Radius of Starlikeness for Classes of Analytic Functions

Theorem 2.1. For the class $K_1$ , the following results hold: - (1) The $S^*(\alpha)$ -radius is the smallest positive real root of the equation $r^4(1+\alpha)-4r^3-2r^2-4r^3-6r^2$ $4r + (1 - \alpha) = 0, \quad 0 \le \alpha < 1.$ - (2) The $S_L$ -radius is $R_{S_L} = (\sqrt{5} 2)/(\sqrt{2} + 1) \approx 0.0977826$ . - (2) The $S_L$ -radius is $RS_L^* = (\sqrt{3} 2)/(\sqrt{2} + 1) \approx 0.0911820$ . (3) The $S_P$ -radius is the smallest positive real root of the equation $3r^4 8r^3 4r^2 8r + 1 = 0$ i.e. $R_{S_P} \approx 0.116675$ . - (4) The $\mathcal{S}_e$ -radius is the smallest positive real root of the equation $(2r^2+4r+4r^3-1-r^4)e=$ $r^4 - 1$ i.e. $R_{\mathcal{S}_a} \approx 0.144684$ . - (5) The $S_c$ -radius is the smallest positive real root of the equation $4r^4-12r^3-6r^2-12r+2=0$ i.e. $R_{S_a} \approx 0.15182$ . - (6) The $\mathcal{S}_{\mathbb{Q}}$ -radius is the smallest positive real root of the equation $4r^3 + 2r^2 + 4r + \sqrt{2}(1-r^4) = 0$ 2 i.e. $R_{S_{\sigma}} \approx 0.134993$ . - (7) The $S_{\sin}$ -radius is $R_{S_{\sin}} = (-2 + \sqrt{4 + \sin 1(2 + \sin 1)})/(2 + \sin 1) \approx 0.185835$ . (8) The $S_{RL}$ -radius is $R_{S_{RL}} \approx 0.0687813$ . (9) The $\mathcal{S}_R$ -radius is the smallest positive real root of the equation $4r^3 + 2r^2 + 4r - r^4 - 1 = 2(1 - \sqrt{2})(1 - r^4)$ i.e. $R_{\mathcal{S}_P} \approx 0.0419413$ . All the radii obtained are sharp. We would use the following lemmas in order to prove our results:

Lemma 2.2. ( [3, Lemma 2.2, p. 4]). For $0 < a < \sqrt{2}$ , let $r_a$ be given by $$r_a = \left\{ \begin{array}{ll} (\sqrt{1-a^2} - (1-a^2))^{1/2}, & 0 < a \le 2\sqrt{2}/3; \\ \sqrt{2} - a, & 2\sqrt{2}/3 \le a < \sqrt{2}. \end{array} \right.$$ Then $\{w : |w - a| < r_a\} \subseteq \{w : |w^2 - 1| < 1\}$

Lemma 2.3. ( [19, Lemma 1, p. 321]). For a > 1/2, let $r_a$ be given by $$r_a = \begin{cases} a - 1/2, & 1/2 < a \le 3/2; \\ \sqrt{2a - 2}, & a \ge 3/2. \end{cases}$$ Then $\{w : |w-a| < r_a\} \subseteq \{w : \text{Re } w > |w-1|\} = \Omega_{1/2}$ . Here, $\Omega_p$ is a parabolic region which is symmetric with respect to the real axis and vertex at (p,0).

Lemma 2.4. ( [11, Lemma 2.2, p. 368]). For $e^{-1} < a < e$ , let $r_a$ be given by $$r_a = \left\{ \begin{array}{ll} a - e^{-1}, & e^{-1} < a \le (e + e^{-1})/2; \\ e - a, & (e + e^{-1})/2 \le a < e. \end{array} \right.$$ Then $\{w : |w - a| < r_a\} \subseteq \{w : |\log w| < 1\} = \Omega_e$ , which is the image of the unit disk $\mathbb{D}$ under the exponential function.

Lemma 2.5. ( [23, Lemma 2.5, p. 926]). For 1/3 < a < 3, let $r_a$ be given by $$r_a = \left\{ \begin{array}{ll} (3a-1)/3, & 1/3 < a \leq 5/3; \\ 3-a, & 5/3 \leq a \leq 3. \end{array} \right.$$ Then $\{w: |w-a| < r_a\} \subseteq \Omega_c$ . Here $\Omega_c$ is the region bounded by the cardioid $\{x+iy: (9x^2+9y^2-18x+5)^2-16(9x^2+9y^2-6x+1)=0\}$ .

