Theorem 3.1
Let \(\pi \in \mathcal{C}^{*}\). If \(f\in \Sigma \) and \(\mathcal{J}\) is convex univalent in \(\mathfrak{D}\) with \(\mathcal{J}(0) = 1\). If f and \(\mathcal{J}\) attain any one of the following pairs of stipulation:
-
1.
$$\begin{aligned} &\Re \biggl\{ {1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}' ( z )}}} \biggr\} > \max \biggl\{ {0,\frac{1}{\xi }\Re \biggl( { \frac{1}{\pi }} \biggr)} \biggr\} , \end{aligned}$$
(3.1)
$$\begin{aligned} & \pi \bigl( {z\mathcal{L}^{\varsigma +1}_{\sigma }(\varrho, \tau )f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \mathcal{J} ( z ) - \pi \vartheta z \mathcal{J}' ( z ),\quad z \in \mathfrak{D}, \end{aligned}$$
(3.2)
-
2.
$$\begin{aligned} & \Re \biggl\{ {1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}' ( z )}}} \biggr\} > \max \biggl\{ {0,\Re \biggl( {\frac{{\varrho - 1}}{\pi }} \biggr)} \biggr\} , \end{aligned}$$
(3.3)
$$\begin{aligned} & \pi \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho -1, \tau )f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \mathcal{J} ( z ) - \frac{\pi }{{\varrho - 1}}z \mathcal{J}' ( z ), \\ &\quad z \in \mathfrak{D}, \end{aligned}$$
(3.4)
-
3.
$$\begin{aligned} &\Re \biggl\{ {1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}' ( z )}}} \biggr\} > \max \biggl\{ {0, - ( {\pi + 1} )\Re \biggl( { \frac{1}{\pi }} \biggr)} \biggr\} , \end{aligned}$$
(3.5)
$$\begin{aligned} &\lambda \bigl( {z\mathcal{L}^{\varsigma }_{\sigma +1}( \varrho,\tau )f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \mathcal{J} ( z ) - \frac{\pi }{{\sigma + 1}}z \mathcal{J}' ( z ), \\ &\quad z \in \mathfrak{D}, \end{aligned}$$
(3.6)
-
4.
$$\begin{aligned} & \Re \biggl\{ {1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}' ( z )}}} \biggr\} > \max \biggl\{ {0,\Re \biggl( {\frac{\tau }{\pi }} \biggr)} \biggr\} , \end{aligned}$$
(3.7)
$$\begin{aligned} & \pi \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho, \tau +1)f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \mathcal{J} ( z ) - \frac{\pi }{\tau }z \mathcal{J}' ( z ),\quad z \in \mathfrak{D}. \end{aligned}$$
(3.8)
Then
$$\begin{aligned} z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z) \prec \mathcal{J} ( z ) \end{aligned}$$
(3.9)
and \(\mathcal{J}\) is the best dominant of (3.9).
