Sharp Bounds of the Fekete–Szegö Problem and Second Hankel Determinant for Certain Bi-Univalent Functions Defined by a Novel q-Differential Operator Associated with q-Limaçon Domain

Abstract

In this present paper, we define a new operator in conjugation with the basic (or q-) calculus. We then make use of this newly defined operator and define a new class of analytic and bi-univalent functions associated with the q-derivative operator. Furthermore, we find the initial Taylor–Maclaurin coefficients for these newly defined function classes of analytic and bi-univalent functions. We also show that these bounds are sharp. The sharp second Hankel determinant is also given for this newly defined function class.

1. Introduction and Definitions

Due to its numerous uses in mathematical analysis and the physical sciences, such as q-difference operators, fractional and q-symmetric fractional q-calculus, optimal control, q-symmetric functions, and q-integral equations, the q-derivative has undergone accelerated development in a variety of scientific fields in recent decades (for more details, see [1,2,3,4,5,6,7,8]). Jackson introduces the q-difference operator and describes several applications of the q-integral and q-derivative in [9] (see also [10]). However, in the context of the geometric function theory of the complex analysis, the usage of hypergeometric functions was first used by Srivastava in [11]. Important applications of the q-calculus theory are discussed by Arif et al. in [12]. Using q-difference operators, Ismail et al. have given the q-deformation of starlike functions in [13]. A subclass of analytical functions with a Ruscheweyh q-differential operator is studied by Sokol et al. in [14]. Kanas and Raducanu established some convolution properties of a few normalized analytic functions in [15] and introduced q-analogues of the Ruscheweyh differential operators. Using q-hypergeometric functions, Darus et al. examined a q-analogue of a certain operator in [16]. Furthermore, the writers of [17,18,19,20] used the q-difference operator’s characteristics and defined different subclasses of complex analytical functions. Univalent functions are employed in electrostatics problems to find solutions to Laplace’s equation in complex geometries. For instance, in the study of electric fields around conductors or dielectrics, univalent functions can be used to map the problem domain onto a simpler domain wherein analytical solutions are available. Univalent functions can be used for data visualization purposes, particularly in the field of complex data visualization. By mapping complex data onto simpler domains using univalent functions, it becomes possible to visualize complex relationships and patterns in the data. It goes without saying that the q-derivative is crucial to modern mathematics and has several uses in physics and engineering. It is well known that they are crucial in approximation theory issues. Both mathematical statistics and the science of differential and integral equations contain them. Additionally known are their uses in the axially symmetric potential theory, automated control, scattering theory, signal analysis, quantum mechanics, and absorption of radio waves in the ionospheric space environment and aeronomy [21,22].

Definition 1

([9,23]). In a subset of $\mathcal{C}$ of complex numbers, the q-derivative (or q-difference) of a function f with the form (2) is denoted by ${\mathcal{D}}_{q}f\left(z\right)$ and defined by

$${\mathcal{D}}_{q}f\left(z\right)=\left\{\begin{array}{cc}\frac{f\left(z\right)-f\left(qz\right)}{(1-q)z}\hfill & forz\ne 0\hfill \\ \\ \\ f^{\prime }\left(0\right)\hfill & forz=0,\hfill \end{array}\right.$$

(1)

given that $f^{\prime }\left(z\right)$ is real. The q-derivative operator exists when $q⟶1^{-}$. ${\mathcal{D}}_q$ approaches the universal derivative operator, and as a result, we have

$$\underset{q⟶1^{-}}{lim}\left[{\mathcal{D}}_{q}f\left(z\right)\right]=f^{\prime }\left(z\right).$$

The family of all analytic functions on the unified open disc $\mathcal{U}=\{z\in \mathbb{C}:|z|<1\}$ is denoted by $\mathcal{H}$. Let $\mathcal{A}$ be the subset of $\mathcal{H}$ that contains all functions $f\left(z\right)$ given by

$$f\left(z\right)=z+\sum _{k=2}^{\infty }{s}_k{z}^k,$$

(2)

that are normalized by $f\left(0\right)=0$ and $f^{\prime }\left(0\right)=1$, and let $\mathcal{S}$ be the set of univalent functions (see, for details, [24]). Salagean introduced a differential operator and looked at some of its applications on a specific subclass of univalent functions in [25] for a function f in the class $\mathcal{S}$. Generalizations of the Salagean differential operator have recently been introduced by multiple writers in a variety of works (for more details, see [26,27,28]). Furthermore, investigated in [29] was the q-analogue of the Salagean operator. For details on the study about this topic, we may refer the readers to [30,31].

Now, using the binomial series

$${(1-\sigma )}^j=\sum _{\kappa =0}^j\left(\begin{array}{c}j\\ \kappa \end{array}\right){(-1)}^{\kappa }{\sigma }^{\kappa },(j\in \mathbb{N}=\{1,2,3,\cdots \},\kappa \in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\})$$

and

$${\mathcal{L}}_{\kappa }^j\left(\sigma \right)=\sum _{\kappa =1}^j\left(\begin{array}{c}j\\ \kappa \end{array}\right){(-1)}^{\kappa +1}{\sigma }^{\kappa }.$$

we define the new q-differential operator ${\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)$ for a function $f\left(z\right)$ in $\mathcal{A}$ as follows:

$$\begin{array}{c}\hfill {\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)={\mathcal{D}}_q\left({\mathcal{T}}_{\sigma ,q}^{n-1,\delta ,\alpha ,j}f\left(z\right)\right)=z+\sum _{m=2}^{\infty }{\left[{{\rm Y}}_m(\sigma ,\delta ,\alpha ,j)\right]}_q^n{a}_m{z}^m\end{array}$$

(3)

where

$${{\rm Y}}_m(\sigma ,\delta ,\alpha ,j)=1+(m+\alpha -\delta -1){\mathcal{L}}_{\kappa }^j\left(\sigma \right),$$

$j\in \mathbb{N}$, $\delta \in [0,\alpha ]$, $\sigma \ge 0$, $n\in {\mathbb{N}}_0$, $0<q<1$ and $z\in \mathbb{U}$.

Using the relation (3), we have the following identity:

$$\begin{array}{c}z{\mathcal{L}}_{\kappa }^j\left(\sigma \right){\mathcal{D}}_q\left({\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)\right)={\mathcal{T}}_{\sigma ,q}^{n+1,\delta ,\alpha ,j}f\left(z\right)-\left(1+(\alpha -\delta -1){\mathcal{L}}_{\kappa }^j\left(\sigma \right)\right){\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)\hfill \\ \hfill -z(\mu -\beta ){\mathcal{L}}_{\kappa }^j\left(\sigma \right).\end{array}$$

(4)

By using certain values for the parameters $\delta$, $\alpha$, $\sigma$, and j, the above newly developed q-analogous differential operators are generated into the following known operators.

1.

Let $q⟶1^{-}$, $\alpha =\delta =0$, we have the Frasin differential operator [27].

2.

Let $q⟶1^{-}$, $j=1$, we have the Opoola differential operator [28].

