Shi et al. Journal of Inequalities and Applications 2013, 2013:336

Theorem 1. Let $f \in \mathcal{MS}_{p}(\alpha, \beta)$ . Then <span id="page-2-2"></span><span id="page-2-1"></span> $$f(z) = z^{-p} \cdot \exp\left(2(\beta - p\cos\alpha)e^{-i\alpha} \int_0^z \frac{\omega(t)}{t(1 - \omega(t))} dt\right) \quad (z \in \mathbb{U}^*), \tag{2.1}$$ where $\omega$ is analytic in $\mathbb{U}$ with $\omega(0) = 0$ and $|\omega(z)| < 1$ . Proof For $f \in \mathcal{MS}_p(\alpha, \beta)$ , we know that (1.6) holds true. It follows that $$-e^{i\alpha}\frac{zf'(z)}{f(z)} = pe^{i\alpha} - \frac{2(\beta - p\cos\alpha)\omega(z)}{1 - \omega(z)},\tag{2.2}$$ where $\omega$ is analytic in $\mathbb{U}$ with $\omega(0) = 0$ and $|\omega(z)| < 1$ . We next find from (2.2) that $$\frac{f'(z)}{f(z)} + \frac{p}{z} = \frac{2(\beta - p\cos\alpha)e^{-i\alpha}\omega(z)}{z(1 - \omega(z))} \quad (z \in \mathbb{U}^*), \tag{2.3}$$ which, upon integration, yields <span id="page-3-0"></span> $$\log(z^p f(z)) = 2(\beta - p\cos\alpha)e^{-i\alpha} \int_0^z \frac{\omega(t)}{t(1-\omega(t))} dt.$$ (2.4) The assertion (2.1) of Theorem 1 can be easily derived from (2.4). Note that $f \in \mathcal{MS}_p(\alpha, \beta)$ if and only if $$-\frac{zf'(z)}{p}\in\mathcal{MC}_p(\alpha,\beta),$$ we get the following result.

Corollary 1. Let $f \in \mathcal{MC}_p(\alpha, \beta)$ . Then $$f(z) = -p \int_{z_0}^z u^{-p-1} \cdot \exp\left(2(\beta - p\cos\alpha)e^{-i\alpha} \int_0^u \frac{\omega(t)}{t(1-\omega(t))} dt\right) du \quad (z \in \mathbb{U}^*),$$ <span id="page-3-4"></span>where $\omega$ is analytic in $\mathbb{U}$ with $\omega(0) = 0$ and $|\omega(z)| < 1$ . Next, we discuss the coefficient estimates of functions belonging to the classes $\mathcal{MS}_p(\alpha,\beta)$ and $\mathcal{MC}_p(\alpha,\beta)$ . The following lemma will be required in the proof of Theorem 2.

Lemma 1. Let $p \in \mathbb{N}$ . Suppose also that the sequence $\{A_{p+m}\}_{m=0}^{\infty}$ is defined by <span id="page-3-1"></span> $$\begin{cases} A_{p} = \frac{\beta - p \cos \alpha}{p} & (m = 0), \\ A_{p+m} = \frac{2(\beta - p \cos \alpha)}{2p + m} (1 + \sum_{k=0}^{m-1} A_{p+k}) & (m \in \mathbb{N}). \end{cases}$$ (2.5) Then <span id="page-3-2"></span> $$A_{p+m} = \frac{2(\beta - p\cos\alpha)}{2\beta + m + 2p - 2p\cos\alpha} \prod_{k=0}^{m} \frac{2\beta + k + 2p - 2p\cos\alpha}{2p + k}$$ $$(m \in \mathbb{N}_0 := \mathbb{N} \cup \{0\}). \tag{2.6}$$ Proof By virtue of (2.5), we get <span id="page-3-3"></span> $$(2p+m+1)A_{p+m+1} = 2(\beta - p\cos\alpha)\left(1 + \sum_{k=0}^{m} A_{p+k}\right),\tag{2.7}$$ and $$(2p+m)A_{p+m} = 2(\beta - p\cos\alpha)\left(1 + \sum_{k=0}^{m-1} A_{p+k}\right).$$ (2.8) Combining (2.7) and (2.8), we find that $$\frac{A_{p+m+1}}{A_{p+m}} = \frac{2\beta + m + 2p - 2p\cos\alpha}{2p + m + 1} \quad (m \in \mathbb{N}_0).$$ (2.9) <span id="page-4-4"></span><span id="page-4-1"></span> Thus, <span id="page-4-0"></span> $$A_{p+m} = \frac{A_{p+m}}{A_{p+m-1}} \cdot \frac{A_{p+m-1}}{A_{p+m-2}} \cdot \cdot \cdot \frac{A_{p+1}}{A_p} \cdot A_p$$ $$= \frac{2\beta + m - 1 + 2p - 2p\cos\alpha}{2p + m} \cdot \cdot \cdot \cdot \frac{2\beta + 2p - 2p\cos\alpha}{2p + 1} \cdot \cdot \frac{2\beta - 2p\cos\alpha}{2p}$$ $$= \frac{2(\beta - p\cos\alpha)}{2\beta + m + 2p - 2p\cos\alpha} \prod_{k=0}^{m} \frac{2\beta + k + 2p - 2p\cos\alpha}{2p + k} \quad (m \in \mathbb{N}).$$ (2.10) The proof of Lemma 1 is thus completed.

