Toeplitz determinants on bounded starlike circular domain in $\mathbb{C}^n$

Lemma 2.1. [20] $\Omega \subset \mathbb{C}^n$ is a bounded starlike circular domain if and only if there exists a unique real continuous functions $\rho: \mathbb{C}^n \to \mathbb{R}$ , called the Minkowski functional of $\Omega$ , such that (i) $$\rho(z) \ge 0$$ , $z \in \mathbb{C}^n$ ; $\rho(z) = 0 \Leftrightarrow z = 0$ ; (ii) $$\rho(tz) = |t|\rho(z), t \in \mathbb{C}, z \in \mathbb{C}^n$$ : (iii) $$\Omega = \{z \in \mathbb{C}^n : \rho(z) < 1\}.$$ Furthermore, if $\rho(z) \in C^1$ in $\mathbb{C}^n \setminus \{0\}$ , then the function $\rho(z)$ has the following properties. <span id="page-3-0"></span> $$2\frac{\partial \rho(z)}{\partial z}z = \rho(z), \quad z \in \mathbb{C}^n,$$ $$2\frac{\partial \rho(z_0)}{\partial z}z_0 = 1, \quad z_0 \in \partial\Omega,$$ $$\frac{\partial \rho(\lambda z)}{\partial z} = \frac{\partial \rho(z)}{\partial z}, \quad \lambda \in (0, \infty),$$ $$\frac{\partial \rho(e^{i\theta}z)}{\partial z} = e^{-i\theta}\frac{\partial \rho(z)}{\partial z}, \quad \theta \in \mathbb{R}.$$ (2.1)

Theorem 3.1. Let $g \in \mathcal{H}(\Omega, \mathbb{C})$ with g(0) = 1 and G(z) = zg(z). If $J_G^{-1}(z)G(z) \in \mathcal{M}_{\Phi}$ such that $\Phi$ satisfies $|\Phi''(0) + 2(\Phi'(0))^2| > 2\Phi'(0) > 0$ , then $$\left| \left( 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right)^2 - \left( 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2! \rho^2(z)} \right)^2 \right| \leq (\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2.$$ The bound is sharp. Proof. Since $J_G^{-1}(z)$ exists, therefore $g(z) \neq 0$ , $z \in \Omega$ . For fix $z \in \Omega \setminus \{0\}$ , let us denote $z_0 = \frac{z}{\rho(z)}$ and define $h : \mathbb{U} \to \mathbb{C}$ such that $$h(\zeta) = \begin{cases} \frac{\zeta}{2^{\frac{\partial p(z_0)}{\partial z}} J_G^{-1}(\zeta z_0) G(\zeta z_0)}, & \zeta \neq 0, \\ 1, & \zeta = 0. \end{cases}$$ Using the property $2\frac{\partial \rho(z_0)}{\partial z}z_0=1$ for $z_0\in\partial\Omega$ of Minkowski functional, we obtain $h\in\mathcal{H}(\mathbb{U})$ and since $J_G^{-1}(z)G(z)\in\mathcal{M}_{\Phi}$ , therefore $$\begin{split} h(\zeta) &= \frac{\zeta}{2\frac{\partial \rho(z_0)}{\partial z}J_G^{-1}(\zeta z_0)G(\zeta z_0)} \\ &= \frac{\rho(\zeta z_0)}{2\frac{\partial \rho(\zeta z_0)}{\partial z}J_G^{-1}(\zeta z_0)G(\zeta z_0)} \in \Phi(\mathbb{U}), \quad \zeta \in \mathbb{U} \setminus \{0\}. \end{split}$$ Applying the same technique as in [23] (also see [8, Theorem 7.1.14]), we obtain $$J_G^{-1}(z) = \frac{1}{g(z)} \left( I - \frac{\frac{zJ_g(z)}{g(z)}}{1 + \frac{J_g(z)z}{g(z)}} \right).$$ Now, using G(z) = zg(z), we have $$J_G^{-1}(z)G(z) = \frac{zg(z)}{g(z) + J_q(z)z}, \quad z \in \Omega \setminus \{0\},$$ which together with (2.1) gives $$\frac{\rho(z)}{2\frac{\partial \rho(z)}{\partial z}J_G^{-1}(z)G(z)} = 1 + \frac{J_g(z)z}{g(z)}, \quad z \in \Omega \setminus \{0\}.$$ In view of the above equation, we obtain $$h(\zeta) = \frac{\rho(\zeta z_0)}{2\frac{\partial \rho(\zeta z_0)}{\partial z} J_0^{-1}(\zeta z_0) G(\zeta z_0)} = 1 + \frac{J_g(\zeta z_0) \zeta z_0}{g(\zeta z_0)},$$ which immediately yields $$h(\zeta)g(\zeta z_0) = g(\zeta z_0) + J_g(\zeta z_0)\zeta z_0.$$ Based on the Taylor series expansions in $\zeta$ , the above equation gives $$\left(1 + h'(0)\zeta + \frac{h''(0)}{2!