**1. Introduction and Definitions**

Let the class of functions, which are analytic in the open unit disk

$$\mathcal{U} = \{ z : z \in \mathbb{C} \quad \text{and} \quad |z| < 1 \} \dots$$

be denoted by <sup>L</sup> (U). Also let <sup>A</sup> denote the class of all functions *<sup>f</sup>* , which are analytic in the open unit disk U and normalized by

$$f\left(0\right) = 0 \quad \text{and} \quad f'\left(0\right) = 1.$$

Then, clearly, each *f* ∈ A has a Taylor–Maclaurin series representation as follows:

$$f\left(z\right) = z + \sum\_{n=2}^{\infty} a\_n z^n \qquad \left(z \in \mathbb{U}\right). \tag{1}$$

Suppose that S is the subclass of the analytic function class A, which consists of all functions which are also univalent in U.

A function *<sup>f</sup>* ∈ A is said to be starlike in <sup>U</sup> if it satisfies the following inequality:

$$\Re\left(\frac{zf'(z)}{f(z)}\right) > 0 \qquad (z \in \mathbb{U})\ .$$

We denote by <sup>S</sup><sup>∗</sup> the class of all such starlike functions in <sup>U</sup>.

For two functions *f* and *g*, analytic in U, we say that the function *f* is subordinate to the function *g* and write this subordination as follows:

$$f \preccurlyeq \mathbf{g} \quad \text{or} \quad f\left(z\right) \preccurlyeq g\left(z\right) \cdot \mathbf{g}$$

if there exists a Schwarz function *w* which is analytic in U, with

$$w\left(0\right) = 0 \qquad \text{and} \qquad \left|w\left(z\right)\right| < 1,$$

such that

$$f\left(z\right) = \lg\left(w\left(z\right)\right).$$

In the case when the function *g* is univalent in U, then we have the following equivalence (see, for example, [1]; see also [2]):

$$f(z) \preccurlyeq g(z) \quad (z \in \mathbb{U}) \iff f(0) = \emptyset(0) \quad \text{and} \quad f(\mathbb{U}) \subset \mathcal{g}(\mathbb{U}).$$

Next, for a function *f* ∈ A given by (1) and another function *g* ∈ A given by

$$g(z) = z + \sum\_{n=2}^{\infty} b\_n z^n \qquad (z \in \mathbb{U})\ \_\nu$$

the convolution (or the Hadamard product) of *f* and *g* is defined here by

$$(f \ast g)\left(z\right) := z + \sum\_{n=2}^{\infty} a\_n b\_n z^n =: \left(g \ast f\right)\left(z\right). \tag{2}$$

Let <sup>P</sup> denote the well-known Carathéodory class of functions *<sup>p</sup>*, analytic in the open unit disk <sup>U</sup>, which are normalized by

$$p\left(z\right) = 1 + \sum\_{n=1}^{\infty} c\_n z^n,\tag{3}$$

such that

$$\Re\left(p\left(z\right)\right) > 0 \qquad \qquad \left(z \in \mathbb{U}\right)\dots$$

Following the works of Kanas et al. (see [3,4]; see also [5]), we introduce the conic domain Ω*<sup>k</sup>* (*k* 0) as follows:

$$
\Omega\_k = \left\{ u + iv : u > k\sqrt{\left(u - 1\right)^2 + v^2} \right\}.\tag{4}
$$

In fact, subjected to the conic domain Ω*<sup>k</sup>* (*k* 0), Kanas and Wi´sniowska (see [3,4]; see also [6]) studied the corresponding class *<sup>k</sup>*-ST of *<sup>k</sup>*-starlike functions in <sup>U</sup> (see Definition <sup>1</sup> below). For fixed *k*, Ω*<sup>k</sup>* represents the conic region bounded successively by the imaginary axis (*k* = 0), by a parabola (*k* = 1), by the right branch of a hyperbola (0 < *k* < 1), and by an ellipse (*k* > 1).

