Noor Journal of Inequalities and Applications 2014, 2014:254

Lemma 2.1. [15] Let q(z) be a convex function in E with q(0) = 1 and let another function $h: E \to \mathbb{C}$ be with $\Re h(z) > 0$ . Let p(z) be analytic in E with p(0) = 1 such that $$(p(z) + h(z)zp'(z)) \prec q(z), \quad z \in E.$$ Then $p(z) \prec q(z)$ , $z \in E$ .

Lemma 2.2. Let N(z), D(z) be analytic in E with $$N(0) = 0 = D(z)$$ and let $D \in S^*$ for $z \in E$ . Then $\frac{N'(z)}{D'(z)} \in P(p_{k,\beta})$ implies that $\frac{N(z)}{D(z)} \in P(p_{k,\beta})$ for $z \in E$ . Proof Let $$\frac{N(z)}{D(z)} = p(z).$$ Then $$\frac{N'(z)}{D'(z)} = p(z) + h(z)(zp'(z)), \quad h(z) = \frac{1}{h_0(z)},$$ where $$h_0(z) = \frac{zD'(z)}{D(z)} \in P.$$ Since $\frac{N'(z)}{D'(z)} \in P(p_{k,\beta})$ , we have $$\frac{N'(z)}{D'(z)} = (p(z) + h(z)(zp'(z))) \prec p_{k,\beta}(z), \quad z \in E.$$ We now use Lemma 2.1 and this implies that $$\frac{N(z)}{D(z)} = p(z) \prec p_{k,\beta}(z) \quad \text{in } E.$$ This proves that $\frac{N(z)}{D(z)} \in P(p_{k,\beta})$ for $z \in E$ . The following lemma is an easy extension of a result proved in [5].

Lemma 2.3. Let $k \in [0, \infty)$ and $\gamma_1$ , $\delta_1$ be any complex numbers with $\gamma_1 \neq 0$ and let $\Re\{\frac{\gamma_1 k}{k+1} + \delta_1\} > \beta$ . If h(z) is analytic in E, h(0) = 1 and it satisfies <span id="page-4-0"></span> $$\left(h(z) + \frac{zh'(z)}{\gamma_1 h(z) + \delta_1}\right) \prec p_{k,\beta}(z),\tag{2.1}$$ and $q_{k,\beta}(z)$ is an analytic solution of $$\left(q_{k,\beta}(z)+\frac{zq_{k,\beta}'(z)}{\gamma_1q_{k,\beta}(z)+\delta_1}\right)=p_{k,\beta}(z),$$ then $q_{k,\beta}$ is univalent and <span id="page-4-1"></span> $$h(z) \prec q_{k,\beta}(z) \prec p_{k,\beta}(z)$$ , and $q_{k,\beta}(z)$ is the best dominant of (2.1).