Lemma 2.6. ( [4, Lemma 3.3, p. 7]). For $1 - \sin 1 < a < 1 + \sin 1$ , let $r_a = \sin 1 - |a - 1|$ . Then $\{w : |w - a| < r_a\} \subseteq \Omega_{sin}$ . Here $\Omega_{sin}$ is the image of the unit disk $\mathbb{D}$ under the function $1 + \sin z$ .

Lemma 2.7. ( [5, Lemma 2.1, p. 3]). For $\sqrt{2} - 1 < a < \sqrt{2} + 1$ , let $r_a = 1 - |\sqrt{2} - a|$ . Then $\{w : |w - a| < r_a\} \subseteq \Omega_{\mathbb{C}} = \{w : |w^2 - 1| < 2|w|\}$ .

Lemma 2.8. ( [6, Lemma 2.2, p. 202]). For $2(\sqrt{2}-1) < a < 2$ , let $r_a$ be given by $$r_a = \begin{cases} a - 2(\sqrt{2} - 1), & 2(\sqrt{2} - 1) < a \le \sqrt{2}; \\ 2 - a, & \sqrt{2} \le a < 2. \end{cases}$$ Then $\{w : |w-a| < r_a\} \subseteq \Omega_R$ , where $\Omega_R$ is the image of the unit disk $\mathbb{D}$ under the function $1 + ((zk+z^2)/(k^2-kz)), k = \sqrt{2} + 1.$

Lemma 2.9. ( [12, Lemma 3.2, p. 10]). For $0 < a < \sqrt{2}$ , let $r_a$ be given by $$r_a = \begin{cases} a, & 0 < a \le \sqrt{2}/3; \\ \left( (1 - (\sqrt{2} - a)^2)^{1/2} - (1 - (\sqrt{2} - a)^2) \right)^{1/2}, & \sqrt{2}/3 \le a < \sqrt{2}. \end{cases}$$ Then $\{w : |w - a| < r_a\} \subseteq \{w : \text{Re } w > 0, |(w - \sqrt{2})^2 - 1| < 1\} = \Omega_{RL}$ .

