On a subfamily of starlike functions related to Hyperbolic Cosine function

Lemma 2.1. Suppose $\sigma \neq 0$ , then $\varrho_{\sigma}(z)$ satisfies the following inclusion $$\{u \in \mathbb{C} : |u - c| < r_{\sigma c}\} \subset \varrho_{\sigma}(\mathbb{D}) =: \Omega_{\rho_{\sigma}} \quad (-\pi/2 \le \sigma \le \pi/2),$$ where $$r_{\sigma c} = \begin{cases} c - \cos \sigma, & \cos \sigma < c \le (\cosh \sigma + \cos \sigma)/2\\ \cosh \sigma - c, & (\cosh \sigma + \cos \sigma)/2 \le c < \cosh \sigma. \end{cases}$$

Theorem 2.6. Let $f \in \mathcal{S}_{\varrho\sigma}^*$ then for each $\sigma \in [-\pi/2, \pi/2] - \{0\}$ , following inclusions hold: - (i) $S_{\varrho_{\sigma}}^ \subset S^(\zeta)$ , where $\zeta = \cos \sigma$ . (ii) $S_{\varrho_{\sigma}}^* \subset \mathcal{M}(\beta)$ , where $\beta = \cosh \sigma$ . - (iii) $S_{q_{\kappa}}^ \subset S_{\varrho_{\sigma}}$ , whenever $\kappa \leq 1 \cos^2 \sigma$ . - (iv) $k \mathcal{ST} \subset \mathcal{S}_{\varrho_{\sigma}}^{}$ , whenever $k \geq \cosh \sigma / (\cosh \sigma 1)$ . (v) $\mathcal{S}_{\varrho_{\sigma}}^{} \subset \mathcal{S}_{hpl}^{*}(s)$ , whenever $\log(\sec \sigma)/\log 2 \leq s \leq 1$ , $\sigma \in [-\pi/3, \pi/3] \{0\}$ . - (vi) $S_{\varrho\sigma}^ \subset S_L^(s)$ , whenever $1 \sqrt{\cos \sigma} \leq s \leq \frac{1}{\sqrt{2}}$ .

Lemma 2.9. [22] If $p \in \mathcal{P}_n[A, B]$ , then for |z| = r $$\left| p(z) - \frac{1 - ABr^{2n}}{1 - B^2r^{2n}} \right| \le \frac{|A - B|r^n}{1 - B^2r^{2n}}.$$ Particularly, if $p \in \mathcal{P}_n(\alpha)$ , then $$\left| p(z) - \frac{1 + (1 - 2\alpha)r^{2n}}{1 - r^{2n}} \right| \le \frac{2(1 - \alpha)r^n}{1 - r^{2n}}.$$

Lemma 2.10. [25] If $p \in \mathcal{P}_n(\alpha)$ , then for |z| = r $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)nr^n}{(1-r^n)(1+(1-2\alpha)r^n)}.$$  <span id="page-6-0"></span>FIGURE 2. Inclusion graphs in context of Corollary 2.7 associated with $\varrho(z)$ .

Theorem 2.11. Let p(z) = (1 + Az)/(1 + Bz), where $-1 < B < A \le 1$ , then $p(z) < \cosh \sqrt{z}$ , if and only if $$A \le \begin{cases} 1 - (1 - B)c_0 & \text{if } 2(1 - AB) \le (c_0 + c_1)(1 - B^2) \\ (1 + B)c_1 - 1 & \text{if } 2(1 - AB) \ge (c_0 + c_1)(1 - B^2). \end{cases}$$ (2.3)

Theorem 3.1. The class $S_{\rho}^* \subset \mathcal{M}(\beta)$ for $|z| < r_{\beta}$ , where $$r_{\beta} = \begin{cases} r(\beta), & 1 < \beta < c_1 \\ 1, & \beta \ge c_1. \end{cases}$$ and $r(\beta) \in (0,1)$ is the smallest root of the equation $\cosh \sqrt{r} = \beta$ . Equality holds when $f(z) = \varphi_{\rho}(z)$ .

