Sufficient conditions for strong starlikeness

Lemma 1.1. [16, Theorem 2.3i, p.35] Let $\Omega \subset \mathbb{C}$ and suppose that $\psi : \mathbb{C}^3 \times \mathbb{D} \to \mathbb{C}$ satisfies the condition $\psi(i\rho, \sigma, \mu + i\nu; z) \notin \Omega$ whenever $\rho, \sigma, \mu$ and $\nu$ are real numbers, $\sigma \leq -n(1+\rho^2)/2, \mu + \sigma \leq 0$ . If $p \in \mathcal{H}[1,n]$ and $\psi(p(z), zp'(z), z^2p''(z); z) \in \Omega$ for $z \in \mathbb{D}$ , then $\operatorname{Re} p(z) > 0$ in $\mathbb{D}$ .

Lemma 1.2. [16, Theorem 3.4i, p.134] Let q be univalent in $\mathbb{D}$ and let $\varphi$ and $\nu$ be analytic in a domain D containing $q(\mathbb{D})$ with $\varphi(w) \neq 0$ when $w \in q(\mathbb{D})$ . Set $Q(z) := zq'(z)\varphi(q(z))$ , $h(z) := \nu(q(z)) + Q(z)$ . Suppose that (i) either h is convex or Q(z) is starlike univalent in $\mathbb{D}$ and (ii) $\operatorname{Re}(zh'(z)/Q(z)) > 0$ for $z \in \mathbb{D}$ . If p is analytic in $\mathbb{D}$ , p(0) = q(0) and satisfies (1.1) $$\nu(p(z)) + zp'(z)\varphi(p(z)) \prec \nu(q(z)) + zq'(z)\varphi(q(z)),$$ then $p \prec q$ and q is the best dominant.

Lemma 1.3. [16, Theorem 3.4a, p.120] Let q be analytic in $\mathbb{D}$ and $\phi$ be analytic in a domain D containing $q(\mathbb{D})$ and suppose that (i) $\operatorname{Re} \phi(q(z)) > 0$ and either (ii) q is convex, or (iii) $Q(z) = zq'(z)\phi(q(z))$ is starlike. If p is analytic in $\mathbb{D}$ , p(0) = q(0), $p(\mathbb{D}) \subset D$ and $p(z) + zp'(z)\phi(p(z)) \prec q(z)$ , then $p \prec q$ .

Theorem 2.1. Let n be a positive integer, $0 < \alpha \le 1, C(z) = C \ge 0$ . Suppose that the functions $D, E : \mathbb{D} \to \mathbb{C}$ satisfy $$(2.1) |\operatorname{Im} E(z)| < n\alpha(\operatorname{Re} D(z) - C).$$ If $p \in \mathcal{H}[1, n]$ satisfies the equation (2.2) $$Cz^{2}p''(z) + D(z)zp'(z) + E(z)p(z) = 0$$ and $p(z) \neq 0$ , then $p(z) \prec ((1+z)/(1-z))^{\alpha}$ .

Corollary 2.2. Let n be a positive integer, $C(z) = C \ge 0$ . Suppose that the functions $D, E : \mathbb{D} \to \mathbb{C}$ satisfy $$|\operatorname{Im} E(z)| < n(\operatorname{Re} D(z) - C).$$ If $p \in \mathcal{H}[1,n]$ satisfies $Cz^2p''(z) + D(z)zp'(z) + E(z)p(z) = 0$ and $p(z) \neq 0$ , then $\operatorname{Re} p(z) > 0$ . By taking C(z) = 0 in Theorem 2.1, we get the following result for first order differential subordination.

