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Abstract

Let $f$ be analytic in the unit disk $\mathbb{D}= \{z \in \mathbb{C}~:~ |z| < 1\}$, and $\mathcal{S}$ be the subclass of normalized univalent functions given by $f(z)=\sum_{n=1}^{\infty}a_{n}z^{n},~a_{1}:=1$ for $z \in\mathbb{D}$. We present the sharp bounds of the third-order Hankel determinant for inverse functions when it belongs to of the class of Ozaki close-to-convex.

Results & Lemmas (8)

Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.

Lemma 1.1 Lemma 1.1. If, is of the form (1.5) with, such that then where, for some, and such that, and.
Lemma 1.1. If $p \in \mathcal{P}$ , is of the form (1.5) with $c_1 \geq 0$ , such that $c_1 \in [0,2]$ then $$2c_2 = c_1^2 + \nu \mu,$$ $$4c_3 = c_1^3 + 2c_1\nu\mu - c_1\nu\mu^2 + 2\nu\left(1 - |\mu|^2\right)\rho,$$ $$8c_4 = c_1^4 + 3c_1^2\nu\mu + (4 - 3c_1^2)\nu\mu^2 + c_1^2\nu\mu^3 + 4\nu(1 - |\mu|^2)(1 - |\rho|^2)\psi + 4\nu(1 - |\mu|^2)(c_1\rho - c\mu\rho - \bar{\mu}\rho^2),$$ where $\nu := 4 - c_1^2$ , for some $\mu$ , $\rho$ and $\psi$ such that $|\mu| \le 1$ , $|\rho| \le 1$ and $|\psi| \le 1$ .
Lemma 1.2 Lemma 1.2. Let be a function definded by Then following are true (a). for (b). for (c). for (d). for (d). for (d). for (e). for Proof(a).…
Lemma 1.2. Let $\psi_1, \psi_2, \psi_3, \psi_4 : [0,2] \to \mathbb{R}$ be a function definded by $$\psi_1(c) := -160c^2 + 16c^3 + 20c^4 - 4c^5 + \frac{5c^6}{4}$$ $$\psi_2(c) := 32c + 48c^2 + 32c^3 + 14c^4 - 10c^5 - \frac{13c^6}{2}$$ $$\psi_3(c) := -256 + 64c + 276c^2 - 48c^3 - 82c^4 + 8c^5 + \frac{29c^6}{4}$$ $$\psi_4(c) := 320 - 32c - 272c^2 - 32c^3 + 76c^4 + 10c^5 - 7c^6$$ $$\psi_5(c) := -64 - 64c + 48c^2 + 32c^3 - 12c^4 - 4c^5 + c^6$$ Then following are true (a) $$\psi_1(c) \leq 0$$ . for $c \in [0, 2]$ (b) $$\psi_1(c) + \psi_2(c) \le 0$$ . for $c \in (\frac{87137}{250000}, 2]$ (c) $$\psi_1(c) + \psi_2(c) = 0$$ . for $c \in (\frac{87137}{250000}, \frac{4511}{4000}]$ (d) $\psi_1(c) + \psi_2(c) + \psi_3(c) = 0$ . for $c \in (\frac{87137}{250000}, \frac{4511}{4000}]$ (d) $\psi_1(c) + \psi_2(c) + \psi_3(c) + 0.6\psi_4(c) \leq 0$ . for $c \in [\frac{4511}{4000}, 2]$ (d) $$\psi_1(c) + \psi_2(c) + \psi_3(c) + 0.6\psi_4(c) \le 0$$ . for $c \in \left[\frac{4511}{4000}, 2\right]$ (e) $$\psi_5(c) \leq 0$$ . for $c \in [0, 2]$ Proof(a). Since $4-c^2>0$ for $c\in[0,2]$ $$\psi_1(c) := -160c^2 + 16c^3 + 20c^4 - 4c^5 + \frac{5c^6}{4}$$ $$= -48c^2 - \frac{3c^6}{4} - 2c^2(4 - c^2)(14 - 2c + c^2) \le 0$$ $$Proof(b)$$ . Let $c = \frac{87137}{250000}t, 1 < t \le \frac{500000}{87137}$ . Then $$\psi_1(c(t)) + \psi_2(c(t)) := 11.1535t - 13.6064t^2 + 2.03249t^3 + 0.501798t^4 - 0.072018t^5 - 0.00941315t^6$$ $$\leq 0.0094t(1-t)(17.189-7.9049t+t^2)(69.0293+16.5646t+t^2) \leq 0$$ $$Proof(c)$$ . Let $c = \frac{87137}{250000}t, 1 < t \le \frac{563875}{174274}$ . Then $$\psi_1(c(t)) + \psi_2(c(t)) + \psi_3(c(t)) := -256. + 33.4606t + 19.9237t^2 - 0.708421t^4$$ $$-0.0308649t^5 + 0.00358596t^6$$ $$\leq 0.00358596(-18.403 + t)(8.71052 + t)$$ $$(15.6059 - 7.89366t + t^2)(28.5372 + 8.97902t + t^2) < 0$$ П Proof(d). Let $c = \frac{4511}{4000}t$ , $1 \le t \le \frac{8000}{4511}$ . Then $$\begin{split} \psi_1 + \psi_2 + \psi_3 + 06\psi_4 = &64 + 72.176t - 137.357t^2 - 45.8974t^3 \\ &+ 45.2907t^4 + 7.29666t^5 - 10.286t^6 \\ \leq &10.286(1-t)(0.511168+t) \\ &(4.07379 - 3.46659t + t^2)(3.06843 + 3.21981t + t^2) \leq 0 \end{split}$$ Proof(e). Since $4-c^2>0$ for $c\in[0,2]$ $$\psi_1 = (4 - c^2)^2 (-4 - 4c + c^2) \le 0$$
Lemma 1.3 Lemma 1.3. Let be a function definded by where define as lemma 1.2 for, Then for and 0 < x < 0.25
Lemma 1.3. Let $\Psi: [0, \frac{87137}{250000}] \times (0, 0.25) \to \mathbb{R}$ be a function definded by $$\Psi(c,x) = 320 + \psi_1(c) + \psi_2(c)x + \psi_3(c)x^2 + \psi_4(c)x^3 + \psi_5(c)x^4$$ where $\psi_1, \psi_2, \psi_3, \psi_4, \psi_5$ define as lemma 1.2 for $c \in [0, \frac{87137}{250000}]$ , Then $\Psi(c, x) \leq 320$ for $0 \leq c \leq \frac{87137}{250000}$ and 0 < x < 0.25
Lemma 1.4 Lemma 1.4. Let be a function definded by where for, Then for and
Lemma 1.4. Let $\Psi : \left[0, \frac{87137}{250000}\right] \times [0.25, 1] \to \mathbb{R}$ be a function definded by $\Phi(c, x) = \phi_1(x) + \phi_2(x)c + \phi_3(x)c^2 + \phi_4(x)c^3 + \phi_5(x)c^4 + \phi_6(x)c^5 + \phi_7(x)c^6$ where for $x \in [0.25, 1]$ , $$\phi_1(x) := -256x^2 + 320x^3 - 64x^4$$ $$\phi_2(x) := 32x + 64x^2 - 32x^3 - 64x^4$$ $$\phi_3(x) := -160 + 48x + 276x^2 - 272x^3 + 48x^4$$ $$\phi_4(x) := 16 + 32x - 48x^2 - 32x^3 + 32x^4$$ $$\phi_5(x) := 20 + 14x - 82x^2 + 76x^3 - 12x^4$$ $$\phi_6(x) := -4 - 10x + 8x^2 + 10x^3 - 4x^4$$ $$\phi_7(x) := \frac{5}{4} - \frac{13x}{2} + \frac{29x^2}{4} - 7x^3 + x^4$$ Then $\Phi(c, x) < 0$ for $0 \le c \le \frac{87137}{250000}$ and $0.25 \le x < 1$
Lemma 1.5 Lemma 1.5. Let be a function definded as in lemma 1.2. Then for and 0 < x < 0.6 Proof. Since,,,, and in for
