Sandwich type results for meromorphic functions with respect to symmetrical points

1. Introduction

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

then \(p\) is called a solution of the differential subordination (1). The univalent function \(q\) is called a dominant of differential subordination (1) if \(p\prec q\) for all \(p\) satisfying \(1.\) A dominant \(\tilde q \prec q \) for all dominants \(q\) of \(1,\) is said to be the best dominant of \(1.\)

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

An analytic function \(q\) is called a subordinant of differential superordination (2) if \(q\prec p\) for all \(p\) satisfying \((2).\) A univalent subordinant \(\tilde q\) that satisfies \(q \prec \tilde q \) for all subordinants \(q\) for (2), is said to be the best subordinant of (2).

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.

2. Preliminaries

We shall use the following lemmas to prove our result.

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

• (i) \(h\) is convex, or
• (ii) \(Q_1\) is starlike.

In addition, assume that

• (iii) \(\displaystyle \Re\left(\frac{zh'(z)}{Q_1(z)}\right)>0\) for all \(z\) in \(\mathbb E\).

If \(p\) is analytic in \(\mathbb E\), with \(p(0)=q(0),~p(\mathbb E)\subset \mathbb D\) and \[\theta[p(z)]+zp'(z)\phi[p(z)]\prec\theta[q(z)]+zq'(z)\phi[q(z)],~z\in\mathbb E,\] then \(p(z)\prec q(z)\) and \(q\) is the best dominant.

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 \(|2ab+1|\leq1.\)

3. Subordination results

Theorem 1. Let \(q\) be univalent in \(\mathbb E,\) with \(q(0)=1,\) and let

where \(\gamma\in\mathbb C^*:=\mathbb C\setminus\{0\}.\) If \(f\in\Sigma\) satisfy the condition and then \[\frac{-2zf'(z)}{f(z)-f(-z)}\prec q(z),\] and \(q\) is the best dominant of (5).

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

It is easy to see that the function \(\displaystyle \chi(\zeta)=\frac{1-\zeta}{1+\zeta}\) with \(|\zeta|< |B|\) is convex in \(E\) and since \(\chi(\overline{\zeta}) =\overline{\chi(\zeta) }\) for all \(|\zeta|< |B|\), it follows that \(\chi(\mathbb E)\) is a convex domain symmetric with respect to the real axis. Hence Thus, the inequality (6) is equivalent to \(\displaystyle\Re\frac {1}{\lambda}\geq \frac{|B|-1}{|B|+1}\), hence we deduce the following corollary:

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

then \[\frac{-2zf'(z)}{f(z)-f(-z)}\prec \frac{1+Az}{1+Bz},\] i.e., \(f\in\mathcal{MS}^{s*}[A,B].\) Moreover, the function \(\displaystyle\frac{1+Az}{1+Bz}\) is the best dominant of (8).

For \(A=1\) and \(B=-1\), the above corollary reduces to the next special case:

Remark 1. Let \(\lambda\in\mathbb C^*\) with \(\Re \frac{1}{\lambda}\geq0\). If \(f\in\Sigma\) satisfy the condition (4) and

then \[\frac{-2zf'(z)}{f(z)-f(-z)}\prec \frac{1+z}{1-z},\] i.e., \(f\in\mathcal{MS}^{s*},\) or \(f\) is meromorphic starlike with respect to symmetrical points in \(\mathbb E\). Moreover, the function \(\displaystyle\frac{1+z}{1-z}\) is the best dominant of (9).

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

where \(\gamma,\mu\in\mathbb C^*\) and \(\nu,\eta\in\mathbb C\) with \(\nu+\eta\neq 0\) and let \(f\in\Sigma\) satisfy the conditions and If then \[-\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\prec q(z)\] and \(q\) is the best dominant of (13) (the power is the principal one).

Proof. Define a function

According to the assumptions (11) and (12), the multivalued power function \(p\) has an analytic branch in \(\mathbb E\) with \(p(0)=1\) and from the (14) it follows that \[\mu\left[1-\frac{\nu z[z(f'(z)+ f'(-z))]' +\eta z(f'(z)+ f'(-z))}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}+\frac{zf''(z)}{f'(z)}\right]=\frac{zp'(z)}{p(z)}.\] To prove our desired result we will use Lemma 1. Thus, let the functions \(\theta(w)=1\) and \(\phi(w)=\displaystyle\frac{\gamma}{w}\). Also, if we let \( Q(z)=zq'(z) \phi(q(z))=\gamma zq'(z)\) and \(h(z)=\theta(q(z))+Q(z)=1+\gamma\displaystyle\frac{z q'(z)}{q(z)}.\)

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

If then \[-\left[\frac{2zf'(z)}{f(z)-f(-z)}\right]^\mu\prec \frac{(1+Az)}{(1+Bz)}, \] and the function \( \displaystyle\frac{(1+Az)}{(1+Bz)}\) is the best dominant of (16) (the power is principal one).

Using \(\nu=0, \eta=1, \gamma=\displaystyle \frac{1}{ab}\) with \(a,b\in\mathbb C^*, \mu=a\) and \( q(z)=\displaystyle \frac{1}{(1-z)^{2ab}}\) in Theorem 2, then using this result together with Lemma 3, we deduce the following result:

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

and the function \(\displaystyle \frac{1}{(1-z)^{2ab}}\) is the best dominant of (17) (the power is the principal one).

