Generalized Briot-Bouquet differential equation based on new differential operator with complex connections

Lemma 1.3. If β ∈ [0, 1], z ∈ U then each function of the form • $$\sigma(z) = \beta + (1-\beta)\sqrt{1+z}$$ , • $$\sigma(z) = \beta + (1 - \beta)e^z$$ , • $$\sigma(z) = \beta + (1 - \beta)(1 + \sin(z)),$$ $$\bullet \ \sigma(z) = \beta + (1 - \beta)e^{e^z - 1},$$ has the upper and lower bound for all r ∈ (0, 1), θ ∈ [0, 2π) as follows: $$\min_{|z|=r} \Re(\sigma(z)) = \sigma(-r) = \min_{|z|=r} |\sigma(\xi)|$$ and $$\max_{|z|=r} \Re(\sigma(z)) = \sigma(r) = \max_{|z|=r} |\sigma(z)|.$$ Proof. The first and second type can be located in [8]. We only need to prove the third type. For β = 0, we have the function σ(z) = 1 + sin(z) (see [3]). It is clear that $$\sin(z) = \sin(re^{i\theta}) = \sin(r\cos(\theta))\cosh(r\sin(\theta)) + i\cos(r\cos(\theta))\sinh(r\sin(\theta))$$ therefore, we have $$\Re(\sigma(z)) = 1 + \sin(r\cos(\theta))\cosh(r\sin(\theta)).$$ Consequently, by taking r → 0, we obtain $$\min_{|z|=r} \Re(\sigma(z)) = 1 - \sin(r) = \min_{|z|=r} |\sigma(z)| = 1.$$ Moreover, we have | sin(rei^θ )| ² = cos² (r cos θ) sinh 2(r sin θ) + sin² 2(r cos θ) cosh 2(r sin r) ≤ sinh² (r); thus, this yields $$\max_{|z|=r} \Re(\sigma(z)) = 1 + \sin(r) = \max_{|z|=r} |\sigma(z)| \le 1 + \sinh^2(r).$$ Extend the above result, for β > 0, we have $$\min_{|z|=r} \Re(\sigma(z)) = \beta + (1-\beta)(1-\sin(r)) = \min_{|z|=r} |\sigma(z)| = 1,$$ and $$\max_{|z|=r} \Re(\sigma(z)) = \beta + (1-\beta)(1+\sin(r)) = \max_{|z|=r} |\sigma(z)| \le \beta + (1-\beta)(1+\sinh^2(r)).$$ Similarly for the last assertion. The next result can be found in [12]

Lemma 1.4. If $\tau > 0$ and $\sigma \in \mathfrak{H}[1,n]$ , then there are constants $\wp > 0$ and $\nu > 0$ with $\nu = \nu(\wp, \tau, n)$ so that $$\sigma(z) + \tau z \sigma'(z) \prec \left[\frac{1+z}{1-z}\right]^{\nu} \Rightarrow \sigma(z) \prec \left[\frac{1+z}{1-z}\right]^{\wp}.$$

Lemma 1.5. Let $\varphi(z)$ be a convex function in U and $h(z) = \varphi(z) + n\nu(z\varphi'(z))$ for $\nu > 0$ and n is a positive integer. If $\varrho \in \mathfrak{H}[\varphi(0), n]$ , and $$\varrho(z) + \nu z \varrho'(z) \prec h(z), \quad z \in U,$$ then $$\rho(z) \prec \varphi(z)$$ and this result is sharp.