For these conic regions, the following functions play the role of extremal functions.

$$\begin{cases} \frac{1+z}{1-z} = 1 + 2z + 2z^2 + \cdots & (k=0) \\\\ 1 + \frac{2}{\pi^2} \left[ \log \left( \frac{1+\sqrt{z}}{1-\sqrt{z}} \right) \right]^2 & (k=1) \\\\ \cdot & \cdot \cdot \cdot \cdot & \cdot \end{cases} \tag{5}$$

$$p\_k(z) = \begin{cases} 1 + \frac{2}{1 - k^2} \sinh^2\left[\left(\frac{2}{\pi} \arccos k\right) \arctan\left(h\sqrt{z}\right) \right] & (0 \le k < 1) \\\\ 1 + \frac{1}{k^2 - 1} \left[1 + \sin\left(\frac{\pi}{2K(\kappa)} \int\_0^{\frac{\mu(z)}{\sqrt{\kappa}}} \frac{dt}{\sqrt{(1 - t^2)(1 - \kappa^2 t^2)}}\right) \right] & (k > 1) \end{cases}$$

$$\left[1+\frac{1}{k^2-1}\left[1+\sin\left(\frac{\pi}{2\overline{\kappa}(\kappa)}\int\_0^{\frac{u(z)}{\sqrt{\kappa}}} \frac{dt}{\sqrt{(1-t^2)(1-\kappa^2t^2)}}\right)\right] \quad (k>1)\right)$$

where

$$
\mu(z) = \frac{z - \sqrt{\kappa}}{1 - \sqrt{\kappa}z} \qquad (z \in \mathbb{U})\ ,
$$

and *κ* ∈ (0, 1) is so chosen that

$$k = \cosh\left(\frac{\pi K'(\kappa)}{4K(\kappa)}\right).$$

Here *K*(*κ*) is Legendre's complete elliptic integral of first kind and

$$K'(\kappa) = K\left(\sqrt{1-\kappa^2}\right),$$

that is, *K* (*κ*) is the complementary integral of *K* (*κ*) (see, for example, ([7], p. 326, Equation 9.4 (209))). Indeed, from (5), we have

*pk*(*z*) = <sup>1</sup> <sup>+</sup> *<sup>p</sup>*1*<sup>z</sup>* <sup>+</sup> *<sup>p</sup>*2*z*<sup>2</sup> <sup>+</sup> *<sup>p</sup>*3*z*<sup>3</sup> <sup>+</sup> ··· . (6)

The class *k*-ST is defined as follows.

**Definition 1.** *A function f* ∈ A *is said to be in the class k-*ST *if and only if*

$$\frac{zf'(z)}{f'(z)} \prec p\_k \left(z\right) \quad \left(\forall \, z \in \mathbb{U}; \ k \ge 0\right).$$

We now recall some basic definitions and concept details of the *q*-calculus which will be used in this paper (see, for example, ([7], p. 346 et seq.)). Throughout the paper, unless otherwise mentioned, we suppose that 0 < *q* < 1 and

$$\mathbb{N} = \{1, 2, 3 \cdots \} = \mathbb{N}^0 \mid \{0\} \qquad \text{( $\mathbb{N}\_0 := \{0, 1, 2, \cdots\}$ ) . ]}$$

**Definition 2.** *Let q* ∈ (0, 1) *and define the q-number* [*λ*]*<sup>q</sup> by*

$$[\lambda]\_q = \begin{cases} \frac{1 - q^{\lambda}}{1 - q} & (\lambda \in \mathbb{C}) \\\\ \sum\_{k=0}^{n-1} q^k = 1 + q + q^2 + \dots + q^{n-1} & (\lambda = n \in \mathbb{N}) \end{cases} \tag{\lambda = n \in \mathbb{N}} \ .$$

**Definition 3.** *Let q* ∈ (0, 1) *and define the q-factorial* [*n*]*q*! *by*

$$[n]\_q! = \begin{cases} 1 & (n=0) \\\\ \prod\_{k=1}^n [k]\_q & (n \in \mathbb{N}) \end{cases}$$

**Definition 4** (see [8,9])**.** *The q-derivative (or q-difference) operator Dq of a function f defined, in a given subset of* C*, by*

$$\left(D\_q f\right)(z) = \begin{cases} \frac{f\left(z\right) - f\left(qz\right)}{\left(1 - q\right)z} & \quad (z \neq 0) \\\\ f'\left(0\right) & \quad (z = 0) \end{cases} \tag{7}$$

