In the present paper, we use the technique of differential subordination and superordination involving meromorphic functions with respect to symmetric points and also derive some sandwich results. As a consequence of main result, we obtain results for meromorphic starlike functions with respect to symmetrical points.
Let \(\Sigma\) denote the class of functions of the form \[f(z)=\frac{1}{z}+\sum_{k=1}^\infty a_kz^{k-1},\] which are analytic in the punctured unit disc \( \mathbb E_0=\mathbb E\setminus\{0\},\) where \(\mathbb E = \{z\in\mathbb C:|z|< 1\}. \) A function \(f \in \Sigma \) is said to be meromorphic starlike of order \(\alpha \) if \(f(z) \neq 0 \) for \(z \in \mathbb E_0 \) and
\[-\Re\left(\frac{zf'(z)}{f(z)}\right)>\alpha,\hspace{1.7cm}(\alpha< 1;z \in \mathbb E).\] The class of such functions is denoted by \(\mathcal {MS}^*(\alpha)\) and write \(\mathcal {MS}^*=\mathcal {MS}^*(0)\)-the class of meromorphic starlike functions.In 1959, Sakaguchi [1] introduced and studied the class of starlike functions with respect to symmetric points in \(\mathbb E\). Further investigations into the class of starlike functions with respect to symmetric points can be found in [2,3].
Recently, Ghaffar et al., [4] introduced and investigated a class of meromorphic starlike functions with respect to symmetric points which satisfies the condition
\[-\frac{2zf'(z)}{f(z)-f(-z)}\prec \frac{1+Az}{1+Bz}, ~z\in\mathbb E_0\] and \(-1\leq B< A\leq1.\) We denote the above class by \(\mathcal {MS}^{s*}[A,B].\) If \(f\) is analytic and \(g\) is analytic univalent in open unit disk \(\mathbb E\), we say that \(f(z)\) is subordinate to \(g(z)\) in \(\mathbb E\) and written as \(f(z)\prec g(z)\) if \(f(0) = g(0)\) and \(f( \mathbb E) \subset g(\mathbb E)\). To derive certain sandwich-type results, we use the dual concept of differential subordination and superordination.Let \(~\Phi:\mathbb C^2 \times \mathbb E\longrightarrow \mathbb C\) (\(\mathbb C\) is the complex plane) and \(h\) be univalent in \(\mathbb E\). If \(p\) is analytic in \(\mathbb E\) and satisfies the differential subordination
Let \(\Psi:\mathbb C^2 \times \mathbb E\longrightarrow \mathbb C\) \((\mathbb C\) is the complex plane) be analytic and univalent in domain \(\mathbb C^2 \times \mathbb E, h\) be analytic in \(\mathbb E,\) \(p\) is analytic and univalent in \(\mathbb E,\) with \((p(z), zp'(z); z)\in \mathbb C^2 \times \mathbb E \) for all \(z\in \mathbb E.\) Then \(p\) is called a solution of first order differential superordination if it satisfies
In this paper we study the concepts of subordination and superordination to obtain meromorphic starlikeness with respect to symmetric points.On the basis of the theory we also investigate some important sandwich results of symmtric meromorphic functions.
Lemma 1. [5] Let \(q\) be univalent in \(\mathbb E\) and let \(\theta\) and \(\phi\) be analytic in a domain \(\mathbb D\) containing \(q(\mathbb E)\), with \(\phi(w)\neq 0\), when \(w\in q (\mathbb E)\). Set \(Q_1(z)=zq'(z)\phi[q(z)],~h(z)=\theta[q(z)]+Q_1(z)\) and suppose that either
Definition 1. We denote by \(Q\) the set of functions \(p\) that are analytic and injective on \(\overline{\mathbb E}\setminus\mathbb B(p),\) where \[\mathbb B(p)=\left\{\zeta\in \partial{\mathbb E}: \lim_{z\rightarrow\zeta} {p(z)=\infty} \right\},\] and are such that \(p'(\zeta)\neq 0\) for \(\zeta \in \partial \mathbb E\setminus\mathbb B(p).\)
Lemma 2. [6] Let \(q\) be the univalent in \(\mathbb E\) and let \(\theta\) and \(\phi\) be analytic in a domain \(\mathbb D\) containing \(q(\mathbb E).\) Set \(Q_1(z)=zq'(z)\phi[q(z)],~h(z)=\theta[q(z)]+Q_1(z)\) and suppose that (i) \( Q_1(z)\) is starlike in \(\mathbb E;\) and (ii) \( \Re\left(\frac{\theta'(q(z))}{\phi(q(z))}\right)> 0,\) for \(z \in \mathbb E.\) If \(p\in \mathcal H[q(0),1]\cap Q,\) with \(p(\mathbb E)\subset \mathbb D\) and \(\theta[p(z)]+zp'(z)\phi[p(z)]\) is univalent in \(\mathbb E\) and \[\theta[q(z)]+zq'(z)\phi[q(z)]\prec \theta[p(z)]+zp'(z)\phi[p(z)],z\in \mathbb E,\] then \(q(z)\prec p(z)\) and \(q\) is the best subordinant.