Theorem 2. Let $f(z) = z^{-p} + \sum_{m=0}^{\infty} a_{p+m} z^{p+m} \in \mathcal{MS}_p(\alpha, \beta)$ . Then $$|a_{p+m}| \le \frac{2(\beta - p\cos\alpha)}{2\beta + m + 2p - 2p\cos\alpha} \prod_{k=0}^{m} \frac{2\beta + k + 2p - 2p\cos\alpha}{2p + k} \quad (m \in \mathbb{N}_0).$$ (2.11) Proof Let $$h(z) := \frac{\beta + e^{i\alpha} \frac{zf'(z)}{f(z)} + ip \sin \alpha}{\beta - p \cos \alpha} \quad (z \in \mathbb{U}; f \in \mathcal{MS}_p(\alpha, \beta)). \tag{2.12}$$ We know that $h \in \mathcal{P}$ . It follows that $$e^{i\alpha}zf'(z) = (\beta - p\cos\alpha)f(z)h(z) - (\beta + ip\sin\alpha)f(z). \tag{2.13}$$ Suppose that $$h(z) = 1 + h_1 z + h_2 z^2 + \cdots$$ (2.14) Then <span id="page-4-2"></span> $$e^{i\alpha} \left( -pz^{-p} + pa_p z^p + (p+1)a_{p+1} z^{p+1} + \dots + (p+m)a_{p+m} z^{p+m} + \dots \right)$$ $$= (\beta - p\cos\alpha) \left( z^{-p} + a_p z^p + a_{p+1} z^{p+1} + \dots \right) \times \left( 1 + h_1 z + h_2 z^2 + \dots \right)$$ $$- (\beta + ip\sin\alpha) \left( z^{-p} + a_p z^p + a_{p+1} z^{p+1} + \dots + a_{p+m} z^{p+m} + \dots \right). \tag{2.15}$$ By evaluating the coefficient of $z^{p+m}$ on both sides of (2.15), we get <span id="page-4-3"></span> $$e^{i\alpha}(p+m)a_{p+m} = (\beta - p\cos\alpha)(h_{2p+m} + a_ph_m + a_{p+1}h_{m-1} + \dots + a_{p+m})$$ $$-(\beta + ip\sin\alpha)a_{p+m}. \tag{2.16}$$ On the other hand, it is well known that $$|h_k| \le 2 \quad (k \in \mathbb{N}). \tag{2.17}$$ From (2.16) and (2.17), we easily get $$|a_p| \le \frac{\beta - p\cos\alpha}{p} \tag{2.18}$$ and <span id="page-5-2"></span><span id="page-5-0"></span> $$|a_{p+m}| \le \frac{2(\beta - p\cos\alpha)}{2p + m} \left(1 + \sum_{k=0}^{m-1} |a_{p+k}|\right). \tag{2.19}$$ Suppose that $p \in \mathbb{N}$ . We define the sequence $\{A_{p+m}\}_{m=0}^{\infty}$ as follows: $$\begin{cases} A_p = \frac{\beta - p \cos \alpha}{p} & (m = 0), \\ A_{p+m} = \frac{2(\beta - p \cos \alpha)}{2p + m} (1 + \sum_{k=0}^{m-1} A_{p+k}) & (m \ge 1). \end{cases}$$ (2.20) In order to prove that $$|a_{p+m}| \le A_{p+m} \quad (m \in \mathbb{N}_0), \tag{2.21}$$ we use the principle of mathematical induction. It is easy to verify that <span id="page-5-1"></span> $$|a_p| \le A_p = \frac{\beta - p\cos\alpha}{p}.\tag{2.22}$$ Thus, assuming that $$|a_{p+j}| \le A_{p+j} \quad (j=0,1,\ldots,m; m \in \mathbb{N}_0),$$ (2.23) we find from (2.19) and (2.23) that <span id="page-5-3"></span> $$|a_{p+m+1}| \leq \frac{2(\beta - p\cos\alpha)}{2p + m + 1} \left( 1 + \sum_{k=0}^{m} |a_{p+k}| \right)$$ $$\leq \frac{2(\beta - p\cos\alpha)}{2p + m + 1} \left( 1 + \sum_{k=0}^{m} |A_{p+k}| \right)$$ $$= A_{p+m+1} \quad (m \in \mathbb{N}_0). \tag{2.24}$$ Therefore, by the principle of mathematical induction, we have <span id="page-5-4"></span> $$|a_{n+m}| \le A_{n+m} \quad (m \in \mathbb{N}_0).$$ (2.25) By means of Lemma 1 and (2.20), we know that $$A_{p+m} = \frac{2(\beta - p\cos\alpha)}{2\beta + m + 2p - 2p\cos\alpha} \prod_{k=0}^{m} \frac{2\beta + k + 2p - 2p\cos\alpha}{2p + k} \quad (m \in \mathbb{N}_0).$$ (2.26) Combining (2.25) and (2.26), we readily get the coefficient estimates (2.11) asserted by Theorem 2. $\Box$ From Theorem 2, we easily get the following result.