*provided that f* (0) *exists.*

From Definition 4, we can observe that

$$\lim\_{q \to 1-} \left( D\_q f \right)(z) = \lim\_{q \to 1-} \frac{f(z) - f(qz)}{(1-q)z} = f'(z)$$

for a differentiable function *f* in a given subset of C. It is also known from (1) and (7) that

$$\left(D\_{\emptyset}f\right)(z) = 1 + \sum\_{n=2}^{\infty} \left[n\right]\_{\emptyset} a\_{\emptyset} z^{n-1}.\tag{8}$$

**Definition 5.** *The q-Pochhammer symbol* [*ξ*]*n*,*<sup>q</sup>* (*<sup>ξ</sup>* <sup>∈</sup> <sup>C</sup>; *<sup>n</sup>* <sup>∈</sup> <sup>N</sup>0) *is defined as follows:*

$$[\xi]\_{n,q} = \frac{\left(q^{\mathbb{Z}};q\right)\_n}{(1-q)^n} = \begin{cases} 1 & (n=0) \\\\ [\xi]\_q \left[\xi+1\right]\_q [\xi+2]\_q \cdots \left[\xi+n-1\right]\_q & (n \in \mathbb{N}) \end{cases}$$

*Moreover, the q-gamma function is defined by the following recurrence relation:*

$$
\Gamma\_q \left( z + 1 \right) = \left[ z \right]\_q \Gamma\_q \left( z \right) \quad \text{and} \quad \Gamma\_q \left( 1 \right) = 1.
$$

**Definition 6** (see [10])**.** *For <sup>f</sup>* ∈ A, *let the q-Ruscheweyh derivative operator* <sup>R</sup>*<sup>λ</sup> <sup>q</sup> be defined*, *in terms of the Hadamard product* (*or convolution*) *given by* (2), *as follows*:

$$\mathcal{R}\_q^{\lambda} f\left(z\right) = f\left(z\right) \* \mathcal{F}\_{q,\lambda+1}\left(z\right) \qquad \left(z \in \mathbb{U}; \lambda > -1\right),$$

*where*

$$\mathcal{F}\_{q,\lambda+1}(z) = z + \sum\_{n=2}^{\infty} \frac{\Gamma\_q\left(\lambda+n\right)}{[n-1]\_q!\Gamma\_q\left(\lambda+1\right)} z^n = z + \sum\_{n=2}^{\infty} \frac{[\lambda+1]\_{q,n-1}}{[n-1]\_q!} z^n.$$

We next define a certain *q*-integral operator by using the same technique as that used by Noor [11].

**Definition 7.** *For f* ∈ A*, let the q-integral operator* F*q*,*<sup>λ</sup> be defined by*

$$
\mathcal{F}\_{q,\lambda+1}^{-1}(z) \ast \mathcal{F}\_{q,\lambda+1}(z) = z \left( D\_q f \right)(z) \dots
$$

*Then*

$$\begin{split} \mathcal{Z}\_{\eta}^{\lambda} f \begin{pmatrix} z \\ \end{pmatrix} &= f \begin{pmatrix} z \\ \end{pmatrix} \* \mathcal{F}\_{\eta,\lambda+1}^{-1} \begin{pmatrix} z \\ \end{pmatrix} \\ &= z + \sum\_{n=2}^{\infty} \psi\_{n-1} a\_n z^n \\ \end{split} \qquad \begin{split} \begin{split} &= \begin{split} &= \begin{split} &= \begin{split} &= \begin{split} &= \begin{split} &= \begin{split} &= \begin{split} &= \begin{split} &= \end{split} \\ &= \begin{split} &= \begin{split} &= \begin{split} &= \begin{split} &= \end{split} \\ \end{split} \end{split} \end{split} \end{split} \end{split} \end{split} \tag{9}$$

*where*

$$\mathcal{F}\_{q,\lambda+1}^{-1}(z) = z + \sum\_{n=2}^{\infty} \psi\_{n-1} z^n$$

