Tang et al. Journal of Inequalities and Applications 2013, 2013:317

Theorem 2.1. Let the function f given by (1.1) be in the class $H^{\mu}_{\sigma}(\lambda, \varphi)$ , $\lambda \geq 1$ and $\mu \geq 0$ . Then <span id="page-2-2"></span> $$|a_2| \le \min\left\{\frac{B_1}{\lambda + \mu}, \sqrt{\frac{2(B_1 + |B_2 - B_1|)}{(1 + \mu)(2\lambda + \mu)}}\right\} \tag{2.5}$$ and <span id="page-2-0"></span> $$|a_3| \le \begin{cases} \min\{\frac{B_1}{2\lambda + \mu} + \frac{B_1^2}{(\lambda + \mu)^2}, \frac{2(B_1 + |B_2 - B_1|)}{(1 + \mu)(2\lambda + \mu)}\}, & 0 \le \mu < 1, \\ \frac{B_1}{2\lambda + \mu} + \frac{2|B_2 - B_1|}{(1 + \mu)(2\lambda + \mu)}, & \mu \ge 1. \end{cases}$$ $$(2.6)$$ Proof Since $f \in H^{\mu}_{\sigma}(\lambda, \varphi)$ , there exist two analytic functions $u, v : D \to D$ , with u(0) = v(0) = 0, such that $$(1-\lambda)\left(\frac{f(z)}{z}\right)^{\mu} + \lambda f'(z)\left(\frac{f(z)}{z}\right)^{\mu-1} = \varphi(u(z))$$ (2.7) and <span id="page-3-5"></span><span id="page-3-0"></span> $$(1-\lambda)\left(\frac{g(w)}{w}\right)^{\mu} + \lambda g'(w)\left(\frac{g(w)}{z}\right)^{\mu-1} = \varphi(\nu(w)). \tag{2.8}$$ Define the functions p and q by <span id="page-3-1"></span> $$p(z) = \frac{1 + u(z)}{1 - u(z)} = 1 + p_1 z + p_2 z^2 + \cdots \quad \text{and}$$ $$q(z) = \frac{1 + v(z)}{1 - v(z)} = 1 + q_1 z + q_2 z^2 + \cdots,$$ (2.9) or, equivalently, <span id="page-3-2"></span> $$u(z) = \frac{p(z) - 1}{p(z) + 1} = \frac{1}{2} \left( p_1 z + \left( p_2 - \frac{p_1^2}{2} \right) z^2 + \cdots \right)$$ (2.10) and <span id="page-3-3"></span> $$\nu(z) = \frac{q(z) - 1}{q(z) + 1} = \frac{1}{2} \left( q_1 z + \left( q_2 - \frac{q_1^2}{2} \right) z^2 + \cdots \right). \tag{2.11}$$ It is clear that p and q are analytic in D and p(0) = q(0) = 1. Since $u, v : D \to D$ , the functions p and q have positive real part in D, and hence $|p_i| \le 2$ and $|q_i| \le 2$ (i = 1, 2, ...). By virtue of (2.7), (2.8), (2.10) and (2.11), we have $$(1-\lambda)\left(\frac{f(z)}{z}\right)^{\mu} + \lambda f'(z)\left(\frac{f(z)}{z}\right)^{\mu-1} = \varphi\left(\frac{p(z)-1}{p(z)+1}\right) \tag{2.12}$$ and <span id="page-3-6"></span> $$(1 - \lambda) \left(\frac{g(w)}{w}\right)^{\mu} + \lambda g'(w) \left(\frac{g(w)}{z}\right)^{\mu - 1} = \varphi\left(\frac{q(w) - 1}{g(w) + 1}\right). \tag{2.13}$$ Using (2.10), (2.11), together with (2.1), we easily obtain <span id="page-3-4"></span> $$\varphi\left(\frac{p(z)-1}{p(z)+1}\right) = 1 + \frac{1}{2}B_1p_1z + \left(\frac{1}{2}B_1\left(p_2 - \frac{1}{2}p_1^2\right) + \frac{1}{4}B_2p_1^2\right)z^2 + \cdots$$ (2.14) and $$\varphi\left(\frac{q(w)-1}{q(w)+1}\right) = 1 + \frac{1}{2}B_1q_1w + \left(\frac{1}{2}B_1\left(q_2 - \frac{1}{2}q_1^2\right) + \frac{1}{4}B_2q_1^2\right)w^2 + \cdots$$ (2.15) Since $f \in \sigma$ has the Maclaurin series given by (1.1), a computation shows that its inverse $g = f^{-1}$ has the expansion given by (1.2). Also, since $$f'(z) = 1 + 2a_2z + 3a_3z^2 + \cdots$$ and $g'(w) = 1 - 2a_2w + 3(2a_2 - a_3)w^2 - \cdots$ it follows from (2.12)-(2.15) that <span id="page-4-2"></span><span id="page-4-1"></span><span id="page-4-0"></span> $$(\lambda + \mu)a_2 = \frac{1}{2}B_1p_1,\tag{2.16}$$ <span id="page-4-3"></span> $$(2\lambda + \mu)a_3 + \frac{(\mu - 1)(2\lambda + \mu)}{2}a_2^2 = \frac{1}{2}B_1\left(p_2 - \frac{1}{2}p_1^2\right) + \frac{1}{4}B_2p_1^2,\tag{2.17}$$ $$-(\lambda + \mu)a_2 = \frac{1}{2}B_1q_1 \tag{2.18}$$ and <span id="page-4-6"></span><span id="page-4-4"></span> $$-(2\lambda + \mu)a_3 + \frac{(3+\mu)(2\lambda + \mu)}{2}a_2^2 = \frac{1}{2}B_1\left(q_2 - \frac{1}{2}q_1^2\right) + \frac{1}{4}B_2q_1^2. \tag{2.19}$$ From (2.16) and (2.18), we get $$p_1 = -q_1 (2.20)$$ and <span id="page-4-5"></span> $$8(\lambda + \mu)^2 a_2^2 = B_1^2 (p_1^2 + q_1^2). \tag{2.21}$$ Also, from (2.17) and (2.19), we obtain $$(1+\mu)(2\lambda+\mu)a_2^2=\frac{1}{2}B_1(p_2+q_2)+\frac{1}{4}(B_2-B_1)\big(p_1^2+q_1^2\big),$$ or $$a_2^2 = \frac{2B_1(p_2 + q_2) + (B_2 - B_1)(p_1^2 + q_1^2)}{4(1 + \mu)(2\lambda + \mu)}.$$ (2.22) Since $|p_i| \le 2$ and $|q_i| \le 2$ (i = 1, 2), it follows from (2.21) and (2.22) that $$|a_2| \le \frac{B_1}{\lambda + \mu} \tag{2.23}$$ and <span id="page-4-7"></span> $$|a_2| \le \sqrt{\frac{2(B_1 + |B_2 - B_1|)}{(1 + \mu)(2\lambda + \mu)}},\tag{2.24}$$ which yields the desired estimate on $|a_2|$ as asserted in (2.5). Next, in order to find the bound on $|a_3|$ , by subtracting (2.19) from (2.17), we get $$2(2\lambda + \mu)(a_3 - a_2^2) = \frac{1}{2}B_1(p_2 - q_2) + \frac{1}{4}(B_2 - B_1)(p_1^2 - q_1^2). \tag{2.25}$$ Using (2.20) and (2.21) in (2.25), we have $$a_3 = \frac{1}{4(2\lambda + \mu)} B_1(p_2 - q_2) + \frac{1}{4(\lambda + \mu)^2} B_1^2 p_1^2,$$ which evidently yields <span id="page-5-2"></span><span id="page-5-0"></span> $$|a_3| \le \frac{B_1}{2\lambda + \mu} + \frac{B_1^2}{(\lambda + \mu)^2}. (2.26)$$ On the other hand, by using (2.20) and (2.22) in (2.25), we obtain <span id="page-5-1"></span> $$a_3 = \frac{B_1[(\mu+3)p_2 + (1-\mu)q_2] + (B_2 - B_1)(p_1^2 + q_1^2)}{4(1+\mu)(2\lambda + \mu)},$$ (2.27) and applying $|p_i| \le 2$ and $|q_i| \le 2$ (i = 1, 2) for (2.27), we get <span id="page-5-3"></span> $$|a_3| \le \frac{B_1}{2(2\lambda + \mu)} \left\lceil \frac{\mu + 3}{1 + \mu} + \frac{|1 - \mu|}{1 + \mu} \right\rceil + \frac{2|B_2 - B_1|}{(1 + \mu)(2\lambda + \mu)}.$$ (2.28) Now, we consider the bounds on $|a_3|$ according to $\mu$ . Case 1. If $0 \le \mu < 1$ , then from (2.28) <span id="page-5-4"></span> $$|a_3| \le \frac{2(B_1 + |B_2 - B_1|)}{(1 + \mu)(2\lambda + \mu)}. (2.29)$$ Case 2. If $\mu \ge 1$ , then from (2.28) $$|a_3| \le \frac{B_1}{2\lambda + \mu} + \frac{2|B_2 - B_1|}{(1 + \mu)(2\lambda + \mu)}. (2.30)$$ <span id="page-5-5"></span>Thus, from (2.26), (2.29) and (2.30), we obtain the desired estimate on $|a_3|$ given in (2.6). This completes the proof of Theorem 2.1. Putting $\mu = 1$ and $\lambda = \mu = 1$ in Theorem 2.1, we respectively get the following Corollaries 2.1 and 2.2.

