Starlike Functions Associated with a Non-Convex Domain

Theorem 1. A function $f \in \mathcal{S}_{nc}^*$ if and only if there exists an analytic function $q(z) \prec \varphi_{nc}(z) = (1+z)/\cos z$ , such that $$f(z) = z \exp \int_0^z \frac{q(t) - 1}{t} dt. \tag{4}$$

Theorem 2. Let $f \in \mathcal{S}_{nc}^*$ and $\tilde{f}$ be the extremal function given by (6). Then the following holds: 1. Growth theorem: For $|z_0| = r < 1$ , we have $$-\tilde{f}(-r) \le |f(z_0)| \le \tilde{f}(r).$$ Equality holds for some $z_0 \neq 0$ if and only if f is a rotation of $\tilde{f}$ . 2. Rotation theorem: For $|z_0| = r < 1$ , we have $$|\arg\{f(z_0)/z_0\}| \le \max_{|z|=r} \arg\{\tilde{f}(z)/z\}.$$ 3. Distortion theorem: For $|z_0| = r < 1$ , we have $$\tilde{f}'(-r) \le |f'(z_0)| \le \tilde{f}'(r).$$ Equality holds for some $z_0 \neq 0$ if and only if f is a rotation of $\tilde{f}$

Theorem 3. For |z| = r, min $\operatorname{Re}\{\varphi_{nc}(z)\} = \varphi(-r)$ and $\max \operatorname{Re}\{\varphi_{nc}(z)\} = \varphi(r)$ .

Theorem 4. (Function's Bounds) Let $\Phi_R$ and $\Phi_I$ denote the real and imaginary part of $\Phi_{nc}(z) = (1+z)/\cos z$ respectively. Then we have - 1. $0 \le \Phi_R \le (4\cos 1)/(1+\cos 2)$ . - 2. $-\gamma_0 \leq \Phi_I \leq \gamma_0$ , where $\gamma_0 \approx 1.6471$ . - 3. $|\arg \varphi_{nc}(z)| \leq \pi/2$ .  Figure 2: Bounds of $\varphi_{nc}$

Theorem 5. If $f \in \mathcal{S}_{nc}^*$ , then $$0 \le \det T_{2,1}(f) \le 1$$ and $\det T_{3,1}(f) \le 1$ . All the estimates are sharp.

Theorem 6. If $f \in \mathcal{S}_{nc}^*$ , then $\det T_{3,1}(f) \geq -1/15$ . The bound is sharp.

Lemma 7. [19] If $p(z) = 1 + \sum_{n=2}^{\infty} p_n z^n \in \mathcal{P}$ , then for a complex number $\nu$ $$|p_2 - \nu p_1^2| \le 2 \max(1, |2\nu - 1|).$$ This result is sharp for the function $p(z) = (1+z^2)/(1-z^2)$ and p(z) = (1+z)/(1-z).

Lemma 8. [27] [5] Let $p(z) = 1 + \sum_{n=2}^{\infty} p_n z^n \in \mathcal{P}$ , then for $n, m \in \mathbb{N}$ $$|p_{n+m} - \nu p_n p_m| \le \begin{cases} 2, & 0 \le \nu \le 1\\ 2|2\nu - 1|, & elsewhere. \end{cases}$$ The next lemma gives the representations of $p_2$ , $p_3$ and $p_4$ in terms of $p_1$ . The expressions for $p_2$ and $p_3$ were given in [14, 15] and for $p_4$ in [13].

Lemma 9. Let $p(z) = 1 + \sum_{n=2}^{\infty} p_n z^n \in \mathcal{P}$ , then $$2p_{2} = p_{1}^{2} + \gamma(4 - p_{1}^{2}),$$ $$4p_{3} = p_{1}^{3} + 2(4 - p_{1}^{2})p_{1}\gamma - (4 - p_{1}^{2})p_{1}\gamma^{2} + 2(4 - p_{1}^{2})(1 - |\gamma|^{2})\eta \quad and$$ $$8p_{4} = p_{1}^{4} + (4 - p_{1}^{2})\gamma(p_{1}^{2}(\gamma^{2} - 3\gamma + 3) + 4\gamma) - 4(4 - p_{1}^{2})(1 - |\gamma|^{2})(p_{1}(\gamma - 1)\eta + \bar{\gamma}\eta^{2} - (1 - |\gamma|^{2})\rho)$$ for some complex numbers $\gamma, \eta, \rho$ with $|\gamma| \leq 1$ , $|\eta| \leq 1$ and $|\rho| \leq 1$ .