3.

Let $q⟶1^{-}$, $\alpha =\delta =0$ and $j=1$, we have the Al-Oboudi differential operator [26].

4.

Let $\alpha =\delta =0$, $\sigma =1$ and $j=1$, we have the q-Salagean differential operator [29].

5.

Let $q⟶1^{-}$, $\alpha =\delta =0$, $\sigma =1$ and $j=1$, we have the Salagean differential operator [25].

According to [32], the Koebe one-quarter theorem assures that every $f\in \mathcal{S}$ function’s image of the unit disk contains a disk with a radius of $1/4$.

The inverse function of $f\left(z\right)$, denoted by $f^{-1}\left(z\right)$, has the following definition if the function $f\left(z\right)$ is in $\mathcal{S}$:

$$f^{-1}\left(f\left(z\right)\right)=z(z\in \mathcal{U})$$

and

$$f\left(f^{-1}\left(\mu \right)\right)=\mu (|\mu |<d_0\left(f\right);d_0\left(f\right)\ge 1/4)$$

where

$$f^{-1}\left(\mu \right)=\mu -s_2{\mu }^2+(2s_2^2-s_3){\mu }^3-(5s_2^3-5s_2{s}_3+s_4){\mu }^4+\cdots .$$

(5)

If both $f^{-1}\left(z\right)$ and $f\left(z\right)$ are univalent in $\mathcal{U}$, then a function f is said to be bi-univalent in $\mathcal{U}$. Let $\Sigma$ denote the class of bi-univalent functions in $\mathcal{U}$ given by (2). See [33] for examples of the functions in the class $\Sigma$.

In [33], the study of analytical and bi-univalent functions is reintroduced; earlier investigations [34,35,36,37,38,39] are among them. New subclasses of bi-univalent function were introduced by a number of authors, and bounds were obtained for the initial coefficients and second Hankel determinant (see [30,31,40,41,42,43]).

One of the fundamental ideas of the geometry function theory (GFT) is the study of univalent functions that map the open unit disc onto a domain symmetric with respect to the real axis in the right-half plane. Considering this, recent times have seen an increase in the study of its subclasses. In order to accomplish this, Ma and Minda [44] offered a generalized classification of these subclasses; for more information, see [45,46,47].

Definition 2

([48]). Let $f\in \mathcal{A}$. Then, $f\in {\mathcal{ST}}_{{\mathcal{L}}_q}\left(t\right)$ if and only if

$$\frac{z{\mathcal{D}}_{q}f\left(z\right)}{f\left(z\right)}\prec {\left(\frac{2(1+tz)}{2+t(1-q)z}\right)}^2,0<q<1,0<t\le \frac{1}{\sqrt{2}},z\in \mathcal{U},$$

where ≺ is called the subordination principle. For more understanding on the subordination principle, see [32].

In this article, we introduce and investigate a subclass of analytical and bi-univalent functions defined by a new q-differential operator associated with the q-limaçon domain, which is inspired by the work of Saliu et al. [48,49].

Definition 3.

$f\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$, suppose the following conditions are met:

$$\frac{z{\mathcal{D}}_q\left({\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)\right)}{{\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)}\prec {\left(\frac{2(1+tz)}{2+t(1-q)z}\right)}^2,0<q<1,0<t\le \frac{1}{\sqrt{2}},z\in \mathcal{U}$$

and

$$\frac{z{\mathcal{D}}_q\left({\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}{f}^{-1}\left(\mu \right)\right)}{{\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}{f}^{-1}\left(\mu \right)}\prec {\left(\frac{2(1+t\mu )}{2+t(1-q)\mu }\right)}^2,0<q<1,0<t\le \frac{1}{\sqrt{2}},\mu \in \mathcal{U},$$

where $z,\mu \in \mathcal{U}$, $0<q<1$, and $f^{-1}\left(\mu \right)$ is given in (5).

These functions are extremal for several problems in the class $\Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$.

$$\begin{array}{c}\hfill f\left(z\right)=z+\frac{t(q+1)}{q{\left[m\right]}_q{\left[{{\rm Y}}_{m+1}(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{z}^{m+1}+\cdots ,m\in \mathbb{N},z\in \mathcal{U}.\end{array}$$

Lemma 1

([32,50]). Suppose that $\mathcal{P}$ is the set of all analytic functions $\mathfrak{p}$ of the form

$$\mathfrak{p}\left(z\right)=1+\sum _{k=1}^{\infty }{\mathfrak{p}}_k{z}^k$$

(6)

satisfying $\Re \left(\mathfrak{p}\right(z\left)\right)>0$, $z\in \mathcal{E}$ and $\mathfrak{p}\left(0\right)=1$. Then,

$$|{\mathfrak{p}}_k| \le 2,k\in \mathbb{N}.$$

For any value of $k\in \mathbb{N}$, this inequality is sharp.

Lemma 2

([32,50]). Suppose that $\mathcal{P}$ is the set of all analytic functions $\mathfrak{p}$ of the form

$$\mathfrak{p}\left(z\right)=1+\sum _{k=1}^{\infty }{\mathfrak{p}}_k{z}^k$$

(7)

satisfying $\Re \left(\mathfrak{p}\right(z\left)\right)>0$, $z\in \mathcal{E}$ and $\mathfrak{p}\left(0\right)=1$. Then,

$$2{\mathfrak{p}}_2={\mathfrak{p}}_1^2+(4-{\mathfrak{p}}_1^2)h$$

$$4{\mathfrak{p}}_3={\mathfrak{p}}_1^3+2(4-{\mathfrak{p}}_1^2){\mathfrak{p}}_{1}h-(4-{\mathfrak{p}}_1^2){\mathfrak{p}}_1{h}^2+2(4-{\mathfrak{p}}_1^2){(1-|h|}^2)z,$$

for some $h,z$ with $|h|\le 1$, $|z|\le 1$.

Lemma 3

([50]). If and only if the Toeplitz determinants

$${\mathcal{H}}_k=\left|\begin{array}{ccccc}2& {\mathfrak{p}}_1& {\mathfrak{p}}_2& \dots & {\mathfrak{p}}_k\\ {\mathfrak{p}}_{-1}& 2& {\mathfrak{p}}_1& \dots & {\mathfrak{p}}_{k-1}\\ ⋮& ⋮& ⋮& \dots & ⋮\\ {\mathfrak{p}}_{-k}& {\mathfrak{p}}_{-k+1}& {\mathfrak{p}}_{-k+2}& \dots & 2\end{array}\right|,k\in \mathbb{N}$$

(8)

and ${\mathfrak{p}}_{-k}=\overline{{}_k}$ are all nonnegative, the power series given in (6) converges in $\mathcal{E}$ to the function $\mathfrak{p}\in \mathcal{P}$. Except for

$$\mathfrak{p}\left(z\right)=\sum _{k=1}^k{\rho }_k{\mathfrak{p}}_0\left({\mathfrak{p}}^{i{x}_{k}z}\right),{\rho }_k>0,x_{k}real$$

and $x_k\ne x_j$ for $k\ne j$; in this example, they are all strictly positive. ${\mathcal{H}}_k>0$ for $k<n-1$ and ${\mathcal{H}}_k=0$ for $k\ge n$.