Corollary 2.3. Let n be a positive integer, $0 < \alpha \le 1$ . Suppose that the functions $D, E : \mathbb{D} \to \mathbb{C}$ satisfy $$|\operatorname{Im} E(z)| < n\alpha \operatorname{Re} D(z).$$ If $p \in \mathcal{H}[1,n]$ satisfies D(z)zp'(z) + E(z)p(z) = 0 and $p(z) \neq 0$ , then $p(z) \prec ((1+z)/(1-z))^{\alpha}$ . Remark 2.4. The Corollary 2.3 for $\alpha = 1$ should be compared with [16, Corollary 4.1a.1, p. 189] The confluent (Kummer) hypergeometric function $\Phi(a, c; z)$ is given by (2.10) $$\Phi(a,c;z) = \frac{\Gamma(c)}{\Gamma(a)} \sum_{n=0}^{\infty} \frac{\Gamma(a+n)}{\Gamma(c+n)} \frac{z^n}{n!} = \sum_{n=0}^{\infty} \frac{(a)_n}{(c)_n} \frac{z^n}{n!},$$ where $a, c \in \mathbb{C}$ , $c \neq 0, -1, -2, \cdots$ , and $(\lambda)_n$ denotes the Pochhammer symbol defined by $(\lambda)_0 = 1$ , $(\lambda)_n = \lambda(\lambda + 1)_{n-1}$ . The function $\Phi \in \mathcal{H}[1, 1]$ is a solution of the differential equation (2.11) $$z\Phi''(a,c;z) + (c-z)\Phi'(a,c;z) - a\Phi(a,c;z) = 0$$ introduced by Kummer in 1837 [29]. The function $\Phi(a, c; z)$ satisfies the following recursive relation $$c\Phi'(a; c; z) = a\Phi(a+1; c+1; z).$$ When $\operatorname{Re} c > \operatorname{Re} a > 0$ , the function $\Phi$ can be expressed in the integral form $$\Phi(a; c; z) = \frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)} \int_0^1 t^{a-1} (1-t)^{c-a-1} e^{tz} dt.$$ There has been several works [1, 4, 14, 22] studying geometric properties of the function $\Phi(a; c; z)$ , such as on its close-to-convexity, starlikeness and convexity. By the use of Theorem 2.1, we obtain the following sufficient conditions for $\Phi(a, c; z) \prec ((1+z)/(1-z))^{\alpha}$ , $(0 < \alpha \le 1)$ .

Corollary 2.5. Let $0 < \alpha \le 1$ . If $a, c \in \mathbb{R}$ satisfy $$|c - 1| > \sqrt{1 + a^2/\alpha^2}$$ and $\Phi(a; c; z) \neq 0$ , then $\Phi(a; c; z) \prec ((1+z)/(1-z))^{\alpha}$ .

Corollary 2.8. Suppose that $0 < \alpha \le 1$ and if $b, p, c \in \mathbb{R}$ satisfy the following condition $$|c| < 4\alpha(k-1)$$ and $u_p(z) \neq 0$ , then $u_p(z) \prec ((1+z)/(1-z))^{\alpha}$ .

Corollary 2.11. Let n be a positive integer, $0 < \alpha \le 1$ and $C(z) = C \ge 0$ . Suppose the functions $D, E : \mathbb{D} \to \mathbb{C}$ satisfy $$|\operatorname{Im} E(z)| < n\alpha(\operatorname{Re} D(z) - C).$$ If $h \in \mathcal{A}_n$ satisfy $$C\left(2\left(\frac{zh'(z)}{h(z)}\right)^{3} - \frac{3z^{3}h'(z)h''(z)}{h^{2}(z)} + \frac{z^{3}h'''(z)}{h(z)}\right) + (2C + D(z))\left(\frac{z^{2}h''(z)}{h(z)} - \left(\frac{zh'(z)}{h(z)}\right)^{2}\right) + (D(z) + E(z))\frac{zh'(z)}{h(z)} = 0,$$ then $h \in \mathcal{S}^*[\alpha]$ .

Corollary 2.12. Let $0 < \alpha \le 1$ . Suppose that the functions $f \in \mathcal{A}$ , $g \in \mathcal{H}[1,1]$ satisfy $$\Big|\operatorname{Im}\Big(1+\frac{zf''(z)g(z)+2zf'(z)g'(z)+zg''(z)f(z)}{f'(z)g(z)+g'(z)f(z)}\Big)-\Big(\frac{zf'(z)}{f(z)}+\frac{zg'(z)}{g(z)}\Big)\Big|<\alpha,$$ then $fa \in \mathcal{S}^*[\alpha]$ . Remark 2.13. For $g(z) \equiv 1$ , Corollary 2.12 reduces to the following result: Let $0 < \alpha \le 1$ . Suppose that the function $f \in \mathcal{A}$ satisfy $$\left| \operatorname{Im} \left( 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right) \right| < \alpha,$$ then $f \in \mathcal{S}^*[\alpha]$ .