Theorem 3.2. Suppose $f \in \mathcal{S}_{\varrho}^*$ , then f(z) is starlike of order $\zeta$ , in $|z| < r_{\zeta}$ , where $r_{\zeta} < 1$ is the least positive root of the equation $\cos \sqrt{r} = \zeta$ . This radius result is sharp.

Theorem 3.3. Let $f \in \mathcal{S}_{\varrho}^*$ , then $f \in \mathcal{C}(\alpha)$ , where $\alpha \in [0,1)$ , provided $|z| \leq r_0$ , where $r_0 \in [0,1)$ is the least positive root of the equation, $2(1-r^2)\cos\sqrt{r} - \sqrt{r}\tan\sqrt{r} = \alpha$ .

Theorem 3.4. For $-1 \le B < A \le 1$ , suppose $f \in \mathcal{S}_n^[A, B]$ , then the sharp $\mathcal{S}_n^(\varrho)$ -radius is given by (i) $$\mathcal{R}_{\mathcal{S}_n^(\varrho)}(\mathcal{S}_n^[A,B]) = \min\{1; ((1-c_0)/(A-Bc_0)^{1/n}\} =: \mathcal{R}_0, \text{ where } 0 \leq B < A \leq 1.$$ $$(ii) \ \mathcal{R}_{\mathcal{S}_n^(\varrho)}(\mathcal{S}_n^[A,B]) = \left\{ \begin{array}{l} \mathcal{R}_0, \quad \mathcal{R}_0 \leq \mathcal{R}_1 \\ \mathcal{R}_2, \quad \mathcal{R}_0 > \mathcal{R}_1, \end{array} \right. \ where \ -1 \leq B < 0 < A \leq 1.$$ where $$\mathcal{R}_1 = \left(\frac{c_0 - 2}{B(c_0 B - 2A)}\right)^{1/2n}, \quad \mathcal{R}_2 = \min\left\{1; \left(\frac{c_1 - 1}{A - Bc_1}\right)^{1/n}\right\}.$$

Theorem 3.5. Let $\beta > 1$ , then the sharp $\mathcal{S}_n^*(\varrho)$ -radius for the class $\mathcal{M}_n(\beta)$ , is given by $$\mathcal{R}_{\mathcal{S}_n^*(\varrho)}(\mathcal{M}_n(\beta)) = \left(\frac{1 - c_0}{2\beta - (1 + c_0)}\right)^{1/n}.$$

Theorem 3.8. The sharp $S_n^*(\varrho)$ – radii for the classes $\mathfrak{F}_1(0)$ , $\mathfrak{F}_1(1/2)$ and $\mathfrak{F}_2$ , are respectively given by (i) $$\mathcal{R}_{\mathcal{S}_n^*(\varrho)}(\mathfrak{F}_1(0)) = \left(\frac{\sqrt{9n^2 - 4(c_0 - 1)(1 + n - c_0)} - 3n}{2(1 + n - c_0)}\right)^{1/n}$$ . (ii) $$\mathcal{R}_{\mathcal{S}_n^*(\varrho)}(\mathfrak{F}_1(1/2)) = \left(\frac{1-c_0}{2n-(c_0-1)}\right)^{1/n}$$ . (iii) $$\mathcal{R}_{\mathcal{S}_n^*(\varrho)}(\mathfrak{F}_2) = \left(\frac{\sqrt{1+n(n+6)+4c_0(c_0-(1+n))}-(1+n)}{2(n-c_0)}\right)^{1/n}$$ .

