A Cardioid Domain and Starlike Functions

Theorem 1.1. Let $f \in \mathscr{S}_{\wp}^*$ and $f_1$ as defined in (1.10). Then - (i) Growth theorem: $-f_1(-|z|) \le |f(z)| \le f_1(|z|)$ . - (ii) Covering theorem: $\{w : |w| \le -f_1(-1) \approx 0.5314\} \subset f(\mathbb{D}).$ (iii) Rotation theorem: $|\arg f(z)/z| \le \max_{|z|=r} \arg (f_1(z)/z).$ - (iv) $f(z)/z \prec f_1(z)/z$ and $|f'(z)| \leq f'_1(|z|)$ . As a consequence of growth theorem, for |z|=r, we obtain $$\log \left| \frac{f(z)}{z} \right| \le \int_0^r e^t dt \le \int_0^1 e^t dt = e - 1,$$ which implies $|f(z)| \le e^{e^{-1}}$ and the bound can not be further improved as $z \exp(e^z - 1)$ acts as an extremal function. In the present work, we discuss the geometric properties of the cardioid domain $\wp(\mathbb{D})$ and the inclusion relationship of $\mathscr{S}_{\wp}$ with the classes $\mathcal{SS}^(\gamma)$ , $\mathcal{S}^(\alpha)$ and many more. We also obtain various sharp radius results associated with $\mathscr{S}_{\wp}$ . Further, we find the coefficient estimates for $f \in \mathscr{S}_{\wp}$ and the sharp bound for the first five coefficients. Conjecture related to the sharp bound of nth coefficient is also posed. We also obtain the estimate for the third Hankel determinant for the class $\mathscr{S}_{\wp}$ using the expression of the carathéodory coefficient $p_4$ in terms of $p_1$ , where the technique has not been exploited much so far. The sharp estimates on third Hankel determinant for the classes of two-fold and three-fold symmetric functions associated with $\mathscr{S}_{\wp}^*$ is also obtained. Further, coefficient related problems are also discussed.

Theorem 2.1. The radius of convexity of the function $\wp(z) = 1 + ze^z$ is the smallest positive root of the equation $r^3 - 4r^2 + 4r - 1 = 0$ , which is given by $$r_c = (3 - \sqrt{5})/2 \approx 0.381966.$$ Proof. Now it is to find the constant $r_c \in (0,1]$ so that <span id="page-3-0"></span> $$\operatorname{Re}\left(1 + \frac{z\wp''(z)}{\wp'(z)}\right) > 0 \quad (|z| < r_c). \tag{2.1}$$ Since $$\operatorname{Re}\left(1 + \frac{z\wp''(z)}{\wp'(z)}\right) = \frac{r^3\cos\theta + r^2(3 + \cos 2\theta) + 4r\cos\theta + 1}{1 + 2r\cos\theta + r^2} =: g(r, \theta)$$ and g is symmetric about the real axis as $g(r,\theta)=g(r,-\theta)$ . Thus we only need to consider $\theta \in [0,\pi]$ . Further we have $1+2\cos\theta r+r^2>0$ for $r\in(0,1)$ and $\theta\in[0,\pi]$ ). So we may consider the numerator of $g(r,\theta)$ as $$g_N(r,\theta) := r^3 \cos \theta + r^2 (3 + \cos 2\theta) + 4r \cos \theta + 1.$$ Now to arrive at (2.1), we only need to show <span id="page-3-1"></span> $$g_N(r,\theta) > 0 \quad (r \in (0, r_c)).$$ (2.2) It is evident that for any fixed $r = r_0$ , $g_N(r_0, \theta)$ attains its minimum at $\theta = \pi$ which is given by $g_N(r_0, \pi) = -r_0^3 + 4r_0^2 - 4r_0 + 1$ . Since $g_N(0, \pi) > 0$ and if $r_c$ is the least positive root of $r^3 - 4r^2 + 4r - 1 = 0$ then (2.2) follows and hence the result. Using elementary calculus, one can easily find the following sharp bounds that are associated with the function $\wp(z) = 1 + ze^z$ , which are used extensively in obtaining our subsequent results.

Lemma 2.2. Let $\wp(z) = 1 + ze^z$ . Then we have (1) $$\{w : |w - a| < r_a\} \subset \wp(\mathbb{D}), \text{ where}$$ $$r_a = \begin{cases} (a-1) + 1/e, & 1 - 1/e < a \le 1 + (e - e^{-1})/2; \\ e - (a-1), & 1 + (e - e^{-1})/2 \le a < 1 + e. \end{cases}$$ (2) $\wp(\mathbb{D}) \subset \{w : |w - a| < R_a\}, \text{ where }$ $$R_a = \begin{cases} 1 + e - a, & 1 - 1/e < a \le (e + e^{-1})/2 \\ \sqrt{d(\theta_a)}, & (e + e^{-1})/2 < a < 1 + e. \end{cases}$$ where $\theta_a \in (0, \pi)$ , is the root of the following equation: <span id="page-4-0"></span> $$\sin(\theta/2) + (1 - a)\sin(3\theta/2 + \sin\theta) = 0. \tag{2.3}$$ Proof. We begin with the first part. The curve $\wp(e^{i\theta}) = 1 + e^{\cos\theta}(\cos(\theta + \sin\theta) + i\sin(\theta + \sin\theta))$ represents the boundary of $\wp(\mathbb{D})$ and is symmetric about the real axis. So it is enough to consider $\theta$ in $[0,\pi]$ . Now, square of the distance of (a,0) from the points on the curve $\wp(e^{i\theta})$ is given by $$d(\theta) := (a - 1 - e^{\cos \theta} \cos(\theta + \sin \theta))^2 + e^{2\cos \theta} \sin^2(\theta + \sin \theta)$$ $$= e^{2\cos \theta} - 2(a - 1)e^{\cos \theta} \cos(\theta + \sin \theta) + (a - 1)^2.$$ Case(i): If $1 - e^{-1} < a \le (e + e^{-1})/2$ , then $d(\theta)$ decreases in $[0, \pi]$ . Therefore, we get $$r_a = \min_{\theta \in [0,\pi]} \sqrt{d(\theta)} = \sqrt{d(\pi)} = (a-1) + 1/e.$$ Now for the range $(e + e^{-1})/2 \le a < 1 + (e - e^{-1})/2$ , it is easy to see that the equation $$d'(\theta) = -4e^{\cos\theta}\cos(\theta/2)(\sin(\theta/2) - (a-1)\sin(3\theta/2 + \sin\theta)) = 0$$ has three real roots $0, \theta_a$ and $\pi$ , where $\theta_a$ is the root of the equation given in (2.3) and we have $\theta_{a_1} < \theta_{a_2}$ whenever $a_1 < a_2$ . Further, we see that $d(\theta)$ increases in $[0, \theta_a]$ and decreases in $[\theta_a, \pi]$ . Also, $$d(\pi) - d(0) = 2(e + e^{-1})(a - (1 + (e - e^{-1})/2)) < 0.$$ Therefore, $\min\{d(0), d(\theta_a), d(\pi)\} = d(\pi)$ and we have $$r_a = \sqrt{d(\pi)} = (a-1) + 1/e.$$ Case(ii): If $1 + (e - e^{-1})/2 \le a < 1 + e$ , we see that $d(\theta)$ is an increasing function for $\theta \in [0, \theta_a]$ and decreasing for $\theta \in [\theta_a, \pi]$ , where $\theta_a$ is the root of the equation defined in (2.3). Also, $$d(\pi) - d(0) = 2(e + e^{-1})(a - (1 + (e - e^{-1})/2)) > 0.$$ Therefore, $\min\{d(0), d(\theta_a), d(\pi)\} = d(0)$ and we have $$r_a = \sqrt{d(0)} = e - (a - 1).$$ This completes the proof of first part. The proof of second part is much akin to the first part so is skipped here. $\Box$ Remark 2.1. We obtain the largest disk $D_L := |w - a| < r_a$ contained in $\wp(\mathbb{D})$ when $a = 1 + (e - e^{-1})/2$ and $r_a = (e + e^{-1})/2$ and the smallest disk $D_S := |w - a| < R_a$ , which contains $\wp(\mathbb{D})$ when $a = (e + e^{-1})/2$ and $R_a = 1 + (e - e^{-1})/2$ . Thus $D_L \subset \wp(\mathbb{D}) \subset D_S$ . Our next result deals with the inclusion relations of the class $\mathscr{S}_{\wp}^*$ involving various classes including the following (see [5, 12, 37]): $$\mathcal{M}(\beta) := \left\{ f \in \mathcal{A} : \operatorname{Re} \frac{zf'(z)}{f(z)} < \beta, \quad \beta > 1 \right\},$$ $$k - \mathcal{ST} := \left\{ f \in \mathcal{A} : \operatorname{Re} \frac{zf'(z)}{f(z)} > k \left| \frac{zf'(z)}{f(z)} - 1 \right|, \quad k \ge 0 \right\}$$ and $$\mathcal{ST}_p(a) := \left\{ f \in \mathcal{A} : \operatorname{Re} \frac{zf'(z)}{f(z)} + a > \left| \frac{zf'(z)}{f(z)} - a \right|, \quad a > 0 \right\}.$$

Lemma 2.3. [32] If $p \in \mathcal{P}_n(\alpha)$ , then for |z| = r, $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)nr^n}{(1-r^n)(1+(1-2\alpha)r^n)}.$$

Lemma 2.4. [28] If $p(z) \in \mathcal{P}_n[A, B]$ , 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 $$\left| p(z) - \frac{1 + (1 - 2\alpha)r^{2n}}{1 - r^{2n}} \right| \le \frac{2(1 - \alpha)r^n}{1 - r^{2n}}.$$