Corollary 2. Let $f(z) = z^{-p} + \sum_{m=0}^{\infty} a_{p+m} z^{p+m} \in \mathcal{MC}_p(\alpha, \beta)$ . Then <span id="page-6-0"></span> $$|a_{p+m}| \leq \frac{2p(\beta - p\cos\alpha)}{(p+m)(2\beta + m + 2p - 2p\cos\alpha)} \prod_{k=0}^{m} \frac{2\beta + k + 2p - 2p\cos\alpha}{2p + k} \quad (m \in \mathbb{N}_0).$$ Remark 3 By setting $\alpha = 0$ in Theorem 2, we get the corresponding result due to Wang et al. [1].

Theorem 3. If $f \in \mathcal{MS}_p(\alpha, \beta)$ , then $$\frac{p\cos\alpha - (2\beta - p\cos\alpha)r}{1 - r} \le \Re\left(-e^{i\alpha}\frac{zf'(z)}{f(z)}\right) \le \frac{p\cos\alpha + (2\beta - p\cos\alpha)r}{1 + r}$$ (2.27) for |z| = r < 1. Proof Consider the function $\varphi$ defined by $$\varphi(z) := \frac{pe^{i\alpha} - (2\beta - pe^{-i\alpha})z}{1 - z} \quad (z \in \mathbb{U}).$$ $$(2.28)$$ Let $z = re^{i\theta}$ (0 < r < 1), we see that $$\Re(\varphi(z)) = p\cos\alpha - \frac{2(\beta - p\cos\alpha)r(\cos\theta - r)}{1 + r^2 - 2r\cos\theta}.$$ (2.29) Suppose $$\psi(t) := p \cos \alpha - \frac{2(\beta - p \cos \alpha)r(t - r)}{1 + r^2 - 2rt} \quad (t := \cos \theta), \tag{2.30}$$ we easily find that $$\psi'(t) = -2(\beta - p\cos\alpha) \cdot \frac{1 - r^2}{(1 + r^2 - 2rt)^2} > 0.$$ (2.31) This implies $$p\cos\alpha - \frac{2(\beta - p\cos\alpha)r}{1 - r} \le \Re(\varphi(z)) \le p\cos\alpha + \frac{2(\beta - p\cos\alpha)r}{1 + r},$$ (2.32) which is equivalent to $$\frac{p\cos\alpha - (2\beta - p\cos\alpha)r}{1 - r} \le \Re(\varphi(z)) \le \frac{p\cos\alpha + (2\beta - p\cos\alpha)r}{1 - r}.$$ (2.33) Noting that $-e^{i\alpha}\frac{zf'(z)}{f(z)}\prec \varphi(z)$ and $\varphi(z)$ is univalent in $\mathbb U$ , we prove the inequality (2.27). Taking $\alpha = 0$ in Theorem 3, we have the following corollary.