Proof
Differentiating the following function:
$$\begin{aligned} \mathcal{H} ( z ) = z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z),\quad z \in \mathfrak{D}, \end{aligned}$$
by using identities (2.1)–(2.6), we consistently gain the following:
$$\begin{aligned} &z\mathcal{L}^{\varsigma +1}_{\sigma }(\varrho,\tau )f(z) = \mathcal{H} ( z ) - \vartheta z\mathcal{H}' ( z ), \end{aligned}$$
(3.10)
$$\begin{aligned} &z\mathcal{L}^{\varsigma }_{\sigma }(\varrho -1,\tau )f(z) = \mathcal{H} ( z ) + \frac{1}{{\varrho - 1}}z\mathcal{H}' ( z ), \end{aligned}$$
(3.11)
$$\begin{aligned} &z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) = \mathcal{H} ( z ) + \frac{{z\mathcal{H}' ( z )}}{{1 + \sigma }}, \end{aligned}$$
(3.12)
and
$$\begin{aligned} z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau +1)f(z) = \mathcal{H} ( z ) + \frac{1}{\tau }z\mathcal{H}' ( z ). \end{aligned}$$
(3.13)
Now, subordination stipulations (3.2), (3.4), (3.6), and (3.8) are consistently equivalent to
$$\begin{aligned} & \mathcal{H} ( z ) - \pi \vartheta z\mathcal{H}' ( z ) \prec \mathcal{J} ( z ) - \pi \vartheta z \mathcal{J}' ( z ), \end{aligned}$$
(3.14)
$$\begin{aligned} &\mathcal{H} ( z ) + \frac{\pi }{{\varrho - 1}}z \mathcal{H}' ( z ) \prec \mathcal{J} ( z ) + \frac{\pi }{{\varrho - 1}}z\mathcal{J}' ( z ), \end{aligned}$$
(3.15)
$$\begin{aligned} &\mathcal{H} ( z ) + \frac{\pi }{{\sigma + 1}}z\mathcal{H}' ( z ) \prec \mathcal{J} ( z ) + \frac{\pi }{{\sigma + 1}}z\mathcal{J}' ( z ), \end{aligned}$$
(3.16)
and
$$\begin{aligned} \mathcal{H} ( z ) + \frac{\pi }{\tau }z\mathcal{H}' ( z ) \prec \mathcal{J} ( z ) + + \frac{\pi }{\tau }z \mathcal{J}' ( z ). \end{aligned}$$
(3.17)
Successively, by employing Lemma 2.3 to each of the subordination stipulations (3.13)–(3.17) with suitable choices of τ and π, we gain assertion (3.9) of Theorem 3.1. □
Theorem 3.2
Let \(\pi \in \mathcal{C}^{*}, { - 1 \le \eta < \varpi \le 1} \), and \(f\in \Sigma \). If any one of the following pairs of stipulations is attained:
-
1.
$$\begin{aligned} &\frac{{ \vert \eta \vert - 1}}{{ \vert \eta \vert + 1}} < \frac{{ - 1}}{\vartheta }\Re \biggl( { \frac{1}{\pi }} \biggr), \end{aligned}$$
(3.18)
$$\begin{aligned} & \pi \bigl( {z\mathcal{L}^{\varsigma +1}_{\sigma }(\varrho, \tau )f(z)} \bigr) +(1- \lambda ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \frac{{1 + \varpi z}}{{1 + \eta z}} + \pi \vartheta \frac{{ ( {\varpi - \eta } )z}}{{{{ ( {1 + \eta z} )}^{2}}}}, \end{aligned}$$
(3.19)
-
2.
$$\begin{aligned} &\frac{{ \vert \eta \vert - 1}}{{ \vert \eta \vert + 1}} < \Re \biggl( {\frac{{\varrho - 1}}{\pi }} \biggr), \end{aligned}$$
(3.20)
$$\begin{aligned} &\pi \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho -1, \tau )f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \\ &\quad \prec \frac{{1 + \varpi z}}{{1 + \eta z}} + \frac{{ ( {\varpi - \eta } )\pi z}}{{ ( {\varrho - 1} ){{ ( {1 + \eta z} )}^{2}}}}, \end{aligned}$$
(3.21)
-
3.
$$\begin{aligned} &\frac{{ \vert \eta \vert - 1}}{{ \vert \eta \vert + 1}} < ( {\sigma + 1} )\Re \biggl( { \frac{1}{\pi }} \biggr), \end{aligned}$$
(3.22)
$$\begin{aligned} & \pi \bigl( {z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho, \tau )f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \frac{{1 + \varpi z}}{{1 + \eta z}} + \frac{{ ( {\varpi - \eta } )\pi z}}{{ ( {\sigma + 1} ){{ ( {1 + \eta z} )}^{2}}}}, \end{aligned}$$
(3.23)
-
4.