*and*

$$\psi\_{n-1} = \frac{[n]\_q!\Gamma\_q\left(\lambda+1\right)}{\Gamma\_q\left(\lambda+n\right)} = \frac{[n]\_q!}{[\lambda+1]\_{q,n-1}}.$$

Clearly, we have

$$
\mathcal{L}\_q^0 f \begin{pmatrix} z \\ \end{pmatrix} = z \begin{pmatrix} D\_q f \\ \end{pmatrix} \begin{pmatrix} z \\ \end{pmatrix} \quad \text{and} \quad \mathcal{L}\_q^1 f \begin{pmatrix} z \\ \end{pmatrix} = f \begin{pmatrix} z \\ \end{pmatrix} \;.
$$

We note also that, in the limit case when *q* → 1−, the *q*-integral operator F*q*,*<sup>λ</sup>* given by Definition 7 would reduce to the integral operator which was studied by Noor [11].

The following identity can be easily verified:

$$zD\_{\eta}\left(\mathcal{Z}\_{\eta}^{\lambda+1}f\left(z\right)\right) = \left(1 + \frac{[\lambda]\_{\eta}}{\eta^{\lambda}}\right)\mathcal{Z}\_{\eta}^{\lambda}f\left(z\right) - \frac{[\lambda]\_{\eta}}{\eta^{\lambda}}\mathcal{Z}\_{\eta}^{\lambda+1}f\left(z\right). \tag{10}$$

When *q* → 1−, this last identity in (10) implies that

$$z\left(\mathcal{I}^{\lambda+1}f\left(z\right)\right)' = \left(1+\lambda\right)\mathcal{I}^{\lambda}f\left(z\right) - \lambda\mathcal{I}^{\lambda+1}f\left(z\right).$$

which is the well-known recurrence relation for the above-mentioned integral operator which was studied by Noor [11].

In geometric function theory, several subclasses belonging to the class of normalized analytic functions class A have already been investigated in different aspects. The above-defined *q*-calculus gives valuable tools that have been extensively used in order to investigate several subclasses of A. Ismail et al. [12] were the first who used the *q*-derivative operator *Dq* to study the *q*-calculus analogous of the class <sup>S</sup><sup>∗</sup> of starlike functions in <sup>U</sup> (see Definition <sup>8</sup> below). However, a firm footing of the *q*-calculus in the context of geometric function theory was presented mainly and basic (or *q*-) hypergeometric functions were first used in geometric function theory in a book chapter by Srivastava (see, for details, ([13], p. 347 et seq.); see also [14]).

**Definition 8** (see [12])**.** *A function f* ∈ A *is said to belong to the class* S<sup>∗</sup> *<sup>q</sup> if*

$$f\left(0\right) = f'\left(0\right) - 1 = 0\tag{11}$$

*and*

$$\left| \frac{z}{f(z)} \left( D\_q f \right) z - \frac{1}{1 - q} \right| \le \frac{1}{1 - q}. \tag{12}$$

It is readily observed that, as *q* → 1−, the closed disk:

$$\left| w - \frac{1}{1-q} \right| \le \frac{1}{1-q}$$

becomes the right-half plane and the class S<sup>∗</sup> *<sup>q</sup>* of *q*-starlike functions reduces to the familiar class S<sup>∗</sup> of normalized starlike functions in U with respect to the origin (*z* = 0). Equivalently, by using the principle of subordination between analytic functions, we can rewrite the conditions in (11) and (12) as follows (see [15]):

$$\frac{z}{f\left(z\right)}\left(D\_{\mathfrak{q}}f\right)\left(z\right) \prec \widehat{p}\left(z\right) \qquad\qquad \left(\widehat{p}\left(z\right) = \frac{1+z}{1-qz}\right).\tag{13}$$

The notation S<sup>∗</sup> *<sup>q</sup>* was used by Sahoo and Sharma [16].

Now, making use of the principle of subordination between analytic functions and the above-mentioned *q*-calculus, we present the following definition.

