On the Third Hankel Determinant of a Certain Subclass of Bi-Univalent Functions Defined by (p,q)-Derivative Operator

Abstract

In this study, the generalized $(p,q)$-derivative operator is used to define a novel class of bi-univalent functions. For this class, we define constraints on the coefficients up to $|{\ell }_5|$. The functions are analyzed using a suitable operational method, which enables us to derive new bounds for the Fekete–Szegö functional, as well as explicit estimates for important coefficients like $|{\ell }_2|$ and $|{\ell }_3|$. In addition, we establish the upper bounds of the second and third Hankel determinants, providing insights into the geometrical and analytical properties of this class of functions.

1. Introduction

In geometric function theory, univalent functions form a subclass of bi-univalent functions, defined by their univalence in both the unit disk and its inverse domain. The Fekete–Szegö functional plays a significant role in providing coefficient bounds for analyzing geometric properties. Additionally, the Hankel determinant is widely applied to study coefficient growth and structural stability, offering insights into the analytic and geometric behavior of bi-univalent functions.

Let A represent the class of analytic functions within the open unit disk $U=\{z\in \mathbb{C}:|z|<1\}$, which can be expressed in the following form:

$$\mathfrak{T}\left(z\right)=z + \sum _{\mathsf{ɩ}=2}^{\infty }{\ell }_{\mathsf{ɩ}}{z}^{\mathsf{ɩ}},$$

(1)

A function $\mathfrak{T}\in A$ is called univalent in U if it is one-to-one. The set of all univalent functions within A is denoted by $\mathbb{S}$. A fundamental result in geometric function theory is the Koebe One-Quarter Theorem [1,2], which guarantees that the image of U under any function in $\mathbb{S}$ contains a disk of radius $1/4$. Consequently, every such function has an inverse ${\mathfrak{T}}^{-1}$ satisfying the following:

$${\mathfrak{T}}^{-1}\left(\mathfrak{T}\left(z\right)\right)=z(z\in U),$$

(2)

where there is an explicit series representation given by the following:

$${\mathfrak{T}}^{-1}\left(w\right)=w-{\ell }_2{w}^2 + \left(2{\ell }_2^2-{\ell }_3\right)w^3-\left(5{\ell }_3^2-5{\ell }_2{\ell }_3 + {\ell }_4\right)w^4 + \cdots .$$

(3)

If $\mathfrak{T}\in A$ is univalent in U and its inverse ${\mathfrak{T}}^{-1}$ is likewise univalent in U, then the function is bi-univalent. The notation $\Sigma$ represents the set of all such functions.

The study of bi-univalent functions has gained attention, following the work of Srivastava et al. [3]. Some well-known examples in $\Sigma$ include the following:

$${\displaystyle \frac{z}{1-z}},-log(1-z),{\displaystyle \frac{1}{2}}log\left({\displaystyle \frac{1 + z}{1-z}}\right)$$

However, the Koebe function and certain other classical univalent functions, such as $z-{\displaystyle \frac{z^2}{2}}$ and ${\displaystyle \frac{z}{1-z^2}}$, do not belong to $\Sigma$; see [3].

Lewin [4] established an upper bound of $1.51$ for $|{\ell }_2|$, while Brannan and Clunie [5] conjectured that $|{\ell }_2| \le \sqrt{2}$. Later, Netanyahu [6] proved that the sharp bound is $|{\ell }_2| \le 4/3$. Further studies introduced bi-starlike and bi-convex subclasses of $\Sigma$ [3,7,8], leading to various coefficient bounds [1,2,9,10].

Another significant functional in geometric function theory is the Fekete–Szegö functional, given by $|{\ell }_3-\mathfrak{M}{\ell }_2^2|$, where $\mathfrak{M}\in \mathbb{C}$. Originally introduced to address a conjecture by Littlewood and Paley, it has since been widely studied for different function classes [1,2,11,12].

In 1976, Noonan and Thomas [13] introduced the concept of the q-th Hankel determinant for the function $\mathfrak{T}$ given in (1).

$${\mathcal{H}}_q(\mathsf{ɩ})=\left|\begin{array}{cccc}{\ell }_{\mathsf{ɩ}}& {\ell }_{\mathsf{ɩ} + 1}& \cdots & {\ell }_{\mathsf{ɩ} + q-1}\\ {\ell }_{\mathsf{ɩ} + 1}& {\ell }_{\mathsf{ɩ} + 2}& \cdots & {\ell }_{\mathsf{ɩ} + q}\\ ⋮& ⋮& \ddots & ⋮\\ {\ell }_{\mathsf{ɩ} + q-1}& {\ell }_{\mathsf{ɩ} + q}& \cdots & {\ell }_{\mathsf{ɩ} + 2q-2}\end{array}\right|(\mathsf{ɩ},q\in \mathbb{N};{\ell }_1=1)$$

(4)

The determinant ${\mathcal{H}}_q(\mathsf{ɩ})$ has been studied by various researchers. Some studies [14,15] explored the growth rate of ${\mathcal{H}}_q(\mathsf{ɩ})$ as $\mathsf{ɩ}\to \infty$ for functions $\mathfrak{T}$ defined by Equation (1) with a bounded boundary. Notably, other recent works [16,17,18] derived sharp upper bounds for ${\mathcal{H}}_2\left(2\right)$ within different classes of functions. The following equations are especially notable:

$${\mathcal{H}}_2\left(1\right)=\left|\begin{array}{cc}{\ell }_1& {\ell }_2\\ {\ell }_2& {\ell }_3\end{array}\right|={\ell }_3-{\ell }_2^2,$$

(5)

and

$${\mathcal{H}}_2\left(2\right)=\left|\begin{array}{cc}{\ell }_2& {\ell }_3\\ {\ell }_3& {\ell }_4\end{array}\right|={\ell }_2{\ell }_4-{\ell }_3^2.$$

(6)

The determinant ${\mathcal{H}}_2\left(1\right)={\ell }_3-{\ell }_2^2$ is commonly recognized as the Fekete–Szegö functional. Recently, Shakir et al. [17] investigated the upper bounds of ${\mathcal{H}}_2\left(2\right)$ for certain classes of analytic functions; see also [14,16,17,19].

