\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 182, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/182\hfil Growth of transcendental solutions]
{Growth of transcendental solutions to higher-order linear
differential equations with entire coefficients}

\author[K. Hamani \hfil EJDE-2016/182\hfilneg]
{Karima Hamani}

\address{Karima Hamani  \newline
Laboratory of Pure and Applied Mathematics,
University of Mostaganem (UMAB),
B. P. 227 Mostaganem, Algeria}
\email{hamanikarima@yahoo.fr, karima.hamani@univ-mosta.dz}

\thanks{Submitted February 15, 2016. Published July 11, 2016.}
\subjclass[2010]{34M10, 30D35}
\keywords{Linear differential equation; transcendental entire function;
\hfill\break\indent  order of growth; hyper-order}

\begin{abstract}
 In this article, we study the growth of transcendental solutions
 of certain higher order linear differential equations with entire coefficients.
 Under some conditions, we prove that every transcendental solution is of infinite
 order. We also give an estimate of its hyper-order. We improve previous
 results by Peng and Chen \cite{PEN}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction and statement of results}

 In this article, we use  fundamental results and the standard
notation of the Nevanlinna's value distribution theory of meromorphic
functions (see \cite{HAY,YAN}). In addition, we use the notation
$\sigma (f)$ to denote the order of growth of a meromorphic function
 $f$ and $\sigma _2(f)$ to denote the hyper-order of $f$ which is defined in
 \cite{YAN}  by
\begin{equation*}
\sigma _2(f)=\limsup_{r\to+\infty}\frac{\log \log T(r,f)}{\log r},
\end{equation*}
where $T(r,f)$ is the Nevanlinna characteristic function of $f$.

 For the second order linear differential equation
\begin{equation}
f''+e^{-z}f'+B(z)f=0,  \label{e1.1}
\end{equation}
where $B(z)$ is an entire function of finite order, it is well known that
every solution of \eqref{e1.1} is an entire function and most
solutions of \eqref{e1.1} have an infinite order. Thus, a natural
question is: what conditions on $B(z)$ will guarantee that
every solution $f(\not\equiv 0) $ of \eqref{e1.1} has an infinite
order?
Ozawa \cite{OZA}, Gundersen \cite{GUN2}, Langley \cite{LAN}.
Amemiya and Ozawa \cite{AME} have studied the
problem, where $B(z) $ is a nonconstant polynomial or a
transcendental entire function with order $\sigma (B) \neq 1$.
In 2002, Chen \cite{CHE1} investigated the growth of solutions of
equation \eqref{e1.1} in the case where $\sigma (B) =1$.

 In 1988, Gundersen \cite{GUN3} studied finite order
solutions of second order linear differential equations, where coefficients
satisfy certain conditions in some angle. This result was generalised to
higher order linear differential equations by Laine and Yang \cite{LAI}.
Recently, the authors \cite{HEI} have studied completely
regular growth solutions of second order linear differential equations and
discussed cases where coefficients and or solutions of these equations are
exponential polynomials.

 Recently, Peng and Chen \cite{PEN} have studied the order and the
hyper-order of solutions of equation \eqref{e1.1} and have proved
the following result:

 \begin{theorem}[\cite{PEN}] \label{thmA}
 Let $A_j(z)$ $(\not\equiv 0) $ $(j=1,2)$ be entire functions with
$\sigma (A_j)<1$, $a_1$, $a_2$ be complex numbers such that
$a_1a_2\neq 0$ and $a_1\neq a_2$ (suppose that $|a_1|\leq |a_2|$).
If $\arg a_1\neq \pi $ or $a_1<-1$, then
every solution $f(\not\equiv 0) $ of the equation
\begin{equation}
f''+e^{-z}f'+(A_1(z) e^{a_1z}+A_2(z) e^{a_2z}) f=0  \label{e1.2}
\end{equation}
is of infinite order and $\sigma _2(f)=1$.
\end{theorem}

 In this article, we continue the research in this type of problem.
We consider the higher order linear differential equation
\begin{equation}
f^{(k)}+h_{k-1}(z)f^{(k-1)}+\dots +h_1(z)f'+h_{0}(z)f=0, \label{e1.3}
\end{equation}
where $k>2$\ is an integer and $h_j(z)$ $(j=0,\dots ,k-1) $
are entire functions. We suppose that there exists
only one coefficient of the form $h_{s}(z)=A_1(z)e^{P_1(z)}+A_2(z)e^{P_2(z)}$,
where $P_l(z)=
\sum_{i=0}^n a_{i,l}z^{i}$ $(l=1,2) $ are
polynomials with degree $n\geq 1$ and $A_l(z)$
 $(\not\equiv 0) $ $(l=1,2)$\ are entire functions with $\sigma (A_l)<n$. The
other coefficients have the form $h_j(z)=B_j(z)e^{Q_j(z) }$
$(j\neq s)$, where $Q_j(z)=\sum_{i=0}^n b_{i,j}z^{i}$
are polynomials with degree $n\geq 1$ and $B_j(z)$
$(\not\equiv 0) $ are entire functions with $\sigma (B_j)<n$. Under some
conditions on the complex numbers $a_{n,l}$ $(l=1,2)$ and $b_{n,j}$
$(j\neq s)$, we will prove that every transcendental solution of equation
\eqref{e1.3} is of infinite order. We also give an estimation of its
hyper-order. We will prove the following results:

\begin{theorem} \label{thm1.1}
Let $P_l(z)=\sum_{i=0}^n a_{i,l}z^{i}$ $(l=1,2)$ be
polynomials with degree $n\geq 1$, where
$a_{0,l},\dots ,a_{n,l}$ $(l=1,2)$  are complex numbers such that
$a_{n,1}\neq a_{n,2}$,  $A_l(z)$ $(\not\equiv 0) $
$(l=1,2)$  be entire functions with
$\sigma (A_l)<n$ and $h_j(z) $ $(j=0,\dots ,k-1) $
be entire functions. Suppose that there exists
$s\in \{ 1,\dots ,k-1\} $  such that
$h_{s}(z) =A_1(z)e^{P_1(z)}+A_2(z)e^{P_2(z)}$ and for
$j\neq s$, $h_j(z) =B_j(z)e^{Q_j(z) }$,  where
$B_j(z)$ $(\not\equiv 0) $ are entire
functions with $\sigma (B_j)<n$,  $Q_j(z) = \sum_{i=0}^n b_{i,j}z^{i}$
 are polynomials with degree $n\geq 1$ and
$b_{0,j},\dots ,b_{n,j}$ $(j\neq s) $  are complex numbers. Let
$I$ and $J$  be two sets satisfying
$I\neq \emptyset$, $J\neq \emptyset$, $I\cap J=\emptyset $
 and $I\cup J=\{0,\dots ,s-1,s+1,\dots ,k-1\}$ such that for
$j\in I$, $b_{n,j}=\alpha _ja_{n,1}$ $(0<\alpha _j<1) $
 and for $j\in J$,  $b_{n,j}=\beta _ja_{n,2}$
$(0<\beta _j<1)$.  Set $a_{n,l}=| a_{n,l}| $
$e^{i\theta _l}$, $\theta _l\in [ 0,2\pi ) $
$(l=1,2)$,  $\alpha =\max \{\alpha _j:j\in I\}$
 and $\beta =\max \{\beta _j:j\in J\}$.

If $\theta _1\neq \theta _2$ or $\theta_1=\theta _2$ and
(i) $| a_{n,1}| <(1-\beta )| a_{n,2}| $ or
(ii) $| a_{n,2}| <(1-\alpha )| a_{n,1}| $,
 then every transcendental solution $f$  of equation
\eqref{e1.3} is of infinite order and satisfies $\sigma_2(f)=n$.
\end{theorem}

 \begin{theorem} \label{thm1.2}
 Let $P_l(z)=\sum_{i=0}^n a_{i,l}z^{i}$ $(l=1,2) $
be polynomials with degree $n\geq 1$, where
$a_{0,l},\dots ,a_{n,l}$ $(l=1,2)$ are complex numbers such that
$a_{n,1}\neq a_{n,2}$ (suppose that $|a_{n,1}|\leq |a_{n,2}|$),
$A_l(z)$ $(\not\equiv 0) $ $(l=1,2)$
 be entire functions with
$\sigma (A_l)<n$ and $h_j(z) $ $(j=0,\dots ,k-1) $ be entire
functions. Suppose that there exists $s\in \{ 1,\dots ,k-1\} $
 such that $h_{s}(z) =A_1(z)e^{P_1(z)}+A_2(z)e^{P_2(z)}$ and for
$j\neq s$, $h_j(z) =B_j(z)e^{Q_j(z) }$, where
$B_j(z)$ $(\not\equiv 0) $  are entire functions with
$\sigma (B_j)<n$, $Q_j(z) =\sum_{i=0}^n b_{i,j}z^{i}$ are polynomials
with degree $n\geq 1$ and $b_{0,j},\dots ,b_{n,j}$
$(j\neq s) $ are complex numbers. Let $I$ and $J$
 be two sets satisfying $I\neq \emptyset$,
$J\neq \emptyset $, $I\cap J=\emptyset $ and
$I\cup J=\{0,\dots ,s-1,s+1,\dots ,k-1\}$ such that for
$j\in I$, $b_{n,j}=\alpha _ja_{n,1}$ $(0<\alpha _j<1) $
 and for $j\in J$,  $b_{n,j}$  are real numbers
satisfying $b_{n,j}<0$.

