In this section, we study the geometric properties of the classes \(S_{q}^{*}(\kappa ,k,h)\) and \(\mathbb{J}_{q}^{\kappa ,b} (A,B,k)\) and the consequences of these classes for recent investigations by researchers.
Theorem 3.1
For
\(\upsilon \in \varLambda \)if one of the following statements is given:
\(\mathcal{S}_{q}^{\kappa ,k}\upsilon (z)\)is of bounded boundary rotation;
υsatisfies the subordination structure
$$ \bigl( \mathcal{S}_{q}^{\kappa ,k}\upsilon (z) \bigr)' \prec \biggl(\frac{1+z}{1-z} \biggr) ^{\flat },\quad \flat >0, z \in {\mathbb{U}}; $$
υfulfills the layout
$$ \Re \biggl(\bigl( \mathcal{S}_{q}^{\kappa ,k}\upsilon (z) \bigr)' \frac{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)}{z} \biggr)> \frac{\varsigma }{2},\quad \varsigma \in [0,1), z \in {\mathbb{U}}, $$
υobeys the relation
$$ \Re \biggl( z \bigl( \mathcal{S}_{q}^{\kappa ,k} \upsilon (z) \bigr)'' -\bigl( \mathcal{S}_{q}^{\kappa ,k} \upsilon (z)\bigr)' + 2 \frac{\mathcal{S}_{q} ^{\kappa ,k} \upsilon (z) }{z} \biggr)> 0, $$
υadmits the relation
$$ \Re \biggl( \frac{z (\mathcal{S}_{q}^{\kappa ,k}\upsilon (z))' }{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z) } + 2 \frac{ \mathcal{S}_{q} ^{\kappa ,k} \upsilon (z)}{z} \biggr)> 1, $$
then
\(\frac{\mathcal{S}_{q}^{\kappa ,k} \upsilon (z)}{z} \in {\mathcal{P}}(\sigma )\)for some
\(\sigma \in [0,1)\).
Proof
Consider a function ρ as follows:
$$ \rho (z)= \frac{\mathcal{S}_{q}^{\kappa ,k} \upsilon (z)}{z}\quad \Rightarrow \quad z\rho '(z)+ \rho (z)= \bigl( \mathcal{S}_{q}^{\kappa ,k} \upsilon (z)\bigr)'. $$
(3)
By the first conclusion, \(\mathcal{S}_{q}^{\kappa ,k}\upsilon (z)\) is of bounded boundary rotation, it implies that \(\Re ( z\rho '(z)+ \rho (z))>0\). Thus, by Lemma 2.1.i, we obtain \(\Re (\rho (z))>0\) which implies the first part of the theorem.
According to the second part, we have the subject subordination layout
$$ \bigl( \mathcal{S}_{q}^{\kappa ,k} \upsilon (z) \bigr)'=z\rho '(z)+ \rho (z) \prec \biggl[ \frac{1+z}{1-z}\biggr]^{\flat }. $$
Now, according to Lemma 2.1.i, there is a constant \(\ell >0\) with \(\flat =\flat (\ell )\) accepting the subordination
$$ \frac{\mathcal{S}_{q}^{\kappa ,k} \upsilon (z)}{z}\prec \biggl(\frac{1+z}{1-z} \biggr) ^{\ell }. $$
This implies that
$$ \Re \biggl(\frac{\mathcal{S}_{q}^{\kappa ,k}\upsilon (z)}{z} \biggr)> \sigma , \quad \sigma \in [0,1). $$
Continuing, we address the third part, which implies that
$$ \Re \bigl(\rho ^{2}(z)+2\rho (z) . z\rho '(z) \bigr)=2\Re \biggl(\bigl( \mathcal{S}_{q}^{\kappa ,k} \upsilon (z)\bigr)' \frac{ \mathcal{S}_{q}^{ \kappa ,k} \upsilon (z)}{z} \biggr) > \varsigma . $$
(4)
In virtue of Lemma 2.1.ii, there is a constant \({\ell }>0\) such that \(\Re (\rho (z))>\ell \) and
$$ \rho (z) = \frac{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z) }{z} \in {\mathcal{P}}(\sigma ), \quad \sigma \in [0,1). $$
It follows from (4) that \(\Re ( \mathcal{S}_{q}^{\kappa ,k} \upsilon (z))' )>0\) and thus by the Noshiro–Warschawski and Kaplan theorems that \(\mathcal{S}_{q}^{\kappa ,k}\upsilon (z)\) is univalent and of bounded boundary rotation in \(\mathbb{U}\).