$$\begin{aligned} &\frac{{ \vert \eta \vert - 1}}{{ \vert \eta \vert + 1}} < \Re \biggl( {\frac{\tau }{\pi }} \biggr), \end{aligned}$$
(3.24)
$$\begin{aligned} & \pi \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho, \tau +1)f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \frac{{1 + \varpi z}}{{1 + \eta z}} + \frac{{ ( {\varpi - \eta } )\pi z}}{{\tau {{ ( {1 + \eta z} )}^{2}}}}. \end{aligned}$$
(3.25)
Then
$$\begin{aligned} z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z) \prec \frac{{1 + \varpi z}}{{1 + \eta z}},\quad z \in \mathfrak{D}, \end{aligned}$$
(3.26)
and \(\frac{{1 + \varpi z}}{{1 + \eta z}}\) is the best dominant of (3.26).
Proof
Taking \(\mathcal{J}(z) = \frac{{1 + \varpi z}}{{1 + \eta z}}\), we see that
$$\begin{aligned} \Re \biggl( {1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}' ( z )}}} \biggr) > \frac{{1 - \vert \eta \vert }}{{1 + \vert \eta \vert }},\quad z \in \mathfrak{D}. \end{aligned}$$
Hence, assumptions (3.18), (3.20), (3.22), and (3.24) imply that stipulations (3.1), (3.3), (3.5), and (3.7) are consistently in Theorem 3.1. Thus from Theorem 3.1 we acquire assertion (3.26). □
Setting \(\varpi = 1\) and \(\eta =- 1\) in Theorem 3.2 yields the following.
Corollary 3.1
Let \(\pi \in \mathcal{C}^{*}\) and \(f\in \Sigma \). If any one of the following pairs of stipulations is achieved:
-
1.
$$\begin{aligned} &\Re \biggl( {\frac{1}{\pi }} \biggr) < 0, \\ &\pi \bigl( {z\mathcal{L}^{\varsigma +1}_{\sigma }(\varrho,\tau )f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \frac{{1 + z}}{{1 - z}} + \pi \vartheta \frac{{2z}}{{{{ ( {1 + z} )}^{2}}}},\quad z \in \mathfrak{D}, \end{aligned}$$
-
2.
$$\begin{aligned} &\Re \biggl( {\frac{{\varrho - 1}}{\pi }} \biggr) > 0, \\ &\pi \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho -1,\tau )f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \frac{{1 + z}}{{1 - z}} + \frac{{2\pi z}}{{ ( {\vartheta - 1} ){{ ( {1 + z} )}^{2}}}}, \end{aligned}$$
-
3.
$$\begin{aligned} &\Re \biggl( {\frac{1}{\pi }} \biggr) > 0, \\ &\pi \bigl( {z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z)} \bigr) +(1- \pi ) \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }( \varrho,\tau )f(z)} \bigr) \prec \frac{{1 + z}}{{1 - z}} + \frac{{2\pi z}}{{ ( {\sigma + 1} ){{ ( {1 + z} )}^{2}}}}, \end{aligned}$$
-
4.
$$\begin{aligned} &\Re \biggl( {\frac{\tau }{\pi }} \biggr) > 0, \\ &\frac{{1 + z}}{{1 - z}} + \frac{{2\pi z}}{{\tau {{ ( {1 + z} )}^{2}}}},\quad z \in \mathfrak{D}. \end{aligned}$$
Then
$$\begin{aligned} z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z) \prec \frac{{1 + z}}{{1 -z}} \end{aligned}$$
and \(\frac{{1 + z}}{{1 -z}}\) is the best dominant.
