Throughout this section, let β and γ be real numbers such that \(-\pi /2 < \beta < \pi /2\) and \(0 \leq \gamma < \cos \beta \) unless we mention it. We define a function \(\varphi _{\beta ,\gamma } :\mathbb{D} \rightarrow \mathbb{C}\) by
$$ \varphi _{\beta ,\gamma }(z) = \frac{ 1+ ({\mathrm{{e}}}^{{\mathrm{{i}}}\beta }-2\gamma ) {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } z }{ 1 - z}. $$
(2.1)
Then it is easy to check that the bilinear function \(\varphi _{\beta ,\gamma }\) maps the unit disk \(\mathbb{D}\) onto the half-plane \(\Omega _{\gamma }(\beta )\). By using the function \(\varphi _{\beta ,\gamma }\) we obtain the following results.
Theorem 2.1
Let \(\alpha \in \mathbb{C}\) with \(\operatorname{Re}(\alpha ) \geq 0\). If \(p \in {\mathcal{H}}_{1}\) satisfies
$$ \biggl\vert p(z) + \alpha \frac{zp'(z)}{p(z)} - 1 \biggr\vert < (\cos \beta - \gamma ) \biggl( 1+\frac{1}{2}\operatorname{Re}( \alpha ) \biggr) \bigl\vert p(z) \bigr\vert , \quad z\in \mathbb{D}, $$
(2.2)
then \(1/p \in {\mathcal{P}}_{\gamma }(-\beta )\). That is, \(\operatorname{Re}\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }/p(z) \} > \gamma \) for all \(z\in \mathbb{D}\).
Proof
Let us define functions q and \(h:\mathbb{D}\rightarrow \mathbb{C}\) by
$$ q(z) = \frac{{\mathrm{{e}}}^{{\mathrm{{i}}}\beta }}{p(z)} $$
(2.3)
and
$$ h(z) = {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }\varphi _{-\beta ,\gamma }(z) = \frac{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }+({\mathrm{{e}}}^{-{\mathrm{{i}}}\beta }-2\gamma )z }{ 1-z }, $$
(2.4)
where \(\varphi _{\beta ,\gamma }\) is the function defined by (2.1). Then the functions q and h are analytic in \(\mathbb{D}\) with
$$ q(0)=h(0)={\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \in \mathbb{C} \quad \text{and} \quad h(\mathbb{D}) = \bigl\{ w\in \mathbb{C}:\operatorname{Re} \{ w \} >\gamma \bigr\} . $$
Suppose now that q is not subordinate to h. Then, by Lemma 1.1, there exist points \(z_{0}\in \mathbb{D}\) and \(\zeta _{0} \in \partial \mathbb{D} \setminus \{ 1 \} \) such that
$$ q(z_{0})=h(\zeta _{0})=\gamma + { \mathrm{{i}}}\rho \quad (\rho \in \mathbb{R}) \quad \text{and} \quad z_{0}q'(z_{0}) = m \zeta _{0} h'(\zeta _{0}) = m \sigma \quad (m\geq 1), $$
(2.5)
where
$$ \sigma = \frac{ -\rho ^{2} +2\rho \sin \beta +2\gamma \cos \beta -1-\gamma ^{2} }{ 2(\cos \beta -\gamma ) }. $$
(2.6)
Since \(\gamma < \cos \beta \), we get \(\sigma <0\). Indeed, we have
$$ 2(\cos \beta -\gamma )\sigma = -(\rho -\sin \beta )^{2} -(\cos \beta - \gamma )^{2} \leq -(\cos \beta -\gamma )^{2}, $$
which implies that
$$ \sigma \leq -\frac{1}{2}(\cos \beta -\gamma ) < 0. $$
Using (2.3) and (2.5), we have
$$ \begin{aligned}[b] \biggl\vert \frac{ p(z_{0}) + \alpha \frac{z_{0} p'(z_{0}) }{p(z_{0})} -1 }{ p(z_{0}) } \biggr\vert &= \bigl\vert \alpha z_{0} q'(z_{0}) + q(z_{0}) -{\mathrm{{e}}}^{{ \mathrm{{i}}}\beta } \bigr\vert \\ &= \bigl\vert \alpha m \sigma + \gamma +{\mathrm{{i}}}\rho -{ \mathrm{{e}}}^{{\mathrm{{i}}} \beta } \bigr\vert . \end{aligned} $$