Notation 1.

$\mathfrak{p}\in \mathcal{P}$, ${\mathcal{H}}_k\ge 0$ and ${\mathfrak{p}}_{-1}=\overline{{\mathfrak{p}}_1}\ge 0$ are true, as stated by Lemma 3. This results in ${\mathcal{H}}_1=\left|\begin{array}{cc}2& {\mathfrak{p}}_1\\ {\mathfrak{p}}_1& 2\end{array}\right|\ge 0$ and ${\mathfrak{p}}_1=\overline{{\mathfrak{p}}_1}={\mathfrak{p}}_{-1}\ge 0$. As a result, $4-{\mathfrak{p}}_1^2\ge 0$ and ${\mathfrak{p}}_1\ge 0$ with ${\mathfrak{p}}_1\in [0,2]$. For these reasons, we will assume throughout our investigation that $|4-{\mathfrak{p}}_1^2|=|4-|{\mathfrak{p}}_1{|}^2|=4-|{\mathfrak{p}}_1{|}^2$ for ${\mathfrak{p}}_1$, the first coefficient in (6).

2. Coefficients Bound Estimates

We establish the following theorem in this section with respect to upper bound estimates for the few initial coefficients of the functions belonging within the class $\Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$.

Theorem 1.

Let $f\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$. Then,

$$|s_2| \le \frac{t(q+1)}{q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n},$$

$$|s_3| \le \frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n},$$

$$|s_4| \le \frac{t(q+1)}{q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

The results are sharp.

Proof. 

Let $f\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$. Then, there are holomorphic functions $b:\mathcal{U}⟶\mathcal{U}$, $c:\mathcal{U}⟶\mathcal{U}$ with $b\left(0\right)=0=c\left(0\right)$, $|b(z)| \le 1$, $|c(w)|<1$ fulfilling the following conditions:

$$\frac{z{\mathcal{D}}_q\left({\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)\right)}{{\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)}={\left(\frac{2(1+tb(z\left)\right)}{2+t(1-q)b\left(z\right)}\right)}^2,z\in \mathcal{U}$$

(9)

and

$$\frac{z{\mathcal{D}}_q\left({\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}{f}^{-1}\left(\mu \right)\right)}{{\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}{f}^{-1}\left(\mu \right)}={\left(\frac{2(1+tc(\mu \left)\right)}{2+t(1-q)c\left(\mu \right)}\right)}^2,\mu \in \mathcal{U}.$$

(10)

The functions $\mathfrak{p},x\in \mathcal{P}$ are defined as follows:

$$\mathfrak{p}\left(z\right)=\frac{1+b\left(z\right)}{1-b\left(z\right)}=1+\sum _{k=1}^{\infty }{\mathfrak{p}}_k{z}^k,z\in \mathcal{U}$$

and

$$\mathfrak{l}\left(\mu \right)=\frac{1+c\left(\mu \right)}{1-c\left(\mu \right)}=1+\sum _{k=1}^{\infty }{\mathfrak{l}}_k{\mu }^k,\mu \in \mathcal{U}.$$

As a result,

$$b\left(z\right)=\frac{\mathfrak{p}\left(z\right)-1}{\mathfrak{p}\left(z\right)+1}=\frac{1}{2}\left[{\mathfrak{p}}_{1}z+\left({\mathfrak{p}}_2-\frac{{\mathfrak{p}}_1^2}{2}\right)z^2+\left({\mathfrak{p}}_3+{\mathfrak{p}}_1{\mathfrak{p}}_2+\frac{{\mathfrak{p}}_1^3}{4}\right)z^3+\cdots \right],z\in \mathcal{U}$$

(11)

and

$$c\left(\mu \right)=\frac{\mathfrak{l}\left(\mu \right)-1}{\mathfrak{l}\left(\mu \right)+1}=\frac{1}{2}\left[{\mathfrak{l}}_1\mu +\left({\mathfrak{l}}_2-\frac{{\mathfrak{l}}_1^2}{2}\right){\mu }^2+\left({\mathfrak{l}}_3+{\mathfrak{l}}_1{\mathfrak{l}}_2+\frac{{\mathfrak{l}}_1^3}{4}\right){\mu }^3+\cdots \right],\mu \in \mathcal{U}.$$

(12)

By substituting the formulation of the functions $b\left(z\right)$ and $c\left(\mu \right)$ in (9) and (10) with those of (11) and (12), we arrive to

$$\begin{array}{cc}\hfill \frac{z{\mathcal{D}}_q\left({\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)\right)}{{\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}f\left(z\right)}& =1+\frac{t(q+1)}{2}{\mathfrak{p}}_{1}z+\frac{t(q+1)}{2}\left({\mathfrak{p}}_2+\frac{{\Lambda }_1(q,t)}{8}{\mathfrak{p}}_1^2\right)z^2\hfill \\ \hfill & +\frac{t(q+1)}{2}\left({\mathfrak{p}}_3+\frac{{\Lambda }_1(q,t)}{4}{\mathfrak{p}}_1{\mathfrak{p}}_2+\frac{{\Lambda }_2(q,t)}{8}{\mathfrak{p}}_1^3\right)z^3+\cdots .\hfill \end{array}$$

(13)

and

$$\begin{array}{cc}\hfill \frac{z{\mathcal{D}}_q\left({\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}{f}^{-1}\left(\mu \right)\right)}{{\mathcal{T}}_{\sigma ,q}^{n,\delta ,\alpha ,j}{f}^{-1}\left(\mu \right)}& =1+\frac{t(q+1)}{2}{\mathfrak{l}}_1\mu +\frac{t(q+1)}{2}\left({\mathfrak{l}}_2+\frac{{\Lambda }_1(q,t)}{8}{\mathfrak{l}}_1^2\right){\mu }^2\hfill \\ \hfill & +\frac{t(q+1)}{2}\left({\mathfrak{l}}_3+\frac{{\Lambda }_1(q,t)}{4}{\mathfrak{l}}_1{\mathfrak{l}}_2+\frac{{\Lambda }_2(q,t)}{8}{\mathfrak{l}}_1^3\right){\mu }^3+\cdots ,\hfill \end{array}$$

(14)

where

$${\Lambda }_1(q,t)=t(3q-1)-4$$

(15)

$${\Lambda }_2(q,t)=2-t(q-1)+q(q-1)t^2.$$

(16)

The following equations are produced for $s_2$, $s_3$, and $s_4$ by performing the operations and simplifications on the left side of Equations (13) and (14) and equating the coefficients of the terms with the same degree.

$$\begin{array}{cc}\hfill q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{s}_2& =\frac{t(q+1)}{2}{\mathfrak{p}}_1,\hfill \end{array}$$