Lemma 2.10. ( [18, Lemma 2, p. 240]). If $p(z) = 1 + b_n z^n + b_{n+1} z^{n+1} + \cdots$ is analytic and satisfies Re $p(z) > \alpha$ , $0 \le \alpha < 1$ , for |z| < 1, then $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2nz^n(1-\alpha)}{(1-|z|^n)(1+(1-2\alpha)|z|^n)}.$$ With all these tools, we are ready to give the proof of our first result. Proof of Theorem 2.1. Let $f \in \mathcal{K}_1$ and the function $g : \mathbb{D} \to \mathbb{C}$ be chosen such that <span id="page-3-0"></span>(2.2) $$\operatorname{Re} \frac{f(z)}{g(z)} > 0 \quad \text{and} \quad \operatorname{Re} \left(\frac{1 - z^2}{z} g(z)\right) > 0 \quad (z \in \mathbb{D}).$$ Let us define $p_1, p_2 : \mathbb{D} \to \mathbb{C}$ as (2.3) $$p_1(z) = \frac{1-z^2}{z}g(z)$$ and $p_2(z) = \frac{f(z)}{g(z)}$ Therefore, by equation (2.2), $p_1$ and $p_2$ are in $\mathcal{P}$ . Equation (2.3) yields <span id="page-3-3"></span><span id="page-3-2"></span><span id="page-3-1"></span> $$f(z) = \frac{z}{(1-z^2)} p_1(z) p_2(z).$$ Take logarithm at both sides and differentiate with respect to z would give (2.4) $$\frac{zf'(z)}{f(z)} = \frac{1+z^2}{1-z^2} + \frac{zp'_1(z)}{p_1(z)} + \frac{zp'_2(z)}{p_2(z)}.$$ It can be easily proved that the bilinear transform $w = (1 + z^2)/(1 - z^2)$ maps the disk $|z| \le r$ onto the disk (2.5) $$\left| \frac{1+z^2}{1-z^2} - \frac{1+r^4}{1-r^4} \right| \le \frac{2r^2}{1-r^4}.$$ Now, by Lemma 2.10, for $p \in \mathcal{P}(\alpha) := \{ p \in \mathcal{P} : \operatorname{Re} p(z) > \alpha, z \in \mathbb{D} \}$ , we have <span id="page-3-4"></span>(2.6) $$\left| \frac{zp'(z)}{p(z)} \right| \le \frac{2(1-\alpha)r}{(1-r)(1+(1-2\alpha)r)} (|z| \le r).$$ By using equations (2.4), (2.5) and (2.6), we can conclude that a function $f \in \mathcal{K}_1$ maps the disk $|z| \leq r$ onto the disk (2.7) $$\left| \frac{zf'(z)}{f(z)} - \frac{1+r^4}{1-r^4} \right| \le \frac{2r(2r^2+r+2)}{1-r^4}.$$ In order to solve radius problems for $f \in \mathcal{K}_1$ , we are interested in computing the value of r for which the disk in (2.7) is contained in the corresponding regions. The classes we are considering here are all subclasses of starlike functions and therefore, we first determine the radius of starlikeness for $f \in \mathcal{K}_1$ . From (2.7), we have <span id="page-3-5"></span>Re $$\frac{zf'(z)}{f(z)} \ge \frac{r^4 - 4r^3 - 2r^2 - 4r + 1}{1 - r^4} \ge 0.$$ Solving the above inequality for r, we get that the function $f \in \mathcal{K}_1$ is starlike in $|z| \leq 0.216845$ . Hence, all the radii that we are going to estimate here, will be less than 0.216845. For the function $f_1$ defined in (2.1), we have $$\frac{zf_1'(z)}{f_1(z)} = \frac{1 + 4iz + 2z^2 - 4iz^3 + z^4}{1 - z^4}$$ $$= \frac{1 + 4iz(1 - z^2) + 2z^2 + z^4}{1 - z^4}$$ At z := ri = (0.216845)i, we have $zf'_1(z)/f_1(z) \approx 0$ , thereby proving that the radius of starlikeness obtained for the class $\mathcal{K}_1$ is sharp. (1) In order to compute $R_{\mathcal{S}^*(\alpha)}$ , we estimate the value of $r \in (0,1)$ satisfying Re $$\frac{zf'(z)}{f(z)} \ge \frac{r^4 - 4r^3 - 2r^2 - 4r + 1}{1 - r^4} \ge \alpha$$ . Therefore, the number $r = R_{\mathcal{S}^*(\alpha)}$ , is the smallest positive real root of the equation $r^4(1+\alpha) - 4r^3 - 2r^2 - 4r + (1-\alpha) = 0$ in (0,1). For the function $f_1 \in \mathcal{K}_1$ given by (2.1), we have <span id="page-4-0"></span>(2.8) $$\frac{zf_1'(z)}{f_1(z)} = \frac{1 + 4iz + 2z^2 - 4iz^3 + z^4}{1 - z^4}$$ At $z := ri = \mathcal{R}_{\mathcal{S}^*(\alpha)}$ , (2.8) reduces to $$\frac{zf_1'(z)}{f_1(z)} = \frac{1 - 4r - 2r^2 - 4r^3 + r^4}{1 - r^4} = \alpha,$$ thereby proving that the radius is sharp. (2) We use lemma 2.2 to compute the lemniscate starlike radius for the function $f \in \mathcal{K}_1$ . Let $a = (1 + r^4)/(1 - r^4)$ . Then for $0 \le r < 1$ , we have $a \ge 1$ . So for $a < \sqrt{2}$ , we get $r < \sqrt[4]{(\sqrt{2} - 1)/(\sqrt{2} + 2)} \approx 0.59018$ . On the other hand, consider $$\frac{2r(2r^2+r+2)}{1-r^4} \le \sqrt{2} - a = \sqrt{2} - \frac{1+r^4}{1-r^4}.$$ From this, let $r^*$ be the smallest positive real roof of the equation $(1 + \sqrt{2})r^4 + 4r^3 + 2r^2 + 4r + (1 - \sqrt{2}) = 0$ . Then the radius of lemniscate starlikeness for $f \in \mathcal{K}_1$ is $$R_{\mathcal{S}_L} = \min\left\{ \left(\frac{\sqrt{2}-1}{\sqrt{2}+2}\right)^{1/4}, r^ \right\} = r^* = \frac{\sqrt{5}-2}{\sqrt{2}+1}.$$ The radius obtained is sharp. Consider the functions $f, g : \mathbb{D} \to \mathbb{C}$ defined by (2.9) $$f(z) = \frac{z(1-z)}{(1+z)^3} \quad \text{and} \quad g(z) = \frac{z}{(1+z)^2}.$$ Then clearly $f \in \mathcal{K}_1$ as <span id="page-4-1"></span>Re $$\frac{f(z)}{g(z)}$$ = Re $\frac{1-z^2}{z}g(z)$ = Re $\frac{1+z}{1-z} > 0$ . Now, for $z := -r^ = -R_{\mathcal{S}_L}$ , we have $(z^2 - 4z + 1)/(1 - z^2) = \sqrt{2}$ and thus $$\left| \left( \frac{zf'(z)}{f(z)} \right)^2 - 1 \right| = \left| \left( \frac{r^2 - 4r + 1}{1 - r^2} \right)^2 - 1 \right| = 1,$$ thereby proving that the radius obtained is sharp by the function f in (2.9). (3) We use Lemma 2.3 to compute the parabolic starlike radius for $f \in \mathcal{K}_1$ . Again, let $a = (1 + r^4)/(1 - r^4)$ , which is larger than or equal to 1 for $0 \le r < 1$ . Note that $$a = \frac{1+r^4}{1-r^4} = \frac{3}{2} \quad \Leftrightarrow \quad r = \left(\frac{1}{5}\right)^{1/4} \approx 0.66874.