Some sufficient problems for strongly close-to-convex of order μ

Lemma 1. Let the function w(z) defined by $$w(z) = a_n z^n + a_{n+1} z^{n+1} + a_{n+2} z^{n+2} + \dots \qquad (n = 1, 2, 3, \dots)$$ ²⁰⁰⁰ Mathematics Subject Classification. 30C45. Key words and phrases. Analytic function, univalent function, close-to-convex, Jack's lemma. be analytic in $\mathbb{U}$ with w(0) = 0. If |w(z)| attains its maximum value on the circle |z| = r at a point $z_0 \in \mathbb{U}$ , then there exists a real number $k \ge n$ such that $$\frac{z_0 w'(z_0)}{w(z_0)} = k.$$

Theorem 1. If $f(z) \in A_{n_1}$ satisfies $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right|^{\beta} \left( \operatorname{Re} \left( \delta + 1 + \frac{zf''(z)}{f'(z)} - \frac{zg'(z)}{g(z)} \right) \right)^{\gamma} < \left( \operatorname{Re}(\delta) + \frac{\mu n}{2} \right)^{\gamma} \ (z \in \mathbb{U})$$ for some real $\beta \geq 0$ , $\gamma \geq 0$ such that $\beta + \gamma > 0$ , some real $0 < \mu \leq 1$ , some complex $\delta$ with $\text{Re}(\delta) > -\frac{\mu n}{2}$ , and for some $g(z) \in \mathcal{A}_{n_2} \cap \mathcal{S}^*$ where $n = \min\{n_1, n_2\}$ , then $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right| < 1 \qquad (z \in \mathbb{U}).$$ This means that $f(z) \in \mathcal{STC}(\mu)$ .

Theorem 2. If $f(z) \in \mathcal{A}_{n_1}$ satisfies $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right|^{\beta} \left( \operatorname{Re} \left( \delta + 1 + \frac{zf''(z)}{f'(z)} \right) \right)^{\gamma} \le \left( \operatorname{Re}(\delta) + \alpha + \frac{\mu n}{2} \right)^{\gamma} \qquad (z \in \mathbb{U})$$ for some real $\beta \geq 0$ , $\gamma \geq 0$ such that $\beta + \gamma > 0$ , some real $0 < \mu \leq 1$ , $0 \leq \alpha < 1$ , some complex $\delta$ with $\text{Re}(\delta) > -\frac{\mu n}{2} - \alpha$ , and for some $g(z) \in \mathcal{A}_{n_2} \cap \mathcal{S}^*(\alpha)$ where $n = \min\{n_1, n_2\}, then$ $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right| < 1 \qquad (z \in \mathbb{U}),$$ which shows that $f(z) \in \mathcal{STC}(\mu)$ . Define w(z) in $\mathbb{U}$ by Proof. $$w(z) = \left(\frac{zf'(z)}{g(z)}\right)^{\frac{1}{\mu}} - 1 \qquad (z \in \mathbb{U})$$ = $b_n z^n + b_{n+1} z^{n+1} + \dots$ (2) where $n = \min\{n_1, n_2\}.