By differentiating (3) and taking the real part, we have
$$\begin{aligned} \Re \bigl( \rho (z)+z\rho '(z)+z^{2} \rho ''(z) \bigr) = \Re \biggl( z\bigl( \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)\bigr)'' -\bigl(\mathcal{S}_{q}^{\kappa ,k} \upsilon (z) \bigr)' + 2 \frac{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z) }{z} \biggr)> 0. \end{aligned}$$
Thus, in view of Lemma 2.1-ii, we attain \(\Re (\frac{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)}{z})>0\).
By logarithmic differentiation (3) and taking the real part, we obtain the following:
$$\begin{aligned} \Re \biggl( \rho (z)+\frac{z\rho '(z)}{\rho (z)}+z^{2} \rho ''(z) \biggr) &=\Re \biggl( \frac{z (\mathcal{S}_{q}^{\kappa ,k}\upsilon (z) )' }{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)} + 2 \frac{ \mathcal{S} _{q}^{\kappa ,k}\upsilon (z) }{z} -1 \biggr)> 0. \end{aligned}$$
Thus, according to Lemma 2.1-iii, where \(\alpha (z)=1\), we get \(\Re ( \frac{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)}{z})>0\). □
Theorem 3.2
Consider
\(\upsilon \in S_{q}^{*}(\kappa ,k,h)\), where
\(h(z)\)is convex univalent function in
\(\mathbb{U}\). Then
$$ \mathcal{S}_{q}^{\kappa ,k} \upsilon (z) \prec z \exp \biggl( \int ^{z} _{0} \frac{h(\eth (w))-1}{w} \,d w \biggr), $$
where
\(\eth (z)\)is analytic in
\(\mathbb{U}\), with
\(\eth (0) = 0\)and
\(|\eth (z)| < 1\). Moreover, for
\(|z|=\chi \), \(\mathcal{S}_{q}^{ \kappa ,k}\upsilon (z)\)fulfills the formula
$$ \exp \biggl( \int ^{1}_{0} \frac{h\eth (-\chi ))-1}{\chi } \biggr) \,d \chi \leq \biggl\vert \frac{ \mathcal{S}_{q}^{\kappa ,k} \upsilon (z)}{z} \biggr\vert \leq \exp \biggl( \int ^{1}_{0} \frac{h(\eth (\chi ))-1}{\chi } \biggr) \,d \chi . $$
Proof
Since \(\psi \in S_{q}^{*}(\kappa ,k,h)\), we get
$$ \biggl(\frac{z( \mathcal{S}_{q}^{\kappa ,k} \upsilon (z))'}{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)} \biggr) \prec h(z), \quad z \in \mathbb{U}, $$
which leads to a Schwarz function with \(\eth (0) = 0\) and \(|\eth (z)| < 1\) satisfying the following equality:
$$ \biggl(\frac{z( \mathcal{S}_{q}^{\kappa ,k} \upsilon (z))'}{ \mathcal{S}_{q}^{\kappa ,k} \upsilon (z)} \biggr) = h\bigl(\eth (z)\bigr),\quad z \in \mathbb{U}. $$
A calculation gives
$$ \biggl(\frac{( \mathcal{S}_{q}^{\kappa ,k} \psi (z))'}{ \mathcal{S} _{q}^{\kappa ,k} \psi (z)} \biggr)- \frac{1}{z} = \frac{ h(\eth (z))-1}{z}. $$
By integrating both sides, we obtain
$$ \log \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)-\log z= \int ^{z}_{0} \frac{h( \eth (w))-1}{w} \,d w. $$
Thus, we have
$$ \log \frac{ \mathcal{S}_{q}^{\kappa ,k} \upsilon (z)}{ z}= \int ^{z} _{0} \frac{h(\eth (\xi ))-1}{w} \,d w. $$
(5)
By utilizing the meaning of subordination, we conclude that
$$ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z) \prec z \exp \biggl( \int ^{z} _{0} \frac{h(\eth (w))-1}{w} \,dw \biggr). $$
