Amal M. Darweesh*
| Waggas G. Atshan | Ali H. Battor | Mohammed S. Mahdi
© 2023 IIETA. This article is published by IIETA and is licensed under the CC BY 4.0 license (http://creativecommons.org/licenses/by/4.0/).
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In this study, we introduce a novel subclass of bi-univalent functions, which are of considerable interest in various fields of mathematics, including complex analysis and geometric function theory. By employing the property of subordination, we define these bi-univalent functions as $\mathcal{R}(t, \gamma, \lambda)$ and impose constraints on the coefficients $\left|a_n\right|$. Our investigation provides the upper bounds for the bi-univalent functions in this newly developed subclass, specifically for n=2, 3, 4, and 5. We then derive the third Hankel determinant for this particular class, which reveals several intriguing scenarios. These findings contribute to the broader understanding of bi-univalent functions and their potential applications in diverse mathematical contexts. Notably, the results obtained may serve as a foundation for future investigations into the properties and applications of bi-univalent functions and their subclasses.
bi-univalent functions, Chebyshev polynomials, subordination, third Hankel
Let N refers to the collection of functions $f$ analytic in the open unit disk $U=\{z: z \in \mathbb{C}$ and $|z|<1\}$. An analytic function $f \in N$ has Taylor series expansion of the form:
$f(z)=z+\sum_{i=2}^{\infty} a_j z^j, \quad(z \in U)$. (1)
The class of all functions in N refers by Nu which are univalent in U. The Koebe One-Quarter Theorem [1] ensures that the image of U under each $f \in N_u$ has a disk of radius ¼. Obviously, each $f \in N_u$ contains an inverse function f-1is satisfying f-1(f(z))=z and $f^{-1}(f(w))=w,\left(\left(|w|<r_0(f), r_0(f) \geq 1 / 4\right)\right)$, where,
$\begin{aligned} g(w)=f^{-1}(w) & =w-a_2 w^2+\left(2 a_2^2-a_3\right) w^3 -\left(5 a_2^3-5 a_2 a_3+a_4\right) w^4+\cdots, (w \in U)\end{aligned}$. (2)
For two $f$ and $\Phi$ be analytic functions, $f$ is told to be subordinate to $\Phi$ in $U$ and written as $f(z) \prec \Phi(z)$, if there exists a Schwarz function $w$ be analytic such that $f(z)=\Phi(w(z))$ with $w(0)=0$ and $|\mathrm{w}(\mathrm{z})| \leq 1,(z \in U)$. A function $f \in \Sigma$ is said to be bi- univalent in $U$ if both $f(z)$ and $f^1(z)$ are univalent in U.
In 1967, Lewin [2], for all function $f \in \Gamma$ of the form (1), and let $\Gamma$ refer to the class of bi-univalent functions in $U$. The second coefficient of $f$ satisfyies the estimate $\left|a_2\right|<1.51$. In 1967, Brannan and Clunie [3] conjectured that $\left|a_2\right| \leq \sqrt{2}$ for $f \in \Gamma$. After that Al-Shaqs [4] proved that $\left|a_2\right|=4 / 3$. In 1985 it was stated that Brannan-Clunie conject for bi-starlike function [5]. Brannan and Taha [6] gained evaluation estimates on the initial coefficients $\left|a_2\right|$ as well as $\left|a_3\right|$ for functions in the classes of bi- starlike functions of order $\rho$ denoted $E_{\Omega}^*(\rho)$ and bi-convex functions of order $\rho$ symbol $Y_{\Omega}(\rho)$. For all of the function classes $E_{\Omega}^*(\rho)$ and YΩ(ρ), non-sharp estimates on the first two Taylor-Maclaurin coefficients were found in studies [6-10]. Many researchers [10-15] have studied numerous curious subclasses of the bi-univalent function class Ω and observed non-sharp bounds on the first two Taylor-Maclaurin coefficient. As well as the coefficient problem for all of the Taylor-Maclaurin coefficient $\left|a_2\right|$, n=3,4,… is as yet an open problem [2, 16].
Noonan and Thomas [3] defined qth Hankel determinant of f, in 1976 for n≥1 and q≥1 by:
$\mathrm{H}_q(n)=\left|\begin{array}{ccc}a_n & a_{n+1} \cdots & a_{n+q-1} \\ a_{n+1} & a_{n+2} \cdots & a_{n+q} \\ \vdots & \vdots & \vdots \\ a_{n+q-1} & a_{n+q} \cdots & a_{n+2 q-2}\end{array}\right|,\left(a_1=1\right)$.
For $q=2$ and $n=1$, we know that the function $\mathrm{H}_2(1)=a_3-$ $a_2^2$. The second Hankel determinant $\mathrm{H}_2(2)$ is given through $\left|\mathrm{H}_2(2)\right|=\left|a_2 a_4-a_3^2\right|$ for the classes of bi- starlike and bi-convex [17, 18] and third Hankel determinant, this functions are studied by Babalola [19] functional given by:
$\mathrm{H}_3(1)=\left|\begin{array}{lll}a_1 & a_2 & a_3 \\ a_2 & a_3 & a_4 \\ a_3 & a_4 & a_5\end{array}\right| \quad\left(a_1=1\right)$ and $(n=1, q=3)$.
By applying triangle inequality for H3(1) we have:
$\begin{gathered}\left|H_3(1)\right| \leq\left|a_3\right|\left|a_2 a_4-a_3^2\right|-\left|a_4\right|\left|a_4-a_2 a_3\right| +\left|a_5\right|\left|a_3-a_2^2\right| .\end{gathered}$ (3)
The Chebyshev polynomials are divided into four kinds. Chebyshev polynomials, which we include in this idea, play an important role in numerical analysis and mathematical physics. The majority of research papers on real orthogonal polynomials of the Chebyshev family include primarily findings of Chebyshev polynomials of the first and second kinds, Tn(x) and Un(x), as well as their numerous uses in diverse applications [20, 21]. Known widely sundry of the Chebyshev polynomials are the first and second kinds. In the case of real variable x on (1), the first and second kinds of Chebyshev polynomials that you are familiar with are: Tn(x)=cos(n arccosx),
$U_n(x)=\frac{\sin [(n+1) \arccos x]}{\sin (\arccos x)}=\frac{\sin [(n+1) \arccos x]}{\sqrt{1-x^2}}$.
The function stated below was taken into consider ion in this work: $N(t, z)=\frac{1}{1-2 t z+z^2} \, , \, t \in\left(\frac{1}{2}, 1\right), z \in U$.
It is familiar that if $t=\cos \rho, \rho \in\left(0, \frac{\pi}{3}\right)$. Then $N(t, z)=1+$ $\sum_{n=1}^{\infty} \frac{\sin [(n+1) \rho]}{\sin \rho} z^n=1+2 \cos \rho z+\left(3 \cos ^2 \rho-\sin ^2 \rho\right) z^2+$ $\left(8 \cos ^3 \rho-4 \cos \rho\right) z^3+\cdots, z \in U$, that is:
$\begin{gathered}N(t, z)=1+U_1(t) z+U_2(t) z^2+U_3(t) z^3+U_4(t) z^4 +\cdots, t \in\left(\frac{1}{2}, 1\right), z \in U, \end{gathered}$ (4)
where,
$U_n(t)=\frac{\sin [(n+1) \arccos t]}{\sqrt{1-t^2}}, n \in \mathbb{N}$,
where, Un(t) are the second kind Chebyshev polynomials. It is understood from the concept of the second kind Chebyshev polynomials that $U_{n+1}(t)=2 t U_n(t)-U_{n-2}(t)$.
We get that
$U_1(t)=2 t, U_2(t)=4 t^2-1, U_3(t)=8 t^3-4 t$, (for each $n \in \mathbb{N}$). (5)
Lemma 1: [1] If P be a class of all analytic functions p(z) of the form:
$p(\mathrm{z})=1+\sum_{n=1}^{\infty} p_n z^n$, (6)
with p(0)=1 and $\mathfrak{R} e\{p(\mathrm{z})\}>0$ for all $\mathrm{z} \in \mathrm{U}$. Then $\left|p_n\right| \leq 2$, for every (n=1,2,3,…). This disparity is sharp for each n.