(17)

$$\begin{array}{cc}\hfill q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n{s}_3-q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2{s}_2^2& =\frac{t(q+1)}{2}\left({\mathfrak{p}}_2+\frac{{\Lambda }_1(q,t)}{8}{\mathfrak{p}}_1^2\right),\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill & q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n{s}_4-q(2+q){\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n{s}_2{s}_3\hfill \\ \hfill & +q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2{s}_2^3=\frac{t(q+1)}{2}\left({\mathfrak{p}}_3+\frac{{\Lambda }_1(q,t)}{4}{\mathfrak{p}}_1{\mathfrak{p}}_2+\frac{{\Lambda }_2(q,t)}{8}{\mathfrak{p}}_1^3\right)\hfill \end{array}$$

(19)

and

$$\begin{array}{cc}\hfill -q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{s}_2& =\frac{t(q+1)}{2}{\mathfrak{l}}_1,\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill (2q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n& -q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2)s_2^2-q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n{s}_3\hfill \end{array}$$

$$\begin{array}{c}\hfill =\frac{t(q+1)}{2}\left({\mathfrak{l}}_2+\frac{{\Lambda }_1(q,t)}{8}{\mathfrak{l}}_1^2\right),\end{array}$$

(21)

$$\begin{array}{cc}\hfill & -q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n{s}_4+(2(2+q){\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n-5q{\left[3\right]}_q\hfill \\ \hfill & {\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n-q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2)s_2^3-(q(2+q){\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\hfill \\ \hfill & +5q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n)s_2{s}_3=\frac{t(q+1)}{2}\left({\mathfrak{l}}_3+\frac{{\Lambda }_1(q,t)}{4}{\mathfrak{l}}_1{\mathfrak{l}}_2+\frac{{\Lambda }_2(q,t)}{8}{\mathfrak{l}}_1^3\right).\hfill \end{array}$$

(22)

On the basis of Equations (17) and (20), we write

$${\mathfrak{p}}_1=-{\mathfrak{l}}_1,{\mathfrak{p}}_1^2={\mathfrak{l}}_1^2,{\mathfrak{p}}_1^3=-{\mathfrak{l}}_1^3.$$

(23)

The first result of the theorem is evident based on this and Lemma 1.

By deducting (21) from (18) and taking into account the equivalence in (23), we obtain

$$s_3=s_2^2+\frac{t(q+1)({\mathfrak{p}}_2-{\mathfrak{l}}_2)}{4q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n};$$

Hence,

$$s_3=\frac{{\mathfrak{p}}_1^2{t}^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}+\frac{t(q+1)({\mathfrak{p}}_2-{\mathfrak{l}}_2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

(24)

According to Lemma 2 and ${\mathfrak{p}}_1=-{\mathfrak{l}}_1$, we may write

$$\begin{array}{c}\hfill {\mathfrak{p}}_2-{\mathfrak{l}}_2=\frac{4-{\mathfrak{p}}_1^2}{2}(h-w),{\mathfrak{p}}_2+{\mathfrak{l}}_2={\mathfrak{p}}_1^2+\frac{4-{\mathfrak{p}}_1^2}{2}(h+w)\end{array}$$

(25)

and

$$\begin{array}{cc}\hfill {\mathfrak{p}}_3-{\mathfrak{l}}_3=& \frac{{\mathfrak{p}}_1^3}{2}+\frac{(4-{\mathfrak{p}}_1^2)(h+w)}{2}{\mathfrak{p}}_1-\frac{(4-{\mathfrak{p}}_1^2)(h^2+w^2)}{4}{\mathfrak{p}}_1\hfill \\ \hfill & +\frac{4-{\mathfrak{p}}_1^2}{2}\left[\left(1-{|h|}^2\right)z-\left(1-{|w|}^2\right)\mu \right]\hfill \end{array}$$

(26)

for some $h,w,z,\kappa$ with $|h|\le 1$, $|w|\le 1$, $|z|\le 1$, $|\mu |\le 1$.

The following expression for the coefficient $s_3$ is written by replacing the first equality (25) in (24).

Take note that $|4-{\mathfrak{p}}_1^2| =|4-|{\mathfrak{p}}_1{|}^2| =4-|{\mathfrak{p}}_1{|}^2=|4-a^2|=4-a^2$ can be written if we take $|{\mathfrak{p}}_1| =a$. (see, also Notation (1)). That is, we may assume without restriction that $\mathfrak{p}\in [0,2]$. In that instance, we can express the inequality for $|s_3|$ by using a triangle inequality and setting $|h|=\phi$ and $|w|=\psi$.

$$|s_3| \le \frac{a^2{t}^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}+\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}(\phi +\psi ),(\phi ,\psi )\in {[0,1]}^2.$$

Let us now define the function $\Delta :{\mathbb{R}}^2⟶\mathbb{R}$ as follows:

$$\Delta (\phi ,\psi )=\frac{a^2{t}^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}+\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}(\phi +\psi ),(\phi ,\psi )\in {[0,1]}^2.$$

The function $\Delta$ on the closed square $\Xi =\{(\phi ,\psi ):(\phi ,\psi )\in {[0,1]}^2\}$ needs to be maximized.

It is evident that the function $\Delta$ reaches its highest value at the closed square’s $\Xi$ boundary. When we differentiate the function $\Delta (\phi ,\psi )$ with regard to parameter $\phi$, we obtain

$${\Delta }_{\phi }(\phi ,\psi )=\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

Since the function $\Delta (\phi ,\psi )$ has a maximum value at $\phi =1$ and ${\Delta }_{\phi }(\phi ,\psi )\ge 0$, it is an increasing function with regard to $\phi$. Hence,

$$max\{\Delta (\phi ,\psi ):\phi \in [0,1]\}=\Delta (1,\psi )=\frac{a^2{t}^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}+\frac{t(q+1)(4-a^2)(1+\psi )}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n},$$

for each $\psi \in [0,1]$ and $a\in [0,2]$.

With the function $\Delta (1,\psi )$ now differentiable, we have

$${\Delta }^{\prime }(1,\psi )=\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

The function $\Delta (1,\psi )$ is an increasing function because ${\Delta }^{\prime }(1,\psi )\ge 0$ and the maximum occurs at $\psi =1$; therefore,

$$max\{\Delta (1,\psi ):\psi \in [0,1]\}=\Delta (1,1)=\frac{a^2{t}^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}+\frac{t(q+1)(4-a^2)}{4q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n},$$

where $a\in [0,2]$.

Thus, we obtain

$$\begin{array}{cc}\hfill \Delta (\phi ,\psi )\le max\{\Delta (\phi ,\psi ):(\phi ,\psi )\in \Xi \}=\Delta (1,1)& =\frac{a^2{t}^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}\hfill \\ \hfill & +\frac{t(q+1)(4-a^2)}{4q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.\hfill \end{array}$$

Since $|s_3| \le \Delta (\phi ,\psi )$, we have

$$|s_3| \le c(\sigma ,\delta ,\alpha ,j,n,q)\times a^2+\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n},a\in [0,2],$$

where

$$c(\sigma ,\delta ,\alpha ,j,n,q)=\frac{t(q+1)}{4q}\left[\frac{t(q+1)}{q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}-\frac{1}{{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\right].$$

Now, let us calculate the maximum of the function $N:\mathbb{R}⟶\mathbb{R}$ defined as follows:

$$N\left(a\right)=c(\sigma ,\delta ,\alpha ,j,n,q)\times a^2+\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}$$

in the range of $[0,2]$.