Theorem 3.9. Let $r \in [0,1)$ , then the sharp $\mathcal{S}_n^*(\varrho)$ -radius for the class $\mathfrak{F}_3$ is given by $$\mathcal{R}_{\mathcal{S}_n^*(\varrho)}(\mathfrak{F}_3) = \left\{ \begin{array}{ll} \mathcal{R}_0, & r \leq \mathcal{R}_0, \\ \mathcal{R}_1, & r \geq \mathcal{R}_0, \end{array} \right.$$ where $$\mathcal{R}_{0} = \begin{cases} \left(\frac{1+A+4n+\sqrt{(1+A+4n)^{2}-4(1-c_{0})(A+c_{0})}}{2(A+c_{0})}\right)^{1/n} & \text{if } -1 \leq A < -c_{0}, \\ \left(\frac{1-c_{0}}{1+4n-c_{0}}\right)^{1/n} & \text{if } A = -c_{0}, \\ \left(\frac{1+A+4n-\sqrt{(1+A+4n)^{2}-4(1-c_{0})(A+c_{0})}}{2(A+c_{0})}\right)^{1/n} & \text{if } -c_{0} < A \leq 1, \end{cases}$$ and $$\mathcal{R}_1 = \left(\frac{\sqrt{(1+A+4n)^2 + 4(A+c_1)(c_1-1)} - (1+A+4n)}{2(A+c_1)}\right)^{1/n}.$$ Proof. Let $f \in \mathfrak{F}_3$ , then $\operatorname{Re} f(z)/g(z) > 0$ and $\operatorname{Re}((1-z^n)^{(1+A)/n}g(z)/z) > 0$ , where $g \in \mathcal{A}_n$ . Define $g(z)/f(z) = p_1(z)$ and $(1-z^n)^{(1+A)/n}g(z)/z = p_2(z)$ , where $p_1(z)$ and $p_2(z)$ are analytic in $\mathbb{D}$ . Since A < 1, then for |z| = r < 1, the inequality $(1 + Ar^{2n}) \ge 1 - r^{2n}$ , holds true. Further on logarithmically differentiating $zp_1(z)p_2(z)(1-z^n)^{-(1+A)/n} = f(z)$ , we get $$\frac{zf'(z)}{f(z)} = \frac{1 + Az^n}{1 - z^n} + \frac{zp'_1(z)}{p_1(z)} + \frac{zp'_2(z)}{p_2(z)}.$$ Due to Lemmas 2.9 - 2.10, for |z|=r, we infer <span id="page-11-0"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1 + Ar^{2n}}{1 - r^{2n}} \right| \le \frac{4nr^n}{1 - r^{2n}} + \frac{(1+A)r^n}{1 - r^{2n}}.$$ (3.6) Assume $c = (1 + Ar^{2n})/(1 - r^{2n})$ . Then $c \le (c_0 + c_1)/2$ leads to $r \le \mathcal{R}$ and vice-versa, where $\mathcal{R} = ((c_0 - 2)/(2A + c_0))^{1/2n}$ . Algebraically, for each $n = 1, 2, 3, \ldots$ , it can be observed that, for the given range of A, we have $\mathcal{R}_0 < \mathcal{R}_1 < \mathcal{R}$ . In particular, if $r \le \mathcal{R}_0$ , then $c \le (c_0 + c_1)/2$ . Further due to Theorem 2.3, inequality (3.6) gives $$\frac{4nr^n}{1-r^{2n}} + \frac{(1+A)r^n}{1-r^{2n}} \le \frac{1+Ar^{2n}}{1-r^{2n}} - c_0,$$ whenever $r \leq \mathcal{R}_0$ . Moreover if $c \geq (c_0 + c_1)/2$ , then $r \geq \mathcal{R}_0$ . Infact, when $r \geq \mathcal{R}_0$ , then we have $c \geq (c_0 + c_1)/2$ . Now inequality (3.6) together with Theorem 2.3 yields $$\frac{4nr^n}{1-r^{2n}} + \frac{(1+A)r^n}{1-r^{2n}} \le c_1 - \frac{1+Ar^{2n}}{1-r^{2n}},$$ provided $r \leq \mathcal{R}_1$ . Thus the following functions, mentioned in Remark 3.7 $$\tilde{f}(z) = \frac{z(1+z^n)^2}{(1-z^n)^{2+(1+A)/n}}$$ and $\tilde{g}(z) = \frac{z(1+z^n)}{(1-z^n)^{1+(1+A)/n}}$ , serve as the extremal function for both the cases.