In this paper, we focus on the Hankel determinant for the case $q=3$ and $\mathsf{ɩ} =1$, represented as ${\mathcal{H}}_3\left(1\right)$, which is defined as follows:

$${\mathcal{H}}_3\left(1\right)=\left|\begin{array}{ccc}{\ell }_1& {\ell }_2& {\ell }_3\\ {\ell }_2& {\ell }_3& {\ell }_4\\ {\ell }_3& {\ell }_4& {\ell }_5\end{array}\right|.$$

For $\mathfrak{T}\in A$, we have ${\ell }_1=1$, resulting in the following:

$${\mathcal{H}}_3\left(1\right)={\ell }_3({\ell }_2{\ell }_4-{\ell }_3^2)-{\ell }_4({\ell }_4-{\ell }_2{\ell }_3) + {\ell }_5({\ell }_3-{\ell }_2^2).$$

Using the triangle inequality, we can derive the following:

$$|{\mathcal{H}}_3\left(1\right)| \le |{\ell }_3||{\ell }_2{\ell }_4-{\ell }_3^2| + |{\ell }_4||{\ell }_4-{\ell }_2{\ell }_3| + |{\ell }_5||{\ell }_3-{\ell }_2^2|.$$

Quantum calculus provides a framework that extends classical calculus by introducing deformation parameters p and q, which are useful in studying various mathematical models, particularly in quantum mechanics and special function theory. One of its key concepts is the $(p,q)$-derivative, which generalizes the classical derivative.

Before proceeding, we will revisit some notations from $(p,q)$-calculus. The $(p,q)$-twin-basic number, represented as ${[\mathsf{ɩ}]}_{p,q}$, is defined as follows:

$${[\mathsf{ɩ}]}_{p,q}={\displaystyle \frac{p^{\mathsf{ɩ}}-q^{\mathsf{ɩ}}}{p-q}},(0<q<p \le 1,\mathsf{ɩ} =0,1,2,\dots ).$$

The $(p,q)$-derivative operator of a function $\mathfrak{T}$ is given by the following:

$$\left({\mathbb{D}}_{p,q}\mathfrak{T}\right)\left(z\right)={\displaystyle \frac{\mathfrak{T}\left(pz\right)-\mathfrak{T}\left(qz\right)}{(p-q)z}},(z\ne 0),$$

(7)

and

$${\mathbb{D}}_{p,q}\mathfrak{T}\left(0\right)={\mathfrak{T}}^{\prime }\left(0\right).$$

(8)

Given that the function $\mathfrak{T}$ is differentiable at $z=0$ (refer to [20]), we can directly deduce the following for a function $\mathfrak{T}$ expressed as in Equation (1):

$${\mathbb{D}}_{p,q}\mathfrak{T}\left(z\right)=1 + \sum _{\mathsf{ɩ}=2}^{\infty }{[\mathsf{ɩ}]}_{p,q}{\ell }_{\mathsf{ɩ}}{z}^{\mathsf{ɩ}-1}.$$

(9)

As mentioned in [21,22,23,24], the $(p,q)$-derivative operator simplifies to the Jackson $\mathfrak{q}$-derivative when $p=1$. This can be found using the following formula:

$${\mathbb{D}}_q\mathfrak{T}\left(z\right)={\displaystyle \frac{\mathfrak{T}\left(z\right)-\mathfrak{T}\left(qz\right)}{(1-q)z}},(z\ne 0).$$

(10)

Additionally, for $p=1$, the q-bracket ${[\mathsf{ɩ}]}_q$ is defined as follows:

$${[\mathsf{ɩ}]}_q={\displaystyle \frac{1-q^{\mathsf{ɩ}}}{1-q}},(\mathsf{ɩ}=0,1,2,\cdots ).$$

In this paper, we employ the $(p,q)$-derivative, since it is a natural generalization of the q-derivative, and it reduces to the q-derivative when $p=1$.

Definition 1. 

A function $\mathfrak{T}\in \Sigma$ belongs to the class ${\mathfrak{A}}_{\Sigma }(p,q,⋉)$ if it holds the following conditions:

$$\Re \left({\mathbb{D}}_{p,q}\mathfrak{T}\left(z\right)\right)>⋉,(0 \le ⋉<1;z\in U),$$

(11)

and

$$\Re \left({\mathbb{D}}_{p,q}g\left(w\right)\right)>⋉,(0 \le ⋉<1;w\in U),$$

(12)

where $g\left(w\right)={\mathfrak{T}}^{-1}$.

We provide an upper bound for the functional ${\mathcal{H}}_2\left(2\right)={\ell }_2{\ell }_4-{\ell }_3^2$ in this paper, and we establish the following:

$$|{\mathcal{H}}_3\left(1\right)| \le |{\ell }_3||{\ell }_2{\ell }_4-{\ell }_3^2| + |{\ell }_4||{\ell }_4-{\ell }_2{\ell }_3| + |{\ell }_5||{\ell }_3-{\ell }_2^2|.$$

(13)

Additionally, we investigate the upper bounds for the following quantities:

$$|{\ell }_3-{\ell }_2^2|,|{\ell }_3|,|{\ell }_4|\mathrm{and}|{\ell }_5|.$$

(14)

2. A Set of Lemmas

The set of a positive real functions $\mathfrak{E}\left(z\right)$ is denoted by $\mathfrak{E}$. All analytic functions $\mathcal{L}:U\to \mathbb{C}$ that meet the following criteria are included in this class:

$$\mathfrak{E}\left(0\right)=1\mathrm{and}\Re \left(\mathfrak{E}\right(z\left)\right)>0.$$

(15)

Lemma 1 

(see [14]). If $\mathcal{L}\in \mathfrak{E}$ is given by

$$\mathcal{L}\left(z\right)=1 + {\mathcal{L}}_{1}z + {\mathcal{L}}_2{z}^2 + {\mathcal{L}}_3{z}^3 + \cdots ,$$

then

$$|{\mathcal{L}}_{\mathsf{ɩ}}| \le 2,(\mathsf{ɩ}\in \mathbb{N}).$$

(16)

Lemma 2 

(see [17]). If $\mathcal{L}\in \mathfrak{E}$ is given by

$$\mathcal{L}\left(z\right)=1 + {\mathcal{L}}_{1}z + {\mathcal{L}}_2{z}^2 + {\mathcal{L}}_3{z}^3 + \cdots ,$$

then

$$2{\mathcal{L}}_2={\mathcal{L}}_1^2 + \gimel (4-{\mathcal{L}}_1^2),$$

(17)

and

$$4{\mathcal{L}}_3={\mathcal{L}}_1^3 + 2(4-{\mathcal{L}}_1^2){\mathcal{L}}_1\gimel -{\mathcal{L}}_1(4-{\mathcal{L}}_1^2){\gimel }^2 + 2(4-{\mathcal{L}}_1^2){(1-|\gimel |}^2)z,$$

(18)

for some ℷ and z with $|\gimel | \le 1$ and $|z| \le 1$.

3. $|{\mathcal{H}}_2\left(2\right)|$ Upper Bound for the Second Hankel Determinant

In this section, we focus on finding upper bounds for the functional ${\mathcal{H}}_2\left(2\right)={\ell }_2{\ell }_4-{\ell }_3^2$ in the class ${\mathfrak{A}}_{\Sigma }(p,q,⋉)$. Our objective is to ascertain its upper bounds and examine the related limitations. Further research on this subject may be found at [7,15,17,19].

Theorem 1. 