Lemma 3. [7] The function \(\displaystyle q(z)=\frac{1}{(1-z)^{2ab}}\) is univalent in \(E\) if and only if \(|2ab-1|\leq1\) or \(|2ab+1|\leq1.\)
Theorem 1. Let \(q\) be univalent in \(\mathbb E,\) with \(q(0)=1,\) and let
Proof. Setting \[p(z)=\frac{-2zf'(z)}{f(z)-f(-z)},~z\in\mathbb E,\] from assumption (4) it follows that \(p\) is analytic in \(\mathbb E,\) with \(p(0)=1.\) A simple computation shows that \[\frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f”(z)}{f(z)-f(-z)}=p(z)+\lambda zp'(z),\] hence, the subordination (5) is equivalent to \(p(z)+\lambda zp'(z)\prec q(z)+\lambda zq'(z)\).
Now, in order to prove our result we will use Lemma 1. Consider the functions \(\theta(w)=w\) and \(\phi(w)=\gamma\) analytic in \(\mathbb C,\) and set
\begin{equation} Q(z)=zq'(z) \phi(q(z))=\gamma zq'(z) \  \  and \  \  h(z)=\theta(q(z))+Q(z)=q(z)+\gamma z q'(z). \end{equation} Since \(Q(0)=0\) and \(Q'(0)=\gamma q'(0)\neq0\), the assumption (3) implies that \(Q\) is starlike in \(\mathbb E\) and \[\Re\left(\frac{zh'(z)}{Q(z)}\right)=\Re \left(1+\frac{zq”(z)}{q'(z)}+\frac{1}{\gamma}\right)>0, ~z \in \mathbb E.\] Therefore, Lemma 1 and assumption (5) imply \(p(z)\prec q(z)\) and the function \(q\) is the best dominant of (5).If we take \(\displaystyle q(z)=\frac{1+Az}{1+Bz},-1\leq B< A\leq 1\) in Theorem 1 then the condition (3) reduces to
Corollary 1. Let \(\lambda\in\mathbb C^*\) and \(-1\leq B< A\leq 1\) with \(\displaystyle\frac{1-|B|}{1+|B|}\geq max\left\{0;~-\Re \frac{1}{\lambda}\right\}\). If \(f\in\Sigma\) satisfy the condition (4) and
Remark 1. Let \(\lambda\in\mathbb C^*\) with \(\Re \frac{1}{\lambda}\geq0\). If \(f\in\Sigma\) satisfy the condition (4) and
Theorem 2. Suppose that \(q\) be univalent in \(\mathbb E\) with \(q(0)=1\) and \(q(z)\neq0\) for all \(z\in\mathbb E\) such that
Proof. Define a function
Since \(Q(0)=0\) and \(Q'(0)=\gamma q'(0)\neq0\), the assumption (10) implies that \(Q\) is starlike in \(\mathbb E\) and
\[\Re\left(\frac{zh'(z)}{Q(z)}\right)=\Re \left(1+\frac{zq”(z)}{q'(z)}+\frac{z q'(z)}{q(z)}\right)>0, ~z \in \mathbb E.\] Therefore, using Lemma 1 and assumption (13) implies \(p(z)\prec q(z)\) and the function \(q\) is the best dominant of (13). In particular, taking \(\nu=0, \eta=\gamma=1\) and \(\displaystyle q(z)=\frac{1+Az}{1+Bz}\) in the above theorem, it is easy to check that the inequality (10) holds whenever \(-1\leq B< A\leq 1\). Hence, we deduce the following corollary:Corollary 2. Let \(-1\leq B< A\leq 1\) and \(\mu\in\mathbb C^*\). Let \(f\in\Sigma\) satisfy the conditions (4) and