Theorem 4.1. Let $f \in \mathcal{A}$ , then $f \in \mathcal{S}_{\rho}^*$ if and only if <span id="page-11-1"></span> $$\frac{1}{z} \left( f(z) * \frac{z - kz^2}{(1 - z)^2} \right) \neq 0 \tag{4.1}$$ where $k = \cosh e^{it/2}/(\cosh e^{it/2} - 1)$ for $t \in [-\pi, \pi]$ . Moreover, $f \in \mathcal{S}_{\varrho}^*$ if and only if <span id="page-11-2"></span> $$1 - \sum_{n=2}^{\infty} \frac{(n - \cosh e^{it/2}) a_n}{\cosh e^{it/2} - 1} z^{n-1} \neq 0.$$ (4.2)

Theorem 4.3. Let $f \in \mathcal{S}_{\varrho}^*$ then the following inequality holds $$c_1^2 - 1 \ge \sum_{k=2}^{\infty} (k^2 - c_1^2)|a_k|^2.$$

Definition 1.1. Let $\mathcal{S}_{q_{\sigma}}^{*}$ be the class of normalized starlike functions, defined as follows: $$\mathcal{S}_{\varrho_{\sigma}}^{*} := \left\{ f \in \mathcal{A} : \frac{zf'(z)}{f(z)} \prec \varrho_{\sigma}(z) := \cosh \sigma \sqrt{z}, z \in \mathbb{D} \right\} \quad (\sigma \in [-\pi/2, \pi/2] - \{0\}),$$ where we choose the branch of the square root function so that $$\cosh \sigma \sqrt{z} = 1 + \frac{\sigma^2 z}{2!} + \frac{\sigma^4 z^2}{4!} + \frac{\sigma^6 z^3}{6!} + \cdots$$ The conformal mapping $\varrho_{\sigma}: \mathbb{D} \to \mathbb{C}$ , maps the unit disc $\mathbb{D}$ onto the region $$\Omega_{\varrho_{\sigma}} := \{ u \in \mathbb{C} : |\log(u + \sqrt{u^2 - 1})|^2 < \sigma^2 \} \quad (\sigma \in [-\pi/2, \pi/2] - \{0\}),$$ defined on the principle branch of logarithm and square root functions. For each $\sigma \leq \hat{\sigma}$ , observe that $\varrho_{\sigma}(\mathbb{D}) \subset \varrho_{\hat{\sigma}}(\mathbb{D})$ . Moreover, for each circle |z| = r < 1, $$\begin{cases} \min_{|z|=r} \operatorname{Re} \varrho_{\sigma}(z) = \min_{|z|=r} |\varrho_{\sigma}(z)| = \varrho_{\sigma}(\sqrt{-r}) \\ \max_{|z|=r} \operatorname{Re} \varrho_{\sigma}(z) = \max_{|z|=r} |\varrho_{\sigma}(z)| = \varrho_{\sigma}(\sqrt{r}). \end{cases}$$ (1.1) Assume $\varrho_1(z) =: \varrho(z)$ , therefore we have $\mathcal{S}_{\varrho}^ = \mathcal{S}_{\varrho_1}$ . In the present investigation we shall restrict our major workings to a subclass of starlike functions, namely $\mathcal{S}_{\varrho}$ , and deduce radii constants along with some inclusion relations. In terms of integral representation, we have $f \in \mathcal{S}_{\varrho}$ if and only if <span id="page-1-2"></span><span id="page-1-1"></span><span id="page-1-0"></span> $$f(z) = z \exp\left(\int_0^z \frac{\hat{\varrho}(t) - 1}{t} dt\right) \tag{1.2}$$ where $\hat{\varrho}(z) \prec \varrho(z)$ . Note that if $\psi_{\hat{\varrho}}(z) = 1 + z/3 + z^2/18$ and $\phi_{\hat{\varrho}}(z) = 1 + \sin{(z/3)}$ , then evidently $\psi_{\hat{\varrho}}(z)$ and $\phi_{\hat{\varrho}}(z)$ are subordinate to $\varrho(z)$ , so the corresponding functions $$f_1(z) = z \exp\left(\frac{z}{3} + \frac{z^2}{36}\right)$$ and $f_2(z) = ze^{Si(z)}$ , where $Si(z) = \int_0^z \frac{\sin t}{t} dt$ lie in $\mathcal{S}_{\varrho}^*$ . Now using the representation in (1.2), we obtain different functions, those work as extremal functions for various results. For instance, $\varphi_{\varrho_n} \in \mathcal{A}$ (n = 2, 3, 4, ...), defined as $$\varphi_{\varrho_n}(z) = z \exp\left(\int_0^z \frac{\varrho(t^{n-1}) - 1}{t} dt\right) = z + \frac{z^n}{2(n-1)} + \frac{z^{2n-1}}{48(n-1)} + \cdots, \tag{1.3}$$ belongs to $\mathcal{S}_{\varrho}$ . We denote $\varphi_{\varrho} := \varphi_{\varrho_2}$ . For completeness of our class $\mathcal{S}_{\varrho}$ , we give below a remark using the results of [12, 15]. Remark 1.2. For $f \in \mathcal{S}_{\rho}^*$ and $\varphi_{\varrho}(z)$ be as defined in (1.3), then for $|z| = r_0 < 1$ , we have - $\begin{array}{ll} \text{(i)} & -\varphi_{\varrho}(-r_0) \leq |f(z)| \leq \varphi_{\varrho}(r_0) \text{ (Growth Theorem)}. \\ \text{(ii)} & \varphi_{\varrho}'(-r_0) \leq |f'(z)| \leq \varphi_{\varrho}'(r_0) \text{ (Distortion Theorem)}. \\ \text{(iii)} & |\arg(f(z)/z)| \leq \max_{|z|=r_0} \arg\left(\varphi_{\varrho}(z)/z\right) \text{ (Rotation Theorem)} \ . \end{array}$ Equality for (i)-(iii) holds for some $z_0 \neq 0$ if and only if f(z) is a rotation of $\varphi_{\varrho}(z)$ . Infact if $f \in \mathcal{S}_{\varrho}^*$ then either f is a rotation of $\varphi_{\varrho}(z)$ or $f(\mathbb{D}) \supset \{v : |v| \leq -\varphi_{\varrho}(-1) \approx 0.619...\}$ . Further, from the results in [15] for each $f \in \mathcal{S}_{\varrho}^*$ , (i) $|a_2| \leq 1/2$ , (ii) $|a_3| \leq 1/4$ , (iii) $|a_4| \leq 1/6$ and (iv) for any complex constant $\mu$ , $|a_3 - \mu a_2^2| \le \frac{1}{4} \max\{1, |\mu - 7/12|\}$ . These estimates are sharp. Equality in (i) holds for the function $\varphi_{\varrho}(z)$ and $\tilde{f}(z) = z + z^3/4$ is an extremal function for (ii) and (iv).