Theorem 2.3. Let $-1 < B < A \le 1$ . If one of the following two conditions hold. $$\begin{array}{l} (i) \ \ 2(e-1)(1-B^2) < 2e(1-AB) \leq (e^2+2e-1)(1-B^2) \ \ and \ B-1 \leq e(1-A); \\ (ii) \ \ (e^2+2e-1)(1-B^2) \leq 2e(1-AB) < 2e(1+e)(1-B^2) \ \ and \ A-B \leq e(1+B). \end{array}$$ Then $\mathcal{S}^[A,B] \subset \mathscr{S}^_{\wp}$ . Proof. If $f \in \mathcal{S}^*[A, B]$ , then $zf'(z)/f(z) \in \mathcal{P}[A, B]$ . Therefore, by Lemma 2.4, we have <span id="page-7-0"></span> $$\left| \frac{zf'(z)}{f(z)} - a \right| \le \frac{(A-B)}{1-B^2},$$ (2.4) where $a:=(1-AB)/(1-B^2)$ . Suppose that the conditions in (i) hold. Now multiplying by (B+1) on both sides of the inequality $(B-1) \leq e(1-A)$ and then dividing by $1-B^2$ , gives $(A-B)/(1-B^2) \leq a-(1-1/e)$ . Similarly, the inequality $2(e-1)(1-B^2) < 2e(1-AB) \leq (e^2+2e-1)(1-B^2)$ is equivalent to $1-1/e < (1-AB)/(1-B^2) \leq 1+(e-e^{-1})/2$ . Therefore from (2.4), we see that $zf'(z)/f(z) \in \{w \in \mathbb{C} : |w-a| < r_a\}$ , where $r_a = a-(1-1/e)$ and $1-1/e < a \leq 1+(e-e^{-1})/2$ . Hence, $f \in \mathscr{S}_{\wp}$ by Lemma 2.2. Similarly, we can show that $f \in \mathscr{S}_{\wp}$ , if the conditions in (ii) hold.

Theorem 3.1. Let $f \in \mathscr{S}_{\wp}^*$ . Then (i) $f \in \mathcal{S}^*(\alpha)$ in $|z| < r_{\alpha}$ , $\alpha \in (\alpha_0, 1)$ , where $\alpha_0 = 1 + \frac{\sqrt{5} - 3}{2}e^{\frac{\sqrt{5} - 3}{2}}$ and $r_{\alpha} \in (0, 1)$ is the smallest root of the equation $$1 - re^{-r} - \alpha = 0.$$ (ii) $f \in \mathcal{M}(\beta)$ in $|z| < r_{\beta}$ , where $$r_{\beta} = \begin{cases} r_0(\beta) & for \ 1 < \beta < 1 + e \\ 1 & for \ \beta \ge 1 + e \end{cases}$$ and $r_0(\beta) \in (0,1)$ is the smallest root of $1 + re^r = \beta$ . (iii) $f \in \mathcal{SS}^*(\gamma)$ in $|z| < r_{\gamma}, \gamma \in (0,1]$ , where $$r_{\gamma} = \min\{1, r_0(\gamma)\}\$$ and $r_0(\gamma) \in (0,1)$ is the smallest root of the following equation: <span id="page-7-1"></span> $$\arcsin\left(\frac{1}{r}\ln\left(\frac{r}{\sin(\gamma\pi/2)}\right)\right) + \sqrt{r^2 + \ln^2\left(\frac{r}{\sin(\gamma\pi/2)}\right)} = \frac{\gamma\pi}{2}.$$ (3.1) Proof. Since $zf'(z)/f(z) \prec \wp$ , it suffices to consider the cardioid domain $\wp(\mathbb{D})$ so that certain geometry can be performed. (i) Since $f \in \mathscr{S}_{\wp}^*$ , there exists a function $\omega(z) \in \Omega$ such that $$\frac{zf'(z)}{f(z)} = 1 + \omega(z)e^{\omega(z)}.$$ Since $|\omega(z)| \leq |z|$ , we can assume $\omega(z) = Re^{i\theta}$ , where $R \leq |z| = r$ and $-\pi \leq \theta \leq \pi$ . A calculation shows that $$|Re^{i\theta}e^{Re^{i\theta}}| = Re^{R\cos\theta} =: T(\theta).$$ Since $T(\theta) = T(-\theta)$ , it is sufficient to consider $\theta \in [0, \pi]$ . Further, $T'(\theta) \leq 0$ implies $$|\omega(z)e^{\omega(z)}| = T(\theta) \le T(0) \le Re^R \le re^r.$$ Therefore, we obtain Re $$\frac{zf'(z)}{f(z)} \ge 1 - |\omega(z)e^{\omega(z)}| \ge 1 - re^r \ge \alpha$$ , whenever $1 - re^r - \alpha \ge 0$ . Hence the result. (ii) Since $f \in \mathscr{S}_{\wp}^*$ . Therefore, using subordination principle and Lemma 2.1, we $$\operatorname{Re} \frac{zf'(z)}{f(z)} \le \operatorname{Re} \wp(\omega(z)) \le |\wp(\omega(z))| \le 1 + re^r \quad (|z| = r), \tag{3.2}$$ where $\omega \in \Omega$ . Thus $f \in \mathcal{M}(\beta)$ in |z| < r, whenever $1 + re^r < \beta$ . (iii) Let $f \in \mathscr{S}_{\wp}$ , then $f \in \mathcal{SS}^(\gamma)$ in |z| < r provided $$\left|\arg \frac{zf'(z)}{f(z)}\right| \le \left|\arg(\wp(z))\right| \le \gamma\pi/2 \quad (|z|=r).$$ Assuming $z = re^{i(\theta + \pi/2)}$ <span id="page-8-0"></span> $$\theta + r\cos\theta = \gamma\pi/2 \text{ and } re^{-r\sin\theta} = \sin(\gamma\pi/2),$$ (3.3) we have $$1 + ze^z = 1 + re^{-r\sin\theta} \left( -\sin(\theta + r\cos\theta) + i\cos(\theta + r\cos\theta) \right)$$ $$= 1 + \sin(\gamma\pi/2) \left( -\sin(\gamma\pi/2) + i\cos(\gamma\pi/2) \right)$$ $$= \cos^2(\gamma\pi/2) + i\sin(\gamma\pi/2)\cos(\gamma\pi/2),$$ which implies $|\arg(\wp(z))| \leq \gamma \pi/2$ . Now we obtain equation (3.1) by eliminating $\theta$ from the equations, given in (3.3), a geometrical observation, ensures existence of the unique root for the equations (3.1). Thus the result now follows by considering the smallest root of (3.1). The result is further sharp as we can find $z_0 = r_0 e^{i(\theta_0 + \pi/2)}$ , for any fixed $\gamma$ , at which, the function $f_1$ , given by (1.10), satisfies $$\left| \arg \frac{z_0 f_1'(z_0)}{f_1(z_0)} \right| = |\arg(1 + ze^z)|_{z=z_0}$$ $$= |\arg(\cos^2(\gamma \pi/2) + i\sin(\gamma \pi/2)\cos(\gamma \pi/2))|$$ $$= |\arctan(\tan(\gamma \pi/2))|$$ $$= \gamma \pi/2.$$ Hence the result.

