New criteria for starlikeness in the unit disc

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

Corollary 1. Let $f \in A$ and $$\left| \arg \left[ \frac{zf''(z)}{f'(z)} + 1 \right] \right| < \frac{\pi}{2} \quad (z \in \mathbb{D})$$ Then $f \in \mathcal{S}^*$ . Further, it is easy to verify that the disc with center $\alpha$ and radius $\frac{\alpha\sqrt{3}}{\sqrt{3+(\alpha-1)^2}}$ is contained in the angle $|\arg z| < \arctan\frac{\sqrt{3}}{\alpha-1}$ , and so by using the result of Theorem 1, we have that $$\left| \frac{zf''(z)}{f'(z)} \right| = \left| \left( \frac{zf''(z)}{f'(z)} + \alpha \right) - \alpha \right| < \frac{\alpha\sqrt{3}}{\sqrt{3 + (\alpha - 1)^2}} \quad (z \in \mathbb{D})$$ implies $f \in \mathcal{S}^*$ . Since the function $\varphi(\alpha) =: \frac{\alpha\sqrt{3}}{\sqrt{3+(\alpha-1)^2}}$ , $\alpha \geq 1$ , attains its maximal value 2 for $\alpha = 4$ , we get

Corollary 2. Let $f \in A$ and $$\left| \frac{zf''(z)}{f'(z)} \right| < 2 \quad (z \in \mathbb{D})$$ Then $f \in \mathcal{S}^*$ . In a similar way as in Theorem 1, we can consider starlikeness problem in connection with the class defined by $$\mathcal{G} = \left[ f \in \mathcal{A} : \operatorname{Re} \left[ 1 + \frac{zf''(z)}{f'(z)} \right] < \frac{3}{2}, z \in \mathbb{D} \right].$$ Ozaki in [4] proved that if $f \in \mathcal{G}$ , then f is univalent in $\mathbb{D}$ . Later, Umezava in [6] showed the functions from $\mathcal{G}$ are convex in one direction. Also, it is shown in the papers [2] and [5] that $\mathcal{G}$ is subclass of $\mathcal{S}^*$ . If Re $\left[1 + \frac{zf''(z)}{f'(z)}\right] < \frac{3}{2}$ is not satisfied for every $z \in \mathbb{D}$ , then we can pose a question if for some $\beta < 1$ , such that (4) $$\operatorname{Re}\left[\beta + \frac{zf''(z)}{f'(z)}\right] < \frac{3}{2} \quad (z \in \mathbb{D}),$$ is it possible obtain sufficient condition for starlikeness, in a similar way as in Theorem 1. The condition (4) is equivalent to $$\operatorname{Re}\left[\frac{3-2\beta}{2} - \frac{zf''(z)}{f'(z)}\right] > 0 \quad (z \in \mathbb{D}),$$ and we can use similar technique. This lead so

Theorem 2. Let $f \in \mathcal{A}$ , $\beta < 1$ , and let $$\left| \arg \left[ \frac{3 - 2\beta}{2} - \frac{zf''(z)}{f'(z)} \right] \right| < \arctan \frac{2\sqrt{3}}{5 - 2\beta} \quad (z \in \mathbb{D}).$$ Then $f \in \mathcal{S}^*$ .

Corollary 3. Let $f \in A$ and $$\left| \arg \left[ \frac{1}{2} - \frac{zf''(z)}{f'(z)} \right] \right| < \operatorname{arctg} \frac{2}{\sqrt{3}} \approx 49.1^{\circ} \quad (z \in \mathbb{D}).$$ Then $f \in \mathcal{S}^*$ . As in the case of Theorem 1, we can show that the disc with center $\frac{3-2\beta}{2}$ and radius $\frac{(3-2\beta)\sqrt{3}}{\sqrt{12+(5-2\beta)^2}}$ is containing in the angle $|\arg z| \le$ arctg $\frac{2\sqrt{3}}{5-2\beta}$ . So, if $$\left| \frac{zf''(z)}{f'(z)} \right| = \left| \left( \frac{3 - 2\beta}{2} - \frac{zf''(z)}{f'(z)} \right) - \frac{3 - 2\beta}{2} \right| < \frac{(3 - 2\beta)\sqrt{3}}{\sqrt{12 + (5 - 2\beta)^2}},$$ for all $z \in \mathbb{D}$ , then Theorem 2 brings that $f \in \mathcal{S}^*$ . Since the function $\psi(\beta) =: \frac{(3-2\beta)\sqrt{3}}{\sqrt{12+(5-2\beta)^2}}$ is decreasing on the interval $(-\infty, 1]$ , and $\psi(\beta) \to \sqrt{3}$ , when $\beta \to -\infty$ , we get