If $a_{n,1}$ is a real number such that
$(1-\alpha )a_{n,1}<b$, where $\alpha =\max \{\alpha _j:j\in I\}$
and $b=\min \{b_{n,j}:j\in J\}$, then every transcendental
solution $f$ of equation $\eqref{e1.3}$ is of
infinite order and satisfies $\sigma _2(f)=n$.
\end{theorem}

 \begin{theorem} \label{thm1.3}
Let $P_l(z)=\sum_{i=0}^n a_{i,l}z^{i}$ $(l=1,2) $  be
polynomials with degree $n\geq 1$, where
$a_{0,l},\dots ,a_{n,l}$ $(l=1,2)$  are complex numbers such that
$a_{n,1}\neq a_{n,2}$ (suppose that $|a_{n,1}|\leq |a_{n,2}|$),
$A_l(z)$ $(\not\equiv 0) $ $(l=1,2)$ be entire
functions with $\sigma (A_l)<n$ and $h_j(z) $
$(j=0,\dots ,k-1) $ be entire functions.
Suppose that there exists $s\in \{ 1,\dots ,k-1\} $
such that $h_{s}(z) =A_1(z)e^{P_1(z)}+A_2(z)e^{P_2(z)}$
and for $j\neq s$, $h_j(z) =B_j(z)e^{Q_j(z) }$, where
 $B_j(z)$ $(\not\equiv 0)$  are entire functions with
$\sigma (B_j)<n$, $Q_j(z) =\sum_{i=0}^n b_{i,j}z^{i}$  are
polynomials with degree $n\geq 1$ and $b_{0,j},\dots ,b_{n,j}$
$(j\neq s) $ are complex numbers. Let $I$ and $J $ be two sets satisfying
$I\neq \emptyset $, $J\neq \emptyset$, $I\cap J=\emptyset $ and
$I\cup J=\{1,\ldots,s-1,s+1,\dots ,k-1\} $ such that for
$j\in I$, $b_{n,j}=\alpha_ja_{n,1}+\beta _ja_{n,2}$ $(0<\alpha _j<1) $,
$(0<\beta_j<1) $  and
for $j\in J$, $b_{n,j}$  are real numbers satisfying
$b_{n,j}<0$.  Set $\alpha =\max \{\alpha _j:j\in I\}$,
$\beta =\max \{\beta _j:j\in I\}$  and $b=\min \{b_{n,j}:j\in J\}$.

If $a_{n,1}$ and $a_{n,2}$  are real numbers such that
(i) $(1-\beta ) a_{n,2}-b<a_{n,1}<0$  or
(ii) $(1-\alpha )a_{n,1}-b<a_{n,2}<0$,
then every transcendental solution $f$
of equation \eqref{e1.3}  is of infinite order and
satisfies $\sigma _2(f)=n$.
\end{theorem}


 \begin{remark} \label{rmk1.1} \rm
In Theorem \ref{thmA}, the authors have considered conditions only on one complex
number $a_1$. But in Theorem \ref{thm1.1} and Theorem \ref{thm1.3}, 
conditions are imposed to the two numbers $a_{n,l}$ $(l=1,2)$.
\end{remark}

\section{Preliminary Lemmas}

 \begin{lemma}[\cite{GUN1}] \label{lem2.1}
 Let $f(z) $ be a transcendental meromorphic function of finite
order $\sigma$.  Let $\Gamma =\{ (k_1,j_1),(k_2,j_2) ,\dots ,(k_m,j_m) \} $
denotes a set of distinct pairs of integers satisfying
$k_{i}>j_{i}\geqslant 0$ $(i=1,2,\dots ,m) $,  and
let $\varepsilon >0$  be a given constant. Then there exists a set
$E_1\subset [ 0,2\pi ) $ that has linear measure zero
such that if $\theta \in [ 0,2\pi ) \backslash E_1$,
then there is a constant $R_1=R_1(\theta ) >1$
such that for all $z$ satisfying $\arg z=\theta $ and
$| z| \geqslant R_1$,  and for all $(k,j) \in \Gamma $, we have
\begin{equation}
| \frac{f^{(k) }(z) }{f^{(j)
}(z) }| \leqslant | z| ^{(
k-j) (\sigma -1+\varepsilon ) }.  \label{e2.1}
\end{equation}
\end{lemma}

 \begin{lemma}[\cite{BAN,MAR}] \label{lem2.2}
Let $P(z) =(\alpha +i\beta) z^{n}+\dots $ ($\alpha $, $\beta $ are real
numbers, $| \alpha | +| \beta | \neq 0 $) be a polynomial with degree
$n\geq 1$,  and $A(z) $ be an entire function with
$\sigma (A) <n$. Set $f(z) =A(z) e^{P(z) }$,
$z=re^{i\theta }$, $\delta (P,\theta )=\alpha \cos (n\theta ) -\beta \sin (n\theta )$.
Then for any given $\varepsilon >0$, there exists a set
$E_2\subset [ 0,2\pi )$ that has linear measure zero such that for any
$\theta \in [ 0,2\pi ) \backslash E_2\cup H$, where
$H=\{ \theta \in [ 0,2\pi ) :\delta (P,\theta ) =0\} $ is a finite set,
there is a constant $R_2>1$ such that for $| z|=r\geqslant R_2$, we have
\begin{itemize}
\item[(i)]  if $\delta (P,\theta ) >0$, then
\begin{equation}
\exp \{ (1-\varepsilon ) \delta (P,\theta )
r^{n}\} \leq | f(re^{i\theta }) | \leq
\exp \{ (1+\varepsilon ) \delta (P,\theta )
r^{n}\} ,  \label{e2.2}
\end{equation}

 \item[(ii)] if $\delta (P,\theta ) <0$, then
\begin{equation}
\exp \{ (1+\varepsilon ) \delta (P,\theta )
r^{n}\} \leq | f(re^{i\theta }) | \leq
\exp \{ (1-\varepsilon ) \delta (P,\theta )
r^{n}\} .  \label{e2.3}
\end{equation}
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{GUN3,HEI}] \label{lem2.3}
Let $d\geq 1$ be an integer, $f(z) $
be an entire function and suppose that $| f^{(d)}(z) | $ 
is unbounded on some ray $\arg z=\theta $.  
Then there exists an infinite sequence of points $
z_m=r_me^{i\theta }$ $(m=1,2,\dots ) $, where
$r_m\to +\infty $, such that $f^{(d) }(z_m) \to \infty $ and
\begin{equation}
\big| \frac{f^{(j) }(z_m) }{f^{(
d) }(z_m) }\big| \leqslant \frac{1}{(d-j) !}(1+o(1) ) | z_m|
^{d-j}\quad (j=0,\dots ,d-1).  \label{e2.4}
\end{equation}
\end{lemma}

 The following Lemma is a trivial consequence of theorems by
Phragm\`{e}n-Lindel\"{o}f and Liouville (see [13, p. 214]):

 \begin{lemma}[\cite{CHE2}] \label{lem2.4}
Let $f(z) $ be an entire function of finite order $\rho$.
Suppose that there exists a set $E_3\subset [ 0,2\pi )$
 that has linear measure zero such that for any ray 
$\arg z=\theta \in [ 0,2\pi ) \backslash E_3$, 
$| f(re^{i\theta }) | \leq Mr^k$, where 
$M=M(\theta ) >0$  is a constant and $k$ 
$(>0) $ is a constant independent of $\theta $. Then 
$f(z) $ is a polynomial with $\deg f\leq k$. 
\end{lemma}

\begin{lemma}[\cite{GUN1}] \label{lem2.5}
Let $f(z)$ be a transcendental meromorphic function. Let 
$\alpha >1$ and $\varepsilon >0$ be given constants. Then there
exist a set $F_1\subset (1,+\infty )$  having finite logarithmic
measure and a constant $B>0$ that depends only on $\alpha $
 and $(i,j) $ ($i,j$ are positive integers with $i>j$), such that for all 
$z$ satisfying $|z|=r\notin [ 0,1] \cup F_1$, we have
\begin{equation}
| \frac{f^{(i) }(z)}{f^{(j) }(z)}
| \leq B\big[ \frac{T(\alpha r,f)}{r}(\log ^{\alpha
}r) \log T(\alpha r,f)\big] ^{i-j}.  \label{e2.5}
\end{equation}
\end{lemma}

 \begin{lemma}[\cite{TU}] \label{lem2.6}
 Let $f(z) $ be a transcendental entire function. For each
sufficiently large $| z| =r$,  let $z_r=r$
 $e^{i\theta _r}$ be a point satisfying $|f(z_r) | =M(r,f) $. Then there
exist a constant $\delta _r$ $(>0) $ and a
set $F_2$ of finite logarithmic measure, such that for all $z$
 satisfying $| z| =r\notin F_2$ and 
$\arg z=\theta \in [ \theta _r-\delta _r,\theta _r+\delta _r] $, we have 
\begin{equation}
| \frac{f(z)}{f^{(d) }(z)}| \leq r^{2d}\quad
(d\geq 1\text{ is an integer}).  \label{e2.6}
\end{equation}
\end{lemma}

\begin{lemma}[\cite{GUN3}] \label{lem2.7}
  Let $\varphi :[ 0,+\infty ) \to \mathbf{\mathbb{R}}$ and 
$\psi :[ 0,+\infty ) \to \mathbf{\mathbb{R}}$  be monotone 
non-decreasing functions such that 
$\varphi (r) \leqslant \psi (r) $ for all $r\notin F_3\cup [ 0,1] $, 
 where $F_3\subset (1,+\infty) $ is a set of finite logarithmic measure. 
Let $\alpha >1$  be a given constant.  Then  there exists  an
$r_{0}=r_{0}(\alpha ) >0$  such that $\varphi (r) \leqslant \psi (\alpha r) $ 
for all $r>r_{0}$.
\end{lemma}

 \begin{lemma}[\cite{CHE2}] \label{lem2.8}
 Let $k\geq 2$ be an integer and $A_j(z)$ $(j=0,1,\dots ,k-1)$ be entire 
functions of finite order. Set $\rho=\max \{\sigma (A_j):j=0,1,\dots ,k-1\}$. 
If $f$ is a solution of equation
\begin{equation}
f^{(k) }+A_{k-1}(z) f^{(k-1)
}+\dots +A_1(z) f'+A_{0}(z) f=0,  \label{e2.7}
\end{equation}
 then $\sigma _2(f)\leq \rho $.
\end{lemma}