Theorem 3.1. Let $f \in k - ST_s(\beta)$ . Then the odd function $$\Psi(z) = \frac{1}{2} [f(z) - f(-z)], \tag{3.1}$$ belongs to $k - ST(\beta)$ in E. In particular $\Psi(z)$ is an odd starlike function of order $\beta_1 = \frac{k+\beta}{k+1}$ in E. Proof Logarithmic differentiation of (3.1) and simple computation yield $$\begin{split} \frac{z\Psi'(z)}{\Psi(z)} &= \frac{1}{2} \left[ \frac{2zf'(z)}{f(z) - f(-z)} + \frac{2(-z)f'(-z)}{f(-z) - f(z)} \right] \\ &= \frac{1}{2} \left[ p_1(z) + p_2(z) \right], \quad \text{for } z \in E, p_1, p_2 \in P(p_{k,\beta}). \end{split}$$ Since $P(p_{k,\beta})$ is a convex set, it follows that $\frac{z\Psi'(z)}{\Psi(z)} \in P(p_{k,\beta})$ and thus $\Psi \in k - ST(\beta)$ in E. As a special case, we note that, for $k=0=\beta$ , $\frac{1}{2}[f(z)-f(-z)]=\Psi(z)\in S^*$ in E, and hence $\frac{zf'}{\Psi}\in P$ . We now discuss a geometric property for $f\in k-ST_s(\beta)$ . Here we investigate the behavior of the inclusion of the tangent at a point $w(\theta)=f(re^{i\theta})$ to the image $\Gamma_r$ of the circle $C_r=\{z:|z|=r\},\ 0\leq r<1,\ \theta\in[0,2\pi]$ , under the mapping by means of a function f from the class $f\in k-ST_s(\beta)$ . Let $$\Phi(\theta) = \frac{\pi}{2} + \theta + \arg f'(re^{i\theta}) = \arg \frac{\partial}{\partial \theta} f(re^{i\theta}),$$ and, for $\theta_2 > \theta_1$ , $\theta_1$ , $\theta_2 \in [0, 2\pi]$ , $$\Phi(\theta_2) - \Phi(\theta_1) = \theta_2 + \arg f'(re^{i\theta_2}) - \theta_1 - \arg f'(re^{i\theta_1}).$$ Now, since $$\theta + \arg f'(re^{i\theta}) = \theta + \Re\{-i \ln f'(re^{i\theta})\},\$$ then $$\frac{\partial}{\partial \theta} \left( \theta + \arg f' \left( r e^{i \theta} \right) \right) = \Re \left\{ 1 + \frac{r e^{i \theta} f'' (r e^{i \theta})}{f' (r e^{i \theta})} \right\}.$$ Hence $$\int_{\theta_1}^{\theta_2} \frac{\partial}{\partial \theta} \left(\theta + \arg f' \left(re^{i\theta}\right)\right) d\theta = \int_{\theta_1}^{\theta_2} \Re \left\{1 + \frac{re^{i\theta} f''(re^{i\theta})}{f'(re^{i\theta})}\right\} d\theta.$$ Also, on the other hand, $$\int_{\theta_1}^{\theta_2} \frac{\partial}{\partial \theta} (\theta + \arg f'(re^{i\theta})) d\theta = \theta_2 + \arg f'(re^{i\theta_2}) - \theta_1 - \arg f'(re^{i\theta_1})$$ $$= \Phi(\theta_2) - \Phi(\theta_1).$$ <span id="page-5-1"></span>So, the integral on the left side of the last inequality characterizes the increment of the angle of the inclination of the tangent to the curve $\Gamma_r$ between the points $w(\theta_2)$ and $w(\theta_1)$ for $\theta_2 > \theta_1$ . We have the following necessary condition for $f \in k - ST_s(\beta)$ .