Corollary 4. Let $f \in A$ and $$\left| \frac{zf''(z)}{f'(z)} \right| < \sqrt{3} \quad (z \in \mathbb{D}).$$ Then $f \in \mathcal{S}^*$ . For a starlike function f it is not necessary that $Re\frac{f(z)}{z} > 0$ , $z \in \mathbb{D}$ . For example, for Koebe function $k(z) = \frac{z}{(1-z)^2}$ we have $$\operatorname{Re} \frac{k(z)}{z} \bigg|_{z=(1+i)/\sqrt{2}} = -(\sqrt{2}+1) < 0,$$ and this also holds for points in $\mathbb{D}$ close enought to $\frac{1+i}{\sqrt{2}}$ . In the next theorem we give a condition which provides $\operatorname{Re} \frac{f(z)}{z} > 0$ , $z \in \mathbb{D}$ .

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 $\frac{f(z)}{z} > 0$ , $z \in \mathbb{D}$ .

Corollary 5. Let $f \in A$ and $$\left|\arg \frac{zf'(z)}{f(z)}\right| < \frac{\pi}{4} \quad (z \in \mathbb{D}).$$ Then Re $\frac{f(z)}{z} > 0$ , $z \in \mathbb{D}$ . In Theorem 3 the function f from $\mathcal{A}$ need not be starlike as the next example shows. Example 2. Let $f \in \mathcal{A}$ is defined by (7) $$\frac{zf'(z)}{f(z)} = \sqrt{3} \left(\frac{1+z}{1-z}\right)^{\frac{1}{3}} + 1 - \sqrt{3},$$ where we use the principal value for square root. For real z close to -1 we have that $\operatorname{Re}\left[\frac{zf'(z)}{f(z)}\right] < 0$ , which means that f is not starlike. But from (7) we have $$\frac{zf'(z)}{f(z)} + (\sqrt{3} - 1) = \sqrt{3} \left(\frac{1+z}{1-z}\right)^{\frac{1}{3}},$$ and from here $$\left| \arg \left[ \frac{zf'(z)}{f(z)} + (\sqrt{3} - 1) \right] \right| \le \frac{1}{3} \left| \arg \left( \frac{1+z}{1-z} \right) \right|$$ $$< \frac{\pi}{6} = \operatorname{arctg} \frac{1}{(\sqrt{3} - 1) + 1},$$ which by Theorem 3 (with $\gamma = \sqrt{3} - 1$ ) implies that $\operatorname{Re} \frac{f(z)}{z} > 0$ , $z \in \mathbb{D}$ . It is possible to show that the disc with center $1 + \gamma$ and radius $\frac{1+\gamma}{\sqrt{1+(1+\gamma)^2}}$ lies in the angle $|\arg z| \leq \arctan\frac{1}{1+\gamma}$ , and so, by using the result of Theorem 3 we have that $$\left| \frac{zf'(z)}{f(z)} - 1 \right| = \left| \left( \frac{zf'(z)}{f(z)} + \gamma \right) - (1+\gamma) \right| < \frac{1+\gamma}{\sqrt{1+(1+\gamma)^2}} \quad (z \in \mathbb{D})$$ implies Re $\frac{f(z)}{z} > 0$ , $z \in \mathbb{D}$ . When $\gamma \to +\infty$ we receive

Corollary 6. Let $f \in A$ and $$\left| \frac{zf'(z)}{f(z)} - 1 \right| < 1 \quad (z \in \mathbb{D}).$$ Then Re $\frac{f(z)}{z} > 0$ , $z \in \mathbb{D}$ .