b$ IM, then $f\equiv f^{(1)}$.
\end{theoA}
In the direction of value sharing and uniqueness problem, Yang and Zhang (\cite{7.1}) were the first authors to consider the uniqueness of a power of a meromorphic (resp. entire) function $F = f^{m}$ and its derivative $F^{(1)}$ as:
\begin{theoB}(\cite{7.1}) Let $f$ be a non-constant entire (resp. meromorphic) function and $m>7$ (resp. $12)$ be an integer. If $F$ and $F^{(1)}$ share $1$ CM, then $F = F^{(1)}$, and $f$ assumes the form
$$f(z) = ce^{\frac{z}{m}},$$ where $c$ is a nonzero constant.\end{theoB}
In this direction, Zhang (\cite{8}) further improved the above result in the following manner:
\begin{theoC}(\cite{8})  Let $f$ be a non-constant entire function, $m$, $k$ be positive integers and $a(z)(\not\equiv 0,\infty)$ be a small function of $f$. Suppose $f^{m} -a$ and $(f^{m})^{(k)}-a$ share the value $0$ CM and $m > k + 4$.
Then $f^m\equiv (f^m)^{(k)}$ and $f$ assumes the form $$f(z) = ce^{\frac{\lambda}{m}z},$$ where $c$ is a nonzero constant and $\lambda^k = 1$.
\end{theoC}
Afterwards, there were many improvements and generalizations concerning the uniqueness of $f^{m}$ and $(f^{m})^{(k)}$. But all authors paid their attention on value sharing or small function sharing, not on set sharing problem. Thus the natural curiosity will be:
\begin{question}\label{q1} Is it possible to change the \enquote{value sharing notion} into \enquote{set sharing notion}  in {\it Theorem C} keeping the conclusions same?
\end{question}
In connection to Question \ref{q1}, recently we considered the uniqueness of $f$ and $f^{(k)}$ when they share a set $S$ instead of a value $a (\not=0,\infty)$. To discuss the results in (\cite{2.1}), we first introduce the polynomial of Lin and Yi (\cite{4.1}).
\be\label{e5.1} P(z)=az^{n}-n(n-1)z^{2}+2n(n-2)b z-(n-1)(n-2)b^{2},\ee where $n\geq 3$ is an integer and $a$ and $b$ are two nonzero complex numbers satisfying $ab^{n-2}\not=2$. Clearly the polynomial $P(z)$ has only simple zeros.\\
In (\cite{2.1}), we considered the uniqueness of $f$ and $f^{(k)}$ when they share a set.
\begin{theoE}(\cite{2.1})  Let $n(\geq 8),~k(\geq1)$ be two positive integers and $f$ be a non-constant meromorphic function.
Suppose that $S=\{z :P(z)=0 \}$ where $P(z)$ is defined by (\ref{e5.1}). If $E_{f}(S,3)=E_{f^{(k)}}(S,3)$, then $f \equiv f^{(k)}$.
\end{theoE}
%In that paper (\cite{2.1}), \emph{it was also asked that can the cardinality of the set $S$ in Theorem E be further reduced without imposing any constraints
%on the functions}?\par
But in this paper, we will see that if we impose some restrictions on $f$, then the cardinality of the set $S$ defined in Theorem E will be reduced remarkably.\par Thus our main goal is \emph{to reduce the cardinality of this particular set $S$ and to establish the uniqueness of the power of a meromorphic function with its derivative counterpart sharing the set $S$}.\par
The method of proving of the main result of this paper is from (\cite{2.1, 2.11}).
\section{Main Result}
\begin{theo}\label{thB1} Let $f$ be a non-constant meromorphic function and $n(\geq 6)$, $k(\geq1)$ and $m(\geq k+1)$ be three positive integers. Suppose that $S=\{z :P(z)=0 \}$ where $P(z)$ is defined by (\ref{e5.1}). If $E_{f^{m}}(S,3)=E_{(f^{m})^{(k)}}(S,3)$, then $f^{m} \equiv (f^{m})^{(k)}$ and hence $f$ takes the form $$f(z)=ce^{\frac{\zeta}{m}z},$$ where $c$ is a non-zero constant and $\zeta^{k}=1$.