Differentiating the function $N\left(a\right)$, we have $N^{\prime }\left(a\right)=2c(\sigma ,\delta ,\alpha ,j,n,q)a,a\in [0,2]$. Since $N^{\prime }\left(a\right)\le 0$ when $c(\sigma ,\delta ,\alpha ,j,n,q)\le 0$, the function $N\left(a\right)$ is a decreasing function and the maximum occurs at $a=0$; therefore,

$$max\{N\left(a\right):a\in [0,2]\}=N\left(0\right)=\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}$$

and $N^{\prime }\left(a\right)\ge 0$ when $c(\sigma ,\delta ,\alpha ,j,n,q)\ge 0$, the function $N\left(a\right)$ is an increasing function and the maximum occurs at $a=2$, which gives

$$max\{N\left(a\right):a\in [0,2]\}=N\left(2\right)=\frac{t^2{(q+1)}^2}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}.$$

As a result, we arrive at the upper bound estimate for $|s_3|$ that is provided below:

$$|s_3| \le max\left\{\frac{t^2{(q+1)}^2}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2},\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\right\}.$$

Additionally, by reducing (22) to (19), taking into account the equalities (23) and (24), we have

$$\begin{array}{cc}\hfill s_4=& \frac{5t^3{(q+1)}^3}{16q^3{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3}{\mathfrak{p}}_1^3-\frac{M_1(q,n)t^3{(q+1)}^3}{16q^4{\left[3\right]}_q\left[{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right]{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3}{\mathfrak{p}}_1^3\hfill \\ \hfill & +\frac{{\Lambda }_2(q,t)t(q+1)}{16q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{\mathfrak{p}}_1^3+\frac{5t^2{(q+1)}^2({\mathfrak{p}}_2-{\mathfrak{l}}_2)}{16q^2{\left[2\right]}_q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{\mathfrak{p}}_1\hfill \\ \hfill & +\frac{{\Lambda }_1(q,t)t(q+1)({\mathfrak{p}}_2+{\mathfrak{l}}_2)}{16q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{\mathfrak{p}}_1+\frac{t(q+1)({\mathfrak{p}}_3-{\mathfrak{l}}_3)}{4q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n},\hfill \end{array}$$

(27)

where

$$\begin{array}{cc}\hfill M_1(q,n)=& 2q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2-2(2+q){\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\hfill \\ \hfill & +5q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n.\hfill \end{array}$$

(28)

The following inequality for $|s_4|$ is obtained from (27), using (25), (26), and triangle inequality.

$$|s_4| \le v_1(a,n)+v_2(a,n)(\phi +\psi )+v_3(a,n)({\phi }^2+{\psi }^2):=G(\phi ,\psi )$$

where

$$\begin{array}{cc}\hfill v_1(a,n)=& \frac{M_2(q,n)t(q+1)}{16q^4{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3}{a}^3+\frac{t(1+q)(4-a^2)}{4q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n},\hfill \\ \hfill v_2(a,n)=& \frac{5t^2{(q+1)}^2(4-a^2)}{32q^2{\left[2\right]}_q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}a+\frac{{\Lambda }_1(q,t)t(q+1)(4-a^2)}{32q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}a\hfill \\ \hfill & +\frac{t(q+1)(4-a^2)}{8q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}a,\hfill \\ \hfill v_3(a,n)=& \frac{t(q+1)(4-a^2)(a-2)}{16q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n},\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill M_2(q,n)=& 5t^2{(q+1)}^{2}q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n-t^2{(q+1)}^2{M}_1(q,n)\hfill \\ \hfill & +q^3{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3(2+{\Lambda }_1(q,t)+{\Lambda }_2(q,t)).\hfill \end{array}$$

The function G for each $a\in [0,2]$ must now be maximized.

We must examine the maximum of the function G for various values of the parameter a since the coefficients $v_1(a,n)$, $v_2(a,n)$, and $v_3(a,n)$ of the function G depend on the parameter a.

Let $a=0$, since $v_2(0,n)=0,$

$$v_1(0,n)=\frac{t(q+1)}{q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\mathrm{and}$$

$$v_3(0,n)=-\frac{t(q+1)}{q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

We then have

$$G(\phi ,\psi )=\frac{t(q+1)}{q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}-\frac{t(q+1)}{q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}({\phi }^2+{\psi }^2),(\phi ,\psi )\in {[0,1]}^2.$$

Hence, we obtain

$$G(\phi ,\psi )\le max\{Q(\phi ,\psi ):(\phi ,\psi )\in \Delta \}=G(0,0)=\frac{t(q+1)}{q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

Let $a=2$. Then, since $v_2(2,n)=v_3(2,n)=0$ and

$$v_1(2,n)=\frac{M_2(q,n)t(q+1)}{2q^4{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3}.$$

The following function G is a constant.

$$G(\phi ,\psi )=v_1(2,n)=\frac{M_2(q,n)t(q+1)}{2q^4{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3}.$$

We can rapidly demonstrate that the function G cannot have a maximum on the $\Xi$ in the case $a\in (0,2)$. Thus, we attain

$$|s_4| \le max\left\{\frac{M_2(q,n)t(q+1)}{2q^4{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3},\frac{t(q+1)}{q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\right\}.$$

□

However, for the subsequent function, the obtained results hold with equalities.

$$\begin{array}{c}\hfill f_m\left(z\right)=z+\frac{t(q+1)}{q{\left[m\right]}_q{\left[{{\rm Y}}_{m+1}(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{z}^{m+1}+\cdots ,m\in \mathbb{N},z\in \mathcal{U}.\end{array}$$

This concludes the proof of Theorem 1.

From the special instances $n=0$, we obtain the following results, respectively.

Corollary 1.

Let $f\in \Lambda \mathcal{M}\left(q\right)$. Then,

$$|s_2| \le \frac{t(q+1)}{q},$$

$$|s_3| \le \frac{t(q+1)}{q{\left[2\right]}_q},$$

$$|s_4| \le \frac{t(q+1)}{q{\left[3\right]}_q}.$$

The results are sharp for

$$\begin{array}{c}\hfill f_m\left(z\right)=z+\frac{t(q+1)}{q{\left[m\right]}_q}{z}^{m+1}+\cdots ,m\in \mathbb{N},z\in \mathcal{U}.\end{array}$$

3. The Second Hankel Determinant and Fekete-Szegö Inequality

For the function belonging to the class $\Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$ described by Definition 3, we provide an upper limit estimate for the second Hankel determinant and Fekete–Szegö inequality in this section.