Theorem 3.2. Let $f \in \mathscr{S}_{\wp}^*$ . Then $f \in \mathcal{C}(\alpha)$ in $|z| < r_{\alpha}$ , where $r_{\alpha} \in (0,1)$ is the smallest root of the equation <span id="page-8-1"></span> $$(1-r)(1-re^r)(1-re^r-\alpha) - re^r = 0. (3.4)$$ Proof. If $f \in \mathscr{S}_{\wp}^*$ , then there exists a function $\omega \in \Omega$ such that $$\frac{zf'(z)}{f(z)} = 1 + \omega(z)e^{\omega(z)}.$$ Now a computation yields <span id="page-9-0"></span> $$1 + \frac{zf''(z)}{f'(z)} = 1 + \omega(z)e^{\omega(z)} + \frac{z\omega'(z)e^{\omega(z)}(1+\omega(z))}{1+\omega(z)e^{\omega(z)}}.$$ (3.5) From (3.5), we obtain <span id="page-9-1"></span> $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge 1 + \operatorname{Re}(\omega(z)e^{\omega(z)}) - \frac{|z\omega'(z)||(\omega(z) + 1)e^{\omega(z)}|}{1 - |\omega(z)e^{\omega(z)}|}.$$ (3.6) Since $|\omega(z)| \leq |z|$ , we can assume that $\omega(z) = Re^{i\theta}$ , where $R \leq |z| = r$ and $-\pi \leq \theta \leq \pi$ . Now using triangle inequality together with the Schwarz-Pick inequality: $$\frac{|\omega'(z)|}{1 - |\omega(z)|^2} \le \frac{1}{1 - |z|^2},$$ we have $|z\omega'(z)e^{\omega(z)}(1+\omega(z))| \leq re^r/(1-r)$ . Also $|\omega(z)e^{\omega(z)}| \leq Re^R \leq re^r$ . Upon using these inequalities in (3.6), we get <span id="page-9-2"></span> $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) \ge 1 - re^r - \frac{re^r}{(1-r)(1-re^r)} \ge \alpha,$$ (3.7) and with the least root of (3.4), the above inequality (3.7) hold and hence the result. By taking $\alpha = 0$ in Theorem 3.2, we obtain the following result: Corollary 3.3. Let $f \in \mathscr{S}_{o}^{*}$ . Then $f \in \mathcal{C}$ whenever $|z| < r_0 \approx 0.256707$ . Remark 3.1. Let $\omega(z) = z = re^{i\theta}$ and $\alpha = 0$ in Theorem 3.2. Then for the function given by (1.10), we have $$\operatorname{Re}\left(1 + \frac{zf_1''(z)}{f_1'(z)}\right) = \operatorname{Re}\left(1 + ze^z + \frac{z(1+z)e^z}{1+ze^z}\right) =: F(r,\theta),$$ where $$F(r,\theta) = -\frac{1 + r\cos\theta + R\cos\theta_1 + rR\cos(\theta_1 - \theta)}{1 + 2R\cos\theta_1 + R^2} + 2 + r\cos\theta + R\cos\theta_1$$ and $$z = re^{i\theta}, R = re^{r\cos\theta}, \theta_1 = \theta + r\sin\theta.$$ Numerically, we note that for all $\theta \in [0, \pi]$ , $F(r, \theta) \ge 0$ whenever $r \le r_0 \approx 0.599547$ , but when $\theta$ approaches to $\pi$ , then $F(r, \theta) < 0$ for $r = r_0 + \epsilon$ , $\epsilon > 0$ . Thus we previse that the sharp radius of convexity for the class $\mathscr{S}_{\wp}^*$ is $r_0$ . For the next theorems 3.4-3.7, we need to recall some classes: Let $f \in \mathcal{A}_n$ , if we set p(z) = zf'(z)/f(z), then the class $\mathcal{P}_n[A, B]$ reduces to $\mathcal{S}_n^[A, B]$ , the class of Janowski starlike functions and $\mathcal{S}_n^(\alpha) := \mathcal{S}_n^*[1 - 2\alpha, -1]$ . Further, let $$\mathscr{S}_{\varrho,n}^ := \mathcal{A}_n \cap \mathscr{S}_{\varrho}^ \quad \text{and} \quad \mathscr{S}_n^(\alpha) := \mathcal{A}_n \cap \mathscr{S}^(\alpha).$$ Ali et al. [3] studied the classes, $\mathcal{F}_n := \{ f \in \mathcal{A}_n : f(z)/z \in \mathcal{P}_n \}, \ \mathcal{S}_n^*[A, B]$ and the subclass consisting of close-to-starlike functions of type $\alpha$ given by $$\mathcal{CS}_n(\alpha) := \left\{ f \in \mathcal{A}_n : \frac{f(z)}{g(z)} \in \mathcal{P}_n, \ g \in \mathcal{S}_n^*(\alpha) \right\}.$$ $\Box$ We find the $\mathscr{S}_{\wp,n}^*$ -radii for the classes defined above.

Theorem 3.4. The sharp $\mathscr{S}_{\omega,n}^*$ -radius of the class $\mathcal{F}_n$ is given by: $$R_{\mathscr{S}_{n,n}^*}(\mathcal{F}_n) = (\sqrt{1 + n^2 e^2} - ne)^{1/n}.$$ Proof. If $f \in \mathcal{F}_n$ , then the function $h(z) := f(z)/z \in \mathcal{P}_n$ and $$\frac{zf'(z)}{f(z)} - 1 = \frac{zh'(z)}{h(z)}.$$ Using Lemmas 2.2 and 2.3, we get $$\left| \frac{zf'(z)}{f(z)} - 1 \right| = \left| \frac{zh'(z)}{h(z)} \right| \le \frac{2nr^n}{1 - r^{2n}} \le \frac{1}{e}.$$ Upon simplifying the last inequality, we get $r^{2n} + 2ner^n - 1 \leq 0$ . Thus, the $\mathscr{S}_{\wp,n}$ -radius of $\mathcal{F}_n$ is the least positive root of $r^{2n} + 2ner^n - 1 = 0$ in (0,1). Since for the function $f_0(z) = z(1+z^n)/(1-z^n)$ , $\operatorname{Re}(f_0(z)/z) > 0$ in $\mathbb{D}$ . We have $f_0 \in \mathcal{F}_n$ and $zf_0'(z)/f_0(z) = 1 + 2nz^n/(1-z^{2n})$ . Moreover, the result is sharp as we have at $z = R_{\mathscr{S}_{\wp,n}}(\mathcal{F}_n)$ : $$\frac{zf_0'(z)}{f_0(z)} - 1 = \frac{2nz^n}{1 - z^{2n}} = \frac{1}{e}.$$ This completes the proof. Let $\mathcal{F} := \mathcal{F}_1$ , which is $\mathcal{F} := \{ f \in \mathcal{A} : f(z)/z \in \mathcal{P} \}$ . MacGregor [22] showed that $r_0 = \sqrt{2} - 1$ is the radius of univalence and starlikeness for the class $\mathcal{F}$ , we here below provide the $\mathscr{S}_{\wp}^*$ -radius for the same: Corollary 3.5. The $\mathscr{S}_{\wp}^*$ -radius of the class $\mathcal{F}$ is given by $$R_{\mathcal{S}_{o}^{*}}(\mathcal{F}) = \sqrt{1 + e^{2}} - e \approx 0.178105.$$

Theorem 3.6. The sharp $\mathscr{S}_{\wp,n}^*$ -radius of the class $\mathcal{CS}_n(\alpha)$ is given by $$R_{\mathscr{S}_{\wp,n}^*}(\mathcal{CS}_n(\alpha)) = \left(\frac{1/e}{\sqrt{(1+n-\alpha)^2 - (1/e)(2(1-\alpha)-1/e)} + 1 + n - \alpha}\right)^{1/n}.$$ Proof. Let $f \in \mathcal{CS}_n(\alpha)$ and $g \in \mathcal{S}_n^*(\alpha)$ . Then, we have $h(z) := f(z)/g(z) \in \mathcal{P}_n$ , which implies: $$\frac{zf'(z)}{f(z)} = \frac{zg'(z)}{g(z)} + \frac{zh'(z)}{h(z)}.$$ Using Lemma 2.3 with $\alpha = 0$ and Lemma 2.4, we have <span id="page-10-1"></span> $$\left| \frac{zf'(z)}{f(z)} - a \right| \le \frac{2(1+n-\alpha)r^n}{1-r^{2n}},\tag{3.8}$$ where $a := (1 + (1 - 2\alpha)r^{2n})/(1 - r^{2n}) \ge 1$ . Note that $a \le 1 + (e - e^{-1})/2$ if and only if $r^{2n} \le (e^2 - 1)/(e^2 - 1 + 4e(1 - \alpha))$ . Let $r \le R_{\mathscr{S}_{e,n}^*}(\mathcal{CS}_n(\alpha))$ . Then $$r^{2n} \le \left(\frac{1}{e(2-\alpha) + \sqrt{e^2(2-\alpha)^2 - e(2-2\alpha - \frac{1}{e})}}\right)^2$$ $$\le \frac{1}{2e^2\alpha^2 - (8e^2 - 2e)\alpha + (8e^2 - 2e + 1)}.$$ Further, the expression on the right of the above inequality is less than or equal to $\frac{e^2-1}{(e^2-1)+4e(1-\alpha)}$ , provided $$T(\alpha) := 2e^{2}(e^{2} - 1)\alpha^{2} - (8e^{4} - 2e^{3} - 8e^{2} - 2e)\alpha + (8e^{4} - 2e^{3} - 8e^{2} - 2e) \ge 0.$$ Since $T'(\alpha) < 0$ and $\min_{0 < \alpha < 1} T(\alpha) = \lim_{\alpha \to 1} T(\alpha) = 2e^2(e^2 - 1) > 0$ . Therefore, $a \le 1 + (e - e^{-1})/2$ . Using Lemma 2.2, it follows that the disk, given by (3.8) is contained in the cardioid $\wp(\mathbb{D})$ , if $$\frac{1 - 2(1 + n - \alpha)r^n + (1 - 2\alpha)r^{2n}}{1 - r^{2n}} \ge 1 - \frac{1}{e},$$ which is equivalent to $(2-2\alpha-1/e)r^{2n}-2(1+n-\alpha)r^n+1/e\geq 0$ , and holds when $r\leq R_{\mathscr{S}_{\varrho,n}^*}(\mathcal{CS}_n(\alpha))$ . For sharpness, we consider the following functions $$f_0(z) := \frac{z(1+z^n)}{(1-z^n)^{(n+2-2\alpha)/n}}$$ and $g_0(z) := \frac{z}{(1-z^n)^{2(1-\alpha)/n}}$ , (3.9) such that $f_0(z)/g_0(z) = (1+z^n)/(1-z^n)$ and $zg_0'(z)/g_0(z) = (1+(1-2\alpha)z^n)/(1-z^n)$ . Moreover, $\text{Re}(f_0(z)/g_0(z)) > 0$ and $\text{Re}(zg_0'(z)/g_0(z)) > \alpha$ in $\mathbb{D}$ . Hence $f_0 \in \mathcal{CS}_n(\alpha)$ and $$\frac{zf_0'(z)}{f_0(z)} = \frac{1 + 2(1 + n - \alpha)z^n + (1 - 2\alpha)z^{2n}}{1 - z^{2n}}.$$ For $z = R_1 e^{i\pi/n}$ , we have $zf'_0(z)/f_0(z) = 1 - 1/e$ .