Definition 3.6. Let $-1 \le A \le 1$ and $g \in \mathcal{A}_n$ , then for each $n = 1, 2, ..., \mathfrak{F}_3 \subset \mathcal{A}_n$ , be defined as: $$\mathfrak{F}_3 := \left\{ f \in \mathcal{A}_n : \operatorname{Re} \frac{f(z)}{g(z)} > 0 \text{ and } \operatorname{Re} \frac{(1-z^n)^{(1+A)/n}g(z)}{z} > 0 \right\}.$$ <span id="page-9-0"></span>Remark 3.7. The functions $\tilde{f}(z)=z(1+(1-2\beta)z^n)$ and $\tilde{g}(z)=z(1+(1-2\beta)z^n)/(1-z^n)$ defined on $\mathbb{D}$ satisfy $|(\tilde{f}(z)/\tilde{g}(z))-1|=|z|^n<1$ and $\operatorname{Re}\tilde{g}(z)/z=\operatorname{Re}(1+(1-2\beta)z^n)/(1-z^n)>\beta$ . Therefore $\tilde{f}\in\mathfrak{F}_1(\beta)$ , where $\beta\in\{0,1/2\}$ . If $\tilde{f}(z)=z(1+z^n)/(1-z^n)^{1/n}$ and $\tilde{g}(z)=z/(1-z^n)^{1/n}$ , then $\tilde{f}\in\mathfrak{F}_2$ . Similarly when $\tilde{f}(z)=z(1+z^n)^2/(1-z^n)^{2+(1+A)/n}$ and $\tilde{g}(z)=z(1+z^n)/(1-z^n)^{1+(1+A)/n}$ , then $\tilde{f}\in\mathfrak{F}_3$ . Therefore the class $\mathfrak{F}_3$ is non-empty.