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Results & Lemmas (5)
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Lemma 2
Lemma 2. Let. Then we have where Using the estimate (2.1), the lemma follows easily. Hence we omit the
Lemma 2. Let $w \in B$ . Then we have
$$\begin{array}{l} (2.2) \ \operatorname{Re} \left\{ \frac{zw'(z)}{(1+sw(z))(1+tw(z))} \right\} \leq -\frac{1}{(s-t)^2} \operatorname{Re} \left\{ sp(z) \ + \ \frac{t}{p(z)} - s - t \right\} \\ + \frac{1}{(s-t)^2} \frac{r^2 |sp(z) - t|^2 - |1-p(z)|^2}{(1-r^2)|p(z)|} \end{array}$$
where $p(z) = (1 + tw(z))/(1 + sw(z)), |z| = r \text{ and } -1 \le t < s \le 1.$
Using the estimate (2.1), the lemma follows easily. Hence we omit the
Lemma 3 · radius
Lemma 3. If p(z) = (1 + tw(z))/(1 + sw(z)),, then for each and s, t satisfying, p(z) lies in the disc, where and r = |z| < 1.
Lemma 3. If p(z) = (1 + tw(z))/(1 + sw(z)), $w \in B$ , then for each $b \in [0, 1]$ and s, t satisfying $-1 \le t < s \le 1$ , p(z) lies in the disc
$$\Delta(z) \equiv \{\zeta \colon |\zeta - A_b| \le D_b\}$$
,
where
$$A_{\scriptscriptstyle b} = \frac{(1+br)^2 - str^2(b+r)^2}{(1+br)^2 - s^2r^2(b+r)^2}; \, D_{\scriptscriptstyle b} = \; \frac{(s-t)r(b+r)(1+br)}{(1+br)^2 - s^2r^2(b+r)^2}$$
and r = |z| < 1.
Lemma 4
Lemma 4. If p(z) = (1 + tw(z))/(1 + sw(z)),, then for |z| = r,, we have where, (2.5; b) and, where, are defined as in Lemma 3 and,.
Lemma 4. If p(z) = (1 + tw(z))/(1 + sw(z)), $w \in B$ , then for |z| = r, $0 \le r < 1$ , we have
$$\operatorname{Re}\left\{kp(z) + \frac{t}{p(z)}\right\} - \frac{r^2|sp(z) - t|^2 - |1 - p(z)|^2}{(1 - r^2)|p(z)|}$$
$$(2.5) \ \ge \begin{bmatrix} \frac{2}{1-r^2} [\sqrt{(1+t)(1-tr^2)(k(1-r^2)+1-s^2r^2)} - (1-str^2)] \ if \ R_b \le R^ \ W/W^ \ if \ R_b \ge R^* \end{cases}$$
where
$$W=t(kt+s^2)r^4+2bt\{(k+s)+(kt+s^2)\}r^3 \ +[b^2(1+t)\{(k+t)+(kt+s^2)\}+2t(k+s)-(s-t)^2]r \ +2b\{(k+t)+t(k+s)\}r+(k+t)$$
,
(2.5; b)
$$W^* = \{1 + rb(1+t) + tr^2\}\{1 + rb(1+s) + sr^2\}$$
and $R^{*2} = (1+t)(1-tr^2)/(k(1-r^2)+1-s^2r^2)$ , $R_b = A_b - D_b$ where $A_b$ , $D_b$ are defined as in Lemma 3 and $k \ge s$ , $-1 \le t < s \le 1$ .
Theorem 1 · radius
Theorem 1. Let, then f is convex in where is the smallest positive root of the equation if and if where, and r = |z| < 1. The result is…
Theorem 1. Let $f \in R_a(\alpha, \beta)$ , then f is convex in $|z| < r_0$ where $r_0$ is the smallest positive root of the equation
$$\begin{aligned} 1 + 4\alpha\beta ar + (4\alpha\beta^2a^2 - 2(1+\beta - 3\alpha\beta))r^2 + 4\beta(2\alpha\beta - 1)ar^3 \ &+ (2\beta - 1)(2\alpha\beta - 1)r^4 = 0 \end{aligned}$$
if $R_a \geq R^*$ and
$$r_{\scriptscriptstyle 0} = [\{-\alpha\beta + \sqrt{\alpha(1-2\alpha\beta+\alpha\beta^2)}\}/(1-2\alpha\beta)]^{1/2}$$
if $R_a \leq R^*$ where
$$R_a = \frac{1 + 2\alpha\beta ar + (2\alpha\beta - 1)r^2}{1 + 2\beta ar + (2\beta - 1)r^2}$$
, $R^* = \left(\frac{\alpha(1 - (2\alpha\beta - 1)r^2)}{1 - (2\beta - 1)r^2}\right)^{\!\scriptscriptstyle 1/2}$
and r = |z| < 1. The result is sharp for each $\alpha$ , $\beta(0 \le \alpha < 1, 0 < \beta \le 1)$ and $0 \le \alpha \le 1$ .
Theorem 2 · radius
Theorem 2. Let, then g is convex in where is the smallest positive root of the equation if and if, where and r = |z| < 1. The result is…
Theorem 2. Let $g \in S_a^*(\alpha, \beta)$ , then g is convex in $|z| < r_0$ where $r_0$ is the smallest positive root of the equation
$$\begin{aligned} 1 + 2\beta(3\alpha - 1)\alpha r + (4\alpha^2\beta^2\alpha^2 + 8\alpha\beta - 2 - 4\beta)r^2 \ & - 2\beta(1 + \alpha - 4\beta\alpha^2)\alpha r^3 + (1 - 2\alpha\beta)^2r^4 = 0 \end{aligned}$$
if $R_a \geq R^*$ and
$$r_{\text{o}} = [(5\alpha - 1)/\{(1 - \alpha + 4\beta\alpha^{2}) + 4\alpha\sqrt{(1 + \beta - 3\alpha\beta + \alpha^{2}\beta^{2})\}}]^{1/2}$$
if $R_a \leq R^*$ , where
$$R_a = \frac{1 + 2\alpha\beta ar + (2\alpha\beta - 1)r^2}{1 + 2\beta ar + (2\beta - 1)r^2} \; , \;\;\;\; R^* = \left(\frac{\alpha(1 + (1 - 2\alpha\beta)r^2)}{(2 - \alpha) - (2\beta - \alpha)r^2}\right)^{\!\scriptscriptstyle 1/2}$$
and r = |z| < 1. The result is sharp for each $\alpha$ , $\beta$ $(0 \le \alpha < 1, 0 < \beta \le 1)$ and $0 \le \alpha \le 1$ .