Theorem 3.2. Let $f \in k - ST_s(\beta)$ . Then, with $z = re^{i\theta}$ and $0 \le \theta_1 < \theta_2 \le 2\pi$ , $0 \le \beta < 1$ and $0 \le k \le 1$ , we have $$\int_{\theta_1}^{\theta_2} \Re \left\{ \frac{(zf'(z))'}{f'(z)} \right\} d\theta > -\sigma \pi + 2 \cos^{-1} \left\{ \frac{2(1-\beta)}{1-(1-2\beta)r^2} \right\} + \beta_1(\theta_2 - \theta_1),$$ where $\sigma$ is given by (1.3) and $\beta_1 = \frac{k+\beta}{k+1}$ . Proof Since $\frac{f'(z)}{\Psi'(z)} \in P(p_{k,\beta})$ , $\Psi(z) = \frac{1}{2}[f(z) - f(-z)]$ and $\Psi \in k - UCV(\beta) \subset C(\beta)$ . We can write <span id="page-5-0"></span> $$f'(z) = (\Psi_1'(z))^{1-\beta_1} h^{\sigma}(z), \quad \Psi_1 \in C, h \in P(\beta),$$ and this gives us, with $z = re^{i\theta}$ , $0 \le r < 1$ , $0 \le \theta_1 < \theta_2 \le 2\pi$ , $$\int_{\theta_1}^{\theta_2} \Re\left\{\frac{(zf'(z))'}{f'(z)}\right\} d\theta = (1 - \beta_1) \int_{\theta_1}^{\theta_2} \Re\left\{\frac{(z\Psi_1'(z))'}{\Psi_1'(z)}\right\} d\theta + \sigma \int_{\theta_1}^{\theta_2} \Re\frac{2h'(z)}{h(z)} d\theta + \beta_1(\theta_2 - \theta_1).$$ $$(3.2)$$ For $h \in P(\beta)$ , we observe that $$\frac{\partial}{\partial \theta} \arg h(re^{i\theta}) = \frac{\partial}{\partial \theta} \Re \left\{ -i \ln h(re^{i\theta}) \right\}$$ $$= \Re \left\{ re^{i\theta} \frac{h'(re^{i\theta})}{h(re^{i\theta})} \right\}.$$ Therefore $$\int_{\theta_1}^{\theta_2} \Re\left\{\frac{re^{i\theta}h'(re^{i\theta})}{h(re^{i\theta})}\right\}d\theta = \arg h(re^{i\theta_2}) - \arg h(re^{i\theta_1}),$$ and $$\max_{h \in P(\beta)} \left| \int_{\theta_1}^{\theta_2} \Re \left\{ \frac{re^{i\theta}h'(re^{i\theta})}{h(re^{i\theta})} \right\} d\theta \right| = \max_{h \in P(\beta)} \left| \arg h(re^{i\theta_2}) - \arg h(re^{i\theta_1}) \right|.$$ We can write $$\frac{1}{1-\beta}[h(z)-\beta]=p(z), \quad p\in P,$$ and for |z| = r < 1, it is well known that $$\left| p(z) - \frac{1+r^2}{1-r^2} \right| \le \frac{2r}{1-r^2}.$$ From this, we have <span id="page-6-0"></span> $$\left| h(z) - \frac{1 + (1 - 2\beta)r^2}{1 - r^2} \right| \le \frac{2(1 - \beta)r}{1 - r^2}.$$ Thus the values of h are contained in the circle of Apollonius whose diameter is the line segment from $\frac{1-(1-2\beta)r}{1+r}$ to $\frac{1+(1-2\beta)r}{1-r}$ and has the radius $\frac{2(1-\beta)r}{1-r^2}$ . So $|\arg h(z)|$ attains its maximum at points where a ray from origin is tangent to the circle, that is, when $$\arg h(z) = \pm \sin^{-1} \left( \frac{2(1-\beta)r}{1-(1-2\beta)r^2} \right). \tag{3.3}$$ From (3.3), we observe that $$\max_{h \in P(\beta)} \left| \int_{\theta_1}^{\theta_2} \Re \left\{ r e^{i\theta} \frac{h'(re^{i\theta})}{h(re^{i\theta})} \right\} d\theta \right| \leq 2 \sin^{-1} \left( \frac{2(1-\beta)r}{1-(1-2\beta)r^2} \right) \\ = \pi - 2 \cos^{-1} \left( \frac{2(1-\beta)r}{1-(1-2\beta)r^2} \right). \tag{3.4}$$ Also, for $\Psi_1 \in C$ $$\int_{\theta_1}^{\theta_2} \Re\left\{1 + re^{i\theta} \frac{\Psi_1''(re^{i\theta})}{\Psi_1'(re^{i\theta})}\right\} d\theta \ge 0. \tag{3.5}$$ <span id="page-6-2"></span><span id="page-6-1"></span> Using (3.4) and (3.5) in (3.2), we obtain the required result. We note the following special cases: 1. For k = 0, $0 \le \theta_1 < \theta_2 \le 2\pi$ , $z = re^{i\theta}$ , it follows from Theorem 3.2 that $$\int_{\theta_1}^{\theta_2} \Re\left\{1 + \frac{zf''(z)}{f'(z)}\right\} d\theta > -\pi \quad (z \in E).$$ This is a necessary and sufficient condition for f to be close-to-convex (hence univalent) in E; see [7]. This also shows that $ST_s(\beta) \subset K$ . - 2. For k=1 $\int_{\theta_1}^{\theta_2} \Re\{1+\frac{zf''(z)}{f'(z)}\} d\theta > -\frac{\pi}{2}$ . 3. When $k \in [0,1]$ , it is obvious that $\sigma \in (0,1]$ . In this case, the class $k-ST_s(\beta)$ consists of strongly close-to-convex functions of order $\sigma$ in the sense of Pommerenke [20, 21].