Theorem 10. If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_{nc}^*$ , then $$\sum_{n=2}^{\infty} (n^2 k_1 - 4) |a_n|^2 \le (4 - k_1),$$ where $k_1 = \cos^2 1$ .

Corollary 10.1. Let $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}_{nc}^*$ and $k_1 = \cos^2 1$ , then $$|a_n| \le \sqrt{\frac{4 - k_1}{n^2 \cos^2 1 - 4}}, \quad n = 2, 3, 4, \cdots.$$ The following sharp bound of Fekete-Szegö functional for $f \in \mathcal{S}_{nc}^*$ can be deduced directly from [1].

Theorem 11. Let $f \in \mathcal{S}_{nc}^*$ , then $$|a_3 - \mu a_2^2| \le \begin{cases} -\mu + \frac{3}{4}, & \mu < \frac{1}{4} \\ \frac{1}{2}, & \frac{1}{4} \le \mu \le \frac{5}{4} \\ \mu - \frac{3}{4}, & \mu > \frac{5}{4}. \end{cases}$$ The bound is sharp.

Theorem 12. If the function $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ belongs to the class $\mathcal{S}_{nc}^*$ , then $$|a_2| \le 1$$ , $|a_3| \le \frac{3}{4}$ , $|a_4| \le \frac{7}{12}$ , $|a_5| \le \frac{1}{3}$ . The bounds of $|a_n|$ for n = 2, 3, 4 are sharp.

Theorem 13. If f ∈ S^∗ nc, then $$|H_2(2)(f)| \le \frac{1}{4}.$$ The result is sharp. Proof. For f ∈ S^∗ nc, from [\(8\)](#page-5-2), we can get $$a_2 a_4 - a_3^2 = \frac{1}{768} (p_1^4 - 8p_1^2 p_2 - 48p_2^2 + 64p_1 p_3).$$ Since the class P is invariant under rotation, we can take p¹ ∈ [0, 2]. Let p := p¹ and substituting the value of p² and p³ from Lemma [9,](#page-4-2) we have $$\begin{split} |a_2a_4-a_3^2| &= \frac{1}{768} |(p^4-4p^2(p^2+(4-p^2)x)-12(p^2+(4-p^2)x)^2\\ &+ 16p(p^3+2p(4-p^2)x-p(4-p^2)x^2+2(4-p^2)(1-|x|^2)z))|.\\ &= \frac{1}{768} |(p^4+12p^2(4-p^2)x+(28p^4-96p^2+192)x^2\\ &+ 32p(4-p^2)(1-|x|^2)z)|. \end{split}$$ Now, applying the triangle inequality and replacing |x| by ϱ, we obtain $$|a_2a_4 - a_3^2| \le \frac{1}{768}(p^4 + 12p^2(4 - p^2)\varrho + (28p^4 + 32p^3 - 96p^2 - 128p + 192)\varrho^2 + 32p(4 - p^2))$$ =: $G(p, \varrho)$ (say). On differentiation G(p, ϱ) with respect to ϱ, we obtain $$\frac{\partial G(p,\varrho)}{\partial \varrho} = \frac{1}{768} (12p^2(4-p^2) + 2(192 - 128p - 96p^2 + 32p^3 + 28p^4)\varrho).$$ Since ∂G(p, ϱ)/∂ϱ > 0, therefore G(p, ϱ) is an increasing function and attains its maximum value at ϱ = 1. Hence $$G(p,1) =: G(p) = \frac{1}{768}(192 - 48p^2 + 17p^4).$$ A simple calculation yields that the maximum value of G(p) occurs at p = 0. Thus, $$|H_2(2)(f)| = |a_2a_4 - a_3^2| \le \frac{192}{768} = \frac{1}{4}.$$ The bound is sharp for the function f3(z) given by [\(5\)](#page-2-2).

