On some inequalities of a certain class of analytic functions

Lemma 1.4. ([14]) If $-1 \le B < A \le 1$ , $\lambda > 0$ and the complex number $\gamma$ satisfies $\text{Re}\{\gamma\} \ge -\lambda(1-A)/(1-B)$ , then the differential equation $$q(z)+\frac{zq'(z)}{\lambda q(z)+\gamma}=\frac{1+Az}{1+Bz},\quad z\in E,$$ has a univalent solution in E given by $$q(z) = \begin{cases} \frac{z^{\lambda+\gamma} (1+Bz)^{\lambda(A-B)/B}}{\lambda \int_0^z t^{\lambda+\gamma-1} (1+Bt)^{\lambda(A-B)/B} dt} - \frac{\gamma}{\lambda}, & B \neq 0, \\ \frac{z^{\lambda+\gamma} e^{\lambda Az}}{\lambda \int_0^z t^{\lambda+\gamma-1} e^{\lambda At} dt} - \frac{\gamma}{\lambda}, & B = 0. \end{cases}$$ If $h(z) = 1 + c_1 z + c_2 z^2 + \cdots$ is analytic in E and satisfies $$h(z)+\frac{zh'(z)}{\lambda h(z)+\gamma}\prec \frac{1+Az}{1+Bz},\quad z\in E,$$ then $$h(z) \prec q(z) \prec \frac{1+Az}{1+Bz}$$ and q(z) is the best dominant.

Lemma 1.5. ([15]) Let $\varepsilon$ be a positive measure on [0,1]. Let g be a complex-valued function defined on $E \times [0,1]$ such that $g(\cdot,t)$ is analytic in E for each $t \in [0,1]$ and $g(z,\cdot)$ is $\varepsilon$ -integrable on [0,1] for all $z \in E$ . In addition, suppose that $\operatorname{Re} g(z,t) > 0$ , g(-r,t) is real and $\operatorname{Re}\{1/g(z,t)\} \geq 1/g(-r,t)$ for $|z| \leq r < 1$ and $t \in [0,1]$ . If $g(z) = \int_0^1 g(z,t) \, d\varepsilon(t)$ , then $\operatorname{Re}\{1/g(z)\} \geq 1/g(-r)$ .

Lemma 1.6. ([16, Chapter 14]) Let $a_1$ , $b_1$ and $c_1 \neq 0, -1, -2, ...$ be complex numbers. Then, for $\text{Re } c_1 > \text{Re } b_1 > 0$ , (i) $${}_{2}F_{1}(a_{1},b_{1},c_{1};z) = \frac{\Gamma(c_{1})}{\Gamma(c_{1}-b_{1})\Gamma(b_{1})} \int_{0}^{1} t^{b_{1}-1} (1-t)^{c_{1}-b_{1}-1} (1-tz)^{-a_{1}} dt,$$ (ii) $$_2F_1(a_1,b_1,c_1;z) = _2F_1(b_1,a_1,c_1;z),$$ $$({\rm iii}) \quad {}_2F_1(a_1,b_1,c_1;z) = (1-z)^{-a_1} {}_2F_1 \left(a_1,c_1-b_1,c_1;\frac{z}{z-1}\right).$$

Lemma 1.7. ([17]) Let $-1 \le B_1 \le B_2 < A_2 \le A_1 \le 1$ . Then $$\frac{1 + A_2 z}{1 + B_2 z} \prec \frac{1 + A_1 z}{1 + B_1 z}$$

Lemma 1.8. ([18]) Let F be analytic and convex in E. If $f,g \in A_p$ and $f,g \prec F$ , then $$\mu f + (1 - \mu)g \prec F$$ , $0 \le \mu \le 1$ .

Lemma 1.9. ([19]) Let $f(z) = \sum_{k=0}^{\infty} a_k z^k$ be analytic in E and $F(z) = \sum_{k=0}^{\infty} b_k z^k$ be analytic and convex in E. If $f \prec F$ , then $$|a_k| \leq |b_1| \quad (k \in \mathbb{N}).$$

Lemma 1.10. ([20]) Let $h(z) = 1 + d_1z + d_2z^2 + \cdots$ be analytic in E and $h(z) \neq 0$ in E. If there exists a point $z_0 \in E$ such that $|\arg h(z)| < \frac{\pi}{2}\eta$ ( $|z| < |z_0|$ ) and $|\arg h(z_0)| = \frac{\pi}{2}\eta$ ( $0 < \eta \leq 1$ ), then we have $\frac{z_0h'(z_0)}{h(z_0)} = ik\eta$ , where $$\begin{cases} k \ge \frac{1}{2}(x + \frac{1}{x}), & \text{when } \arg h(z_0) = \frac{\pi}{2}\eta, \\ k \le -\frac{1}{2}(x + \frac{1}{x}), & \text{when } \arg h(z_0) = -\frac{\pi}{2}\eta, \end{cases}$$ and $(h(z_0))^{1/\eta} = \pm ix (x > 0)$ .