**Definition 9.** *A function p is said to be in the class k-*P*<sup>q</sup> if and only if*

$$p\left(z\right) \prec \frac{2p\_k\left(z\right)}{\left(1+q\right) + \left(1-q\right)p\_k\left(z\right)}\gamma$$

*where pk* (*z*) *is defined by* (5)*.*

Geometrically, the function *p* (*z*) ∈ *k*-P*<sup>q</sup>* takes on all values from the domain Ω*k*,*<sup>q</sup>* (*k* 0) which is defined as follows:

$$\Omega\_{k,q} = \left\{ w : \Re \left( \frac{(1+q)\,w}{(q-1)\,w+2} \right) > k \left| \frac{(1+q)\,w}{(q-1)\,w+2} - 1 \right| \right\}.$$

The domain Ω*k*,*<sup>q</sup>* represents a generalized conic region. It can be seen that

$$\lim\_{\eta \to 1-} \Omega\_{k,\eta} = \Omega\_{k,\eta}$$

where Ω*<sup>k</sup>* is the conic domain considered by Kanas and Wi´sniowska [3]. Below, we give some basic facts about the class *k*-P*q*.

**Remark 1.** *First of all, we see that*

$$k \cdot \mathcal{P}\_q \subseteq \mathcal{P} \left[ \frac{2k}{2k+1+q} \right] \text{ .}$$

*where* P <sup>2</sup>*<sup>k</sup>* <sup>2</sup>*k*+1+*<sup>q</sup> is the well-known class of functions with real part greater than* <sup>2</sup>*<sup>k</sup>* <sup>2</sup>*k*+1+*<sup>q</sup>* . *Secondly, we have*

$$\lim\_{q \to 1-} k \text{-} \mathcal{P}\_q = \mathcal{P} \left( p\_k \right) \text{-} k$$

*where* P (*pk*) *is the well-known function class introduced by Kanas and Wi´sniowska [3]. Thirdly, we have*

$$\lim\_{q \to 1-} 0 \text{-} \mathcal{P}\_q = \mathcal{P}\_r$$

*where* P *is the well-known class of analytic functions with positive real part.*

**Definition 10.** *A function f is said to be in the class* ST (*k*, *λ*, *q*) *if and only if*

$$\frac{z\left(D\_{\mathfrak{q}}\mathcal{Z}\_{\mathfrak{q}}^{\lambda}f\right)(z)}{f\left(z\right)} \in k\cdot\mathcal{P}\_{\mathfrak{q}} \qquad \qquad \left(k \ge 0; \ \lambda \ge 0\right),$$

*or, equivalently,*

$$\left| \Re \left( \frac{(1+q)\frac{z\left(D\_qT\_q^\lambda f\right)(z)}{f(z)}}{(q-1)\frac{z\left(D\_qT\_q^\lambda f\right)(z)}{f(z)}+2} \right) > k \left| \frac{(1+q)\frac{z\left(D\_qT\_q^\lambda f\right)(z)}{f(z)}}{(q-1)\frac{z\left(D\_qT\_q^\lambda f\right)(z)}{f(z)}+2} - 1 \right| \right| $$

.

**Remark 2.** *First of all, it is easily seen that*

$$\mathcal{ST}\left(0,1,q\right) = \mathcal{S}\_{\emptyset}^\*.$$

*where* S<sup>∗</sup> *<sup>q</sup> is the function class introduced and studied by Ismail et al. [12]. Secondly, we have*

$$\lim\_{q \to 1-} \mathcal{ST}\left(k, 1, q\right) = k \text{-} \mathcal{ST}\_{\prime\prime}$$

*where k-*ST *is a function class introduced and studied by Kanas and Wi´sniowska [4]. Finally, we have*

$$\lim\_{q \to 1-} \mathcal{ST}\left(0, 1, q\right) = \mathcal{S}^\*\,,$$

*where* <sup>S</sup><sup>∗</sup> *is the well-known class of starlike functions in* <sup>U</sup> *with respect to the origin* (*<sup>z</sup>* <sup>=</sup> <sup>0</sup>)*.*

**Remark 3.** *Further studies of the new q-starlike function class* ST (*k*, *λ*, *q*), *as well as of its more consequences*, *can next be determined and investigated in future papers.*