Corollary 2.1. If $f \in H_{\sigma}(\lambda, \varphi)$ $(\lambda \ge 1)$ , then $$|a_2| \le \min \left\{ \frac{B_1}{\lambda + 1}, \sqrt{\frac{B_1 + |B_2 - B_1|}{2\lambda + 1}} \right\}$$ <span id="page-5-6"></span>and $$|a_3| \leq \begin{cases} \min\{\frac{B_1}{2\lambda+1} + \frac{B_1^2}{(\lambda+1)^2}, \frac{B_1+|B_2-B_1|}{2\lambda+1}\}, & 0 \leq \mu < 1, \\ \frac{B_1+|B_2-B_1|}{2\lambda+1}, & \mu \geq 1. \end{cases}$$

Corollary 2.2. If $f \in H_{\sigma}(\varphi)$ , then $$|a_2| \le \min\left\{\frac{B_1}{2}, \sqrt{\frac{B_1 + |B_2 - B_1|}{3}}\right\}$$ and $$|a_3| \leq \begin{cases} \min\{\frac{B_1}{3} + \frac{B_1^2}{4}, \frac{B_1 + |B_2 - B_1|}{3}\}, & 0 \leq \mu < 1, \\ \frac{B_1 + |B_2 - B_1|}{3}, & \mu \geq 1. \end{cases}$$ Remark 2.1 The estimates of the coefficients $|a_2|$ and $|a_3|$ of Corollaries 2.1 and 2.2 are the improvement of the estimates obtained in [14, Theorem 2.1] and [13, Theorem 2.1], respectively.

Theorem 2.2. Let $\gamma \in C^*$ , $\lambda \geq 0$ and $\mu \geq 0$ . If $f \in M_{\sigma}^{\gamma}(\lambda, \mu, \varphi)$ , then <span id="page-6-2"></span> $$|a_{2}| \leq \frac{|\gamma|B_{1}\sqrt{B_{1}}}{\sqrt{|[2(6\lambda\mu + 2\lambda - 2\mu + 1) - (2\lambda\mu + \lambda - \mu + 1)^{2}]B_{1}^{2}\gamma + 2(2\lambda\mu + \lambda - \mu + 1)^{2}(B_{1} - B_{2})|}}$$ (2.31) and <span id="page-6-0"></span> $$|a_3| \le \frac{|\gamma|(B_1 + |B_2 - B_1|)}{|2(6\lambda\mu + 2\lambda - 2\mu + 1) - (2\lambda\mu + \lambda - \mu + 1)^2|}.$$ (2.32) Proof If $f \in M_{\sigma}^{\gamma}(\lambda, \mu, \varphi)$ , then there are analytic functions $u, v : D \to D$ , with u(0) = v(0) = 00, satisfying $$1 + \frac{1}{\gamma} \left( \frac{zf'(z) + (2\lambda\mu + \lambda - \mu)z^2 f''(z) + \lambda\mu z^3 f'''(z)}{(1 - \lambda + \mu)f(z) + (\lambda - \mu)z f'(z) + \lambda\mu z^2 f''(z)} - 1 \right) = \varphi(u(z))$$ (2.33) and <span id="page-7-1"></span><span id="page-7-0"></span> $$1 + \frac{1}{\gamma} \left( \frac{wg'(w) + (2\lambda\mu + \lambda - \mu)w^2 g''(w) + \lambda\mu w^3 g'''(w)}{(1 - \lambda + \mu)g(w) + (\lambda - \mu)wg'(w) + \lambda\mu w^2 g''(w)} - 1 \right) = \varphi(\nu(w)). \tag{2.34}$$ Let p and q be defined as in (2.8), then it is clear from (2.33), (2.34), (2.9) and (2.10) that <span id="page-7-2"></span> $$1 + \frac{1}{\gamma} \left( \frac{zf'(z) + (2\lambda\mu + \lambda - \mu)z^2 f''(z) + \lambda\mu z^3 f'''(z)}{(1 - \lambda + \mu)f(z) + (\lambda - \mu)z f'(z) + \lambda\mu z^2 f''(z)} - 1 \right)$$ $$= \varphi \left( \frac{p(z) - 1}{p(z) + 1} \right)$$ (2.35) and <span id="page-7-3"></span> $$1 + \frac{1}{\gamma} \left( \frac{wg'(w) + (2\lambda\mu + \lambda - \mu)w^2 g''(w) + \lambda\mu w^3 g'''(w)}{(1 - \lambda + \mu)g(w) + (\lambda - \mu)wg'(w) + \lambda\mu w^2 g''(w)} - 1 \right)$$ $$= \varphi \left( \frac{q(w) - 1}{q(w) + 1} \right). \tag{2.36}$$ It follows from (2.35), (2.36), (2.14) and (2.15) that $$(2\lambda\mu + \lambda - \mu + 1)a_2 = \frac{1}{2}B_1p_1\gamma, \tag{2.37}$$ $$-(2\lambda\mu + \lambda - \mu + 1)^2a_2^2 + 2(6\lambda\mu + 2\lambda - 2\mu + 1)a_3$$ <span id="page-7-6"></span><span id="page-7-5"></span><span id="page-7-4"></span> $$= \gamma \left[ \frac{1}{2} B_1 \left( p_2 - \frac{1}{2} p_1^2 \right) + \frac{1}{4} B_2 p_1^2 \right], \tag{2.38}$$ $$-(2\lambda\mu + \lambda - \mu + 1)a_2 = \frac{1}{2}B_1q_1\gamma \tag{2.39}$$ and <span id="page-7-8"></span><span id="page-7-7"></span> $$\left[4(6\lambda\mu + 2\lambda - 2\mu + 1) - (2\lambda\mu + \lambda - \mu + 1)^{2}\right]a_{2}^{2} - 2(6\lambda\mu + 2\lambda - 2\mu + 1)a_{3}$$ $$= \gamma \left[\frac{1}{2}B_{1}\left(q_{2} - \frac{1}{2}q_{1}^{2}\right) + \frac{1}{4}B_{2}q_{1}^{2}\right].$$ (2.40) Equations (2.37) and (2.39) yield $$p_1 = -q_1 (2.41)$$ and $$8(2\lambda\mu + \lambda - \mu + 1)^2 a_2^2 = B_1^2 \gamma^2 (p_1^2 + q_1^2). \tag{2.42}$$ From (2.38), (2.40), (2.41) and (2.42), it follows that $$a_2^2 = \frac{\gamma^2 B_1^3 (p_2 + q_2)}{4[(2(6\lambda\mu + 2\lambda - 2\mu + 1) - (2\lambda\mu + \lambda - \mu + 1)^2)B_1^2\gamma + (2\lambda\mu + \lambda - \mu + 1)^2(B_1 - B_2)]}$$ which yields the desired estimate on $|a_2|$ as described in (2.31). Similarly, it can be obtained from (2.38), (2.40) and (2.41) that <span id="page-8-0"></span> $$a_3 = \frac{\gamma B_1[p_2(4(6\lambda\mu + 2\lambda - 2\mu + 1) - (2\lambda\mu + \lambda - \mu + 1)^2) + q_2(2\lambda\mu + \lambda - \mu + 1)^2]}{8[2(6\lambda\mu + 2\lambda - 2\mu + 1) - (2\lambda\mu + \lambda - \mu + 1)^2](6\lambda\mu + 2\lambda - 2\mu + 1)} + \frac{2\gamma (B_2 - B_1)(6\lambda\mu + 2\lambda - 2\mu + 1)p_1^2}{8[2(6\lambda\mu + 2\lambda - 2\mu + 1) - (2\lambda\mu + \lambda - \mu + 1)^2](6\lambda\mu + 2\lambda - 2\mu + 1)}$$ which easily leads to the desired estimate (2.32) on $|a_3|$ . Taking $\mu = 0$ in Theorem 2.2, we obtain the following corollary.