Theorem 3.3. (Integral representation) Let $f \in k - ST_s(\beta)$ . Then $$f'(z) = \frac{1}{2}p(z) \exp \int_0^z \frac{1}{t} [p(t) + p(-t) - 2] dt,$$ where $p \in P(p_{k,\beta})$ , $z \in E$ . Proof Since $f \in k - ST_s(\beta)$ , we can write $$\frac{2zf'(z)}{f(z)-f(-z)}=p(z), \quad p\in P(p_{k,\beta}).$$ This gives us <span id="page-7-0"></span> $$\frac{2[f(z) - f(-z)]'}{f(z) - f(-z)} - \frac{1}{z} = \frac{1}{2} [p(z) - p(-z) - 2]$$ <span id="page-7-1"></span>and the result follows when we integrate. When k = 0, $\beta = 0$ , we obtain the result for the class $S_s^*$ given in [5]. We now study the class $k - ST_s(\beta)$ under a certain integral operator.

Theorem 3.4. Let $g \in k - ST_s(\beta)$ and let for m = 1, 2, 3, ..., G be defined by $$G(z) = \frac{m+1}{2z^m} \int_0^z t^{m-1} \{g(t) - g(-t)\} dt.$$ (3.6) Then G(z) belongs to $k - ST_s(\beta)$ in E. Proof Let $$J(z) = \int_0^z t^{m-1} \frac{g(t) - g(-t)}{2} dt.$$ Since $g \in k - ST_s(\beta)$ , $\frac{1}{2}\{g(z) - g(-z)\} \in k - ST(\beta) \subset S^(\beta_1) \subset S$ , and $\beta_1 = \frac{k+\beta}{k+1}$ . Therefore it can easily be verified that J(z) is (m + 1)-valently starlike in E. We can write (3.6) as $$z^m G(z) = (m+1)J(z),$$ and, differentiating logarithmically, we have $$\frac{zG'(z)}{G(z)} = \frac{zJ'(z) - mJ(z)}{J(z)} = \frac{N(z)}{D(z)},$$ say, where N(0) = D(0) = 0 and D is (m + 1)-valently starlike. Let $$\frac{N(z)}{D(z)} = h(z).$$ Then $$\frac{N'(z)}{D'(z)} = h(z) + \frac{zh'(z)}{h_0(z)}, \quad h_0(z) = \frac{zD'(z)}{D(z)} \in P$$ $$= h(z) + H_0(z)(zh'(z)), \quad H_0 = \frac{1}{h_0} \in P.$$ (3.7) Since $$\frac{N'(z)}{D'(z)} = \frac{(zh'(z))' - mJ'(z)}{J'(z)}$$ $$= \left\{ \frac{(zJ'(z))'}{J'(z)} - m \right\} \in P(p_{k,\beta}).$$ We now apply Lemma 2.2 to obtain <span id="page-8-0"></span> $$\frac{N(z)}{D(z)} = \frac{zG'(z)}{G(z)} \in P(p_{k,\beta}), \quad z \in E.$$ This proves that $G \in k - ST(\beta)$ in E.