Lemma 1.11. Let $g \in S^[A, B]$ . Then the function* $$G(z) = \left[ \frac{c + \alpha + i\beta}{z^{c+i\beta}} \int_0^z t^{c+i\beta - 1} g^{\alpha}(t) dt \right]^{\frac{1}{\alpha}}$$ (1.5) belongs to $S^*[A,B]$ for $c \ge -\alpha \frac{1-A}{1-B}$ . Proof is straightforward by using Lemma 1.4. Throughout this paper, $\alpha \ge 0$ , $\beta \in \mathbb{R}$ , $\mu > 0$ , and $-1 \le B < A \le 1$ unless otherwise stated.

Theorem 2.1. If $f \in M_p(\alpha, \beta, \mu, A, B)$ , then $$\frac{zf'(z)}{pf(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^p}\right)^{i\beta} \prec q(z),\tag{2.1}$$ where $q(z) = \frac{\mu}{pO(z)}$ and $$Q(z) = \begin{cases} \int_0^1 t^{\frac{p}{\mu} - 1} (\frac{1 + Bzt}{1 + Bz})^{\frac{p}{\mu}(A - B)/B} dt, & B \neq 0, \\ \int_0^1 t^{\frac{p}{\mu} - 1} e^{\frac{p}{\mu}(t - 1)Az} dt, & B = 0. \end{cases}$$ (2.2) In hypergeometric function form, $$q(z) = \begin{cases} \left[ {}_{2}F_{1}(1, \frac{p}{\mu}(1 - \frac{A}{B}); \frac{p}{\mu} + 1; \frac{Bz}{Bz+1}) \right]^{-1}, & B \neq 0, \\ \left[ {}_{1}F_{1}(1, \frac{p}{\mu} + 1; -\frac{p}{\mu}Az) \right]^{-1}, & B = 0, \end{cases}$$ (2.3) and if $A < -\frac{\mu B}{p}$ , $-1 \le B < 0$ , then $M_p(\alpha, \beta, \mu, A, B) \subset B_p(\alpha, \beta, \rho)$ , where $$\rho = p \left\{ {}_{2}F_{1} \left( 1, \frac{p}{\mu} \left( 1 - \frac{A}{B} \right); \frac{p}{\mu} + 1; \frac{B}{B - 1} \right) \right\}^{-1}. \tag{2.4}$$ This result is best possible. Proof Let $$h(z) = \frac{zf'(z)}{pf(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^p}\right)^{i\beta},$$ where h(z) is analytic in E with h(0) = 1. Differentiating logarithmically, we obtain $$\frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^{p}}\right)^{i\beta} + \mu \left\{1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1)\frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - ip\beta\right\}$$ $$= ph(z) + \frac{\mu z h'(z)}{h(z)} \prec p \frac{1 + Az}{1 + Bz}.$$ (2.5) Using Lemma 1.4 for $\lambda = \frac{p}{\mu}$ and $\gamma = 0$ , we have $$h(z) \prec q(z) \prec \frac{1 + Az}{1 + Bz}$$ where q(z) is given in (2.3) and is the best dominant of (2.5). Next, in order to prove $M_p(\alpha,\beta,\mu,A,B)\subset B_p(\alpha,\beta,\rho)$ , we show that $\inf_{|z|<1}\{\operatorname{Re} q(z)\}=q(-1)$ . Now, we set $a=\frac{p}{\mu}(B-A)/B$ , $b=\frac{p}{\mu}$ and $c=\frac{p}{\mu}+1$ , then it is clear that c>b>0; therefore, for $B\neq 0$ it follows from (2.2) by using Lemma 1.6 that $$Q(z) = (1 + Bz)^{a} \int_{0}^{1} t^{b-1} (1 + Btz)^{-a} dt$$ $$= \frac{\Gamma(b)}{\Gamma(c)} {}_{2}F_{1}\left(1, a, c; \frac{Bz}{Bz+1}\right). \tag{2.6}$$ To prove that $\inf_{|z|<1} \{\operatorname{Re} q(z)\} = q(-1)$ , we need to show that $$\text{Re}\{1/Q(z)\} \ge 1/Q(-1).$$ Since $A < -\frac{\mu B}{p}$ with $-1 \le B < 0$ implies that c > a > 0, therefore, by using Lemma 1.6, (2.6) yields $$Q(z) = \int_0^1 g(z,t) \, d\varepsilon(t),$$ where $$g(z,t) = \frac{1+Bz}{1+(1-t)Bz} \quad (0 \le t \le 1),$$ $$d\varepsilon(t) = \frac{\Gamma(b)}{\Gamma(a)\Gamma(c-a)} t^{a-1} (1-t)^{c-a-1} dt,$$ which is a positive measure on [0,1]. For $-1 \le B < 0$ it is clear that $\operatorname{Re} g(z,t) > 0$ and g(-r,t) is real for $0 \le |z| \le r < 1$ and $t \in [0,1]$ . Also, $$\operatorname{Re}\left\{\frac{1}{g(z,t)}\right\} = \operatorname{Re}\left\{\frac{1 + (1-t)Bz}{1 + Bz}\right\} \ge \frac{1 - (1-t)Br}{1 - Br} = \frac{1}{g(-r,t)}$$ for $|z| \le r < 1$ . Therefore, using Lemma 1.5, we have $$\operatorname{Re}\left\{1/Q(z)\right\} \ge 1/Q(-r).$$ Now, letting $r \to 1^-$ , it follows $$\text{Re}\{1/Q(z)\} \ge 1/Q(-1).$$ Therefore, $$M_p(\alpha, \beta, \mu, A, B) \subset B_p(\alpha, \beta, \rho)$$ . For $\beta = 0$ , we have the following result proved in [12].