\section{Proof of Theorem \ref{thm1.1}}

 Assume $f$ is a transcendental solution of  \eqref{e1.3}.
\smallskip

\noindent\textbf{First step.} 
We prove that $\sigma (f)=+\infty $. Suppose that $\sigma (f) =\rho <+\infty $. 
By Lemma \ref{lem2.1}, there exists a set $E_1\subset [ 0,2\pi )$ that has linear measure 
zero such that if $\theta \in [0,2\pi ) -E_1$, then there is a constant 
$R_1=R_1(\theta ) >1$\ such that for all $z$ satisfying $\arg z=\theta $ and 
$| z| \geqslant R_1$, we have
\begin{equation}
\big| \frac{f^{(j) }(z)}{f^{(i) }(z)}\big| 
\leq | z| ^{k\rho }\text{ }(0\leq i<j\leq k). \label{e3.1}
\end{equation}
By Lemma \ref{lem2.2}, for any given $\varepsilon >0$, there exists a set 
$E_2\subset [ 0,2\pi ) $ that has linear measure zero such that
if $z=re^{i\theta }$, $\theta \in [ 0,2\pi ) /E_2\cup H_1$
and $r$ is sufficiently large, then $A_l(z)e^{P_l(z) }$
$(l=1,2) $ and $B_j(z)e^{Q_j(z) }$ $(j\neq s) $ satisfy \eqref{e2.2}  or 
\eqref{e2.3},  where $H_1=\{ \theta \in [ 0,2\pi ) :\delta
(P_1,\theta ) \text{ or }\delta (P_2,\theta )=0\} $.
\smallskip

\noindent\textbf{Case 1.}
 Suppose that $\theta _1$ $\neq \theta _2$.
Set $H_2=\{ \theta \in [ 0,2\pi ) :\delta (P_1,\theta ) =\delta (P_2,\theta )\} $. 
Since $\theta _1\neq \theta _2$, it follows that $H_2$ has linear measure zero. 
For any given $\theta \in [ 0,2\pi ) \backslash E_1\cup E_2\cup H_1\cup H_2$, 
we have
\begin{equation}
\begin{gathered}
\delta (P_1,\theta ) \neq 0,\quad 
\delta (P_2,\theta ) \neq 0\quad\text{and}\\
\delta (P_1,\theta ) >\delta (P_2,\theta ) \quad\text{or}\quad
\delta (P_1,\theta ) <\delta (P_2,\theta ) . 
\end{gathered} \label{e3.2}
\end{equation}
Set $\delta _1=\delta (P_1,\theta ) $ and 
$\delta_2=\delta (P_2,\theta ) $.
\smallskip

\noindent\textbf{Subcase 1.1: $\delta _1>\delta _2$.}
Here we also divide our proof in three subcases:

\noindent \textbf{(a): $\delta _1>\delta _2>0$.} Set 
$\delta _3=\max \{ \delta _2,\delta (Q_j,\theta ):j\in I\} $. Then 
$0<\delta _3<\delta _1$. Thus for any given $\varepsilon $ 
$(0<\varepsilon <\frac{\delta _1-\delta _3}{2(\delta _1+\delta _3) }) $ and all 
$z$ satisfying $\arg z=\theta $ and $|z| =r$ sufficiently large, we have
\begin{gather}
| A_1(z)e^{P_1(z) }| \geq \exp
\{(1-\varepsilon )\delta _1r^{n}\},  \label{e3.3} \\
| A_2(z)e^{P_2(z) }| \leq \exp
\{(1+\varepsilon )\delta _3r^{n}\}  \label{e3.4} \\
| B_j(z)e^{Q_j(z) }| \leq \exp \{(1+\varepsilon )\delta _3r^{n}\}\text{ }(j\neq s) .
\label{e3.5}
\end{gather}
 Now we prove that $| f^{(s) }(z)| $\ is bounded on the ray $\arg z=\theta $. 
If $|f^{(s) }(z) | $ is unbounded on the ray $\arg z=\theta $, then by
 Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$ $(m=1,2,\dots ) $, where
$r_m\to +\infty $ such that $f^{(s) }(z_m) \to \infty $ and
\begin{equation}
\big| \frac{f^{(j) }(z_m) }{f^{(
s) }(z_m) }\big| \leqslant \frac{1}{(
s-j) !}(1+o(1) ) | z_m|
^{s-j}\quad (j=0,\dots ,s-1) .  \label{e3.6}
\end{equation}
By \eqref{e1.3}, \eqref{e3.1} and \eqref{e3.3}--\eqref{e3.6}, 
for the above $z_m$, we obtain
\begin{equation}
\exp \{(1-\varepsilon )\delta _1r_m^{n}\}
\leq M_1r_m^{d_1}\exp \{(1+\varepsilon )\delta _3r_m^{n}\},  \label{e3.7}
\end{equation}
where $M_1$, $d_1(>0) $ are constants. This is a
contradiction. Hence 
$| f^{(s) }(z)| \leq M$ on $\arg z=\theta $, where $M$ $(>0) $ is a constant. 
We can easily obtain
\begin{equation}
| f(z) | \leq M| z| ^{s}
\label{e3.8}
\end{equation}
on $\arg z=\theta $. By Lemma \ref{lem2.4}, \eqref{e3.8}  and the
fact that $E_1\cup E_2\cup H_1\cup H_2$ has linear measure zero, we
obtain that $f(z) $ is a polynomial with $\deg f\leq s$,
 which contradicts our assumption. Therefore $\sigma (f)=+\infty $.
\smallskip


\noindent\textbf{(b): $\delta _1>0>\delta _2$.} Thus for any given 
$\varepsilon $ $(0<\varepsilon <\frac{1-\alpha }{2(1+\alpha )})$ and all $z$
satisfying $\arg z=\theta $ and $| z| =r$ sufficiently
large, we have \eqref{e3.3},
\begin{gather}
| A_2(z)e^{P_2(z) }| \leq \exp
\{(1-\varepsilon )\delta _2r^{n}\}<1,  \label{e3.9} \\
| B_j(z)e^{Q_j(z)}| \leq \exp \{(1+\varepsilon
)\alpha \delta _1r^{n}\}\text{ }(j\in I) , \label{e3.10} \\
| B_j(z)e^{Q_j(z)}| \leq \exp \{(1-\varepsilon
)\delta (Q_j,\theta )r^{n}\}<1\text{ }(j\in J) .  \label{e3.11}
\end{gather}
 Now we prove that $| f^{(s) }(z)| $ is bounded on the ray $\arg z=\theta $.
 If $|f^{(s) }(z) | $ is unbounded on the ray $\arg z=\theta $, then by
 Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$ $(m=1,2,\dots ) $, where 
$r_m\to +\infty $ such that $f^{(s) }(z_m) \to \infty $ and 
\eqref{e3.6} holds.

 By \eqref{e1.3}, \eqref{e3.1}, \eqref{e3.3}, \eqref{e3.6}
and \eqref{e3.9}-\eqref{e3.11}, for the above $z_m$, we obtain
\begin{equation}
\exp \{(1-\varepsilon )\delta _1r_m^{n}\}\leq M_2r_m^{d_2}\exp
\{(1+\varepsilon )\alpha \delta _1r_m^{n}\},  \label{e3.12}
\end{equation}
where $M_2$, $d_2$  $(>0) $ are constants. This is a
contradiction. Hence $| f^{(s) }(z)| \leq M$ on $\arg z=\theta $, where 
$M$ $(>0) $ is a constant. We can easily obtain \eqref{e3.8} 
on $\arg z=\theta $. Using similar arguments as above, we deduce
that $\sigma (f) =+\infty $.
\smallskip


\noindent\textbf{(c): $0>\delta _1>\delta _2$.} Thus for any given $
\varepsilon $ $(0<2\varepsilon <1) $ and all $z$ satisfying 
$\arg z=\theta $ and $| z| =r$ sufficiently large, we have
\begin{gather}
| A_l(z)e^{P_l(z) }| \leq \exp
\{(1-\varepsilon )\delta (P_l,\theta )r^{n}\}\quad (l=1,2),
\label{e3.13} \\
| B_j(z)e^{Q_j(z) }| \leq \exp
\{(1-\varepsilon )\delta (Q_j,\theta )r^{n}\}(j\neq s) .
\label{e3.14}
\end{gather}
 By \eqref{e1.3}, we obtain
\begin{equation}
-1=h_{k-1}(z)\frac{f^{(k-1)}(z)}{f^{(k) }(z)}+\dots +h_{s}(z)
\frac{f^{(s)}(z)}{f^{(k) }(z)}+\dots +h_{0}(z)\frac{f(z)}{
f^{(k) }(z)}.  \label{e3.15}
\end{equation}

Now we prove that $| f^{(k) }(z)| $ is bounded on the ray $\arg z=\theta $. 
If $|f^{(k) }(z) | $\ is unbounded on the ray 
$\arg z=\theta $, then by Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$\ $(m=1,2,\dots ) $, where 
$r_m\to +\infty $ such that $f^{(k) }(z_m) \to \infty $ and
\begin{equation}
\big| \frac{f^{(j) }(z_m) }{f^{(k) }(z_m) }\big| 
\leqslant 
\frac{1}{(k-j) !}(1+o(1) ) | z_m|^{k-j}\quad (j=0,\dots ,k-1).  \label{e3.16}
\end{equation}
Substituting \eqref{e3.1}, \eqref{e3.13}, \eqref{e3.14} and \eqref{e3.16} into 
\eqref{e3.15},  as $r_m\to +\infty $, we obtain
$1\leq 0$.
This  contradiction implies $| f^{(k) }(z) | \leq M'$ on 
$\arg z=\theta$, where $M'$ $(>0) $ is a constant. 
We can easily obtain that $| f(z) | \leq M'| z| ^k$ on $\arg z=\theta $.
From this and the fact $E_1\cup E_2\cup H_1\cup H_2$ has linear measure zero, we
obtain by Lemma \ref{lem2.4} that $f(z) $ is a polynomial with $\deg f\leq k$, 
which contradicts our assumption. Therefore $\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{Subcase 1.2: $\delta _1<\delta _2$.}
Using the same reasoning as in subcase 1.1 , we can also obtain that 
$f(z) $ is a polynomial, which contradicts our assumption.
Therefore $\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{Case 2.} 
Suppose that $\theta _1=\theta _2$. For any
given $\theta \in [ 0,2\pi ) /E_1\cup E_2\cup H_1$, where $
E_1$, $E_2$ and $H_1$ are defined above, we have
\begin{equation}
\delta (P_1,\theta ) >0\quad \text{or}\quad 
\delta (P_1,\theta) <0.  \label{e3.17}
\end{equation}
\smallskip