Let *<sup>n</sup>* <sup>∈</sup> <sup>N</sup><sup>0</sup> and *<sup>j</sup>* <sup>∈</sup> <sup>N</sup>. The following *<sup>j</sup>*th Hankel determinant was considered by Noonan and Thomas [17]:


,

where *a*<sup>1</sup> = 1. In fact, this determinant has been studied by several authors, and sharp upper bounds on H<sup>2</sup> (2) were obtained by several authors (see [18–20]) for various classes of functions. It is well-known that the Fekete–Szegö functional *a*<sup>3</sup> <sup>−</sup> *<sup>a</sup>*<sup>2</sup> 2 can be represented in terms of the Hankel determinant as <sup>H</sup><sup>2</sup> (1). This functional has been further generalized as *a*<sup>3</sup> <sup>−</sup> *<sup>μ</sup>a*<sup>2</sup> 2 for some real or complex *μ*. Fekete and Szegö gave sharp estimates of *a*<sup>3</sup> <sup>−</sup> *<sup>μ</sup>a*<sup>2</sup> 2 for *μ* real and *f* ∈ S, the class of normalized univalent functions in U. It is also known that the functional *a*2*a*<sup>4</sup> <sup>−</sup> *<sup>a</sup>*<sup>2</sup> 3 is equivalent to H<sup>2</sup> (2) (see [18]). Babalola [21] studied the Hankel determinant H<sup>3</sup> (1) for some subclasses of normalized analytic functions in <sup>U</sup>. The symmetric Toeplitz determinant <sup>T</sup>*<sup>j</sup>* (*n*) is defined by

$$\mathcal{T}\_{\vec{l}}(n) = \begin{vmatrix} a\_{n} & a\_{n+1} & \cdot & \cdot & \cdot & a\_{n+j-1} \\ a\_{n+1} & \cdot & & \cdot & & \cdot \\ \cdot & \cdot & & & \cdot & \\ \cdot & \cdot & & & \cdot & \\ \cdot & \cdot & \cdot & & \cdot & \\ a\_{n+j-1} & \cdot & \cdot & \cdot & \cdot & a\_{n} \end{vmatrix} \prime$$

so that

T<sup>2</sup> (2) = *a*<sup>2</sup> *a*<sup>3</sup> *a*<sup>3</sup> *a*<sup>2</sup> , T<sup>2</sup> (3) = *a*<sup>3</sup> *a*<sup>4</sup> *a*<sup>4</sup> *a*<sup>3</sup> , T<sup>3</sup> (2) = *a*<sup>2</sup> *a*<sup>3</sup> *a*<sup>4</sup> *a*<sup>3</sup> *a*<sup>2</sup> *a*<sup>3</sup> *a*<sup>4</sup> *a*<sup>3</sup> *a*<sup>2</sup> ,

and so on.

For *f* ∈ S, the problem of finding the best possible bounds for ||*an*+1| − |*an*|| has a long history (see, for details, [22]). It is a known fact from [22] that

$$\left| \left| \left| a\_{n+1} \right| - \left| a\_{n} \right| \right| \right| < \infty$$

for a constant *c*. However, the problem of finding exact values of the constant *c* for S and its various subclasses has proved to be difficult. In a very recent investigation, Thomas and Abdul-Halim [23] succeeded in obtaining some sharp estimates for T*<sup>j</sup>* (*n*) for the first few values of *n* and *j* involving symmetric Toeplitz determinants whose entries are the coefficients *an* of starlike and close-toconvex functions.

In the present investigation, our focus is on the Hankel determinant and the Toeplitz matrices for the function class ST (*k*, *λ*, *q*) given by Definition 10.

#### **2. A Set of Lemmas**

In order to prove our main results in this paper, we need each of the following lemmas.

**Lemma 1** (see [20])**.** *If the function p* (*z*) *given by* (3) *is in the Carathéodory class* P *of analytic functions with positive real part in* U, *then*

$$2c\_2 = c\_1^2 + x\left(4 - c\_1^2\right)$$

*and*

$$4c\_3 = c\_1^3 + 2\left(4 - c\_1^2\right)c\_1x - c\_1\left(4 - c\_1^2\right)x^2 + 2\left(4 - c\_1^2\right)\left(1 - \left|x^2\right|\right)z^2$$

*for some x*, *<sup>z</sup>* <sup>∈</sup> <sup>C</sup> *with* <sup>|</sup>*x*<sup>|</sup> - 1 *and* |*z*| -1.