Corollary 2.2. If $f \in M_p(\alpha, \mu, A, B)$ , then $$\frac{zf'(z)}{pf(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \prec \begin{cases} \left[ {}_{2}F_{1}(1, \frac{p}{\mu}(1-\frac{A}{B}); \frac{p}{\mu}+1; \frac{Bz}{Bz+1}) \right]^{-1}, & B \neq 0, \\ \left[ {}_{1}F_{1}(1, \frac{p}{\mu}+1; -\frac{p}{\mu}Az) \right]^{-1}, & B = 0, \end{cases}$$ and if $A < -\frac{\mu B}{p}$ , $-1 \le B < 0$ , then $M_p(\alpha, \mu, A, B) \subset B_p(\alpha, \rho)$ , where $$\rho = p \left\{ {}_{2}F_{1}\left(1,\frac{p}{\mu}\left(1-\frac{A}{B}\right);\frac{p}{\mu}+1;\frac{B}{B-1}\right)\right\}^{-1}.$$ This result is best possible. For p = 1, we have the class $M_1(\alpha, \beta, \mu, A, B) = M(\alpha, \beta, \mu, A, B)$ . We denote the class of functions $f \in A$ , having Taylor series representation of the form $$f(z) = z + \sum_{k=n+1}^{\infty} a_k z^k$$ and satisfying the condition $$\frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z}\right)^{i\beta} + \mu \left\{ 1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1) \frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - i\beta \right\} < \frac{1 + Az}{1 + Bz}, \quad z \in E,$$ (2.7) by $M^(\alpha, \beta, \mu, A, B)$ , where $g(z) = z + \sum_{k=n+1}^{\infty} b_k z^k$ such that $\text{Re } \frac{zg'(z)}{g(z)} > 0$ . Now, we derive the following result for the class $M^(\alpha, \beta, \mu, A, B)$ .

Theorem 2.3. Let $f \in M^(\alpha, \beta, \mu, A, B)$ . Then* $$|a_{n+1}| \le \frac{(A-B) + \alpha(n+1)(1+\mu n)}{|\alpha + i\beta + n|(1+\mu n)}.$$ (2.8) Proof Since $f \in M^*(\alpha, \beta, \mu, A, B)$ , therefore, $$\begin{split} &\frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z}\right)^{i\beta} \\ &+ \mu \left\{1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1)\frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - i\beta\right\} < \frac{1 + Az}{1 + Bz}. \end{split}$$ Now, using the fact that $f(z) = z + \sum_{k=n+1}^{\infty} a_k z^k$ and $g(z) = z + \sum_{k=n+1}^{\infty} b_k z^k$ , we obtain $$1 + \left\{ (\alpha + i\beta + n)(1 + \mu n)a_{n+1} - \alpha(1 + \mu n)b_{n+1} \right\} z^n + \dots < \frac{1 + Az}{1 + Bz}.$$ By a well-known result due to Janowski and Lemma 1.9, we have $$\left| (\alpha + i\beta + n)(1 + \mu n)a_{n+1} - \alpha(1 + \mu n)b_{n+1} \right| \leq A - B.$$ By the triangle inequality, we obtain $$|(\alpha + i\beta + n)(1 + \mu n)a_{n+1}| - |\alpha(1 + \mu n)b_{n+1}| \le A - B.$$ Using the coefficient bound for the class $S^*$ , we have the required result. For $\beta = 0$ and g(z) = z, we have the following result proved in [13].

Corollary 2.4. Let $f \in M(\alpha, \mu, A, B)$ . Then $$|a_{n+1}| \leq \frac{(A-B)}{(\alpha+n)(1+\mu n)}.$$ For $\beta = 0$ , p = 1, g(z) = z, $A = 1 - 2\rho$ and B = -1, we have the following result proved in [21].