\noindent\textbf{Subcase 2.1: $\delta (P_1,\theta ) >0$.}
\smallskip

\noindent\textbf{(i)} If $| a_{n,1}| <(1-\beta) | a_{n,2}| $, 
for any given $\varepsilon $ $(0<\varepsilon <\frac{(1-\beta ) | a_{n,2}|
-| a_{n,1}| }{2[ (1+\beta ) |a_{n,2}| +| a_{n,1}| ] })$ and all $z$
satisfying $\arg z=\theta $ and $| z| =r$ sufficiently large, we have
\begin{gather}
| A_1(z)e^{P_1(z) }| \leq \exp \{(1+\varepsilon )\delta (P_1,\theta ) r^{n}\}, 
 \label{e3.18} \\
| A_2(z)e^{P_2(z) }| \geq \exp
\{(1-\varepsilon )\delta (P_2,\theta )r^{n}\},  \label{e3.19} \\
| B_j(z)e^{Q_j(z) }| \leq \exp
\{(1+\varepsilon )\alpha \delta (P_1,\theta )r^{n}\}\text{ }(j\in
I) ,  \label{e3.20} \\
| B_j(z)e^{Q_j(z) }| \leq \exp
\{(1+\varepsilon )\beta \delta (P_2,\theta )r^{n}\}\text{ }(j\in
J) .  \label{e3.21}
\end{gather}
 Now we prove that $| f^{(s) }(z)| $ is bounded on the ray $\arg z=\theta $. 
If $|f^{(s) }(z) | $ is unbounded on the ray $\arg z=\theta $, 
then by Lemma \ref{lem2.3}, 
there exists an infinite sequence of points $z_m=r_me^{i\theta }$
 $(m=1,2,\dots ) $, where $r_m\to +\infty $ such that $f^{(s) }(z_m) \to \infty $
 and \eqref{e3.6} holds.

 By \eqref{e1.3}, \eqref{e3.1}, \eqref{e3.6} and \eqref{e3.18}-\eqref{e3.21},
 for the above $z_m$, we obtain
\begin{equation}
\begin{aligned}
&\exp \{(1-\varepsilon )\delta (P_2,\theta )r_m^{n}\}   \\
&\leq M_3r_m^{d_3}\exp \{(1+\varepsilon )\delta (P_1,\theta
)r_m^{n}\}\exp \{(1+\varepsilon )\beta \delta (P_2,\theta )r_m^{n}\},
\end{aligned} \label{e3.22}
\end{equation} 
where $M_3$, $d_3$ $(>0) $ are constants. By \eqref{e3.22},
 we have
\begin{equation}
\exp \{\gamma _1r_m^{n}\}\leq M_3r_m^{d_3},  \label{e3.23}
\end{equation}
where
\begin{equation*}
\gamma _1=(1-\varepsilon )\delta (P_2,\theta )-(1+\varepsilon )\delta
(P_1,\theta )-(1+\varepsilon )\beta \delta (P_2,\theta ).
\end{equation*}
Since
\[
0<\varepsilon <\frac{(1-\beta ) |a_{n,2}| -| a_{n,1}| }{2[ (1+\beta
)| a_{n,2}| +| a_{n,1}| ] },
\]
$\theta _1=\theta _2$ and $\cos (\theta _1+n\theta )>0$, we have
\begin{align*}
\gamma _1
&=\{ (1-\beta ) | a_{n,2}|
-| a_{n,1}| -\varepsilon [ (1+\beta )|
a_{n,2}| +| a_{n,1}| ] \} \cos(\theta _1+n\theta ) \\
&> \frac{(1-\beta ) | a_{n,2}| -|a_{n,1}| }{2}\cos (\theta _1+n\theta )>0.
\end{align*}
Hence \eqref{e3.23} is a contradiction. Hence
$| f^{(s) }(z) | \leq M$ on
$\arg z=\theta $, where $M$ $(>0) $ is a constant.
 We can easily obtain \eqref{e3.8} on
$\arg z=\theta $. By Lemma \ref{lem2.4}, \eqref{e3.8} and the fact that
$E_1\cup E_2\cup H_1$ has linear
measure zero, we obtain that $f(z) $ is a polynomial with
$\deg f\leq s$, which contradicts our assumption. Therefore
$\sigma (f) =+\infty $.
\smallskip


\noindent\textbf{(ii)} 
If $| a_{n,2}| <(1-\alpha ) | a_{n,1}| $, for any given $\varepsilon$ 
$(0<\varepsilon <\frac{(1-\alpha ) | a_{n,1}| -| a_{n,2}| }{2[ (1+\alpha ) |
a_{n,1}| +| a_{n,2}| ] })$ and all $z$
satisfying $\arg z=\theta $ and $| z| =r$ sufficiently
large, we have
\begin{gather}
| A_1(z)e^{P_1(z) }| 
\geq \exp \{(1-\varepsilon )\delta (P_1,\theta ) r^{n}\} , \label{e3.24} \\
| A_2(z)e^{P_2(z) }| \leq \exp
\{(1+\varepsilon )\delta (P_2,\theta ) r^{n}\}.  \label{e3.25}
\end{gather}
 Now we prove that $| f^{(s) }(z)| $ is bounded on the ray 
$\arg z=\theta $. If $|f^{(s) }(z) | $ is unbounded on the ray 
$\arg z=\theta $, then by Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$\ $(m=1,2,\dots ) $, where 
$r_m\to +\infty $ such that $f^{(s) }(z_m) \to \infty $ and \eqref{e3.6} holds.

 By \eqref{e1.3}, \eqref{e3.1}, \eqref{e3.6}, \eqref{e3.20}, 
\eqref{e3.21}, \eqref{e3.24} and \eqref{e3.25}, for the above $z_m$, we obtain
\begin{equation}
\begin{aligned}
&\exp \{(1-\varepsilon )\delta (P_1,\theta )r_m^{n}\}   \\
&\leq M_4r_m^{d_4}\exp \{(1+\varepsilon )\alpha \delta (P_1,\theta
)r_m^{n}\}\exp \{(1+\varepsilon )\delta (P_2,\theta )
r_m^{n}\},
\end{aligned} \label{e3.26}
\end{equation}
where $M_4$, $d_4$ $(>0) $ are constants. By \eqref{e3.26}, we have
\begin{equation}
\exp \{\gamma _2r_m^{n}\}\leq M_4r_m^{d_4},  \label{e3.27}
\end{equation}
where
\begin{equation*}
\gamma _2=(1-\varepsilon )\delta (P_1,\theta )-(1+\varepsilon )\delta
(P_2,\theta ) -(1+\varepsilon )\alpha \delta (P_1,\theta )>0.
\end{equation*}
Since \eqref{e3.27} is a contradiction, $| f^{(s) }(z) | \leq M$ on
$\arg z=\theta $, where $M$ $(>0) $ is a constant. We can easily
obtain \eqref{e3.8}  on $\arg z=\theta $. Using similar
arguments as above, we conclude that $\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{Subcase 2.2: $\delta (P_1,\theta ) <0$.}
Using the same reasoning as in subcase 1.1(c), we can also
conclude that $\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{Second step.} Now we prove that 
$\sigma _2(f) =n$. By Lemma \ref{lem2.5}, there exists a constant $B>0$ and a set 
$F_1\subset (1,+\infty )$ having finite logarithmic measure, such that for
all $z$ satisfying $|z|=r\notin [ 0,1]\cup F_1$, we have
\begin{equation}
\big| \frac{f^{(j)}(z)}{f^{(i) }(z)}\big| 
\leq Br [ T(2r,f)] ^{j+1}(0\leq i<j\leq k).  \label{e3.28}
\end{equation}
For each sufficiently large $| z| =r$, we take a point 
$z_r=re^{i\theta _r}$ satisfying $| f(z_r)| =M(r,f) $. By Lemma \ref{lem2.6}, 
there  exists a constant $\delta _r$ $(>0)$ and a set $F_2$ of finite
logarithmic measure such that for all all $z$ satisfying 
$|z| =r\notin F_2$ and $\arg z=\theta \in [ \theta_r-\delta _r,\theta _r
+\delta _r] $, we have
\begin{equation}
| \frac{f(z)}{f^{(d) }(z)}| \leq r^{2d}\quad (d=s,k).  \label{e3.29}
\end{equation}
\smallskip

\noindent\textbf{Case 1.} 
Suppose that $\theta _1$ $\neq \theta _2$. For any
given 
$\theta \in [ \theta _r-\delta _r,\theta _r+\delta _r
] \backslash E_2\cup H_1\cup H_2$, 
we have \eqref{e3.2},
where $E_2$, $H_1$ and $H_2$ are defined above. Set 
$\delta_1=\delta (P_1,\theta ) $ and $\delta _2=\delta (P_2,\theta ) $.
\smallskip

\noindent\textbf{Subcase 1.1: $\delta _1>\delta _2$.}
Here we also divide our proof in three subcases:

\noindent\textbf{(a): $\delta _1>\delta _2>0$.} Thus for any given 
$\varepsilon $ $(0<\varepsilon <\frac{\delta _1-\delta _3}{2(\delta _1+\delta _3) }) $
 and all $z$ satisfying $\arg z=\theta $ and $| z| =r$ sufficiently large, we have 
\eqref{e3.3}--\eqref{e3.5}, where $\delta _3$ is defined
above. By \eqref{e1.3}, \eqref{e3.3}--\eqref{e3.5}, 
\eqref{e3.28} and \eqref{e3.29}, for all $z$ satisfying 
$|z|=r\notin [ 0,1]\cup F_1\cup F_2$ and
$\arg z\in [ \theta _r-\delta _r,\theta _r+\delta _r] \backslash E_2\cup
H_1\cup H_2$, we obtain
\begin{equation}
\exp \{(1-\varepsilon )\delta _1r^{n}\}\leq M_5r^{2s+1}\exp \{
(1+\varepsilon ) \delta _3r^{n}\} [ T(2r,f)]
^{k+1},  \label{e3.30}
\end{equation}
where $M_5(>0) $ is a constant. Hence by using Lemma \ref{lem2.7} and 
\eqref{e3.30}, we obtain $\sigma _2(f) \geq n$. From
this and Lemma \ref{lem2.8}, we have $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{(b): $\delta _1>0>\delta _2$.}
 Thus for any given 
$\varepsilon $ $(0<\varepsilon <\frac{1-\alpha }{2(1+\alpha )})$ and all $z$
satisfying $\arg z=\theta $ and $|z| =r$ sufficiently
large, we have \eqref{e3.3}  and \eqref{e3.9}--\eqref{e3.11}. 
By \eqref{e1.3}, \eqref{e3.3}, \eqref{e3.9}---\eqref{e3.11}, \eqref{e3.28}
and \eqref{e3.29}, for all $z$ satisfying
 $|z|=r\notin [ 0,1]\cup F_1\cup F_2$ and
$\arg z\in [ \theta _r-\delta _r,\theta
_r+\delta _r] \backslash E_2\cup H_1\cup H_2$, we obtain
\begin{equation}
\exp \{(1-\varepsilon )\delta _1r^{n}\}\leq M_6r^{2s+1}\exp
\{(1+\varepsilon )\alpha \delta _1r^{n}\}[ T(2r,f)] ^{k+1},
\label{e3.31}
\end{equation}
where $M_6(>0)$ is a constant. Hence by using Lemma \ref{lem2.7} and 
\eqref{e3.31}, we obtain $\sigma _2(f) \geq n$. From
this and Lemma \ref{lem2.8}, we have $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{(c): $0>\delta _1>\delta _2$.} Set 
$\gamma =\min \{ \alpha _j,\beta _j:j\neq s\}$. Thus for any given 
$\varepsilon $ $(0<2\varepsilon <1) $ and all $z$ satisfying 
$\arg z=\theta $ and $| z| =r$ sufficiently large, we have
\begin{gather}
| A_l(z)e^{P_l(z) }| \leq \exp
\{(1-\varepsilon )\gamma \delta _1r^{n}\}\quad (l=1,2), \label{e3.32} \\
| B_j(z)e^{Q_j(z) }| \leq \exp
\{(1-\varepsilon )\gamma \delta _1r^{n}\}\text{ }(j\neq s) .
\label{e3.33}
\end{gather}
By \eqref{e1.3}, \eqref{e3.28}, \eqref{e3.29}, 
\eqref{e3.32} and \eqref{e3.33}, for all $z$ satisfying 
$|z|=r\notin [ 0,1]\cup F_1\cup F_2$ and
 $\arg z\in [ \theta _r-\delta _r,\theta _r+\delta _r] \backslash E_2\cup
H_1\cup H_2$, we obtain
\begin{equation}
1\leq M_7r^{2k+1}\exp \{ (1-\varepsilon ) \gamma \delta
_1r^{n}\} [ T(2r,f)] ^{k+1},  \label{e3.34}
\end{equation}
where $M_7$ $(>0) $ is a constant. Hence by using Lemma \ref{lem2.7}
and \eqref{e3.34}, we obtain $\sigma _2(f) \geq n$.
From this and Lemma \ref{lem2.8}, we have $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{Subcase 1.2: $\delta _1<\delta _2$.}
Using the same reasoning as in subcase 1.1 of the second step, we can also obtain
that $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{Case 2.} Suppose that $\theta _1=\theta _2$. For any
given $\theta \in [ \theta _r-\delta _r,\theta _r+\delta _r
] \backslash E_2\cup H_1$, where $E_2$ and $H_1$ are defined
above, we have \eqref{e3.17}.
\smallskip

\noindent\textbf{Subcase 2.1: $\delta (P_1,\theta ) >0$.}
\smallskip

\noindent\textbf{(i)} 
If $| a_{n,1}| <(1-\beta ) | a_{n,2}| $, for any given $\varepsilon $,
\[
0<\varepsilon <\frac{(1-\beta ) | a_{n,2}|-| a_{n,1}| }{2[ (1+\beta ) |
a_{n,2}| +| a_{n,1}| ] },
\]
 and all $z$ satisfying $\arg z=\theta $ and $| z| =r$ sufficiently
large, we have \eqref{e3.18} -\eqref{e3.21}. By
\eqref{e1.3}, \eqref{e3.18}--\eqref{e3.21}, \eqref{e3.28}
 and \eqref{e3.29}, for all $z$ satisfying $|z|=r\notin [0,1]\cup F_1\cup F_2$ 
and $\arg z\in [ \theta _r-\delta _r,\theta _r+\delta _r] \backslash E_2\cup H_1$, 
we obtain
\begin{equation}
\exp \{\gamma _1r^{n}\}\leq M_8r^{2s+1}[ T(2r,f)] ^{k+1},
\label{e3.35}
\end{equation}
where $M_8$ $(>0) $ is a constant and $\gamma _1$ is defined
above. Hence by using Lemma \ref{lem2.7} and \eqref{e3.35}, we obtain 
$\sigma_2(f) \geq n$. From this and Lemma \ref{lem2.8}, we have 
$\sigma_2(f) =n$.
\smallskip

\noindent\textbf{(ii)} If $| a_{n,2}| <(1-\alpha) | a_{n,1}| $, for any 
given $\varepsilon$ $(0<\varepsilon <\frac{(1-\alpha ) | a_{n,1}|
-| a_{n,2}| }{2[ (1+\alpha ) | a_{n,1}| +| a_{n,2}| ] })$ and all 
$z$ satisfying $\arg z=\theta $ and $| z| =r$ sufficiently
large, we have \eqref{e3.20} $, \eqref{e3.21} $, 
\eqref{e3.24} and \eqref{e3.25}. By \eqref{e1.3}, 
\eqref{e3.20}, \eqref{e3.21}, \eqref{e3.24}, \eqref{e3.25}, \eqref{e3.28}
 and \eqref{e3.29}  for all $z$ satisfying $|z|=r\notin [ 0,1]\cup F_1\cup F_2$ 
and $\arg z\in[ \theta _r-\delta _r,\theta _r+\delta _r] \backslash
E_2\cup H_1$, we obtain
\begin{equation}
\exp \{\gamma _2r^{n}\}\leq M_9r^{2s+1}[ T(2r,f)] ^{k+1},
\label{e3.36}
\end{equation}
where $M_9$ $(>0) $ is a constant and $\gamma _2$ is defined
above. Hence by using Lemma \ref{lem2.7} and \eqref{e3.36}, we obtain 
$\sigma_2(f) \geq n$. From this and Lemma \ref{lem2.8}, we have $\sigma_2(f) =n$.
\smallskip

\noindent\textbf{Subcase 2.2: $\delta (P_1,\theta ) <0$.}
Using the same reasoning as in subcase 1.1(c) of the second
step, we can also obtain that $\sigma _2(f) =n$.

\section{Proof of Theorem \ref{thm1.2}}

Assume $f$ is a transcendental solution of  \eqref{e1.3}.
\smallskip

\noindent\textbf{First step.} 
We prove that $\sigma (f)=+\infty $. Suppose that
 $\sigma (f) =\rho <+\infty $. By Lemma \ref{lem2.1}, there exists a set 
$E_1\subset [ 0,2\pi )$ that has linear measure zero such that if
$\theta \in [ 0,2\pi ) \backslash E_1$, then there is a constant 
$R_1=R_1(\theta ) >1$ such that for all $z$ satisfying 
$\arg z=\theta $ and $| z| \geqslant R_1$, we have 
\eqref{e3.1}. Set $a_{n,l}=| a_{_{n,l}}|e^{i\theta _l}$,
$\theta _l\in [ 0,2\pi )$ $(l=1,2) $.