$$ Since the radius we are looking for would be less than 0.216845, we only consider the case $1/2 < a \le 3/2$ in Lemma 2.3. So when considering $$\frac{2r(2r^2+r+2)}{1-r^4} \le \frac{1+r^4}{1-r^4} - \frac{1}{2},$$ let $r^*$ be the smallest positive real root of the equation $3r^4 - 8r^3 - 4r^2 - 8r + 1 = 0$ . Then the radius of parabolic starlikeness for $f \in \mathcal{K}_1$ is $$R_{\mathcal{S}_P} = \min\left\{ \left(\frac{1}{5}\right)^{1/4}, r^ \right\} = r^ \approx 0.116675.$$ We see that the sharpness follows for the function $f_1 \in \mathcal{K}_1$ defined in (2.1). At z = ir, we have $$F(r) = \frac{zf_1'(z)}{f_1(z)}\bigg|_{z=ir} = \frac{1 - 4r - 2r^2 - 4r^3 + r^4}{1 - r^4}.$$ Then. $$|F(r) - 1| = \left| \frac{2r(r^3 - 2r^2 - r - 2)}{1 - r^4} \right|.$$ For $z := ir^* = iR_{\mathcal{S}_P}$ , we have Re $$\frac{zf_1'(z)}{f_1(z)} = \frac{1 + r^4 - 4r^3 - 2r^2 - 4r}{1 - r^4} (\approx 0.5)$$ = $\frac{2r(2 + r + 2r^2 - r^3)}{1 - r^4} = \left| \frac{zf_1'(z)}{f_1(z)} - 1 \right|$ . Thus the radius obtained is sharp for the function $f_1$ (4) By using Lemma 2.4 and the argument similar to the above, we get that the exponential starlike radius $R_{\mathcal{S}_{e}^{*}}$ for the class $\mathcal{K}_{1}$ is the smallest positive real root of the equation $(4r^{3} + 2r^{2} + 4r - 1 - r^{4})e = r^{4} - 1$ . The radius is sharp for the function $f_1$ defined in (2.1). For $z := ir = iR_{\mathcal{S}_e^*}$ , we have $$\left|\log \frac{zf_1'(z)}{f_1(z)}\right| = \left|\log \frac{1 + r^4 - 4r^3 - 2r^2 - 4r}{1 - r^4}\right| = 1.$$ (5) By using Lemma 2.5, and similar argument as before, the $S_c^*$ -radius for the class $K_1$ is the smallest positive real root of the equation $2r^4 - 6r^3 - 3r^2 - 6r + 1 = 0$ . The radius is sharp for the function $f_1$ defined in (2.1). Indeed, for the function $f_1$ defined in (2.1), we have at $z := ir = i\mathcal{R}_{\mathcal{S}_{}^{}}$ , $$\frac{zf_1'(z)}{f_1(z)} = \frac{1 + r^4 - 4r^3 - 2r^2 - 4r}{1 - r^4} = \frac{1}{3} = h_c(-1) \in \partial h_c(\mathbb{D}),$$ where $h_c(z) = 1 + (4/3)z + (2/3)z^2$ . This shows that the result is sharp. (6) To determine the $\mathcal{S}_{\mathbb{Q}}$ -radius, $R_{\mathcal{S}_{\mathbb{Q}}}$ , we will use Lemma 2.7. After some computations following the idea above, it can be shown that $R_{\mathcal{S}_{\mathbb{Q}}^*}$ is the smallest positive real root of the equation $4r^3 + 2r^2 + 4r = 2 - \sqrt{2}(1 - r^4)$ . The radius is sharp for the function $f_1$ defined in (2.1), since at $z := ir = iR_{\mathcal{S}_{\mathbb{Q}}^*}$ , we have $$\left| \left( \frac{zf_1'(z)}{f_1(z)} \right)^2 - 1 \right| = \left| \left( \frac{1 + r^4 - 4r^3 - 2r^2 - 4r}{1 - r^4} \right)^2 - 1 \right| (\approx 0.134993)$$ $$= 2 \left| \frac{1 + r^4 - 4r^3 - 2r^2 - 4r}{1 - r^4} \right| = 2 \left| \frac{zf_1'(z)}{f_1(z)} \right|.$$ (7) In order to find the $\mathcal{S}_{\sin}$ -radius for function $f \in \mathcal{K}_1$ , we make use of Lemma 2.6. Similarly as above, with $a = (1 + r^4)/(1 - r^4) > 1$ , it can be shown by arguing similarly as above that the $\mathcal{S}_{\sin}$ -radius is the smallest positive real root of the equation $(2 + \sin 1)r^4 + 4r^3 + 2r^2 + 4r - \sin 1 = 0$ . The radius is sharp for the function $f_1$ defined in (2.1). (8) In order to compute the $S_{RL}$ - radius for the class $K_1$ , we use Lemma 2.9. As $\sqrt{2}/3 \le a = (1+r^4)/(1-r^4) < \sqrt{2}$ , a computation using Lemma 2.9 shows that the $S_{RL}$ - radius is the smallest positive real root of the equation $$4r^{2}(2r^{2}+r+2)^{2}=(1-r^{4})\sqrt{\left(\sqrt{2}-1\right)+\left(\sqrt{2}-2\right)r^{4}}-2\left(\sqrt{2}-1+(\sqrt{2}-2)r^{4}\right).$$ The radius obtained is sharp for the function $f \in \mathcal{K}_1$ given by (2.9). At $z := -r = -R_{\mathcal{S}_{RL}^*}$ , we have $(z^2 - 4z + 1)/(1 - z^2) = \sqrt{2}$ and therefore, $$\left| \left( \frac{zf'(z)}{f(z)} - \sqrt{2} \right)^2 - 1 \right| = \left| \left( \frac{1 - 4z + z^2}{1 - z^2} - \sqrt{2} \right)^2 - 1 \right| = 1.$$ Hence the result. (9) Since $2(\sqrt{2}-1) < a = (1+r^4)/(1-r^4) \le \sqrt{2}$ , by using Lemma 2.8, it can be shown that the $S_R$ - radius is obtained by solving the equation $$(2\sqrt{2} - 1)r^4 - 4r^3 - 2r^2 - 4r + (3 - 2\sqrt{2}) = 0.$$ The radius is sharp for the function $f_1$ defined in (2.1). Indeed, for the function $f_1$ defined in (2.1), we have at $z := ir = iR_{\mathcal{S}_R^*}$ that $$\frac{zf_1'(z)}{f_1(z)} = \frac{1 + r^4 - 4r^3 - 2r^2 - 4r}{1 - r^4} = 2\sqrt{2} - 2 = h_R(-1) \in \partial h_R(\mathbb{D}).$$ Here, $$h_R = 1 + (zk + z^2)/(k^2 - kz)$$ , and $k = \sqrt{2} + 1$ . Our next result gives various radii of starlikeness for the $K_2$ , which consists of functions $f \in \mathcal{A}$ satisfying |(f(z)/g(z)) - 1| < 1 for some $g \in \mathcal{A}$ and $\text{Re}((1-z^2)g(z)/z) > 0$ . Consider the functions $f_2, g_2 : \mathbb{D} \to \mathbb{C}$ defined by (2.10) $$f_2(z) = \frac{z(1+iz)^2}{(1-z^2)(1-iz)} \quad \text{and} \quad g_2(z) = \frac{z(1+iz)}{(1-z^2)(1-iz)}.$$ Clearly, <span id="page-6-2"></span> $$\left| \frac{f_2(z)}{g_2(z)} - 1 \right| = |iz| = |z| < 1$$ and $\operatorname{Re} \frac{1 - z^2}{z} g_2(z) = \operatorname{Re} \frac{1 + iz}{1 - iz} > 0$ . Therefore, the function $f_2$ is in $\mathcal{K}_2$ and this shows $\mathcal{K}_2 \neq \phi$ . Note that this function $f_2$ would serve as an extremal function for several radii-problems that we study here.