Corollary 3. If $f \in \mathcal{MS}_{p}(0, \beta)$ , then $$\frac{p - (2\beta - p)r}{1 - r} \le \Re\left(\frac{zf'(z)}{f(z)}\right) \le \frac{p + (2\beta - p)r}{1 + r}$$ for |z| = r < 1. Similar to the proof of Theorem 3, we get the following result.

Corollary 4. If $f \in \mathcal{MC}_p(\alpha, \beta)$ , then $$\frac{p\cos\alpha - (2\beta - p\cos\alpha)r}{1 - r} \le \Re\left(-e^{i\alpha}\left(1 + \frac{zf''(z)}{f'(z)}\right)\right) \le \frac{p\cos\alpha + (2\beta - p\cos\alpha)r}{1 + r}$$ for |z| = r < 1.

Corollary 5. If $f \in \mathcal{MC}_p(0, \beta)$ , then <span id="page-7-0"></span> $$\frac{p - (2\beta - p)r}{1 - r} \le \Re\left(1 + \frac{zf''(z)}{f'(z)}\right) \le \frac{p + (2\beta - p)r}{1 + r}$$ <span id="page-7-3"></span>for |z| = r < 1. Now, we present some sufficient conditions for functions belonging to the classes $\mathcal{MS}_p(\alpha, \beta)$ and $\mathcal{MC}_p(\alpha, \beta)$ .

Theorem 4. If $f \in \mathcal{MS}_p(\alpha, \beta)$ satisfies the condition <span id="page-7-2"></span> $$\sum_{n=1-p}^{\infty} (\left| ne^{i\alpha} + \lambda \right| + \left| ne^{i\alpha} + 2\beta - \lambda \right|) |a_n| \le \left| pe^{i\alpha} - 2\beta + \lambda \right| - \left| pe^{i\alpha} - \lambda \right|$$ (2.34) for some real $\alpha$ , $\beta$ and $\lambda$ ( $0 \le \lambda \le p \cos \alpha$ ), then $f \in \mathcal{MS}_p(\alpha, \beta)$ . Proof To prove $f \in \mathcal{MS}_p(\alpha, \beta)$ , it suffices to show that <span id="page-7-1"></span> $$\left| \frac{e^{i\alpha} \frac{zf'(z)}{f(z)} + \lambda}{e^{i\alpha} \frac{zf'(z)}{f(z)} + (2\beta - \lambda)} \right| < 1 \quad (z \in \mathbb{U}; 0 \le \lambda \le p \cos \alpha). \tag{2.35}$$ From (2.34), we know that $$\left| p e^{i\alpha} - 2\beta + \lambda \right| - \sum_{n=1-n}^{\infty} \left| n e^{i\alpha} + 2\beta - \lambda \right| |a_n| \ge \left| p e^{i\alpha} - \lambda \right| + \sum_{n=1-n}^{\infty} \left| n e^{i\alpha} + \lambda \right| |a_n| > 0. \quad (2.36)$$ Now, by the maximum modulus principle, we deduce from (1.1) and (2.36) that $$\left| \frac{e^{i\alpha} \frac{zf'(z)}{f(z)} + \lambda}{e^{i\alpha} \frac{zf'(z)}{f(z)} + (2\beta - \lambda)} \right| = \left| \frac{(-pe^{i\alpha} + \lambda) + \sum_{n=1-p}^{\infty} (ne^{i\alpha} + \lambda) a_n z^{n+p}}{(-pe^{i\alpha} + 2\beta - \lambda) + \sum_{n=1-p}^{\infty} (ne^{i\alpha} + 2\beta - \lambda) a_n z^{n+p}} \right| < \frac{|pe^{i\alpha} - \lambda| + \sum_{n=1-p}^{\infty} |ne^{i\alpha} + \lambda| |a_n|}{|pe^{i\alpha} - 2\beta + \lambda| - \sum_{n=1-p}^{\infty} |ne^{i\alpha} + 2\beta - \lambda| |a_n|} \leq 1.$$ (2.37) Therefore, if f satisfies the coefficient estimate (2.34), then we know that f satisfies the inequality (2.35). This completes the proof of Theorem 4.