Theorem 14. If f ∈ S^∗ nc, then $$|H_3(1)(f)| \le \frac{1}{9}.$$ The bound is sharp. Proof. Let f(z) = z + P^∞ ⁿ=2 anz ⁿ ∈ S^∗ nc. Then from [\(2\)](#page-1-1), we have $$H_3(1)(f) = a_3(a_2a_4 - a_3^2) - a_4(a_4 - a_2a_3) + a_5(a_3 - a_2^2).$$ Using the values of a2, a3, a⁴ and a⁵ from [\(8\)](#page-5-2) together with Lemma [9](#page-4-2) and p¹ = p ∈ [0, 2], we obtain $$H_3(1)(f) = \frac{1}{36864} \bigg( \vartheta_1(p,\gamma) + \vartheta_2(p,\gamma)\eta + \vartheta_3(p,\gamma)\eta^2 + \psi(p,\gamma,\eta)\rho \bigg),$$ where $$\begin{split} \vartheta_1(p,\gamma) &= 5p^6 - 26\gamma(4-p^2)p^4 - 144\gamma^2(4-p^2)p^2 + 56\gamma^2(4-p^2)p^4 + 68\gamma^2(4-p^2)^2p^2 \\ &- 36\gamma^3(4-p^2)p^4 - 40\gamma^3(4-p^2)^2p^2 + 8\gamma^4(4-p^2)^2p^2, \\ \vartheta_2(p,\gamma) &= (4-p^2)(1-|\gamma|^2)(-40p^3 + 144p^3\gamma + 80p(4-p^2)\gamma - 32p(4-p^2)\gamma^2), \\ \vartheta_3(p,\gamma) &= (4-p^2)(1-|\gamma|^2)(-256(4-p^2) - 32(4-p^2)|\gamma|^2 + 144p^2\bar{\gamma}), \\ \psi(p,\gamma,\eta) &= (4-p^2)(-144p^2 + 288(4-p^2)\gamma)(1-|\gamma|^2)(1-|\eta|^2), \end{split}$$ |γ| ≤ 1, |η| ≤ 1 and |ρ| ≤ 1 and p = p¹ ∈ [0, 2] by the rotationally invariant property of the class P. Now, by taking |γ| := x, |η| := y and keeping in mind |ρ| ≤ 1, we get $$|H_3(1)(f)| \leq \frac{1}{36864} \bigg( |\vartheta_1(p,\gamma)| + |\vartheta_2(p,\gamma)| y + |\vartheta_3(p,\gamma)| y^2 + |\psi(p,\gamma,\eta)| \bigg) =: G(p,x,y),$$ where $$G(p, x, y) = \frac{1}{36864} \left( g_1(p, x) + g_2(p, x)y + g_3(p, x)y^2 + g_4(p, x) \right)$$ such that $$\begin{split} g_1(p,x) &= 5p^6 + 26p^4(4-p^2)x + 144p^2(4-p^2)x^2 + 56p^4(4-p^2)x^2 + 68p^2(4-p^2)^2x^2 \\ &\quad + 36p^4(4-p^2)x^3 + 40p^2(4-p^2)^2x^3 + 8p^2(4-p^2)^2x^4, \\ g_2(p,x) &= (4-p^2)(1-x^2)(40p^3 + 144p^3x + 80p(4-p^2)x + 32p(4-p^2)x^2), \\ g_3(p,x) &= (4-p^2)(1-x^2)(256(4-p^2) + 32(4-p^2)x^2 + 144p^2x), \\ g_4(p,x) &= (4-p^2)(1-y^2)(144p^2 + 288(4-p^2)x)(1-x^2). \end{split}$$ Thus, we obtain $$\begin{split} G(p,x,y) &= \frac{1}{36864} \bigg( 5p^6 + 26p^4(4-p^2)x + 144p^2(4-p^2)x^2 + 56p^4(4-p^2)x^2 \\ &\quad + 36p^4(4-p^2)x^3 + 68p^2(4-p^2)^2x^2 + 40p^2(4-p^2)^2x^3 + 8p^2(4-p^2)^2x^4 \\ &\quad + (4-p^2)(1-x^2)(40p^3 + 144p^3x + 80p(4-p^2)x + 32p(4-p^2)x^2)y \\ &\quad + (4-p^2)(1-x^2)(256(4-p^2) + 144p^2x + 32(4-p^2)x^2)y^2 \\ &\quad + (4-p^2)(144p^2 + 288(4-p^2)x)(1-x^2)(1-y^2) \bigg). \end{split}$$ Clearly, the maximum value of G(p, x, y) in the closed cuboid $\Gamma : [0, 2] \times [0, 1] \times [0, 1]$ is the required bound of $|H_3(1)(f)|$ . We establish this by finding the maximum value of G(p, x, y) in the interior of six faces, on the twelve edges and in the interior of $\Gamma$ . Case I. Firstly, we proceed with interior points of $\Gamma$ . Let $(p, x, y) \in (0, 2) \times (0, 1) \times (0, 1)$ . A simple computation yields $$\frac{\partial G}{\partial y} = \frac{(4-p^2)}{2304} \left( 9p^2(x(1-x^2)(1+2y) - 2(4-p^2)y) + 2(4-p^2)(-18(4-p^2)xy + 8(1-x^2+2(1-x^2)y) + x^2(1-x^2+2(1-x^2)y)) \right).