First, we establish the following theorem on the second Hankel determinant’s upper-bound estimate.

Theorem 2.

Let $f\left(z\right)\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$. Then:

$$|s_2{s}_4-s_3^2| \le {\left(\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\right)}^2.$$

The results obtained here are sharp.

Proof. 

Let $f\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$. The following equality for $s_2{s}_4-s_3^2$ is therefore written from Equations (23), (24), and (27):

$$\begin{array}{cc}\hfill s_2{s}_4-s_3^2=& \frac{t(q+1)M_3(q,n)}{32q^5{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^4}{\mathfrak{p}}_1^4-\frac{t^4{(q+1)}^4}{16q^4{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^4}{\mathfrak{p}}_1^4\hfill \\ \hfill & -\frac{3t^3{(q+1)}^3({\mathfrak{p}}_2-{\mathfrak{p}}_2)}{64q^3{\left[2\right]}_q{\left({\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{\mathfrak{p}}_1^2\hfill \\ \hfill & +\frac{{\Lambda }_1(q,t)t^2{(q+1)}^2({\mathfrak{p}}_2+{\mathfrak{l}}_2)}{32q^2{\left[3\right]}_q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{\mathfrak{p}}_1^2-\frac{t^2{(q+1)}^2{({\mathfrak{p}}_2-{\mathfrak{l}}_2)}^2}{16q^2{\left[2\right]}_q^2{\left({\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}\hfill \\ \hfill & +\frac{t^3{(q+1)}^2({\mathfrak{p}}_3-{\mathfrak{l}}_3)}{8q^3{\left[3\right]}_q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{\mathfrak{p}}_1,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill M_3(q,n)=& t^3{(q+1)}^3[5q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n-M_1(q,n)]\hfill \\ \hfill & +t{q}^3(q+1){\Lambda }_2(q,t){\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3.\hfill \end{array}$$

(29)

We obtain the following estimate for $|s_2{s}_4-s_3^2|$ by using equalities (25) and (26), followed by triangle inequality and setting $|{\mathfrak{p}}_1| =a$, $|h|=\phi$, and $|w|=\psi$.

$$|s_2{s}_4-s_3^2| \le T_1(a,n)+T_2(a,n)(\phi +\psi )+T_3(a,n)({\phi }^2+{\psi }^2)+T_4(a,n){(\phi +\psi )}^2$$

(30)

where

$$\begin{array}{cc}\hfill T_1(a,n)=& \frac{M_4(q,n)}{32q^5{\left[3\right]}_q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^4{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{a}^4\hfill \\ \hfill & +\frac{t^2{(q+1)}^2(4-a^2)}{8q^2{\left[3\right]}_q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}a\ge 0\hfill \\ \hfill T_2(a,n)=& \frac{3t^3(q+1)(4-a^2)}{128q^3{\left[2\right]}_q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{a}^2\hfill \\ \hfill & +\frac{{\Lambda }_1(q,t)t^2{(q+1)}^2(4-a^2)}{64q^2{\left[3\right]}_q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{a}^2\hfill \\ \hfill & +\frac{t^2{(q+1)}^2(4-a^2)}{16q^3{\left[3\right]}_q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{a}^2\ge 0\hfill \\ \hfill T_3(a,n)& =\frac{t^2{(q+1)}^2(4-a^2)a(a-2)}{32q^3{\left[3\right]}_q{\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\le 0\hfill \\ \hfill T_4(a,n)& =\frac{t^2{(q+1)}^2{(4-a^2)}^2}{64q^2{\left[2\right]}_q^2{\left({\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}\ge 0,\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill M_4(q,n)=& t(q+1)[M_3(q,n)-2q{\left[3\right]}_q{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n{t}^3{(q+1)}^3]\hfill \\ \hfill & +q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^3{t}^3{(q+1)}^2[{\Lambda }_1(q,t)q+2].\hfill \end{array}$$

(31)

Let us now define the function $\Psi :{\mathbb{R}}^2⟶\mathbb{R}$ as follows:

$$\begin{array}{c}\hfill \Psi (\phi ,\psi )=T_1(a,n)+T_2(a,n)(\phi +\psi )+T_3(a,n)({\phi }^2+{\psi }^2)+T_4(a,n){(\phi +\psi )}^2,\end{array}$$

for each $(\phi ,\psi )\in {[0,1]}^2$ and $a\in [0,2]$.

The function $\Psi$ on $\Xi$ for each $a\in [0,2]$ must now be maximized.

We must examine the maximum of the function $\Psi$ for various values of the parameter a since the coefficients $T_1(a,n)$, $T_2(a,n)$, $T_3(a,n)$, and $T_4(a,n)$ of the function $\Psi$ depend on the parameter a.

1.

Let $a=0$. Since $T_1(0,n)=T_2(0,n)=T_3(0,n)=0$ and

$$T_4(0,n)=\frac{t^2{(q+1)}^2}{4q^2{\left[2\right]}_q^2{\left({\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}$$

the function $\Psi (\phi ,\psi )$ is written as follows:

$$\Psi (\phi ,\psi )=\frac{t^2{(q+1)}^2}{4q^2{\left[2\right]}_q^2{\left({\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{(\phi +\psi )}^2,(\phi ,\psi )\in \Delta .$$

It is obvious that the function $\Psi$ reaches its maximum near the closed-square boundary $\Delta$. Now, applying some differentiation on the function $\Psi (\phi ,\psi )$ with respect to $\phi$, we have

$${\Psi }_{\phi }(\phi ,\psi )=\frac{t^2{(q+1)}^2}{2q^2{\left[2\right]}_q^2{\left({\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}(\phi +\psi )$$

for each $\psi \in [0,1]$.

The function $\Psi (\phi ,\psi )$ is an increasing function with regard to $\phi$ and reaches its maximum at $\phi =1$ since ${\Psi }_{\phi }(\phi ,\psi )\ge 0$. Therefore,

$$max\{\Psi (\phi ,\psi ):\psi \in [0,1]\}=\Psi (1,\psi )=\frac{t^2{(q+1)}^2}{4q^2{\left[2\right]}_q^2{\left({\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{(1+\psi )}^2,\psi \in [0,1].$$

Taking the differentiation of the function $\Psi (1,\psi )$, we obtain

$${\Psi }^{\prime }(1,\psi )=\frac{t^2{(q+1)}^2}{2q^2{\left[2\right]}_q^2{\left({\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}(1+\psi ),\psi \in [0,1].$$

Since ${\Psi }^{\prime }(1,\psi )>0$, the function $\Psi (1,\psi )$ is an increasing function and the maximum occurs at $\psi =1$. Hence,

$$max\{\Psi (1,\psi ):\psi \in [0,1]\}=\Psi (1,1)={\left(\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\right)}^2.$$

Thus, in the instance of $a=0$, we obtain

$$\Psi (\phi ,\psi )\le max\{\Psi (\phi ,\psi ):(\phi ,\psi )\in {[0,1]}^2\}=\Psi (1,1)={\left(\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\right)}^2.$$

We know that $|s_2{s}_4-s_3^2| \le \Psi (\phi ,\psi )$, so we can have

$$|s_2{s}_4-s_3^2| \le {\left(\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\right)}^2.$$

2.