Lemma 2: [22] If $p \in P$, then
$\begin{gathered}2 p_2=p_1^2+\left(4-p_1^2\right) x, \\ 4 p_3=p_1^3+2\left(4-p_1^2\right) p_1 x-\left(4-p_1^2\right) p_1 x^2+2\left(4-p_1^2\right)\left(1-|x|^2\right) z,\end{gathered}$
for some $x, z$, with $|x| \leq 1,|z| \leq 1$.
The aim of this idea is to evaluate the third Hankel determinant for class of bi-univalent functions.
Definition 1: A function $f \in \Sigma$, given by (1) is said to be in the class $\mathcal{R}(t, \gamma, \lambda)$, if the following conditions are satisfied holds:
$\frac{1}{\gamma}\left\{(1-\lambda) \frac{f(z)}{z}+\lambda f^{\prime}(z)+z f^{\prime \prime}(z)\right\} \prec N(t, z)$, (7)
and
$\frac{1}{\gamma}\left\{(1-\lambda) \frac{g(w)}{w}+\lambda g^{\prime}(w)+w g^{\prime \prime}(w)\right\} \prec N(t, w)$, (8)
where, $\gamma \in \mathbb{C} \backslash\{0\}$ and $\lambda \geq 1, z, w \in U$ and $g=f^{-1}$, where the functional $g$ is given by (2).
Remark 1: A function $f \in \Sigma$, given by (1) is said to be in the class $\mathcal{R}(t, \gamma, \lambda)(\lambda=1)$, if the following conditions are satisfied holds:
$\frac{1}{\gamma}\left\{f^{\prime}(z)+z f^{\prime \prime}(z)\right\}<N(t, z)$,
and
$\frac{1}{\gamma}\left\{g^{\prime}(w)+w g^{\prime \prime}(w)\right\} \prec N(t, w)$.
Remark 2: A function $f \in \Sigma$, given by (1) is said to be in the class $\mathcal{R}(t, \gamma, \lambda),(\gamma=1)$, if the following conditions are satisfied holds:
$\left\{(1-\lambda) \frac{f(z)}{z}+\lambda f^{\prime}(z)+z f^{\prime \prime}(z)\right\} \prec N(t, z)$,
and
$\left\{(1-\lambda) \frac{g(w)}{w}+\lambda g^{\prime}(w)+w g^{\prime \prime}(w\} \prec N(t, w)\right.$.
Theorem 1: If f is given by (1) belongs to the subclass $\mathcal{R}(t, \gamma, \lambda), t \in\left(\frac{1}{2}, 1\right)$, then
$\left|a_2 a_4-a_3{ }^2\right| \leq \begin{cases}D(t, 2-), & \text { if } \Lambda(\xi, t) \geq 0 \text { and } \mathcal{C}(\xi, t) \geq 0 \\ \max \left\{\frac{4 \gamma^2 t^2}{(7+2 \lambda)^2}, D(t, 2-)\right\}, & \text { if } \Lambda(\xi, t)>0 \text { and } \mathcal{C}(\xi, t)<0, \\ \frac{4 \gamma^2 t^2}{(7+2 \lambda)^2} & \text { if } \Lambda(\xi, t) \leq 0 \text { and } \mathcal{C}(\xi, t) \leq 0, \\ \max \left\{D\left(t, \tau_0\right), D(t, 2-)\right\}, & \text { if } \Lambda(\xi, t)<0 \text { and } \mathcal{C}(\xi, t)>0,\end{cases}$ (9)
where
$D(t, 2-)=\frac{4 \gamma^2 t^2}{(7+2 \lambda)^2}+\frac{\Lambda(\xi, t)+\mathcal{C}(\xi, t)}{2(3+\lambda)^4(7+2 \gamma)^2(13+3 \lambda)}$,
$D\left(t, \tau_0\right)=\frac{4 \gamma^2 t^2}{(7+2 \lambda)^2}-\frac{\mathcal{c}^2(\xi, t)}{8 \Lambda(\xi, t)(3+\lambda)^4(7+2 \lambda)^2(13+3 \lambda)}$, $\tau_0=\sqrt{\frac{-2 \mathcal{C}(\xi, t)}{\Lambda(\xi, t)}}$
$\begin{gathered}\Lambda(\xi, t)=2 \gamma^2 \mathrm{t}\left\{4(3+\lambda)^3(7+2 \lambda)^2\left[4 t^3+4 t^2-3 \mathrm{t}-1\right]+\right. \\ 2 \mathrm{t}(13+3 \lambda)\left[2(3+\lambda)^4-10 t(3+\lambda)^2(7+2 \lambda)^2-\right. \left.\left.8 \gamma^2 t^2(7+2 \lambda)^2\right]\right\},\end{gathered}$
$\begin{gathered}\mathcal{C}(\xi, t)=2 \gamma^2 \mathrm{t}(3+\lambda)^2\left\{4(3+\lambda)(7+2 \lambda)^2\left[\mathrm{t}+4 t^2-1\right]+\right. \\ \left.2 \mathrm{t}(13+3 \lambda)\left[10 t-4(3+\lambda)^2\right]\right\} .\end{gathered}$
Proof: Since $f \in \mathcal{R}(t, \gamma, \lambda)$. Then, we have
$\begin{aligned} \frac{1}{\gamma}\left\{(1-\lambda) \frac{f(z)}{z}\right. & \left.+\lambda f^{\prime}(z)+z f^{\prime \prime}(z)\right\}=N(t, m(z)),\end{aligned}$ (10)
and
$\begin{aligned} \frac{1}{\gamma}\left\{(1-\lambda) \frac{g(w)}{w}\right. & +\lambda g^{\prime}(w)+w g^{\prime \prime}(w\} =N(t, n(w)) .\end{aligned}$ (11)
Let $p, q \in P$ is defined as by follows
$p(z)=\frac{1+\mathcal{S}(z)}{1-\mathcal{S}(z)}=1+p_1 z+p_2 z^2+p_3 z^3+\cdots$,
and
$q(w)=\frac{1+\vartheta(w)}{1-\vartheta(w)}=1+q_1 w+q_2 w^2+q_3 w^3+\cdots$.