Corollary 2.3. [14, Theorem 2.3] If $f \in N_{\sigma,\nu}^{\lambda}(\varphi)$ , then <span id="page-8-1"></span> $$|a_2| \leq \frac{|\gamma|B_1\sqrt{B_1}}{\sqrt{|(1+2\lambda-\lambda^2)B_1^2\gamma + (1+\lambda)^2(B_1-B_2)|}} \quad and \quad |a_3| \leq \frac{|\gamma|(B_1+|B_2-B_1|)}{|1+2\lambda-\lambda^2|}.$$ Further, for $\gamma = 1$ , putting $\lambda = 0$ and $\lambda = 1$ in Corollary 2.3, respectively, we have the following Corollaries 2.4 and 2.5.

Corollary 2.4. [13, Corollary 2.1] If $f \in M^1_{\sigma}(0,0,\varphi) = ST_{\sigma}(\varphi)$ , then <span id="page-8-2"></span> $$|a_2| \le \frac{B_1\sqrt{B_1}}{\sqrt{|B_1^2 + B_1 - B_2|}}$$ and $|a_3| \le B_1 + |B_2 - B_1|$ .

Corollary 2.5. [13, Corollary 2.2] If $f \in M^1_{\sigma}(1,0,\varphi) = CV_{\sigma}(\varphi)$ , then $$|a_2| \le \frac{B_1\sqrt{B_1}}{\sqrt{2|B_1^2 + 2B_1 - 2B_2|}}$$ and $|a_3| \le \frac{1}{2}(B_1 + |B_2 - B_1|).$ Remark 2.3 If we set $$\varphi(z) = \left(\frac{1+z}{1-z}\right)^{\alpha} = 1 + 2\alpha z + 2\alpha^2 z^2 + \cdots \quad (0 < \alpha \le 1)$$ and $$\varphi(z) = \frac{1 + (1 - 2\beta)z}{1 - z} = 1 + 2(1 - \beta)z + 2(1 - \beta)z^2 + \cdots \quad (0 \le \beta < 1)$$ in Corollaries 2.4 and 2.5, we obtain the results of Brannan and Taha [6, Theorems 2.1, 3.1 and 4.1, respectively].

Definition 2.1. A function $f \in \sigma$ given by (1.1) is said to be in the class $H^{\mu}_{\sigma}(\lambda, \varphi)$ if it satisfies $$(1-\lambda)\left(\frac{f(z)}{z}\right)^{\mu} + \lambda f'(z)\left(\frac{f(z)}{z}\right)^{\mu-1} \prec \varphi(z) \quad (\lambda \ge 1, \mu \ge 1, z \in D)$$ (2.2) and <span id="page-2-4"></span> $$(1-\lambda)\left(\frac{g(w)}{w}\right)^{\mu} + \lambda g'(w)\left(\frac{g(w)}{z}\right)^{\mu-1} \prec \varphi(w) \quad (\lambda \ge 1, \mu \ge 1, w \in D),\tag{2.3}$$ where the function g is given by $$g(w) = f^{-1}(w) = w - a_2 w^2 + (2a_2^2 - a_3)w^3 - (5a_2^3 - 5a_2 a_3 + a_4)w^4 + \cdots$$ (2.4) We note that, for suitable choices $\lambda$ , $\mu$ and $\varphi$ , the class $H^{\mu}_{\sigma}(\lambda, \varphi)$ reduces to the following known classes. - (1) $H^{\mu}_{\sigma}(\lambda, (\frac{1+z}{1-z})^{\alpha}) = H^{\mu}_{\sigma}(\lambda, \alpha)$ ( $\lambda \ge 1$ , $0 < \alpha \le 1$ , $\mu \ge 0$ ) (see Caglar et al. [12, Definition 2.1]); - (2) $H_{\sigma}^{\mu}(\lambda, \frac{1+(1-2\beta)z}{1-z}) = H_{\sigma}^{\mu}(\lambda, \beta)$ ( $\lambda \ge 1$ , $0 \le \beta < 1$ , $\mu \ge 0$ ) (see Caglar et al. [12, Definition 3.1]); - (3) $H^1_{\sigma}(\lambda, \varphi) = H_{\sigma}(\lambda, \varphi)$ ( $\lambda \ge 1$ ) (see Kumar et al. [14, Definition 