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Abstract

It is well-known that the condition ${\operatorname{Re}} \left[1+\frac{zf''(z)}{f'(z)}\right]>0$, $z\in{\mathbb D}$, implies that $f$ is starlike function (i.e. convexity implies starlikeness). If the previous condition is not satisfied for every $z\in {\mathbb D}$, then it is possible to get new criteria for starlikeness by using $\left|\arg\left[α+\frac{zf''(z)}{f'(z)}\right]\right|$, $z\in{\mathbb D}$, where $α>1.$

Results & Lemmas (2)

Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.

Theorem 1 Theorem 1. Let,, and let <span id="page-1-1"></span>(1) Then.
Theorem 1. Let $f \in \mathcal{A}$ , $\alpha > 1$ , and let <span id="page-1-1"></span>(1) $$\left| \arg \left[ \alpha + \frac{zf''(z)}{f'(z)} \right] \right| < \arctan \frac{\sqrt{3}}{\alpha - 1} \quad (z \in \mathbb{D}).$$ Then $f \in \mathcal{S}^*$ .
Theorem 3 Theorem 3. Let, and let Then Re,.
Theorem 3. Let $f \in \mathcal{A}, \gamma \geq 0$ , and let $$\left| \arg \left[ \frac{zf'(z)}{f(z)} + \gamma \right] \right| < \arctan \frac{1}{1+\gamma} \quad (z \in \mathbb{D}).$$ Then Re $\frac{f(z)}{z} > 0$ , $z \in \mathbb{D}$ .

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