It follows that:
$\begin{array}{r}\mathcal{S}(z)=\frac{p(z)-1}{p(z)+1}=\frac{1}{2} p_1 z+\frac{1}{2}\left(p_2-\frac{p_1^2}{2}\right) z^2 +\frac{1}{2}\left(p_3-p_1 p_2+\frac{p_1^3}{4}\right) z^3+\cdots,\end{array}$ (12)
and
$\begin{aligned} & \vartheta(w)=\left(\frac{q(w)-1}{q(w)+1}\right) \quad=\frac{1}{2} q_1 w+\frac{1}{2}\left(q_2-\frac{q_1^2}{2}\right) w^2 \, +\frac{1}{2}\left(q_3-q_1 q_2+\frac{q_1^3}{4}\right) w^3+\cdots\end{aligned}$. (13)
From (12) and (13), we take into account (5), we can easy to conclude
$N(t, \mathcal{S}(z))=1+\frac{U_1(t)}{2} p_1 z^{+}\left[\frac{U_1(t)}{2}\left(p_2-\frac{p_1^2}{2}\right)+\frac{U_2(t)}{4} p_1^2\right] z^2+\left[\frac{U_1(t)}{2}\left(p_3-p_1 p_2+\frac{p_1^3}{4}\right)+\frac{U_2(t)}{2} p_1\left(p_2-\frac{p_1^2}{2}\right)+\frac{U_3(t)}{8} p_1^3\right]$
$z^3+\left[\frac{U_1(t)}{2}\left(p_4-\frac{p_2^2}{2}+p_1\left(\frac{p_1^3}{4}+\frac{3}{4} p_1 p_2-p_3\right)\right)+\frac{U_2(t)}{2}\left(p_2^2+p_1\left(\frac{3 p_1^3}{8}+\frac{3}{2} p_1 p_2+\frac{p_1}{2}+\frac{p_3}{2}\right) \frac{3}{8} U_3(t) p_1\left(p_1^3+p_1 p_2\right)+U_4(t) p_1^4\right] z^4+\cdots\right.$, (14)
and
$\begin{aligned} N(t, \vartheta(w))=1 & +\frac{U_1(t)}{2} q_1 w+\left[\frac{U_1(t)}{2}\left(q_2-\frac{q_1^2}{2}\right)+\frac{U_2(t)}{4} q_1^2\right] w^2 +\left[\frac{U_1(t)}{2}\left(q_3-q_1 q_2+\frac{q_1^3}{4}\right)+\frac{U_2(t)}{2} q_1\left(q_2-\frac{q_1^2}{2}\right)+\frac{U_3(t)}{8} q_1^3\right]\end{aligned}$
$w^3+\left[\frac{U_1(t)}{2}\left(q_4-\frac{q_2^2}{2}+q_1\left(\frac{q_1^3}{4}+\frac{3}{4} q_1 q_2-q_3\right)\right)+\frac{U_2(t)}{2}\left(q_2^2+q_1\left(\frac{3 q_1^3}{8}+\frac{3}{2} q_1 q_2+\frac{q_1}{2}+\frac{q_3}{2}\right)-\frac{3}{8} U_3(t) q_1\left(q_1^3+q_1 q_2\right)+U_4(t) q_1^4\right] w^4+\cdots\right.$. (15)
From (10) and (14), we obtain that
$\frac{3+\lambda}{\gamma} a_2=\frac{U_1(t)}{2} p_1$, (16)
$\frac{7+2 \lambda}{\gamma} a_3=\frac{U_1(t)}{2}\left(p_2-\frac{p_1^2}{2}\right)+\frac{U_2(t)}{4} p_1^2$ (17)
$\begin{aligned} \frac{13+3 \lambda}{\gamma} a_4 & =\frac{U_1(t)}{2}\left(p_3-p_1 p_2+\frac{p_1^3}{4}\right) +\frac{U_2(t)}{2} p_1\left(p_2-\frac{p_1^2}{2}\right) +\frac{U_3(t)}{8} p_1^3 .\end{aligned}$ (18)
Also, from (11) and (15), we obtain that
$-\frac{3+\lambda}{\gamma} a_2=\frac{U_1(t)}{2} q_1$, (19)
$\frac{7+2 \lambda}{\gamma}\left(2 a_2^2-a_3\right)=\frac{U_1(t)}{2}\left(q_2-\frac{q_1^2}{2}\right)+\frac{U_2(t)}{4} q_1^2$, (20)
$\begin{aligned} & -\frac{13+3 \lambda}{\gamma}\left(5 a_2^3-5 a_2 a_3+a_4\right) \, =\frac{U_1(t)}{2}\left(q_3-q_1 q_2+\frac{q_1^3}{4}\right) \, +\frac{U_2(t)}{2} q_1\left(q_2-\frac{q_1^2}{2}\right)+\frac{U_3(t)}{8} q_1^3 .\end{aligned}$ (21)
From (16) and (19), we get
$\mathrm{a}_2=\frac{\gamma \mathrm{U}_1(\mathrm{t})}{2(3+\lambda)} \mathrm{p}_1=\frac{-\gamma \mathrm{U}_1(\mathrm{t})}{2(3+\lambda)} \mathrm{q}_1$, (22)
It follows that its
$p_{1=-} q_1$. (23)
Subtracting (20) and (17) and considering (22), we get
$a_3=\frac{\gamma^2 U_1^2(t)}{4(3+\lambda)^2} \,\, p_1^2+\frac{\gamma U_1(t)}{4(7+2 \lambda)}\left(p_2-q_2\right)$. (24)
Also, subtracting (21) from (18) and considering (22) and (24), we find that
$\begin{aligned} \frac{1}{\gamma}[(16(13+3 \lambda) & \left.\left.a_4+8(13+3 \lambda)\right)\left(5 a_2^3-5 a_2 a_3\right)\right] =4 U_1(t)\left(p_3-q_3\right) \\ & +4\left(U_2(t)-U_1(t)\right) \mathrm{p}_1\left(\mathrm{p}_2+\mathrm{q}_2\right) +2\left(\left(\mathrm{U}_1(\mathrm{t})-2 \mathrm{U}_2(\mathrm{t})+\mathrm{U}_3(\mathrm{t})\right)\right) \mathrm{p}_1^3\end{aligned}$,
then
$\begin{aligned} & a_4=\frac{-5}{2}\left(a_2^3-a_2 a_3\right)+\frac{2 \gamma U_1(t)\left(p_3-q_3\right)}{8(13+3 \lambda)}+\frac{2 \gamma\left(U_2(t)-U_1(t)\right)}{8(13+3 \lambda)} \mathrm{p}_1\left(\mathrm{p}_2+\mathrm{q}_2\right) +\frac{\gamma\left(\mathrm{U}_1(\mathrm{t})-2 \mathrm{U}_2(\mathrm{t})+\mathrm{U}_3(\mathrm{t})\right)}{8(13+3 \lambda)} p_3\end{aligned}$,
since
$\begin{aligned} \frac{-5}{2}\left(a_2^3-a_2 a_3\right) & =-\frac{5 \gamma^3 \mathrm{U}_1^3(\mathrm{t})}{16(3+\lambda)^3} \mathrm{p}_1^3+\frac{5 \gamma^3 \mathrm{U}_1^3(\mathrm{t})}{16(3+\lambda)^3} \mathrm{p}_1^3 +\frac{5 \gamma^2 \mathrm{U}_1^2(\mathrm{t}) \mathrm{p}_1\left(\mathrm{p}_2-\mathrm{q}_2\right)}{16(3+\lambda)(7+2 \lambda)}\end{aligned}$,
Thus
$\begin{array}{r}a_4=\frac{5 \gamma^2 \mathrm{U}_1^2(\mathrm{t}) \mathrm{p}_1\left(\mathrm{p}_2-\mathrm{q}_2\right)}{16(3+\lambda)}+\frac{\gamma \mathrm{U}_1(\mathrm{t})\left(\mathrm{p}_3-\mathrm{q}_3\right)}{4(13+3 \lambda)} +\frac{\gamma\left(\mathrm{U}_2(\mathrm{t})-\mathrm{U}_1(\mathrm{t})\right)}{4(13+3 \lambda)} \mathrm{p}_1\left(\mathrm{p}_2+\mathrm{q}_2\right) +\frac{\gamma\left(\mathrm{U}_1(\mathrm{t})-2 \mathrm{U}_2(\mathrm{t})+\mathrm{U}_3(\mathrm{t})\right)}{8(13+3 \lambda)} \mathrm{p}_1^3\end{array}$ (25)
Thus, form (22), (24) and (25), we get
$a_2 a_4-a_3^2=\frac{5 \gamma^3 \mathrm{U}_1^3(\mathrm{t}) \mathrm{p}_1^2\left(\mathrm{p}_2-\mathrm{q}_2\right)}{32(3+\lambda)^2(7+2 \lambda)}+\frac{\gamma^2 \mathrm{U}_1^2(\mathrm{t}) \mathrm{p}_1\left(\mathrm{p}_3-\mathrm{q}_3\right)}{8(3+\lambda)(13+3 \lambda)}+\frac{\gamma^2 \mathrm{~b}_1(\mathrm{t})\left(\mathrm{U}_2(\mathrm{t})-\mathrm{U}_1(\mathrm{t})\right)}{8(3+\lambda)(13+3 \lambda)} \mathrm{p}_1^2\left(\mathrm{p}_2+\mathrm{q}_2\right)-\frac{\gamma^2 \mathrm{U}_1^2(\mathrm{t})\left(\mathrm{p}_2-\mathrm{q}_2\right)^2}{16(7+2 \lambda)^2}+$
$\frac{\gamma^2 \mathrm{U}_1(\mathrm{t})}{16(3+\lambda)^4(13+3 \lambda)} \mathrm{p}_1^4\left[\left(\mathrm{U}_1(\mathrm{t})-2 \mathrm{U}_2(\mathrm{t})+\mathrm{U}_3(\mathrm{t})\right)(3+\lambda)^3-\gamma^2 \mathrm{U}_1^3(\mathrm{t})(13+3 \lambda)\right]$. (26)
According to Lemma (2) we get
$2 p_2=p_1^2+\left(4-p_1^2\right) x$ and $2 q_2=q_1^2+\left(4-q_1^2\right) y$, (27)
and
$\begin{array}{r}4 \mathrm{p}_3=p_1^3+2\left(4-p_1^2\right) \mathrm{p}_1 x-\left(4-p_1^2\right) \mathrm{p}_1 x^2 +2\left(4-p_1^2\right)\left(1-|x|^2\right) z\end{array}$, (28)
and
$\begin{gathered}4 q_3=q_1^3+2\left(4-q_1^2\right) q_1 y-\left(4-q_1^2\right) q_1 y^2+2(4- \left.q_1^2\right)\left(1-|y|^2\right) w,\end{gathered}$
for $x, y, z, w$ with $|x| \leq 1,|y| \leq 1,|z| \leq 1,|w| \leq 1$.