\end{theo}
The following example shows that for a non-constant entire function the set $S$ in Theorem \ref{thB1} can not be replaced by an arbitrary set containing six distinct elements.
\begin{exm} (\cite{2.1})
For a non-zero complex number $a$, let $S=\{a\omega, a\sqrt{\omega},a, \frac{a}{\sqrt{\omega}},\frac{a}{\omega}, \frac{a}{\omega \sqrt{\omega}}\}$, where $\omega$ is the non-real cubic root of unity. Choosing $f(z)=e^{^{\frac{{\omega}^{^{\frac{1}{2k}}}}{m}}z}$ (taking the principal branch when $m\geq 2$), it is easy to verify that $f^{m}$ and $(f^{m})^{(k)}$ share $(S,\infty)$, but $f^{m}\not \equiv (f^m)^{(k)}$
\end{exm}
\begin{rem} However the following questions are still unknown to us:
\begin{enumerate}
\item [i)] Is it possible to omit the condition $m\geq k+1$ keeping the condition $n(\geq 6)$ same in Theorem \ref{thB1}?
\item [ii)] Under the same conditions of Theorem \ref{thB1}, is it possible to further reduce the cardinality of $S$?
\end{enumerate}
\end{rem}
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\section{Auxiliary Lemmas}
Before going to discuss the necessary lemmas, we recall a well known auxiliary function as
\bea\label{h} H=\left(\frac{F''}{F'}-\frac{2F'}{F-1}\right)-\left(\frac{G''}{G'}-\frac{2G'}{G-1}\right),\eea
where $F:=R(f^{m})$ , $G:=R\left((f^{m})^{(k)}\right)$ and the expression $\frac{a(\ast)^{n}}{n(n-1)(\ast-\alpha_{1})(\ast-\alpha_{2})}$ is denoted by $R(\ast)$.\par  In addition, in the expression of $R(\ast)$, we choose $\alpha_{1}$ and $\alpha_{2}$ as the distinct roots of the equation $n(n-1)z^{2}-2n(n-2)bz+(n-1)(n-2)b^{2}=0.$
\begin{lem}\label{abc212}(\cite{1}) Let $Q(z)=(n-1)^{2}(z^{n}-1)(z^{n-2}-1)-n(n-2)(z^{n-1}-1)^{2}.$ Then $$Q(z)=(z-1)^{4}\prod\limits_{i=1}^{2n-6}(z-\beta_{i}),$$
where $\beta_{i} \in \mathbb{C}\setminus\{0,1\} (i=1,2,...,2n-6),$ which are distinct.
\end{lem}
\begin{lem}\label{l.n.1}
 Let $F$ and $G$ share $(1,l)$ where $F$ and $G$ defined as earlier, then
\begin{enumerate}
\item [i)] $\overline{N}_{L}(r,1;F)\leq \mu\left(\overline{N}(r,0;f)+\overline{N}(r,\infty;f)\right)+S(r,f),$
\item [ii)] $\overline{N}_{L}(r,1;G)\leq \mu \left(\overline{N}(r,0; (f^{m})^{k})+\overline{N}(r,\infty;f)\right)+S(r,f)$,
\end{enumerate}
where $\mu=\min\{\frac{1}{l},1\}$.
\end{lem}
\begin{proof} The proofs are similar to the proof of Lemma 2.2 of (\cite{2.1}). So we omit the details.
\end{proof}
\begin{lem}\label{bc121} Suppose that $F$ and $G$ share $(1,l)$ and $F \not\equiv G$. If $m\geq k+1$, then
\beas\label{bc111} \ol{N}(r,0;f)\leq \ol{N}(r,0,(f^m)^{(k)}) &\leq& \frac{2\mu+1}{\eta-2\mu}\overline{N}(r,\infty;f)+\frac{2}{\eta-2\mu}T(r)+S(r),\eeas
where  $T(r)=T(r,f^m)+T\left(r,(f^m)^{(k)}\right)$, $S(r)=S(r,f)$ and $\eta=(m-k)n-1.$
\end{lem}
\begin{proof} For the proof, we define $U:=\frac{F'}{(F-1)}-\frac{G'}{(G-1)}$ and consider two cases:\\
\textbf{Case - 1} Assume $U\equiv 0$. Then by integration, we get $$F-1=B(G-1).$$
If $z_{0}$ is a zero of $f$, then $B=1$ which is impossible, thus $\ol{N}(r,0;f)=S(r,f)$. Hence the result holds.\\
\textbf{Case - 2} Next we assume that $U\not\equiv 0$.\par
If $z_{0}$ is a zero of $f$ of order $t$, then it is a zero of $F$ of order $mtn$ and that of $G$ is of order $(mt-k)n$. Hence $z_{0}$ is a zero of $U$ of order at least $\eta=(m-k)n-1.$ Thus