Corollary 2.5. Let f satisfy the condition $$\operatorname{Re}\left\{\frac{zf'(z)}{f(z)}\left(\frac{f(z)}{z}\right)^{\alpha} + \mu\left[1 + \frac{zf''(z)}{f'(z)} + (1 - \alpha)\left(1 - \frac{zf'(z)}{f(z)}\right)\right]\right\} > \rho.$$ Then $$|a_{n+1}| \le \frac{2(1-\rho)}{(n+\alpha)(1+\mu n)}.$$

Theorem 2.6. For $\mu_2 \ge \mu_1 \ge 0$ and $-1 \le B_1 \le B_2 < A_2 \le A_1 \le 1$ , $$M_p(\alpha, \beta, \mu_2, A_2, B_2) \subset M_p(\alpha, \beta, \mu_1, A_1, B_1).$$ Proof Let $f \in M_p(\alpha, \beta, \mu_2, A_2, B_2)$ . Then $$\begin{split} &\frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^{p}}\right)^{i\beta} \\ &+ \mu_{2} \left\{ 1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1) \frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - ip\beta \right\} \prec p \frac{1 + A_{2}z}{1 + B_{2}z}. \end{split}$$ Since $-1 \le B_1 \le B_2 < A_2 \le A_1 \le 1$ , therefore by Lemma 1.7, we have $$\frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^{p}}\right)^{i\beta} + \mu_{2} \left\{ 1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1) \frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - ip\beta \right\} \prec p \frac{1 + A_{1}z}{1 + B_{1}z}.$$ Hence, we have $f \in M_p(\alpha, \beta, \mu_2, A_1, B_1)$ . For $\mu_2 = \mu_1 \ge 0$ , we have the required result. When $\mu_2 > \mu_1 \ge 0$ , Theorem 2.1 implies that $$\frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^p}\right)^{i\beta} \prec p \frac{1 + A_1 z}{1 + B_1 z}.$$ Now $$\begin{split} & \frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^{p}}\right)^{i\beta} \\ & + \mu_{1} \left\{ 1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1) \frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - ip\beta \right\} \\ & = \left( 1 - \frac{\mu_{1}}{\mu_{2}} \right) \frac{zf'(z)}{f(z)} \left( \frac{f(z)}{g(z)} \right)^{\alpha} \left( \frac{f(z)}{z^{p}} \right)^{i\beta} + \frac{\mu_{1}}{\mu_{2}} \left\{ \frac{zf'(z)}{f(z)} \left( \frac{f(z)}{g(z)} \right)^{\alpha} \left( \frac{f(z)}{z^{p}} \right)^{i\beta} \right. \\ & + \mu_{2} \left\{ 1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1) \frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - ip\beta \right\} \right\}. \end{split}$$ Using Lemma 1.8, we have the required result. For $\beta = 0$ , we have the following result.

Corollary 2.7. For $\mu_2 \ge \mu_1 \ge 0$ and $-1 \le B_1 \le B_2 < A_2 \le A_1 \le 1$ , $$M_p(\alpha, \mu_2, A_2, B_2) \subset M_p(\alpha, \mu_1, A_1, B_1).$$ This result is proved in [13]. For $\beta = 0$ , p = 1, g(z) = z, $A = 1 - 2\rho$ and B = -1, we have the class $M(\alpha, \mu, \rho)$ defined as $$\operatorname{Re}\left[\frac{zf'(z)}{f(z)}\left(\frac{f(z)}{z}\right)^{\alpha} + \mu\left\{1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} + \alpha\left(\frac{zf'(z)}{f(z)} - 1\right)\right\}\right] > \rho \tag{2.9}$$ for $z \in E$ . Now have the following result for the class $M(\alpha, \mu, \rho)$ proved in [21].

Corollary 2.8. For $\alpha \ge 0$ , $\mu_2 \ge \mu_1 \ge 0$ and $1 > \rho_2 \ge \rho_1 \ge 0$ , $$M(\alpha, \mu_2, \rho_2) \subset M(\alpha, \mu_1, \rho_1).$$