Theorem 3.5. Let $f, g \in k - ST_s(\beta)$ and let F be defined by the following integral operator: $$F(z) = \left(\gamma + \frac{1}{\delta}\right) z^{1 - \frac{1}{\delta}} \int_0^z t^{\frac{1}{\delta} - 2} \left[ \frac{f(t) - f(-t)}{2} \right]^{\frac{1}{1 + \gamma}} \left[ \frac{g(t) - g(-t)}{2} \right] dt, \tag{3.8}$$ where $z \in E$ , $\delta > 0$ , $\gamma \ge 0$ and $\left[\frac{k(1+\gamma)}{k+1} + \left(\frac{1}{\delta} - 1\right)\right] > \beta$ . Then F(z) belongs to $k - ST(\beta)$ for $z \in E$ . When g(z) = z, $\gamma = 0$ , we obtain a generalized form of the Bernardi operator; see [1]. Also for g(z) = z, $\gamma = 0$ , and $\delta = \frac{1}{2}$ , we have the well-known integral operator studied by Libera [11] who showed that it preserves the geometric properties of convexity, starlikeness, and close-to-convexity. Proof Let $\frac{f(z)-f(-z)}{2} = \Psi_1(z)$ , $\frac{g(z)-g(-z)}{2} = \Psi_2(z)$ . Then $\Psi_1, \Psi_2 \in k - ST(\beta)$ in E. We can write (3.8) as <span id="page-8-1"></span> $$F(z) = \left(\gamma + \frac{1}{\delta}\right) z^{1 - \frac{1}{\delta}} \int_0^z t^{\frac{1}{\delta} - 2} \left(\Psi_1(t)\right)^{\frac{1}{1 + \gamma}} \left(\Psi_2(t)\right) dt. \tag{3.9}$$ Differentiating (3.9) logarithmically, and with $p(z) = \frac{zF'(z)}{F(z)}$ , we have <span id="page-9-1"></span><span id="page-9-0"></span> $$\frac{\gamma}{1+\gamma} \frac{z\Psi_1'}{\Psi_1(z)} + \frac{1}{1+\gamma} \frac{z\Psi_2'}{\Psi_2(z)} = p(z) + \frac{zp'(z)}{(1+\gamma)p(z) + (\frac{1}{\delta} - 1)}.$$ (3.10) Since, for $i=1,2,\ \Psi_i\in k-ST(\beta),\ \frac{z\Psi_1'(z)}{\Psi_1}=h_1(z),\ \frac{z\Psi_2'(z)}{\Psi_2}=h_2(z)$ both belong to $P(p_{k,\beta})$ in E, and $P(p_{k,\beta})$ is a convex set. Therefore $$\left(\frac{\gamma}{1+\gamma}h_1(z) + \frac{1}{1+\gamma}h_2(z)\right) \in P(p_{k,\beta}), \quad z \in E.$$ (3.11) From (3.10) and (3.11), it follows that $$\left(p(z) + \frac{zp'(z)}{(1+\gamma)p(z) + (\frac{1}{\delta} - 1)}\right) \prec p_{k,\beta}(z).$$ We now apply Lemma 2.3 which gives us $$p(z) \prec q_{k,\beta}(z) \prec p_{k,\beta}(z)$$ . Thus $F \in k - ST(\beta)$ and the proof is complete.