Since (23) $p_{1=-} q_1$, from (27) and (28) we get
$\mathrm{p}_2-\mathrm{q}_2=\frac{4-p_1^2}{2}(x-y), \mathrm{p}_2+\mathrm{q}_{2=} p_1^2+\frac{4-p_1^2}{2}(x+y)$, (29)
and
$\begin{array}{r}\mathrm{p}_3-\mathrm{q}_3=\frac{p_1^3}{2}+\frac{\left(4-p_1^2\right) \mathrm{p}_1}{2}(x+y)-\frac{\left(4-p_1^2\right) \mathrm{p}_1}{4}\left(x^2+\right. \left.y^2\right)+\frac{4-p_1^2}{2}\left[\left(1-|x|^2\right) z-\left(1-|y|^2\right) w\right] .\end{array}$ (30)
According to Lemma (2), we can draw assumptions without any restrictions that $\tau \in[0,2]$, where $\tau=\left|\mathrm{p}_1\right|$.
Thus, substituting the expressions (29) and (30) in (26) and utilize triangle inequality, letting $|x|=\mathcal{J}$, $|y|=\Upsilon$, we can easily obtain that
$\begin{array}{r}\left|a_2 a_4-a_3{ }^2\right| \leq b_1(t, \tau)(\mathcal{J}+\Upsilon)^2+b_2(t, \tau)\left(\mathcal{J}^2+\right. \left.\Upsilon^2\right)+b_3(t, \tau)(\mathcal{J}+\Upsilon)+b_4(t, \tau):=F(\mathcal{J}, \Upsilon),\end{array}$ (31)
where,
$b_1(t, \tau)=\frac{\gamma^2 \mathrm{U}_1^2(\mathrm{t})\left(4-t^2\right)^2}{64(7+2 \lambda)^2} \geq 0, b_2(t, \tau)=\frac{\gamma^2 \mathrm{U}_1^2(\mathrm{t}) \mathrm{t}(\mathrm{t}-2)\left(4-t^2\right)}{32(3+\lambda)(13+3 \lambda)} \leq 0$,
$b_3(t, \tau)=\frac{5 \gamma^3 \mathrm{U}_1^3(\mathrm{t}) t^2\left(4-t^2\right)}{64(3+\lambda)^2(7+2 \lambda)}+\frac{\gamma^2 \mathrm{U}_1(\mathrm{t}) \mathrm{U}_2(\mathrm{t}) t^2\left(4-t^2\right)}{16(3+\lambda)(13+3 \lambda)} \geq 0$,
$b_4(t, \tau)=\frac{\gamma^2 \mathrm{U}_1(\mathrm{t})\left[(3+\lambda)^3 \mathrm{U}_3(\mathrm{t})-U_1^3(t)(13+3 \lambda)\right] t^4}{16(3+\lambda)^4(13+3 \lambda)}$
$+\frac{\gamma^2 \mathrm{U}_1^2(\mathrm{t}) \mathrm{t}\left(4-t^2\right)}{8(3+\lambda)(13+3 \lambda)} \geq 0, t \in\left(\frac{1}{2}, 1\right), \tau \in[0,2]$.
Therefore, we must maximize the function. $F(\mathcal{J}+\Upsilon)$ on the closed square.
$\Psi=\{(\mathcal{J}+\Upsilon): \mathcal{J}, \Upsilon \in[0,1]\}$ for $\tau \in[0,2]$. Since the coefficients of the function $F(\mathcal{J}, \curlyvee)$ is dependent to variable $\tau$ for fixed value of $\mathrm{t}$, we must investigate the maximum of $F(\mathcal{J}, \Upsilon)$ respect to $\tau$ taking account these cases $\tau=0, \tau=2$ and $\tau \in(0,2)$.
Let $\tau=0$. Then, we write
$F(\mathcal{J}, \Upsilon)=b_1(t, 0)=\frac{\gamma^2 U_1^2(\mathrm{t})}{4(7+2 \lambda)^2}(\mathcal{J}+\Upsilon)^2$.
It is obvious that the maximum of the function is $F(\mathcal{J}, \Upsilon)$ occurs at $(\mathcal{J}, \Upsilon)=(1,1)$, and
$\max \{F(\mathcal{J}, \Upsilon): \mathcal{J}, \Upsilon \in[0,1]\}=F(1,1)=\frac{\gamma^2 \mathrm{U}_1^2(\mathrm{t})}{(7+2 \lambda)^2}$ (32)
Now, let $\tau=2$. In this case, $F(\mathcal{J}, \curlyvee)$ is constant function as follows:
$\begin{gathered}F(\mathcal{J}, \Upsilon)=b_4(t, 2)=\frac{\gamma^2 \mathrm{U}_1(\mathrm{t})\left[(3+\lambda)^3 \mathrm{U}_3(\mathrm{t})-U_1^3(t)(13+3 \lambda)\right]}{(3+\lambda)^4(13+3 \lambda)}\end{gathered}$ (33)
In the case $\tau \in(0,2)$, we will examine the maximum of the function $F(\mathcal{J}+\eta)$ taking into account the sing of $\Gamma(\mathcal{J}, \Upsilon)=F_{\mathcal{J J}}(\mathcal{J}, \Upsilon) F_{\Upsilon \Upsilon}(\mathcal{J}, \Upsilon)-\left[F_{\mathcal{J} \Upsilon}(\mathcal{J}, \Upsilon)\right]^2$. We can easily see that by simple computation, we can easily see that
$\Gamma(\mathcal{J}, \Upsilon)=4 b_2(t, \tau)\left[2 b_1(t, \tau)+b_2(t, \tau)\right]$.
Since $b_2(t, \tau)<0$ for all $t \in\left(\frac{1}{2}, 1\right), \tau \in[0,2]$ and $2 b_1(t, \tau)+b_2(t, \tau)>0$, we conclude that $\Gamma(\mathcal{J}, \curlyvee)<0$. Therefore, the function F cannot have a local maximum in interior of the closed square $\Gamma$. Let
$\begin{aligned} & \partial \Gamma=\{(0, \Upsilon): \Upsilon \in[0,1]\} \cup\{(\mathcal{J}, 0): \mathcal{J} \in[0,1]\} \cup\{(1, \Upsilon): \Upsilon \in \\ & [0,1]\} \cup\{(\mathcal{J}, 1): \mathcal{J} \in[0,1]\} \text {. } \\ & \end{aligned}$
We may actually demonstrate that the function's maximum $F(\mathcal{J}, \Upsilon)$ on the boundary $\partial \Gamma$ of the square $\Gamma$ value is occurs at $(\mathcal{J}, Y)=(1,1)$, and
$\begin{aligned} & \max \quad\{F(\mathcal{J}, \Upsilon):(\mathcal{J}, \Upsilon) \in \partial \Gamma\}=F(1,1)= 4 b_1(t, \tau)\left[2 b_2(t, \tau)+b_3(t, \tau)\right]+b_4(t, \tau) \, , \, t \in \left(\frac{1}{2}, 1\right), \tau \in(0,2) .\end{aligned}$ (34)
Let us now define the function $D:(0,2) \rightarrow \mathbb{R}$ as follows:
$D(t, \tau)=4 b_1(t, \tau)\left[2 b_2(t, \tau)+b_3(t, \tau)\right]+b_4(t, \tau)$, (35)
for fixed value of t.