Theorem 3.7. The $\mathscr{S}_{\wp,n}$ -radius of the class $\mathscr{S}_n^[A,B]$ is given by (i) $$R_{\mathscr{S}_{\wp,n}}(\mathcal{S}_n^[A,B]) = \min \left\{ 1, \left( \frac{1/e}{A - (1-1/e)B} \right)^{\frac{1}{n}} \right\}, \text{ when } 0 \le B < A \le 1.$$ (ii) $$R_{\mathscr{S}_{\wp,n}}(\mathcal{S}_n^[A,B]) = \begin{cases} R_1, & \text{if } R_1 \le r_1 \\ R_2, & \text{if } R_1 > r_1 \end{cases} \text{ when } -1 \le B < 0 \le A \le 1.$$ $$R_1 := R_{\mathscr{S}_{\wp,n}}(\mathcal{S}_n^[A,B])$$ as defined in part (i), $$R_2 = \min\{1, (e/(A - (e+1)B))^{1/n}\}\$$ and $$r_1 = \left(\frac{(e^2 - 1)/2e}{((e^2 + 2e - 1)/2e)B^2 - AB}\right)^{1/2n}.$$ In particular, for the class $S$ , we have $R_{\mathscr{S}_{\wp}}(S^*) = 1/(2e-1)$ . Proof. Let $f \in \mathcal{S}_n^*[A, B]$ . Using Lemma 2.4, we have <span id="page-11-1"></span> $$\left| \frac{zf'(z)}{f(z)} - \frac{1 - ABr^{2n}}{1 - B^2r^{2n}} \right| \le \frac{(A - B)r^n}{1 - B^2r^{2n}}.$$ (3.10) (i) If $0 \le B < A \le 1$ , then $$a := \frac{1 - ABr^{2n}}{1 - B^2r^{2n}} \le 1.$$ Further, by Lemma 2.2 and equation (3.10), we see that $f \in \mathscr{S}_{\wp,n}^*$ if $$\frac{ABr^{2n} + (A-B)r^n - 1}{1 - R^2r^{2n}} \le \frac{1}{e} - 1,$$ which upon simplification, yields $$r \le \left(\frac{1/e}{A - (1 - 1/e)B}\right)^{1/n}.$$ The result is sharp due to the function $f_0(z)$ , given by <span id="page-12-0"></span> $$f_0(z) = \begin{cases} z(1 + Bz^n)^{\frac{A-B}{nB}}; & B \neq 0, \\ z \exp\left(\frac{Az^n}{n}\right); & B = 0. \end{cases}$$ (3.11) (ii) If $-1 \le B < 0 < A \le 1$ , then $$a := \frac{1 - ABr^{2n}}{1 - B^2r^{2n}} \ge 1.$$ Let us first assume that $R_1 \leq r_1$ . Note that $r \leq r_1$ if and only if $a \leq 1 + (e - e^{-1})/2$ . In particular, for $0 \leq r \leq R_1$ , we have $a \leq 1 + (e - e^{-1})/2$ . Further, from Lemma 2.2, we have $f \in \mathscr{S}_{\omega,n}^*$ in $|z| \leq r$ , if $$\frac{(A-B)r^n}{1-B^2r^{2n}} \le (a-1) + \frac{1}{e},$$ which holds whenever $r \leq R_1$ . Let us now assume that $R_1 > r_1$ . Thus $r \geq r_1$ if and only if $a \geq 1 + (e - e^{-1})/2$ . In particular, for $r \geq R_1$ , we have $a \geq 1 + (e - e^{-1})/2$ . Further, from Lemma 2.2, we have $f \in \mathscr{S}_{\wp,n}^*$ in $|z| \leq r$ whenever $$\frac{(A-B)r^n}{1-B^2r^{2n}} \le e - (a-1),$$ which holds when $r \leq R_2$ . Hence, the result follows with sharpness due to $f_0(z)$ given in (3.11).

Theorem 3.8. The $\mathscr{S}_{\wp,n}$ -radius of the class $\mathcal{M}_n^(\beta)$ , $(\beta > 1)$ is given by $$R_{\mathscr{S}_{n,n}}(\mathcal{M}_n^(\beta)) = (2e(\beta-1)+1)^{-1/n}.$$ Proof. If $f \in \mathcal{M}_n(\beta)$ , then $zf'(z)/f(z) \prec (1 + (2\beta - 1)z)/(1 + z)$ . Now using Lemma 2.4, we have $$\left| \frac{zf'(z)}{f(z)} - \frac{1 + (1 - 2\beta)r^{2n}}{1 - r^{2n}} \right| \le \frac{(\beta - 1)2r^n}{1 - r^{2n}}.$$ Note that for $\beta > 1$ , we have $(1 + (1 - 2\beta)r^{2n})/(1 - r^{2n}) < 1$ . Therefore, by Lemma 2.2, we get $f \in \mathscr{S}_{\alpha,n}^*$ in |z| < r, provided $$\frac{(\beta-1)2r^n}{1-r^{2n}} - \frac{1+(1-2\beta)r^{2n}}{1-r^{2n}} \le \frac{1}{e} - 1,$$ which holds whenever $r \leq R_{\mathscr{L}_n}(\mathcal{M}_n^(\beta))$ . The result is sharp due to $$f_0(z) := \frac{z}{(1-z^n)^{2(1-\beta)/n}},$$ as we see that $zf_0'(z)/f_0(z) = (1 + (1 - 2\beta)z^n)/(1 - z^n) = 1 - 1/e$ when $z = R_{\mathscr{S}_{0,n}}(\mathcal{M}_n^(\beta))$ . In the following theorem, we attempt to find the sharp $\mathscr{S}_{\wp}$ -radii of the class $\mathcal{S}^(\psi)$ , for different choices of $\psi$ such as $1 + \sin(z)$ , $\sqrt{2} - c\sqrt{(1-z)/(1+2cz)}$ , $1 + 4z/3 + 2z^2/3$ , $z + \sqrt{1+z^2}$ , $e^z$ and $\sqrt{1+z}$ , where $c := \sqrt{2} - 1$ . Authors in [8, 23, 24, 27, 33, 35] introduced and studied these subclasses of starlike functions which we denote by $\mathcal{S}_s$ , $\mathcal{S}_{RL}$ , $\mathcal{S}_e$ , $\mathcal{S}_C$ and $\mathcal{S}_L^*$ , respectively.

Theorem 3.9. The sharp $\mathscr{S}_{\wp}$ -radii of $\mathcal{S}_L$ , $\mathcal{S}_{RL}$ , $\mathcal{S}_e$ , $\mathcal{S}_C$ , $\mathcal{S}_s$ and $\Delta^*$ are: (i) $$R_{\mathscr{S}_{n}^{}}(\mathcal{S}_{L}^{}) = (2e-1)/e^{2} \approx 0.600423.$$ (ii) $$R_{\mathscr{S}_{\wp}}(\mathcal{S}_{RL}^) = \frac{1+2(\sqrt{2}-1)e}{e^2(\sqrt{2}-1)(\sqrt{2}-1+2(\sqrt{2}-1+e^{-1})^2)} \approx 0.648826.$$ (iii) $R_{\mathscr{S}_{\wp}}(\mathcal{S}_e^) = 1 - \ln(e-1) \approx 0.458675.$ (iii) $$R_{\mathscr{S}_{e}^{}}(\mathcal{S}_{e}^{}) = 1 - \ln(e - 1) \approx 0.458675.$$ (iv) $$R_{\mathscr{S}_{\wp}}(\mathcal{S}_C^) = 1 - \sqrt{1 - 3/2e} \approx 0.330536.$$ (v) $R_{\mathscr{S}_{\wp}}(\mathcal{S}_s^) = \arcsin(1/e) \approx 0.376727.$ $$(v) \ R_{\mathscr{S}_{s}^{}}(\mathcal{S}_{s}^{}) = \arcsin(1/e) \approx 0.376727$$ (vi) $$R_{\mathscr{S}_{\alpha}}(\Delta^) = (2e-1)/(2e(e-1)) \approx 0.474928.$$