Theorem 2.9. Let $f \in A_n$ satisfy $$\operatorname{Re}\left\{\frac{f(z)}{z^{p}}\right\} > 0 \quad and \quad \left|\frac{zf'(z)}{f(z)}\left(\frac{f(z)}{g(z)}\right)^{\alpha}\left(\frac{f(z)}{z^{p}}\right)^{i\beta} - p\right| < \sigma p, \quad 0 < \sigma \leq 1,$$ for $g \in S_p^*$ . Then f is p-valent convex in $|z| < R_{\alpha,\beta,\sigma}$ , where $$R_{\alpha,\beta,\sigma} = \left( \left( 2|1 - \alpha - i\beta| + 2\alpha p + \sigma \right) - \sqrt{\left( 2|1 - \alpha - i\beta| + 2\alpha p + \sigma \right)^2 - 4p(2\alpha p - p - \sigma)} \right) / \left( 2(2\alpha p - p - \sigma) \right). \tag{2.10}$$ Proof Let $$h(z) = \frac{zf'(z)}{pf(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^p}\right)^{i\beta} - 1,$$ where h(z) is analytic in E with h(0) = 0 and |h(z)| < 1. By using the Schwarz lemma, we get $$h(z) = \sigma z \psi(z),$$ where $\psi(z)$ is analytic in E with $|\psi(z)| < 1$ . Differentiating logarithmically, we have $$1 + \frac{zf''(z)}{f'(z)} = (1 - \alpha - i\beta)\frac{zf'(z)}{f(z)} + \alpha\frac{zg'(z)}{g(z)} + \frac{\sigma z(z\psi'(z) + \psi(z))}{1 + \sigma z\psi(z)} + ip\beta.$$ Since Re{ $\frac{f(z)}{z^p}$ } > 0, therefore, $$\frac{zf'(z)}{f(z)} = p + \frac{z\varphi'(z)}{\varphi(z)}, \quad \operatorname{Re}\varphi(z) > 0.$$ This implies that $$\operatorname{Re}\left\{1+\frac{zf''(z)}{f'(z)}\right\} \geq (1-\alpha)p + \alpha\operatorname{Re}\frac{zg'(z)}{g(z)} - |1-\alpha-i\beta| \left|\frac{z\varphi'(z)}{\varphi(z)}\right| - \sigma \left|\frac{z(z\psi'(z)+\psi(z))}{1+\sigma z\psi(z)}\right|.$$ Now, using the well-known results for classes $S_n^*$ , P and the Schwarz function [22], we have $$\operatorname{Re}\left\{1 + \frac{zf''(z)}{f'(z)}\right\} \ge (1 - \alpha)p + \alpha p \frac{1 - r}{1 + r} - |1 - \alpha - i\beta| \frac{2r}{(1 - r^2)} - \frac{\sigma r}{(1 - r)} \quad (0 < \sigma \le 1)$$ $$= \frac{(2\alpha p - p - \sigma)r^2 - (2|1 - \alpha - i\beta| + 2\alpha p + \sigma)r + p}{(1 - r^2)}.$$ Let $P(r) = (2\alpha p - p - \sigma)r^2 - (2|1 - \alpha - i\beta| + 2\alpha p + \sigma)r + p$ . Since $p \in \mathbb{N}$ and $0 < \sigma \le 1$ , therefore, P(0) = p > 0 and $P(1) = -2(|1 - \alpha - i\beta| + \sigma) < 0$ . It follows that the root lies in (0,1). This implies that $\text{Re}\{1 + \frac{zf''(z)}{f'(z)}\} > 0$ if $r < R_{\alpha,\beta,\sigma}$ , where $R_{\alpha,\beta,\sigma}$ is given by (2.10). For $\sigma = 1$ and $\beta = 0$ , we have the following result which is proved in [12].

Corollary 2.10. Let $f \in A_p$ satisfy $$\operatorname{Re}\left\{\frac{f(z)}{z^p}\right\} > 0 \quad and \quad \left|\frac{zf'(z)}{f(z)}\left(\frac{f(z)}{g(z)}\right)^{\alpha} - p\right| < p,$$ $g \in S_n^*$ . Then f is p-valent convex in $|z| < R_\alpha$ , where $$R_{\alpha} = \frac{3 + 2\alpha(p-1) - \sqrt{(3 + 2\alpha(p-1))^2 - 4p(2\alpha p - p - 1)}}{2(2\alpha p - p - 1)}.$$

Theorem 2.11. Let $f \in A_p$ satisfy $$\left| \frac{zf'(z)}{f(z)} \left( \frac{f(z)}{g(z)} \right)^{\alpha} - p \right| < \sigma p,$$ for $g \in S_p^*$ . Then, for $\alpha > 0$ , f is p-valent $\frac{1}{\alpha}$ -convex in $|z| < R_{\alpha,\sigma}$ , where $$R_{\alpha,\sigma} = \frac{2\alpha p + \sigma - \sqrt{(2\alpha p + \sigma)^2 - 4\alpha p(\alpha p - \sigma)}}{2(\alpha p - \sigma)}.$$ (2.11) Proof Let $$h(z) = \frac{zf'(z)}{pf(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} - 1,$$ where h(z) is analytic in E with h(0) = 0 and |h(z)| < 1. By using the Schwarz lemma, we get $$h(z) = \sigma z \psi(z),$$ where $\psi(z)$ is analytic in E with $|\psi(z)| < 1$ . Differentiating logarithmically, we have $$\frac{1}{\alpha}\left(1+\frac{zf''(z)}{f'(z)}\right)+\left(1-\frac{1}{\alpha}\right)\frac{zf'(z)}{f(z)}=\frac{zg'(z)}{g(z)}+\frac{\sigma}{\alpha}\frac{z(z\psi'(z)+\psi(z))}{1+\sigma z\psi(z)}.$$ This implies that $$\operatorname{Re}\left\{\frac{1}{\alpha}\left(1+\frac{zf''(z)}{f'(z)}\right)+\left(1-\frac{1}{\alpha}\right)\frac{zf'(z)}{f(z)}\right\} \geq \operatorname{Re}\left(\frac{zg'(z)}{g(z)}-\frac{\sigma}{\alpha}\right|\frac{(z\psi'(z)+\psi(z))}{1+\sigma z\psi(z)}\right|.$$ Now, using the well-known results for classes $S_p^*$ , and the Schwarz function, we have $$\operatorname{Re} \frac{1}{p} \left\{ \frac{1}{\alpha} \left( 1 + \frac{zf''(z)}{f'(z)} \right) + \left( 1 - \frac{1}{\alpha} \right) \frac{zf'(z)}{f(z)} \right\} \ge \frac{1 - r}{1 + r} - \frac{\sigma r}{\alpha p (1 - r)} \quad (0 < \sigma \le 1)$$ $$= \frac{(\alpha p - \sigma)r^2 - (2\alpha p + \sigma)r + \alpha p}{\alpha (1 - r^2)}.$$ Let $Q(r)=(\alpha p-\sigma)r^2-(2\alpha p+\sigma)r+\alpha p$ . Then for $p\in\mathbb{N}$ , $\alpha>0$ and $0<\sigma\leq 1$ , $Q(0)=\alpha p>0$ and $Q(1)=-2\sigma<0$ . It shows that the root lies in (0,1). This implies that $\operatorname{Re}\{\frac{1}{\alpha}(1+\frac{zf''(z)}{f'(z)})+(1-\frac{1}{\alpha})\frac{zf'(z)}{f(z)}\}>0$ if $r< R_{\alpha,\sigma}$ , where $R_{\alpha,\sigma}$ is given by (2.11).