 Assume that $a_{n,1}$ is a real number such that 
$(1-\alpha)a_{n,1}<b$, which is $\theta _1=\pi $. By Lemmas \ref{lem2.2}, for any given 
$\varepsilon >0$, there exist a set $E_2\subset [ 0,2\pi ) $
that has linear measure zero, such that if $z=re^{i\theta }$, 
$\theta \in[ 0,2\pi ) /E_2\cup H_1$ and $r$ is sufficiently large,
then $A_l(z)e^{P_l(z) }$ $(l=1,2) $ and 
$B_j(z)e^{Q_j(z) }$ $(j\neq s) $ satisfy \eqref{e2.2} or \eqref{e2.3}, 
where $H_1$ is defined as in the proof of Theorem \ref{thm1.1}.
\smallskip

\noindent\textbf{Case 1.} 
Suppose that $\theta _1$ $\neq \theta _2$.
For any given $\theta \in [ 0,2\pi ] /E_1\cup E_2\cup
H_1\cup H_2$, we have \eqref{e3.2}, where $H_2$ is defined as
in the proof of Theorem \ref{thm1.1}. Since $(1-\alpha )a_{n,s}<b$, it follows that
 $| b_{_{n,j}}| <| a_{_{n,1}}| $ $(j\in J) $. Set 
$\delta _1=\delta (P_1,\theta ) $ and $
\delta _2=\delta (P_2,\theta ) $.
\smallskip

\noindent\textbf{Subcase 1.1: $\delta _1>\delta _2$.}
If (a): $\delta _1>\delta _2>0$ or 
(b): $\delta _1>0>\delta _2$, it follows that 
$0<\delta (Q_j,\theta )<\delta _1$  $(j\in J) $.
 Hence by using the same reasoning as in subcase
1.1(a) of the first step in the proof of Theorem \ref{thm1.1}, 
we can also obtain that $\sigma (f) =+\infty $. If 
(c): $0>\delta_1>\delta _2$, by using similar reasoning as in subcase
 1.1(c) of the first step in the proof of Theorem \ref{thm1.1}, we can also obtain 
$\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{Subcase 1.2: $\delta _2>\delta _1$.}
Here we also divide our proof in three subcases:

\noindent\textbf{(a): $\delta _2>\delta _1>0$.}
 Thus for any given  $\varepsilon $ 
$(0<\varepsilon <\frac{\delta _2-\delta _1}{2(
\delta _2+\delta _1) }) $ and all $z$ satisfying 
$\arg z=\theta $ and $| z| =r$ sufficiently large, we have
\begin{gather}
| A_2(z)e^{P_2(z) }| \geq \exp
\{(1-\varepsilon )\delta _2r^{n}\},  \label{e4.1} \\
| A_1(z)e^{P_1(z) }| \leq \exp
\{(1+\varepsilon )\delta _1r^{n}\},  \label{e4.2} \\
| B_j(z)e^{Q_j(z) }| \leq \exp
\{(1+\varepsilon )\delta _1r^{n}\}\text{ }(j\neq s) .
\label{e4.3}
\end{gather}
 Now we prove that $| f^{(s) }(z)| $ is bounded on the ray 
$\arg z=\theta $. If $|f^{(s) }(z) | $ is unbounded on the ray 
$\arg z=\theta $, then by Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$\ $(m=1,2,\dots ) $, where 
$r_m\to +\infty $ such that $f^{(s) }(z_m) \to \infty $ and \eqref{e3.6} holds.

 By \eqref{e1.3}, \eqref{e3.1}, \eqref{e3.6}
and \eqref{e4.1}--\eqref{e4.3}, for the above $z_m$, we obtain
\begin{equation}
\exp \{(1-\varepsilon )\delta _2r_m^{n}\}\leq M_1r_m^{d_1}\exp
\{(1+\varepsilon )\delta _1r_m^{n}\},  \label{e4.4}
\end{equation}
where $M_1$, $d_1(>0) $ are constants. This is a
contradiction. Hence $| f^{(s) }(z)| \leq M$ on 
$\arg z=\theta $, where $M$ $(>0) $ is a constant. 
We can easily obtain \eqref{e3.8} 
on $\arg z=\theta $. By Lemma \ref{lem2.4}, \eqref{e3.8} and the
fact that $E_1\cup E_2\cup H_1\cup H_2$ has linear measure zero, we
obtain that $f(z) $ is a polynomial with $\deg f\leq s$
which contradicts our assumption. Therefore $\sigma (f) =+\infty$.
\smallskip

\noindent\textbf{(b): $\delta _2>0>\delta _1$.} Thus for any given 
$\varepsilon $ $(0<2\varepsilon <1) $ and all $z$ satisfying 
$\arg z=\theta $ and $| z| =r$ sufficiently large, we have
\eqref{e4.1},
\begin{gather}
| A_1(z)e^{P_1(z) }| \leq \exp
\{(1-\varepsilon )\delta _1r^{n}\}<1,  \label{e4.5} \\
| B_j(z)e^{Q_j(z) }| \leq \exp
\{(1-\varepsilon )\delta (Q_j,\theta )r^{n}\}<1\text{ }(j\neq s) .  \label{e4.6}
\end{gather}

 Now we prove that $| f^{(s) }(z) | $ is bounded on the ray $\arg z=\theta $. 
If $|f^{(s) }(z) | $\ is unbounded on the ray 
$\arg z=\theta $, then by Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$\ $(m=1,2,\dots ) $, where 
$r_m\to +\infty $ such that $f^{(s) }(z_m) \to \infty $
 and \eqref{e3.6} holds.

 By \eqref{e1.3}, \eqref{e3.1}, \eqref{e3.6}, \eqref{e4.1}, \eqref{e4.5}
 and \eqref{e4.6}, for the above $z_m$, we obtain
\begin{equation}
\exp \{(1-\varepsilon )\delta _2r_m^{n}\}\leq M_2r_m^{d_2},
\label{e4.7}
\end{equation}
where $M_2$, $d_2(>0) $ are constants. This is a
contradiction. Hence $| f^{(s) }(z) | \leq M$ on $\arg z=\theta $, 
where $M$ $(>0) $ is a constant. We can easily obtain \eqref{e3.8} 
on $\arg z=\theta $. Using similar arguments as above, we conclude
that $\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{(c): $0>\delta _2>\delta _1$.} Using similar reasoning
as in subcase 1.1.(c) of the first step in the proof of Theorem \ref{thm1.1}, 
we can also obtain that $\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{Case 2.} Assume that $\theta _1=\theta _2$. Then
 $ \theta _1=\theta _2=\pi $. For any given $\theta \in [ 0,2\pi
) /E_1\cup E_2\cup H_1$, we have \eqref{e3.17}.
\smallskip

\noindent\textbf{Subcase 2.1: $\delta (P_1,\theta ) >0$.}
 Since $| a_{n,1}| \leq |a_{n,2}| ,a_{n,1}\neq a_{n,2}$ and 
$\theta _1=\theta _2$, it follows that
$| a_{n,1}| <| a_{n,2}| $. Thus for
any given $\varepsilon $ $(0<\varepsilon <\frac{|
a_{n,2}| -| a_{n,1}| }{2(|a_{n,2}| +| a_{n,1}| ) })$ and all $z$
satisfying $\arg z=\theta $ and $| z| =r$ sufficiently
large, we have \eqref{e3.18}--\eqref{e3.20} and
\begin{equation}
\begin{aligned}
| B_j(z) e^{Q_j(z) }|
&\leq \exp \{(1+\varepsilon )\delta (Q_j,\theta )r^{n}\}   \\
&\leq \exp \{(1+\varepsilon )br^{n}\cos (n\theta ) \}(j\in J) .  \label{e4.8}
\end{aligned}
\end{equation}
By \eqref{e3.18} and \eqref{e3.19}, we obtain
\begin{equation}
| A_1(z)e^{P_1(z)}+A_2(z)e^{P_2(z)}| \geq \exp
\{(1+\varepsilon )\delta (P_1,\theta )r^{n}\}[ \exp \{\gamma
_1r^{n}\}-1] ,  \label{e4.9}
\end{equation}
where
\begin{equation*}
\gamma _1=(1-\varepsilon )\delta (P_2,\theta )-(1+\varepsilon )\delta
(P_1,\theta )>0.
\end{equation*}
Hence from \eqref{e4.9}, we obtain
\begin{equation}
\begin{aligned}
| A_1(z)e^{P_1(z)}+A_2(z)e^{P_2(z)}|
\geq (1-o(1)) \exp \{(1+\varepsilon )\delta (P_1,\theta )r^{n}\}\exp
\{\gamma _1r^{n}\}.
\end{aligned}  \label{e4.10}
\end{equation}
Now we prove that $| f^{(s) }(z)| $ is bounded on the ray $\arg z=\theta $.
If $|f^{(s) }(z) | $ is unbounded on the ray
$\arg z=\theta $, then by Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$ $(m=1,2,\dots ) $, where $r_m\to +\infty $ such that
$f^{(s) }(z_m) \to \infty $\ and \eqref{e3.6} holds.

 By \eqref{e1.3}, \eqref{e3.1}, \eqref{e3.6}, \eqref{e3.20}, 
\eqref{e4.8} and \eqref{e4.10}, we obtain
\begin{equation}
\begin{aligned}
&(1-o(1)) \exp \{(1+\varepsilon )\delta (P_1,\theta
)r_m^{n}\}\exp \{\gamma _1r_m^{n}\}   \\
&\leq M_3r_m^{d_3}\exp \{(1+\varepsilon )\alpha \delta (P_1,\theta
)r_m^{n}\}\exp \{(1+\varepsilon )br_m^{n}\cos (n\theta ) \},
\end{aligned} \label{e4.11}
\end{equation}
where $M_3$, $d_3$ $(>0) $ are constants. Hence
\begin{equation}
(1-o(1)) \exp \{\gamma _2r_m^{n}\}\leq M_3r_m^{d_3}, \label{e4.12}
\end{equation}
where
\begin{equation*}
\gamma _2=(1+\varepsilon )[ (1-\alpha )\delta (P_1,\theta )-b\cos
(n\theta ) ] +\gamma _1.
\end{equation*}
Since $\gamma _1>0$, $\cos (n\theta ) <0$,
$\delta(P_1,\theta )=-| a_{n,1}| \cos (n\theta ) $
and $(1-\alpha ) a_{n,1}<b<0$, we have
\begin{equation*}
\gamma _2=-(1+\varepsilon )[ (1-\alpha )| a_{n,1}|
+b] \cos (n\theta ) +\gamma _1>\gamma _1>0.
\end{equation*}
Hence \eqref{e4.11} is a contradiction. Hence
$| f^{(s) }(z) | \leq M$ on $\arg z=\theta $,
where $M$ $(>0) $ is a constant. We can easily obtain
\eqref{e3.8} on $\arg z=\theta $. By Lemma \ref{lem2.4}, \eqref{e3.8}
 and the fact that $E_1\cup E_2\cup H_1$ has linear measure zero, we
obtain that $f(z) $ is a polynomial with $\deg f\leq s$
which contradicts our assumption. Therefore $\sigma (f) =+\infty$.
\smallskip