Corollary 3. Let \(a,b\in\mathbb C^*\) such that \[|2ab-1|\leq1 \hspace{1cm}or \hspace{1cm}|2ab+1|\leq1\] and suppose that \(f\in\Sigma\) satisfy the conditions (4) and (15). If \[1+\frac{1}{b}\left[1-\frac{z(f'(z)+f'(-z))}{f(z)-f(-z)}+\frac{zf”(z)}{f'(z)}\right]\prec\frac{1+z}{1-z},\] then
Corollary 4. Let \(-1\leq B< A\leq 1\) with \(B\neq0\), and suppose that \[\left|\frac{\mu(A-B)}{B-1}\right|\leq1 \hspace{1cm}or \hspace{1cm}\left|\frac{\mu(A-B)}{B+1}\right|\leq1,\] where \(\mu\in\mathbb C^*\). If \(f\in\Sigma\) satisfy the conditions (4), (15) and
Corollary 5. Let \(a,b\in\mathbb C^*\) and \(|\lambda|< \frac{\pi}{2}\) and suppose that \[ |2abe^{-\iota\lambda} \cos \lambda-1|\leq1 \hspace{1cm} \text{or} \hspace{1cm} |2abe^{-\iota\lambda} \cos \lambda+1|\leq1 .\] If \(f\in\Sigma\) satisfy the conditions (4), (15) and \[1+\frac{e^{\iota\lambda}}{b \cos \lambda}\left[1-\frac{z(f'(z)+f'(-z))}{f(z)-f(-z)}+\frac{zf''(z)}{f'(z)}\right]\prec\frac{1+z}{1-z},\] then
Theorem 3. Suppose that \(q\) be univalent in \(\mathbb E\) with \(q(0)=1\) and \(q(z)\neq0\) for all \(z\in\mathbb E\), such that
Proof. Define a function
Since \(Q(0)=0\) and \(Q'(0)=\gamma q'(0)\neq0\), the assumption (20) implies that \(Q\) is starlike in \(\mathbb E\) and
\[\Re\left(\frac{zh'(z)}{Q(z)}\right)=\Re \left(\frac{\delta}{\gamma}+1+\frac{zq”(z)}{q'(z)}+\frac{z q'(z)}{q(z)}\right)>0, ~z \in \mathbb E.\] Therefore, using Lemma 1 and assumption (22) implies \(p(z)\prec q(z)\) and the function \(q\) is the best dominant of (22). Taking \(\displaystyle q(z)=\frac{1+Az}{1+Bz}\) in Theorem 3, where \(-1\leq B< A\leq1\), according to (7) the condition (20) becomes \[max\left\{0;~-\Re \frac{\delta}{\gamma}\right\}\leq\frac{1-|B|}{1+|B|}.\] Hence, for the particular cases \(\eta=0\) and \(\mu=\gamma=1\), we have the following result:Corollary 6. Let \(-1\leq B< A\leq 1, \mu\in\mathbb C^*\) and \(\delta\in\mathbb C\) with \[max\left\{0;~-\Re \frac{\delta}{\gamma}\right\}\leq\frac{1-|B|}{1+|B|}.\] If \(f\in\Sigma\) satisfy the conditions (4), (15) and
Corollary 7. Let \(\mu\in\mathbb C^*\) and \(\delta\in\mathbb C\) with \(\Re \delta \geq 0.\) If \(f\in\Sigma\) satisfy the conditions (4), (15) and
Theorem 4. Let \(q\) be convex in \(\mathbb E\) with \(q(0)=1\) and \(\lambda\in\mathbb C\) with \(\Re~ \lambda>0\). Let \(f\in\Sigma\) satisfy the condition (4) such that \(\displaystyle-\frac{2zf'(z)}{f(z)-f(-z)}\in\mathbb Q\) and suppose that the function \[ \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f”(z)}{f(z)-f(-z)}\] is univalent in \(\mathbb E\). If