Theorem 2.11. For $f \in \mathcal{K}_2$ , the following results hold: - (1) The sharp $S^*(\alpha)$ radius is the smallest positive real root of the equation $\alpha r^4 3r(r^2 + r + 1) + (1 \alpha) = 0$ , $0 \le \alpha < 1$ . - (2) The $S_L$ radius is $R_{S_L} = (\sqrt{2} 1)/(\sqrt{2} + 2) \approx 0.12132$ . - (3) The sharp $S_P$ radius is the smallest positive real root of the equation $6r^3 + 6r^2 + 6r 1 r^4 = 0$ i.e., $R_{S_P} \approx 0.1432698$ . - (4) The sharp $S_e$ radius is the smallest positive real root of the equation $(3r^3 + 3r^2 + 3r 1)e + 1 r^4 = 0$ i.e., $R_{S_e} \approx 0.174887$ . - (5) The sharp $S_c$ radius is the smallest positive real root of the equation $9r^3 + 9r^2 + 9r 2 r^4 = 0$ i.e., $R_{S_c} \approx 0.182815$ . - (6) The sharp $\mathcal{S}_{\mathbb{Q}}$ radius is the smallest positive real root of the equation $r^4(1-\sqrt{2})+3r^3+3r^2+3r=2-\sqrt{2}$ i.e., $R_{\mathcal{S}_{\mathcal{Q}}}\approx 0.164039$ . - (7) The sharp $S_{\sin}$ radius is $R_{S_{\sin}}^* = \sin 1/(3 + \sin 1) \approx 0.219049$ . - (8) The sharp $S_R$ radius is the smallest positive real root of the equation $2r^4 + 3r^3 + 3r^2 + 3r 3 + 2\sqrt{2}(1 r^4) = 0$ i.e., $R_{S_R} \approx 0.0541073$ . - <span id="page-6-0"></span>(9) The $\mathcal{S}_{RL}$ radius is $R_{\mathcal{S}_{RL}} \approx 0.0870259$ .