Substituting the value bj(t,τ), j=1,2,3,4 in the (35), we get
$D(t, \tau)=\frac{\Lambda(\xi, t) \tau^4+4 \mathcal{C}(\xi, t) \tau^2}{32(3+\lambda)^4(7+2 \lambda)^2(13+3 \lambda)}+\frac{\delta^2 U_1^2(\mathrm{t})}{(7+2 \lambda)^2}$,
where,
$\begin{gathered}\Lambda(\xi, t)=\gamma^2 \mathrm{U}_1(\mathrm{t})\left\{2(3+\lambda)^3(7+2 \lambda)^2\left[-\mathrm{U}_1(\mathrm{t})+2 \mathrm{U}_2(\mathrm{t})+\right.\right. \left.\mathrm{U}_3(\mathrm{t})\right]+\mathrm{U}_1(\mathrm{t})(13+3 \lambda)\left[2(3+\lambda)^4-5 \mathrm{U}_1(\mathrm{t})(3+\lambda)^2(7+\right. \left.\left.2 \lambda)^2-2 \gamma^2 U_1^2(t)(7+2 \lambda)^2\right]\right\},\end{gathered}$
$\begin{aligned} \mathcal{C}(\xi, t)=\gamma^2 \mathrm{U}_1(\mathrm{t}) & (3+\lambda)^2\left\{2(3+\lambda)(7+2 \lambda)^2\left[\mathrm{U}_1(\mathrm{t})\right.\right. \left.+2 \mathrm{U}_2(\mathrm{t})\right] \left.+\mathrm{U}_1(\mathrm{t})(13+3 \lambda)\left[5 \mathrm{U}_1(\mathrm{t})-4(3+\lambda)^2\right]\right\}\end{aligned}$,
where,
$U_{n+1}(t)=2 t U_n(t)-U_{n-2}(t)$,
We get that,
$\begin{gathered}U_1(t)=2 t, U_2(t)=4 t^2-1, U_3(t)=8 t^3-4 t \text {. Then } \Lambda(\xi, t)=2 \gamma^2 t\left\{4(3+\lambda)^3(7+2 \lambda)^2\left[4 t^3+4 t^2-3 t-1\right]+\right. 2 t(13+3 \lambda)\left[2(3+\lambda)^4-10 t(3+\lambda)^2(7+2 \lambda)^2-\right. \left.\left.8 \gamma^2 t^2(7+2 \lambda)^2\right]\right\},\end{gathered}$
$\begin{gathered}\mathcal{C}(\xi, t)=2 \gamma^2 \mathrm{t}(3+\lambda)^2\left\{4(3+\lambda)(7+2 \lambda)^2\left[\mathrm{t}+4 t^2-1\right]+\right.\left.2 \mathrm{t}(13+3 \lambda)\left[10 t-4(3+\lambda)^2\right]\right\} .\end{gathered}$
Now, suppose that H(t,τ) has maximum value in an interior of $\tau \in[0,2]$, then
$D^{\prime}(t, \tau)=\frac{\Lambda(\xi, t) \tau^2+\mathcal{C}(\xi, t)}{8(3+\lambda)^4(7+2 \lambda)^2(13+3 \lambda)} \quad \tau$.
After some calculations, we take into account the following cases:
Let $\Lambda(B, t) \geq 0$ and $\mathcal{C}(B, t) \geq 0$. Then $D^{\prime}(t, \tau) \geq 0$,so $D(t, \tau)$ is an increasing function. Therefore,
$\begin{gathered}\operatorname{Max}\{D(t, \tau): \tau \in(0,2)\}=D(t, 2-)= \frac{\Lambda(\xi, t)+\mathcal{C}(\xi, t)}{2(3+\lambda)^4(7+2 \lambda)^2(13+3 \lambda)}+\frac{\gamma^2 \mathrm{U}_1^2(\mathrm{t})}{(7+2 \lambda)^2} .\end{gathered}$ (36)
$\begin{aligned} & \text { That is, } \quad \max \quad\{\max \{F(\mathcal{J}, \Upsilon):(\mathcal{J}, \Upsilon) \in[0,1]\}: \tau \in (0,2)\}=H(t, 2-) .\end{aligned}$
1. Let $\Lambda(\xi, t)>0$ and $\mathcal{C}(\xi, t)<0$. Then $\tau_0=\sqrt{\frac{-2 \mathcal{C}(\xi, t)}{\Lambda(B, t)}}$ is a critical point of the function $D(t, \tau)$. We suppose that $\tau_0 \in(0,2)$. Since $D^{\prime \prime}(t, \tau)>0, \tau_0$ is a local minimum point of the function $D(t, \tau)$. That is the function $D(t, \tau)$ cannot have a local maximum.
2. Let $\Lambda(\xi, t) \leq 0$ and $\mathcal{C}(\xi, t) \leq 0$. Then $D^{\prime}(t, \tau) \leq 0$ Thus, $D(t, \tau)$ is a decreasing function on the interval $(0,2)$. Therefore,
$\begin{gathered}\max \{D(t, \tau): \tau \in(0,2)\}=D(t, 0+)=4 b_1(t, 0)=\frac{\gamma^2 U_1^2(\mathrm{t})}{(7+2 \lambda)^2} .\end{gathered}$ (37)
1. Let $\Lambda(\xi, t)<0$ and $\mathcal{C}(\xi, t)>0$. Then $\tau_0$ is a critical point of the function $D(t, \tau)$.
2. We suppose that $\tau_0 \in(0,2)$. Since $D^{\prime \prime}(t, \tau)<0, \tau_0$ is a local maximum of the function $D(t, \tau)$ and maximum value occurs at $\tau=\tau_0$. Therefore,
$\max \{D(t, \tau): \tau \in(0,2)\}=D\left(t, \tau_0\right)$, (38)
where
$D\left(t, \tau_0\right)=\frac{4 \gamma^2 t^2}{(7+2 \lambda)^2}-\frac{\mathcal{e}^2(\xi, t)}{8 \Lambda(\xi, t)(3+\lambda)^4(7+2 \lambda)^2(13+3 \lambda)}$.
Thus from (32) to (38), the proof of Theorem (1) is complete.
Theorem 2: If f is given by (1) belongs to the subclass $\mathcal{R}(t, \gamma, \lambda), t \in\left(\frac{1}{2}, 1\right)$, then
$\begin{aligned} & \left|a_2 a_3-a_4\right| \leq\left\{\begin{array}{ll}\frac{t^3 \gamma^3(13+3 \lambda)-\gamma(3+\lambda)^3\left(8 t^3-8 t^2-2 t-4\right)}{(3+\lambda)^3(13+3 \lambda)} & \omega \leq \tau \leq 2 \\ \frac{t \gamma}{2(13+3 \lambda)} & 0 \leq \tau \leq \omega\end{array},\right.\end{aligned}$ (39)
where
$\omega=\frac{d_2+\sqrt{d_2{ }^2+12\left(d_3-d_1\right)}}{3\left(d_3-d_1\right)}$.