$ Evidently, w(z) is analytic in $\mathbb{U}$ and w(0) = 0. Differentiating (2) logarithmically and simplyfing, we have $$1 + \frac{zf''(z)}{f'(z)} = \frac{zg'(z)}{g(z)} + \frac{\mu zw'(z)}{w(z) + 1},$$ and hence, $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right|^{\beta} \left( \operatorname{Re} \left( \delta + 1 + \frac{zf''(z)}{f'(z)} \right) \right)^{\gamma}$$ $$= |w(z)|^{\beta} \left( \operatorname{Re} \left( \delta + \frac{zg'(z)}{g(z)} + \frac{\mu zw'(z)}{w(z) + 1} \right) \right)^{\gamma} \leq \left( \operatorname{Re}(\delta) + \alpha + \frac{\mu n}{2} \right)^{\gamma} \qquad (z \in \mathbb{U}).$$ If there exists a point $z_0 \in \mathbb{U}$ such that $$\max_{|z| \le |z_0|} |w(z)| = |w(z_0)| = 1,$$ then Lemma 1 gives us that $w(z_0) = e^{i\theta}$ and $z_0 w'(z_0) = k w(z_0)$ $(k \ge n)$ . For such a point $z_0$ , we have $$\left| \left( \frac{z_0 f'(z_0)}{g(z_0)} \right)^{\frac{1}{\mu}} - 1 \right|^{\beta} \left( \operatorname{Re} \left( \delta + 1 + \frac{z_0 f''(z_0)}{f'(z_0)} \right) \right)^{\gamma}$$ $$= |w(z_0)|^{\beta} \left( \operatorname{Re} \left( \delta + \frac{z_0 g'(z_0)}{g(z_0)} + \frac{\mu z_0 w'(z_0)}{w(z_0) + 1} \right) \right)^{\gamma}$$ $$= \left( \operatorname{Re} \left( \delta + \frac{z_0 g'(z_0)}{g(z_0)} + \frac{\mu k w(z_0)}{w(z_0) + 1} \right) \right)^{\gamma}$$ $$= \left( \operatorname{Re} \left( \delta + \frac{z_0 g'(z_0)}{g(z_0)} + \frac{\mu k}{2} \left( 1 + i \tan \frac{\theta}{2} \right) \right) \right)^{\gamma}$$ $$= \left( \operatorname{Re}(\delta) + \operatorname{Re} \left( \frac{z_0 g'(z_0)}{g(z_0)} \right) + \frac{\mu k}{2} \right)^{\gamma}$$ $$> \left( \operatorname{Re}(\delta) + \alpha + \frac{\mu n}{2} \right)^{\gamma}.$$ This contradicts our condition in the theorem. Therefore, there is no $z_0 \in \mathbb{U}$ such that $|w(z_0)| = 1$ . This means that |w(z)| < 1 for all $z \in \mathbb{U}$ . This implies that $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right| < 1 \qquad (z \in \mathbb{U})$$ so that $f(z) \in \mathcal{STC}(\mu)$ . We consider a new application for Lemma 1. Our new application is follows.