(2.7)
Let \(\alpha =\alpha _{1}+{\mathrm{{i}}}\alpha _{2}\) with \(\alpha _{1} \geq 0\) and \(\alpha _{2}\in \mathbb{R}\). Then we have
$$ \begin{aligned}[b] & \bigl\vert \alpha m \sigma + \gamma +{\mathrm{{i}}}\rho -{\mathrm{{e}}}^{{ \mathrm{{i}}}\beta } \bigr\vert ^{2} \\ &\quad = \bigl( \alpha m \sigma + \gamma +{\mathrm{{i}}}\rho -{\mathrm{{e}}}^{{\mathrm{{i}}} \beta } \bigr) \bigl( \overline{\alpha } m \sigma + \gamma -{\mathrm{{i}}}\rho -{ \mathrm{{e}}}^{-{ \mathrm{{i}}}\beta } \bigr) \\ &\quad = \vert \alpha \vert ^{2} m^{2} \sigma ^{2} +(\gamma -\cos \beta )^{2} +2 \alpha _{1} m\sigma (\gamma -\cos \beta ) + \kappa , \end{aligned} $$
(2.8)
where
$$ \kappa = (\rho -\sin \beta )^{2} +2\alpha _{2} m\sigma ( \rho -\sin \beta ). $$
Furthermore it is easy to see that
$$ \kappa = ( \rho -\sin \beta + m \alpha _{2} \sigma )^{2} - (m \alpha _{2} \sigma )^{2} \geq - (m \alpha _{2} \sigma )^{2}. $$
Since \(m \geq 1\), from (2.8), we have
$$ \begin{aligned}[b] & \bigl\vert \alpha m \sigma + \gamma +{\mathrm{{i}}}\rho -{\mathrm{{e}}}^{{ \mathrm{{i}}}\beta } \bigr\vert ^{2} \\ &\quad \geq \vert \alpha \vert ^{2} m^{2} \sigma ^{2} +(\gamma -\cos \beta )^{2} +2 \alpha _{1} m\sigma (\gamma -\cos \beta ) - \alpha _{2}^{2} m^{2} \sigma ^{2} \\ &\quad = \alpha _{1}^{2} m^{2} \sigma ^{2} +(\gamma -\cos \beta )^{2} +2 \alpha _{1} m\sigma (\gamma -\cos \beta ) \\ &\quad \geq \alpha _{1}^{2} \sigma ^{2} +(\gamma -\cos \beta )^{2} +2 \alpha _{1} \sigma (\gamma -\cos \beta ) \\ &\quad = [ \alpha _{1} \sigma +\gamma - \cos \beta ]^{2}. \end{aligned} $$
(2.9)
Since \(\sigma <0\), \(\alpha _{1}\geq 0\), and \(\cos \beta >\gamma \), inequality (2.9) implies
$$ \bigl\vert \alpha m \sigma + \gamma +{\mathrm{{i}}}\rho -{ \mathrm{{e}}}^{{\mathrm{{i}}}\beta } \bigr\vert \geq -\alpha _{1}\sigma + \cos \beta - \gamma . $$
(2.10)
Furthermore, since \(\sigma \leq -(\cos \beta -\gamma )/2\), we have
$$ -\alpha _{1}\sigma +\cos \beta -\gamma \geq (\cos \beta -\gamma ) \biggl( 1+\frac{1}{2}\alpha _{1} \biggr). $$
(2.11)
Finally, from (2.7), (2.10), and (2.11), we obtain
$$ \biggl\vert p(z_{0}) + \alpha \frac{z_{0}p'(z_{0})}{p(z_{0})} - 1 \biggr\vert \geq (\cos \beta -\gamma ) \biggl( 1+\frac{1}{2}\alpha _{1} \biggr) \bigl\vert p(z_{0}) \bigr\vert . $$
This inequality contradicts hypothesis (2.2). Therefore, we obtain \(q \prec h\) in \(\mathbb{D}\) and the inequality \(\operatorname{Re}\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }/p(z) \} > \gamma \) holds for all \(z\in \mathbb{D}\). □
We remark that the hypothesis in Theorem 2.1 implies also \(1/p \in {\mathcal{P}}_{\gamma }(\beta )\). And we also remark that Theorem 2.1 reduces the result [13] when \(\alpha =1\).