Besides, we find that the function \(h(z)\) maps the disk \(0< |z|< \chi <1\) onto a domain which is convex and symmetric with respect to the real axis, which means
$$ h\bigl(-\chi \vert z \vert \bigr) \leq \Re \bigl(h\bigl(\eth (\chi z)\bigr) \bigr) \leq h\bigl(\chi \vert z \vert \bigr), \quad \chi \in (0,1), $$
then we obtain the next relations:
$$ h(-\chi ) \leq h\bigl(-\chi \vert z \vert \bigr),\qquad h\bigl(\chi \vert z \vert \bigr) \leq h(\chi ), $$
and
$$ \int ^{1}_{0} \frac{h(\eth (-\chi \vert z \vert ))-1}{\chi }\,d\chi \leq \Re \biggl( \int ^{1}_{0} \frac{h(\eth (\chi ))-1}{\chi } \,d\chi \biggr)\leq \int ^{1}_{0} \frac{h(\eth (\chi \vert z \vert ))-1}{\chi }\,d\chi . $$
By employing Eq. (5), we deduce that
$$ \int ^{1}_{0} \frac{h(\eth (-\chi \vert z \vert ))-1}{\chi }\,d \chi \leq \log \biggl\vert \frac{ \mathcal{S}_{q}^{\kappa ,k} \upsilon (z)}{ z} \biggr\vert \leq \int ^{1}_{0} \frac{h(\eth (\chi \vert z \vert ))-1}{\chi }\,d\chi , $$
which leads to
$$ \exp \biggl( \int ^{1}_{0} \frac{h(\eth (-\chi \vert z \vert ))-1}{\chi }\,d\chi \biggr) \leq \biggl\vert \frac{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)}{ z} \biggr\vert \leq \exp \biggl( \int ^{1}_{0} \frac{h(\eth (\chi \vert z \vert ))-1}{ \chi }\,d\chi \biggr). $$
Hence, we have
$$ \exp \biggl( \int ^{1}_{0} \frac{h(\eth (-\chi ))-1}{\eta } \biggr) \,d \chi \leq \biggl\vert \frac{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)}{z} \biggr\vert \leq \exp \biggl( \int ^{1}_{0} \frac{h(\eth (\chi ))-1}{\chi } \biggr) \,d \chi . $$
□
Corollary 3.1
([8])
Let
\(q \longrightarrow 1\)in Theorem
3.2. Then
$$ \mathcal{S}_{1}^{\kappa ,k}\upsilon (z) \prec z \exp \biggl( \int ^{z} _{0} \frac{h(\eth (w))-1}{w} \,d w \biggr). $$
Note that all the special cases of the class \(S_{q}^{*}(\kappa ,k,h)\) can be considered as consequences of Theorem 3.2.
Theorem 3.3
If
\(\upsilon \in \mathbb{J}_{q}^{\kappa ,b} (A,B,k)\)
then the odd function
$$ \mathfrak{B}(z)=\frac{1}{2} \bigl[\upsilon (z)-\upsilon (-z)\bigr],\quad z \in \mathbb{U,} $$
attains the subordination inequalities
$$ 1+\frac{1}{b} \biggl( \frac{ \mathcal{S}_{q}^{\kappa ,k+1} \mathfrak{B}(z)}{ \mathcal{S}_{q}^{\kappa ,k} \mathfrak{B}(z)}-1 \biggr) \prec \frac{1+Az}{1+Bz} $$
and
$$\begin{aligned} &\Re \biggl( \frac{z \mathfrak{B}(z)'}{ \mathfrak{B}(z)} \biggr) \geq \frac{1- \varrho ^{2}}{1+\varrho ^{2}},\quad \vert z \vert =\varrho < 1, \\ &\quad \bigl(z \in \mathbb{U}, -1\leq B< A\leq 1, k=1,2,\ldots, b \in \mathbb{C} \setminus \{0\}, \kappa \in \mathbb{R} \bigr). \end{aligned}$$
Proof
Let \(\upsilon \in \mathbb{J}_{q}^{\kappa ,b} (A,B,k)\). Then there exists a function \(P \in \mathbb{J}(A,B) \) with the layout
$$ b \bigl(P(z)-1\bigr)= \biggl( \frac{2 \mathcal{S}_{q}^{\kappa ,k+1}\upsilon (z)}{ \mathcal{S}_{q}^{\kappa ,k}\upsilon (z)- \mathcal{S}_{q}^{\kappa ,k} \upsilon (-z)} \biggr) $$