Corollary 6. If $f \in \mathcal{MC}_p(\alpha, \beta)$ satisfies the inequality <span id="page-8-4"></span> $$\sum_{n=1-p}^{\infty} |n| (\left| ne^{i\alpha} + \lambda \right| + \left| ne^{i\alpha} + 2\beta - \lambda \right|) |a_n| \leq p (\left| pe^{i\alpha} - 2\beta + \lambda \right| - \left| pe^{i\alpha} - \lambda \right|)$$ for some real $\alpha$ , $\beta$ and $\lambda$ ( $0 \le \lambda \le p \cos \alpha$ ), then $f \in \mathcal{MC}_p(\alpha, \beta)$ . We need the following lemma to prove our next theorem.

Lemma 2. (See [11]) Let $\varphi$ be a nonconstant regular function in $\mathbb{U}$ . If $|\varphi|$ attains its maximum value on the circle |z| = r < 1 at $z_0$ , then <span id="page-8-2"></span> $$z_0\varphi'(z_0)=k\varphi(z_0),$$ where $k \ge 1$ is a real number.

Theorem 5. If $f \in \mathcal{MS}_{p}(0, \beta)$ satisfies <span id="page-8-0"></span> $$\left|1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)}\right| < \frac{\beta - p}{2\beta} \quad (z \in \mathbb{U})$$ for some real $\beta > p$ , then $f \in \mathcal{MS}_p(0, \beta)$ . Proof Let us define the function $\phi$ by <span id="page-8-1"></span> $$\phi(z) := \frac{\frac{zf'(z)}{f(z)} + p}{\frac{zf'(z)}{f(z)} + 2\beta - p} \quad (z \in \mathbb{U}), \tag{2.39}$$ then we see that $\phi$ is analytic in $\mathbb{U}$ and $\phi(0) = 0$ . It follows from (2.39) that <span id="page-8-3"></span> $$\frac{zf'(z)}{f(z)} = \frac{-p + (2\beta - p)\phi(z)}{1 - \phi(z)}. (2.40)$$ Differentiating both sides of (2.40) logarithmically, we obtain $$1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} = \frac{(2\beta - p)z\phi'(z)}{-p + (2\beta - p)\phi(z)} + \frac{z\phi'(z)}{1 - \phi(z)}.$$ (2.41) <span id="page-9-0"></span> By virtue of (2.38) and (2.41), we find that $$\left| 1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} \right| = \left| \frac{2(\beta - p)z\phi'(z)}{[-p + (2\beta - p)\phi(z)][1 - \phi(z)]} \right| < \frac{\beta - p}{2\beta}. \tag{2.42}$$ Suppose that there exists a point $z_0 \in \mathbb{U}$ such that $$\max_{|z| \le |z_0|} \left| \phi(z) \right| = \left| \phi(z_0) \right| = 1.$$ Then, Lemma 2 gives us that $\phi(z_0) = e^{i\theta}$ and $z_0\phi'(z_0) = ke^{i\theta}$ $(k \ge 1)$ . For such a point $z_0$ , we have that $$\left| 1 + \frac{z_0 f''(z_0)}{f'(z_0)} - \frac{z_0 f'(z_0)}{f(z_0)} \right| = \left| \frac{2(\beta - p)k e^{i\theta}}{[-p + (2\beta - p)e^{i\theta}][1 - e^{i\theta}]} \right| = \frac{2(\beta - p)k}{\sqrt{p^2 + (2\beta - p)^2 - 2p(2\beta - p)\cos\theta} \sqrt{2 - 2\cos\theta}} \ge \frac{\beta - p}{2\beta}.$$ (2.43) <span id="page-9-1"></span>This contradicts our condition (2.38). Therefore, there is no $z_0 \in \mathbb{U}$ such that $|\phi(z_0)| = 1$ . This implies that $|\phi(z)| < 1$ ( $z \in \mathbb{U}^*$ ), that is, $$\left|\frac{\frac{zf'(z)}{f(z)}+p}{\frac{zf'(z)}{f(z)}+(2\beta-p)}\right|<1\quad (z\in\mathbb{U}).$$ Thus, we conclude that $f \in \mathcal{MS}_p(0, \beta)$ .