\beas \ol{N}(r,0;f)&\leq& \ol{N}\left(r,0,(f^m)^{(k)}\right)\\
&\leq& \frac{1}{\eta}N(r,0;U)\leq \frac{1}{\eta}N(r,\infty;U)+S(r,f)\\
&\leq& \frac{1}{\eta}\{\ol{N}_{L}(r,1;F)+\ol{N}_{L}(r,1;G)+\ol{N}_{L}(r,\infty;F)+\ol{N}_{L}(r,\infty;G)\\
&&+ \ol{N}(r,\infty;G|F\neq \infty)+\ol{N}(r,\infty;F|G\neq \infty)\}+S(r,f)\\
&\leq& \frac{1}{\eta}\{\mu\left(\overline{N}(r,0;f)+\overline{N}(r,0;(f^{m})^{(k)})+2\overline{N}(r,\infty;f)\right)\\
&&+ \overline{N}(r,\infty;f)+\overline{N}(r,\alpha_{1};f^m)+\overline{N}(r,\alpha_{2};f^m)\\
&&+ \overline{N}(r,\alpha_{1};(f^m)^{(k)})+\overline{N}(r,\alpha_{2};(f^m)^{(k)}\}+S(r,f)\\
&\leq& \frac{1}{\eta}\{2\mu\overline{N}(r,0;(f^{m})^{(k)})+(2\mu+1)\overline{N}(r,\infty;f)+2T(r)\}+S(r).
\eeas
Thus
\beas \ol{N}(r,0;f)\leq \ol{N}(r,0,(f^m)^{(k)}) &\leq& \frac{2\mu+1}{\eta-2\mu}\overline{N}(r,\infty;f)+\frac{2}{\eta-2\mu}T(r)+S(r).\eeas
Hence the proof of the lemma is completed.
\end{proof}
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\begin{lem}\label{bc123} Let $F$ and $G$ share $(1,l)$ and $F \not\equiv G$. If $m\geq k+1$, then
\bea\label{el.1}
\ol{N}(r,\infty;f)\leq \frac{\mu(\eta-2\mu)+2}{(\lambda-2\mu)(\eta-2\mu)-(2\mu+1)}T(r)+S(r),\eea

where $T(r)=T(r,f^m)+T\left(r,(f^m)^{(k)}\right)$ and $S(r)=S(r,f)$, $\lambda=m(n-2)-1$ and $\mu=\min\{\frac{1}{l},1\}$.
\end{lem}
\begin{proof} For the proof, we define $V:=\frac{F'}{F(F-1)}-\frac{G'}{G(G-1)}$ and consider two cases:\\
\textbf{Case - 1} Assume that $V\equiv 0$. Then by integration, we get $$(1-\frac{1}{F})=A(1-\frac{1}{G}).$$
As $f^{m}$ and $(f^{m})^{(k)}$ share $(\infty,0)$, so if $\overline{N}(r,\infty;f)\neq S(r,f)$, then $A=1$, i.e., $F=G$, which is not possible. So $\overline{N}(r,\infty;f)=S(r,f).$ Thus the lemma holds.\\
\textbf{Case - 2} Next we assume that $V\not\equiv 0$.\par
If $z_{0}$ is a pole of $f$ of order $p$, then it is a pole of $(f^{m})^{(k)}$ of order $(pm+k)$ and that of $F$ and $G$ are $pm(n-2)$ and $(pm+k)(n-2)$ respectively.\par
Hence $z_{0}$ is a zero of $(\frac{F'}{F-1}-\frac{F'}{F})$ of order at least $pm(n-2)-1$ and a zero of $V$ of order at least $\lambda$ where $\lambda=m(n-2)-1$.\par
Since the zeros of $F$ comes from zeros of $f^{m}$ and that of $G$ comes from zeros of $(f^{m})^{(k)}$, so for the points where $f=0$, each zero of $F$ will be of larger multiplicities than that of $G$. Consequently
\beas &&\ol N_{*}(r,0;F,G)+\ol N(r,0;G\mid F\not=0)\\
&\leq& \ol N_{L}(r,0;F)+\ol N_{L}(r,0;G)+\ol N(r,0;G\mid F\not=0)\\
&\leq& \ol N_{L}(r,0;F)+\ol N(r,0;G\mid F\not=0)\leq\ol{N}(r,0;G).\eeas
Thus \beas \overline{N}(r,\infty;f) &\leq& \frac{1}{\lambda}N(r,0;V)\\
&\leq& \frac{1}{\lambda}N(r,\infty;V)+S(r,f)\\
&\leq& \frac{1}{\lambda}\{\overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)+\overline{N}_{L}(r,0;F)\\
&&+ \overline{N}_{L}(r,0;G)+\ol{N}(r,0;G| F\neq 0\}+S(r,f)\\
&\leq& \frac{1}{\lambda}\{\overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)+\overline{N}(r,0;G)\}+S(r,f)\\
&\leq& \frac{1}{\lambda}\{\mu\left(\overline{N}(r,0;f)+\overline{N}(r,0;(f^{m})^{(k)})+2\overline{N}(r,\infty;f)\right)\\
&&+\overline{N}(r,0;(f^{m})^{(k)})\}+S(r,f).