Theorem 2.12. Let $f \in M(\alpha, \beta, \mu, A, B)$ . Then E is mapped by f on a domain that contains the disc $|w| < R_{\alpha,\beta,\mu} \in (0,1)$ , where $$R_{\alpha,\beta,\mu} = \frac{|\alpha + i\beta + 1|(1+\mu)}{2|\alpha + i\beta + 1|(1+\mu) + (A-B) + 2\alpha(1+\mu)}.$$ (2.12) Proof Let $w_0$ be any complex number such that $f(z) \neq w_0$ . Then $$\frac{w_0 f(z)}{w_0 - f(z)} = z + \left(a_2 + \frac{1}{w_0}\right) z^2 + \cdots,$$ is univalent in E, so that $$\left|a_2 + \frac{1}{w_0}\right| \le 2.$$ Therefore, $$\left|\frac{1}{w_0}\right| - |a_2| \le 2.$$ Hence, $$w_0 \ge \frac{|\alpha + i\beta + 1|(1 + \mu)}{2|\alpha + i\beta + 1|(1 + \mu) + (A - B) + 2\alpha(1 + \mu)} = R_{\alpha,\beta,\mu}.$$ For $\beta = 0$ , g(z) = z, $A = 1 - 2\rho_1$ , B = -1, we have the following result proved in [13].

Corollary 2.13. Let $f \in M(\alpha, \mu, \rho)$ . Then E is mapped by f on a domain that contains the disc $|w| < R_{\alpha,\mu}$ , where $$R_{\alpha,\beta,\mu} = \frac{(1+\alpha)(1+\mu)}{2(1+\alpha)(1+\mu) + 2(1-\rho_1)}.$$