Theorem 6. If $f \in \mathcal{MS}_p(0,\beta)$ for some real $p < \beta \leq p + \frac{1}{2}$ , then $$\Re\left(\frac{1}{z^p f(z)}\right) > \frac{1}{1 - 2\beta + 2p} \quad (z \in \mathbb{U}). \tag{2.44}$$ Proof Consider the function $\eta$ such that $$\frac{1}{z^p f(z)} = \frac{1 + (1 - 2\gamma)\eta(z)}{1 - \eta(z)} \tag{2.45}$$ for $\gamma = \frac{1}{1-2\beta+2p}$ and $f(z) \in \mathcal{MS}_p(0,\beta)$ . Then we know that $$\Re\left(-\frac{zf'(z)}{f(z)}\right) = \Re\left(p + \frac{(1 - 2\gamma)z\eta'(z)}{1 + (1 - 2\gamma)\eta(z)} + \frac{z\eta'(z)}{1 - \eta(z)}\right) < \beta. \tag{2.46}$$ Since $\eta(z)$ is analytic in $\mathbb{U}$ and $\eta(0) = 0$ , we suppose that there exists a point $z_0 \in \mathbb{U}$ such that $$\max_{|z| \le |z_0|} \left| \eta(z) \right| = \left| \eta(z_0) \right| = 1.$$ Then, applying Lemma 2, we can write that $\eta(z_0) = e^{i\theta}$ and $z_0 \eta'(z_0) = k e^{i\theta}$ $(k \ge 1)$ . This gives us that $$\Re\left(-\frac{z_0 f'(z_0)}{f(z_0)}\right) = \Re\left(p + \frac{(1 - 2\gamma)ke^{i\theta}}{1 + (1 - 2\gamma)e^{i\theta}} + \frac{ke^{i\theta}}{1 - e^{i\theta}}\right)$$ $$\geq p - \frac{(1 - 2\gamma)k}{2\gamma} - \frac{k}{2}$$ $$\geq p + \frac{\gamma - 1}{2\gamma} = \beta,$$ (2.47) which contradicts the inequality (2.46). Therefore, there is no $z_0 \in \mathbb{U}$ such that $|\eta(z_0)| = 1$ . This means that $|\eta(z)| < 1$ , and that $$\Re\left(\frac{1}{z^p f(z)}\right) > \frac{1}{1 - 2\beta + 2p} \quad (z \in \mathbb{U}). \tag{2.48}$$ The proof of Theorem 6 is thus completed. In view of Theorem 6, we get the following result. Corollary 7 If $f \in \mathcal{MC}_p(0,\beta)$ for some real $p < \beta \leq p + \frac{1}{2}$ , then $$\Re\left(\frac{p}{z^{p+1}f'(z)}\right) > \frac{1}{1 - 2\beta + 2p} \quad (z \in \mathbb{U}).$$

Definition 1. A function $f \in \Sigma_p$ is said to be in the class $\mathcal{MS}_p(\alpha, \beta)$ if it satisfies the condition $$\Re\left(e^{i\alpha}\frac{zf'(z)}{f(z)}\right) > -\beta \quad (z \in \mathbb{U})$$ (1.4) for some real $\alpha$ and $\beta$ , where (and throughout this paper unless otherwise mentioned) the parameters $\alpha$ and $\beta$ are constrained as follows: $$|\alpha| < \frac{\pi}{2}$$ and $\beta > p \cos \alpha$ . Furthermore, a function $f \in \Sigma_p$ is said to be in the class $\mathcal{MC}_p(\alpha, \beta)$ if it satisfies the inequality <span id="page-2-0"></span> $$\Re\left(e^{i\alpha}\left(1+\frac{zf''(z)}{f'(z)}\right)\right) > -\beta \quad (z \in \mathbb{U}). \tag{1.5}$$ Remark 1 Taking $\alpha = 0$ , we get the function classes introduced by Wang et al. [1]. Remark 2 We note that $f \in \mathcal{MS}_p(\alpha, \beta)$ if and only if $$-e^{i\alpha}\frac{zf'(z)}{f(z)} < \frac{pe^{i\alpha} - (2\beta - pe^{-i\alpha})z}{1 - z}.$$ (1.6) Also, $f \in \mathcal{MC}_p(\alpha, \beta)$ if and only if $$-e^{i\alpha}\left(1+\frac{zf''(z)}{f'(z)}\right) \prec \frac{pe^{i\alpha}-2(\beta-pe^{-i\alpha})z}{1-z}.$$ (1.7) For some investigations of meromorphic functions, see (for example) the works [1, 3-10] and the references cited in. <span id="page-2-3"></span>In the present paper, we aim at proving some interesting properties such as integral representations and coefficient inequalities of the function classes $\mathcal{MS}_p(\alpha, \beta)$ and $\mathcal{MC}_p(\alpha, \beta)$ .