}\zeta^2 + \cdots\right) \left(1 + J_g(0)(z_0)\zeta + \frac{D^2g(0)(z_0^2)}{2!}\zeta^2 + \cdots\right) = \left(1 + J_g(0)(z_0)\zeta + \frac{D^2g(0)(z_0^2)}{2!}\zeta^2 + \cdots\right) \left(J_g(0)(z_0)\zeta + D^2g(0)(z_0^2)\zeta^2 + \cdots\right).$$ By the comparison of homogeneous expansions, we get <span id="page-4-0"></span> $$h'(0) = J_a(0)(z_0).$$ Further, using $z_0 = \frac{z}{\rho(z)}$ in the above relation, we have $$h'(0)\rho(z) = J_q(0)(z). \tag{3.1}$$ Since, we also have G(z) = zg(z), therefore $$\frac{D^2G(0)z^2}{2!} = J_g(0)(z)z,$$ which together with (2.1) leads to <span id="page-5-0"></span> $$2\frac{\partial\rho}{\partial z}\frac{D^2G(0)z^2}{2!} = J_g(0)(z)\rho(z). \tag{3.2}$$ Thus, from (3.1) and (3.2), we obtain <span id="page-5-2"></span> $$2\frac{\partial \rho}{\partial z} \frac{D^2 G(0) z^2}{2! \rho^2(z)} = h'(0).$$ Since $h \prec \Phi$ , therefore $|h'(0)| \leq |\Phi'(0)|$ , using this fact, we get <span id="page-5-3"></span><span id="page-5-1"></span> $$\left| 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0) z^2}{2! \rho^2(z)} \right| \le |\Phi'(0)|. \tag{3.3}$$ For $\lambda \in \mathbb{C}$ , Xu et al. [28, Theorem 1] proved that $$\left| 2 \frac{\partial \rho}{\partial z} \frac{D^{3}G(0)(z^{3})}{3! \rho^{3}(z)} - \lambda \left( 2 \frac{\partial \rho}{\partial z} \frac{D^{2}G(0)(z^{2})}{2! \rho^{2}(z)} \right)^{2} \right| \\ \leq \frac{|\Phi'(0)|}{2} \max \left\{ 1, \left| \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + (1 - 2\lambda)\Phi'(0) \right| \right\}, \quad z \in \Omega \setminus \{0\}.$$ (3.4) Thus, when $|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0)$ , the equation (3.4) readily yields $$\left| 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right| \le \frac{\Phi'(0)}{2} \left| \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right|. \tag{3.5}$$ Using the bounds given in (3.3) and (3.5), together with the following inequality $$\begin{split} \left| \left( 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right)^2 - \left( 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2! \rho^2(z)} \right)^2 \right| \\ & \leq \left| 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right|^2 + \left| 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2! \rho^2(z)} \right|^2, \end{split}$$ we find the required bound. To see the sharpness of the bound consider the function <span id="page-5-5"></span> $$G(z) = z \exp \int_0^{\frac{z_1}{r}} \frac{(\Phi(it) - 1)}{t} dt, \quad z \in \Omega,$$ (3.6) where $r = \sup\{|z_1| : z = (z_1, z_2, \dots, z_n)' \in \Omega\}$ . It can be easily showed that $J_G^{-1}(z)G(z) \in \mathcal{M}_{\Phi}$ and $$\frac{D^2G(0)(z^2)}{2!} = i\Phi'(0)(\frac{z_1}{r})z \text{ and } \frac{D^3G(0)(z^3)}{3!} = -\frac{1}{2}\left(\frac{\Phi''(0)}{2} + (\Phi'(0))^2\right)(\frac{z_1}{r})^2z.$$ By applying (2.1) in the above relations, we get $$2\frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2!} = i\Phi'(0)(\frac{z_1}{r})\rho(z)$$ and $$2\frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3!} \rho(z) = -\frac{1}{2} \left( \frac{\Phi''(0)}{2} + (\Phi'(0))^2 \right) (\frac{z_1}{r})^2 \rho^2(z).$$ Setting z = Ru (0 < R < 1), where $u = (u_1, u_2, \dots, u_n)' \in \partial \Omega$ and $u_1 = r$ , we obtain <span id="page-5-4"></span> $$2\frac{\partial\rho}{\partial z}\frac{D^2G(0)(z^2)}{2!