Theorem 3.3
Let \(\mathcal{J}(z)\) be a nonzero univalent function in \(\mathfrak{D}\) with \(\mathcal{J}(0) = 1\), \(\gamma \in {\mathcal{C}^{*}}, \alpha, \beta \in \mathcal{C}\), while \(\alpha + \beta \ne 0\) and \(f\in \Sigma \). Consider that f and \(\mathcal{J}\) attain the stipulations
$$\begin{aligned} \frac{{\alpha z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}{{\alpha + \beta }} \ne 0\quad ( {z \in \mathfrak{D}} ) \end{aligned}$$
and
$$\begin{aligned} \Re \biggl\{ {1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}' ( z )}} - \frac{{z\mathcal{J}' ( z )}}{{\mathcal{J} ( z )}}} \biggr\} > 0 \quad ( {z \in \mathfrak{D}} ). \end{aligned}$$
(3.27)
If
$$\begin{aligned} \gamma \biggl[ {1 + \frac{{\alpha z{{ ( {\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z)} )}^{\prime }} + \beta z{{ ( {\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} )}^{\prime }}}}{{\alpha \mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta \mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}} \biggr] \prec \frac{{z\mathcal{J}' ( z )}}{{\mathcal{J} ( z )}}, \end{aligned}$$
(3.28)
then
$$\begin{aligned} \frac{{\alpha z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}{{\alpha + \beta }} \prec \mathcal{J} ( z ) \end{aligned}$$
(3.29)
and \(\mathcal{J}\) is the best dominant of (3.29).
Proof
From Lemma 2.2, we consider
$$\begin{aligned} &\varphi ( \mathfrak{a} ) = 0, \qquad\rho ( \mathfrak{a} ) = \frac{1}{\mathfrak{a}}, \\ &\mathfrak{P} ( z ) = z\mathcal{J}' ( z )\pi \bigl( {\mathcal{J} ( z )} \bigr) = \frac{{z\mathcal{J}' ( z )}}{{\mathcal{J} ( z )}}, \end{aligned}$$
and
$$\begin{aligned} \mathcal{H} ( z ) = \mathfrak{P} ( z ). \end{aligned}$$
By utilizing (3.27), \(\mathfrak{P} ( z )\) is univalent starlike in \(\mathfrak{D}\). Further, we obtain
$$\begin{aligned} \Re \biggl( { \frac{{z\mathcal{H}' ( z )}}{{\mathfrak{P} ( z )}}} \biggr) > 0 \quad ( {z \in \mathfrak{D}} ). \end{aligned}$$
Next, let \(\mathfrak{p}\) be a function formulated as follows:
$$\begin{aligned} \mathfrak{p}(z) = { \biggl[ { \frac{{\alpha z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}{{\alpha + \beta }}} \biggr]^{\gamma }} \quad ( {z \in \mathfrak{D}} ). \end{aligned}$$
(3.30)
Then \(\mathfrak{p}\) is regular in \(\mathfrak{D}\) with \(\mathfrak{p}(0) = \mathcal{J}(0) = 1\) and
$$\begin{aligned} \frac{{z\mathfrak{p}' ( z )}}{{\mathfrak{p} ( z )}} = \gamma \biggl[ {1 + \frac{{\alpha z{{ ( {\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z)} )}^{\prime }} + \beta z{{ ( {\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} )}^{\prime }}}}{{\alpha \mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta \mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}} \biggr]. \end{aligned}$$
(3.31)
Using (3.31) in (3.28), we gain
$$\begin{aligned} \frac{{z\mathfrak{p}' ( z )}}{{\mathfrak{p} ( z )}} \prec \frac{{z\mathcal{J}' ( z )}}{{\mathcal{J} ( z )}}. \end{aligned}$$
This means that
$$\begin{aligned} \theta \bigl( {\mathfrak{p} ( z )} \bigr) + z \mathfrak{p}' ( z ) \rho \bigl( {\mathfrak{p} ( z )} \bigr) \prec \varphi \bigl( {\mathcal{J} ( z )} \bigr) + z\mathcal{J}' ( z )\pi \bigl( { \mathcal{J} ( z )} \bigr). \end{aligned}$$
So, from Lemma 2.2, we deduce
$$\begin{aligned} \mathfrak{p} ( z ) \prec \mathcal{J} ( z ) \quad ( {z \in \mathfrak{D}} ), \end{aligned}$$
and \(\mathcal{J}(z)\) is the best dominant. This is quite the assertion in (3.29). □
Setting \(\alpha = 0, \beta = 1\), and \(\mathcal{J} ( z ) = \frac{{1 + \varpi z}}{{1 + \eta z}}\) in Theorem 3.3, it is clear to find out that hypothesis (3.27) holds whenever \({ - 1 \le \eta < \varpi \le 1} \), and therefore we acquire the next outcome.