Now, taking $a=2$. Since $T_2(2,n)=T_3(2,n)=T_4(2,n)=0$ and

$$T_1(2,n)=\frac{M_4(q,n)}{2q^5{\left[3\right]}_q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^4{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}$$

the function $\Psi (\phi ,\psi )$ is a constant, as follows:

$$\Psi (\phi ,\psi )=T_1\left(2\right)=\frac{M_4(q,n)}{2q^5{\left[3\right]}_q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^4{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

Thus, we obtain

$$|s_2{s}_4-s_3^2|\le \frac{M_4(q,n)}{2q^5{\left[3\right]}_q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^4{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n},$$

in the case of $a=2$.

3.

Let us say $a\in (0,2)$. In this instance, we must look into the maximum of the function $\Psi$ while accounting for the sign of $\Lambda (\Psi )={\Psi }_{\phi \phi }(\phi ,\psi ){\Psi }_{\psi \psi }(\phi ,\psi )-{\left({\Psi }_{\phi \psi }(\phi ,\psi )\right)}^2$.

The equation $\Lambda (\Psi )=4T_3(a,n)[T_3(a,n)+2T_4(a,n)]$ is clear to see. We will look into two instances of the sign $\Lambda (\Psi )$.

(a)

Let $T_3(a,n)+2T_4(a,n)\le 0$ for the same $a\in (0,2)$. In this case, since ${\Psi }_{\phi ,\psi }(\phi ,\psi )={\Psi }_{\psi ,\phi }(\phi ,\psi )=2T_4(a,n)\ge 0$ and $\Lambda (\Psi )\ge 0$, the function $\Psi$ (having a minimum) cannot have a maximum on the square $\Delta$ according to basic calculus.

(b)

Now, let $T_3(a,n)+2T_4(a,n)\ge 0$ for some $a\in (0,2)$. In this case, since $\Lambda (\Psi )\le 0$, the function $\Psi$ cannot have a maximum on the square $\Delta$.

Consequently, in light of all three instances, we write

$$|s_2{s}_4-s_3^2| \le max\left\{\frac{M_4(q,n)}{2q^5{\left[3\right]}_q{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^4{\left[{{\rm Y}}_4(\sigma ,\delta ,\alpha ,j)\right]}_q^n},{\left(\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}\right)}^2\right\}.$$

However, for the subsequent function, the obtained results hold with equalities.

$$\begin{array}{c}\hfill f_2\left(z\right)=z+\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{z}^3+\cdots ,m\in \mathbb{N},z\in \mathcal{U}.\end{array}$$

Theorem 2 has now been successfully proved. □

From the special instances $n=1$, we obtain the following results, respectively.

Corollary 2.

Let $f\left(z\right)\in \Lambda \mathcal{M}\left(q\right)$. Then,

$$|s_2{s}_4-s_3^2| \le {\left(\frac{t(q+1)}{q{\left[2\right]}_q}\right)}^2.$$

The results obtained here are sharp for

$$\begin{array}{c}\hfill f_2\left(z\right)=z+\frac{t(q+1)}{q{\left[2\right]}_q}{z}^3+\cdots ,m\in \mathbb{N},z\in \mathcal{U}.\end{array}$$

We now provide the subsequent theorem on the Fekete–Szegö inequality.

Theorem 3.

Let $f\left(z\right)\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$, $\chi \in \mathcal{C}$. Then,

$$\left|s_3-\chi s_2^2\right|\le \left\{\begin{array}{ccc}\frac{L(q,n)}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}\hfill & \hfill & |1-\chi |\le L(q,n)\hfill \\ \hfill & \hfill & \\ \frac{|1-\chi |}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}\hfill & \hfill & |1-\chi |\ge L(q,n),\hfill \end{array}\right.$$

where

$$L(q,n)=\frac{t(q+1)q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

The results obtained here are sharp.

Proof. 

Let $f\left(z\right)\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$ and $\chi \in \mathcal{C}$. Then, from (23), (24), (25), and (26), we have the expression $s_3-\chi s_2^2$ be

$$s_3-\chi s_2^2=(1-\chi )\frac{{\mathfrak{p}}_1^2{t}^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}+\frac{t(q+1)(4-{\mathfrak{p}}_1^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}(h-w)$$

(32)

for some $h,w$ with $|h|\le 1$ and $|w|\le 1$.

With $|h|=\phi$, $|w|=\psi$, $|{\mathfrak{p}}_1|=a$ and the triangle inequality to the equality (32), we can estimate the upper bound of $|s_3-\chi s_2^2|$ as follows:

$$|s_3-\chi s_2^2| \le \frac{|1-\chi |t^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{a}^2+\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}(\phi +\psi ),(\phi ,\psi )\in \Delta ,$$

(33)

for each $a\in [0,2]$.

Let us now define the function $\Omega :{\mathbb{R}}^2⟶\mathbb{R}$ as follows:

$$\Omega (\phi ,\psi )=\frac{|1-\chi |t^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{a}^2+\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}(\phi +\psi ),(\phi ,\psi )\in \Delta ,$$

for each $a\in [0,2]$. The function $\Omega$ on $\Delta$ for each $a\in [0,2]$ must now be maximized.

The function $\Omega$ clearly reaches its maximum value along the boundary of the closed square $\Delta$.

Applying the concept of differentiation on the function $\Omega (\phi ,\psi )$ with respect to $\phi$, we have

$${\Omega }_{\phi }(\phi ,\psi )=\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}$$

(34)

for each $a\in [0,2]$.

Since ${\Omega }_{\phi }(\phi ,\psi )>0$, the function $\Omega (\phi ,\psi )$ is an increasing function with respect to $\phi$ and the maximum occurs at $\phi =1$. Hence,

$$max\{\Omega (\phi ,\psi ):\psi \in [0,1]\}=\Omega (1,\psi )=\frac{|1-\chi |t^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{a}^2+\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}(1+\psi )$$

for each $\psi \in [0,1]$ and $a\in [0,2]$.

Now, differentiating the function $\Omega (1,\psi )$, we obtain

$${\Omega }^{\prime }(1,\psi )=\frac{t(q+1)(4-a^2)}{8q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}$$

for each $a\in [0,2]$.