Theorem 3. If $f(z) \in A_{n_1}$ satisfies $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right|^{\beta} \left| \delta + 1 + \frac{zf''(z)}{f'(z)} - \frac{zg'(z)}{g(z)} \right|^{\gamma} < \rho^{\beta} \left( \delta + \frac{\mu \rho n}{1 + \rho} \right)^{\gamma} \qquad (z \in \mathbb{U})$$ $\textit{for so} \ \underline{\textit{me real}} \ \beta \geqq 0, \ \gamma \geqq 0 \ \textit{such that} \ \beta + \gamma > 0, \ \textit{some real} \ 0 < \mu \leqq 1, \ \delta > 0, \ \rho \ \textit{with}$ $\rho > \sqrt{\frac{\delta}{\delta + \mu n}}$ , and for some $g(z) \in \mathcal{A}_{n_2} \cap \mathcal{S}^*$ where $n = \min\{n_1, n_2\}$ , then $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right| < \rho \qquad (z \in \mathbb{U}).$$ In addition for $\rho < 1$ , we have $f(z) \in \mathcal{STC}(\mu)$ . Proof. Defining the function w(z) by $$w(z) = \left(\frac{zf'(z)}{g(z)}\right)^{\frac{1}{\mu}} - 1 \qquad (z \in \mathbb{U})$$ $$= b_n z^n + b_{n+1} z^{n+1} + \dots$$ where $n = \min\{n_1, n_2\}$ , we have that w(z) is analytic in $\mathbb{U}$ with w(0) = 0. Since, $$1 + \frac{zf''(z)}{f'(z)} - \frac{zg'(z)}{g(z)} = \frac{\mu zw'(z)}{w(z) + 1},$$ we obtain that $$\left| \left( \frac{zf'(z)}{g(z)} \right)^{\frac{1}{\mu}} - 1 \right|^{\beta} \left| \delta + 1 + \frac{zf''(z)}{f'(z)} - \frac{zg'(z)}{g(z)} \right|^{\gamma} = |w(z)|^{\beta} \left| \delta + \frac{\mu zw'(z)}{w(z) + 1} \right|^{\gamma}$$ $$< \rho^{\beta} \left( \delta + \frac{\mu \rho n}{1 + \rho} \right)^{\gamma}$$ If there exists a point $z_0 \in \mathbb{U}$ such that $$\max_{|z| \le |z_0|} |w(z)| = |w(z_0)| = \rho,$$ then Lemma 1 gives us that $w(z_0) = \rho e^{i\theta}$ and $z_0 w'(z_0) = k w(z_0)$ $(k \ge n)$ . Thus we have $$\left| \left( \frac{z_0 f'(z_0)}{g(z_0)} \right)^{\frac{1}{\mu}} - 1 \right|^{\beta} \left| \delta + 1 + \frac{z_0 f''(z_0)}{f'(z_0)} - \frac{z_0 g'(z_0)}{g(z_0)} \right|^{\gamma}$$ $$= |w(z_0)|^{\beta} \left| \delta + \frac{\mu z_0 w'(z_0)}{w(z_0) + 1} \right|^{\gamma}$$ $$= \rho^{\beta} \left| \delta + \frac{\mu k w(z_0)}{w(z_0) + 1} \right|^{\gamma}$$ $$= \rho^{\beta} \left| \delta + \frac{\mu \rho k(\rho + \cos \theta)}{\rho^2 + 1 + 2\rho \cos \theta} + i \frac{\mu \rho k \sin \theta}{\rho^2 + 1 + 2\rho \cos \theta} \right|^{\gamma}$$ $$= \rho^{\beta} \left( \delta^2 + \mu \delta k + \frac{\mu \delta k(\rho^2 - 1) + \mu^2 \rho^2 k^2}{\rho^2 + 1 + 2\rho \cos \theta} \right)^{\frac{\gamma}{2}}$$ $$\geq \rho^{\beta} \left( \delta^2 + \mu \delta k + \frac{\mu \delta k(\rho^2 - 1) + \mu^2 \rho^2 k^2}{\rho^2 + 1 + 2\rho} \right)^{\frac{\gamma}{2}}$$ $$= \rho^{\beta} \left( \delta + \frac{\mu \rho k}{1 + \rho} \right)^{\gamma}$$ $$\geq \rho^{\beta} \left( \delta + \frac{\mu \rho n}{1 + \rho} \right)^{\gamma}.$$ This contradicts our condition in the theorem. Therefore, there is no $z_0 \in \mathbb{U}$ such that $|w(z_0)| = \rho$ . This means that $|w(z)| < \rho$ for all $z \in \mathbb{U}$ . We also consider a new aplocation for Lemma 1.

Theorem 4. If $f(z) \in A_{n_1}$ satisfies $$\left| \left( \frac{g(z)}{zf'(z)} \right)^{\frac{1}{\mu}} - 1 \right|^{\beta} \left( \operatorname{Re} \left( \delta - 1 - \frac{zf''(z)}{f'(z)} + \frac{zg'(z)}{g(z)} \right) \right)^{\gamma} < \left( \operatorname{Re}(\delta) + \frac{\mu n}{2} \right)^{\gamma} \ (z \in \mathbb{U})$$ for some real $\beta \geq 0$ , $\gamma \geq 0$ such that $\beta + \gamma > 0$ , some real $0 < \mu \leq 1$ , some complex $\delta$ with $\text{Re}(\delta) > -\frac{\mu n}{2}$ , and for some $g(z) \in \mathcal{A}_{n_2} \cap \mathcal{S}^*$ where $n = \min\{n_1, n_2\}$ , then $$\left| \left( \frac{g(z)}{zf'(z)} \right)^{\frac{1}{\mu}} - 1 \right| < 1 \qquad (z \in \mathbb{U}).$$ This means that $f(z) \in \mathcal{STC}(\mu)$ .