Let $\mathfrak{T}$ belong to the class ${\mathfrak{A}}_{\Sigma }(p,q,⋉)$ and be defined by (1).

$$\left|{\ell }_2{\ell }_4-{\ell }_3^2\right| \le \left\{\begin{array}{c}4{(1-⋉)}^2({\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}} + {\displaystyle \frac{1}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}),\hfill \\ \\ \mathit{if}(0 \le ⋉ \le 1-{\displaystyle \frac{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}^2-32{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}{16{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}})\hfill \\ \\ {\displaystyle \frac{4{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2}}-{\displaystyle \frac{{\left({\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉) + 3{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2-2{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}\right)}^2{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}(4{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}{(1-⋉)}^2-{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉)-2{\left[2\right]}_{p,q}^3{\left[3\right]}_{p,q}^2 + {\left[2\right]}_{p,q}^4{\left[4\right]}_{p,q})}},\hfill \\ \\ \mathit{if}(1-{\displaystyle \frac{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}^2-32{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}{16{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}<⋉<1).\hfill \end{array}\right.$$

(19)

Proof. 

Suppose that $\mathfrak{T}\in {\mathfrak{A}}_{\Sigma }(p,q,⋉)$. Then, by referring to Definition (1), we obtain the following:

$$D_{p,q}f\left(z\right)=⋉ + (1-⋉)\mathfrak{v}\left(z\right),$$

(20)

and

$$D_{p,q}g\left(w\right)=⋉ + (1-⋉)d\left(w\right).$$

(21)

where $\mathfrak{v}\left(z\right)$ and $d\left(w\right)$ are in the class $\mathfrak{E}$ of functions that are provided by

$$\mathfrak{v}\left(z\right)=1 + {\mathfrak{v}}_{1}z + {\mathfrak{v}}_2{z}^2 + {\mathfrak{v}}_3{z}^3 + \cdots ,$$

(22)

and

$$d\left(w\right)=1 + d_{1}w + d_2{w}^2 + d_3{w}^3 + \cdots .$$

(23)

It follows from (20) and (21), together with (3) and (9), that

$${\left[2\right]}_{p,q}{\ell }_2=(1-⋉){\mathfrak{v}}_1,$$

(24)

$${\left[3\right]}_{p,q}{\ell }_3=(1-⋉){\mathfrak{v}}_2,$$

(25)

$${\left[4\right]}_{p,q}{\ell }_4=(1-⋉){\mathfrak{v}}_3,$$

(26)

$${\left[5\right]}_{p,q}{\ell }_5=(1-⋉){\mathfrak{v}}_4,$$

(27)

and

$$-{\left[2\right]}_{p,q}{\ell }_2=(1-⋉)d_1,$$

(28)

$${\left[3\right]}_{p,q}\left(2{\ell }_2^2-{\ell }_3\right)=(1-⋉)d_2,$$

(29)

$$-{\left[4\right]}_{p,q}\left({\ell }_4 + 5{\ell }_3^2-5{\ell }_2{\ell }_3\right)=(1-⋉)d_3.$$

(30)

$${\left[5\right]}_{p,q}(14{\ell }_2^4-21{\ell }_2^2{\ell }_3 + 3{\ell }_3^2 + 6{\ell }_2{\ell }_4-{\ell }_5)=(1-⋉)d_4.$$

(31)

From (24) and (28), we obtain the following:

$${\mathfrak{v}}_1=-d_1,$$

(32)

and

$${\ell }_2={\displaystyle \frac{1-⋉}{{\left[2\right]}_{p,q}}}{\mathfrak{v}}_1.$$

(33)

Upon subtracting (25) from (29), we have the following:

$${\ell }_3={\displaystyle \frac{{(1-⋉)}^2}{{\left[2\right]}_{p,q}^2}}{\mathfrak{v}}_1^2 + {\displaystyle \frac{1-⋉}{2{\left[3\right]}_{p,q}}}({\mathfrak{v}}_2-d_2).$$

(34)

Also, if we subtract (26) from (30), we achieve the following:

$${\ell }_4={\displaystyle \frac{5{(1-⋉)}^2}{4{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}}}{\mathfrak{v}}_1({\mathfrak{v}}_2-d_2) + {\displaystyle \frac{1-⋉}{2{\left[4\right]}_{p,q}}}({\mathfrak{v}}_3-d_3).$$

(35)

Thus, by applying (33)–(35), we conclude that

$$\left|{\ell }_2{\ell }_4-{\ell }_3^2\right|=\left|{\displaystyle \frac{-{(1-⋉)}^4}{{\left[2\right]}_{p,q}^4}}{\mathfrak{v}}_1^4 + {\displaystyle \frac{{(1-⋉)}^3}{4{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}}{\mathfrak{v}}_1^2({\mathfrak{v}}_2-d_2)\right.$$

$$\left. + {\displaystyle \frac{{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}{\mathfrak{v}}_1({\mathfrak{v}}_3-d_3)-{\displaystyle \frac{{(1-⋉)}^2}{4{\left[3\right]}_{p,q}^2}}{({\mathfrak{v}}_2-d_2)}^2\right|.$$

(36)

Next, by applying Lemma 2 along with Equation (32), we obtain the following:

$$2{\mathfrak{v}}_2={\mathfrak{v}}_1^2 + \gimel \left(4-{\mathfrak{v}}_1^2\right),2d_2=d_1^2 + \zeta \left(4-d_1^2\right)$$

$$\Rightarrow {\mathfrak{v}}_2-d_2=\left({\displaystyle \frac{4-{\mathfrak{v}}_1^2}{2}}\right)(\gimel -\zeta )$$

(37)

and

$$4{\mathfrak{v}}_3={\mathfrak{v}}_1^3 + 2\left(4-{\mathfrak{v}}_1^2\right){\mathfrak{v}}_1\gimel -{\mathfrak{v}}_1\left(4-{\mathfrak{v}}_1^2\right){\gimel }^2 + 2\left(4-{\mathfrak{v}}_1^2\right)\left(1-{|\gimel |}^2\right)z,$$

(38)

$$4d_3=d_1^3 + 2\left(4-d_1^2\right)d_1\zeta -d_1\left(4-d_1^2\right){\zeta }^2 + 2\left(4-d_1^2\right)\left(1-{|\zeta |}^2\right)w,$$

(39)

and

$$\begin{array}{cc}\hfill {\mathfrak{v}}_3-d_3& ={\displaystyle \frac{{\mathfrak{v}}_1^3}{2}} + {\displaystyle \frac{1}{2}}{\mathfrak{v}}_1\left(4-{\mathfrak{v}}_1^2\right)\left(\gimel + \zeta \right)\hfill \\ & -{\displaystyle \frac{1}{4}}{\mathfrak{v}}_1\left(4-{\mathfrak{v}}_1^2\right)\left({\gimel }^2 + {\zeta }^2\right)\hfill \\ & + {\displaystyle \frac{4-{\mathfrak{v}}_1^2}{2}}\left[\left(1-{|\gimel |}^2\right)z-\left(1-{|\zeta |}^2\right)w\right].\hfill \end{array}$$