Proof. Setting \[p(z)=\frac{-2zf'(z)}{f(z)-f(-z)},~z\in\mathbb E,\] then \(p\) is analytic in \(\mathbb E\) with \(p(0) = 1.\) Taking logarithmic differentiation of the above relation with respect to \(z,\) we have \[\frac{zp'(z)}{p(z)}=1-z\left(\frac{f'(z)+f'(-z)}{f(z)-f(-z)}-\frac{f”(z)}{f'(z)}\right)\] and a simple calculation yields that the assumption (26) is equivalent to \[ q(z)+\lambda zq'(z)\prec p(z)+\lambda zp'(z).\] Now, in order to prove our result we will use Lemma 2. Consider the functions \(\theta(w)=w\) and \(\phi(w)=\lambda\) analytic in \(\mathbb C\) and set \[h(z)=zq'(z)\phi(q(z))=\lambda z q'(z).\] Since \(h(0)=0\), \(h'(0)=\lambda q'(0)\neq0\) and \(q\) is convex in \(\mathbb E\), it follows that \(h\) is starlike in \(\mathbb E\) and \[\Re \frac{\theta’q(z)}{\phi(q(z))}=\Re \frac{1}{\gamma}>0, ~z \in \mathbb E.\] Therefore, Lemma 2 and assumption (26) imply \(q(z)\prec p(z)\) and the function \(q\) is the best subordinant of (26).
Taking \(\displaystyle q(z)=\frac{1+Az}{1+Bz}\) in Theorem 4, where \(-1\leq B< A\leq1\), we get the following corollary:Corollary 8. Let \(q\) be convex in \(\mathbb E\) with \(q(0)=1\) and \(\lambda\in\mathbb C\) with \(\Re~ \lambda>0\). Let \(f\in\Sigma\) satisfy the condition (4) such that \(\displaystyle-\frac{2zf'(z)}{f(z)-f(-z)}\in\mathbb Q\) and suppose that the function \[ \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f”(z)}{f(z)-f(-z)}\] is univalent in \(\mathbb E\). If
Theorem 5. Let \(\gamma,\mu\in\mathbb C^*\) and \(\delta,\nu,\eta\in\mathbb C\) with \(\nu+\eta\neq 0\) and \(\Re\left (\displaystyle\frac{\delta}{\gamma}\right)>0\). Suppose that \(q\) is convex in \(\mathbb E\), with \(q(0)=1\) and let \(f\in\Sigma\) satisfy the conditions (11), (12) and \[-\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\in\mathbb Q. \] If the function \(\phi \) given by (21) is univalent in \(\mathbb E\) and
Theorem 6. Let \(q_1\) and \(q_2\) be two convex functions in \(\mathbb E\) with \(q_1(0)=q_2(0)=1\) and \(\lambda\in\mathbb C\) with \(\Re~ \lambda>0\). Let \(f\in\Sigma\) satisfy the condition (4), such that \(\displaystyle-\frac{2zf'(z)}{f(z)-f(-z)}\in\mathbb Q\) and suppose that the function \[ \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f”(z)}{f(z)-f(-z)}\] is univalent in \(\mathbb E\). If \[ q_1(z)+\lambda zq_1′(z)\prec \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f”(z)}{f(z)-f(-z)}\,.\]
Theorem 7. Let \(q_1\) and \(q_2\) be two convex functions in \(\mathbb E\) with \(q_1(0)=q_2(0)=1\) and let \(\gamma,\mu\in\mathbb C^*\) and \(\nu,\eta\in\mathbb C\) with \(\nu+\eta\neq 0\) and \(\Re\displaystyle \frac{\delta}{\gamma}>0\). Let \(f\in\Sigma\) satisfy the condition (11), (12) and such that \(\displaystyle-\frac{2zf'(z)}{f(z)-f(-z)}\in\mathbb Q\) and \[-\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)-f(-z))}\right]^\mu\in\mathbb Q. \] If the function \(\phi \) given by (21) is univalent in \(\mathbb E\) and