\rho^2(z)} = i\Phi'(0)$$ (3.7) and <span id="page-6-0"></span> $$2\frac{\partial\rho}{\partial z}\frac{D^3G(0)(z^3)}{3!\rho^3(z)} = -\frac{1}{2}\left(\frac{\Phi''(0)}{2} + (\Phi'(0))^2\right). \tag{3.8}$$ Thus, from (3.7) and (3.8), we have $$\left| \left( 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right)^2 - \left( 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2! \rho^2(z)} \right)^2 \right| \\ \leq \frac{(\Phi'(0))^2}{4} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2 + (\Phi'(0))^2,$$ which shows the sharpness of the bound and completes the

Theorem 3.2. Let $g \in \mathcal{H}(\Omega, \mathbb{C})$ with g(0) = 1 and G(z) = zg(z). If $J_G^{-1}(z)G(z) \in \mathscr{M}_{\Phi}$ such that $\Phi$ satisfies $2\Phi'(0) - 2(\Phi'(0))^2 < \Phi''(0) < 6(\Phi'(0))^2 - 2\Phi'(0)$ . then $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 1 + 2(\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left(3\Phi'(0) - \frac{\Phi''(0)}{2\Phi'(0)}\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right),$$ where $$b_3 = 2 \frac{\partial \rho}{\partial z} \frac{2D^3 G(0)(z^3)}{3! \rho^3(z)} \quad and \quad b_2 = 2 \frac{\partial \rho}{\partial z} \frac{2D^2 G(0)(z^2)}{2! \rho^2(z)}. \tag{3.9}$$ The bound is sharp. Proof. Since $2\Phi'(0) < \Phi''(0) + 2(\Phi'(0))^2$ , the inequality (3.4) gives <span id="page-6-4"></span><span id="page-6-2"></span><span id="page-6-1"></span> $$\left| 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right| \le \frac{\Phi'(0)}{2} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right). \tag{3.10}$$ Also, since $2\Phi'(0) + \Phi''(0) \le 6(\Phi'(0))^2$ , the inequality (3.4) for $\lambda = 2$ gives $$\left| 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} - 2 \left( 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2! \rho^2(z)} \right)^2 \right| \le \frac{\Phi'(0)}{2} \left( 3\Phi'(0) - \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} \right). \tag{3.11}$$ Using the estimates given in (3.3) and (3.10), and the bound given by (3.11) in the following inequality $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| < 1 + 2|b_2|^2 + |b_3||b_3 - 2b_2^2|$$ the required bound is established. The result is sharp for the function G(z) given by (3.6). As for this function, we have $b_2 = i\Phi'(0)$ and $b_3 = -(\Phi''(0) + 2(\Phi'(0))^2)/4$ from (3.7) and (3.8), respectively. Therefore $$1 - b_3(b_3 - 2b_2^2) - 2b_2^2 = 1 + 2(\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left(3\Phi'(0) - \frac{\Phi''(0)}{2\Phi'(0)}\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right),$$ which proves the sharpness of the bound. In case of $\Omega = \mathbb{U}^n$ , Theorem 3.1 and Theorem 3.2 directly give the following results, which we state here without

Theorem 3.3. Let $g \in \mathcal{H}(\mathbb{U}^n, \mathbb{C})$ with g(0) = 1 and G(z) = zg(z). If $J_G^{-1}(z)G(z) \in \mathscr{M}_{\Phi}$ such that $\Phi$ satisfies $$|\Phi''(0) + 2(\Phi'(0))^2| \ge 2\Phi'(0) > 0,$$ then $$\left| \left( \frac{1}{\|z\|^4} \frac{D^3 G(0)(z^3)}{3!} \bar{z} \right)^2 - \left( \frac{1}{\|z\|^3} \frac{D^2 G(0)(z^2)}{2!} \bar{z} \right)^2 \right| \\ \leq (\Phi'(0))^2 + \frac{(\Phi'(0))^2}{4} \left( \frac{1}{2} \frac{\Phi''(0)}{\Phi'(0)} + \Phi'(0) \right)^2.$$ The bound is sharp.