Theorem 2.1. The class $\mathfrak{S}_m^*(\alpha, \kappa, \sigma)$ achieves the following inclusion: $$\mathfrak{S}_{m}^{}(\alpha,\kappa,\sigma)\subset\mathfrak{S}_{m}^{}(\alpha,\kappa,\gamma)\subset\mathfrak{S}_{m}^{*}(\alpha,\kappa),$$ where $\sigma$ is one of the type in Lemma 1.3 and $$\mathfrak{S}_{m}^{*}(\alpha,\kappa,\gamma) := \{ f \in \Lambda \Re \left( \frac{z(J_{\alpha,\kappa}^{m}f(z)))'}{J_{\alpha,\kappa}^{m}f(z)} \right) > \gamma \};$$ $$\mathfrak{S}_{m}^{*}(\alpha,\kappa) := \{ f \in \Lambda \Re \left( \frac{z(J_{\alpha,\kappa}^{m} f(z)))'}{J_{\alpha,\kappa}^{m} f(z)} \right) > 0 \}.$$ Proof. Let $Y \in \mathfrak{S}_m^*(\alpha, \kappa, \sigma)$ and let $\sigma(z) = \beta + (1 - \beta)\sqrt{1 + z}$ , then we have the inequality $$\frac{z(J^m_{\alpha,\kappa}f(z)))'}{J^m_{\alpha,\kappa}f(z)} \prec \beta + (1-\beta)\sqrt{1+z}, \quad z \in U.$$ In view of Lemma 1.3, we obtain $$\min_{|z|=1^{-}} \Re(\beta + (1-\beta)\sqrt{1+z}) < \Re\Big(\frac{z(J_{\alpha,\kappa}^{m}f(z)))'}{J_{\alpha,\kappa}^{m}f(z)}\Big) < \max_{|z|=1^{+}} \Re(\beta + (1-\beta)\sqrt{1+z}),$$ which yields $$\beta < \Re \Big( dfracz(J_{\alpha,\kappa}^m f(z)))' J_{\alpha,\kappa}^m f(z) \Big) < \beta + (1-\beta)\sqrt{2}.$$ Hence, we have $$\Re\left(\frac{z(J_{\alpha,\kappa}^mf(z)))'}{J_{\alpha,\kappa}^mf(z)}\right) > \beta := \gamma \ge 0,$$ consequently, we get the requested result. Consider $\sigma(z) = \beta + (1 - \beta)e^z$ , the we have $$\min_{|z|=1} \Re(\beta+(1-\beta)e^z) < \Re\left(\frac{z(J_{\alpha,\kappa}^m f(z)))'}{J_{\alpha,\kappa}^m f(z)}\right) < \max_{|z|=1} \Re(\beta+(1-\beta)e^z),$$ which implies $$(\beta + (1-\beta)\frac{1}{e}) < \Re\left(\frac{z(J_{\alpha,\kappa}^m f(z)))'}{J_{\alpha,\kappa}^m f(z)}\right) < (\beta + (1-\beta)e),$$ that is $$\Re\Big(\frac{z(J^m_{\alpha,\kappa}f(z)))'}{J^m_{\alpha,\kappa}f(z)}\Big) > (\beta + (1-\beta)\frac{1}{e}) := \gamma \geq 0.