\documentclass[10pt]{studiamnew} \usepackage{graphicx} \usepackage{amsmath} \usepackage{amssymb} \sloppy \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \theoremstyle{definition} \newtheorem{remark}[theorem]{Remark} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{problem}[theorem]{Problem} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{solution}[theorem]{Solution} \newtheorem{notation}[theorem]{Notation} \renewcommand{\theequation}{\thesection.\arabic{equation}} \numberwithin{equation}{section} \begin{document} % \setcounter{page}{1} \setcounter{firstpage}{1} \setcounter{lastpage}{4} \renewcommand{\currentvolume}{??} \renewcommand{\currentyear}{??} \renewcommand{\currentissue}{??} % \title[Some classes of Janowski functions associated with conic domains]{Some Classes Of Janowski Functions Associated With Conic Domain And A Shell-Like Curve Involving Ruscheweyh Derivative} \author{Karthikeyan K. R} \address{Department of Mathematics and Statistics,\\ National University of Science \& Technology (By Merger of Caledonian College of Engineering and Oman Medical College),\\ Sultanate of Oman.} \email{kr\_karthikeyan1979@yahoo.com} % \author{Varadharajan S} \address{Mathematics Section,\\ Department of Information Technology,\\ University of Technology and Applied Sciences - Al Mussanah,\\ Sultanate of Oman.} \email{svrajanram@gmail.com} % \author{Lakshmi S} \address{Independent Researcher, \\Former Doctoral Student, Department of Mathematics, Presidency College (Autonomous), Chennai-600005, Tamilnadu, India.} \email{laxmirmk@gmail.com} % \subjclass{30C45} \keywords{analytic function, Schwarz function, starlike, convex, shell-like functions, Janowski functions, subordination, Fekete-Szeg\"{o} inequality, Ruscheweyh derivative} \begin{abstract} Making use of Ruscheweyh derivative, we define a new class of starlike functions of complex order subordinate to a conic domain impacted by Janowski functions. Coefficient estimates and Fekete-Szeg\"{o} inequalities for the defined class are our main results. Some of our results generalize the related work of some authors. \end{abstract} \maketitle \section{Introduction}\label{sec1} Let $\mathcal{A}$ denote the class of functions $f$ analytic in the open unit disk $$\mathcal{U}=\left\lbrace\ z \in \mathbb{C}:|z|< 1\right\rbrace$$ and satisfying the normalization condition $$ f(0)=0 \quad \textrm{and} \quad f^{\prime}(0)=1.$$ Thus, the functions in $\mathcal{A}$ are represented by the Taylor-Maclaurin series expansion given by \begin{equation}\label{eq1.1} f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n}, \quad (z \in \mathcal{U}). \end{equation} Let $\mathcal{S}\subset \mathcal{A}$ be the class of functions which are univalent. We let $\mathcal{S}^{*},\,\mathcal{C}$ and $\mathcal{K}$ to denote the well known classes of starlike, convex and close-to-convex (normalized) function respectively. For $0\leq\alpha<1$, $\mathcal{S}^{*}(\alpha)$ and $\mathcal{C}(\alpha)$ symbolize the classes of starlike functions of order $\alpha$ and convex functions of order $\alpha$ respectively. Also let $\mathcal{P}$ denote the class of functions of the form $p(z)=1+p_{1}z+p_{2}z^{2}+p_{3}z^{3}+\cdots$ that are analytic in $\mathcal{U}$ and such that $Re\left(p(z)\right)>0$ for all $z$ in $\mathcal{U}$. For arbitrary fixed numbers $A,\, B$, $-1k\sqrt{(u-1)^{2}+v^2}\right\}$ and is given by \begin{equation}\label{eqadditional} \hat{p}_{k,\alpha}(z)=\begin{cases} \frac{1+(1-2\alpha)z}{1-z}, & \mbox{if } k=0, \\ 1+\frac{2(1-\alpha)}{\pi^2}\left(\log\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)^{2}, & \mbox{if }k=1, \\ 1+\frac{2(1-\alpha)}{1-k^2}\sinh^2\left[\left(\frac{2}{\pi}\arccos k\right) arc\tanh\sqrt{z}\right], & \mbox{if }01, \end{cases} \end{equation} where $u(z)=\frac{z-\sqrt{t}}{1-\sqrt{tz}},\, t\in (0,\,1)$ and $t$ is chosen such that $k=\cosh\left(\frac{\pi R^{\prime}(t)}{4R(t)}\right)$, with $R(t)$ is Legendre’s complete elliptic integral of the first kind and $R^{\prime}(t)$ is complementary integral of $R(t)$. Clearly, $\hat{p}_{k,\alpha}(z)$ is in $\mathcal{P}$ with the expansion of the form \begin{equation}\label{eqpow} \hat{p}_{k,\alpha}(z)=1+\delta_{1}z+\delta_{2}z^{2}+\cdots,\qquad (\delta_{j}=p_{j}(k,\,\alpha),\,j=1,\,2,\,3,\ldots), \end{equation} we get \begin{equation}\label{eql2} \delta_{1}=\begin{cases} \frac{8(1-\alpha)(\arccos k)^{2}}{\pi^2(1-k^2)}, & \mbox{if } 0\leq k<1,\\ \frac{8(1-\alpha)}{\pi^2}, & \mbox{if } k=1, \\ \frac{\pi^2(1-\alpha)}{4\sqrt{t}(k^{2}-1)R^2(t)(1+t)}, & \mbox{if } k>1. \end{cases}\end{equation} Noor in \cite{noor1,noor} replaced $p(z)$ in (\ref{Janowski}) with $\hat{p}_{k,\alpha}(z)$ and studied the impact of Janowski function on conic regions. Let $f(z)$ and $g(z)$ be analytic in $\mathcal{U}$. Then we say that the function $f(z)$ is subordinate to $g(z)$ in $\mathcal{U}$, if there exists an Schwartz function $w(z)$ in $\mathcal{U}$ such that $|w(z)|<|z|$ and $f(z) = g(w(z))$, denoted by $f(z) \prec g(z)$. If $g(z)$ is univalent in $\mathcal{U}$, then the subordination is equivalent to $f(0) = g(0)$ and $f(\mathcal{U})\subset g(\mathcal{U})$. Using the concept of subordination for holomorphic functions, Ma and Minda \cite{Ma} introduced the classes $$\mathcal{S}^{*}(\phi)=\left\{f\in \mathcal{A}:\,\frac{zf^{'}(z)}{f(z)}\prec \phi\right\}\quad \textrm{and}\quad \mathcal{C}(\phi)=\left\{f\in \mathcal{A}:\,1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\prec \phi\right\} $$ where $\phi \in \mathcal{P}$ with $\phi^{'}(0)>0$ maps $\mathcal{U}$ onto a region starlike with respect to $1$ and symmetric with respect to real axis. By choosing $\phi$ to map unit disc on to some specific regions like parabolas, cardioid, lemniscate of Bernoulli, booth lemniscate in the right-half plane of the complex plane, various interesting subclasses of starlike and convex functions can be obtained. Raina\ and\ Sok\'{o}\l \,\cite{Raina} studied the class $\mathcal{S}^{*}(\phi)$ for $\phi(z) =z+\sqrt{1+z^2}$ and found some interesting coefficient inequalities. The function $\phi(z) =z+\sqrt{1+z^2}$ maps the unit disc $\mathcal{U}$ onto a shell shaped region on the right half plane and it is analytic and univalent on $\mathcal{U}$. For detailed study of starlike functions related to shell shaped region, refer to a recent work of Murugusundaramoorthy and Bulboac\u{a} \cite{murugu}. Khatter et al. \cite{Khatter} studied the convex combination of constant function $f(z)=1$ with $e^{z}$ and $\sqrt{1+z}$. Recently, Gandhi in \cite{Gandhi} studied a class $\mathcal{S}^{*}(\phi)$ with $\phi=\beta e^{z}+(1-\beta)(1+z)$, $0\leq \delta\leq 1$ a convex combination of two starlike functions. \begin{definition}\label{defnRuscheyh}\cite{Ruscheweyh} For $f \in \mathcal{A}$ of the form (\ref{eq1.1}) and $\lambda \in \mathbb{N}_0= \mathbb{N} \cup \{0\}$, the operator $R^{\lambda}$ is defined by $R^{\lambda}: \mathcal{A} \rightarrow \mathcal{A}$, \begin{align*} R^{0} f(z) &= f(z),\\ R^{1} f(z) &= zf^{\prime}(z),\\ &\vdots\\ (\lambda+1)R^{\lambda+1} f(z) &= z\left(R^{\lambda} f(z)\right)^{\prime}+\lambda R^{\lambda}f(z), \quad z \in \mathcal{U}. \end{align*} \end{definition} \begin{remark} If $f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n}$, then for $\lambda > -1$ \begin{align*} R^{\lambda} f(z) = \frac{z}{(1-z)^{\lambda+1}} \ast f(z)=z+ \sum\limits_{n=2}^{\infty} \varphi_{n}(\lambda) a_n z^n, \end{align*} where \begin{equation}\label{Ruscheweyh} \varphi_{n}(\lambda)= \frac{[\lambda+1]_{n-1}}{(n-1)!}, \end{equation} \begin{align*} [t]_{n} = \begin{cases} 1, &\quad n=0, \\ (t) (t + 1) (t + 2) \ldots (t + n - 1), &\quad n \in \mathbb{N}. \end{cases} \end{align*} is a Pochhammer symbol, \begin{align*} \Gamma (t + 1) = \begin{cases} 1, &\quad t=1, \\ [t] \Gamma (t), &\quad t >0. \end{cases} \end{align*} is a gamma function. The symbol $``\ast"$ stands for Hadamard product. \end{remark} Motivated by Gandhi \cite{Gandhi}, we introduce the following new subclasses of analytic functions using Ruscheweyh differential operator. \begin{definition}\label{defn1} For $\hat{p}_{k,\alpha}(z),\,(k\geq 0,\, 0\leq \alpha <1)$ is defined as in (\ref{eqadditional}), $-1\leq B < A \leq 1$, $\lambda >-1$, $\mid t \mid \leq 1, t \neq 1$ and for some $b \in \mathbb{C} \backslash \{0\}$, we let $k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, t, b)$ to be the class of functions $f \in \mathcal{A}$ satisfying the inequality \begin{equation}\label{eq1.6} 1+ \frac{1}{b}\left(\frac{(1-t)R^{\lambda+1}f(z)}{R^{\lambda}f(z)-R^{\lambda}f(tz)}-1\right) \prec \frac{(A+1)h(z)-(A-1)}{(B+1)h(z)-(B-1)}, \quad (z \in \mathcal{U}) \end{equation} where \begin{equation}\label{h}h(z)=\beta\left[\hat{p}_{k,\alpha}(z)\right]+(1-\beta)\left[z+\sqrt{1+z^2}\right],\,0\leq \beta\leq 1. \end{equation} \end{definition} \begin{remark} Note that $\hat{p}_{k,\alpha}(z)$ is not univalent but belongs to $\mathcal{P}$, whereas $z+\sqrt{1+z^2}$ is univalent in $\mathcal{U}$. Since the linear combination of two convex function is not convex in $|z| <1$, $h(z)$ is not convex univalent in $\mathcal{U}$. \end{remark} The following definition is motivated by the Alexander transform relationship between convex and starlike functions. \begin{definition}\label{defn2} For $\hat{p}_{k,\alpha}(z),\,(k\geq 0,\, 0\leq \alpha <1)$ is defined as in (\ref{eqadditional}), $-1\leq B < A \leq 1$, $\lambda >-1$, $\mid t \mid \leq 1, t \neq 1$ and for some $b \in \mathbb{C} \backslash \{0\}$, we let $k-\mathcal{CL}(A, B, \alpha, \beta, \lambda, t, b)$ to be the class of functions $f \in \mathcal{A}$ satisfying the inequality \begin{equation}\label{eq1.7} 1+ \frac{1}{b}\left(\frac{(1-t)\left(R^{\lambda+1}f(z)\right)^{\prime}}{\left(R^{\lambda}f(z)-R^{\lambda}f(tz)\right)^{\prime}}-1\right) \prec \frac{(A+1)h(z)-(A-1)}{(B+1)h(z)-(B-1)}, \quad (z \in \mathcal{U}) \end{equation} where $h(z)$ is defined as in (\ref{h}). \end{definition} \medskip We let $k-\mathcal{CL}(A, B, \lambda, t, b)$ and $k-\mathcal{CL}(A, B, \alpha, 1,\lambda, t, b)$ to denote the special cases of the function class $k-\mathcal{CL}(A, B, \alpha, \beta, \lambda, t, b)$ obtained by letting $\beta=0$ and $\beta=1$ respectively. \begin{remark} The versatility of classes $k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, t, b)$ and $k-\mathcal{CL}(A, B, \alpha, \beta, \lambda, t, b)$ is that it unifies the study of starlike and convex functions with respect to symmetric points. Here we list just a few special cases. \begin{enumerate} \item[1. ]If we let $b=1$, $t=0$, $\alpha=0$, $\beta=1$ and $\lambda=0$ in the definition of the function class $k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, t, b)$ and $k-\mathcal{CL}(A, B, \alpha, \beta, \lambda, t, b)$, we get the classes $k-\mathcal{SL} (A,B)$ and $k-\mathcal{CL} (A,B)$ introduced and studied by Noor and Malik in \cite{noor}. \item[2. ] For $b=1$, $\beta=1$ and $\lambda=0$, the class $k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, t, b)$ reduces to the respective classes $k-\mathcal{SL}(A,B,\alpha, 1, 0, t, 1)$ studied by Arif et al. in \cite{arif}. \end{enumerate} \end{remark} Unless otherwise mentioned, we assume throughout this paper that the function \newline $0\leq \alpha <1$, $0\leq \beta\leq 1$, $\lambda >-1$, $k\geq 0$, $-1\leq B < A \leq 1$, $ | t | \leq 1$, $t \neq 1$, $b \in \mathbb{C} \backslash \{0\}$ and $z \in \mathcal{U}$. \section{ Fekete-Szeg\"{o} inequalities for the starlike class $k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, t, b)$}\label{sec3} Many extremal problems within the class of univalent functions are solved by the Koebe function. On the other hand, the Koebe function satisfies $$\left|a_{3}-\lambda a_{2}^{2}\right|=\left|3-4\lambda\right|$$ whereas Fekete and Szeg\"{o} showed $$\max_{f\in \mathcal{S}}\left|a_{3}-\lambda a_{2}^{2}\right|=\left|3-4\lambda\right|=1+2e^{-2\lambda/(1-\lambda)}$$ for $\lambda\in [0,1]$. In this section, we obtain the Fekete-Szeg\"{o} for the class $k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, t, b)$. We need the following lemma to establish our main result. \begin{lemma}\label{lem3}\cite{Ma} Let $p(z)\in \mathcal{P}$ and also let $v$ be a complex number, then \begin{equation}\label{eq3.1} |c_{2}-vc_{1}^2|\leq 2\:\max \left\lbrace 1,|2v-1|\right\rbrace, \end{equation} the result is sharp for functions given by $$p(z)= \frac{1+z^2}{1-z^2}, \hspace{35pt} p(z)= \frac{1+z}{1-z}.$$ \end{lemma} \begin{theorem}\label{th3} If $f(z) \in k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, t, b)$ then for $\mu \in \mathbb{C}$ we have \begin{equation}\label{eq3.2} \mid a_3 - \mu a_2^2 \mid \leq \frac{\mid b \mid |\beta(\delta_{1}-1)+1|(A-B)}{2 [\varphi_{3}(\lambda+1) - u_{3} \varphi_{3}(\lambda)]} \:\max \left\lbrace 1,|2v-1|\right\rbrace, \end{equation} where \begin{align}\label{eq3.3} v & = \frac{1}{2}-\frac{\beta(2\delta_{2}-1)+1}{4\left[\beta(\delta_{1}-1)+1\right]}+ \frac{\left[\beta(\delta_{1}-1)+1\right](B+1)}{4} \nonumber \\ &\hspace{0.5in}- \frac{b \varphi_{2}(\lambda)[\beta(\delta_{1}-1)+1](A-B)}{4 [\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]} \left(u_2- \mu \frac{\varphi_{3}(\lambda +1) - u_{3} \varphi_{3}(\lambda)}{\varphi_{2}(\lambda) [\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]}\right) \end{align} and $u_n=1+t+t^2+\cdots+t^{n-1}.