$$ Now $\partial G/\partial y = 0$ gives $$y_0 = \frac{p^3(-4x^2 + 8x + 5) + 8px(2x + 5)}{4(x - 1)(p^2(2x - 25) - 8(x - 8))}.$$ The critical point $y_0$ lie in the interval (0,1) whenever $$p^{3}(4x^{2} - 8x - 5) + 4p^{2}(2x^{2} - 27x + 25) - 8px(2x + 5) - 32(x^{2} - 9x + 8) > 0$$ $$(9)$$ and $$p^2(25 - 2x) > 8(8 - x) \tag{10}$$ hold. It can be simply observe that as $p \to 2$ , there exists at least one $x \in (0,1)$ such that the inequality (9) is not true. Similarly, when $x \to 0$ , there exists at least one $p \in (0,2)$ such that inequality (9) does not remain valid. The inequality (9) is true only for p = 2 and $x \in (0,13/36)$ , which is outside the considered domain. Hence, we deduce that G(p,x,y) has no critical point. Case II. Secondly, we consider the interior of all the six faces of the cuboid $\Gamma$ . On the face p = 0, let $G(0, x, y) = h_1(x, y)$ , where $$h_1(x,y) = \frac{1}{72}(1-x^2)(x^2y^2 - 9x(y^2-1) + 8y^2)$$ for $x, y \in (0, 1)$ . A simple calculation shows that $$\frac{\partial h_1}{\partial y} = \frac{1}{72} (1 - x^2) (2x^2y - 18xy + 16y) \neq 0 \quad \forall \ x, y \in (0, 1).$$ Hence $h_1(x,y)$ has no critical point in $(0,1) \times (0,1)$ . On the face p = 2, we simply get $$G(2, x, y) = \frac{5}{576}. (11)$$ On the face x = 0, let G(p, 0, y) = h2(p, y), where $$h_2(p,y) = \frac{5p^6 + 144(4-p^2)p^2(1-y^2) + 256(4-p^2)^2y^2 + 40(4-p^2)p^3y}{36864},$$ (12) p ∈ (0, 2) and x ∈ (0, 1). The critical points of h2(p, y) are given by the solution of ∂h2(p, y)/∂p = 0 and ∂h2(p, y)/∂y = 0. The equation ∂h2(p, y)/∂y = 0 satisfies for $$y_0 = \frac{5p^3}{4(25p^2 - 64)}.$$ The root y⁰ lies in the range (0, 1) whenever p ∈ (1.67136, 2). Also, ∂h2(p, y)/∂p = 0 implies $$\frac{p(625p^8 - 18050p^6 + 110320p^4 - 251904p^2 + 196608)}{1536(64 - 25p^2)^2} = 0,$$ which has no solution for the range p ∈ (1.67136, 2). Consequently, h2(p, y) has no critical point for p ∈ (0, 2) and x ∈ (0, 1). On the face x = 1, let G(p, 1, y) = h3(p, y) where $$h_3(p,y) = \frac{5p^6 + 144p^2(4-p^2) + 118p^4(4-p^2) + 116p^2(4-p^2)^2}{36864}, \quad p \in (0,2).$$ (13) A straightforward computation reveals that the maximum of h3(p, y) is attained at p = 2 q 2(25 − √ 587)/3 given by $$\max h_3(p,y) = \frac{587\sqrt{587} - 14200}{324}.$$ (14) On the face y = 0, let G(p, x, 0) = h4(p, x), where $$h_4(p,x) = \frac{1}{36864} \left( 5p^6 + 26p^4(4-p^2)x + 144p^2(4-p^2)x^2 + 56p^4(4-p^2)x^2 + 68p^2(4-p^2)^2x^2 + 36p^4(4-p^2)x^3 + 40p^2(4-p^2)^2x^3 + 8p^2(4-p^2)^2x^4 + (4-p^2)(144p^2 + 288(4-p^2)x)(1-x^2) \right).