\eeas
Now using Lemma \ref{bc121}, we get
\beas (\lambda-2\mu)\ol{N}(r,\infty;f) &\leq& \mu T(r)+\overline{N}(r,0;(f^{m})^{(k)})+S(r)\\
 &\leq& \mu T(r)+\frac{2\mu+1}{\eta-2\mu}\overline{N}(r,\infty;f)+\frac{2}{\eta-2\mu}T(r)+S(r).
\eeas
Thus
\beas \ol{N}(r,\infty;f)\leq \frac{\mu(\eta-2\mu)+2}{(\lambda-2\mu)(\eta-2\mu)-(2\mu+1)}T(r)+S(r).\eeas
Hence the proof is completed.
\end{proof}
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\begin{lem}\label{bc1234} If $H \not\equiv0$ and $F$ and $G$ share $(1,l)$ then
\bea\label{el.2} &&N(r,\infty;H)\\
\nonumber&\leq& \overline{N}(r,\infty;f)+\overline{N}(r,0;(f^m)^{(k)})+\overline{N}(r,b;f^m)+\overline{N}(r,b;(f^m)^{(k)})\\
\nonumber&&+\overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)+\overline{N}_{0}(r,0;(f^{m})')+\overline{N}_{0}(r,0;(f^m)^{(k+1)}),\eea
where $\overline{N}_{0}(r,0;(f^{m})')$ denotes the counting function of the zeros of $(f^{m})'$ which are not the zeros of $f(f^{m}-b)$ and $F-1$. Similarly $\overline{N}_{0}(r,0;(f^{m})^{(k+1)})$ is defined.
\end{lem}
\begin{proof} The proof is obvious if we are keeping the followings in our mind:\par
As zeros of $F$ come from the zeros of $f^{m}$ and that of $G$ come from the zeros of $(f^{m})^{(k)}$, so $\ol{N}(r,0;F)\leq \ol{N}(r,0;G)$ when $m\geq k+1$.
Also $$\overline{N}(r,\infty;F)\leq\overline{N}(r,\infty;f^m)+\overline{N}(r,\alpha_{1};f^m)+\overline{N}(r,\alpha_{2};f^m).$$
Again simple zeros of $f^m-\alpha_{i}$ are not poles of $H$ and multiple zeros of  $f^m-\alpha_{i}$ are zeros of $(f^{m})'$.\par
Similar explanation for $G$ also holds.
\end{proof}
\begin{lem}\label{abc121} If $F \equiv G$ and $n> 5$, then $f^m=(f^{m})^{(k)},$ i.e., $f$ takes the form  $$f(z)=ce^{\frac{\zeta}{m}z},$$
where $c$ is a non zero constant and $\zeta^{k}=1$.