$\begin{gathered}d_1=\frac{14 \gamma^2}{16(3+\lambda)(7+2 \lambda)}-\frac{\gamma\left(4 t^2-2 t-1\right)}{4(13+3 \lambda)}, d_2 =\frac{t \gamma}{8(13+3 \lambda)}\end{gathered}$
and
$d_3=\frac{t^3 \gamma^3(13+3 \lambda)-\gamma(3+\lambda)^3\left(8 t^3-8 t^2-2 t-4\right)}{(3+\lambda)^3(13+3 \lambda)} \quad .$
Proof: From (22), (24) and (25), we get
$\begin{aligned} & \left|a_2 a_3-a_4\right|=\mid \frac{\gamma^3 \mathrm{U}_1{ }^3(\mathrm{t})(13+3 \lambda)-\gamma(3+\lambda)^3\left(\mathrm{U}_1(\mathrm{t})-2 \mathrm{U}_2(\mathrm{t})+\mathrm{U}_3(\mathrm{t})\right)}{8(3+\lambda)^3(13+3 \lambda)} \mathrm{p}_1^3 \\ & +\frac{7 \gamma^2 \mathrm{U}_1^2(\mathrm{t}) \mathrm{p}_1\left(\mathrm{p}_2-\mathrm{q}_2\right)}{8(3+\lambda)(7+2 \lambda)}+\frac{\gamma \mathrm{U}_1(\mathrm{t})\left(\mathrm{p}_3-\mathrm{q}_3\right)}{4(13+3 \lambda)} -\frac{\gamma\left(\mathrm{U}_2(\mathrm{t})-\mathrm{U}_1(\mathrm{t})\right)}{4(13+3 \lambda)} \mathrm{p}_1^2\left(\mathrm{p}_2+\mathrm{q}_2\right) \mid . \\ & \end{aligned}$
According to Lemma 2, we assume without any restriction that $\tau \in[0,2]$, where $\tau=\left|\mathrm{p}_1\right|$ thus for $\eta_1=|\mathrm{x}| \leq 1, \eta_2=$ $|y| \leq 1$, we get
$\begin{aligned}\left|a_2 a_3-a_4\right| & \leq d_1\left(\eta_1+\eta_2\right)+d_2\left(\eta_1^2+\eta_2^2\right)+d_3 \\ & =R\left(\eta_1, \eta_2\right)\end{aligned}$,
where,
$\begin{gathered}d_1=\frac{7 \gamma^2 \mathrm{U}_1^2(\mathrm{t}) \tau\left(4-\tau^2\right)}{32(3+\lambda)(7+2 \lambda)} \geq 0, d_2=\frac{\gamma \mathrm{U}_1(\mathrm{t})(\tau-2)\left(4-\tau^2\right)}{16(13+3 \lambda)} \geq 0,\end{gathered}$
$\begin{aligned} & d_3= \frac{\gamma^3 \mathrm{U}_1{ }^3(\mathrm{t})(13+3 \lambda)-\gamma(3+\lambda)^3\left(\mathrm{U}_1(\mathrm{t})-2 \mathrm{U}_2(\mathrm{t})+\mathrm{U}_3(\mathrm{t})\right)}{8(3+\lambda)^3(13+3 \lambda)} \tau^3 \\ & -\frac{\gamma\left(\mathrm{U}_2(\mathrm{t})-\mathrm{U}_1(\mathrm{t})\right) \tau\left(4-\tau^2\right)}{4(13+3 \lambda)} \geq 0, t\left(\frac{1}{2}, 1\right), \tau \in[0,2] .\end{aligned}$
By using the same method of Theorem 1, thus maximize occur at η1=1 and η2=1 in closed square [0,2],
$R(\tau)=4 d_1+2\left[d_2+d_3\right]$.
Substituting the value of d1, d2, d3 in R(τ), we get that
$R(\tau)=d_1 \tau\left(4-\tau^2\right)-d_2\left(4-\tau^2\right)+d_3 \tau^3$,
so that
$R(\tau)=4 d_1 \tau-d_1 \tau^3-4 d_2+d_2 \tau^2+d_3 \tau^3$,
where
$\begin{gathered}d_1=\frac{7 \gamma^2 \mathrm{U}_1^2(\mathrm{t})}{32(3+\lambda)(7+2 \lambda)}-\frac{\gamma\left(\mathrm{U}_2(\mathrm{t})-\mathrm{U}_1(\mathrm{t})\right)}{4(13+3 \lambda)}, d_2=\frac{\gamma \mathrm{U}_1(\mathrm{t})}{16(13+3 \lambda)}\end{gathered}$
$d_3=\frac{\gamma^3 \mathrm{U}_1^3(\mathrm{t})(13+3 \lambda)-\gamma(3+\lambda)^3\left(\mathrm{U}_1(\mathrm{t})-2 \mathrm{U}_2(\mathrm{t})+\mathrm{U}_3(\mathrm{t})\right)}{8(3+\lambda)^3(13+3 \lambda)}$
Since
$U_{n+1}(t)=2 t U_n(t)-U_{n-2}(t)$,
We get that,
$U_1(t)=2 t, U_2(t)=4 t^2-1, U_3(t)=8 t^3-4 t$.
Then
$\begin{gathered}d_1=\frac{14 \gamma^2}{16(3+\lambda)(7+2 \lambda)}-\frac{\gamma\left(4 t^2-2 t-1\right)}{4(13+3 \lambda)}, d_2=\frac{t \gamma}{8(13+3 \lambda)}\end{gathered}$
and
$d_3=\frac{t^3 \gamma^3(13+3 \lambda)-\gamma(3+\lambda)^3\left(8 t^3-8 t^2-2 t-4\right)}{(3+\lambda)^3(13+3 \lambda)}$.
We obtain
$\begin{gathered}R^{\prime}(\tau)=3\left(d_3-d_1\right) \tau^2+2 d_2 \tau+4 d_1, \quad R^{\prime \prime}(\tau) =6\left(d_3-d_1\right) \tau+2 d_2,\end{gathered}$
If $d_3-d_1>0$, that is $d_3>d_1$, then we have $R^{\prime}(\tau)>0$. Thus R(τ) is an increasing function on the closed interval [0,2] and so the function R(τ) get the maximum value at τ=2, that is
$\begin{aligned} & \left|a_2 a_3-a_4\right| \leq R(2)=\frac{t^3 \gamma^3(13+3 \lambda)-\gamma(3+\lambda)^3\left(8 t^3-8 t^2-2 t-4\right)}{(3+\lambda)^3(13+3 \lambda)}\end{aligned}$.
If $d_3-d_1<0$, let $R^{\prime}(\tau)=0$, then we have
$\tau=\omega=\frac{d_2+\sqrt{d_2^2+12\left(d_3-d_1\right)}}{3\left(d_3-d_1\right)}$,
when ω<τ≤2. Then we get which means the function on the closed interval [0,2], thus R(τ) gets the maximum value at R(2), which means the function R(τ) is an increasing function the closed interval [0,2], thus R(τ) gets the maximum value at τ=0, we have
$\left|a_2 a_3-a_4\right| \leq R(0)=\frac{t \gamma}{2(13+3 \lambda)}$.
The proof of Theorem 2 is complete.
Theorem 3: If $f$ is given by (1) belongs to the subclass $\mathcal{R}(t, \gamma, \lambda), t \in\left(\frac{1}{2}, 1\right)$, then
$\left|a_3-a_2^2\right| \leq \frac{2 \gamma \mathrm{t}}{(7+2 \lambda)}$, (40)
$\left|a_3\right| \leq \frac{4 t^2 \gamma^2}{(3+\lambda)^2}$. (41)
Proof: By using (24) and Lemma 1, we get that
$\left|a_3\right| \leq \frac{\gamma^2 \mathrm{U}_1^2(\mathrm{t})}{(3+\lambda)^2}$ (42)
since
$U_1(t)=2 t$ (43)
substituting (43) in (42), we get (41).
The following Fekete-Szegö functional, for $\mu \in \mathbb{C}, f \in$ $\mathcal{R}(t, \gamma, \lambda)$
$a_3-\mu a_2^2=\frac{(1-\mu) \gamma^2 \mathrm{U}_1^2(\mathrm{t}) \mathrm{p}_1^2}{4(3+\lambda)^2}+\frac{\gamma \mathrm{U}_1(\mathrm{t})}{4(7+2 \lambda)}\left(\mathrm{p}_2-\mathrm{q}_2\right)=\frac{\gamma \mathrm{U}_1(\mathrm{t})}{4}\left[\frac{(1-\mu) \gamma \mathrm{U}_1(\mathrm{t})}{(3+\lambda)^2} \mathrm{p}_1^2+\frac{1}{(7+2 \lambda)} \mathrm{p}_2-\frac{1}{(7+2 \lambda)} \mathrm{q}_2\right]$,
since U1(t)=2t, then
$a_3-\mu a_2^2=\frac{\gamma t}{2}\left[\frac{1}{(7+2 \lambda)} \mathrm{p}_2+s(\mu) \mathrm{p}_1^2-\frac{1}{(7+2 \lambda)} \mathrm{q}_2\right]$,
where
$s(\mu)=\frac{(1-\mu) 2 \gamma \mathrm{t}}{(3+\lambda)^2}$.