1.1]); - (4) $H_{\sigma}^{\mu}(1,\varphi) = H_{\sigma}^{\mu}(\varphi) \ (\mu \ge 0)$ (see Kumar et al. [14, Definition 2.1]); - (5) $H_{\sigma}^{1}(1,\varphi) = H_{\sigma}(\varphi)$ (see Ali et al. [13, p.345]); - (6) $H_{\sigma}^{1}(\lambda,(\frac{1+z}{1-z})^{\alpha}) = H_{\sigma}(\lambda,\alpha)$ ( $\lambda \geq 1, 0 < \alpha \leq 1$ ) (see Frasin and Aouf [11, Definition 2.1]); - <span id="page-2-3"></span>(7) $H^1_{\sigma}(\lambda, \frac{1+(1-2\beta)z}{1-z}) = H_{\sigma}(\lambda, \beta)$ ( $\lambda \ge 1$ , $0 \le \beta < 1$ ) (see Frasin and Aouf [11, Definition 3.1]); - (8) $H^1_{\sigma}(1,(\frac{1+z}{1-z})^{\alpha}) = H_{\sigma}(\alpha)$ (0 < $\alpha \le 1$ ) (see Srivastava et al. [10, Definition 1]); - (9) $H_{\sigma}^{1}(1, \frac{1+(1-2\beta)z}{1-z}) = H_{\sigma}(\beta)$ (0 $\leq \beta < 1$ ) (see Srivastava et al. [10, Definition 2]). <span id="page-2-1"></span>For functions in the class $H^{\mu}_{\sigma}(\lambda,\varphi)$ , the following estimates are obtained.

Definition 2.2. Let $\gamma \in C^* = C \setminus \{0\}, \lambda > 0$ and $\mu > 0$ . A function $f \in \sigma$ given by (1.1) is said to be in the class $M_{\sigma}^{\gamma}(\lambda, \mu, \varphi)$ , if the following subordinations hold: $$1 + \frac{1}{\gamma} \left( \frac{zf'(z) + (2\lambda\mu + \lambda - \mu)z^2 f''(z) + \lambda\mu z^3 f'''(z)}{(1 - \lambda + \mu)f(z) + (\lambda - \mu)z f'(z) + \lambda\mu z^2 f''(z)} - 1 \right) \prec \varphi(z)$$ and $$1 + \frac{1}{\gamma} \left( \frac{wg'(w) + (2\lambda\mu + \lambda - \mu)w^2g''(w) + \lambda\mu w^3g'''(w)}{(1 - \lambda + \mu)g(w) + (\lambda - \mu)wg'(w) + \lambda\mu w^2g''(w)} - 1 \right) < \varphi(w),$$ where the function g is defined by (2.4). We note that, by choosing appropriate values for $\lambda$ , $\mu$ , $\gamma$ and $\varphi$ , the class $M_{\sigma}^{\gamma}(\lambda, \mu, \varphi)$ reduces to several earlier known classes. - <span id="page-6-3"></span>(1) $M_{\sigma}^{\gamma}(\lambda, 0, \varphi) = N_{\sigma, \gamma}^{\lambda}(\varphi)$ ( $\lambda \ge 0, \gamma \in C$ ) (see Kumar et al.* [14, Definition 2.2]); - <span id="page-6-1"></span>(2) $M_{\sigma}^{1}(0,0,\frac{1+(1-2\beta)z}{1-z}) = S_{\sigma}^{*}(\beta)$ ( $0 \le \beta < 1$ ) (see Brannan and Taha [6, Definition 3.1]); (3) $M_{\sigma}^{1}(1,0,\frac{1+(1-2\beta)z}{1-z}) = K_{\sigma}(\beta)$ ( $0 \le \beta < 1$ ) (see Brannan and Taha [6, Definition 4.1]); - (4) $M_{\sigma}^{1}(0,0,(\frac{1+z}{1-z})^{\alpha}) = S_{\sigma}^{*}(\alpha) \ (0 < \alpha \le 1)$ (see Taha [7]). For functions in the class $M_{\sigma}^{\gamma}(\lambda, \mu, \varphi)$ , the following estimates are derived.