$ The result is sharp. \end{theorem} \begin{proof} Let $p(z) \in \mathcal{P}$ be of the form $1+ \sum\limits_{n=1}^{\infty} p_n z^n$, we consider $$p(z) = \frac{1+w(z)}{1-w(z)},$$ where $w(z)$ is such that $w(0)=0$ and $\mid w(z)\mid <1$. On simple computation, we have \begin{align}\label{eq3.4} w(z)&=\frac{p(z)-1}{p(z)+1}=\frac{p_1z+p_2z^2+p_3z^3+\cdots}{2+p_1z+p_2z^2+p_3z^3+\cdots}\nonumber\\ &= \frac{1}{2}p_1z+ \frac{1}{2}\left(p_2- \frac{1}{2}p_1^2\right)z^2+\frac{1}{2}\left(p_3-p_1p_2+ \frac{1}{4}p_1^3 \right)z^3+ \cdots. \end{align} Using (\ref{eq3.4}) in $h(z)=1+\left[\beta(\delta_{1}-1)+1\right]z+\displaystyle \frac{1}{2}\left[\beta(2\delta_{2}-1)+1\right]z^2+\cdots$, we have{\scriptsize \begin{align*} h(w(z))&=1+\left[\beta(\delta_{1}-1)+1\right]w(z)+\frac{1}{2}\left[\beta(2\delta_{2}-1)+1\right][w(z)]^2+\cdots\\ &= 1+ \left[\beta(\delta_{1}-1)+1\right] \left[\frac{1}{2}p_1z+ \frac{1}{2}\left(p_2- \frac{1}{2}p_1^2\right)z^2+\frac{1}{2}\left(p_3-p_1p_2+ \frac{1}{4}p_1^3 \right)z^3+ \cdots \right]\\& \quad\quad + \frac{1}{2}\left[\beta(2\delta_{2}-1)+1\right] \left[\frac{1}{2}p_1z+ \frac{1}{2}\left(p_2- \frac{1}{2}p_1^2\right)z^2+\frac{1}{2}\left(p_3-p_1p_2+ \frac{1}{4}p_1^3 \right)z^3+ \cdots \right]^2 + \cdots\\ &= 1+ \frac{\left[\beta(\delta_{1}-1)+1\right]p_1}{2}z+\frac{\left[\beta(\delta_{1}-1)+1\right]}{2}\left[p_2-\frac{p_1^2}{2}\left(1- \frac{\beta(2\delta_{2}-1)+1}{2 \left[\beta(\delta_{1}-1)+1\right]}\right) \right] z^2+\cdots . \end{align*} } As $f(z) \in k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, t, b)$, by (\ref{eq1.6}) we have \begin{equation}\label{eq3.5} 1+ \frac{1}{b}\left(\frac{(1-t)R^{\lambda+1}f(z)}{R^{\lambda}f(z)-R^{\lambda}f(tz)}-1\right) = p(z), \end{equation} where \begin{align}\label{eq3.6} p(z) &= \frac{(A+1)h(w(z))-(A-1)}{(B+1)h(w(z))-(B-1)}\nonumber \\ &= {\footnotesize \frac{2+ \frac{(A+1)\left[\beta(\delta_{1}-1)+1\right]p_1}{2}z+ \frac{(A+1)\left[\beta(\delta_{1}-1)+1\right]}{2}\left[p_2- \frac{p_1^2}{2}\left(1- \frac{\beta(2\delta_{2}-1)+1}{2 \left[\beta(\delta_{1}-1)+1\right]}\right) \right] z^2+\cdots}{2+ \frac{(B+1)\left[\beta(\delta_{1}-1)+1\right]p_1}{2}z+ \frac{(B+1)\left[\beta(\delta_{1}-1)+1\right]}{2}\left[p_2- \frac{p_1^2}{2}\left(1- \frac{\beta(2\delta_{2}-1)+1}{2 \left[\beta(\delta_{1}-1)+1\right]}\right) \right] z^2+\cdots}}\nonumber \\ &=1+ \frac{\left[\beta(\delta_{1}-1)+1\right](A-B)p_1}{4}z+\frac{\left[\beta(\delta_{1}-1)+1\right](A-B)}{4}\nonumber \\ &\hspace{0.5in}\Bigg[p_2 -\frac{p_1^2}{2}\Bigg(1- \frac{\beta(2\delta_{2}-1)+1}{2 \left[\beta(\delta_{1}-1)+1\right]} + \frac{\left[\beta(\delta_{1}-1)+1\right](B+1)}{2}\Bigg) \Bigg] z^2+\cdots . \end{align} From (\ref{eq3.5}), we obtain \begin{align}\label{eq3.7} 1+ \frac{1}{b}&\left(\frac{(1-t)R^{\lambda+1}f(z)}{R^{\lambda}f(z)-R^{\lambda}f(tz)}-1\right)= 1+ \frac{1}{b}\Bigg([\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]a_2 z \nonumber \\ &+ \bigg[[\varphi_{3}(\lambda +1) - u_{3} \varphi_{3}(\lambda)]a_3 -[\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]u_{2} \varphi_{2}(\lambda)a_2^2\bigg]z^2+ \cdots \Bigg). \end{align} From (\ref{eq3.6}) and (\ref{eq3.7}), the coefficients of $z$ and $z^2$ are given by \begin{align*} a_2 &= \frac{b\left[\beta(\delta_{1}-1)+1\right](A-B)p_1}{4[\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]} \end{align*} and {\scriptsize \begin{align*} a_3&= \frac{b\left[\beta(\delta_{1}-1)+1\right](A-B)}{4[\varphi_{3}(\lambda +1) - u_{3} \varphi_{3}(\lambda)]}\Bigg[p_2-\frac{p_1^2}{2}\Bigg(1- \frac{\beta(2\delta_{2}-1)+1}{2 \left[\beta(\delta_{1}-1)+1\right]}+ \frac{\left[\beta(\delta_{1}-1)+1\right](B+1)}{2} \\& \hspace{2in}- \frac{b u_2\varphi_{2}(\lambda)\left[\beta(\delta_{1}-1)+1\right](A-B)}{2[\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]}\Bigg) \Bigg]. \end{align*}} Therefore, we have \begin{equation}\label{eq3.8} \mid a_3 - \mu a_2^2 \mid \leq \frac{\mid b \mid|\beta(\delta_{1}-1)+1|(A-B)}{2 [\varphi_{3}(\lambda+1) - u_{3} \varphi_{3}(\lambda)]} \mid p_2-v p_1^2 \mid, \end{equation} where \begin{align}\label{eq3.9} v & = \frac{1}{2}-\frac{\beta(2\delta_{2}-1)+1}{4\left[\beta(\delta_{1}-1)+1\right]}+ \frac{\left[\beta(\delta_{1}-1)+1\right](B+1)}{4} \nonumber \\ &\hspace{0.3in}- \frac{b \varphi_{2}(\lambda)[\beta(\delta_{1}-1)+1](A-B)}{4 [\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]} \left(u_2- \mu \frac{\varphi_{3}(\lambda +1) - u_{3} \varphi_{3}(\lambda)}{\varphi_{2}(\lambda) [\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]}\right). \end{align} Taking the modules for both sides of the above relation, with the aid of the inequality (\ref{eq3.1}) of Lemma \ref{lem3}, we easily get the required estimate. The result is sharp for the functions $$1+ \frac{1}{b}\left(\frac{(1-t)R^{\lambda+1}f(z)}{R^{\lambda}f(z)-R^{\lambda}f(tz)}-1\right)= p(z)$$ and $$1+ \frac{1}{b}\left(\frac{(1-t)R^{\lambda+1}f(z)}{R^{\lambda}f(z)-R^{\lambda}f(tz)}-1\right) = p(z^2).$$ where $p(z)$ is given by the equation (\ref{eq3.6}). Hence the proof of the Theorem \ref{th3} is complete. \end{proof} If $\beta=0$ in the Theorem \ref{th3}, we get the following corollary. \begin{corollary}\label{th7} If $f(z) \in k-\mathcal{CL}(A, B, \lambda, t, b)$ then for $\mu \in \mathbb{C}$ we have \begin{equation}\label{eq1th7} \mid a_3 - \mu a_2^2 \mid \leq \frac{\mid b \mid (A-B)}{12 [\varphi_{3}(\lambda+1) - u_{3} \varphi_{3}(\lambda)]} \:\max \left\lbrace 1,|2v-1|\right\rbrace, \end{equation} where \begin{align}\label{eq3.10} v & = \frac{B+1}{4} - \frac{b \varphi_{2}(\lambda)(A-B)}{4 [\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]} \left(u_2- \mu \frac{3[\varphi_{3}(\lambda +1) - u_{3} \varphi_{3}(\lambda)]}{4\varphi_{2}(\lambda) [\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]}\right) \end{align} and $u_n=1+t+t^2+\cdots+t^{n-1}.$ The result is sharp. \end{corollary} If $\beta=1$ in the Theorem \ref{th3}, we get the following corollary. \begin{corollary}\label{th8} If $f(z) \in k-\mathcal{CL}(A, B, \alpha, 1, \lambda, t, b)$ then for $\mu \in \mathbb{C}$ we have \begin{equation}\label{eq3.11} \mid a_3 - \mu a_2^2 \mid \leq \frac{\mid b \mid |\delta_{1}|(A-B)}{12 [\varphi_{3}(\lambda+1) - u_{3} \varphi_{3}(\lambda)]} \:\max \left\lbrace 1,|2v-1|\right\rbrace, \end{equation} where \begin{align}\label{eq3.12} v & = \frac{1}{2}-\frac{\delta_{2}}{2\delta_{1}}+ \frac{\delta_{1}(B+1)}{4} \nonumber \\ &\hspace{0.5in}- \frac{b \varphi_{2}(\lambda)\delta_{1}(A-B)}{4 [\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]} \left(u_2- \mu \frac{3[\varphi_{3}(\lambda +1) - u_{3} \varphi_{3}(\lambda)]}{4\varphi_{2}(\lambda) [\varphi_{2}(\lambda +1) - u_{2} \varphi_{2}(\lambda)]}\right), \end{align} $\delta_{1}$ is defined as in (\ref{eql2}) and $u_n=1+t+t^2+\cdots+t^{n-1}.