By the above remark, taking \(\gamma =1/2\) in Theorem 2.1 gives the following corollary.
Corollary 2.1
Let α and \(\beta \in \mathbb{R}\) with \(\alpha \geq 0\) and \(\beta \in [0,\pi /3)\). If \(p \in {\mathcal{H}}_{1}\) satisfies (2.2) with \(\gamma =1/2\), then \(p(\mathbb{D}) \subset \Xi _{\beta }\), where
$$ \Xi _{\beta }= \bigl\{ w\in \mathbb{C}: \bigl\vert { \mathrm{{e}}}^{-{\mathrm{{i}}}\beta } w -1 \bigr\vert < 1 \textit{ and } \bigl\vert {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } w -1 \bigr\vert < 1 \bigr\} , $$
and we have \(\operatorname{Re}\{ p(z) \} > 0\) for all \(z\in \mathbb{D}\). Furthermore, if \(\beta \neq0\), then \(|\arg \{ p(z) \}| < \cot \beta \) for all \(z\in \mathbb{D}\).
Taking \(p(z)=zf'(z)/f(z)\), \(f\in {\mathcal{A}}\), in Corollary 2.1 gives the following result.
Corollary 2.2
Let \(\alpha \in \mathbb{R}\) with \(\alpha \geq 0\). If \(\beta \in (0,\pi /3)\) and \(f\in {\mathcal{A}}\) satisfies
$$ \biggl\vert (1-\alpha ) \frac{zf'(z)}{f(z)} + \alpha \biggl( 1+ \frac{zf''(z)}{f'(z)} \biggr) -1 \biggr\vert < \sqrt{\alpha +1} \biggl( \cos \beta -\frac{1}{2} \biggr) \biggl\vert \frac{zf'(z)}{f(z)} \biggr\vert , \quad z\in \mathbb{D}, $$
(2.12)
then \(f \in {\mathcal{SS}}_{0}^{*}(\cot \beta )\), i.e., f is strongly starlike of order \(2(\cot \beta )/\pi \) in \(\mathbb{D}\). If \(f\in {\mathcal{A}}\) satisfies (2.12) with \(\beta =0\), then \(f \in {\mathcal{S}}_{0}^{*}(0)\), i.e., f is a starlike function in \(\mathbb{D}\).
Example 2.1
Let \(a\in \mathbb{C}\) be given, and let \(f_{a}(z)=z/(1-az)\), \(z\in \mathbb{D}\). Then a computation shows that
$$ \frac{1}{2} \biggl\vert \frac{zf'(z)}{f(z)} + \frac{zf''(z)}{f'(z)} -1 \biggr\vert = \frac{3 \vert a \vert \vert z \vert }{2 \vert 1-az \vert } < \frac{3 \vert a \vert }{2 \vert 1-az \vert } = \frac{3 \vert a \vert }{2} \biggl\vert \frac{zf'(z)}{f(z)} \biggr\vert , \quad z\in \mathbb{D}. $$
Hence if
$$ \vert a \vert < \frac{ \sqrt{6} }{3} \biggl( \cos \beta - \frac{1}{2} \biggr), $$
(2.13)
then inequality (2.12) with \(\alpha =1/2\) holds. Thus, by Corollary 2.2 with \(\alpha =1/2\), we conclude that \(f_{a}\) is strongly starlike of order \(2(\cot \beta )/\pi \) in \(\mathbb{D}\) provided inequality (2.13) holds.
Example 2.2
Let \(g_{a}(z)=z/(1-az)^{2}\), \(z\in \mathbb{D}\), with \(a\in \mathbb{C}\). Then a similar computation with Example 2.1 and Corollary 2.2 gives that if \(a\in \mathbb{C}\) satisfies
$$ \frac{2 \vert a \vert (3+2 \vert a \vert )}{ (1- \vert a \vert )^{2} } \leq \frac{ \sqrt{6} }{2} \biggl( \cos \beta - \frac{1}{2} \biggr), $$
then \(g_{a}\) is strongly starlike of order \(2(\cot \beta )/\pi \) in \(\mathbb{D}\).