Theorem 2.12. For $f \in \mathcal{K}_3$ , the following results hold: - (1) The sharp $S^*(\alpha)$ radius is the smallest positive real root of the equation $(1 + \alpha)r^4 2r(r^2 + r + 1) + (1 \alpha) = 0$ , $0 \le \alpha < 1$ . - (2) The sharp $S_L^*$ radius is $R_{S_L} = (\sqrt{2} 1)/(\sqrt{2} + 1) \approx 0.171573$ . - (3) The sharp $S_P$ radius is the smallest positive real root of the equation $4r^3 + 4r^2 + 4r 1 3r^4 = 0$ i.e. $R_{S_P} \approx 0.2021347$ . - (4) The sharp $S_e$ radius is the smallest positive real root of the equation $(2r^3 + 2r^2 + 2r 1 r^4)e + 1 r^4 = 0$ i.e. $R_{S_e} \approx 0.244259$ . - (5) The sharp $S_c$ radius is the smallest positive real root of the equation $3r^3 + 3r^2 + 3r 1 2r^4 = 0$ i.e. $R_{S_c} \approx 0.254726$ . - (6) The sharp $\mathcal{S}_{\mathbb{Q}}$ radius is the smallest positive real root of the equation $2r^3 + 2r^2 + 2r \sqrt{2}r^4 = 2 \sqrt{2}$ i.e. $R_{\mathcal{S}_{\mathbb{Q}}} \approx 0.229877$ . - (7) The sharp $S_{\sin}$ radius is $R_{S_{\sin}} = \sin 1/(2 + \sin 1) \approx 0.296139$ . - (8) The sharp $S_R$ radius is the smallest positive real root of the equation $r^4 + 2r^3 + 2r^2 + 2r 3 + 2\sqrt{2}(1 r^4) = 0$ i.e. $R_{S_R} \approx 0.0790749$ . - (9) The $S_{RL}$ radius is $R_{S_{RL}} \approx 0.125145$ .