Theorem 3.1. Let p be an analytic function defined on $\mathbb{D}$ with p(0) = 1 satisfying $$1 + \beta \frac{zp'(z)}{p(z)} \prec \varphi_{CAR}(z).$$ If $$|\beta| \ge (\sqrt{(4\sqrt{3}+8)/(3\sqrt{3})})/\alpha \simeq 1.6947/\alpha \ (0 < \alpha \le 1)$$ , then $p(z) \prec ((1+z)/(1-z))^{\alpha}$ .

Theorem 3.2. Let p be an analytic function defined on $\mathbb{D}$ with p(0) = 1 satisfying $$1 + \beta \frac{zp'(z)}{p^2(z)} \prec \varphi_{CAR}(z),$$ then the following results hold: (a) If $\beta \ge 4$ or $\beta \le -4/3$ , then p(z) < (1+z)/(1-z). (b) If $\beta > 0$ and $0 < \alpha < 1$ satisfy $$(3.5) \qquad 9\alpha^2\beta^2 \left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha} (1-\alpha^2)^{-2} \left(-8+\alpha^2\left(8+3\beta^2\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}\right)\right) \\ -64\alpha\beta(1-\alpha^2)^{-\frac{1}{2}} \left(\sqrt{\frac{1-\alpha}{1+\alpha}}\right)^{\alpha} \sin\left(\frac{\alpha\pi}{2}\right) \ge 16\beta^2$$ then $$p(z) \prec ((1+z)/(1-z))^{\alpha}$$ .