Theorem 3.4. Let $g \in \mathcal{H}(\mathbb{U}^n,\mathbb{C})$ with g(0) = 1 and G(z) = zg(z). If $J_G^{-1}(z)G(z) \in \mathscr{M}_{\Phi}$ such that $\Phi$ satisfies $$2\Phi'(0) - 2(\Phi'(0))^2 \le \Phi''(0) \le 6(\Phi'(0))^2 - 2\Phi'(0),$$ then $$|2d_2^2d_3 - d_3^2 - 2d_2^2 + 1| \le \frac{(\Phi'(0))^2}{4} \left(3\Phi'(0) - \frac{\Phi''(0)}{2\Phi'(0)}\right) \left(\frac{\Phi''(0)}{2\Phi'(0)} + \Phi'(0)\right) + 2(\Phi'(0))^2 + 1,$$ where $$d_3 = \frac{1}{\|z\|^4} \frac{D^3 G(0)(z^3)}{3!} \bar{z} \quad and \quad d_2 = \frac{1}{\|z\|^3} \frac{D^2 G(0)(z^2)}{2!} \bar{z}. \tag{3.12}$$ The bound is sharp. In sight of remark 2.1, various choices of $\Phi$ in Theorem 3.1 to Theorem 3.4 lead to the following results for different subclasses of $S(\Omega)$ . Corollary 3.5. If $g: \Omega \to \mathbb{C}$ and $G(z) = zg(z) \in \mathcal{S}^*(\Omega)$ , then $$\left| \left( 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right)^2 - \left( 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2! \rho^2(z)} \right)^2 \right| \le 13$$ and <span id="page-7-1"></span> $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 24,$$ where $b_2$ and $b_3$ are given by (3.9). All these estimations are sharp. Corollary 3.6. If $g: \Omega \to \mathbb{C}$ and $G(z) = zg(z) \in \mathcal{S}^*_{\alpha}(\Omega)$ , then $$\left| \left( 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right)^2 - \left( 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2! \rho^2(z)} \right)^2 \right| \le (1 - \alpha)^2 (4\alpha^2 - 12\alpha + 13)$$ and for $\alpha \in [0, 2/3]$ , $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 12\alpha^4 - 52\alpha^3 + 91\alpha^2 - 74\alpha + 24,$$ where $b_2$ and $b_3$ are given by (3.9). All these estimations are sharp.

Corollary 3.7. If $g: \Omega \to \mathbb{C}$ and $G(z) = zg(z) \in \mathcal{SS}^*_{\beta}(\Omega)$ , then for $\beta \in [1/3, 1]$ , the following sharp inequalities hold: $$\left| \left( 2 \frac{\partial \rho}{\partial z} \frac{D^3 G(0)(z^3)}{3! \rho^3(z)} \right)^2 - \left( 2 \frac{\partial \rho}{\partial z} \frac{D^2 G(0)(z^2)}{2! \rho^2(z)} \right)^2 \right| \le 9\beta^4 + 4\beta^2$$ and $$|2b_2^2b_3 - b_3^2 - 2b_2^2 + 1| \le 15\beta^4 + 8\beta^2 + 1,$$ where $b_2$ and $b_3$ are given by (3.9). If $\Omega = \mathbb{U}^n$ , we obtain the following bounds. Corollary 3.8. If $g: \mathbb{U}^n \to \mathbb{C}$ and $G(z) = zg(z) \in \mathcal{S}^*(\mathbb{U}^n)$ , then $$\left| \left( \frac{1}{\|z\|^4} \frac{D^3 G(0)(z^3)}{3!} \bar{z} \right)^2 - \left( \frac{1}{\|z\|^3} \frac{D^2 G(0)(z^2)}{2!} \bar{z} \right)^2 \right| \le 13 \tag{3.13}$$ and <span id="page-8-1"></span><span id="page-8-0"></span> $$|2d_2^2d_3 - d_3^2 - 2d_2^2 + 1| \le 24, (3.14)$$ where $d_2$ and $d_3$ are given by (3.12). All these bounds are sharp. Remark 3.1. When n = 1, (3.13) and (3.14) reduce to the following: $$\left| \left( \frac{G^{(3)}(0)}{3!} \right)^2 - \left( \frac{G''(0)}{2!} \right)^2 \right| \le 13$$ and $$|2d_2^2d_3 - d_3^2 - 2d_2^2 + 1| \le 24,$$ where <span id="page-8-3"></span><span id="page-8-2"></span> $$d_3 = \frac{G^{(3)}(0)}{3!}$$ and $d_2 = \frac{G''(0)}{2!}$ . which are equivalent to the bounds given in Theorem A. Corollary 3.9. If $g: \mathbb{U}^n \to \mathbb{C}$ and $G(z) = zg(z) \in \mathcal{S}^*_{\alpha}(\mathbb{U}^n)$ , then $$\left| \left( \frac{1}{\|z\|^4} \frac{D^3 G(0)(z^3)}{3!} \bar{z} \right)^2 - \left( \frac{1}{\|z\|^3} \frac{D^2 G(0)(z^2)}{2!} \bar{z} \right)^2 \right| \le (1 - \alpha)^2 (4\alpha^2 - 12\alpha + 13) \tag{3.15}$$ and for $\alpha \in [0, 2/3]$ , $$|2d_2^2d_3 - d_3^2 - 2d_2^2 + 1| \le 12\alpha^4 - 52\alpha^3 + 91\alpha^2 - 74\alpha + 24,\tag{3.16}$$ where $d_2$ and $d_3$ are given by (3.12). All these estimations are sharp. Remark 3.2. When n=1, (3.15) and (3.16) reduce to the bounds given in Theorem B.