\end{lem}
\begin{proof}
Given $F\equiv G$, that is,
\beas && n(n-1)f^{2m}\big((f^m)^{(k)}\big)^{2}\big\{f^{(n-2)m}-((f^m)^{(k)})^{n-2}\big\}\\
&&-2n(n-2)bf^m(f^m)^{(k)}\big\{f^{(n-1)m}-\big((f^m)^{(k)}\big)^{n-1}\big\}\\
&&+(n-1)(n-2)b^{2}\big\{(f^m)^{n}-\big((f^m)^{(k)}\big)^{n}\big\}=0. \eeas
By substituting $h=\frac{(f^m)^{(k)}}{f^m}$ in above, we get
\bea\label{pe1.5} n(n-1)h^{2}f^{2m}(h^{n-2}-1)&-&2n(n-2)bhf^{m}(h^{n-1}-1)\\
\nonumber&+&(n-1)(n-2)b^{2}(h^{n}-1)=0. \eea
If $h$ is a non-constant meromorphic function, then by Lemma \ref{abc212}, we get
\beas&&\{n(n-1)hf^m(h^{n-2}-1)-n(n-2)b(h^{n-1}-1)\}^{2}\\&&=-n(n-2)b^{2}(h-1)^{4}\prod\limits_{i=1}^{2n-6}(h-\beta_{i}).\eeas
Then by the second fundamental theorem, we get
\beas &&(2n-6)T(r,h)\\ &\leq& \overline{N}(r,\infty;h)+\overline{N}(r,0;h)+\sum\limits_{i=1}^{2n-6}\overline{N}(r,0;h-\beta_{i})+S(r,h) \\
&\leq& \overline{N}(r,\infty;h+\overline{N}(r,0;h)+\frac{1}{2}\sum\limits_{i=1}^{2n-6}N(r,0;h-\beta_{i})+S(r,h)\\
&\leq& (n-1) T(r,h)+S(r,h), \eeas
which is a contradiction as $n> 5$.\\
Thus $h$ is a constant. Hence as $f$ is non-constant and $b\neq0$, we get from (\ref{pe1.5}) that $$(h^{n-2}-1)=0,~~(h^{n-1}-1)=0~~\text{and}~~(h^{n}-1)=0.$$ That is, $h=1$. Consequently $f^m=(f^{m})^{(k)}.$\\
\medbreak
If $f^m=(f^{m})^{(k)}$, then we claim that $0$ and $\infty$ are the Picard exceptional value of $f$.\medbreak
For the proof, if $z_{0}$ is a zero of $f$ of order $t$, then it is a zero of $f^m$ and $(f^m)^{(k)}$ of order $mt$ and $(mt-k)$ respectively, which is impossible.\par
Again if $z_{0}$ is a pole of $f$ of order $s$, then it is a pole of $f^m$ and $(f^m)^{(k)}$ of order $ms$ and $(ms+k)$ respectively, which is impossible.\par
Thus our claim is true and hence $f$ takes the form of $$f(z)=ce^{\frac{\zeta}{m}z},$$
where $c$ is a non zero constant and $\zeta^{k}=1$.
\end{proof}
\begin{lem}\label{rjm}
If $H \equiv0$ and $n>5$, then  $f^m=(f^{m})^{(k)}$.
\end{lem}
\begin{proof} Since $H \equiv0$, on integration, we have
\bea\label{pe1.1} F=\frac{AG+B}{CG+D},\eea
where $A,B,C,D$ are constant satisfying $AD-BC\neq0$, and $F$ and $G$ share $(1,\infty)$.\par
Thus applying Mokhon'ko's Lemma (\cite{5}) in (\ref{pe1.1}), we get
\bea\label{pe1.2} T(r,f^m)=T(r,(f^m)^{(k)})+S(r,f).\eea
Again from (\ref{pe1.1}), we get $\overline{N}(r,\infty;f)=S(r,f)$ if $C\neq0$, otherwise $f^{m}$ and $(f^m)^{(k)}$ share $(\infty,\infty)$ if $C=0$ .\par
As $AD-BC\neq0$, so $A=C=0$ never occurs. Thus we consider the following cases:\\
\textbf{Case - 1} If $AC\neq0$, then
\bea F-\frac{A}{C}=\frac{BC-AD}{C(CG+D)}.\eea
So, $$\overline{N}\left(r,\frac{A}{C};F\right)=\overline{N}(r,\infty;G).$$
Now by using the second fundamental theorem, we get
\beas &&T(r,F)\\ &\leq& \overline{N}(r,\infty;F)+\overline{N}(r,0;F)+\overline{N}(r,\frac{A}{C};F)+S(r,F)\\
&\leq& \overline{N}(r,\infty;f)+\overline{N}(r,\alpha_{1};f^m)+\overline{N}(r,\alpha_{2};f^m)+\overline{N}(r,0;f^m)\\
&&+ \overline{N}(r,\infty;(f^m)^{(k)})+\overline{N}(r,\alpha_{1};(f^m)^{(k)})+\overline{N}(r,\alpha_{2};(f^m)^{(k)})+S(r,f)\\
&\leq& \frac{5}{n}T(r,F)+S(r,F),\eeas
which is a contradiction as $n> 5$.\\
\textbf{Case - 2} Next we consider $AC=0$. Then obviously $A=0$ and $C=0$ can't occur. Thus we consider the following two subcases:\\