and
$$ b \bigl(P(-z)-1\bigr)= \biggl( \frac{-2 \mathcal{S}_{q}^{\kappa ,k+1}\upsilon (-z)}{ \mathcal{S}_{q}^{\kappa ,k} \upsilon (z)- \mathcal{S}_{q}^{\kappa ,k}\upsilon (-z)} \biggr). $$
This yields
$$ 1+\frac{1}{b} \biggl( \frac{ \mathcal{S}_{q}^{\kappa ,k+1} \mathfrak{B}(z)}{ \mathcal{S}_{q}^{\kappa ,k} \mathfrak{B}(z)}-1 \biggr)= \frac{P(z)+P(-z)}{2}. $$
In addition, since P fulfills the inequality
$$ P(z) \prec \frac{1+Az}{1+Bz}, $$
where \(\frac{1+Az}{1+Bz}\) is univalent, by the idea of subordination, we obtain
$$ 1+\frac{1}{b} \biggl( \frac{ \mathcal{S}_{q}^{\kappa ,k+1} \mathfrak{B}(z)}{ \mathcal{S}_{q}^{\kappa ,k} \mathfrak{B}(z)}-1 \biggr) \prec \frac{1+Az}{1+Bz}. $$
Also, the odd function \(\mathfrak{B}(z)\) is starlike in \(\mathbb{U,}\) which produces the subordination inequality
$$ \frac{z \mathfrak{B}(z)'}{ \mathfrak{B}(z)} \prec \frac{1-z^{2}}{1+z ^{2}}, $$
that is, there exists a Schwarz function \(\gamma \in \mathbb{U}, | \gamma (z)| \leq |z|<1, \gamma (0)=0\) with the property
$$ \varXi (z):=\frac{z \mathfrak{B}(z)'}{ \mathfrak{B}(z)} \prec \frac{1- \gamma (z)^{2}}{1+\gamma (z)^{2}}, $$
which leads to
$$ \gamma ^{2}(\zeta )=\frac{1-\varXi (\zeta )}{1+\varXi (\zeta )}, \quad \zeta \in \mathbb{U}, \zeta , \vert \zeta \vert =r< 1. $$
A computation implies that
$$ \biggl\vert \frac{1-\varXi (\zeta )}{1+\varXi (\zeta )} \biggr\vert = \bigl\vert \gamma (\zeta ) \bigr\vert ^{2} \leq \vert \zeta \vert ^{2}. $$
Therefore, we get the following inequality:
$$ \biggl\vert \varXi (\zeta )- \frac{1+ \vert \zeta \vert ^{4}}{1- \vert \zeta \vert ^{4}} \biggr\vert ^{2} \leq \frac{ 4 \vert \zeta \vert ^{4}}{(1- \vert \zeta \vert ^{4})^{2}} $$
or
$$ \biggl\vert \varXi (z)- \frac{1+ \vert \zeta \vert ^{4}}{1- \vert \zeta \vert ^{4}} \biggr\vert \leq \frac{2 \zeta |^{2}}{(1- \vert \zeta \vert ^{4})}. $$
Consequently, we obtain the result
$$ \Re \bigl(\varXi (z)\bigr) \geq \frac{1-\varrho ^{2}}{1+\varrho ^{2}}, \qquad \vert \zeta \vert = \varrho < 1. $$
□
The following consequences of Theorem 3.3 can be found in [26, 27] and [8], respectively.
Corollary 3.2
Let
\(\lambda =1\)in Theorem
3.3. Then
$$ 1+\frac{1}{b} \biggl( \frac{\mathcal{S}_{q}^{0,k+1}\mathfrak{B}(z)}{ \mathcal{S}_{q}^{0,k} \mathfrak{B}(z)}-1 \biggr) \prec \frac{1+Az}{1+Bz}. $$
Corollary 3.3
Let
\(\kappa =0, k=1\)and
\(q \longrightarrow 1\)in Theorem
3.3. Then
$$ 1+\frac{1}{b} \biggl( \frac{ \mathcal{S}_{q}^{0,2}\mathfrak{B}(z)}{ \mathcal{S}_{q}^{0,1} \mathfrak{B}(z)}-1 \biggr) \prec \frac{1+Az}{1+Bz}. $$
Corollary 3.4
Let
\(q \longrightarrow 1\)in Theorem
3.3. Then
$$ 1+\frac{1}{b} \biggl( \frac{ \mathcal{S}_{q}^{\kappa ,k+1} \mathfrak{B}(z)}{\mathcal{S}_{q}^{\kappa ,k} \mathfrak{B}(z)}-1 \biggr) \prec \frac{1+Az}{1+Bz}. $$