(40)

Then, by using (37), (38), (39) and (36) in (40), we can achieve the following equation:

$$\begin{array}{cc}\hfill |{\ell }_2{\ell }_4-{\ell }_3^2|=& | -{\displaystyle \frac{{(1-⋉)}^4}{{\left[2\right]}_{p,q}^4}}{\mathfrak{v}}_1^4 + {\displaystyle \frac{{(1-⋉)}^3}{4{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}}{\mathfrak{v}}_1^2{\displaystyle \frac{\left(4-{\mathfrak{v}}_1^2\right)}{2}}(\gimel -\zeta ) + {\displaystyle \frac{{(1-⋉)}^2}{4{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}}}{\mathfrak{v}}_1^4\hfill \\ & + {\displaystyle \frac{{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}{\mathfrak{v}}_1^2{\displaystyle \frac{\left(4-{\mathfrak{v}}_1^2\right)}{2}}(\gimel + \zeta )-{\displaystyle \frac{{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}{\mathfrak{v}}_1^2{\displaystyle \frac{\left(4-{\mathfrak{v}}_1^2\right)}{4}}({\gimel }^2 + {\zeta }^2)\hfill \\ & + {\displaystyle \frac{{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}{\mathfrak{v}}_1^2{\displaystyle \frac{\left(4-{\mathfrak{v}}_1^2\right)}{2}}\left[\left(1-{|\gimel |}^2\right)z-\left(1-{|\zeta |}^2\right)w\right]\hfill \\ & -{\displaystyle \frac{{(1-⋉)}^2}{4{\left[3\right]}_{p,q}}}{\displaystyle \frac{{\left(4-{\mathfrak{v}}_1^2\right)}^2}{4}}{(\gimel -\zeta )}^2|.\hfill \end{array}$$

(41)

Since $\mathfrak{v}\left(z\right)\in \mathfrak{E}$, it follows from Lemma (1) that $|{\mathfrak{v}}_1| \le 2$. Thus, by setting $|{\mathfrak{v}}_1| = \mathfrak{v}$, we can consider that, $0 \le \mathfrak{v} \le 2$. Based on Equation (41), we can obtain the following:

$$\begin{array}{cc}\hfill |{\ell }_2{\ell }_4-{\ell }_3^2| \le & {\displaystyle \frac{{(1-⋉)}^4}{{\left[2\right]}_{p,q}^4}}{\mathfrak{v}}^4 + {\displaystyle \frac{{(1-⋉)}^2}{4{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}}}{\mathfrak{v}}^4 + {\displaystyle \frac{{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}\mathfrak{v}(4-{\mathfrak{v}}^2)\hfill \\ & + \left[{\displaystyle \frac{{(1-⋉)}^3}{4{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}}{\mathfrak{v}}^2{\displaystyle \frac{\left(4-{\mathfrak{v}}^2\right)}{2}} + {\displaystyle \frac{{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}{\mathfrak{v}}^2{\displaystyle \frac{\left(4-{\mathfrak{v}}_1^2\right)}{2}}\right](\gimel + \zeta )\hfill \\ & + \left[{\displaystyle \frac{{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}}}{\mathfrak{v}}^2{\displaystyle \frac{\left(4-{\mathfrak{v}}^2\right)}{4}}-{\displaystyle \frac{{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[4\right]}_{\wp ,q}}}\mathfrak{v}{\displaystyle \frac{\left(4-{\mathfrak{v}}^2\right)}{2}}\right]{(|\gimel |}^2 + {|\zeta |}^2)\hfill \\ & + {\displaystyle \frac{{(1-⋉)}^2}{4{\left[3\right]}_{p,q}}}{\displaystyle \frac{{\left(4-{\mathfrak{v}}^2\right)}^2}{4}}(|\gimel | + {|\zeta |)}^2.\hfill \end{array}$$

(42)

Moreover, for $\mathbb{k}=|\gimel | \le 1$ and $\varrho =|\mathbf{y}| \le 1$, Equation (42) can be rewritten as follows:

$$|{\ell }_2{\ell }_4-{\ell }_3^2| \le {\top }_1 + (\mathbb{k} + \varrho ){\top }_2 + ({\mathbb{k}}^2 + {\varrho }^2){\top }_3 + {(\mathbb{k} + \varrho )}^2{\top }_4=:\mathfrak{G}(\mathbb{k},\varrho ),$$

(43)

where

$${\top }_1={\top }_1\left(\mathfrak{v}\right):=$$

$${\displaystyle \frac{{(1-⋉)}^2}{4}}\left[\left({\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}} + {\displaystyle \frac{1}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}\right){\mathfrak{v}}^4 + {\displaystyle \frac{8\mathfrak{v}-2{\mathfrak{v}}^3}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}\right]\ge 0,$$

(44)

$${\top }_2={\top }_2\left(\mathfrak{v}\right):={\displaystyle \frac{{(1-⋉)}^2}{8}}{\mathfrak{v}}^2(4-{\mathfrak{v}}^2)\left[{\displaystyle \frac{(1-⋉)}{{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}} + {\displaystyle \frac{2}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}\right]\ge 0,$$

(45)

$${\top }_3={\top }_3\left(\mathfrak{v}\right):={\displaystyle \frac{{(1-⋉)}^2}{8{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}\mathfrak{v}(4-{\mathfrak{v}}^2)(\mathfrak{v}-2) \le 0,$$

(46)

and

$${\top }_4={\top }_4\left(\mathfrak{v}\right):={\displaystyle \frac{{(1-⋉)}^2}{16{\left[3\right]}_{p,q}^2}}{(4-{\mathfrak{v}}^2)}^2\ge 0.$$

(47)

Our goal is to determine the maximum value of $\mathfrak{G}(\mathbb{k},\varrho )$, as given in Equation (43) within the closed square $[0,1]\times [0,1]$. To accomplish this, we examine its behavior under the conditions $\mathfrak{v}\in (0,2)$, $\mathfrak{v}=0$, and $\mathfrak{v}=2$, while also considering the sign of the subsequent phrase, as follows:

$${\mathfrak{G}}_{\mathbb{k},\varrho }:=\mathfrak{G}(\mathbb{k},\mathbb{k})·\mathfrak{G}(\varrho ,\varrho )-{\left[\mathfrak{G}(\mathbb{k},\varrho )\right]}^2.$$

(48)

First, let us consider the case when $\mathfrak{v}\in (0,2)$. Since ${\top }_3<0$ and ${\top }_3 + 2{\top }_4>0$ for $\mathfrak{v}\in (0,2)$, it follows that

$${\mathfrak{G}}_{\mathbb{k},\varrho }=\mathfrak{G}(\mathbb{k},\mathbb{k})·\mathfrak{G}(\varrho ,\varrho )-{\left[\mathfrak{G}(\mathbb{k},\varrho )\right]}^2<0.$$

(49)

This shows that $\mathfrak{G}(\mathbb{k},\varrho )$ does not reach a local maximum inside the square $[0,1]\times [0,1]$.