Corollary 3.2
Let \({ - 1 \le \eta < \varpi \le 1} \) and \(\gamma \in {\mathcal{C}^{*}}\). Let \(f\in \Sigma \) and assume that
$$\begin{aligned} z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z) \ne 0\quad ( {z \in \mathfrak{D}; \varsigma \in {\mathcal{N}_{0}}; \sigma > - 1 } ). \end{aligned}$$
If
$$\begin{aligned} \gamma \biggl[ {1 + \frac{{z{{ ( {\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} )}^{\prime }}}}{{\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}} \biggr] \prec \frac{{ ( {\varpi - \eta } )z}}{{ ( {1 + \varpi z} ) ( {1 + \eta z} )}}, \end{aligned}$$
then
$$\begin{aligned} { \bigl[ {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} \bigr]^{\gamma }} \prec \frac{{1 + \varpi z}}{{1 + \eta z}}, \end{aligned}$$
(3.32)
and \(\frac{{1 + \varpi z}}{{1 + \eta z}}\) is the best dominant of (3.32).
Theorem 3.4
If f is univalent meromorphic starlike of order \(\delta ( {0 \le \delta < 1} )\) in \(\mathfrak{D}^{*}\) and if \(( {1 - \delta } ) = \frac{\varepsilon }{\gamma }, 0 \le \varepsilon \le 1\), then
$$\begin{aligned} { \bigl( {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} \bigr)^{\gamma }} \prec { ( {1 - z} )^{2\varepsilon }}. \end{aligned}$$
(3.33)
The function \({ ( {1 - z} )^{2\varepsilon }}\) is the best dominant of (3.33). Especially, \(\vert {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} \vert \) is bounded by \({2^{2 ( {1 - \delta } )}}\) in \(\mathfrak{D}\).
Proof
Since
$$\begin{aligned} - \frac{{z{{ ( {\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} )}^{\prime }}}}{{\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}} \prec \frac{{1 + ( {1 - 2\delta } )z}}{{1 - z}}\quad ( {z \in \mathfrak{D}} ), \end{aligned}$$
we have
$$\begin{aligned} \gamma \biggl[ {1 + \frac{{z{{ ( {\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} )}^{\prime }}}}{{\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}} \biggr] \prec \frac{{2\varepsilon z}}{{1 - z}}. \end{aligned}$$
Therefore, taking \(\alpha = \sigma = \varsigma = 0, \beta = 1, \varrho = \tau \), and \(\mathcal{J} ( z ) = { ( {1 - z} )^{2 \varepsilon }}\) in Theorem 3.3, we gain (3.33). Thus, it is clear to achieve that \({ ( {1 - z} )^{2\varepsilon }}\) is univalent in \(\mathfrak{D}\) if \(0 \le \varepsilon < 1\). □
Theorem 3.5
Let \(\gamma \in {\mathcal{C}^{*}}\) and \(\chi, \alpha, \beta \in \mathcal{C}\) with \(\alpha + \beta \ne 0\). Let \(\mathcal{J}\) be univalent in \(\mathfrak{D}\) with \(\mathcal{J}(0) = 1\) and
$$\begin{aligned} \Re \biggl( {1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}'(z)}}} \biggr) > \max \bigl\{ {0, - \Re ( \chi )} \bigr\} \quad ( {z \in \mathfrak{D}} ). \end{aligned}$$
(3.34)
Assume that \(f\in \Sigma \) attains
$$\begin{aligned} { \frac{{\alpha z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}{{ \alpha + \beta }}} \ne 0 \quad ( {z \in \mathfrak{D}} ). \end{aligned}$$
Set