Corollary 3.10. If $g: \mathbb{U}^n \to \mathbb{C}$ and $G(z) = zg(z) \in \mathcal{SS}^*_{\beta}(\mathbb{U}^n)$ , then for $\beta \in [1/3, 1]$ , the following sharp inequalities hold: $$\left| \left( \frac{1}{\|z\|^4} \frac{D^3 G(0)(z^3)}{3!} \bar{z} \right)^2 - \left( \frac{1}{\|z\|^3} \frac{D^2 G(0)(z^2)}{2!} \bar{z} \right)^2 \right| \le 9\beta^4 + 4\beta^2 \tag{3.17}$$ and <span id="page-8-5"></span><span id="page-8-4"></span> $$|2d_2^2d_3 - d_3^2 - 2d_2^2 + 1| \le 15\beta^4 + 8\beta^2 + 1, (3.18)$$ where $d_2$ and $d_3$ are given by (3.12). Remark 3.3. When n=1, (3.17) and (3.18) reduce to the bounds given in Theorem C.

Definition 2.1. Let $\Omega$ be a bounded starlike circular domain in $\mathbb{C}^n$ with $0 \in \Omega$ and its Minkowski functional $\rho \in C^1$ in $\mathbb{C}^n \setminus \{0\}$ . A normalized locally biholomorphic mapping $g: \Omega \to \mathbb{C}^n$ is said to be starlike of order $\alpha$ $(0 \le \alpha < 1)$ if $$\left| \frac{2}{\rho(z)} \frac{\partial \rho}{\partial z} J_g^{-1}(z) g(z) - \frac{1}{2\alpha} \right| < \frac{1}{2\alpha}, \quad \forall z \in \Omega \setminus \{0\}.$$ Equivalently, the above equation can be written as $$\operatorname{Re}\left\{\frac{\rho(z)}{2\frac{\partial\rho(z)}{\partial z}J_g^{-1}(z)g(z)}\right\} > \alpha, \quad \forall z \in \Omega \setminus \{0\}.$$ Clearly, when $\Omega = \mathbb{U}^n$ , the aforementioned inequality is equivalent to $$\operatorname{Re}\left\{\frac{\|z\|^2}{\langle J_g^{-1}(z)g(z), z\rangle}\right\} > \alpha, \quad \forall z \in \mathbb{U}^n \setminus \{0\}.$$ In case of $n=1, \Omega=\mathbb{U}$ and the above relation is equivalent to $$\operatorname{Re} \frac{zg'(z)}{g(z)} > 0, \quad z \in \mathbb{U}.$$ We denote by $\mathcal{S}_{\alpha}^{*}(\Omega)$ the set of all starlike mappings of order $\alpha$ on $\Omega$ .