We conclude that
$\begin{aligned} & \left|a_3-\mu a_2{ }^2\right| \leq\left\{\begin{array}{lr}\{2 \gamma \mathrm{t}|s(\mu)| & |s(\mu)| \geq \frac{1}{(7+2 \lambda)}, \\ \frac{2 \gamma \mathrm{t}}{(7+2 \lambda)} & 0 \leq|s(\mu)| \leq \frac{1}{(7+2 \lambda)},\end{array}\right. \\ & \end{aligned}$
for μ=1, we get (40).
Theorem 4: If f is given by (1) belongs to the subclass $\mathcal{R}(t, \gamma, \lambda), t \in\left(\frac{1}{2}, 1\right)$, then
$\left|a_4\right| \leq \frac{\gamma\left(8 t^3-4 t^2-5 t+2\right)}{(13+3 \lambda)}$ (44)
$\begin{aligned} & \left|a_5\right| \leq \frac{6 \gamma^2\left[(3+\lambda)\left(4 t^3-2 t^2-t\right)+\left(8 t^4-8 t^3-2 t^2-2 t\right)\right]}{(3+\lambda)^2(13+3 \lambda)} \\ & -\frac{4 \gamma\left[3(3+\lambda)^4\left(8 t^3-16 t^2+4\right)+16 \gamma t(21+4 \lambda)\right]}{(3+\lambda)^4(21+4 \lambda)} .\end{aligned}$ (45)
Proof: By (25), we have
$\begin{gathered}a_4=\frac{5 \gamma^2 \mathrm{U}_1^2(\mathrm{t}) \mathrm{p}_1\left(\mathrm{p}_2-\mathrm{q}_2\right)}{16(3+\lambda)(7+2 \lambda)}+\frac{\gamma \mathrm{U}_1(\mathrm{t})\left(\mathrm{p}_3-\mathrm{q}_3\right)}{4(13+3 \lambda)} +\frac{\gamma\left(\mathrm{U}_2(\mathrm{t})-\mathrm{U}_1(\mathrm{t})\right)}{4(13+3 \lambda)} \mathrm{p}_1\left(\mathrm{p}_2+\mathrm{q}_2\right) +\frac{\gamma\left(\mathrm{U}_1(\mathrm{t})-2 \mathrm{U}_2(\mathrm{t})+\mathrm{U}_3(\mathrm{t})\right)}{8(13+3 \lambda)} \mathrm{p}_1^3 .\end{gathered}$
Since
$U_{n+1}(t)=2 t U_n(t)-U_{n-2}(t)$,
we get that,
$U_1(t)=2 t, \, U_2(t)=4 t^2-1, U_3(t)=8 t^3-4 t$, and Lemma 1, we get (44).
From (10) and (14), we obtain that
$\frac{(21+4 \lambda)}{\gamma} a_5=\frac{U_1(t)}{2}\left(p_4-\frac{p_2^2}{2}+p_1\left(\frac{p_1^3}{4}+\frac{3}{4} p_1 p_2-p_3\right)\right)$
$+\frac{U_2(t)}{2}\left(p_2^2+p_1\left(\frac{3 p_1^3}{8}+\frac{3}{2} p_1 p_2+\frac{p_1}{2}+\frac{p_3}{2}\right)-\frac{3}{8} U_3(t) p_1\left(p_1^3+p_1 p_2\right)+U_4(t) p_1^4\right.$. (46)
also, from (11) and (15), we have
$\frac{(21+4 \lambda)}{\gamma}\left(14 a_2^4+3 a_3^2-21 a_2^2 a_3+6 a_2 a_4-a_5\right)=\frac{U_1(t)}{2}\left(q_4-\frac{q_2^2}{2}+q_1\left(\frac{q_1^3}{4}+\frac{3}{4} q_1 q_2-q_3\right)\right)$
$+\frac{U_2(t)}{2}\left(q_2^2+q_1\left(\frac{3 q_1^3}{8}+\frac{3}{2} q_1 q_2+\frac{q_1}{2}+\frac{q_3}{2}\right)-\frac{3}{8} U_3(t) q_1\left(q_1^3+q_1 q_2\right)+U_4(t) q_1^4\right.$ (47)
Subtracting (47) from (46), we get that
$a_5=\frac{\gamma U_1(t)}{(21+4 \lambda)}\left(p_4-q_4\right)+\frac{3 \gamma(3+\lambda)(13+3 \lambda)\left(2 U_2(t)-U_1(t)\right)+3 \gamma^2 U_1(t)(21+4 \lambda)\left(U_2(t)-U_1(t)\right) p_1^2}{4(3+\lambda)(13+3 \lambda)(21+4 \lambda)}\left(p_2+q_2\right)$
$+\frac{2 \gamma(13+3 \lambda)(21+4 \lambda)\left(U_2(t)-2 U_1(t)\right)+3 \gamma^2(21+4 \lambda) U_1(t) p_1}{4(3+\lambda)(13+3 \lambda)(21+4 \lambda)}\left(p_3-q_3\right)-\frac{3 \gamma U_3(t) p_1}{4(21+3 \lambda)}\left(p_2+q_2\right)$
$-\frac{63 \gamma^3 U_1^3(t) p_1^2}{16(3+\lambda)^2(7+2 \lambda)}\left(p_2-q_2\right)+\frac{3 \gamma^2 U_1^2(t)}{16(7+2 \lambda)}\left(p_2-q_2\right)^2$
$+\frac{3 \gamma^2(3+\lambda)^3 U_1(t)\left(U_1(t)-2 U_2(t)+U_3(t)\right)-4 \gamma^4(13+3 \lambda) U_1^4(t)}{16(3+\lambda)^2(7+2 \lambda)} p_1^4$.
Since
$U_{n+1}(t)=2 t U_n(t)-U_{n-2}(t)$,
we get that,
$U_1(t)=2 t, \, U_2(t)=4 t^2-1, \, U_3(t)=8 t^3-4 t$, and Lemma 1, we get (45).
Theorem 5: If f is given by (1) belongs to the subclass $\mathcal{R}(t, \gamma, \lambda), t \in\left(\frac{1}{2}, 1\right), \gamma \in \mathbb{C} \backslash\{0\}$ and $\lambda \geq 1$, then we have
$H_3(1) \leq\left\{\begin{array}{c}N(D(t, 2-))-N_1 N_2-N_3 N_4, \quad \text { if } \Lambda(\xi, t) \geq 0 \text { and } \mathcal{C}(\xi, t) \geq 0, \\ N\left(\max \left\{\frac{4 \gamma^2 t^2}{(7+2 \lambda)^2}, D(t, 2-)\right\}\right)-N_1 N_2-N_3 N_4, \text { if } \Lambda(\xi, t)>0 \text { and } \mathcal{C}(\xi, t)<0, \\ N\left(\frac{4 \gamma^2 t^2}{(7+2 \lambda)^2}\right)-N_1 N_2-N_3 N_4, \quad \text { if } \Lambda(\xi, t) \leq 0 \text { and } \mathcal{C}(\xi, t) \leq 0, \\ N\left(\max \left\{D\left(t, \tau_0\right), D(t, 2-)\right\}\right)-N_1 N_2-N_3 N_4, \text { if } \Lambda(\xi, t)<0 \text { and } \mathcal{C}(\xi, t)>0,\end{array}\right.$ (48)
where, $N, N_1, N_2, N_3, N_4$ and $(\Lambda(\xi, t), \mathcal{C}(\xi, t))$ are given by (41), (44), (39), (45), (40) and (9) respectively.
Proof: Since
$H_3(1)=a_3\left(a_2 a_4-a_3^2\right)-a_4\left(a_4-a_2 a_3\right)+a_5\left(a_3-\right. a_2{ }^2$), by applying the triangle inequality, we get
$\begin{gathered}\left|\mathrm{H}_3(1)\right| \leq\left|a_3\right|\left|a_2 a_4-a_3{ }^2\right|-\left|a_4\right|\left|a_4-a_2 a_3\right| +\left|a_5\right|\left|a_3-a_2{ }^2\right| .\end{gathered}$ (49)
Substituting (41), (44), (39), (45), (40) and (9) in (49), we get (48).