We now look at its maximum along the domain’s edge. Specifically, for $\mathbb{k}=0$ with $0 \le \varrho \le 1$ (and similarly for $\varrho =0$ with $0 \le \mathbb{k} \le 1$), we obtain the following result:

$$\mathfrak{G}(0,\varrho )=\mathcal{H}\left(\varrho \right)=({\top }_3 + {\top }_4){\varrho }^2 + {\top }_2\varrho + {\top }_1.$$

(50)

(i) When ${\top }_3 + {\top }_4=0$: In this case, for $0<\varrho <1$ and any fixed $\mathfrak{v}$ with $0<\mathfrak{v}<2$, we clearly have the following:

$${\mathcal{H}}^{\prime }\left(\varrho \right)=2({\top }_3 + {\top }_4)\varrho + {\top }_2>0.$$

(51)

This indicates that the function $\mathcal{H}\left(\varrho \right)$ is rising. Consequently, the highest value of $\mathcal{H}\left(\varrho \right)$ for fixed $\mathfrak{v}\in (0,2)$ occurs at $\varrho =1$ and

$$max\left\{\mathcal{H}\left(\varrho \right)\right\}=\mathcal{H}\left(1\right)={\top }_1 + {\top }_2 + {\top }_3 + {\top }_4.$$

(52)

(ii) When ${\top }_3 + {\top }_4<0$: Since ${\top }_2 + 2({\top }_3 + {\top }_4)\ge 0$ for $0<\varrho <1$ and any fixed $\mathfrak{v}$ with $0<\mathfrak{v}<2$, we observe the following:

$${\top }_2 + 2({\top }_3 + {\top }_4)<2({\top }_3 + {\top }_4)\varrho + {\top }_2<{\top }_2,$$

(53)

which implies that ${\mathcal{H}}^{\prime }\left(\varrho \right)>0$. Thus, the maximum value of $\mathcal{H}\left(\varrho \right)$ for fixed $\mathfrak{v}\in (0,2)$, occurs at $\varrho =1$.

For $\mathbb{k}=1$ and $0 \le \varrho \le 1$ (and similarly for $\varrho =1$ and $0 \le \mathbb{k} \le 1$), we obtain

$$\mathfrak{G}(1,\varrho )=F\left(\varrho \right)=({\top }_3 + {\top }_4){\varrho }^2 + ({\top }_2 + 2{\top }_4)\varrho + {\top }_1 + {\top }_2 + {\top }_3 + {\top }_4.$$

(54)

The identical method used in the earlier instances with ${\top }_3 + {\top }_4$, yields

$$max\left\{F\left(\varrho \right)\right\}=F\left(1\right)={\top }_1 + 2{\top }_2 + 2{\top }_3 + 4{\top }_4.$$

(55)

Since $\mathbb{H}\left(1\right) \le F\left(1\right)$ for $\mathfrak{v}\in (0,2)$, we conclude the following:

$$max\left\{\mathfrak{G}\right(\mathbb{k},\varrho \left)\right\}=G(1,1).$$

(56)

The function $\mathfrak{G}(\mathbb{k},\varrho )$ reaches its maximum value on the boundary of the square $[0,1]\times [0,1]$ at $\mathbb{k}=1$ and $\varrho =1$, inside the closed region $[0,1]\times [0,1]$.

Let the function $K:(0,2)\to \mathbb{R}$ be defined as follows:

$$K\left(\mathfrak{v}\right):=max\left\{\mathfrak{G}(\mathbb{k},\varrho )\right\}=G(1,1)={\top }_1 + 2{\top }_2 + 2{\top }_3 + 4{\top }_4.$$

(57)

By substituting the expressions for ${\top }_1$, ${\top }_2$, ${\top }_3$, and ${\top }_4$ into $K\left(\mathfrak{v}\right)$ from Equation (57), we obtain the following:

$$\begin{array}{cc}\hfill K\left(\mathfrak{v}\right)={(1-⋉)}^2& \left[\left({\displaystyle \frac{{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}}-{\displaystyle \frac{(1-⋉)}{4{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}}-{\displaystyle \frac{1}{2{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}} + {\displaystyle \frac{1}{4{\left[3\right]}_{p,q}^2}}\right){\mathfrak{v}}^4\right.\hfill \\ & + \left({\displaystyle \frac{(1-⋉)}{{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}} + {\displaystyle \frac{3}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}-{\displaystyle \frac{2}{{\left[3\right]}_{p,q}^2}}\right){\mathfrak{v}}^2\left. + {\displaystyle \frac{4}{{\left[3\right]}_{p,q}^2}}\right].\hfill \end{array}$$

(58)

Assuming that $K\left(\mathfrak{v}\right)$ reaches its maximum at some interior point $\mathfrak{v}\in (0,2)$, differentiating Equation (58) results in the following:

$$\begin{array}{cc}\hfill K^{\prime }\left(\mathfrak{v}\right)={(1-⋉)}^2& \left[\left({\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}}-{\displaystyle \frac{(1-⋉)}{{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}}-{\displaystyle \frac{2}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}} + {\displaystyle \frac{1}{{\left[3\right]}_{p,q}^2}}\right){\mathfrak{v}}^3\right.\hfill \\ & \left. + \left({\displaystyle \frac{2(1-⋉)}{{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}} + {\displaystyle \frac{6}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}-{\displaystyle \frac{4}{{\left[3\right]}_{p,q}^2}}\right)\mathfrak{v}\right].\hfill \end{array}$$

(59)

Case 1. Consider the following condition:

$${\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}}-{\displaystyle \frac{(1-⋉)}{{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}}-{\displaystyle \frac{2}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}} + {\displaystyle \frac{1}{{\left[3\right]}_{p,q}^2}}\ge 0.$$

(60)

Thus, we obtain

$$0 \le ⋉ \le 1-{\displaystyle \frac{\begin{array}{c}{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}(32{\left[3\right]}_{p,q}^2-15{\left[2\right]}_{p,q}{\left[4\right]}_{p,q})}\end{array}}{8{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}},$$

(61)

which ensures that $K^{\prime }\left(\mathfrak{v}\right)>0$ for $\mathfrak{v}\in (0,2)$. Since $K\left(\mathfrak{v}\right)$ is increasing within $(0,2)$, it does not attain a maximum in this interval.