Theorem 4.1. Let $f \in k - UK_s(\beta)$ . Then, for 0 < r < 1, $k \in [0,1]$ , $$L(r,f) = O(1) \left(\frac{1}{1-r}\right)^{\sigma-\beta_1}, \quad \beta_1 < \frac{\sigma}{2},$$ where $\beta_1 = \frac{k+\beta}{k+1}$ and $\sigma$ is given by (1.3), and O(1) is a constant depending only on k, $\beta$ . Proof For $f \in k - UK_s(\beta)$ , we can write $$zf'(z) = \Psi(z)h^{\sigma}(z), \quad h \in P, \Psi \in S^*(\beta_1), \tag{4.1}$$ and $\Psi(z) = \{g(z) - g(-z)\}, g \in k - ST_s(\beta)$ . Since $\Psi \in S^*(\beta_1)$ and is odd, there exists an odd starlike function $\Psi_1(z)$ such that $$\Psi(z) = z \left(\frac{\Psi_1(z)}{z}\right)^{1-\beta_1} = z \left(\frac{\Psi_1(z)}{z}\right)^{\frac{1-\beta_1}{k+1}}.$$ Thus, with $z = re^{i\theta}$ , $$L(r,f) = \int_0^{2\pi} \left| zf'(z) \right| d\theta = \int_0^{2\pi} \left| z^{\beta_1} \left( \Psi_1(z) \right)^{1-\beta_1} h^{\sigma}(z) \right| d\theta,$$ and using Hölder's inequality, we have <span id="page-10-1"></span><span id="page-10-0"></span> $$L(r,f) \le 2\pi r^{\beta_1} \left( \frac{1}{2\pi} \int_0^{2\pi} \left| \Psi_1(z) \right|^{(1-\beta)(\frac{z}{z-\sigma})} d\theta \right)^{\frac{2-\sigma}{z}} \left( \frac{1}{2\pi} \int_0^{2\pi} \left| h(z) \right|^2 d\theta \right)^{\frac{\sigma}{2}}. \tag{4.2}$$ For $h \in P$ , it is well known [20] that $$\frac{1}{2\pi} \int_0^{2\pi} \left| h(z) \right|^2 d\theta \le \frac{1 + 3r^2}{1 - r^2}.$$ (4.3) Using (4.3) and subordination for odd starlike functions in (4.2), it follows that $$\begin{split} L(r,f) &\leq C(\beta_1,\sigma) \bigg(\frac{1}{1-r^2}\bigg)^{[(1-\beta_1)(\frac{2}{2-\sigma})-1][\frac{1+3r^2}{1-r}]^{\frac{\sigma}{2}}} \\ &= O(1) \bigg(\frac{1}{1-r}\bigg)^{\sigma-\beta_1}, \end{split}$$ where C and O(1) are constants depending only on $\beta_1$ and $\sigma$ . This completes the proof. $\square$ We now discuss the growth rate of coefficients of $f \in k - UK_s(\beta)$ .

Theorem 4.2. Let $f \in k - UK_s(\beta)$ and be given by (1.1). Then $$a_n = O(1)n^{\sigma - \beta_1 - 1}, \quad n \ge 1, \beta_1 < \frac{\sigma}{2},$$ where O(1) is a constant depending only on $\sigma$ and $\beta_1$ and $\sigma$ , $\beta_1$ are as given in Theorem 4.1. Proof For $z = re^{i\theta}$ , $n \ge 1$ , Cauchy's Theorem gives us <span id="page-10-2"></span> $$\begin{aligned} n|a_n| &= \frac{1}{2\pi r^{n+1}} \left| \int_0^{2\pi} z f'(z) e^{-in\theta} d\theta \right| \\ &\leq \frac{1}{2\pi r^{n+1}} \int_0^{2\pi} |z f'(z)| d\theta \\ &= \frac{1}{2\pi r^n} L(r, f). \end{aligned}$$ With $r = (1 - \frac{1}{n})$ , we use Theorem 4.1 and obtain the required result.