Theorem 2.2
Let \(\alpha \in \mathbb{R}\) with \(\alpha \geq 0\). Assume that
$$ (2\lambda +\gamma ) \vert \sin \beta \vert < 2\sqrt{\Delta }, $$
(2.14)
where
$$ \Delta =\lambda (\lambda + \cos \beta ) \bigl(-2\gamma \cos \beta +1 + \gamma ^{2} \bigr), $$
(2.15)
with
$$ \lambda = \frac{\alpha }{ 2(\cos \beta -\gamma ) } \geq 0. $$
Let \(p \in {\mathcal{H}}_{1}\) with \(\gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } \notin p(\mathbb{D})\). If
$$ \biggl\vert \operatorname{Im} \biggl\{ p(z) + \frac{\alpha zp'(z)}{p(z) - \gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } } + { \mathrm{{i}}}(2 \lambda +\gamma )\sin \beta \biggr\} \biggr\vert < 2\sqrt{ \Delta }, \quad z\in \mathbb{D}, $$
(2.16)
then \(p \in {\mathcal{P}}_{\gamma }(-\beta )\). That is, \(\operatorname{Re}\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }p(z) \} > \gamma \) for all \(z\in \mathbb{D}\).
Proof
We first note that, since \(p(0)=1\), (2.14) implies that inequality (2.16) is well-defined. Next we define functions q and h by
$$ q(z) = {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }p(z) $$
(2.17)
and (2.4), respectively. If q is not subordinate to h, then there exist points \(z_{0} \in \mathbb{D}\) and \(\zeta _{0} \in \partial \mathbb{D} \setminus \{ 1 \}\) satisfying (2.5) with \(\rho \in \mathbb{R}\). We note that \(\rho \neq0\). Indeed, if \(\rho =0\), then \({\mathrm{{e}}}^{{\mathrm{{i}}}\beta }p(z_{0}) = q(z_{0}) = \gamma \). Therefore we have \(p(z_{0}) = \gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta }\), which contradicts the condition \(\gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } \notin p(\mathbb{D})\).
Simple computations give
$$ \begin{aligned} &p(z_{0}) + \frac{ \alpha z_{0} p'(z_{0}) }{ p(z_{0}) - \gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } } \\ &\quad = {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta }q(z_{0}) + \frac{ \alpha z_{0} q'(z_{0}) }{ q(z_{0}) -\gamma } \\ &\quad = \gamma \cos \beta + \rho \sin \beta +{\mathrm{{i}}} \biggl( \rho \cos \beta - \gamma \sin \beta - \frac{\alpha m \sigma }{\rho } \biggr), \end{aligned} $$
where σ is given by (2.6). Therefore we get
$$ \begin{aligned}[b] &\operatorname{Im} \biggl\{ p(z_{0}) + \frac{\alpha z_{0}p'(z_{0})}{p(z_{0}) - \gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } } + {\mathrm{{i}}}(2 \lambda +\gamma )\sin \beta \biggr\} \\ &\quad = \rho \cos \beta - \frac{\alpha m \sigma }{\rho } + 2 \lambda \sin \beta \\ &\quad = \rho \cos \beta +m \lambda \biggl[ \rho - 2\sin \beta + \frac{-2\gamma \cos \beta +1+\gamma ^{2}}{\rho } \biggr] +2 \lambda \sin \beta . \end{aligned} $$
(2.18)
Assume that \(\rho >0\), and put
$$ \tilde{ \lambda } = \rho - 2\sin \beta + \frac{-2\gamma \cos \beta +1+\gamma ^{2}}{\rho }. $$
Note that
$$ -2\gamma \cos \beta +1+\gamma ^{2} = (\gamma -\cos \beta )^{2} +\sin ^{2} \beta \geq 0 $$
and
$$ -2\gamma \cos \beta +1+\gamma ^{2} - \sin ^{2}\beta = ( \cos \beta - \gamma )^{2} \geq 0. $$
Since \(\rho >0\), these inequalities yield that
$$ \tilde{ \lambda } \geq 2\sqrt{-2\gamma \cos \beta +1+\gamma ^{2}} -2 \sin \beta \geq 0. $$
Therefore, since \(m \geq 1\) and \(\lambda \geq 0\), from (2.18), we obtain
$$ \begin{aligned}[b] &\operatorname{Im} \biggl\{ p(z_{0}) + \frac{\alpha z_{0}p'(z_{0})}{p(z_{0}) - \gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } } + {\mathrm{{i}}}(2 \lambda +\gamma )\sin \beta \biggr\} \\ &\quad \geq \rho \cos \beta + \lambda \tilde{ \lambda } +2 \lambda \sin \beta \\ &\quad = (\cos \beta + \lambda )\rho + \frac{ \lambda (-2\gamma \cos \beta +1+\gamma ^{2})}{\rho } \\ &\quad \geq 2\sqrt{\Delta }, \end{aligned} $$
(2.19)
where Δ is given by (2.15). This contradicts condition (2.16).