$$ Similarly, by letting $\sigma(z) = \beta + (1 - \beta)(1 + \sin(z))$ then we have $$\min_{|z|=1} \Re(\beta + (1-\beta)(1+\sin(z))) < \Re\Big(\frac{z(J_{\alpha,\kappa}^m f(z)))'}{J_{\alpha,\kappa}^m f(z)}\Big) < \max_{|z|=1} \Re(\beta + (1-\beta)(1+\sin(z))),$$ which leads to $$(\beta + 0.158(1 - \beta)) < \Re\left(\frac{z(J_{\alpha,\kappa}^m f(z)))'}{J_{\alpha,\kappa}^m f(z)}\right) < (\beta + 1.841(1 - \beta)),$$ this brings the inequality $$\Re\left(\frac{z(J_{\alpha,\kappa}^mf(z)))'}{J_{\alpha,\kappa}^mf(z)}\right) > (\beta + 0.158(1-\beta)) := \gamma \ge 0.$$ Remark 2.2. In Theorem 2.1, - $m = 0, \beta = 0, \sigma(z) = 1 + \sin z \Longrightarrow [3]$ : - $m=0 \Longrightarrow [8]$ ; - $m = 0, \beta = 0, \sigma(z) = e^z \Longrightarrow [7]$ - $m = 0, \beta = 0, \sigma(z) = \sqrt{1+z} \Longrightarrow [7].$

Theorem 2.3. The class $\mathfrak{S}_m^*(\alpha, \kappa, \sigma)$ achieves the following inclusion: $$\mathfrak{S}_{m}^{*}(\alpha,\kappa,\sigma)\subset\mathfrak{M}_{m}(\alpha,\kappa,\gamma):=\{f\in\Lambda\Re\Big(\frac{z(J_{\alpha,\kappa}^{m}f(z)))'}{J_{\alpha,\kappa}^{m}f(z)}\Big)<\gamma,\,\gamma>1\}.$$ where $\sigma$ is given in Lemma 1.3. The set $\mathfrak{M}_m(\alpha, \lambda, \gamma)$ is a generalization of the set $$\mathfrak{M}(\gamma) := \{ f \in \Lambda \Re \left( \frac{z(J_{\alpha,\lambda}^m f(z)))'}{J_{\alpha,\lambda}^m f(z)} \right) < \gamma, \, \gamma > 1 \}$$ given by Uralegaddi et al. [15]. Proof. Let $f \in \mathfrak{S}_m^*(\alpha, \lambda, \sigma)$ , where $\sigma$ is given in Lemma 1.3. By the proof of Theorem 2.1, we have $$\Re\Big(\frac{z(J_{\alpha,\lambda}^mf(z)))'}{J_{\alpha,\lambda}^mf(z)}\Big)<\beta+(1-\beta)\sqrt{2}:=\gamma,$$ $$\Re\left(\frac{z(J_{\alpha,\lambda}^m f(z)))'}{J_{\alpha,\lambda}^m f(z)}\right) < \beta + (1-\beta)e := \gamma$$ and $$\Re\left(\frac{z(J_{\alpha,\kappa}^mf(z)))'}{J_{\alpha,\kappa}^mf(z)}\right)<(\beta+1.841(1-\beta)):=\gamma,$$ Hence, $f \in \mathfrak{M}_m(\alpha, f(z), \gamma)$ , where the value of $\gamma$ based on the function $\sigma$ , which completes the proof. Remark 2.4. In Theorem 2.3, - $m = 0, \beta = 0, \sigma(z) = 1 + \sin z \Longrightarrow [3];$ - $m = 0, \sigma(z) = \beta + (1 \beta)e^z \Longrightarrow [8]$ , Theorem 2.5; - $m = 0, \sigma(z) = \beta + (1 \beta)(\sqrt{1 + z}) \Longrightarrow [8]$ , Theorem 2.6. - $m = 0, \beta = 0, \sigma(z) = (\sqrt{1+z}) \Longrightarrow [8]$ , Corollary 2.7.