Case 2. Assume the following:

$${\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}}-{\displaystyle \frac{(1-⋉)}{{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}}-{\displaystyle \frac{2}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}} + {\displaystyle \frac{1}{{\left[3\right]}_{p,q}^2}}<0.$$

(62)

Then, it follows that

$$1-{\displaystyle \frac{\begin{array}{c}{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}(32{\left[3\right]}_{p,q}^2-15{\left[2\right]}_{p,q}{\left[4\right]}_{p,q})}\end{array}}{8{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}<⋉<1.$$

(63)

Under these circumstances, solving $K^{\prime }\left(\mathfrak{v}\right)=0$ gives the real critical point $\mathfrak{v}={\mathfrak{v}}_0^{\left(1\right)}$, where

$${\mathfrak{v}}_0^{\left(1\right)}=\sqrt{{\displaystyle \frac{-2{\left[2\right]}_{p,q}^2\left({\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉) + 3{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2-2{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}\right)}{\begin{array}{c}4{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}{(1-⋉)}^2-{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉)-2{\left[2\right]}_{p,q}^3{\left[3\right]}_{p,q}^2 + {\left[2\right]}_{p,q}^4{\left[4\right]}_{p,q}\end{array}}}}.$$

(64)

If the parameter ⋉ is limited as follows:

$$\begin{array}{cc}\hfill 1& -{\displaystyle \frac{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}(32{\left[3\right]}_{p,q}^2-15{\left[2\right]}_{p,q}{\left[4\right]}_{p,q})}}{8{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}<⋉\hfill \\ & \le 1-{\displaystyle \frac{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}^2 + 32{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}}}{16{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}.\hfill \end{array}$$

(65)

It is evident that ${\mathfrak{v}}_0^{\left(1\right)}\ge 2$, implying that ${\mathfrak{v}}_0^{\left(1\right)}$ lies outside the interval $(0,2)$. Conversely, when the parameter ⋉ is restricted by the following:

$$1-{\displaystyle \frac{\begin{array}{c}{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}^2 + 32{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}}\end{array}}{16{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}<⋉<1.$$

(66)

It is observed that ${\mathfrak{v}}_0^{\left(1\right)}<2$, meaning that ${\mathfrak{v}}_0^{\left(2\right)}$ lies within the interior of the closed interval $[0,2]$. Since $K^{″}\left({\mathfrak{v}}_0^{\left(2\right)}\right)<0$, it follows that the function $K\left(\mathfrak{v}\right)$ reaches its maximum at $\mathfrak{v}={\mathfrak{v}}_0^{\left(2\right)}$. Thus, we have the following:

$$\begin{array}{cc}\hfill K\left({\mathfrak{v}}_0^{\left(2\right)}\right)& ={(1-⋉)}^2·\left({\displaystyle \frac{4}{{\left[3\right]}_{p,q}^2}}\right.\hfill \\ & \left.-{\displaystyle \frac{{\left({\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉) + 3{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2-2{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}\right)}^2}{{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}\left(4{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}{(1-⋉)}^2-{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉)-2{\left[2\right]}_{p,q}^3{\left[3\right]}_{p,q}^2 + {\left[2\right]}_{p,q}^4{\left[4\right]}_{p,q}\right)}}\right).\hfill \end{array}$$

(67)

Secondly, in the case when $\mathfrak{v}=2$, we obtain the following:

$$\mathfrak{G}(\mathbb{k},\varrho )=4{(1-⋉)}^2·\left({\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}} + {\displaystyle \frac{1}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}\right).$$

(68)

Finally, in the case when $\mathfrak{v}=0$, we find that

$$\mathfrak{G}(\mathbb{k},\varrho )={(1-⋉)}^2·{\displaystyle \frac{{(\kappa + \varrho )}^2}{{\left[3\right]}_{p,q}^2}}.$$

(69)

It is clear that the function $\mathfrak{G}(\mathbb{k},\varrho )$ attains its maximum value when both $\mathbb{k}=1$ and $\varrho =1$, as follows:

$$max\left\{\mathfrak{G}(\mathbb{k},\varrho )\right\}=G(1,1)={\displaystyle \frac{4{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2}}.$$

(70)

We thus find the following from (67)–(69):

$$\begin{array}{c}{\displaystyle \frac{4{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2}}<{\displaystyle \frac{16{(1-⋉)}^4}{{\left[2\right]}_{p,q}^4}} + {\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}<{\displaystyle \frac{4{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2}}-\hfill \\ \hfill {\displaystyle \frac{{\left({\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉) + 3{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2-2{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}\right)}^2{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}\left(\begin{array}{c}4{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}{(1-⋉)}^2-{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉)\\ -2{\left[2\right]}_{p,q}^3{\left[3\right]}_{p,q}^2 + {\left[2\right]}_{p,q}^4{\left[4\right]}_{p,q}\end{array}\right)}}.\end{array}$$

(71)

For the parameter ⋉ constrained by the following:

$$1-{\displaystyle \frac{\begin{array}{c}{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}^2 + 32q{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}}\end{array}}{16{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}<⋉<1,$$

(72)

which leads us to the second inequality of (19).

Alternatively, considering the left-hand side of inequality (71), we obtain the following:

$${\displaystyle \frac{4{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2}}<{\displaystyle \frac{16{(1-⋉)}^4}{{\left[2\right]}_{p,q}^4}} + {\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}},$$

(73)

which leads to the first inequality in (19), with the parameter $\alpha$ being restricted by the following:

$$0 \le ⋉ \le 1-{\displaystyle \frac{\begin{array}{c}{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q} + {\left[2\right]}_{p,q}\sqrt{{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}^2 + 32q{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}}\end{array}}{16{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}}}.$$

(74)

□

4. Some Results on Third Hankel Determinants and Their Upper Bounds

In this section, we investigate the upper bounds for several important coefficients. We focus on proving a key inequality involving the third Hankel determinant $|{\mathcal{H}}_3\left(1\right)|$, and we also discuss the estimation of the upper bounds for $|{\ell }_4|$ and $|{\ell }_5|$. It is worth mentioning that the coefficients $|{\ell }_2|$ and $|{\ell }_3|$ are essential when studying the Fekete–Szegö inequality. As for $|{\ell }_4|$ and $|{\ell }_5|$, their estimation can become highly complicated for some subclasses of bi-univalent functions. In fact, many researchers have classified these cases as open problems.

Theorem 2. 