Now assume that \(\rho <0\). From (2.18), we have
$$ \begin{aligned} &\operatorname{Im} \biggl\{ p(z_{0}) + \frac{\alpha z_{0}p'(z_{0})}{p(z_{0}) - \gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } } + {\mathrm{{i}}}(2 \lambda +\gamma )\sin \beta \biggr\} \\ &\quad = - \biggl[ \tilde{\rho } \cos \beta +m \lambda \biggl( \tilde{\rho } +2 \sin \beta + \frac{-2\gamma \cos \beta +1+\gamma ^{2}}{\tilde{\rho }} \biggr) \biggr] + 2 \lambda \sin \beta , \end{aligned} $$
where \(\tilde{\rho } = -\rho >0\). A similar calculation with (2.19) gives us to get
$$ \operatorname{Im} \biggl\{ p(z_{0}) + \frac{\alpha z_{0}p'(z_{0})}{p(z_{0}) - \gamma {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } } + { \mathrm{{i}}}(2 \lambda +\gamma )\sin \beta \biggr\} \leq -2\sqrt{ \Delta }, $$
where Δ is given by (2.15). This also contradicts condition (2.16). Therefore we get \(q \prec h\) in \(\mathbb{D}\), and the inequality \(\operatorname{Re}\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }p(z) \} > \gamma \), \(z\in \mathbb{D}\), follows. □
We remark that Theorem 2.2 reduces the result [13] when \(\alpha =1\).
Theorem 2.3
Let \(\alpha \in \mathbb{C}\) with \(\operatorname{Re}(\alpha ) \geq 0\). Assume that \(\Psi (\alpha ,\beta ,\gamma ) < \cos \beta \), where
$$ \Psi (\alpha ,\beta ,\gamma ) = \frac{ \sin ^{2}\beta (\gamma +s)^{2} }{ \cos \beta +s } + \gamma ^{2} \cos \beta +s \bigl(2\gamma \cos \beta -1-\gamma ^{2} \bigr), $$
(2.20)
with
$$ s= \frac{ \operatorname{Re}(\alpha ) }{ 2(\cos \beta -\gamma ) }. $$
If \(p \in {\mathcal{H}}_{1}\) satisfies
$$ \operatorname{Re} \bigl\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \bigl[ \bigl(p(z) \bigr)^{2} + \alpha z p'(z) \bigr] \bigr\} > \Psi (\alpha ,\beta ,\gamma ), \quad z \in \mathbb{D}, $$
(2.21)
then \(\operatorname{Re}\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }p(z) \} > \gamma \) for all \(z\in \mathbb{D}\).
Proof
We first note that, since \(p(0)=1\), the hypothesis \(\Psi (\alpha ,\beta ,\gamma ) < \cos \beta \) implies that inequality (2.21) is well defined. Now we define the functions q and h by (2.17) and (2.4), respectively. If q is not subordinate to h, then there exist points \(z_{0} \in \mathbb{D}\) and \(\zeta _{0} \in \partial \mathbb{D} \setminus \{ 1 \}\) satisfying (2.5) with \(\rho \in \mathbb{R}\).