Theorem 2.5. If $f \in \Lambda$ satisfies the subordination $$\left(\frac{z(J_{\alpha,\kappa}^m f(z))'}{J_{\alpha,\kappa}^m f(z)}\right) \left(2 + \frac{z(J_{\alpha,\kappa}^m f(z))''}{(J_{\alpha,\kappa}^m f(z))'} - \frac{z(J_{\alpha,\kappa}^m f(z))'}{J_{\alpha,\kappa}^m f(z)}\right) \prec \left[\frac{1+z}{1-z}\right]^{\tau}$$ then $$f \in \mathfrak{S}_m^*(\alpha, \kappa, \sigma)$$ , where $\sigma(z) = \left\lceil \frac{1+z}{1-z} \right\rceil^{\wp}$ for $\wp > 0, \tau > 0$ . Proof. We aim to apply Lemma 1.4, a calculation implies that $$\begin{split} &\left(\frac{z(J_{\alpha,\kappa}^{m}f(z))'}{J_{\alpha,\kappa}^{m}f(z)}\right) + z\left(\frac{z(J_{\alpha,\kappa}^{m}f(z))'}{J_{\alpha,\kappa}^{m}f(z)}\right)' \\ &= \left(\frac{z(J_{\alpha,\kappa}^{m}f(z))'}{J_{\alpha,\kappa}^{m}f(z)}\right) \left(2 + \frac{z(J_{\alpha,\kappa}^{m}f(z))''}{(J_{\alpha,\kappa}^{m}f(z))'} - \frac{z(J_{\alpha,\kappa}^{m}f(z))'}{J_{\alpha,\kappa}^{m}f(z)}\right) \\ &\prec \left[\frac{1+z}{1-z}\right]^{\tau}. \end{split}$$ Thus, in view of Lemma 1.4, we have $$\left(\frac{z(J_{\alpha,\kappa}^m f(z))'}{J_{\alpha,\kappa}^m f(z)}\right) \prec \left[\frac{1+z}{1-z}\right]^{\wp} := \sigma(z),$$ which implies that $f \in \mathfrak{S}_m^*(\alpha, \kappa, \sigma)$ .

Theorem 2.6. Let $\varphi$ be a convex function such that $\varphi(0) = 0$ and let $\hbar$ be the function $$hbar (z) = \varphi(z) + \frac{z}{1-\ell} \varphi'(z), \quad z \in U, \ell \in (0,1).$$ If for a function $f \in \Lambda$ satisfies the subordination $$\left(\frac{z}{J_{\alpha,\kappa}^{m+1}f(z)}\right)^{\ell}\frac{J_{\alpha,\kappa}^{m}f(z)}{1-\ell}\left(\frac{\left(J_{\alpha,\kappa}^{m+1}f(z)\right)'}{J_{\alpha,\kappa}^{m+1}f(z)}-\ell\frac{\left(J_{\alpha,\kappa}^{m}f(z)\right)'}{J_{\alpha,\kappa}^{m}f(z)}\right)\prec \hbar(z)$$ then $$\left(\frac{J_{\alpha,\kappa}^{m+1}f(z)}{z}\right)\left(\frac{z}{J_{\alpha,\kappa}^{m+1}f(z)}\right)^{\ell} \prec \varphi(z), \quad z \in U.$$ The outcome is sharp. Proof. We aim to apply Lemma 1.5. Let $$\varrho(z) = \left(\frac{J_{\alpha,\kappa}^{m+1} f(z)}{z}\right) \left(\frac{z}{J_{\alpha,\kappa}^{m+1} f(z)}\right)^{\ell}.$$ A differentiation implies that $$\left(\frac{\xi}{J_{\alpha,\kappa}^{m+1}f(z)}\right)^{\ell}\frac{J_{\alpha,\kappa}^{m}f(z)}{1-\ell}\left(\frac{\left(J_{\alpha,\kappa}^{m+1}f(z)\right)'}{J_{\alpha,\kappa}^{m+1}f(z)}-\ell\frac{\left(J_{\alpha,\kappa}^{m}\right)f(z)\right)'}{J_{\alpha,\kappa}^{m}f(z)}\right)=\varrho(z)+\left(\frac{1}{1-\ell}\right)z\varrho'(z)$$ Thus, by the assumption, we have $$\varrho(z) + (\frac{1}{1-\ell})z\varrho'(z) \prec \hbar(z) = \varphi(z) + \frac{z}{1-\ell}\varphi'(z), \quad z \in U.