Let $\mathfrak{T}\left(z\right)$ be defined by (1) and belong to the class ${\mathfrak{A}}_{\Sigma }(p,q,⋉)$.

$$|{\ell }_2{\ell }_3-{\ell }_4| \le \left\{\begin{array}{cc}8(1-⋉)\left[{\displaystyle \frac{{(1-⋉)}^2}{{\left[2\right]}_{p,q}^3}} + {\displaystyle \frac{1}{4{\left[4\right]}_{p,q}}}\right],\hfill & \mathsf{ɩ} \le \mathfrak{v} \le 2,\hfill \\ {\displaystyle \frac{2(1-⋉)}{{\left[4\right]}_{p,q}}},\hfill & 0 \le \mathfrak{v} \le \mathsf{ɩ}.\hfill \end{array}\right.$$

(75)

where

$$\mathsf{ɩ} ={\displaystyle \frac{{\mathcal{M}}_3\pm \sqrt{{\mathcal{M}}_3^2-12{\mathcal{M}}_2({\mathcal{M}}_1-{\mathcal{M}}_2)}}{3({\mathcal{M}}_1-{\mathcal{M}}_2)}},$$

(76)

with

$$\begin{array}{cc}\hfill {\mathcal{M}}_1& =(1-⋉)\left[{\displaystyle \frac{{(1-⋉)}^2}{{\left[2\right]}_{p,q}^3}} + {\displaystyle \frac{1}{4{\left[4\right]}_{p,q}}}\right],\hfill \end{array}$$

(77)

$$\begin{array}{cc}\hfill {\mathcal{M}}_2& =(1-⋉)\left[{\displaystyle \frac{3(1-⋉)}{4{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}}} + {\displaystyle \frac{3}{4{\left[4\right]}_{p,q}}}\right],\hfill \end{array}$$

(78)

$$\begin{array}{cc}\hfill {\mathcal{M}}_3& ={\displaystyle \frac{(1-⋉)}{2{\left[4\right]}_{p,q}}}.\hfill \end{array}$$

(79)

Proof. 

From Equations (33)–(35), we obtain the following:

$$|{\ell }_2{\ell }_3-{\ell }_4| =\left|{\displaystyle \frac{{(1-⋉)}^3{\mathfrak{v}}_1^3}{{\left[2\right]}_{p,q}^3}}-{\displaystyle \frac{3{(1-⋉)}^2{\mathfrak{v}}_1({\mathfrak{v}}_2-d_2)}{4{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}}}-{\displaystyle \frac{(1-⋉)({\mathfrak{v}}_3-d_3)}{2{\left[4\right]}_{p,q}}}\right|.$$

(80)

Using Lemma 2 and assuming $\mathfrak{v}\in [0,2]$, with ${\mathfrak{v}}_1=\mathfrak{v}$, we define the following:

$$|{\ell }_2{\ell }_3-{\ell }_4| \le {\mathcal{Q}}_1 + {\mathcal{Q}}_2(\delta + \gamma ) + {\mathcal{Q}}_3({\delta }^2 + {\gamma }^2):=\mathcal{Q}(\delta ,\gamma ),$$

(81)

where $\delta =|x| \le 1$ and $\gamma =|y| \le 1$, and

$$\begin{array}{cc}\hfill {\mathcal{Q}}_1(⋉,\mathfrak{v})& =(1-⋉){\mathfrak{v}}^3\left[{\displaystyle \frac{{(1-⋉)}^2}{{\left[2\right]}_{p,q}^3}} + {\displaystyle \frac{1}{4{\left[4\right]}_{p,q}}}\right]\ge 0,\hfill \\ \hfill {\mathcal{Q}}_2(\alpha ,\mathfrak{v})& =(1-⋉)(4-{\mathfrak{v}}^2)\mathfrak{v}\left[{\displaystyle \frac{3(1-⋉)}{8{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}}} + {\displaystyle \frac{1}{4{\left[4\right]}_{p,q}}}\right]\ge 0,\hfill \\ \hfill {\mathcal{Q}}_3(\alpha ,\mathfrak{v})& =(1-⋉)(4-{\mathfrak{v}}^2)\left[{\displaystyle \frac{\mathfrak{v} + 2}{8{\left[4\right]}_{p,q}}}\right]\ge 0.\hfill \end{array}$$

By the method of Theorem 1, the maximum at $\delta =1$ and $\gamma =1$ within the closed interval $[0,2]$. We define the following:

$$\psi \left(\mathfrak{v}\right)=max\left(\mathcal{Q}(\delta ,\gamma )\right)={\mathcal{Q}}_1 + 2{\mathcal{Q}}_2 + 2{\mathcal{Q}}_3.$$

(82)

Substituting the values of ${\mathcal{Q}}_1$, ${\mathcal{Q}}_2$, and ${\mathcal{Q}}_3$ in $\psi \left(\mathfrak{c}\right)$, we obtain the following:

$$\psi \left(\mathfrak{v}\right)={\mathcal{M}}_1{\mathfrak{v}}^3 + {\mathcal{M}}_2\mathfrak{v}(4-{\mathfrak{v}}^2) + {\mathcal{M}}_3(4-{\mathfrak{v}}^2),$$

(83)

where ${\mathcal{M}}_1$, ${\mathcal{M}}_2$, and ${\mathcal{M}}_3$ are as defined earlier.

Using the derivative,

$$\begin{array}{cc}\hfill {\psi }^{\prime }\left(\mathfrak{v}\right)& =3({\mathcal{M}}_1-{\mathcal{M}}_2){\mathfrak{v}}^2-2{\mathcal{M}}_3\mathfrak{v} + 4{\mathcal{M}}_2,\hfill \\ \hfill {\psi }^{\prime \prime }\left(\mathfrak{v}\right)& =6({\mathcal{M}}_1-{\mathcal{M}}_2)\mathfrak{v}-2{\mathcal{M}}_3.\hfill \end{array}$$

Also, ${\psi }^{\prime }\left(\mathfrak{v}\right)>0$ indicates that $\psi \left(\mathfrak{v}\right)$ is an increasing function on $[0,2]$, reaching its maximum at $\mathfrak{v}=2$, if ${\mathcal{M}}_1-{\mathcal{M}}_2>0$:

$$|{\ell }_2{\ell }_3-{\ell }_4| \le \psi \left(2\right)=8(1-⋉)\left[{\displaystyle \frac{{(1-⋉)}^2}{{\left[2\right]}_{p,q}^3}} + {\displaystyle \frac{1}{4{\left[4\right]}_{p,q}}}\right].$$

(84)

If ${\mathcal{M}}_1-{\mathcal{M}}_2<0$, solving ${\psi }^{\prime }\left(\mathfrak{v}\right)=0$ gives $\mathfrak{v}=n$. For $n<\mathfrak{v} \le 2$, ${\psi }^{\prime }\left(\mathfrak{v}\right)>0$, so $\psi \left(\mathfrak{v}\right)$ is decreasing on $[0,2]$ and reaches its maximum at $\mathfrak{v}=0$:

$$|{\ell }_2{\ell }_3-{\ell }_4| \le \psi \left(0\right)={\displaystyle \frac{2(1-⋉)}{{\left[4\right]}_{p,q}}}.$$

(85)

Then, the proof is complete. □

Theorem 3. 

Let $\mathfrak{T}\left(z\right)$ given by (1) be in the class ${\mathfrak{A}}_{\Sigma }(\wp ,q,⋉)$.

$$|{\ell }_3-{\ell }_2^2| \le {\displaystyle \frac{2(1-⋉)}{{\left[3\right]}_{p,q}}},$$

(86)

$$|{\ell }_3| \le {\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^2}} + {\displaystyle \frac{2(1-⋉)}{{\left[3\right]}_{p,q}}}.$$

(87)

Proof. 