$ The result is sharp. \end{corollary} If $t=-1$ in the Theorem \ref{th3}, we get the following corollary. \begin{corollary}\label{cor3.1} If $f\in k-\mathcal{SL}(A, B, \alpha, \beta, \lambda, -1, b)$ then for $\mu \in \mathbb{C}$ we have \begin{equation}\label{eq1cor3.1} \mid a_3 - \mu a_2^2 \mid \leq \frac{\mid b \mid |\beta(\delta_{1}-1)+1|(A-B)}{2[\varphi_{3}(\lambda +1) - \varphi_{3}(\lambda)]} \:\max \left\lbrace 1,|2v-1|\right\rbrace, \end{equation} where \begin{equation}\label{eq2cor3.1} \begin{aligned} v & = \frac{1}{2}-\frac{\beta(2\delta_{2}-1)+1}{4\left[\beta(\delta_{1}-1)+1\right]}+ \frac{\left[\beta(\delta_{1}-1)+1\right](B+1)}{4} \\&\hspace{1.5in}+ \frac{\mu b [\varphi_{3}(\lambda +1) - \varphi_{3}(\lambda)]}{[\varphi_{2}(\lambda +1)]^{2}}\frac{\left[\beta(\delta_{1}-1)+1\right](A-B)}{4}. \end{aligned} \end{equation} The result is sharp. \end{corollary} If $t=-1$, $A=1$, $B=-1$, $\alpha=0$, $\beta=1$, $\lambda=0$ and $b=1$ in the Theorem \ref{th3}, we get the following corollary of \cite{Kavitha}. \begin{corollary}\label{cor3.3} If $f(z) \in \mathcal{M}_s(p_k)$ then we have $$ a_2= \frac{\delta_{1} p_1}{4}, \quad a_3= \frac{\delta_{1}}{4} \left[p_2-\frac{p_1^2}{2}\left(1-\frac{\delta_2}{\delta_1}\right)\right]$$ and for any complex number $\mu$, \begin{align*} \mid a_3 - \mu a_2^2 \mid \leq \frac{\delta_{1}}{2} \:\max \left\lbrace 1,\left|\frac{\delta_{2}}{\delta_{1}}-\frac{\mu \delta_{1}}{2}\right|\right\rbrace. \end{align*} \end{corollary} If $t=0$, $A=1$, $B=-1$, $\alpha=0$, $\beta=0$, $\lambda=0$ and $b=1$ in the Theorem \ref{th3}, we get the following corollary. \begin{corollary}\label{cor3.4}\cite{Raina} If $f(z) \in \mathcal{SL}_q$ then $| a_2 | \leq 1 $, $| a_3 | \leq \frac{3}{4}$ and $| a_3 - \mu a_2^2 | \leq \max \left\lbrace \frac{1}{2}, | \mu - \frac{3}{4} | \right\rbrace $. \end{corollary} \section{Coefficient estimates for the convex classes \fontsize{9}{10} $k-\mathcal{CL}(A, B, \lambda, t, b)$ and $k-\mathcal{CL}(A, B, \alpha, 1, \lambda, t, b)$}\label{sec2} To find the coefficient estimates, we need the following lemmas. \begin{lemma}\label{lem1}\cite{Rogosinski} Let $f(z) = \sum\limits_{n=1}^{\infty}a_n z^n$ be an analytic and $g(z)= \sum\limits_{n=1}^{\infty}b_n z^n$ is an analytic and convex in $\mathcal{U}$. If $f(z) \prec g(z)$, then $|a_n| \leq |b_1|,$ for $n=1,2,\dotso$ . \end{lemma} \begin{remark}\label{rem3.1} Since Lemma \ref{lem1} can be applied only if $g(z)$ is convex in $\mathcal{U}$. But the right hand side in (\ref{eq1.6}) namely $\frac{(A+1)h(z)-(A-1)}{(B+1)h(z)-(B-1)}$ (where $h(z)$ is given as in (\ref{h})) is not convex in $\mathcal{U}$. So we find the coefficient inequalities for the fixed values of $\beta=0$ and $\beta=1$. \end{remark} The following result was obtained by Noor\ and Malik in \cite{noor}. \begin{lemma}\label{lem5}\cite{noor} Let the function $\hat{p}_{k,\alpha}(z)$ be defined as in (\ref{eqpow}) and let $p(z)\in \mathcal{P}$ satisfy the condition \begin{equation}\label{eq1lem5} p(z) \prec \frac{(A+1)\hat{p}_{k,\alpha}(z)-(A-1)}{(B+1)\hat{p}_{k,\alpha}(z)-(B-1)}. \end{equation} Then \begin{equation}\label{eq2lem5} \mid p_n \mid \leq \frac{|\delta_{1}|(A-B)}{2}, \quad (n \geq 1). \end{equation} \end{lemma} \begin{remark} Similar result fails if $\frac{(A+1)k(z)-(A-1)}{(B+1)k(z)-(B-1)}$, as $k(z)=z+\sqrt{1+z^{2}}$ is starlike but not convex. \end{remark} \begin{theorem}\label{th6} Let $k-\mathcal{CL}(A, B, \alpha, 1, \lambda, t, b)$, then for $n \geq 2$ \begin{align}\label{eq2.7th6} \mid a_n \mid &\leq \displaystyle\prod_{j=1}^{n-1} \frac{\Big|b \,j\,u_j \varphi_{j}(\lambda)\delta_{1}(A-B)-2j\left[\varphi_{j}(\lambda+1)-u_j\varphi_{j}(\lambda)\right]B\Big|}{2(j+1) [\varphi_{j+1}(\lambda+1) - u_{j+1} \varphi_{j+1}(\lambda)]}, \end{align} where $\delta_{1}$ is defined as in (\ref{eql2}) and $u_n=1+t+t^2+\cdots+t^{n-1}.$ \end{theorem} \begin{proof} By the definition of $k-\mathcal{CL}(A, B, \alpha, \lambda, t, b)$, we have \begin{equation}\label{eq2.8} 1+ \frac{1}{b}\left(\frac{(1-t)\left(R^{\lambda+1}f(z)\right)^{\prime}}{\left(R^{\lambda}f(z)-R^{\lambda}f(tz)\right)^{\prime}}-1\right) = p(z), \end{equation} where $p(z) \in \mathcal{P}$ and satisfies the subordination condition $$ p(z) \prec \frac{(A+1)\hat{p}_{k,\alpha}(z)-(A-1)}{(B+1)\hat{p}_{k,\alpha}(z)-(B-1)}.$$ Equivalently (\ref{eq2.8}) can be rewritten as {\footnotesize \begin{align*} 1+ \frac{1}{b}\left(\frac{\sum\limits_{n=2}^{\infty}n \left[\varphi_{n}(\lambda+1) - u_n \varphi_{n}(\lambda)\right] a_n z^{n-1}}{1+\sum\limits_{n=2}^{\infty}nu_n \varphi_{n}(\lambda) a_n z^{n-1}}-1\right) &= 1+ \sum\limits_{n=1}^{\infty} p_n z^n \\ \sum\limits_{n=2}^{\infty}n\left[\varphi_{n}(\lambda+1) - u_n \varphi_{n}(\lambda)\right] a_n z^{n-1} &= b \left(1+\sum\limits_{n=2}^{\infty}n u_n \varphi_{n}(\lambda)a_n z^{n-1}\right) \sum\limits_{n=1}^{\infty} p_n z^n . \end{align*}} Equating the coefficients of $z^{n-1}$ on both sides of the above equation, we have $$n\left[\varphi_{n}(\lambda+1) - u_n \varphi_{n}(\lambda)\right] a_n = b \sum\limits_{j=1}^{n-1}(n-j)u_{n-j} \varphi_{n-j}(\lambda)\, a_{n-j}p_j$$ which implies that \begin{equation}\label{eq2.9} n\left[\varphi_{n}(\lambda+1) - u_n \varphi_{n}(\lambda)\right]\mid a_n \mid \leq b \sum\limits_{j=1}^{n-1} (n-j)u_{n-j}\varphi_{n-j}(\lambda) \mid a_{n-j} \mid \,\, \mid p_j \mid. \end{equation} Since $ p \in \mathcal{P}$, by Lemma \ref{lem5}, we obtain $$ \mid p_j \mid \leq \frac{|\delta_{1}|A-B}{2}. $$ Following the steps as in Theorem 2.6 of Noor and Malik \cite{noor}, we can establish the assertion of the Theorem \ref{th6}. \end{proof} If $b=1$, $t=0$, $\alpha=0$ and $\lambda=0$ in the Theorem \ref{th6}, we get the following result. \begin{corollary}\label{cor2.12}\cite{noor} Let $f\in k-\mathcal{CL}(A,B)$, then \begin{align*} \mid a_n \mid &\leq \frac{1}{n}\prod_{j=0}^{n-2} \frac{\left|\delta_1(A-B)-2 jB\right|}{2 (j+1)} \quad (n\geq 2). \end{align*} \end{corollary} \begin{thebibliography}{99} \bibitem{arif} Arif, M., et al., Coefficient inequalities for Janowski-Sakaguchi type functions associated with conic regions, Hacet. 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