By using \(\nu=0, \eta=\gamma=1\) and \( q(z)= \displaystyle(1+Bz)^{\frac{\mu(A-B)}{B}}\), \(-1\leq B< A\leq 1\), \(B\neq0\) in Theorem 2, then using Lemma 3 we have the next result:

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

then \[-\left[\frac{2zf'(z)}{f(z)-f(-z)}\right]^\mu\prec (1+Bz)^{\frac{\mu(A-B)}{B}} \] and the function \((1+Bz)^{\frac{\mu(A-B)}{B}}\) is the best dominant of (18) (the power is the principal one).

By taking \(\nu=0, \eta=1\), \(\gamma=\frac{e^{\iota\lambda}}{ab \cos \lambda}, a,b\in\mathbb C^*, |\lambda|< \displaystyle\frac{\pi}{2}, \mu=a\) and \(q(z)=\displaystyle\frac{1}{(1-z)^{2abe^{-\iota\lambda} \cos \lambda}}\) in Theorem 2, we obtain the following:

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

and the function \(\displaystyle \frac{1}{(1-z)^{2abe^{-\iota\lambda} \cos \lambda}}\) is the best dominant of (19) (the power is the principal one).

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

where \(\gamma,\mu\in\mathbb C^*\) and \(\delta,\nu,\eta\in\mathbb C\) with \(\nu+\eta\neq 0\) and let \(f\in\Sigma\) satisfy the conditions (11), (12) and \[\left[\delta+\gamma\mu\left(1-\frac{\nu z[z(f'(z)+ f'(-z))]' +\eta z(f'(z)+ f'(-z))}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)-f(-z))}+\frac{zf''(z)}{f'(z)}\right)\right]\,.\] If then \[-\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\prec q(z)\] and \(q\) is the best dominant of (13) (the power is the principal one).

Proof. Define a function

From the assumptions (11) and (12) it follows that the multivalued power function \(p\) has an analytic branch in \(\mathbb E\) with \(p(0)=1\) and from the (23) it follows that \[\mu p(z) \left[1-\frac{\nu z[z(f'(z)+ f'(-z))]' +\eta z(f'(z)+ f'(-z))}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}+\frac{zf''(z)}{f'(z)}\right]=zp'(z).\] Consider the functions \(\theta(w)=\delta w\) and \(\phi(w)=\gamma\) that are analytic in \(\mathbb C^*\). Also, if we let \( Q(z)=zq'(z) \phi(q(z))=\gamma zq'(z)\) and \(h(z)=\theta(q(z))+Q(z)=\delta q(z)+\gamma\frac{z q'(z)}{q(z)}.\)

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

then \[ -\left[\frac{2f'(z)}{f'(z)+ f'(-z)}\right]^\mu \prec \frac{1+Az}{1+Bz}\] and the function \(\displaystyle \frac{1+Az}{1+Bz}\) is the best dominant of (24) (all the powers are principal ones).

Taking \(\eta=\gamma=1,\mu=0\) and \(q(z)=\displaystyle\frac{1+z}{1-z}\) in Theorem 3, we get the following corollary:

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

then \[ -\left[\frac{2f'(z)}{f'(z)+ f'(-z)}\right]^\mu \prec \frac{1+z}{1-z},\] and the function \(\displaystyle \frac{1+z}{1-z}\) is the best dominant of (25) (all the powers are principal ones).

4. Superordination and sandwich theorems

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

then \[q(z)\prec-\frac{2zf'(z)}{f(z)-f(-z)}\] and \(q\) is the best subordinant of (26).

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

then \[\displaystyle\frac{1+Az}{1+Bz}\prec-\frac{2zf'(z)}{f(z)-f(-z)}\] and \(q\) is the best subordinant of (27).

Using the same techniques as in proof of Theorem 3 and then applying Lemma 2, we could prove the next theorem:

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

then \[q(z)\prec -\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\] and \(q\) is the best subordinant of (28) (the power is the principal one).

If we combine Theorem 1 with Theorem 4 and Theorem 3 with Theorem 5, we deduce the following sandwich results, respectively:

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 \[ q_1(z)+\lambda zq_1'(z)\prec \frac{-2(1+\lambda)zf'(z)}{f(z)-f(-z)}+\frac{2\lambda z^2 f'(z)(f'(z)+f'(-z))}{(f(z)-f(-z))^2}-\frac{2\lambda z^2 f''(z)}{f(z)-f(-z)}\,.\]

then \[q_1(z)\prec\frac{-2zf'(z)}{f(z)-f(-z)}\prec q_2(z),\] where \(q_1\) and \(q_2\) are respectively the best subordinant and the best dominant of (29).

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

then \[q_1(z)\prec -\left[\frac{2(\nu+\eta) zf'(z)}{\nu z(f'(z)+ f'(-z)) +\eta (f(z)- f(-z))}\right]^\mu\prec q_2(z)\] and \(q_1\) and \(q_2\) are respectively the best subordinant and the best dominant of (29) (the power is the principal one).

Author Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

Conflicts of Interest

''The authors declare no conflict of interest.''