Theorem 3.10. The sharp $\mathscr{S}_{\wp,n}^*$ -radii of functions in the classes $\mathcal{F}_1(\alpha)$ , $\mathcal{F}_2$ and $\mathcal{F}_3$ , respectively, are: (i) $$R_{\mathscr{S}_{n,n}^*}(\mathcal{F}_1(0)) = \left(\sqrt{4n^2e^2 + 1} - 2ne\right)^{1/n}$$ (ii) $$R_{\mathscr{S}_{\wp,n}^*}(\mathcal{F}_1(1/2)) = \left(2/(\sqrt{(3ne+2)^2 - 8ne} + 3ne)\right)^{1/n}$$ . (iii) $$R_{\mathscr{S}_{\wp,n}^*}(\mathcal{F}_2) = \left(2/(\sqrt{(3ne+2)^2 - 8ne} + 3ne)\right)^{1/n}$$ . $$(iv) R_{\mathscr{S}_{\wp,n}^*}(\mathcal{F}_3) = \left(\frac{\sqrt{(n+1)^2 + 4(n-1+1/e)/e} - (1+n)}}{2(n-1+1/e)}\right)^{1/n}.$$ Proof. Let us consider the functions $p, h : \mathbb{D} \to \mathbb{C}$ , defined by p(z) = g(z)/z and h(z) = f(z)/g(z). We write $p_0(z) = g_0(z)/z$ and $h_0(z) = f_0(z)/g_0(z)$ . (i) If $f \in \mathcal{F}_1(0)$ , then $p, h \in \mathcal{P}_n$ such that f(z) = zp(z)h(z). Thus it follows from Lemma 2.3 that $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{4nr^n}{1 - r^{2n}} \le \frac{1}{e},$$ provided $r \leq (\sqrt{4n^2e^2+1}-2ne)^{1/n} =: R_{\mathscr{S}_{\wp,n}}(\mathcal{F}_1(0))$ . Thus $f \in \mathscr{S}_{\wp,n}$ whenever $r \leq R_{\mathscr{S}_{\wp,n}^*}(\mathcal{F}_1(0))$ . Now for the functions $$f_0(z) = z \left(\frac{1+z^n}{1-z^n}\right)^2$$ and $g_0(z) = z \left(\frac{1+z^n}{1-z^n}\right)$ , we have, $\operatorname{Re} h_0(z) > 0$ and $\operatorname{Re} p_0(z) > 0$ . Hence $f_0 \in \mathcal{F}_1(0)$ . For $z = R_{\mathscr{S}_{0,n}^*}(\mathcal{F}_1(0))e^{i\pi/n}$ , we see that $$\frac{zf_0'(z)}{f_0(z)} = 1 + \frac{4nz^n}{1 - z^{2n}} = 1 - \frac{1}{e}.$$ Thus the result is sharp. (ii) Let $f \in \mathcal{F}_1(1/2)$ . Then $h \in \mathcal{P}_n$ and $p \in \mathcal{P}_n(1/2)$ . Since f(z) = zp(z)h(z), it follows from Lemma 2.3 that $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{2nr^n}{1 - r^{2n}} + \frac{nr^n}{1 - r^n} = \frac{3nr^n + nr^{2n}}{1 - r^{2n}} \le \frac{1}{e},$$ provided $$r \le \left(\frac{\sqrt{9n^2e^2 + 4(ne+1)} - 3ne}{2(ne+1)}\right)^{1/n} =: R_{\mathscr{S}_{\wp,n}^*}(\mathcal{F}_1(1/2)).$$ Thus $f \in \mathscr{S}_{\wp,n}$ whenever $r \leq R_{\mathscr{S}_{\wp,n}}(\mathcal{F}_1(1/2))$ . For the functions $$f_0(z) = \frac{z(1+z^n)}{(1-z^n)^2}$$ and $g_0(z) = \frac{z}{1-z^n}$ , we have, $\operatorname{Re} h_0(z) > 0$ and $\operatorname{Re} p_0(z) > 1/2$ . Hence $f \in \mathcal{F}_1(1/2)$ . The result is sharp, since for $z = R_{\mathscr{S}_{\mathfrak{o},n}}^*(\mathcal{F}_1(1/2))$ , we have $$\frac{zf_0'(z)}{f_0(z)} - 1 = \frac{3nz^n + nz^{2n}}{1 - z^{2n}} = \frac{1}{e}.$$ (iii) Let $f \in \mathcal{F}_2$ . Then $p \in \mathcal{P}_n$ . Since |h(z) - 1| < 1 if and only if Re(1/h(z)) > 1/2. Therefore, $1/h \in \mathcal{P}_n(1/2)$ . Since f(z)/h(z) = zp(z), using Lemma 2.3, we have $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{3nr^n + nr^{2n}}{1 - r^{2n}} \le \frac{1}{e},$$ provided $r^n \leq 2/(\sqrt{(3ne+2)^2-8ne}+3ne)$ . For the sharpness, consider $$f_0(z) := \frac{z(1+z^n)^2}{1-z^n}$$ and $g_0(z) := \frac{z(1+z^n)}{1-z^n}$ . Since $$|h_0(z) - 1| = |z^n| < 1 \text{ and } \operatorname{Re} p_0(z) = \operatorname{Re} \frac{1 + z^n}{1 - z^n} > 0.$$ Therefore, $f_0 \in \mathcal{F}_2$ and for $z = R_{\mathscr{S}_{\wp,n}^*}(\mathcal{F}_2)e^{i\pi/n}$ , we have $$\left| \frac{zf_0'(z)}{f_0(z)} - 1 \right| = \left| \frac{3nz^n - nz^{2n}}{1 - z^{2n}} \right| = \frac{1}{e}.$$ (iv) Let $f \in \mathcal{F}_3$ . Then $1/h(z) = g(z)/f(z) \in \mathcal{P}_n(1/2)$ and $$\frac{zf'(z)}{f(z)} = \frac{zg'(z)}{g(z)} - \frac{zh'(z)}{h(z)}. (3.12)$$ Using a result due to Marx-Strohhäcker that every convex function is starlike of order 1/2, it follows from Lemma 2.4 that <span id="page-16-0"></span> $$\left| \frac{zg'(z)}{g(z)} - \frac{1}{1 - r^{2n}} \right| \le \frac{r^n}{1 - r^{2n}}.$$ (3.13) Now using Lemma 2.3 and equation (3.13), we have $$\left| \frac{zf'(z)}{f(z)} - \frac{1}{1 - r^{2n}} \right| \le \frac{r^n}{1 - r^{2n}} + \frac{nr^n}{1 - r^n} = \frac{(n+1)r^n + nr^{2n}}{1 - r^{2n}}.$$ Thus using Lemma 2.2, we have $f \in \mathscr{S}_{\varrho,n}^*$ , provided $$\frac{(n+1)r^n + nr^{2n}}{1 - r^{2n}} \le \left(\frac{1}{1 - r^{2n}} - 1\right) + \frac{1}{e},$$ which implies $r \leq R_{\mathscr{S}_{n,n}^*}(\mathcal{F}_3)$ . Now consider the functions $$f_0(z) = \frac{z(1+z^n)}{(1-z^n)^{1/n}}$$ and $g_0(z) = \frac{z}{(1-z^n)^{1/n}}$ . Since $g_0 \in \mathcal{C}$ and $|h_0(z) - 1| = |z^n| < 1$ . Therefore, $f_0 \in \mathcal{F}_3$ and for $z = R_{\mathcal{S}_{\wp,n}^*}(\mathcal{F}_3)e^{i\pi/n}$ , we have $zf_0'(z)/f_0(z) = 1 - 1/e$ , which confirms the sharpness of the result.

Lemma 4.1. [21] Let $p \in \mathcal{P}$ be of the form (1.2). Then for a complex number $\tau$ , we have $$|p_2 - \tau p_1^2| \le 2 \max(1, |2\tau - 1|).$$

Lemma 4.2. [29] Let $p \in \mathcal{P}$ be of the form (1.2). Then for $n, m \in \mathbb{N}$ , $$|p_{n+m} - \gamma p_n p_m| \le \begin{cases} 2, & 0 \le \gamma \le 1; \\ 2|2\gamma - 1|, & elsewhere. \end{cases}$$ Here below, we partially disclose the lemma given in [4], which is required in sequel.

Lemma 4.3. [4] If $\omega \in \Omega$ be of the form (1.3), then $$|c_3 + \mu c_1 c_2 + \nu c_1^3| \le \Psi(\mu, \nu),$$ where $$\Psi(\mu,\nu) = \frac{2}{3}(|\mu|+1)\left(\frac{|\mu|+1}{3(1+\nu+|\mu|)}\right)^{\frac{1}{2}} \quad for \quad (\mu,\nu) \in D_8 \cup D_9$$ and $$D_8 := \left\{ (\mu, \nu) : \frac{1}{2} \le |\mu| \le 2, -\frac{2}{3} (|\mu| + 1) \le \nu \le \frac{4}{27} (|\mu| + 1)^3 - (|\mu| + 1) \right\}$$ $$D_9 := \left\{ (\mu, \nu) : |\mu| \ge 2, -\frac{2}{3} (|\mu| + 1) \le \nu \le \frac{2|\mu| (|\mu| + 1)}{\mu^2 + 2|\mu| + 4} \right\}.$$ The following lemma, carries the expression for $p_2$ and $p_3$ in terms of $p_1$ , derived in [19, 20] and $p_4$ in terms of $p_1$ obtained in [16].