Theorem 3.3. Let p be an analytic function defined on D with p(0) = 1. Then following are the sufficient conditions for p(z) ≺ ((1 + z)/(1 − z))^α (0 < α ≤ 1). (a) The function p satisfies the subordination $$p(z) + \beta z p'(z) \prec \left(\frac{1+z}{1-z}\right)^{\alpha} \quad (\beta > 0).$$ (b) The function p satisfies the subordination $$p(z) + \beta \frac{zp'(z)}{p(z)} \prec \left(\frac{1+z}{1-z}\right)^{\alpha} \quad (\beta > 0).$$ (c) The function p satisfies the subordination $$p(z) + \beta \frac{zp'(z)}{p^2(z)} \prec \left(\frac{1+z}{1-z}\right)^{\alpha} \quad (either \quad 0 < \alpha < \frac{1}{2}, \beta > 0 \quad or \quad \frac{1}{2} < \alpha \le 1, \beta < 0).$$ Proof. Define the function q : D → C by q(z) = ((1 + z)/(1 − z))^α (0 < α ≤ 1), with q(0) = 1. Clearly, the function q is convex in D. (a) Let us define φ(w) = β (β > 0). Since Re φ(q(z)) = β > 0, the result now follows from Lemma [1.3.](#page-2-10) (b) Define the function $\phi$ as $\phi(w) = \beta/w$ ( $\beta > 0$ ). Consider $$\operatorname{Re} \phi(q(z)) = \beta \operatorname{Re} \left(\frac{1}{q(z)}\right) > 0.$$ The result now follows from Lemma 1.3. (c) We define the function $\phi$ as $\phi(w) = \beta/w^2$ . A simple computation shows that $$\phi(q(z)) = \beta\left(\frac{1}{q^2(z)}\right) = \beta\left(\frac{1-z}{1+z}\right)^{2\alpha} =: M(z).$$ and for $-\pi \leq t \leq \pi$ , $$\operatorname{Re}(M(e^{it})) = \beta \cos(\alpha \pi) \left(\frac{\sin t/2}{\cos t/2}\right)^{2\alpha}.$$ It is easy to see that $\text{Re}(M(e^{it})) > 0$ if either $0 < \alpha < 1/2, \beta > 0$ or $1/2 < \alpha \le 1, \beta < 0$ . This completes the proof by the use of Lemma 1.3. We will illustrate the above three theorems by the following examples. By taking p(z) = zf'(z)/f(z) in Theorems 3.1, 3.2 and 3.3 gives the following example: Example 3.4. Let $f \in \mathcal{A}$ and $0 < \alpha \leq 1$ . The following are some sufficient conditions for the function $f \in \mathcal{S}^*[\alpha]$ : (1) The function f satisfies the subordination $$1 + \beta \left( 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right) \prec \varphi_{CAR}(z) \quad \left( |\beta| \ge \frac{1}{\alpha} \sqrt{\frac{4\sqrt{3} + 8}{3\sqrt{3}}} \simeq \frac{1.6947}{\alpha} \right).$$ (2) The function f satisfies the subordination $$1 - \beta + \beta \frac{1 + \frac{zf''(z)}{f'(z)}}{\frac{zf'(z)}{f(z)}} \prec \varphi_{CAR}(z).$$ where $0 < \alpha < 1$ and $\beta > 0$ satisfy $$9\alpha^{2}\beta^{2}\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}(1-\alpha^{2})^{-2}\left(-8+\alpha^{2}\left(8+3\beta^{2}\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}\right)\right)$$ $$-64\alpha\beta(1-\alpha^{2})^{-\frac{1}{2}}\left(\sqrt{\frac{1-\alpha}{1+\alpha}}\right)^{\alpha}\sin\left(\frac{\alpha\pi}{2}\right) \ge 16.$$ (3) The function f satisfies the subordination $$\frac{zf'(z)}{f(z)} + \beta \frac{zf'(z)}{f(z)} \left( 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right) \prec \left( \frac{1+z}{1-z} \right)^{\alpha} \quad (\beta > 0).$$ (4) The function f satisfies the subordination $$(1-\beta)\frac{zf'(z)}{f(z)} + \beta\left(1 + \frac{zf''(z)}{f'(z)}\right) \prec \left(\frac{1+z}{1-z}\right)^{\alpha} \quad (\beta > 0).$$ (5) The function f satisfies the subordination $$\frac{zf'(z)}{f(z)} - \beta + \beta \frac{1 + \frac{zf''(z)}{f'(z)}}{\frac{zf'(z)}{f(z)}} \prec \left(\frac{1+z}{1-z}\right)^{\alpha}$$ whenever either 0 < α < ¹ 2 , β > 0 or ¹ ² < α ≤ 1, β < 0. By taking p(z) = f 0 (z) in Theorems [3.1,](#page-8-4) [3.2](#page-9-3) and [3.3](#page-10-1) gives the following example: Example 3.5. Let f ∈ A and 0 < α ≤ 1. The following are the sufficient conditions for the function f 0 (z) ≺ ((1 + z)/(1 − z))^α : (1) The function f satisfies the subordination $$1 + \beta \left( \frac{zf''(z)}{f'(z)} \right) \prec \varphi_{CAR}(z) \quad \left( |\beta| \ge \frac{1}{\alpha} \sqrt{\frac{4\sqrt{3} + 8}{3\sqrt{3}}} \simeq \frac{1.6947}{\alpha} \right).