Theorem 4.3. Let $f \in k - UK_s(\beta)$ and let F be defined by $$F(z) = \frac{m+1}{2z^m} \int_0^z t^{m-1} \{ f(t) - f(-t) \} dt.$$ (4.4) Then $F \in k - UK_s(\beta)$ in E. That is, the class $k - UK_s(\beta)$ is preserved under the integral operator (4.4). Proof Since $f \in k - UK_s(\beta)$ , we can write $$\left\{\frac{2zf'(z)}{g(z)-g(-z)}\right\}\in P(p_{k,\beta}), \quad g\in k-ST_s(\beta)\subset S_S^*(\beta_1).$$ Let $G(z) = \frac{1}{2} \{g_1(z) - g_1(-z)\}$ and be defined by (3.5). By Theorem 3.4, $g_1 \in k - ST(\beta)$ and $G \in k - S_sT(\beta) \subset S_s^*(\beta_1)$ . Let $G = zG_1'$ . Then we can write $$G'_1(z) = \frac{1}{2} [zg_1(z) - g_1(-z)]', \quad G_1 \in k - UCV_s(\beta).$$ Thus, from (4.4) and $g = zg'_1, g_1 \in C_s(\beta_1)$ , we have $$\begin{split} \frac{2F'(z)}{[g_1(z)-g_1(-z)]'} &= \frac{z^m \{f(z)-f(-z)\} - m \int_0^z t^{m-1} \{f(t)-f(-t)\} \, dt}{z^m \{g_1(z)-g_1(-z)\} - m \int_0^z t^{m-1} \{g_1(t)-g_1(-t)\} \, dt} \\ &= \frac{N(z)}{D(z)}, \end{split}$$ say. We note that N(0) = D(0) = 0, and for $g_1 \in C_S(\beta_1)$ , $$\frac{(zD'(z))'}{D'(z)} = m + \frac{\{z[g_1(z) - g_1(-z)]'\}'}{\{g_1(z) - g_1(-z)\}'}$$ $$= m + h_1(z), \quad h_1 \in P(\beta_1).$$ Since $P(\beta_1)$ is a convex set, $D \in C_s(\beta_1) \subset S^*$ in E. We thus have $$\frac{N'(z)}{D'(z)} = \frac{1}{2} \left[ \frac{2zf'(z)}{[g_1(z) - g_1(-z)]'} + \frac{2(-z)f'(-z)}{[g_1(-z) - g_1(z)]'} \right] \in P(p_{k,\beta}).$$ Now, using Lemma 2.2, it follows that <span id="page-11-0"></span> $$\frac{N(z)}{D(z)} = \frac{2F'(z)}{(g_1(z) - g_1(-z))'} \in P(p_{k,\beta}) \quad \text{for } z \in E.$$ This proves that $F \in k - UK_S(\beta)$ in E. We study a partial converse of the above result as follows.