$$\begin{aligned} {\mathrm{E}} ( z ) ={}& { \biggl[ { \frac{{\alpha z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z)+ \beta z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}{{ \alpha + \beta }}} \biggr]^{\gamma }} \\ &{}\times \biggl[ {\chi + \gamma \biggl( { \frac{{\alpha z{{ ( {\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z)} )}^{\prime }} + \beta z{{ ( {\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} )}^{\prime }}}}{{\alpha \mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta \mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}} + 1} \biggr)} \biggr]. \end{aligned}$$
If
$$\begin{aligned} {\mathrm{E}} ( z ) \prec \chi \mathcal{J} ( z ) + z \mathcal{J}' ( z ), \end{aligned}$$
(3.35)
then
$$\begin{aligned} { \biggl[ { \frac{{\alpha z\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}{{ \alpha + \beta }}} \biggr]^{\gamma }} \prec \mathcal{J} ( z ), \end{aligned}$$
(3.36)
and \(\mathcal{J}\) is the best dominant of (3.36).
Proof
The evidence for this theorem is similar to that of Theorem 3.3, and we draw major steps. Let \(\mathfrak{p}(z)\) be a function given in (3.30). So, (3.31) leads to
$$\begin{aligned} z\mathfrak{p}'(z) = \gamma \mathfrak{p} ( z ) \biggl[ {1 + \frac{{\alpha z{{ ( {\mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z)} )}^{\prime }} + \beta z{{ ( {\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} )}^{\prime }}}}{{\alpha \mathcal{L}^{\varsigma }_{\sigma +1}(\varrho,\tau )f(z) + \beta \mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}} \biggr]. \end{aligned}$$
(3.37)
In this case, we take
$$\begin{aligned} &\varphi (\mathfrak{a}) = \chi w,\qquad \rho (\mathfrak{a}) = 1 , \quad \mathfrak{a} \in \mathcal{C} \\ &\mathfrak{P} ( z ) = z\mathcal{J}'(z), \qquad\rho \bigl( \mathcal{J} ( z ) \bigr) = z\mathcal{J}'(z) \quad ( {z \in \mathfrak{D}} ), \end{aligned}$$
and
$$\begin{aligned} \mathfrak{p} ( z ) = \varphi \bigl( {\mathfrak{P} ( z )} \bigr) + \mathcal{J} ( z ) = \chi \mathcal{J} ( z ) + z\mathcal{J}'(z),\quad ( {z \in \mathfrak{D}} ). \end{aligned}$$
From (3.34), it yields \({\mathfrak{P} ( z )}\) is starlike in \(\mathfrak{D}\) and that
$$\begin{aligned} \Re \biggl\{ { \frac{{z\mathfrak{p}'(z)}}{{\mathfrak{P} ( z )}}} \biggr\} = \Re \biggl\{ {\chi + 1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}'(z)}}} \biggr\} > 0. \end{aligned}$$
Besides, by substituting the expressions for \(\mathfrak{p}(z)\) and \(z\mathfrak{p}'(z)\) from (3.30) and from (3.37) consistently, we obtain
$$\begin{aligned} \varphi \bigl( {\mathfrak{p} ( z )} \bigr) + z \mathfrak{p}'(z)\rho \bigl( {\mathfrak{p} ( z )} \bigr) = \chi \mathfrak{p} ( z ) + z \mathfrak{p}'(z) = {\mathrm{E}} ( z ). \end{aligned}$$
Assumption (3.35) is equivalent to
$$\begin{aligned} \varphi \bigl( {\mathfrak{p} ( z )} \bigr) + z \mathfrak{p}'(z)\rho \bigl( {\mathfrak{p} ( z )} \bigr) \prec \varphi \bigl( {\mathcal{J} ( z )} \bigr) + z \mathcal{J}'(z) \rho \bigl( {\mathcal{J} ( z )} \bigr). \end{aligned}$$
So Lemma 2.2 gains
$$\begin{aligned} \mathfrak{p} ( z ) \prec \mathcal{J} ( z ). \end{aligned}$$
This last statement gets the assertion of (3.37). □
Taking \(\mathcal{J} ( z ) = \frac{{1 + \varpi z}}{{1 + \eta z}}, ( { - 1 \le \eta < \varpi \le 1} ), \alpha = 0\), and \(\beta = 1\) in Theorem 3.5, we deduce the following.