Theorem 2.14. Let $\alpha > 0$ , $c \ge -\alpha \frac{1-A}{1-B}$ and let $f \in A$ . If $$\left|\arg \frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z}\right)^{i\beta} - \rho \right| < \frac{\pi}{2} \upsilon \quad (0 < \upsilon \le 1, 0 \le \rho < 1), \tag{2.13}$$ for some $g \in S$ , then* $$\left|\arg\frac{zF'(z)}{F(z)}\left(\frac{F(z)}{G(z)}\right)^{\alpha}\left(\frac{F(z)}{z}\right)^{i\beta} - \rho\right| < \frac{\pi}{2}\eta \quad (0 < \eta \le 1), \tag{2.14}$$ where $$F(z) = \left[ \frac{c + \alpha + i\beta}{z^c} \int_0^z t^{c-1} f^{\alpha + i\beta}(t) dt \right]^{\frac{1}{\alpha + i\beta}}, \tag{2.15}$$ with $$\upsilon = \begin{cases} \eta + \frac{2}{\pi} \tan^{-1} \left( \frac{(1+B)\eta \sin(\pi(1-t(\alpha,\beta,c,A,B))/2)}{|c+i\beta|(1+B)+\alpha(1+A)+\eta(1+B)\cos(\pi(1-t(\alpha,\beta,c,A,B))/2)} \right), & B \neq -1, \\ \eta, & B = -1, \end{cases}$$ (2.16) and $$t(\alpha, \beta, c, A, B) = \frac{2}{\pi} \sin^{-1} \frac{\alpha(A - B) + \beta(1 - B^2)}{|c + i\beta|(1 - B^2) + \alpha(1 - AB)}.$$ (2.17) Proof Since $$F^{\alpha+i\beta}(z) = \frac{c+\alpha+i\beta}{z^c} \int_0^z t^{c-1} f^{\alpha+i\beta}(t) dt,$$ therefore, $$z^{1-i\beta}F^{\alpha+i\beta-1}(z)F'(z) = \frac{1}{\alpha+i\beta} \left\{ (c+\alpha+i\beta)z^{-i\beta}f^{\alpha+i\beta}(z) - cz^{-i\beta}F^{\alpha+i\beta}(z) \right\}.$$ Now using (1.5), we have $$\frac{zF'(z)}{F(z)} \left(\frac{F(z)}{G(z)}\right)^{\alpha} \left(\frac{F(z)}{z}\right)^{i\beta} = \frac{\frac{1}{\alpha+i\beta} \left\{ (c+\alpha+i\beta)z^c f^{\alpha+i\beta}(z) - cz^c F^{\alpha+i\beta}(z) \right\}}{(c+\alpha+i\beta)\int_0^z t^{c+i\beta-1} g^{\alpha}(t) \, dt}.$$ Let $$h(z) = \frac{1}{1 - \rho} \left( \frac{zF'(z)}{F(z)} \left( \frac{F(z)}{G(z)} \right)^{\alpha} \left( \frac{F(z)}{z} \right)^{i\beta} - \rho \right) = \frac{N(z)}{D(z)},$$ where h(z) is analytic with h(0) = 1. Now $$\frac{N'(z)}{D'(z)} = \frac{1}{1-\rho} \left( \frac{zf'(z)}{f(z)} \left( \frac{f(z)}{g(z)} \right)^{\alpha} \left( \frac{f(z)}{z} \right)^{i\beta} - \rho \right)$$ $$= h(z) \left\{ 1 + \frac{D(z)}{zD'(z)} \cdot \frac{zh'(z)}{h(z)} \right\},$$ where $$\begin{split} N(z) &= \frac{1}{1-\rho} \left( \frac{1}{\alpha+i\beta} \left\{ (c+\alpha+i\beta) z^c f^{\alpha+i\beta}(z) - c z^c F^{\alpha+i\beta}(z) \right\} - \rho z^{c+i\beta-1} G^{\alpha}(z) \right), \\ D(z) &= (c+\alpha+i\beta) \int_0^z t^{c+i\beta-1} g^{\alpha}(t) \, dt. \end{split}$$ Since $G \in S^*[A, B]$ , therefore, we can write $$\frac{zD'(z)}{D(z)} = c + i\beta + \alpha \frac{zG'(z)}{G(z)} = r_1 e^{i\frac{\pi}{2}\theta},$$ where $$\begin{cases} |c + i\beta| + \alpha \frac{1-A}{1-B} < r_1 < |c + i\beta| + \alpha \frac{1+A}{1+B}, & B \neq -1, \\ -t(\alpha, \beta, c, A, B) < \theta < t(\alpha, \beta, c, A, B), & B \neq -1, \end{cases}$$ (2.18) similarly $$\begin{cases} |c + i\beta| + \alpha \frac{1-A}{2} < r_1 < \infty, & B = -1, \\ -1 < \theta < 1, & B = -1. \end{cases}$$ (2.19) Suppose that $h(z) \neq 0$ in E, there exists a point $z_0 \in E$ such that $|\arg h(z)| < \frac{\pi}{2}\eta$ ( $|z| < |z_0|$ ) and $|\arg h(z_0)| = \frac{\pi}{2}\eta$ . Now, using Lemma 1.10, we have $\frac{zh'(z_0)}{h(z_0)} = ik\eta$ . At first suppose that $h(z_0) = (ix)^{\eta}$ (x > 0) for the case $B \neq -1$ , we obtain $$\arg\left(\frac{z_{0}f'(z_{0})}{f(z_{0})}\left(\frac{f(z_{0})}{g(z_{0})}\right)^{\alpha}\left(\frac{f(z_{0})}{z_{0}}\right)^{i\beta} - \rho\right)$$ $$= \arg h(z_{0}) + \arg\left(1 + \frac{1}{c + i\beta + \alpha \frac{zG'(z_{0})}{G(z_{0})}} \cdot \frac{z_{0}h'(z_{0})}{h(z_{0})}\right)$$ $$= \frac{\pi}{2}\eta + \arg\left(1 + \left(r_{1}e^{i\frac{\pi}{2}\theta}\right)^{-1}ik\eta\right)$$ $$= \frac{\pi}{2}\eta + \tan^{-1}\left(\frac{k\eta\sin\pi(1 - \theta)/2}{r_{1} + k\eta\cos\pi(1 - \theta)/2}\right)$$ $$\geq \frac{\pi}{2}\eta + \tan^{-1}\left(\frac{\eta\sin\pi(1 - t(\alpha, \beta, c, A, B))/2}{|c + i\beta| + \alpha \frac{1+A}{1+B}} + \eta\cos\pi(1 - t(\alpha, \beta, c, A, B))/2\right)$$ $$= \frac{\pi}{2}\upsilon,$$ where v and $t(\alpha, \beta, c, A, B)$ are given by (2.16) and (2.17) respectively. For B = -1, we have $$\arg \left(\frac{z_0f'(z_0)}{f(z_0)}\left(\frac{f(z_0)}{g(z_0)}\right)^{\alpha}\left(\frac{f(z_0)}{z_0}\right)^{i\beta}-\rho\right) \geq \frac{\pi}{2}\eta,$$ which is a contradiction to the assumption of our theorem. Now we suppose that $h(z_0) = (-ix)^{\eta}$ . For the case $B \neq -1$ using a similar method, we obtain $$\arg\left(\frac{z_0f'(z_0)}{f(z_0)}\left(\frac{f(z_0)}{g(z_0)}\right)^{\alpha}\left(\frac{f(z_0)}{z_0}\right)^{i\beta}-\rho\right)\leq -\frac{\pi}{2}\upsilon,$$ and for B = -1, we have $$\arg\left(\frac{z_0f'(z_0)}{f(z_0)}\left(\frac{f(z_0)}{g(z_0)}\right)^{\alpha}\left(\frac{f(z_0)}{z_0}\right)^{i\beta}-\rho\right)\leq -\frac{\pi}{2}\eta,$$ which is a contradiction to the assumption of our theorem. Hence, we have the proof. This kind of problem is also considered in [23]. For $\beta = 0$ , we have the following result proved in [24].