Theorem 4.4. Let $(\frac{2zf'(z)}{g(z)-g(-z)}) \prec p_k(z)$ in E and let <span id="page-11-3"></span> $$F_1(z) = \frac{1}{1+m} z^{1-m} \left( z^m f(z) \right)', \quad m = 1, 2, 3, \dots$$ (4.5) Then $F_1 \in K_s$ for $|z| < r_1$ , where <span id="page-11-2"></span><span id="page-11-1"></span> $$r_1 = \left\{ \frac{m+1}{(2-\beta_1) + \sqrt{(z-\beta_1)^2 + (m+1)(m-1+2\beta_1)}} \right\}, \quad \beta_1 = \frac{k+\beta}{k+1}.$$ (4.6) Proof We shall need the following well-known results for $p \in P(\alpha)$ , $0 \le \alpha < 1$ ; see [4]: $$\frac{1 - (1 - 2\alpha)r}{1 + r} \le |p(z)| \le \frac{1 + (1 - 2\alpha)r}{1 - r},\tag{4.7}$$ $$\left| p'(z) \right| \le \frac{2[\Re p(z) - \alpha]}{1 - r^2}.\tag{4.8}$$ Since $f \in k - UK_s(\beta)$ , there exists $g \in S_s^*(\beta_1)$ such that, for $z \in E$ . $$\left(\frac{2zf'(z)}{g(z)-g(-z)}\right)=p(z), \quad p\in P(p_k)\subset P(\alpha), \alpha=\frac{k}{k+1}.$$ From (4.5), we have $$F_1(z) = \frac{1}{1+m} \left[ mf(z) + zf'(z) \right],$$ and this gives us $$\frac{2zF_1'(z)}{g(z) - g(-z)} = \frac{1}{m+1} \left[ \frac{2mf'(z)}{g(z) - g(-z)} + \frac{2z(zf'(z))'}{g(z) - g(-z)} \right]$$ $$= \frac{1}{m+1} \left[ mp(z) + zp'(z) + p(z)h(z) \right],$$ where <span id="page-12-0"></span> $$h(z) = \frac{z\Psi'(z)}{\Psi(z)} \in P(\beta_1), \quad \Psi(z) = g(z) - g(-z).$$ Now, using (4.7) and (4.8), we have $$\Re\left\{\frac{2zF_{1}'(z)}{g(z)-g(-z)}\right\} \ge \frac{(\Re p(z)-\alpha)}{1+m} \left\{m + \frac{1-(1-2\beta_{1})r}{1+r} - \frac{2r}{1-r^{2}}\right\}$$ $$= \frac{\Re p(z)-\alpha}{1+m} \left[\frac{T(r)}{1-r^{2}}\right],\tag{4.9}$$ where $$T(r) = (m+1) - 2(2-\beta_1)r + (-m-2\beta_1+1)r^2$$ . We note that T(0) = 1 + m > 0 and T(1) = -3 < 0. So there exists $r_1 \in (0,1)$ . The right hand side of (4.9) is positive for $|z| < r_1$ , where $r_1$ is given by (4.6). This implies that $F \in K_s$ for $|z| < r_1$ and the proof is complete. We have the following special cases. - 1. For $k = 0 = \beta$ , $f \in K_s$ . Then $F_1$ , defined by (4.5) belongs to $K_s$ for $|z| < r_0 = \frac{1+m}{2+\sqrt{3+m^2}}$ . - 2. When m = 1 and $\beta_1 = 0$ (that is, $k = 0 = \beta$ ), then $F_1(z) = \frac{(zf(z))'}{2}$ belongs to the same class for $|z| < \frac{1}{2}$ . This result has been proved by Livingston [12] for convex and starlike functions.

Definition 1.1. Let $f \in A$ . The f is said to be in the class $k - ST_s(\beta)$ if, and only if, $$\frac{2zf'(z)}{(f(z)-f(-z))} \in P(p_{k,\beta}), \quad z \in E.$$ It can easily be seen that $$k - ST_s(\beta) \subset S_s^ \subset S_s^, \qquad \beta_1 = \frac{k + \beta}{k + 1}.$$ Also, for $\beta = 0 = k$ , the class $k - ST_s(\beta)$ reduces to $S_s^*$ . The class $k - UCV_s(\beta)$ is defined as follows.

Definition 1.2. Let $f \in A$ . Then $f \in k - UCV_s(\beta)$ if, and only if $zf' \in k - ST_s(\beta)$ for $z \in E$ . We note that $$k - UCV_s(\beta) \subset C_s(\beta_1) \subset C_s$$ , $\beta_1 = \frac{k + \beta}{k + 1}$ .

Definition 1.3. Let $f \in A$ . Then $f \in k - UK_s(\beta)$ if, and only if, there exists $g \in k - ST_s(\beta)$ such that $$\left(\frac{2zf'(z)}{g(z)-g(-z)}\right)\in P(p_{k,\beta}),\quad z\in E.$$ Since $P(p_{k,\beta}) \subset P(\beta_1) \subset P$ , $\beta_1 = \frac{k+\beta}{k+1}$ , and $k - ST_s(\beta) \subset S_s^*$ , we note that $$k - UK_s(\beta) \subset K_s \subset K$$ , where $k_S$ consists of close-to-convex functions with respect to symmetrical starlike functions <span id="page-3-0"></span>From the definition, it is clear that $k - UK_s(\beta)$ consists of univalent functions. For k = 0, $\beta = 0$ and f(z) = g(z), $k - UK_s(\beta)$ reduces to the class $S_s^*$ .