Put \(\alpha = \alpha _{1} + {\mathrm{{i}}}\alpha _{2}\) with \(\alpha _{1} \geq 0\) and \(\alpha _{2}\in \mathbb{R}\). By (2.17) and (2.5), we obtain
$$ \begin{aligned} & {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \bigl[ \bigl(p(z_{0}) \bigr)^{2} + \alpha z_{0} p'(z_{0}) \bigr] \\ &\quad = {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \bigl[ {\mathrm{{e}}}^{-2{\mathrm{{i}}}\beta } \bigl(q(z_{0}) \bigr)^{2} + \alpha {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } z_{0}q'(z_{0}) \bigr] \\ &\quad = {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } (\gamma + {\mathrm{{i}}}\rho )^{2} + ( \alpha _{1} + {\mathrm{{i}}}\alpha _{2} ) m\sigma \\ &\quad = \bigl(\gamma ^{2}-\rho ^{2} \bigr) \cos \beta + 2 \gamma \rho \sin \beta + m \sigma \alpha _{1} \\ &\qquad{} + {\mathrm{{i}}} \bigl[ 2\gamma \rho \cos \beta - \bigl( \gamma ^{2}-\rho ^{2} \bigr)\sin \beta +m \sigma \alpha _{2} \bigr]. \end{aligned} $$
Hence taking real parts in the above, and from \(\sigma \alpha _{1} \leq 0\) and \(m \geq 1\), we have
$$ \begin{aligned}[b] \operatorname{Re} \bigl\{ { \mathrm{{e}}}^{{\mathrm{{i}}}\beta } \bigl[ \bigl(p(z_{0}) \bigr)^{2} + \alpha z_{0} p'(z_{0}) \bigr] \bigr\} &= \bigl(\gamma ^{2}-\rho ^{2} \bigr) \cos \beta + 2\gamma \rho \sin \beta + m \sigma \alpha _{1} \\ &\leq \bigl(\gamma ^{2}-\rho ^{2} \bigr) \cos \beta + 2 \gamma \rho \sin \beta + \sigma \alpha _{1}. \end{aligned} $$
(2.22)
Now equation (2.6) gives
$$ \bigl(\gamma ^{2}-\rho ^{2} \bigr) \cos \beta + 2\gamma \rho \sin \beta + \sigma \alpha _{1} = - a_{2} \rho ^{2} + a_{1} \rho + a_{0}, $$
(2.23)
where
$$ a_{2} = \cos \beta + \frac{\alpha _{1}}{2\mu }, \qquad a_{1} = \biggl( 2 \gamma + \frac{ \alpha _{1} }{\mu } \biggr) \sin \beta $$
and
$$ a_{0} = \gamma ^{2} \cos \beta + \frac{ \alpha _{1}(2\gamma \cos \beta -1-\gamma ^{2}) }{ 2\mu }, $$
with \(\mu = \cos \beta -\gamma \).
Clearly, \(a_{2}>0\). Thus we have
$$ -a_{2} \rho ^{2} + a_{1} \rho +a_{0} \leq \frac{ a_{1}^{2} }{ 4a_{2} } + a_{0}, \quad \rho \in \mathbb{R}. $$
(2.24)
Consequently, by (2.22), (2.23), and (2.24), we obtain
$$ \operatorname{Re} \bigl\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \bigl[ \bigl(p(z_{0}) \bigr)^{2} + \alpha z_{0} p'(z_{0}) \bigr] \bigr\} \leq \frac{ a_{1}^{2} }{ 4a_{2} } + a_{0} = \Psi (\alpha ,\beta ,\gamma ). $$
This contradicts (2.21). Therefore we obtain \(q \prec h\) in \(\mathbb{D}\), and it follows that the inequality \(\operatorname{Re}\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta }p(z) \} > \gamma \) holds for all \(z\in \mathbb{D}\). □
Since the condition
$$ \bigl\vert \arg ( w-a\sec \beta ) \bigr\vert < \frac{\pi }{2}-\beta $$
implies
$$ \operatorname{Re} \bigl\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } w \bigr\} > a \quad \text{and}\quad \operatorname{Re} \bigl\{ {\mathrm{{e}}}^{-{\mathrm{{i}}}\beta } w \bigr\} > a, $$
for \(w\in \mathbb{C}\), \(a \in \mathbb{R}\) and \(\beta \in (-\pi /2,\pi /2)\), by noting that \(\Psi (\alpha ,\beta ,\gamma ) = \Psi (\alpha ,-\beta ,\gamma )\), the following result can be obtained from Theorem 2.3.
Theorem 2.4
Let \(\alpha \in \mathbb{C}\) with \(\operatorname{Re}(\alpha )\geq 0\). Assume that \(\Psi (\alpha ,\beta ,\gamma ) < \cos \beta \), where Ψ is given by (2.20). If \(p \in {\mathcal{H}}_{1}\) satisfies
$$ \bigl\vert \arg \bigl\{ \bigl(p(z) \bigr)^{2} + \alpha zp'(z) - \Psi (\alpha , \beta ,\gamma ) \sec \beta \bigr\} \bigr\vert < \frac{\pi }{2} -\beta , \quad z\in \mathbb{D}, $$
then
$$ \bigl\vert \arg \bigl(p(z)-\gamma \bigr) \bigr\vert < \frac{\pi }{2} -\beta , \quad z \in \mathbb{D}. $$
Taking \(\alpha =1\) and \(p(z)=zf'(z)/f(z)\), \(f \in {\mathcal{A}}\), in Theorems 2.3 and 2.4 we have the following corollary.