$$ for x ∈ (0, 1) and p ∈ (0, 2). A simple calculation yields that ∂h4/∂p = 0 and ∂h4/∂x = 0 have no solutions for p ∈ (0, 2) and x ∈ (0, 1). Hence, h4(p, x) does not have any critical point. On the face y = 1, let G(p, x, 1) = h5(p, x) such that $$h_5(p,x) = \frac{1}{36864} \left( 5p^6 + 26p^4(4-p^2)x + 144p^2(4-p^2)x^2 + 56p^4(4-p^2)x^2 + 68p^2(4-p^2)^2x^2 + 36p^4(4-p^2)x^3 + 40p^2(4-p^2)^2x^3 + 8p^2(4-p^2)^2x^4 + (4-p^2)(1-x^2)(32(4-p^2)x^2 + 144p^2x + 256(4-p^2)) + (4-p^2)(1-x^2)(144p^3x^2 + 40p^3 + 32(4-p^2)px^2 + 80(4-p^2)px) \right).$$ Proceeding as in the above case, we obtain that h5(p, x) has no critical point for p ∈ (0, 2) and x ∈ (0, 1). Case III. Now, we find the maximum value of G(p, x, y) on the eight edges of the cuboid Γ. In view of the function G(p, x, y), we have $$G(p, 0, 0) = 5p^6 + 144(4 - p^2)p^2/36864 =: k_1(p).$$ Clearly, the maximum value of ^k1(p) is attained at ^p = 2^q 2(6 − √ 21)/5 and given by $$\max k_1(p) = \frac{1}{300} (7\sqrt{21} - 27) = 0.0169268.$$ Again considering the function G(p, x, y), we obtain G(p, 0, 1) = (5p ⁶ + 256(4 − p 2 ) ² + 40(4 − p 2 )p 3 )/36864 =: k2(p) for p ∈ (0, 2). Since k2(p) is a decreasing function, therefore maximum will attain at p = 0. Thus $$\max k_2(p) = \frac{1}{9}.$$ From [\(13\)](#page-9-0), it can be observed that G(p, 1, y) does not depend on y, hence we directly obtain G(p, 1, 0) = G(p, 1, 1) = h3(p, y) =: k3(p). Thus from [\(14\)](#page-9-1), we get that max k3(p) = max h3(p, y) = (587^√ 587 − 14200)/324. On substituting p = 0 in [\(13\)](#page-9-0), we get G(0, 1, y) = 0. Also for p = 2, from [\(11\)](#page-9-2) note that G(2, x, y) is independent from of all the variables x and y. Therefore the value of G(p, x, y) on the edges p = 2, x = 0; p = 2, x = 1; p = 2, y = 0 and p = 2, y = 1 given by $$G(2,0,y) = G(2,1,y) = G(2,x,0) = G(2,x,1) = \frac{5}{576}$$ respectively. From [\(12\)](#page-9-3) at p = 0, we obtain G(0, 0, y) = y ²/9 =: k4(y). It is easy to verify that $$\max k_4(y) = \frac{1}{9}.$$ Again, for p = 0 and y = 1, we have G(0, x, 1) = (1 − x 2 )(x ² + 8)/72 =: k5(x). Since k5(x) is a decreasing function of x, hence $$\max k_5(x) = \frac{1}{9}.$$ Substituting p = y = 0 in G(p, x, y), we get G(0, x, 0) = x(1−x 2 )/8 =: k6(x). It is a simple exercise to check that the maximum value of k6(x) is attained at x = 1/ √ 3 given by $$\max k_6(x) = \frac{1}{12\sqrt{3}}.$$ In view of the above all three cases, the maximum value 1/9 of G(p, x, y) is attained at the edge p = x = 0 and y = 1, which is the required bound. The sharpness of the bound follows from the function f4(z) given by [\(5\)](#page-2-2).