Lemma 4.4. Let $p \in \mathcal{P}$ . Then for some complex numbers $\zeta$ , $\eta$ and $\xi$ with $|\zeta| \leq 1$ , $|\eta| \leq 1$ and $|\xi| \leq 1$ , we have $$2p_{2} = p_{1}^{2} + \zeta(4 - p_{1}^{2}),$$ $$4p_{3} = p_{1}^{3} + 2p_{1}\zeta(4 - p_{1}^{2}) - p_{1}\zeta^{2}(4 - p_{1}^{2}) + 2(4 - p_{1}^{2})(1 - |\zeta|^{2})\eta$$ and $$8p_{4} = p_{1}^{4} + (4 - p_{1}^{2})\zeta(p_{1}^{2}(\zeta^{2} - 3\zeta + 3) + 4\zeta)$$ $$-4(4 - p_{1}^{2})(1 - |\zeta|^{2})(p_{1}(\zeta - 1)\eta + \bar{\zeta}\eta^{2} - (1 - |\eta|^{2})\xi).$$ We now define the function fⁿ such that fn(0) = f 0 n (0) − 1 = 0 and $$\frac{zf_n'(z)}{f_n(z)} = \wp(z^n) \qquad (n = 1, 2, 3, \dots),$$ which acts as an extremal function for many subsequent results and we have <span id="page-18-0"></span> $$f_n(z) = z \exp((e^{z^n} - 1)/n)$$ (4.1) Theorem 4.1. Let f(z) = z + P^∞ ^k=2 bkz ^k ∈ S ^∗ ^℘ and α = (1 + e) 2 , then $$\sum_{k=2}^{\infty} (k^2 - \alpha)|b_k|^2 \le \alpha - 1.$$ (4.2) Proof. If f ∈ S ^∗ ℘ , then zf⁰ (z)/f(z) = ℘(ω(z)), where ω ∈ Ω. Now for 0 ≤ r < 1, we have $$\begin{split} 2\pi \sum_{k=1}^{\infty} k^2 |b_k|^2 r^{2k} &= \int_0^{2\pi} |re^{i\theta} f'(re^{i\theta})|^2 d\theta \\ &= \int_0^{2\pi} |f(re^{i\theta}) + f(re^{i\theta}) \omega(re^{i\theta}) e^{\omega(re^{i\theta})}|^2 d\theta \\ &\leq \int_0^{2\pi} \left( |f(re^{i\theta})| + |f(re^{i\theta}) \omega(re^{i\theta}) e^{\omega(re^{i\theta})}| \right)^2 d\theta \\ &= \int_0^{2\pi} |f(re^{i\theta})|^2 d\theta + \int_0^{2\pi} |f(re^{i\theta}) \omega(re^{i\theta}) e^{\omega(re^{i\theta})}|^2 d\theta \\ &+ 2 \int_0^{2\pi} |f(re^{i\theta})|^2 |\omega(re^{i\theta}) e^{\omega(re^{i\theta})}| d\theta \\ &\leq \int_0^{2\pi} |f(re^{i\theta})|^2 d\theta + \int_0^{2\pi} |f(re^{i\theta}) e^{\omega(re^{i\theta})}|^2 d\theta \\ &+ 2 \int_0^{2\pi} |f(re^{i\theta})|^2 |e^{\omega(e^{i\theta})}| d\theta \\ &\leq \int_0^{2\pi} |f(re^{i\theta})|^2 d\theta + e^{2\tau} \int_0^{2\pi} |f(re^{i\theta})|^2 d\theta \\ &\leq \int_0^{2\pi} |f(re^{i\theta})|^2 d\theta + e^{2\tau} \int_0^{2\pi} |f(re^{i\theta})|^2 d\theta \\ &\leq 2\pi (1 + e^{\tau})^2 \sum_{k=1}^{\infty} |b_k|^2 r^{2k}, \end{split}$$ which finally yields $$\sum_{k=1}^{\infty} (k^2 - (1 + e^r)^2) |b_k|^2 r^{2k} \le 0, \quad 0 \le r < 1.$$ Letting r → 1 ⁻, we get the desired result. Corollary 4.2. Let f(z) = z + P^∞ ^k=4 bkz ^k ∈ S ^∗ ^℘ and α = (1 + e) 2 , then ^|bk| ≤ ^r α − 1 k ² − α , for all k ≥ 4. Example 4.3. (i) z/(1 − Az) ² ∈ S ^∗ ℘ if and only if |A| ≤ 1/(2e − 1). (ii) $f(z) = z + b_k z^k \in \mathscr{S}^_{\wp}$ if and only if $|b_k| \le 1/(e(k-1)+1)$ , where $k \in \mathbb{N} - \{1\}$ . (iii) $f(z) = z \exp(Az) \in \mathscr{S}^_{\wp}$ if and only if $|A| \le 1/e$ . Proof. (i) If A=1, then $z/(1-z)^2 \notin \mathscr{S}_{\wp}$ , since $|f(z)| \leq e^{e-1}$ for $f \in \mathscr{S}_{\wp}$ . Now let $K(z) = z/(1 - Az)^2$ . If $A \neq 1$ , then the disk <span id="page-19-0"></span> $$\left| w - \frac{1 + |A|^2}{1 - |A|^2} \right| < \frac{2|A|}{1 - |A|^2},\tag{4.3}$$ is the image of $\mathbb{D}$ under the bilinear transformation w = zK'(z)/K(z) = (1 +Az)/(1-Az) with diameter's end points $x_L:=(1-|A|)/(1+|A|)$ and $x_R:=(1+|A|)/(1+|A|)$ |A|)/(1-|A|). Now for the disk (4.3) to be inside the cardioid $\wp(\mathbb{D})$ , it is necessary that $x_L \ge 1 - 1/e$ which gives $|A| \le 1/(2e - 1)$ . Conversely, let $|A| \le 1/(2e - 1)$ . Then we have $$a := \frac{1+|A|^2}{1-|A|^2} \le \frac{2e-e^{-1}+2}{2(e-1)}$$ and $r := \frac{2|A|}{1-|A|^2} \le \frac{2e-1}{2e(e-1)}$ . Since $r_a > r$ , thus Lemma 2.2 ensures that disk $\{w : |w - a| < r\} \subset \wp(\mathbb{D})$ . Hence, $K \in \mathscr{S}_{\wp}^*$ . (ii) Since $zf'(z)/f(z) = (1 + kb_k z^{k-1})/(1 + b_k z^{k-1})$ maps $\mathbb{D}$ onto the disk $\{w \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C} : x \in \mathbb{C}$ |w-a| < r, where $$a := \frac{1 - k|b_k|^2}{1 - |b_k|^2}$$ and $r := \frac{(k - 1)|b_k|}{1 - |b_k|^2}$ . Further $f(z) = z + b_k z^k \in \mathcal{S}^*$ if and only if $|b_k| \le 1/k$ , which ensures $(1 - k|b_k|^2)/(1 - k|b_k|^2)$ $|b_k|^2 \le 1$ . Therefore, in view of Lemma 2.2, $\{w \in \mathbb{C} : |w-a| < r\} \subset \wp(\mathbb{D})$ if and only if $$\frac{(k-1)|b_k|}{1-|b_k|^2} \le \frac{1-k|b_k|^2}{1-|b_k|^2} - 1 + \frac{1}{e},$$ which is equivalent to $(ke-e+1)|b_k|^2+(ke-e)|b_k|-1\leq 0$ . Hence, $|b_k|\leq$ 1/(e(k-1)+1). (iii) Since zf'(z)/f(z) = 1 + Az maps $\mathbb{D}$ onto the disk $\{w \in \mathbb{C} : |w-1| < |A|\}$ . Therefore, in view of Lemma 2.2, the inequality $$|w - 1| < |A| \le 1/e$$ , yields the necessary and sufficient condition |A| < 1/e for $1 + Az < \wp(z)$ . Remark 4.1. Note that when $k = \sqrt{2} + 1$ , we have $q_0(z) = 1 + (z/k)((k+z)/(k-z)) \prec$ $\wp(z)$ . Therefore, the class of starlike functions $\mathcal{S}^(q_0)$ introduced in [15] is contained in $\mathscr{S}_{\wp}$ . Further the sharp $\mathcal{S}^(\psi)$ -radius for the class $\mathcal{S}$ is also given by the relation <span id="page-19-1"></span> $$R_{\mathcal{S}^(\psi)}(\mathcal{S}^) = \max|A|, \tag{4.4}$$ where A is defined in such a way that $z/(1-Az)^2 \in \mathcal{S}^(\psi)$ . Thus if $z/(1-Az)^2 \in$ $\mathcal{S}^(q_0) \subset \mathscr{S}_{\wp}^*$ , then by Example 4.3, we see that $|A| \leq 1/(2e-1)$ and therefore, in view of (4.4), we now state a result ([15], theorem 2.3, pg 203) in its correct form using the result ([15], theorem 3.2, pg 206): $$z/(1-Az)^2 \in \mathcal{S}^*(q_0)$$ if and only if $|A| \leq \frac{3-2\sqrt{2}}{2\sqrt{2}-1} < \frac{1}{2e-1}$ . The authors proved that $|A| \leq 1/3$ .