\textbf{Subcase - 2.1} If $A=0$ and $C\neq0$, then obviously $B\neq0$ and
$$F=\frac{1}{\gamma G+\delta},$$
where $\gamma=\frac{C}{B}$ and $\delta=\frac{D}{B}$.\\
If $F$ has no 1-point, then by using the second fundamental theorem, we get
\beas &&T(r,F)\\ &\leq& \overline{N}(r,\infty;F)+\overline{N}(r,0;F)+\overline{N}(r,1;F)+S(r,F)\\
&\leq& \overline{N}(r,\infty;f^m)+\overline{N}(r,\alpha_{1};f^m)+\overline{N}(r,\alpha_{2};f^m)+\overline{N}(r,0;f^m)+S(r,f)\\
&\leq& \frac{3}{n} T(r,F)+S(r,F),\eeas
which is a contradiction as $n> 5$. Thus $\gamma+\delta=1$ and $\gamma\neq0$. So, $$F=\frac{1}{\gamma G+1-\gamma}.$$
Consequently, $\overline{N}(r,0;G+\frac{1-\gamma}{\gamma})=\overline{N}(r,\infty;F)$.\par
Now if $\gamma\neq1$, then applying the second fundamental theorem and \ref{pe1.2}, we get
\beas &&T(r,G)\\ &\leq& \overline{N}(r,\infty;G)+\overline{N}(r,0;G)+\overline{N}(r,0;G+\frac{1-\gamma}{\gamma})+S(r,G)\\
&\leq& \overline{N}(r,\infty;(f^m)^{(k)})+\overline{N}(r,\alpha_{1};(f^m)^{(k)})+\overline{N}(r,\alpha_{2};(f^m)^{(k)})\\
&+&\overline{N}(r,0;(f^m)^{(k)})+ \overline{N}(r,\infty;(f^m))+\overline{N}(r,\alpha_{1};(f^m))\\
&+&\overline{N}(r,\alpha_{2};(f^m))+S(r,f)\\
&\leq& \frac{5}{n}T(r,G)+S(r,G),\eeas
which is a contradiction as $n> 5$. Thus $\gamma=1$ and hence $FG\equiv 1$ which give
$$f^{mn}\left((f^m)^{(k)}\right)^{n}=\frac{n^{2}(n-1)^{2}}{a^{2}}(f^m-\alpha_{1})(f^m-\alpha_{2})((f^m)^{(k)}-\alpha_{1})((f^m)^{(k)}-\alpha_{2}).$$
It is clear from the above equation that $f$ has no pole, because $n>5$. Now let $z_{0}$ be a $\alpha_{1i}$ point of $f$ of order $s$, where $(\alpha_{1i})^{m}=\alpha_{1}$, then it can't be a pole of $(f^m)^{(k)}$ as $f$ has no pole, so $z_{0}$ is a zero of $(f^m)^{(k)}$ of order $q$ satisfying $n\leq nq =s$. Thus
\beas\overline{N}(r,\infty;f)&=&S(r,f),\\
\overline{N}(f,\alpha_{1i};f) &\leq& \frac{1}{n}{N}(f,\alpha_{1i};f)~~~~\text{and}\\
\overline{N}(f,\alpha_{2j};f) &\leq& \frac{1}{n}{N}(f,\alpha_{2j};f).\eeas
Thus by the second fundamental theorem, we get
\beas &&(2m-1)T(r,f)\\&\leq& \overline{N}(r,\infty;f)+\sum\limits_{i=1}^{m}\overline{N}(r,\alpha_{1i};f)+\sum\limits_{j=1}^{m}\overline{N}(r,\alpha_{2j};f)+S(r,f)\\
&\leq& \frac{2m}{n}T(r,f)+S(r,f),\eeas
which is not possible as $n > 5$.\\
\textbf{Subcase - 2.2} If $A\neq0$ and $C=0$, then obviously $D\neq0$ and
$$F=\lambda G+\mu,$$
where $\lambda=\frac{A}{D}$ and $\mu=\frac{B}{D}$. If $F$ has no $1$-point, then we arrive at a contradiction as the previous case. Thus $\lambda+\mu=1$ with $\lambda\neq0$. Also $$\overline{N}(r,0;G+\frac{1-\lambda}{\lambda})=\overline{N}(r,0;F).$$
Now if $\lambda \neq1$, then by using the second fundamental theorem and (\ref{pe1.2}), we get
\beas &&T(r,G)\\ &\leq& \overline{N}(r,\infty;G)+\overline{N}(r,0;G)+\overline{N}(r,0;G+\frac{1-\lambda}{\lambda})+S(r,G)\\
&\leq& \overline{N}(r,\infty;(f^m)^{(k)})+\overline{N}(r,\alpha_{1};(f^m)^{(k)})+\overline{N}(r,\alpha_{2};(f^m)^{(k)})\\
&+&\overline{N}(r,0;(f^m)^{(k)})+ \overline{N}(r,0;(f^m))+S(r,f)\\
&\leq& \frac{5}{n}T(r,G)+S(r,G),\eeas
which is a contradiction as $n> 5$. Thus $\lambda=1$ and hence $F\equiv G$.\par
Now in view of Lemma \ref{abc121}, we get $f^m=(f^{m})^{(k)}$ as $n> 5$, i.e. $f$ takes the form  $$f(z)=ce^{\frac{\zeta}{m}z},$$
where $c$ is a non zero constant and $\zeta^{k}=1$.