Theorem 15. $k - \mathcal{ST} \subset \mathcal{S}_{nc}^*$ for $k \ge (4\cos 1)/(4\cos 1 - \cos 2 - 1)$ .

Theorem 16. $S_{nc}^* \subset ST_p(a)$ whenever $a \ge a_0 \approx 0.402301$ .

Theorem 17. The class S ∗ nc satisfies the following relations: - 1. S ∗ nc ⊂ S^∗ . - 2. S ∗ nc ̸⊂ S^∗ (α) for all α ∈ (0, 1). - 3. S ∗ nc ⊂ µ(β) whenever β ≥ 2 sec 1.  Figure 4: Inclusion graphs

Theorem 18. Let f ∈ S^∗ nc, then the following hold: 1. $$f \in S^*(\alpha)$$ in $|z| \le r_\alpha$ , where $r_\alpha$ is the root of $(1-r) - \alpha \cos r = 0$ . 2. $f \in \mu(\beta)$ in $|z| < r_{\beta}$ , where $$r_{\beta} = \begin{cases} r_0(\beta), & 1 < \beta < 2/(\cos 1), \\ 1, & 2/(\cos 1) \le \beta \end{cases}$$ (15) and $r_0(\beta)$ is the smallest root of $$1 + r = \beta \cos r.$$ The radii are sharp.

Theorem 19. If $f \in \mathcal{S}_{nc}^*$ , then f is convex function of order $\alpha$ for $|z| < r_c$ , where $r_c$ is the root of $$(1-r)^2 - (r+\alpha(1-r))\cos r - r(1-r)\sin r = 0.$$ (16)

Corollary 19.1. The C-radius, for the class $S_{nc}^*$ is $r_c \approx 0.454$ .

Theorem 20. If $f \in \mathcal{S}_{nc}^*$ then f is M-starlike in $|z| < r_3(M)$ , where $r_3(M)$ is the root of the equation $$1 - r - 2M\cos r = 0.$$ when $0 < M \le 1/2$ and $r_3(M) = 1$ when M > 1/2.

Lemma 21. [21, p.24] Let $q \in \mathcal{Q}$ with q(0) = a, and let $p(z) = a + a_n z^n + \cdots$ be analytic in $\mathbb{D}$ with $p(z) \neq a$ and $n \geq 1$ . If p is not subordinate to q, then there exist points $z_0 = r_0 e^{i\theta_0} \in \mathbb{D}$ and $\zeta_0 \in \partial \mathbb{D} \setminus E(q)$ and an $m \geq n \geq 1$ for which $p(\mathbb{D}_{r_0}) \subset q(\mathbb{D})$ , - 1. $p(z_0) = q(\zeta_0)$ , - 2. $z_0 p'(z_0) = m\zeta_0 q'(\zeta_0),$ - 3. $\operatorname{Re}\left\{\frac{z_0 p''(z_0)}{p'(z_0)} + 1\right\} \ge m \operatorname{Re}\left\{\frac{\zeta_0 q''(\zeta_0)}{q'(\zeta_0)} + 1\right\}$

Lemma 22. [21, Corollary 3.4h, p.132] Let q be univalent in $\mathbb{D}$ , and let $\theta$ and $\psi$ be analytic in a domain D containing $q(\mathbb{D})$ with $\psi(w) \neq 0$ , when $w \in q(\mathbb{D})$ . Set $Q(z) := zq'(z)\psi(q(z))$ and $h(z) := \theta(q(z)) + Q(z)$ . Suppose that either h is convex or Q(z) is starlike. In addition, assume that Re(zh'(z)/Q(z)) > 0 $(z \in \mathbb{D})$ . If p is analytic in $\mathbb{D}$ with p(0) = q(0), $p(\mathbb{D}) \subset D$ and $$\theta(p(z)) + zp'(z)\psi(p(z)) \prec \theta(q(z)) + zq'(z)\psi(q(z)),$$ then $p \prec q$ and q is the best dominant.

Theorem 23. If $p \in \mathcal{H}$ such that p(0) = 1 and $$\operatorname{Re}\left\{\frac{zp'(z)}{p(z)}\right\} < \frac{1}{2} + k_2,\tag{18}$$ where $k_2 = 1/\cosh 2$ , then $$p(z) \prec \frac{1+z}{\cos z}$$ .

Corollary 23.1. If $f \in A$ and Re $$\left\{ 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right\} < \frac{1}{2} + k_2 \quad (z \in \mathbb{D}),$$ then $f \in \mathcal{S}_{nc}^*$ . The next theorem provides an additional sufficient condition for a function f to belong to the class $\mathcal{S}_{nc}^*$ .

Theorem 24. If $f \in A$ and satisfies $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) > \frac{1}{2} + \frac{2 + \sinh 1}{\cos 1} \quad (z \in \mathbb{D}),$$ then $f \in \mathcal{S}_{nc}^*$ .

