ANGULAR ESTIMATIONS OF CERTAIN ANALYTIC FUNCTIONS

Corollary 2. Let $f \in A$ . If $$|\arg\left((1-\gamma)\frac{f(z)}{z}+\gamma f'(z)\right)|<\frac{\pi\delta}{2}\ (\gamma\geq 0, 0<\delta\leq 1),$$ then $$\left|\arg\frac{f(z)}{z}\right|<\frac{\pi\eta}{2}$$ where $\eta(0 < \eta \le 1)$ is the solution of the equation $$\delta = \eta + \frac{2}{\pi} Tan^{-1} \gamma \eta.$$ Similarly, we have

Theorem 2. Let $f \in A$ and $g \in S(a, \alpha)$ . If $$|\arg\left(\beta-\left((1-\gamma)\frac{f(z)}{g(z)}+\gamma\frac{f'(z)}{g'(z)}\right)\right)|<\frac{\pi\delta}{2}(\gamma\geq0,\beta>1,0<\delta\leq1),$$ then $$|\arg\left(\beta-\frac{f(z)}{g(z)}\right)|<\frac{\pi\eta}{2},$$ where $\eta(0 < \eta \le 1)$ is the solution of the equaiton (2.1). Remark 2. Putting $\alpha = 0$ , $a \to \infty$ and $\gamma = 1$ in Theorem 1 and Teorem 2, we have the corresponding results of Nunokawa[6]. Next, we prove

Theorem 3. Let $f \in A$ . If $$|\arg\frac{zf'(z)}{f^{1-m}(z)z^m}|<\frac{\pi\delta}{2}\ (0<\delta\leq 1, m\in N),$$ then $$|\arg\left(\frac{f(z)}{z}\right)^m|<\frac{\pi\eta}{2},$$ where $\eta(0 < \eta \le 1)$ is the solution of the equation $$\delta = \eta + \frac{2}{\pi} Tan^{-1} \frac{\eta}{m}$$ Proof. Let us put $$p(z) = \left(\frac{f(z)}{z}\right)^m.$$ Then p(z) is analytic in U with p(0) = 1 and we have $$\frac{zf'(z)}{f^{1-m}(z)z^m} = p(z) + \frac{1}{m}zp'(z).$$ By the assumption and Lemma 1, we can see that $p(z) \neq 0$ in U. The remaining part of the proof is similar to that of Theorem 1 and so we omit it. Finally, we have

Theorem 4. Let $f \in A$ . If $$\begin{split} |\arg\frac{zf'(z)}{f^{1-u}(z)z^u}| &< \frac{\pi\delta}{2} \ (u > 0, 0 < \delta \le 1), \\ |\arg\frac{zF'(z)}{F^{1-u}(z)z^u}| &< \frac{\pi\eta}{2}, \end{split}$$ where F is given by $$F^{u}(z) = \frac{u+c}{z^{c}} \int_{0}^{z} t^{c-1} f^{u}(t) dt \quad (c>0)$$ (2.5) with $F(z) \neq 0$ in $U - \{0\}$ and $\eta(0 < \eta \leq 1)$ is the solution of the equation $$\delta = \eta + \frac{2}{\pi} T a n^{-1} \frac{\eta}{u+c} \tag{2.6}$$ Proof. From the definition of (2.5), we have $$u\frac{zF'(z)}{F^{1-u}(z)} + cF^{u}(z) = (u+c)f^{u}(z).$$ Putting $$p(z) = \frac{T(z)}{S(z)},$$ where $T(z) = F'(z)z^{u+c}/(F(z)/z)^{1-u}$ and $S(z) = z^{u+c}$ , we have $$\frac{T'(z)}{S'(z)} = p(z) + \frac{1}{u+c}zp'(z) = \frac{zf'(z)}{f^{1-u}(z)z^u}$$ Applying Lemma 1 and the proof of Theorem 1, we obtain $$|\arg\frac{zF'(z)}{F^{1-u}(z)z^u}|<\frac{\pi\eta}{2},$$ where $\eta(0 < \eta \le 1)$ is the solution of the equation (2.6).