\end{proof}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section {Proof of Main Result}
\begin{proof} [\textbf{Proof of Theorem \ref{thB1} }]
Let $H$ be defined by (\ref{h}). Now we consider two cases:\\
\textbf{Case - 1} First we assume $H \not\equiv 0$. Then clearly $F \not\equiv G$ and
$$\overline{N}(r,1;F|=1)=\overline{N}(r,1;G|=1)\leq N(r,\infty;H).$$
So by the second fundamental theorem and Lemma \ref{bc1234},  we get
\bea\label{pe1.1.1} &&(n+1)T(r,f^m)\\ \nonumber &\leq& \overline{N}(r,\infty;f)+\overline{N}(r,0;f)+\overline{N}(r,b;f^m)\\
\nonumber &&+ \overline{N}(r,1;F)-N_{0}(r,0,(f^m)')+S(r,f)\\
\nonumber &\leq& 2\{\overline{N}(r,\infty;f)+\overline{N}(r,b;f^m)\}+\overline{N}(r,0;(f^m)^{(k)})\\
\nonumber &&+ \overline{N}(r,0;f)+\overline{N}(r,b;(f^{m})^{(k)})+\overline{N}(r,1;F|\geq2)\\
\nonumber &&+ \overline{N}_{L}(r,1;F)+\overline{N}_{L}(r,1;G)+\overline{N}_{0}\left(r,0;(f^m)^{(k+1)}\right)+S(r,f).
\eea
Now \bea\label{pe1.1.2} &&\overline{N}(r,1;F|\geq2)+\overline{N}_{\ast}(r,1;F,G)+\overline{N}_{0}\left(r,0;(f^m)^{(k+1)}\right)\\
\nonumber &\leq& \overline{N}(r,1;G|\geq2)+\overline{N}(r,1;G|\geq3)+\overline{N}_{0}\left(r,0;(f^m)^{(k+1)}\right)\\
\nonumber &\leq& N\left(r,0;(f^m)^{(k+1)}~|~(f^m)^{(k)}\neq0\right)+S(r,f)\\
\nonumber &\leq& \overline{N}\left(r,0;(f^m)^{(k)}\right)+\overline{N}(r,\infty;f)+S(r,f).\eea
Thus
\bea\label{pe1.1.3} &&(n+1)T(r,f^{m})\\
\nonumber &\leq& 2\{\overline{N}(r,\infty;f)+\overline{N}(r,b;f^m)\}+\overline{N}(r,0;f)\\
\nonumber &&+ 2\overline{N}(r,0;(f^m)^{(k)})+\overline{N}(r,b;(f^m)^{(k)})+\overline{N}(r,\infty;f)+S(r,f).
\eea
Similarly for $(f^m)^{(k)}$, we get
\bea\label{pe1.1.4} &&(n+1)T(r,(f^m)^{(k)})\\ &\leq& 2\{\overline{N}(r,\infty;f)+\overline{N}(r,b;(f^m)^{(k)})\}+\overline{N}(r,0;(f^m)^{(k)})\\
\nonumber &&+ 2\overline{N}(r,0;f)+\overline{N}(r,b;f^m)+\overline{N}(r,\infty;f)+S(r,f).