Theorem 4.4. Let $f(z) = z + \sum_{k=2}^{\infty} b_k z^k \in \mathscr{S}_{\wp}^*$ , then $|b_2| < 1$ , $|b_3| < 1$ , $|b_4| < 5/6$ and $|b_5| < 5/8$ . (4.5) The bounds are sharp. Proof. Let $p(z) \in \mathcal{P}$ . Since there exists one-one correspondence between the classes $\Omega$ and $\mathcal{P}$ via the following functions: <span id="page-20-4"></span> $$\omega(z) = \frac{p(z) - 1}{p(z) + 1} \quad \text{and} \quad p(z) = \frac{1 + \omega(z)}{1 - \omega(z)}.$$ Therefore, for $f \in \mathscr{S}_{\wp}^*$ , we have <span id="page-20-1"></span> $$\frac{zf'(z)}{f(z)} = \wp(\omega(z)) = \wp\left(\frac{p(z) - 1}{p(z) + 1}\right),$$ where <span id="page-20-0"></span> $$\wp\left(\frac{p(z)-1}{p(z)+1}\right) = 1 + \frac{p_1}{2}z + \frac{p_2}{2}z^2 + \left(-\frac{p_1^3}{16} + \frac{p_3}{2}\right)z^3 + \left(\frac{p_1^4}{24} - \frac{3p_1^2p_2}{16} + \frac{p_4}{2}\right)z^4 + \cdots$$ (4.6) and $$\frac{zf'(z)}{f(z)} = 1 + b_2 z + (2b_3 - b_2^2)z^2 + (3b_4 - 3b_2 b_3 + b_2^3)z^3 + (-b_2^4 + 2b_3(2b_2^2 - b_3) - 4b_2 b_4 + 4b_5)z^4 + \cdots$$ (4.7) On comparing the coefficients of $z^k$ (k = 1, 2, 3, 4) in (4.6) and (4.7), we get <span id="page-20-3"></span> $$b_2 = \frac{p_1}{2}, \ b_3 = \frac{1}{4} \left( p_2 + \frac{p_1^2}{2} \right), \ b_4 = \frac{1}{6} \left( p_3 + \frac{3}{4} p_1 p_2 \right)$$ and $$b_5 = \frac{1}{8} \left( \frac{1}{48} p_1^4 + \frac{1}{4} p_2^2 + \frac{2}{3} p_1 p_3 - \frac{1}{8} p_1^2 p_2 + p_4 \right). \tag{4.8}$$ Now using the fact $|p_k| \le 2$ , Lemma 4.1 with $\tau = -1/2$ and Lemma 4.2 with $\gamma = -3/4$ , we obtain $|b_2| \le 1$ , $|b_3| \le 1$ and $|b_4| \le 5/6$ respectively. For $b_5$ , using proper rearrangement of terms and then applying triangle inequality, we see that $$|b_{5}| = \frac{1}{8} \left| \frac{1}{48} p_{1}^{4} + \frac{1}{4} p_{2}^{2} + \frac{2}{3} p_{1} p_{3} - \frac{1}{8} p_{1}^{2} p_{2} + p_{4} \right|$$ $$= \frac{1}{8} \left| \frac{1}{48} p_{1}^{4} + (p_{4} + \frac{2}{3} p_{1} p_{3}) + \frac{1}{4} p_{2} (p_{2} - \frac{1}{2} p_{1}^{2}) \right|$$ $$\leq \frac{1}{8} \left( \frac{1}{48} |p_{1}|^{4} + |p_{4} + \frac{2}{3} p_{1} p_{3}| + \frac{1}{4} |p_{2}| |p_{2} - \frac{1}{2} p_{1}^{2}| \right)$$ $$\leq \frac{1}{8} \left( \frac{1}{48} |p_{1}|^{4} - \frac{1}{4} |p_{1}|^{2} + \frac{4}{3} |p_{1}| + 3 \right)$$ $$=: G(p_{1}).$$ Now to maximize the above expression, without loss of generality, we write $$G(p) = \frac{1}{48}p^4 - \frac{1}{4}p^2 + \frac{4}{3}p + 3 \quad (p \in [0, 2]),$$ then $G'(p) \ge 0$ . Thus $G(p) \le 5$ , which implies that $|b_5| \le 5/8$ . The bounds for $b_k$ (k=1,2,3,4) are sharp with the extremal function $f_1$ defined in (4.1). Now in view of Theorem 4.4, we conjecture the following: Conjecture. Let $f(z) \in \mathscr{S}_{\wp}^*$ . Then the following sharp estimates hold: $$|b_k| \le \frac{B_{k-1}}{(k-1)!}$$ for all $k \ge 1$ , where $B_k$ are Bell numbers satisfying the recurrence relation defined in (1.11) and the extremal function $f_1$ is given by (4.1). Remark 4.2. The logarithmic coefficients $d_k$ for $f \in \mathcal{S}$ are defined by the following series expansion: <span id="page-21-0"></span> $$\log \frac{f(z)}{z} = 2\sum_{k=1}^{\infty} d_k z^k, \quad z \in \mathbb{D}.$$ (4.9) Recently, Cho [1] obtained the sharp logarithmic coefficient bounds for the class $\mathcal{S}^*(\psi)$ given by (1.4). Consequently, we have the following sharp result: Let $f \in \mathscr{S}_{\wp}^*$ . Then the logarithmic coefficients of f given by (4.9) satisfies $$|d_k| \le 1/2.$$ Remark 4.3. Now if $f(z) \in \mathscr{S}_{\wp}^*$ , then from (4.8), using triangle inequality together with Lemma 4.1, we obtain the following estimates for the Fekete-Szegö functional: <span id="page-21-1"></span> $$|b_3 - \mu b_2^2| = \frac{1}{4} \left| p_2 - \left(\mu - \frac{1}{2}\right) p_1^2 \right| \le \frac{1}{2} \max(1, 2|\mu - 1|).$$ (4.10) Equality cases holds for the functions $f_1(z) = z \exp(e^z - 1)$ , when $\mu \in [1/2, 3/2]$ and $f_2(z) = z \exp((e^{z^2} - 1)/2)$ , when $\mu \le 1/2$ or $\mu \ge 3/2$ given by (4.1). In particular for $\mu = 1$ , we have $|H_2(1)| = |b_3 - b_2^2| \le 1/2$ . Now the Covering Theorem stated in Theorem 1.1 ensures that for every f in $\mathscr{S}_{\wp}$ , $f(\mathbb{D})$ contains a disk of radius $e^{1/e-1}$ centered at the origin. Hence, every function $f \in \mathscr{S}_{\wp}$ has an inverse $f^{-1}$ which given by $$f^{-1}(w) = w + \sum_{k=2}^{\infty} A_k w^k = w - b_2 w^2 + (2b_2^2 - b_3)w^3 - (5b_2^3 - 5b_2b_3 + b_4)w^4 + \cdots,$$ then we have $f^{-1}(f(z)) = z$ and $f(f^{-1}(w)) = w$ for $|w| < r_0(f)$ and $r_0 > e^{1/e-1}$ . Thus using Theorem 4.4 and equation 4.10, we easily obtain $$|A_2| \le 1 \quad \text{and} \quad |A_3| \le 1.$$ The bounds are sharp with extremal function $f_1^{-1}$ , where $f_1$ is defined in (4.1).

Theorem 4.5. Let $f \in \mathscr{S}_{\wp}^*$ . Then for $f^{-1}(\omega) = \omega + \sum_{k=2}^{\infty} A_k \omega^k$ , we have $$|A_4| \le \frac{5}{6}$$ and $|A_3 - \mu A_2^2| \le \begin{cases} 3 - \mu, & \mu \le 5/2; \\ 1/2, & 5/2 \le \mu \le 7/2; \\ \mu - 3, & \mu \ge 7/2. \end{cases}$ The bounds are sharp. Proof. Consider the inverse function $f^{-1}(\omega) = \omega + \sum_{k=2}^{\infty} A_k \omega^k$ , where we have $A_4 =$ $-5b_2^3 + 5b_2b_3 - b_4$ , which can be now rewritten in terms of Carathéodory coefficients using (4.8) as $$A_4 = -\frac{1}{6} \left( p_3 - 3p_1 p_2 + \frac{15}{8} p_1^3 \right).$$ Now using Lemma 4.1 with $\tau = 5/8$ and $|p_k| \le 2$ , $$|b_4| = \frac{1}{6} \left| p_3 - 3p_1 \left( p_2 - \frac{5}{8} p_1^2 \right) \right| \le \frac{1}{6} (|p_3| + 3|p_1||p_2 - \frac{5}{8} p_1^2|) \le \frac{5}{6}.$$ The bound is sharp with extremal function $f_1^{-1}$ , where $f_1$ is defined in (4.1). Now for the Fekete-Szegö type inequality for the inverse function $f^{-1}$ , we have $$|A_3 - \mu A_2^2| = |b_3 - tb_2^2|, \quad t = \mu - 2.$$ Thus using (4.10), the desired sharp result follows.

Theorem 4.6. Let $f(z) = z + \sum_{k=2}^{\infty} b_k z^k \in \mathscr{S}_{\wp}^*$ , then $$|b_2b_3 - b_4| \le \frac{2}{3}\sqrt{\frac{2}{5}}.$$ The bound is sharp. Proof. Let $f \in \mathscr{S}_{\wp}^*$ . Then <span id="page-22-0"></span> $$\frac{zf'(z)}{f(z)} = \wp(\omega(z)),\tag{4.11}$$ where $\omega \in \Omega$ . Then proceeding as in Theorem 4.4, from (4.11), we have $$b_2 = c_1, \quad b_2 = \frac{1}{2}(c_2 + 2c_1^2) \quad \text{and} \quad b_4 = \frac{1}{6}(2c_3 + 7c_1c_2 + 5c_1^3).$$ (4.12) Therefore, with $\mu = 2, \nu = -1/2$ and $\psi(\mu, \nu) = |c_3 + \mu c_1 c_2 + \nu c_1^3|$ , we have $$|b_2b_3 - b_4| = \frac{1}{3}|c_3 + 2c_1c_2 - c_1^3/2| = \frac{1}{3}\psi(\mu, \nu).$$ Now using Lemma 4.3, we obtain $$|b_2b_3 - b_4| \le \frac{2}{3}\sqrt{\frac{2}{5}}.$$ The bound is sharp as there is an extremal function $$f(z) = z \exp \int_0^z \frac{\wp(\omega(t)) - 1}{t} dt,$$ where $$w(z) = z(\sqrt{2/5} - z)/(1 - \sqrt{2/5}z)$$ . We now enlist below in the remark, certain special cases of earlier known results pertaining to our class $\mathscr{S}^*_{\wp}$ : <span id="page-22-2"></span>Remark 4.4. We obtain the following result by using a result ([2], theorem 2.2, pg 230): Let $f(z) = z + \sum_{k=2}^{\infty} b_k z^k \in \mathscr{S}_{\wp}^*$ , then $$|H_2(2)| = |b_2b_4 - b_3^2| \le 1/4,$$ where equality is attained for the function $f_2$ given by (4.1). <span id="page-22-3"></span>Remark 4.5. Now using Theorems 4.4, 4.6 and Remark 4.4 together with the estimate given in (4.10) and triangle inequality, we obtain the following result: Let the function $f(z) = z + \sum_{k=2}^{\infty} b_k z^k \in \mathscr{S}_{\wp}^*$ , then $$|H_3(1)| \le 0.913864 \cdots$$ Remark 4.6. Until now, the bound on third Hankel determinant is obtained using triangle inequality approach, but note that using the method applied in theorem 4.8, we can substantially improve the known bounds for many subclasses of starlike functions such as $\mathcal{S}_s$ , $\mathcal{S}_C$ and $\mathcal{S}_e^*$ . We know that $|H_3(1)| \leq 1$ [38] for $\mathcal{S}$ , the class of starlike functions. Since, $\mathscr{S}^_{\wp} \subset \mathcal{S}$ , it seems reasonable that the bound on $|H_3(1)|$ for $\mathscr{S}^_{\wp}$ can be further improved. A function f in $\mathcal{A}$ is called n-fold symmetric if $f(e^{2\pi i/n}z) = e^{2\pi i/n}f(z)$ holds for all $z \in \mathbb{D}$ , where n is a natural number. We denote the set of n-fold symmetric functions by $\mathcal{A}^{(n)}$ . Let $f \in \mathcal{A}^{(n)}$ , then f has power series expansion $f(z) = z + b_{n+1}z^{n+1} + b_{2n+2}z^{2n+2} + \cdots$ . Therefore, for $f \in \mathcal{A}^{(3)}$ and $f \in \mathcal{A}^{(2)}$ , respectively, we have <span id="page-23-0"></span> $$H_3(1) = -b_4^2$$ and $H_3(1) = b_3(b_5 - b_3^2)$ . (4.13) Thus we can now find estimates on the third Hankel determinant $|H_3(1)|$ in the classes $\mathscr{S}_{\wp}^{(2)}$ and $\mathscr{S}_{\wp}^{(3)}$ .