By using Lemma 1 and (34) and, we obtain (87). The Fekete–Szegö functional for $\mathfrak{M}\in \mathbb{C}$ and ${\mathfrak{A}}_{\Sigma }(p,q,⋉)$ is as follows:

$${\ell }_3-\mathfrak{M}{\ell }_2^2={\displaystyle \frac{{(1-⋉)}^2{\mathfrak{v}}_1^2}{{\left[2\right]}_{p,q}^2}}(1-\mathfrak{M}) + {\displaystyle \frac{(1-⋉)({\mathfrak{v}}_2-d_2)}{2{\left[3\right]}_{p,q}}}.$$

By Lemma 1, we obtain the following:

$$|{\ell }_3-\mathfrak{M}{\ell }_2^2| \le {\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^2}}(1-\mathfrak{M}) + {\displaystyle \frac{(1-⋉)}{{\left[3\right]}_{p,q}}}.$$

For $\mathfrak{M}=1$, we obtain (86). □

Theorem 4. 

Let $\mathfrak{T}\left(z\right)$ given by (1) be in the class ${\mathfrak{A}}_{\Sigma }(p,q,⋉)$.

$$|{\ell }_4| \le {\displaystyle \frac{10{(1-⋉)}^2}{{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}}} + {\displaystyle \frac{2(1-⋉)}{{\left[4\right]}_{p,q}}},$$

(88)

$$|{\ell }_5{| \le 2(1-⋉)}^2\left[{\displaystyle \frac{100{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}} + {\displaystyle \frac{24(1-⋉)}{{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}} + {\displaystyle \frac{3}{{\left[3\right]}_{p,q}^2}} + {\displaystyle \frac{6}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}\right] + {\displaystyle \frac{2(1-⋉)}{{\left[5\right]}_{p,q}}}.$$

(89)

Proof. 

By subtracting (31) from (27), we achieve the following:

$$2{\left[5\right]}_{p,q}{\ell }_5={\left[5\right]}_{p,q}(14{\ell }_2^4-21{\ell }_2^2{\ell }_3 + 3{\ell }_3^2 + 6{\ell }_2{\ell }_4) + (1-⋉)({\mathfrak{v}}_4-d_4).$$

(90)

By substituting properly (33)–(35), we have the following:

$${\ell }_5={\displaystyle \frac{-25{(1-⋉)}^4}{2{\left[2\right]}_{p,q}^4}}{\mathfrak{v}}_1^4-{\displaystyle \frac{12{(1-⋉)}^3}{4{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}}}{\mathfrak{v}}_1^2({\mathfrak{v}}_2-d_2) + {\displaystyle \frac{3{(1-⋉)}^2}{8{\left[3\right]}_{p,q}^2}}{({\mathfrak{v}}_2-d_2)}^2$$

$$+ {\displaystyle \frac{3{(1-⋉)}^2}{2{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}{\mathfrak{v}}_1({\mathfrak{v}}_3-d_3) + {\displaystyle \frac{(1-⋉)}{2{\left[5\right]}_{p,q}}}({\mathfrak{v}}_4-d_4).$$

(91)

By Lemma 1, we obtain (89). □

Theorem 5. 

Let $\mathfrak{T}\left(z\right)\in {\mathfrak{A}}_{\Sigma }(\wp ,q,⋉)$, $0<⋉ \le 1$ to obtain

$$\begin{array}{c}\hfill |{\mathcal{H}}_3\left(1\right)| \le \left\{\begin{array}{cc}{\mathcal{R}}_1\mathcal{X} + {\mathcal{R}}_2\left(8(1-⋉)\left[{\displaystyle \frac{{(1-⋉)}^2}{{\left[2\right]}_{p,q}^3}} + {\displaystyle \frac{1}{4{\left[4\right]}_{p,q}}}\right]\right) + {\mathcal{R}}_3{\mathcal{R}}_4,\hfill & \mathsf{ɩ} \le \mathfrak{v} \le 2,\hfill \\ {\mathcal{R}}_1\mathcal{Y} + {\mathcal{R}}_2\left({\displaystyle \frac{2(1-⋉)}{{\left[4\right]}_{p,q}}}\right) + {\mathcal{R}}_3{\mathcal{R}}_4,\hfill & 0 \le \mathfrak{v} \le \mathsf{ɩ}.\hfill \end{array}\right.\end{array}$$

(92)

where

$$\mathcal{X}=4{(1-⋉)}^2({\displaystyle \frac{4{(1-⋉)}^2}{{\left[2\right]}_{p,q}^4}} + {\displaystyle \frac{1}{{\left[2\right]}_{p,q}{\left[4\right]}_{p,q}}}),$$

(93)

$$\mathcal{Y}={\displaystyle \frac{4{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2}}-{\displaystyle \frac{{\left({\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉) + 3{\left[2\right]}_{p,q}{\left[3\right]}_{p,q}^2-2{\left[2\right]}_{p,q}^2{\left[4\right]}_{p,q}\right)}^2{(1-⋉)}^2}{{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}(4{\left[3\right]}_{p,q}^2{\left[4\right]}_{p,q}{(1-⋉)}^2-{\left[2\right]}_{p,q}^2{\left[3\right]}_{p,q}{\left[4\right]}_{p,q}(1-⋉)-2{\left[2\right]}_{p,q}^3{\left[3\right]}_{p,q}^2 + {\left[2\right]}_{p,q}^4{\left[4\right]}_{p,q})}}.$$

(94)

and ${\mathcal{R}}_1,{\mathcal{R}}_2,{\mathcal{R}}_3$, and ${\mathcal{R}}_4$ are given by (87), (88), (90), and (86), respectively.

Proof. 

Since we start from the equation

$$\begin{array}{c}\hfill {\mathcal{H}}_3\left(1\right)={\ell }_3({\ell }_2{\ell }_4-{\ell }_3^2)-{\ell }_4({\ell }_4-{\ell }_2{\ell }_3) + {\ell }_5({\ell }_3-{\ell }_2^2),\end{array}$$

(95)

by applying the triangle inequality, we have (13). After collecting the Equations (87), (88), (90) and (86) in (13), we obtain (92). This allows us to obtain the third Hankel determinant $|{\mathcal{H}}_3\left(1\right)|$. □

5. Conclusions

In this paper, we explored some mathematical properties related to analytic functions, focusing on the use of the $(p,q)$-derivative operator to determine coefficient estimates and Hankel determinants. The results obtained contribute to deepening the understanding of specific classes of bi-univalent functions. We provided coefficient estimates up to $|{\ell }_5|$, as well as calculations for the Hankel determinants. These findings are significant in advancing the study of bi-univalent functions and paving the way for further research in this area. Additionally, researchers can extend or generalize similar classes of functions using the $(p,q)$-derivative operator. They can also compute coefficient estimates and Hankel determinants within these classes, possibly incorporating differential or integral operators to gain deeper insights into their structures.