\eea
Adding (\ref{pe1.1.3}) and (\ref{pe1.1.4}), we get
\bea\label{pe1.1.5} (n+1)T(r) &\leq& 6\overline{N}(r,\infty;f)+3\{\overline{N}(r,0;f)+\overline{N}(r,0;(f^m)^{(k)})\}\\
\nonumber &&+ 3\{\overline{N}(r,b;f^m)+\overline{N}(r,b;(f^m)^{(k)})\}+S(r,f)
\eea
and
\bea\label{pe1.1.6} (n-5)T(r) &\leq& 6\overline{N}(r,\infty;f)+S(r).
\eea
Thus using Lemma \ref{bc123}, we get
\beas (n-5)T(r) &\leq& \frac{6\mu(\eta-2\mu)+12}{(\lambda-2\mu)(\eta-2\mu)-(2\mu+1)}T(r)+S(r),
\eeas
which is a contradiction as $n\geq6$ and $l\geq3$.\\
\textbf{Case - 2} Next we assume $H\equiv 0$. Then for $n> 5$, applying Lemma \ref{rjm}, we have $f^m=(f^{m})^{(k)}$. Thus by the same arguments using in Lemma \ref{abc121}, we see that $f$ takes the form  $$f(z)=ce^{\frac{\zeta}{m}z},$$
where $c$ is a non zero constant and $\zeta^{k}=1$. Thus the proof is completed.
\end{proof}
\begin{center} {\bf Acknowledgement} \end{center}
The author is grateful to the referees for his/her valuable suggestions which considerably improved the presentation of the paper.

\begin{thebibliography}{99}
\bibitem{1} T. C. Alzahary,  \emph{Meromorphic functions with weighted sharing of one set}, Kyungpook Math. J., \textbf{47}(2007), no. 1, 57-68.
\bibitem{2} A. Banerjee and B. Chakraborty, \emph{A new type of unique range set with deficient values}, Afr. Mat., \textbf{26}(2015), no. 7-8, 1561-1572.
\bibitem{2.1} A. Banerjee and B. Chakraborty, \emph{Uniqueness of the power of meromorphic functions with its differential polynomial sharing a set}, Math. Morav., \textbf{20}(2016), no. 2, 1-14.
\bibitem{2.11} A. Banerjee and B. Chakraborty, \emph{On the uniqueness of power of a meromorphic function sharing a set with its $k$-th derivative}, J. Indian Math. Soc. (N.S.), \textbf{85}(2018), no. 1-2, 1-15.
\bibitem{th} B. Chakraborty, \emph{Some aspects of uniqueness theory of entire and meromorphic functions}, Ph.D. thesis, arXiv: 1711.08808 [math.CV], 2017.
\bibitem{3} W. K. Hayman, \emph{Meromorphic Functions}, Clarendon Press, Oxford, 1964.
\bibitem{4} I. Lahiri, \emph{Weighted sharing and uniqueness of meromorphic functions}, Nagoya Math. J., \textbf{161}(2001), 193-206.
\bibitem{4.1} W. C. Lin and H. X. Yi, \emph{Uniqueness Theorems for meromorphic functions that share three sets}, Complex Var. Elliptic Equ., \textbf{48}(2003), no. 4, 315-327.
\bibitem{5} A. Z. Mokhon'ko, \emph{On the Nevanlinna characteristics of some meromorphic functions}, \enquote{Theory of Functions, functional analysis and their applications}, Izd-vo Khar'kovsk, Un-ta, \textbf{14}(1971), 83-87.
\bibitem{6} E. Mues and N. Steinmetz, \emph{Meromorphe Funktionen, die mit ihrer Ableitung Werte teilen}, Manuscripta Math., \textbf{29}(1979), no. 2-4, 195-206.
\bibitem{7} L. A. Rubel and C. C. Yang, \emph{Values shared by an entire function and its derivative}, Complex analysis (Proc. Conf., Univ. Kentucky, Lexington, Ky., 1976), 101-103; Lecture Notes in Math., 599, Springer, Berlin, 1977.
\bibitem{7.1} L. Z. Yang and J. L. Zhang, \emph{Non-existence of meromorphic solutions of a Fermat type functional equation}, Aequationes Math., \textbf{76}(2008), no. 1-2, 140-150.
\bibitem{8} J. L. Zhang, \emph{Meromorphic functions sharing a small function with their derivatives}, Kyungpook Math. J., \textbf{49}(2009), no. 1, 143-154.
\bibitem{9} J. L. Zhang and L. Z. Yang, \emph{A power of a meromorphic function sharing a small function with its derivative}, Ann. Acad. Sci. Fenn. Math., \textbf{34}(2009), no. 1,  249-260.
\end{thebibliography}

\end{document}