Corollary 2.15. Let $\alpha > 0$ , $c \ge -\alpha \frac{1-A}{1-B}$ and let $f \in A$ . If $$\left|\arg\left(\frac{zf'(z)}{f(z)}\left(\frac{f(z)}{g(z)}\right)^{\alpha}-\rho\right)\right|<\frac{\pi}{2}\upsilon\quad(0<\upsilon\leq1),$$ then $$\left| \arg \left( \frac{zF'(z)}{F(z)} \left( \frac{F(z)}{G(z)} \right)^{\alpha} - \rho \right) \right| < \frac{\pi}{2} \eta \quad (0 < \eta \le 1),$$ where $$F(z) = \left[\frac{c + \alpha}{z^c} \int_0^z t^{c-1} f^{\alpha}(t) dt\right]^{\frac{1}{\alpha}},$$ with $$\upsilon = \begin{cases} \eta + \frac{2}{\pi} \tan^{-1} \left( \frac{(1+B)\eta \sin(\pi(1-t(\alpha,\beta,c,A,B))/2)}{c(1+B)+\alpha(1+A)+\eta(1+B)\cos(\pi(1-t(\alpha,\beta,c,A,B))/2)} \right), & B \neq -1, \\ \eta, & B = -1, \end{cases}$$ and $$t(\alpha, \beta, c, A, B) = \frac{2}{\pi} \sin^{-1} \frac{\alpha(A - B)}{c(1 - B^2) + \alpha(1 - AB)}.$$ Remark 2.16 By using the suitable choices of parameters c, $\alpha$ , A and B, we can find many results proved in the literature.

Definition 1.1. A function $f \in M_p(\alpha, \beta, \mu, A, B)$ if it satisfies the condition $$\begin{split} &\frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^{p}}\right)^{i\beta} \\ &+ \mu \left\{ 1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1) \frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - ip\beta \right\} \prec p \frac{1 + Az}{1 + Bz}, \quad z \in E, \end{split}$$ where $\alpha \ge 0$ , $\mu > 0$ , $g \in S_p^*$ , $-1 \le B < A \le 1$ and $\beta$ is any real. We have the following special cases. - (i) For $\beta = 0$ , we have the subclass of Bazilevic functions defined by Patel [12]. - (ii) For $\beta = 0$ , p = 1, g(z) = z, $A = 1 2\rho$ , B = -1, we obtain the subclass of Bazilevic functions defined in [13]. For $A = 1 - 2\rho$ , B = -1, we have the following subclass of analytic functions.

Definition 1.2. A function $f \in B_p(\alpha, \beta, \rho)$ if it satisfies the condition $$\frac{zf'(z)}{f(z)} \left(\frac{f(z)}{g(z)}\right)^{\alpha} \left(\frac{f(z)}{z^{p}}\right)^{i\beta} + \mu \left\{1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1)\frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - ip\beta\right\} \prec p \frac{1 + (1 - 2\rho)z}{1 - z},$$ where $\alpha \ge 0$ , $\mu > 0$ , $g \in S_p^*$ , $0 \le \rho < 1$ , $\beta$ is any real and $z \in E$ . In other words, a function $f \in B_p(\alpha, \beta, \rho)$ if it satisfies the condition $$\operatorname{Re} \frac{1}{p} \left[ \frac{zf'(z)}{f(z)} \left( \frac{f(z)}{g(z)} \right)^{\alpha} \left( \frac{f(z)}{z^{p}} \right)^{i\beta} + \mu \left\{ 1 + \frac{zf''(z)}{f'(z)} + (\alpha + i\beta - 1) \frac{zf'(z)}{f(z)} - \alpha \frac{zg'(z)}{g(z)} - ip\beta \right\} \right] > \rho, \quad z \in E.$$ We need the following definition and lemmas which will be used in our main results.

Definition 1.3. Let $\Psi: \mathbb{C}^2 \times E \to \mathbb{C}$ be analytic in a domain D and h be univalent in E. If p is analytic in E with $(p(z), zp'(z); z) \in D$ when $z \in E$ , then we say that p satisfies a first-order differential subordination if $$\Psi(p(z), zp'(z); z) \prec h(z), \quad z \in E. \tag{1.4}$$ The univalent function q is called dominant of the differential subordination (1.4) if $p \prec q$ for all p satisfies (1.4). If $\tilde{q} \prec q$ for all dominants of (1.4), then we say that $\tilde{q}$ is the best dominant of (1.4).