Since ${\Omega }^{\prime }(1,\psi )>0$, the function $\Omega (1,\psi )$ is an increasing function and the maximum occurs at $\psi =1$. Hence,

$$max\{\Omega (1,\psi ):\psi \in [0,1]\}=\Omega (1,1)=\frac{|1-\chi |t^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{a}^2+\frac{t(q+1)(4-a^2)}{4q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

Thus, we have

$$\begin{array}{cc}\hfill \Omega (\phi ,\psi )& \le max\{\Omega (\phi ,\psi ):\chi \in [0,1\left]\right\}=\Omega (1,1)\hfill \\ \hfill & =\frac{|1-\chi |t^2{(q+1)}^2}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{a}^2+\frac{t(q+1)(4-a^2)}{4q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.\hfill \end{array}$$

Since $|s_3-\chi s_2^2| \le \Omega (\phi ,\psi )$, we have the following estimate

$$|s_3-\chi s_2^2| \le \frac{|1-\chi |-L(q,n)}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{a}^2+\frac{L(q,n)}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}.$$

where

$$L(q,n)=\frac{t(q+1)q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

In such instance, finding the maximum of the following function, $\vartheta :[0,2]⟶\mathbb{R}$, would be appropriate

$$\vartheta (a,n)=\frac{|1-\chi |-L(q,n)}{4q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{a}^2+\frac{L(q,n)}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}.$$

Differentiating the function $\vartheta (a,n)$, we have

$${\vartheta }^{\prime }(a,n)=\frac{|1-\chi |-L(q,n)}{2q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}a,a\in [0,2].$$

If $|1-\chi |\le L(q,n)$ and the maximum occurs at $a=0$, then the function $\vartheta (a,n)$ is a decreasing function since ${\vartheta }^{\prime }(a,n)\le 0$.

$$max\{\vartheta (a,n):a\in [0,2]\}=\vartheta (0,n)=\frac{L(q,n)}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}$$

and ${\vartheta }^{\prime }(a,n)\ge 0$, the function $\vartheta (a,n)$ being an increasing function. If $|1-\chi |\ge L(a,n)$ and the maximum occurs at $a=2$,

$$max\{\vartheta (a,n):a\in [0,2]\}=\vartheta (2,n)=\frac{|1-\chi |}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}.$$

Consequently, we achieve

$$\left|s_3-\chi s_2^2\right|\le \left\{\begin{array}{ccc}\frac{L(q,n)}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}\hfill & \hfill & |1-\chi |\le L(q,n)\hfill \\ \hfill & \hfill & \\ \frac{|1-\chi |}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}\hfill & \hfill & |1-\chi |\ge L(q,n).\hfill \end{array}\right.$$

The result reached in this instance is sharp for $|1-\chi |\ge L(q,n)$.

If we set the function $f\left(z\right)$ as follows:

$$\begin{array}{c}\hfill f_2\left(z\right)=z+\frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}{z}^3+\cdots ,m\in \mathbb{N},z\in \mathcal{U}.\end{array}$$

□

From the special instances $n=0$, we obtain the following results, respectively.

Corollary 3.

Let $f\left(z\right)\in \Lambda \mathcal{M}\left(q\right)$, $\chi \in \mathcal{C}$. Then,

$$\left|s_3-\chi s_2^2\right|\le \left\{\begin{array}{ccc}\frac{L(q,n)}{q^2}\hfill & \hfill & |1-\chi |\le L(q,n)\hfill \\ \hfill & \hfill & \\ \frac{|1-\chi |}{q^2}\hfill & \hfill & |1-\chi |\ge L(q,n),\hfill \end{array}\right.$$

where

$$L(q,n)=\frac{t(q+1)q^2}{q{\left[2\right]}_q}.$$

The results obtained here are sharp.

Theorem 3 is presented in the following manner for the case $\chi \in \mathbb{R}$.

Theorem 4.

Let $f\left(z\right)\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$, $\chi \in \mathbb{R}$. Then,

$$|s_3-\chi s_2^2|\le \left\{\begin{array}{cc}\hfill \frac{1-\chi }{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}& if \chi \le 1-L(q,n)\hfill \\ \\ \hfill \frac{L(q,n)}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}& if 1-L(q,n)\le \chi \le 1+L(q,n)\hfill \\ \\ \hfill \frac{\chi -1}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}& if 1+L(q,n)\le \chi ,\hfill \end{array}\right.$$

(35)

where

$$L(q,n)=\frac{t(q+1)q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

Proof. 

Let $f\left(z\right)\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$ and $\chi \in \mathbb{R}$. Since in the case $\chi \in \mathbb{R}$, the inequalities $|1-\chi |\ge L(q,n)$ and $|1-\chi |\le L(q,n)$ are equivalent to

$$\chi \le 1-L(q,n)\mathrm{or}\chi \ge 1+L(q,n)$$

and

$$1-L(q,n)\le \chi \le 1+L(q,n),$$

respectively. The conclusion of the theorem is derived from Theorem 3. □

From the special instances $\chi =1$, we obtain the following results, respectively.

Corollary 4.

Let $f\left(z\right)\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$. Then,

$$|s_3-s_2^2| \le \frac{t(q+1)}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

From the special instances $\chi =0$, we obtain the following results, respectively.

Corollary 5.

Let $f\left(z\right)\in \Lambda {\mathcal{M}}_{\alpha ,\delta }^{\sigma ,j,n}\left(q\right)$. Then,

$$|s_3|\le \left\{\begin{array}{cc}\hfill \frac{1}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}& if L(q,n)\le 1,\hfill \\ \\ \hfill \frac{L(q,n)}{q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}& if L(q,n)\ge 1\hfill \end{array}\right.$$

(36)

where

$$L(q,n)=\frac{t(q+1)q^2{\left({\left[{{\rm Y}}_2(\sigma ,\delta ,\alpha ,j)\right]}_q^n\right)}^2}{q{\left[2\right]}_q{\left[{{\rm Y}}_3(\sigma ,\delta ,\alpha ,j)\right]}_q^n}.$$

From the special instances $n=0$ in Theorem 4, we get the following results, respectively.

Corollary 6.

Let $f\left(z\right)\in \Lambda \mathcal{M}\left(q\right)$, $\chi \in \mathbb{R}$. Then:

$$|s_3-\chi s_2^2|\le \left\{\begin{array}{cc}\hfill \frac{1-\chi }{q^2}& if \chi \le 1-L(q,n)\hfill \\ \\ \hfill \frac{L(q,n)}{q^2}& if 1-L(q,n)\le \chi \le 1+L(q,n)\hfill \\ \\ \hfill \frac{\chi -1}{q^2}& if 1+L(q,n)\le \chi ,\hfill \end{array}\right.$$

(37)

where

$$L(q,n)=\frac{t(q+1)q^2}{q{\left[2\right]}_q}.$$

Remark 1.

By varying the parameters in Theorem 1–4, we also have some new results that are associated with the existing operators and were generated by the new q-analogous differential operator discussed in this paper.

4. Conclusions

In this present paper, we have defined a new operator in conjugation with the basic (or q-) calculus. We have then made use of this newly defined operator and have defined a new class of analytic and bi-univalent functions associated with the q-derivative operator. Furthermore, we have found the initial Taylor–Maclaurin coefficients for these newly defined function classes of analytic and bi-univalent functions.We have also shown that these bounds are sharp. The sharp second Hankel determinants have also been given for this newly defined function class. Further research can be conducted based on this paper by using the novel q-differential operator on some other subclasses of analytic and bi-univalent functions to create some other new classes in geometric function theory.