Theorem 3.6
Let \(\gamma \in {\mathcal{C}^{*}}, - 1 \le \eta < \varpi \le 1\), and
$$\begin{aligned} \mathfrak{a} = \frac{{ \vert \eta \vert - 1}}{{ \vert \eta \vert + 1}}. \end{aligned}$$
(3.38)
If \(f\in \Sigma \) attains
$$\begin{aligned} z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z) \ne 0\quad ( {z \in \mathfrak{D}} ) \end{aligned}$$
and
$$\begin{aligned} { \bigl[ {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} \bigr]^{\gamma }} \biggl[ {\chi + \gamma \biggl( {1 + \frac{{z{{ ( {\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} )}^{\prime }}}}{{\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)}}} \biggr)} \biggr] \prec \chi \frac{{1 + \varpi z}}{{1 + \eta z}} + \frac{{ ( {\varpi - \eta } )z}}{{{{ ( {1 + \eta z} )}^{2}}}}, \end{aligned}$$
(3.39)
then
$$\begin{aligned} { \bigl[ {z\mathcal{L}^{\varsigma }_{\sigma }(\varrho,\tau )f(z)} \bigr]^{\gamma }} \prec \frac{{1 + \varpi z}}{{1 + \eta z}}, \end{aligned}$$
(3.40)
and \(\frac{{1 + \varpi z}}{{1 + \eta z}}\) is the best dominant of (3.40).
Proof
In this case
$$\begin{aligned} \Re \biggl( {1 + \frac{{z\mathcal{J}'' ( z )}}{{\mathcal{J}'(z)}}} \biggr) = \Re \biggl\{ {\frac{{1 - \eta z}}{{1 + \eta z}}} \biggr\} > \frac{{1 - \vert \eta \vert }}{{1 + \vert \eta \vert }} \quad ( {z \in \mathfrak{D}} ). \end{aligned}$$
Therefore, stipulation (3.34) gives (3.38). By utilizing Theorem 3.5, we conclude that assertion (3.40) holds. □
Again setting \(\alpha = 1, \beta = \sigma = \varsigma =\chi = 0, a = \tau \), and \(\mathcal{J} ( z ) = \frac{{1 + z}}{{1 - z}}\) in Theorem 3.6, we get the following.
Corollary 3.3
Let \(f\in \Sigma \) be such that \(zf ( z ) \ne 0\) for \(z \in \mathfrak{D}\), and let \(\gamma \in {\mathcal{C}^{*}}\). If
$$\begin{aligned} { \bigl( {zf ( z )} \bigr)^{\gamma }} \biggl[ {\gamma \biggl( {1 + \frac{{zf' ( z )}}{{f ( z )}}} \biggr)} \biggr] \prec \frac{{2z}}{{{{ ( {1 - z} )}^{2}}}}, \end{aligned}$$
then
$$\begin{aligned} { \bigl( {zf ( z )} \bigr)^{\gamma }} \prec \frac{{1 + z}}{{1 - z}}, \end{aligned}$$
(3.41)
and \(\frac{{1 + z}}{{1 - z}}\) is the best dominant of (3.41).