Lemma 1.5. Let $\Psi: \left(\frac{87137}{250000}, \frac{4511}{4000}\right] \times (0, 0.6) \to \mathbb{R}$ be a function definded as in lemma 1.2. Then $\Psi(c, x) \leq 320$ for $\frac{87137}{250000} < c \leq \frac{4511}{4000}$ and 0 < x < 0.6 Proof. Since $\phi_4(c) > 0$ , $\phi_1(c)$ , $\phi_1(c) + \phi_2(c) < 0$ , $\phi_1(c) + \phi_2(c) + \phi_3(c) < 0$ , $\phi_1(c) + \phi_2(c) + \phi_3(c) < 0$ and $\phi_5(c) < 0$ in for $\frac{87137}{250000} < c \le \frac{4511}{4000}$ $$\Psi(c,x) \le 320 + (\phi_1(c) + \phi_2(c) + \phi_3(c) + 0.6\phi_4(c))x^2 + \phi_5(c)x^4 < 320.$$
Lemma 1.6 Lemma 1.6. Let be a function definded by where for, Then for and
Lemma 1.6. Let $\Gamma: \left(\frac{87137}{250000}, 1\right] \times [0.6, 1] \to \mathbb{R}$ be a function definded by $\Gamma(c,x) = \gamma_1(x) + \gamma_2(x)c + \gamma_3(x)c^2 + \gamma_4(x)c^3 + \gamma_5(x)c^4 + \gamma_6(x)c^5 + \phi_7(x)c^6$ where for $x \in [0.6, 1]$ , $$\gamma_1(x) := -256x^2 + 320x^3 - 64x^4$$ $$\gamma_2(x) := 96x^2 - 32x^3 - 64x^4$$ $$\gamma_3(x) := 164x^2 - 272x^3 + 48x^4$$ $$\gamma_4(x) := -32x^3 + 32x^4$$ $$\gamma_5(x) := -48x^2 + 76x^3 - 12x^4$$ $$\gamma_6(x) := -6x^2 + 10x^3 - 4x^4$$ $$\gamma_7(x) := 2x^2 - 7x^3 + x^4$$ Then $\Phi(c,x) < 0$ for $\frac{87137}{250000} < c \le 1$ and $0.6 \le x \le 1$
Lemma 1.7 Lemma 1.7. Let be a function definded as in lemma 1.2. Then for and Proof. Since,, for
Lemma 1.7. Let $\Psi: \left(1, \frac{4511}{4000}\right] \times [0.6, 1] \to \mathbb{R}$ be a function definded as in lemma 1.2. Then $\Psi(c, x) \leq 320$ for $1 < c \leq \frac{4511}{4000}$ and $0.6 \leq x \leq 1$ Proof. Since $\phi_1(c)$ , $\phi_1(c) + \phi_2(c) < 0$ , $\phi_1(c) + \phi_2(c) + \phi_3(c) < 0$ for $1 < c \le 1.12775$ $\Psi(c, x) \le 320 + (\phi_1(c) + \phi_2(c) + \phi_3(c))x^2 + \phi_4(c)x^3 + \phi_5(c)x^4$ $= 320 + (-256 + 96c + 164c^2 - 48c^4 - 6c^5 + 2c^6)x^2$ $+ (320 - 32c - 272c^2 - 32c^3 + 76c^4 + 10c^5 - 7c^6)x^3$ $+ (-64 - 64c + 48c^2 + 32c^3 - 12c^4 - 4c^5 + c^6)x^4$ $\le 320 - 23x^2 + 63x^3 - 53x^4 < 318.459.$
Lemma 1.8 Lemma 1.8. Let be a function definded as in lemma 1.2. Then for and
Lemma 1.8. Let $\Psi: (\frac{4511}{4000}, 2] \times [0, 1] \to \mathbb{R}$ be a function definded as in lemma 1.2. Then $\Psi(c, x) \leq 320$ for $\frac{4511}{4000} < c \leq 2$ and $0 \leq x \leq 1$
Function classes studied:

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