$$ (2) The function f satisfies the subordination $$1 + \beta \left( \frac{zf''(z)}{(f'(z))^2} \right) \prec \varphi_{CAR}(z),$$ where 0 < α < 1 and β > 0 satisfy $$9\alpha^{2}\beta^{2}\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}(1-\alpha^{2})^{-2}\left(-8+\alpha^{2}\left(8+3\beta^{2}\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}\right)\right)$$ $$-64\alpha\beta(1-\alpha^{2})^{-\frac{1}{2}}\left(\sqrt{\frac{1-\alpha}{1+\alpha}}\right)^{\alpha}\sin\left(\frac{\alpha\pi}{2}\right) \geq 16.$$ (3) The function f satisfies the subordination $$f'(z) + \beta z f''(z) \prec \left(\frac{1+z}{1-z}\right)^{\alpha} \quad (\beta > 0).$$ (4) The function f satisfies the subordination $$f'(z) + \beta \left(\frac{zf''(z)}{f'(z)}\right) \prec \left(\frac{1+z}{1-z}\right)^{\alpha} \quad (\beta > 0).$$ (5) The function f satisfies the subordination $$f'(z) + \beta \left(\frac{zf''(z)}{(f'(z))^2}\right) \prec \left(\frac{1+z}{1-z}\right)^{\alpha} \quad (either \quad 0 < \alpha < \frac{1}{2}, \beta > 0 \quad or \quad \frac{1}{2} < \alpha \leq 1, \beta < 0).$$ The five parts of the next example are obtained by taking p(z) = z 2 f 0 (z)/f ² (z) in Theorems [3.1,](#page-8-4) [3.2](#page-9-3) and [3.3](#page-10-1) respectively. Example 3.6. Let f ∈ A and 0 < α ≤ 1. The following are the sufficient conditions for the function z 2 f 0 (z)/f ² (z) ≺ ((1 + z)/(1 − z))^α : (1) The function f satisfies $$1 + \beta \left( \frac{(zf(z))''}{f'(z)} - \frac{2zf'(z)}{f(z)} \right) \prec \varphi_{CAR}(z) \quad \left( |\beta| \ge \frac{1}{\alpha} \sqrt{\frac{4\sqrt{3} + 8}{3\sqrt{3}}} \simeq \frac{1.6947}{\alpha} \right).$$ (2) The function f satisfies the subordination $$1 + \beta \frac{f^2(z)}{z^2 f'(z)} \left( \frac{(zf(z))''}{f'(z)} - \frac{2zf'(z)}{f(z)} \right) \prec \varphi_{CAR}(z).$$ where $0 < \alpha < 1$ and $\beta > 0$ satisfy $$9\alpha^{2}\beta^{2}\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}(1-\alpha^{2})^{-2}\left(-8+\alpha^{2}\left(8+3\beta^{2}\left(\frac{1-\alpha}{1+\alpha}\right)^{\alpha}\right)\right)$$ $$-64\alpha\beta(1-\alpha^{2})^{-\frac{1}{2}}\left(\sqrt{\frac{1-\alpha}{1+\alpha}}\right)^{\alpha}\sin\left(\frac{\alpha\pi}{2}\right) \geq 16.$$ (3) The function f satisfies $$\frac{z^2 f'(z)}{f^2(z)} + \beta \frac{z^2 f'(z)}{f^2(z)} \left( \frac{(zf(z))''}{f'(z)} - \frac{2zf'(z)}{f(z)} \right) \prec \left( \frac{1+z}{1-z} \right)^{\alpha} \quad (\beta > 0).$$ (4) The function f satisfies $$\frac{z^2 f'(z)}{f^2(z)} + \beta \left( \frac{(zf(z))''}{f'(z)} - \frac{2zf'(z)}{f(z)} \right) \prec \left( \frac{1+z}{1-z} \right)^{\alpha} \quad (\beta > 0).$$ (5) Either $0 < \alpha < 1/2, \beta > 0$ or $1/2 < \alpha \le 1, \beta < 0$ and the function f satisfies the subordination $$\frac{z^2 f'(z)}{f^2(z)} + \beta \frac{f^2(z)}{z^2 f'(z)} \left( \frac{(zf(z))''}{f'(z)} - \frac{2zf'(z)}{f(z)} \right) \prec \left( \frac{1+z}{1-z} \right)^{\alpha}.$$ Let p be an analytic function in $\mathbb{D}$ with p(0) = 1. In the next result, condition on $\beta$ is obtained so that the subordination $p(z) + \beta z p'(z) \prec \varphi_{CAR}(z)$ implies $p(z) \prec (1+z)/(1-z)$ .

Theorem 3.7. Let p be an analytic function defined on $\mathbb{D}$ with p(0) = 1 satisfying $$p(z) + \beta z p'(z) \prec \varphi_{CAR}(z)$$ for $\beta \geq 0$ . Then $p(z) \prec (1+z)/(1-z)$ .

Theorem 3.9. Let p be an analytic function defined on $\mathbb{D}$ with p(0) = 1 satisfying $$p(z) + \beta \frac{zp'(z)}{p^2(z)} \prec \varphi_{CAR}(z) \quad for \quad \beta \leq 0.$$ Then $p(z) \prec (1+z)/(1-z)$ .