$$ By employing Lemma 1.5 yields $\varrho(z) \prec \hbar(z)$ , which means $$\left(\frac{J_{\alpha,\kappa}^{m+1}f(z)}{z}\right)\left(\frac{z}{J_{\alpha,\kappa}^{m+1}f(z)}\right)^{\ell} \prec \varphi(z), \quad z \in U.$$ This result is sharp. Remark 2.7. In Theorem 2.5, $\lambda = 0 \Longrightarrow$ [2] Theorem 2.14. Example 2.8. Consider the following data: $$f(z) = \frac{z}{1-z}, \quad \alpha = \kappa = m = 1$$ then, we have $$\int_{0.5}^{m} f(z) = z + 2z^2 + 4z^3 + 4z^4 + 6z^5 + 6z^6 + O(z^7), |z| < 1$$ and $$P(z) = \frac{z(J^m_{\alpha,\lambda}f(z)))'}{J^m_{\alpha,\lambda}f(z)} = 1 + 2^z + 4z^2 - 4z^3 + 8z^4 + 2z^5 + O(z^6), |z| < 1.$$ By using Wolram Alpha, we have $$1 \le \Re(P(z)) \le 1.83929 : \gamma, \beta \to 0.$$ Moreover, the operator $J_{\alpha,\lambda}^m f(z)$ is univalent for $-0.0001 < \kappa \le 0.0001$ , where (Biberbach Conjecture [5]) $$\int_{a_n}^m f(z) = z + 2z^2 + 3.000z^3 + 4z^4 + 5.000z^5 + 6z^6 + O(z^7), |a_n| \le n.$$ Then in view of Theorem 2.3, we have $J_{\alpha,\lambda}^m f(z) \in \mathfrak{M}_m(\alpha,\kappa,\gamma)$ , with $\sigma(z) = \beta + (1-\beta)(1+\sin(z))$ . The roots of P(z) in U $$R_{1,2} = -0.41 \pm 0.32i, R_{3,4} = -0.49 \pm 0.46i, R_{5,6} = -0.11 \pm 0.47i, R_{7,8} \approx \pm 1.$$

Definition 1.2. Let $\epsilon \in [0,1)$ , $\alpha \geq 0$ , $\kappa \geq 0$ , and $m \in \mathbb{N}$ . A function $f \in \Lambda$ is said to be in the set $\mathfrak{S}_m(\alpha,\kappa,\sigma)$ if and only if $$\frac{(J_{\alpha,\kappa}^m f(z))'}{J_{\alpha,\kappa}^m f(z)} \prec \sigma(z), \quad z \in U,$$ where $\sigma$ is univalent function with a positive real part in U satisfying $\sigma(0) = 1$ , $\Re(\sigma'(z)) > 0$ . For functions f and g in $\Lambda$ we say that f is subordinate to g, denoted by $f \prec g$ , if there exists a Schwarz function $\omega \in U$ with $\omega(0) = 0$ and $|\omega(z)| < 1$ , $z \in U$ so that $f(z) = g(\omega(z))$ for all $z \in U$ (see [12]). Evidently $f(z) \prec g(z)$ is equivalent to f(0) = g(0) and $f(U) \subset g(U)$ . Note that the class $\mathfrak{S}_m^(\alpha, \lambda, \sigma)$ is a generalization of some classes of analytic functions. Moreover, this class is a specialist of Ma and Minda class [10] given as follows $(\mathfrak{S}^(\sigma))$ : $$\frac{zf'(z)}{f(z)} \prec \sigma(z).$$ Moreover, when $\sigma(z) = 1 + \sin(z)$ and m = 0, the class $$\frac{zf'(z)}{f(z)} \prec 1 + \sin(z)$$ is studied by Cho et al. [3]. Our class is generalization of two classes given by Khatter et al [8] as follows: $$\frac{zf'(z)}{f(z)} \prec \beta + (1 - \beta)\sqrt{1 + z}$$ and $$\frac{zf'(z)}{f(z)} \prec \beta + (1 - \beta)e^z,$$ where $\beta = 0$ introduces the class [11] $$\frac{zf'(z)}{f(z)} \prec e^z.$$ Kumar et al. [9] defined the class by using Bell numbers as follows: $$\frac{zf'(z)}{f(z)} \prec e^{e^z - 1}.$$