\noindent\textbf{Subcase 2.2: $\delta (P_1,\theta ) <0$.} 
Using similar reasoning as in subcase 1.1(c) of the first step
in the proof of Theorem \ref{thm1.1}, we can also obtain that $\sigma (f)=+\infty $.
\smallskip

\noindent\textbf{Second step.} Now we prove that 
$\sigma _2(f) =n$. By Lemma \ref{lem2.5}, there exist a constant $B>0$ and a set 
$F_1\subset (1,+\infty )$ having finite logarithmic measure such that for
all $z$ satisfying $|z|=r\notin [ 0,1]\cup F_1$, we have \eqref{e3.28}. 
For each sufficiently large $| z| =r$, we take a point $z_r=re^{i\theta _r}$ 
satisfying $| f( z_r) | =M(r,f) $. By Lemma \ref{lem2.6}, there
exists a constant $\delta _r$\ $(>0) $ and a set $F_2$ of
finite logarithmic measure such that for all all $z$ satisfying 
$|z| =r\notin F_2$ and 
$\arg z=\theta \in [ \theta _r-\delta _r,\theta _r+\delta _r] $, we have 
\eqref{e3.29}.
\smallskip

\noindent\textbf{Case 1} Suppose that $\theta _1 \neq \theta _2$.
For any given $\theta \in [ \theta _r-\delta _r,\theta _r+\delta
_r] \backslash E_2\cup H_1\cup H_2$, we have \eqref{e3.2}, where 
$E_2$, $H_1$ and $H_2$ are defined above. Set $\delta _1=\delta (P_1,\theta ) $ 
and $\delta _2=\delta (P_2,\theta ) $.
\smallskip

\noindent\textbf{Subcase 1.1: $\delta _1>\delta _2$.} 
Using the same reasoning as in subcase 1.1(a) of the second step in the proof of
Theorem \ref{thm1.1}, we can also obtain that $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{Subcase 1.2: $\delta _2>\delta _1$.}
Here we also divide our proof in three subcases:
\smallskip

\noindent\textbf{(a): $\delta _2>\delta _1>0$.} 
Thus for any given $\varepsilon $ 
$(0<\varepsilon <\frac{\delta _2-\delta _1}{2(\delta _2+\delta _1) }) $ and all 
$z$ satisfying $\arg z=\theta $ and $|z| =r$ sufficiently large, we have 
\eqref{e4.1}--\eqref{e4.3}. By \eqref{e1.3}, \eqref{e3.28}, \eqref{e3.29} and 
\eqref{e4.1}--\eqref{e4.3}, for all $z$ satisfying 
$|z|=r\notin [ 0,1]\cup F_1\cup F_2$ and 
$\arg z\in [ \theta _r-\delta _r,\theta _r+\delta _r
] \backslash E_2\cup H_1\cup H_2$, we obtain
\begin{equation}
\exp \{(1-\varepsilon )\delta _2r^{n}\}\leq M_4r^{2s+1}\exp
\{(1+\varepsilon )\delta _1r^{n}\}[ T(2r,f)] ^{k+1},  \label{e4.13}
\end{equation}
where $M_4$ $(>0) $ is a constant. Hence by using Lemma \ref{lem2.7} and 
\eqref{e4.13}, we obtain $\sigma _2(f) \geq n$. From
this and Lemma \ref{lem2.8}, we have $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{(b): $\delta _2>0>\delta _1$.}
 Thus for any given $\varepsilon $ $(0<2\varepsilon <1) $ and all $z$ 
satisfying $\arg z=\theta $ and $| z| =r$ sufficiently large, we have
\eqref{e4.1}, \eqref{e4.5} and \eqref{e4.6}. By 
\eqref{e1.3}, \eqref{e3.28}, \eqref{e3.29}, \eqref{e4.1}, \eqref{e4.5}
 and \eqref{e4.6}, for all $z$
satisfying $|z|=r\notin [ 0,1]\cup F_1\cup F_2$ and 
$\arg z\in [ \theta _r-\delta _r,\theta _r+\delta _r] \backslash
E_2\cup H_1\cup H_2$, we obtain
\begin{equation}
\exp \{(1-\varepsilon )\delta _2r^{n}\}\leq M_5r^{2s+1}[ T(2r,f)
] ^{k+1},  \label{e4.14}
\end{equation}
where $M_5(>0) $ is a constant. Hence by using Lemma \ref{lem2.7} and 
\eqref{e4.14}, we obtain $\sigma _2(f) \geq n$. From
this and Lemma \ref{lem2.8}, we have $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{(c): $0>\delta _2>\delta _1$.} Using similar reasoning
as in subcase 1.1(c) of the second step in the proof of Theorem \ref{thm1.1}, we can
also obtain that $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{Case 2.} Assume that $\theta _1=\theta _2$. For any
given $\theta \in [ \theta _r-\delta _r,\theta _r+\delta _r
] \backslash E_2\cup H_1$, where $E_2$ and $H_1$ are defined
above, we have \eqref{e3.17}.
\smallskip

\noindent\textbf{Subcase 2.1: $\delta (P_1,\theta ) >0$.} 
For any given $\varepsilon $,
\[
0<\varepsilon <\frac{|a_{n,2}| -| a_{n,1}| }{2(|
a_{n,2}| +| a_{n,1}| ) },
\]
and all $z$ satisfying $\arg z=\theta $ and $|z| =r$ sufficiently
large, we have \eqref{e3.20}, \eqref{e4.8} and \eqref{e4.10}. By 
\eqref{e1.3}, \eqref{e3.20}, \eqref{e3.28}, \eqref{e3.29}, \eqref{e4.8} and 
\eqref{e4.10}, for all $z$
satisfying $|z|=r\notin [ 0,1]\cup F_1\cup F_2$ and $\arg z\in
[ \theta _r-\delta _r,\theta _r+\delta _r] \backslash
E_2\cup H_1$, we obtain
\begin{equation}
(1-o(1) ) \exp \{\gamma _2r^{n}\}\leq
M_6r^{2s+1} [ T(2r,f)] ^{k+1},  \label{e4.15}
\end{equation}
where $M_6$ $(>0) $ is a constant. Hence by using Lemma \ref{lem2.7}
and \eqref{e4.15}, we obtain $\sigma _2(f) \geq n$.
From this and Lemma \ref{lem2.8}, we have $\sigma _2(f) =n$.
\smallskip

\noindent\textbf{Subcase 2.2: $\delta (P_1,\theta ) <0$.}
Using similar reasoning as in subcase 1.1(c) of the second step
in the proof of Theorem \ref{thm1.1}, we can also obtain that $\sigma _2(f) =n$.

\section{Proof of Theorem \ref{thm1.3}}

 Assume $f$ is a transcendental solution of equation \eqref{e1.3}.
\smallskip

\noindent\textbf{First step.} We prove that $\sigma (f)
=+\infty $. Suppose that $\sigma (f) =\rho <+\infty $. By Lemma
\ref{lem2.1}, there exists a set $E_1\subset [ 0,2\pi )$ that has linear measure 
zero such that if $\theta \in [0,2\pi ) -E_1$, then there is a constant 
$R_1=R_1(\theta ) >1$\ such that for all $z$\ satisfying $\arg z=\theta $ and 
$| z| \geqslant R_1$, we have \eqref{e3.1}. Set 
$a_{n,l}=| a_{_{n,l}}| e^{i\theta _l}$,
$\theta _l\in [ 0,2\pi ) $ $(l=1,2) $.

 Assume that $a_{n,1}$ and $a_{n,2}$ are real numbers such that 
$(1-\beta ) a_{n,2}-b<a_{n,1}<0$ or 
$(1-\alpha ) a_{n,1}-b<a_{n,2}<0$, which is $\theta _1=\theta _2=\pi $. 
By Lemma \ref{lem2.2}, for any given $\varepsilon >0$, there exists a set
 $E_2\subset [ 0,2\pi ) $ that has linear measure zero, such that if 
$z=re^{i\theta }$, $\theta \in [ 0,2\pi ) /E_2\cup H_3$ and $r$ is
sufficiently large, then $A_l(z)e^{P_l(z) }$ 
$(l=1,2) $ and $B_j(z)e^{Q_j(z) }$ $(j\neq s) $ satisfy \eqref{e2.2}
 or \eqref{e2.3}, where $H_3=\{ \theta \in [ 0,2\pi ) :\cos (n\theta )
=0\} $.