Theorem 4.7. Let $f \in \mathscr{S}_{\alpha}^*$ . Then - (i) $\hat{f} \in \mathscr{S}_{\wp}^{*(3)}$ implies that $|H_3(1)| \leq 1/9$ . - (ii) $\hat{f} \in \mathscr{S}_{\wp}^{*(2)}$ implies that $|H_3(1)| \leq 1/16$ . The result is sharp. Proof. (i) Since $f(z) = z + b_2 z^2 + \cdots \in \mathscr{S}_{\wp}$ if and only if $\hat{f}(z) = (f(z^3))^{1/3} = z + \beta_4 z^4 + \cdots \in \mathscr{S}_{\wp}^{(3)}$ . We have $\beta_4 = b_2/3$ . Hence for $\hat{f} \in \mathscr{S}_{\wp}^{*(3)}$ , from (4.5) and (4.13), we obtain $$|H_3(1)| = |\beta_4|^2 = \frac{1}{9}|b_2|^2 \le \frac{1}{9}.$$ The above estimate is sharp for $\hat{f}_1$ , where $f_1$ is given by (4.1). (ii) Since $f(z) = z + b_2 z^2 + \cdots \in \mathscr{S}_{\wp}$ if and only if $\hat{f}(z) = (f(z^2))^{1/2} = z + \alpha_3 z^3 + \alpha_5 z^5 + \cdots \in \mathscr{S}_{\wp}^{(2)}$ . Upon comparing the coefficients in the following: $$z^{2} + b_{2}z^{4} + b_{3}z^{6} + \dots = (z + \alpha_{3}z^{3} + \alpha_{5}z^{5} + \dots)^{2},$$ we obtain <span id="page-23-1"></span> $$\alpha_3 = \frac{1}{2}b_2$$ and $\alpha_5 = \frac{1}{2}b_3 - \frac{1}{8}b_2^2$ . (4.14) If $\hat{f} \in \mathscr{S}_{\wp}^{*(2)}$ , then from (4.13), we have $$H_3(1) = \alpha_3(\alpha_5 - \alpha_3^2).$$ Now using (4.8), (4.14) and Lemma 4.4, we obtain $$|H_3(1)| = \frac{1}{4} \left| b_2 \left( b_3 - \frac{3}{4} b_2^2 \right) \right| = \frac{1}{64} |p_1| |(p_1^2 - p_1) + \xi(4 - p_1^2)|,$$ where $|\xi| \leq 1$ . Since $H_3(1) = \alpha_3(\alpha_5 - {\alpha_3}^2)$ is rotationally invariant, so we may assume $p_1 := p \in [0,2]$ . Thus using triangle inequality, we easily get $|H_3(1)| \leq (3p^3 - 4p^2 + 4p)/256 =: g(p)$ . Since g'(p) > 0 for all $p \in [0,2]$ . Therefore, $\max_{0 \leq p \leq 2} g(p) = g(2)$ . Hence $$|H_3(1)| \le \frac{1}{16}.$$ The above estimate is sharp for $\hat{f}_1$ , where $f_1$ is given by (4.1). In the following result, the bound obtained in the Remark 4.5 is improved.

Theorem 4.8. Let $f \in \mathscr{S}_{\omega}^*$ . Then $|H_3(1)| \leq 0.150627$ . Proof. From (1.7) and (4.8), we have $$H_3(1) = \frac{1}{9216} (-21p_1^6 + 60p_1^4p_2 + 96p_1^3p_3 + 192p_1p_2p_3 - 144p_1^2p_2^2 - 144p_1^2p_4 - 72p_2^3 - 256p_3^2 + 288p_2p_4)$$ and using Lemma 4.4 and writing $p_1$ as p and $t = 4 - p_1^2$ , we have <span id="page-24-1"></span> $$H_3(1) = \frac{1}{9216} \left( \Upsilon_1(p,\zeta) + \Upsilon_2(p,\zeta)\eta + \Upsilon_3(p,\zeta)\eta^2 + \Upsilon_4(p,\zeta,\eta)\xi \right), \tag{4.15}$$ where $\zeta, \eta, \xi \in \overline{\mathbb{D}}$ and $$\Upsilon_1(p,\zeta) = -4p^6 + t(t(-25p^2\zeta^2 + 19p^2\zeta^3 + 2p^2\zeta^4 + 36\zeta^3) + 5p^4\zeta - 16p^4\zeta^2 - 24p^2\zeta^3),$$ $$\Upsilon_2(p,\zeta) = t(1 - |\zeta|^2)(t(64p\zeta^2 - 80p\zeta) + 32p^3),$$ $$\Upsilon_3(p,\zeta) = -t^2(1 - |\zeta|^2)(64 + 8|\zeta|^2)$$ $$\Upsilon_4(p,\zeta,\eta) = 72t^2(1 - |\zeta|^2)^2\zeta.$$ Let $x = |\zeta| \in [0, 1]$ and $y = |\eta| \in [0, 1]$ . Now using $|\xi| \le 1$ and triangle inequality, from (4.15) we obtain $$|H_3(1)| \le \frac{1}{9216} \left( f_1(p, x) + f_2(p, x)y + f_3(p, x)y^2 + f_4(p, x) \right)$$ $$=: \frac{F(p, x, y)}{9216},$$ (4.16) where $$f_1(p,x) = 4p^6 + t(t(25p^2x^2 + 19p^2x^3 + 2p^2x^4 + 36x^3) + 5p^4x + 16p^4x^2 + 24p^2x^3),$$ $$f_2(p,x) = t(1-x^2)(t(80px + 64px^2) + 32p^3),$$ $$f_3(p,x) = t^2(1-x^2)(64+8x^2)$$ and $$f_4(p,x) = 72t^2x(1-x^2)^2.$$ Since $f_2(p, x)$ and $f_3(p, x)$ are non-negative functions over $[0, 2] \times [0, 1]$ . Therefore, from (4.16) together with $y = |\eta| \in [0, 1]$ , we obtain <span id="page-24-2"></span> $$F(p, x, y) \le F(p, x, 1).$$ Thus, $F(p, x, 1) = f_1(p, x) + f_2(p, x) + f_3(p, x) + f_4(p, x) =: G(p, x)$ . Now we shall maximize G(p, x) over $[0, 2] \times [0, 1]$ . For this we consider the following possible cases: (i) when x = 0, we have $$G(p,0) = 1024 - 512p^2 + 128p^3 + 64p^4 - 32p^5 + 4p^6 =: g_1(p).$$ Since $g'_1(p) < 0$ on [0, 2]. Therefore, $g_1(p)$ is an decreasing function over [0, 2]. Thus, the function $g_1(p)$ attains its maximum value at p = 0 which is equal to 1024. (ii) when x = 1, we have $$G(p, 1) = 576 + 544p^2 - 272p^4 + 29p^6 =: g_2(p).$$ Since $g_2'(p) = 0$ has a critical point at $p_0 = 2\sqrt{(68 - 7\sqrt{34})/87} \approx 1.11795$ . Therefore, it is easy to see that $g_2(p)$ is an increasing function for $p \leq p_0$ and decreasing for $p_0 \leq p$ . Thus, the function $g_2(p)$ attains its maximum at $p := p_0$ , which is approximately equal to 887.674. (iii) when p = 0, we have $$G(0,x) = 1024 - 896x^2 + 576x^3 - 128x^4 =: g_3(x).$$ Since $g_3'(x) < 0$ on [0, 1]. Therefore, the function $g_3(x)$ attains its maximum at x = 0, which is equal to 1024 and for the case, when p = 2, we easily obtain $G(p, x) \le 256$ . (iv) when $(p, x) \in (0, 2) \times (0, 1)$ , a numerical computation shows that there exists a unique real solution for the system of equations $$\partial G(p,x)/\partial x = 0$$ and $\partial G(p,x)/\partial p = 0$ inside the rectangular region: $[0,2] \times [0,1]$ , at $(p,x) \approx (0.531621, 0.482768)$ . Consequently, we obtain $G(p,x) \leq 1388.18$ . Hence, from the above cases we conclude that $$F(p, x, y) \le 1388.18$$ on $[0, 2] \times [0, 1] \times [0, 1]$ , which implies that $$H_3(1) \le \frac{1}{9216} F(p, x, y) \le 0.150627.$$ Hence the result. Conjecture. If $f \in \mathscr{S}_{\wp}^*$ , then the sharp bound for the third Hankel determinant is given by $$|H_3(1)| \le \frac{1}{9} \approx 0.1111 \cdots,$$ with the extremal function $f(z) = z \exp\left(\frac{1}{3}(e^{z^3} - 1)\right) = z + \frac{1}{3}z^4 + \frac{2}{9}z^7 + \cdots$