Theorem 25. Let p be analytic function in D such that p(0) = 1 and satisfies $$1 + \gamma z p'(z) \prec \frac{1+z}{\cos z}.$$ Let $$g(z) = \int_0^z \frac{1 + t - \cos t}{t \cos t} dt. \tag{20}$$ Then the following hold: 1. p(z) ≺ (1 + Az)/(1 + Bz), whenever $$\gamma \ge \max\left\{\frac{\gamma_2(1-B)}{A-B}, \left(\frac{1+B^2}{A-B-1-B^2}\right)\operatorname{Im}g(i)\right\},$$ where −1 ≤ B < A ≤ 1 and Im g(i) ≈ 0.862897. - 2. p(z) ≺ e ^z whenever γ ≥ (eγ1)/(1 − e) ≈ 1.4308, - 3. p(z) ≺ 1 + ze^z whenever γ ≥ γ² ≈ 2.45796, 4. $p(z) \prec 1 + \sin z$ whenever $\gamma \geq \gamma_2 \csc 1 \approx 1.82614$ , where $\gamma_1 = g(-1) \approx -0.904233$ and $\gamma_2 = g(1) \approx 1.53664$ .

Theorem 26. If $p \in \mathcal{H}$ such that $$p(z) + \frac{zp'(z)}{p(z)} \in \Omega_b, \tag{25}$$ then $p(z) \prec \varphi_{nc}(z)$ for $b < b_0 \approx -0.005796$ .

Corollary 26.1. Let f ∈ A and let 1 + zf′′(z)/f′ (z) ∈ Ωb, where Ω^b is given in (3.9) and b ≥ b⁰ ≈ −0.005796. Then $$\frac{zf'(z)}{f(z)} \prec \frac{1+z}{\cos z}$$ or equivalently f ∈ S^∗ P . Proof. Assume that 1 + zf′′(z)/f′ (z) ∈ Ωb. Setting p(z) = zf′ (z)/f(z), we have p(0) = 1, and the above condition can be rewritten as $$p(z) + \frac{zp'(z)}{p(z)} \in \Omega_b,$$ in D. Now, applying Theorem 3.5, we conclude the assertion.

Theorem 27. A function f ∈ S^∗ nc if and only if $$\frac{1}{z} \left[ f(z) * \frac{z - \varphi(e^{i\theta})(z - z^2)}{(1 - z)^2} \right] \neq 0, \quad \theta \in [-\pi, \pi]$$ Proof. Let f ∈ S^∗ nc then by the definition $$\frac{zf'(z)}{f(z)} \prec \frac{1+z}{\cos z}.$$ According to the definition of subordination, there exists a Schwarz function ω(z) with ω(0) = 0 and |ω(z)| < 1, such that $$\frac{zf'(z)}{f(z)} = \frac{1 + \omega(z)}{\cos(\omega(z))}.$$ Again by the definition of subordination we can get $$\frac{zf'(z)}{f(z)} \neq \frac{1 + e^{i\theta}}{\cos e^{i\theta}},$$ which is equivalent to $$\frac{1}{z} \left[ zf'(z) - f(z)\varphi(e^{i\theta}) \right] \neq 0.$$ (27) Now using the following basic convolution property in (4.3) $$f(z) = f(z) * \left(\frac{z}{1-z}\right)$$ and $zf'(z) = f(z) * \left(\frac{z}{(1-z)^2}\right)$ (28) we can get desired result. If we put f(z) = z + P^∞ ⁿ=2 anz ⁿ in Theorem [27,](#page-21-0) we get the condition in terms of coefficients. Corollary 27.1. If $f \in \mathcal{A}$ , then a necessary and sufficient condition for $f \in \mathcal{S}_{nc}^*$ is that $$\sum_{n=2}^{\infty} (\varphi(e^{i\theta}) - n)a_n z^{n-1} + \varphi(e^{i\theta}) \neq 1,$$ where $\varphi(z) = (1+z)/\cos z$ .

Corollary 27.2. If $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in A$ such that $$\sum_{n=2}^{\infty} |n - \varphi(e^{i\theta})| |a_n| + M < 1, \quad z \in \mathbb{D}$$ then $f \in \mathcal{S}_{nc}^*$ , where $M = 4\cos 1/(1+\cos 2)$ .