Corollary 1: If $f$ is given by (1) belongs to the subclass $\mathcal{R}(t, \gamma, 1), t \in\left(\frac{1}{2}, 1\right)$ and $\gamma \in \mathbb{C} \backslash\{0\}$, then we have
$H_3(1) \leq\left\{\begin{array}{c}N(D(t, 2-))-N_1 N_2-N_3 N_4, \quad \text { if } \Lambda(\xi, t) \geq 0 \text { and } \mathcal{C}(\xi, t) \geq 0, \\ N\left(\max \left\{\frac{4 \gamma^2 t^2}{(7+2 \lambda)^2}, D(t, 2-)\right\}\right)-N_1 N_2-N_3 N_4, \quad \text { if } \Lambda(\xi, t)>0 \text { and } \mathcal{C}(\xi, t)<0, \\ N\left(\frac{4 \gamma^2 t^2}{(7+2 \lambda)^2}\right)-N_1 N_2-N_3 N_4, \quad \text { if } \Lambda(\xi, t) \leq 0 \text { and } \mathcal{C}(\xi, t) \leq 0, \\ N\left(\max \left\{D\left(t, \tau_0\right), D(t, 2-)\right\}\right)-N_1 N_2-N_3 N_4, \quad \text { if } \Lambda(\xi, t)<0 \text { and } \mathcal{C}(\xi, t)>0,\end{array}\right.$ (50)
where, $N, N_1, N_2, N_3, N_4$ and $(\Lambda(\xi, t), \mathcal{C}(\xi, t))$ are given by (41), (44), (39), (45), (40) and (9) respectively with $(\lambda=1)$.
This article presented a comprehensive investigation of the third Hankel determinant H3(1) for a novel subclass of bi-univalent functions, $\mathcal{R}(t, \gamma, \lambda)$. This subclass is of significant interest in various mathematical fields, including complex analysis and geometric function theory. Utilizing the property of subordination, we defined the bi-univalent functions $\mathcal{R}(t, \gamma, \lambda)$ and imposed constraints on the coefficients $\left|a_n\right|$. Our findings provided the upper bounds for the bi-univalent functions in this newly developed subclass, specifically for n=2, 3, 4, and 5. Furthermore, we advanced the understanding of these functions by deriving the third Hankel determinant for this particular class, which revealed several intriguing scenarios. This achievement led to the improvement of the bound of the third Hankel determinant for the class of bi-univalent functions $\mathcal{R}(t, \gamma, \lambda)$. Our study contributes to the broader understanding of bi-univalent functions, their subclasses, and their potential applications in diverse mathematical contexts. The results obtained may serve as a foundation for future investigations into the properties and applications of bi-univalent functions and their subclasses. Future research endeavors could explore further refinements of the bounds, as well as examine other subclasses of bi-univalent functions to uncover novel insights into their characteristics and potential applications. Ultimately, this study paves the way for a deeper exploration of the fascinating world of bi-univalent functions and their role in the realm of mathematics.
[1] Duren P.L. (1983). Univalent Functions, In: Grundlehren der Mathematischen Wissenschaften, Band 259, Springer, New York, Berlin, Hidelberg and Tokyo.
[2] Lewin M. (1967). On a coefficient problem for bi-univalent function. Proceedings of the American Mathematical Society, 18: 63-68. https://doi.org/10.2307/2035225
[3] Brannan D.A., Clunie J.G. (1980). Aspects of contemporary complex analysis. Proceedings of the NATO Advanced Study Institute, University of Durham, Durham, Academic Press, New York and London.
[4] Al-Shaqs, K. (2014). Strong differential subordinations obtained with new integral operator defined by polylogarithm function. International Journal of Mathematics and Mathematical Sciences, 2014: 260198. https://doi.org/10.1155/2014/260198
[5] Çağlar, M., M., Deniz, E. (2017). Initial coefficients for a subclass of bi-univalent functions defined by Salagean differential operator. Communications Faculty of Sciences University of Ankara Series A1 Mathematics and Statistics, 85-91. https://doi.org/10.1501/Commua1_0000000777
[6] Brannan, D.A., Taha, T.S. (1988). On some classes of bi-univalent functions. Mathematical Analysis and Its Applications, in Proceedings of the International Conference on Mathematical Analysis and its Applications, Kuwait, pp. 70-77. https://doi.org/10.1016/B978-0-08-031636-9.50012-7
[7] Al-Ameedee, S.A., Atshan, W.G., Al-Maamori, F.A. (2020). Second Hankel determinant for certain Subclasses of bi-univalent functions. Journal of Physics: Conference Series, Volume 1664, 1st International Virtual Conference on Pure Science 10th-11th June 2020, College of Science, University of Al-Qadisiyah, Iraq, 1664: 012044. https://doi.org/10.1088/1742-6596/1664/1/012044
[8] Al-Ameedee, S.A., Atshan, W.G., Al-Maamori, F.A. (2020). Coefficients estimates of biunivalent functions defined by new subclass function. Journal of Physics: Conference Series, Volume 1530, Imam Al-Kadhum International Conference for Modern Applications of Information and Communication Technology (MAICT) 4-5 December, 2019, Baghdad, Iraq, 1530: 012105. https://doi.org/10.1088/1742-6596/1530/1/012105
[9] Cantor, D.G. (1963). Power series with integral coefficients. Bulletin of the American Mathematical Society, 69(3): 362-366.
[10] Alhily, S.S. (2018). Some results on Koebe one- quarter theorem and Koebe distortion theorem. Journal of Iraqi Al-Khwarizmi Society (JIKhS), 2: 29-38.
[11] Atshan, W.G., Badawi, E.I. (2019). Results on coefficients estimates for subclasses of analytic and bi-univalent functions. Journal of Physics: Conference Series, 1294(3): 1-10. https://doi.org/10.1088/1742-6596/1294/3/032025
[12] Atshan W.G., Yalcin S., Hadi R.A. (2020). Coefficients estimates for special subclass of k-fold symmetric bi-univalent functions. Mathematics for Applications, 9(2): 83-90. https://doi.org/10.13164/ma.2020.07
[13] Srivastava, H.M., Mishra, A.K., Gochhayat, P. (2010). Certain subclasses analytic and biunivalent functions. Applied Mathematics Letters, 23(10): 1188-1192. https://doi.org/10.1016/j.aml.2010.05.009
[14] Xu, Q.H., Gui, Y.C., Srivastava, H.M. (2012). Coefficient estimates for a certain subclass of analytic and bi-univalent functions. Applied Mathematics Letters, 25(6): 990-994. https://doi.org/10.1016/j.aml.2011.11.013
[15] Xu, Q.H., Xiao, H.G., Srivastava, H.M. (2012). A certain general subclass of analytic and biunivalent functions and associated coefficient estimate problems. Applied Mathematics and Computation, 218(23): 11461-11465. https://doi.org/10.1016/j.amc.2012.05.034
[16] Nehari, Z. (1952). Conformal Mappings, McGraw–Hill, New York.
[17] Deniz, E., Çağlar, M., Orhan, H. (2015). Second Hankel determinant for bi-stalike and biconvex functions of order β. Applied Mathematics and Computation, 271: 301-307. https://doi.org/10.1016/j.amc.2015.09.010
[18] Orhan, H., Magesh, N., Yamini, J. (2016). Bounds for the second Hankel determinant of certain bi-univalent functions, Turkish Journal of Mathematics, 40(3): 679-687. https://doi.org/10.3906/mat-1505-3
[19] Babalola, K.O. (2010). On H3(1) Hankel determinant for some classes of univalent functions. arXiv:0910.3779. https://doi.org/10.48550/arXiv.0910.3779
[20] Doha, E.H. (1994). The first and second kind Chebyshev coefficients of the moments of the general-order derivative of an infinitely differentiable function. International Journal of Computer Mathematics, 51(1-2): 21-35. https://doi.org/10.1080/00207169408804263
[21] Mason, J.C. (1967). Chebyshev polynomials approximations for the embrane eigenvalue problem, SIAM Journal on Applied Mathematics, 15(1): 172-186.
[22] Grenander, U., Szego, G. (1958). Toeplitz Forms and Their Applications. California Monographs in Mathematical Sciences. Berkeley, CA, USA, University California Press.