 For any given $\theta \in [ 0,2\pi ) \backslash
E_1\cup E_2\cup H_3$, we have \eqref{e3.17}.
\smallskip

\noindent\textbf{Case 1: $\delta (P_1,\theta ) >0$.}

\noindent\textbf{(i)} If $(1-\beta ) a_{n,2}-b<a_{n,1}<0$, for
any given $\varepsilon $ $(0<\varepsilon <\frac{(1-\beta )
| a_{n,2}| -| a_{n,1}| +b}{2[
(1+\beta ) | a_{n,2}| +|a_{n,1}| -b] })$ and all $z$ satisfying 
$\arg z=\theta $ and $| z| =r$ sufficiently large, we have \eqref{e3.18}, 
\eqref{e3.19},  \eqref{e4.8}, and
\begin{equation}
| B_j(z)e^{Q_j(z)}| \leq \exp \{(1+\varepsilon
)\alpha \delta (P_1,\theta )r^{n}\}\exp \{(1+\varepsilon )\beta \delta
(P_2,\theta )r^{n}\}\quad (j\in I).  \label{e5.1}
\end{equation}
Now we prove that $| f^{(s) }(z)| $ is bounded on the ray $\arg z=\theta $. 
If $|f^{(s) }(z) | $ is unbounded on the ray $\arg z=\theta $, then 
by Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$\ $(m=1,2,\dots ) $, where 
$r_m\to +\infty $ such that $f^{(s) }(z_m) \to \infty $ and \eqref{e3.6}
holds.
 By \eqref{e1.3}, \eqref{e3.1}, \eqref{e3.6}, \eqref{e3.18}, \eqref{e3.19}, 
\eqref{e4.8} and \eqref{e5.1}, for
the above $z_m$, we obtain
\begin{equation} 
\begin{aligned}
&\exp \{(1-\varepsilon )\delta (P_2,\theta )r_m^{n}\}   \\
&\leq M_1r_m^{d_1}\exp \{(1+\varepsilon )[ \delta (P_1,\theta
)+\beta \delta (P_2,\theta )+b\cos (n\theta ) ]
r_m^{n}\},
\end{aligned}\label{e5.2}
\end{equation}
where $M_1$, $d_1$ $(>0) $ are constants. By \eqref{e5.2}, we have
\begin{equation}
\exp \{\gamma _1r_m^{n}\}\leq M_1r_m^{d_1},  \label{e5.3}
\end{equation}
where
\begin{equation*}
\gamma _1=(1-\varepsilon ) \delta (P_2,\theta
)-(1+\varepsilon )[ \delta (P_1,\theta )+\beta \delta (P_2,\theta
)+b\cos (n\theta ) ] .
\end{equation*}
From
\[
0<\varepsilon <\frac{(1-\beta ) |
a_{n,2}| -| a_{n,1}| +b}{2[ (1+\beta
)| a_{n,2}| +| a_{n,1}| -b] }
\]
and $\cos (n\theta ) <0$, we obtain
\begin{align*}
\gamma _1
&=-\{(1-\beta ) | a_{n,2}|
-| a_{n,1}| +b-\varepsilon [ (1+\beta )
| a_{n,2}| +| a_{n,1}| -b]
\}\cos (n\theta ) \\
&>-\frac{[ (1-\beta ) | a_{n,2}|
-| a_{n,1}| +b] }{2}\cos (n\theta ) >0.
\end{align*}
Thus \eqref{e5.3} is a contradiction. Hence $|f^{(s) }(z) | \leq M$ on
$\arg z=\theta $, where $M$ $(>0) $ is a constant. We can easily
obtain \eqref{e3.8} on
$\arg z=\theta $. By Lemma \ref{lem2.4},
\eqref{e3.8} and the fact that $E_1\cup E_2\cup H_1$ has linear
measure zero, we obtain that $f(z) $ is a polynomial
with $\deg f\leq s$, which contradicts our assumption.
Therefore $\sigma(f) =+\infty $.
\smallskip

\noindent\textbf{(ii)} 
If $(1-\alpha )a_{n,1}-b<a_{n,2}<0$, for any given 
$\varepsilon $,
\[
0<\varepsilon <\frac{(1-\alpha ) |
a_{n,1}| -| a_{n,2}| +b}{2[ (1+\alpha ) | a_{n,1}| +|
a_{n,2}| -b)] }),\]
and all $z$ satisfying $\arg z=\theta $
and $| z| =r$ sufficiently large, we have \eqref{e3.24} and \eqref{e3.25}.

 Now we prove that $| f^{(s) }(z)| $ is bounded on the ray $\arg z=\theta $. 
If $|f^{(s) }(z) | $ is unbounded on the ray 
$\arg z=\theta $, then by Lemma \ref{lem2.3}, there exists an infinite sequence of
points $z_m=r_me^{i\theta }$ $(m=1,2,\dots ) $, where 
$r_m\to +\infty $ such that $f^{(s) }(z_m) \to \infty $
 and \eqref{e3.6} holds.

 By \eqref{e1.3}, \eqref{e3.1}, \eqref{e3.6},
 \eqref{e3.24}, \eqref{e3.25}, \eqref{e4.8} and \eqref{e5.1}, for
the above $z_m$, we obtain
\begin{equation}
\begin{aligned}
&\exp \{(1-\varepsilon )\delta (P_1,\theta )r_m^{n}\}   \\
&\leq M_2r_m^{d_2}\exp \{(1+\varepsilon )[ \delta (P_2,\theta
)+\alpha \delta (P_1,\theta )+b\cos (n\theta ) ]
r_m^{n}\}, 
\end{aligned} \label{e5.4}
\end{equation}
where $M_2$, $d_2$ $(>0) $ are constants. By \eqref{e5.4}, we have
\begin{equation}
\exp \{\gamma _2r_m^{n}\}\leq M_2r_m^{d_2},  \label{e5.5}
\end{equation}
where
\begin{equation*}
\gamma _2=(1-\varepsilon ) \delta (P_1,\theta
)-(1+\varepsilon )[ \delta (P_2,\theta ) +\alpha \delta
(P_1,\theta ) +b\cos (n\theta ) ] >0.
\end{equation*}
 Thus \eqref{e5.5} is a contradiction. Hence $|
f^{(s) }(z) | \leq M$\ on $\arg z=\theta $
, where $M$ $(>0) $ is a constant. We can easily obtain 
\eqref{e3.8} on $\arg z=\theta $. Using similar arguments as
above, we deduce that $\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{Case 2: $\delta (P_1,\theta ) <0$.}
Using similar reasoning as in subcase 1.1(c) of the first step in the proof
of Theorem \ref{thm1.1}, we can also obtain that $\sigma (f) =+\infty $.
\smallskip

\noindent\textbf{Second step.} We prove that $\sigma _2(f)=n$.
 By Lemma \ref{lem2.5}, there exist a constant $B>0$ and a set $F_1\subset(1,+\infty )$ 
having finite logarithmic measure such that for all $z$
satisfying $|z|=r\notin [ 0,1]\cup E_1$, we have \eqref{e3.28}. For each 
sufficiently large $| z| =r$, we take a point
$z_r=re^{i\theta _r}$ satisfying $| f(z_r)| =M(r,f) $. 
By Lemma \ref{lem2.6}, there exists
 a constant $\delta _r$ $(>0) $ and a set $F_2$ of finite
logarithmic measure such that for all all $z$ satisfying 
$|z| =r\notin F_2$ and 
$\arg z=\theta \in [ \theta _r-\delta _r,\theta _r+\delta _r] $, we have 
\eqref{e3.29}. For any given 
$\theta \in [ \theta _r-\delta_r,\theta _r+\delta _r] \backslash E_2\cup H_3$, 
we have \eqref{e3.17}, where $E_2$ and $H_3$ are defined above.
\smallskip

\noindent\textbf{Case 1: $\delta (P_1,\theta ) >0$.}

\noindent\textbf{(i)} If $(1-\beta ) a_{n,2}-b<a_{n,1}<0$, for
any given $\varepsilon $ $(0<\varepsilon <\frac{(1-\beta )
| a_{n,2}| -| a_{n,1}| +b}{2[
(1+\beta ) | a_{n,2}| +|a_{n,1}| -b] })$ and all $z$ satisfying 
$\arg z=\theta $ and $| z| =r$ sufficiently large, we have \eqref{e3.18}, 
\eqref{e3.19}, \eqref{e4.8} and \eqref{e5.1}. By \eqref{e1.3}, \eqref{e3.18}, 
\eqref{e3.19}, \eqref{e3.28}, \eqref{e3.29}, \eqref{e4.8} and \eqref{e5.1}, 
for all $z$ satisfying $|z|=r\notin [ 0,1]\cup F_1\cup F_2$ and 
$\arg z\in [ \theta _r-\delta _r,\theta _r+\delta _r
] \backslash E_2\cup H_3$, we obtain
\begin{equation}
\exp \{\gamma _1r^{n}\}\leq M_3r^{2s+1}[ T(2r,f)] ^{k+1},\label{e5.6}
\end{equation}
where $M_3$ $(>0) $ is a constant and $\gamma _1$ is defined
above. Hence by using Lemma \ref{lem2.7} and \eqref{e5.6}, we obtain 
$\sigma_2(f) \geq n$. From this and Lemma \ref{lem2.8}, we have $\sigma_2(f) =n$.
\smallskip

\noindent\textbf{(ii)} If $(1-\alpha )a_{n,1}-b<a_{n,2}<0$, for any given 
$\varepsilon $,
\[
(0<\varepsilon <\frac{(1-\alpha ) |a_{n,1}| -| a_{n,2}| +b}{2[ (
1+\alpha ) | a_{n,1}| +|a_{n,2}| -b)] }),
\] 
and all $z$ satisfying $\arg z=\theta $
and $| z| =r$ sufficiently large, we have \eqref{e3.23} and \eqref{e3.24}. 
By \eqref{e1.3}, \eqref{e3.24}, \eqref{e3.25}, \eqref{e3.28}, 
\eqref{e3.29} and \eqref{e4.8}, for all $z$ satisfying 
$|z|=r\notin [ 0,1]\cup F_1\cup F_2$ and 
$\arg z\in [ \theta _r-\delta_r,\theta _r+\delta _r] \backslash E_2\cup H_3$, 
we obtain
\begin{equation}
\exp \{\gamma _2r^{n}\}\leq M_4r^{2s+1}[ T(2r,f)] ^{k+1},
\label{e5.7}
\end{equation}
where $M_4$ $(>0) $ is a constant and $\gamma _2$ is defined
above. Hence by using Lemma \ref{lem2.7} and \eqref{e5.7}, we obtain 
$\sigma_2(f) \geq n$. From this and Lemma \ref{lem2.8}, we have $\sigma_2(f) =n$.


\subsection*{Acknowledgements} 
The author is grateful to the referee
for his/her valuable comments which lead to the improvement of this paper. 
I also would like to thank Professor B. Bela\"{\i}di for several discussions
and for advising me about the paper of Heittokangas, Janne; Laine, Ilpo;
Tohge, Kazuya; Wen, Zhi-Tao.

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\end{document}