Corollary 2.3
Assume that \(\Psi (1,\beta ,\gamma ) < \cos \beta \), where Ψ is given by (2.20). If \(f \in {\mathcal{A}}\) satisfies
$$ \operatorname{Re} \biggl\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \biggl( \frac{zf'(z)}{f(z)} \biggr) \biggl( 1+ \frac{zf''(z)}{f'(z)} \biggr) \biggr\} > \Psi (1,\beta ,\gamma ), \quad z\in \mathbb{D}, $$
then f is a β-spirallike function of order γ in \(\mathbb{D}\). If \(f \in {\mathcal{A}}\) satisfies
$$ \biggl\vert \arg \biggl\{ \biggl( \frac{zf'(z)}{f(z)} \biggr) \biggl( 1+ \frac{zf''(z)}{f'(z)} \biggr) - \Psi (1,\beta ,\gamma ) \sec \beta \biggr\} \biggr\vert < \frac{\pi }{2}-\beta , \quad z\in \mathbb{D}, $$
then f is strongly starlike of order \(1-(2/\pi )\beta \) and type γ in \(\mathbb{D}\).
Example 2.3
Let \(a\in \mathbb{C}\), and define a function \(f_{a}:\mathbb{D}\rightarrow \mathbb{C}\) by \(f_{a}(z)=z/(1-az)\). Then, since \(|z|<1\), we have
$$ \begin{aligned} &\operatorname{Re} \biggl\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \biggl( \frac{zf'(z)}{f(z)} \biggr) \biggl( 1+ \frac{zf''(z)}{f'(z)} \biggr) - \cos \beta \biggr\} \\ &\quad = \operatorname{Re} \biggl\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \frac{ az(3-az) }{ (1-az)^{2} } \biggr\} \geq - \frac{ \vert a \vert \vert z \vert \vert 3-az \vert }{ \vert 1-az \vert ^{2} } > - \frac{ \vert a \vert (3+ \vert a \vert ) }{ (1- \vert a \vert )^{2} }, \end{aligned} $$
or, equivalently,
$$ \operatorname{Re} \biggl\{ {\mathrm{{e}}}^{{\mathrm{{i}}}\beta } \biggl( \frac{zf'(z)}{f(z)} \biggr) \biggl( 1+ \frac{zf''(z)}{f'(z)} \biggr) \biggr\} > \cos \beta - \frac{ \vert a \vert (3+ \vert a \vert ) }{ (1- \vert a \vert )^{2} }. $$
By Corollary 2.3, \(f_{a}\) is a β-spirallike function of order γ provided
$$ \frac{ \vert a \vert (3+ \vert a \vert ) }{ (1- \vert a \vert )^{2} } \leq \cos \beta - \Psi (1,\beta , \gamma ), $$
where Ψ is given by (2.20). In particular, if
$$ \vert a \vert \leq \frac{1}{76}(-239+3\sqrt{6769}) =: \tau = 0.1029 \cdots , $$
then \(f_{a}\) is \((\pi /3)\)-spirallike function of order \(1/3\). Indeed, when \(\beta =\pi /3\) and \(\gamma =1/3\), we have \(\cos \beta - \Psi (1,\beta ,\gamma ) = 25/63\). Solving the inequality \(|a|(3+|a|)/(1-|a|)^{2} \leq 25/63\) gives us to get \(|a| \leq \tau \).
Example 2.4
Let \(a\in \mathbb{C}\) be given, and let \(g_{a}(z) = z/(1-az)^{2}\), \(z\in \mathbb{D}\). Then, from a similar computation with Example 2.3 and Corollary 2.3, we have that \(g_{a}\) is a β-spirallike function of order γ, if
$$ \frac{1+4 \vert a \vert + \vert a \vert ^{2}}{(1- \vert a \vert )^{2}} \leq \cos \beta - \Psi (1,\beta , \gamma ). $$