Definition 2.2. [8](also see [16, 12]) Let $\Omega$ be a bounded starlike circular domain in $\mathbb{C}^n$ with $0 \in \Omega$ and its Minkowski functional $\rho \in C^1$ in $\mathbb{C}^n \setminus \{0\}$ . A normalized locally biholomorphic mapping $g: \Omega \to \mathbb{C}^n$ is said to be strongly starlike of order $\beta$ ( $0 < \beta \le 1$ ) if $$\left|\arg\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}J_g^{-1}(z)g(z)\right| < \frac{\pi}{2}\beta, \quad \forall z \in \Omega \setminus \{0\}.$$ Clearly, when $\Omega = \mathbb{U}^n$ , the aforementioned inequality is equivalent to $$|\arg\langle J_g^{-1}(z)g(z),z\rangle| < \frac{\pi}{2}\beta, \quad \forall z \in \mathbb{U}^n \setminus \{0\}.$$ In case of n = 1, $\Omega = \mathbb{U}$ and the above relation is equivalent to $$\left| \arg \frac{zg'(z)}{g(z)} \right| < \frac{\pi}{2}\beta, \quad \forall z \in \mathbb{U}.$$ We denote by $\mathcal{SS}^*_{\beta}(\Omega)$ the set of all strongly starlike mappings of order $\beta$ on $\Omega$ . Next, we recall the class $\mathcal{M}$ , which plays a fundamental role in the study of Loewner chains and Loewner differential equation in several complex variables (see [8, 22]. $$\mathcal{M} = \left\{ p \in \mathcal{H}(\Omega) : p(0) = 0, J_p(0) = I, \operatorname{Re} \frac{\partial \rho}{\partial z} p(z) > 0, z \in \Omega \setminus \{0\} \right\},\,$$ where $\partial \rho(z)/\partial z = (\partial \rho(z)/\partial z_1, \partial \rho(z)/\partial z_2, \cdots, \partial \rho(z)/\partial z_n)$ . Kohr [15] introduced the class $\mathcal{M}_{\Phi}$ on $\mathbb{U}^n$ , which is studied by Graham et al. [7] (see also [6]), where $\Phi: \mathbb{U} \to \mathbb{C}$ is a biholomorphic function such that $\Phi(0) = 1$ and $\operatorname{Re} \Phi(z) > 0$ on $\mathbb{U}$ . Recently, Xu et al. [28] considered the class $\mathcal{M}_{\Phi}$ on $\Omega \subset \mathbb{C}^n$ . Here, we add some more conditions on $\Phi$ and define the following subsets of $\mathcal{M}$ . Assumption 2.3. Let $\Phi : \mathbb{U} \to \mathbb{C}$ be a biholomorphic function such that $\Phi(0) = 1$ , $\Phi'(0) > 0$ , $\Phi''(0) \in \mathbb{R}$ and $\operatorname{Re} \Phi(z) > 0$ on $\mathbb{U}$ . Obviously, there are many functions which satisfy this assumption. Let $$\mathcal{M}_{\Phi} = \left\{ p \in \mathcal{H}(\Omega) : p(0) = 0, J_p(0) = I, \frac{\rho(z)}{2\frac{\partial \rho}{\partial z}p(z)} \in \Phi(\mathbb{U}), z \in \Omega \setminus \{0\} \right\}.$$ The class $\mathcal{M}_{\Phi}$ coincides with $\mathcal{M}$ for $\Phi(z) = (1+z)/(1-z), z \in \mathbb{U}$ . Also, if $\Omega = \mathbb{U}^n$ , then $$\mathcal{M}_{\Phi} = \left\{ p \in \mathcal{H}(\mathbb{U}^n) : p(0) = 0, J_p(0) = I, \frac{\|z\|^2}{\langle p(z), z \rangle} \in \Phi(\mathbb{U}), z \in \mathbb{U}^n \setminus \{0\} \right\}.$$ <span id="page-3-2"></span>Remark 2.1. Let $g \in \mathcal{H}(\mathbb{U})$ be a normalized locally biholomorphic function. If $J_g^{-1}(z)g(z) \in \mathcal{M}_{\Phi}$ , then for different choices of $\Phi$ , we obtain different important classes of $\mathcal{S}(\Omega)$ . For instance, if we take $\Phi(z) = (1+z)/(1-z)$ , $\Phi(z) = (1+(1-2\alpha)z)/(1-z)$ and $\Phi(z) = ((1+z)/(1-z))^{\beta}$ (where the branch point is chosen such that $((1+z)/(1-z))^{\beta} = 1$ at z = 0), then we easily obtain $g \in \mathcal{S}^(\Omega)$ , $g \in \mathcal{S}^_{\alpha}(\Omega)$ and $g \in \mathcal{S}\mathcal{S}^*_{\beta}(\Omega)$ , respectively. The following lemma helps us to prove the main results.