**Lemma 2** (see [24])**.** *Let the function p*(*z*) *given by* (3) *be in the Carathéodory class* P *of analytic functions with positive real part in* <sup>U</sup>*. Also let <sup>μ</sup>* <sup>∈</sup> <sup>C</sup>*. Then*

$$|\mathcal{c}\_n - \mu \mathcal{c}\_k \mathcal{c}\_{n-k}| \stackrel{<}{=} 2 \max\left(1, |2\mu - 1|\right) \qquad (1 \le k \le n - 1) \dots$$

**Lemma 3** (see [22])**.** *Let the function p*(*z*) *given by* (3) *be in the Carathéodory class* P *of analytic functions with positive real part in* U*. Then*

$$|c\_n| \le 2 \qquad (n \in \mathbb{N})\dots$$

*This last inequality is sharp.*

#### **3. Main Results**

Throughout this section, unless otherwise mentioned, we suppose that

$$q \in (0,1), \ \lambda > -1 \quad \text{and} \quad k \in [0,1] \dots$$

**Theorem 1.** *If the function f* (*z*) *given by* (1) *belongs to the class* ST (*k*, *λ*, *q*), *where k* ∈ [0, 1] , *then*

$$|a\_2| \le \frac{(1+q)}{2q\psi\_1},$$

$$a\_3 \le \frac{1}{2q\psi\_2} \left(p\_1 + \left|p\_2 - p\_1 + \frac{(q^2+1)}{2q}\right|^2\right).$$

*and*

$$\begin{split} a\_4 &\le \frac{(1+q)}{4\left(q+q^2+q^3\right)} \left(2p\_1+4\left|p\_2-p\_1+\frac{\left(2+q^2\right)p\_1^2}{4q}\right|\right. \\ &\left.+ \left|2p\_3+2p\_1-4p\_2-\frac{\left(2\left(1+q^2\right)-q\right)p\_1^2}{q}+\frac{\left(4q^2-3q+2\right)}{q}p\_1p\_2\right|\right. \\ &\left.+ \frac{\left(q^2+2q-1\right)}{2q^2}|p\_1^3|\right), \end{split} \tag{14}$$

*where pj* (*j* = 1, 2, 3) *are positive and are the coefficients of the functions pk* (*z*) *defined by* (6). *Each of the above results is sharp for the function g* (*z*) *given by*

$$\lg(z) = \frac{2p\_k(z)}{(1+q) + (1-q)\ p\_k(z)}.$$

**Proof.** Let *f* (*z*) ∈ ST (*k*, *λ*, *q*). Then, we have

$$\frac{z\left(D\_{\mathfrak{q}}f\right)(z)}{f\left(z\right)} = \mathfrak{q}\left(z\right) \prec \mathcal{S}\_{k}\left(z\right),\tag{15}$$

where

$$S\_k\left(z\right) = \frac{2p\_k\left(z\right)}{\left(1+q\right) + \left(1-q\right)p\_k\left(z\right)},$$

and the functions *pk* (*z*) are defined by (6).

We now define the function *p* (*z*) with *p* (0) = 1 and with a positive real part in U as follows:

$$p\left(z\right) = \frac{1 + S\_k^{-1}\left(\mathfrak{q}\left(z\right)\right)}{1 - S\_k^{-1}\left(\mathfrak{q}\left(z\right)\right)} = 1 + c\_1 z + c\_2 z^2 + \cdots \ . \tag{16}$$

After some simple computation involving (16), we get

$$\mathfrak{q}\left(z\right) = S\_k \left( \frac{p\left(z\right) + 1}{p\left(z\right) - 1} \right).$$

We thus find that

$$\begin{split} S\_{k} &\left( \frac{p\left(z\right) + 1}{p\left(z\right) - 1} \right) \\ &= 1 + \left( \frac{q + 1}{2} \right) \left[ \frac{p\_{1}c\_{1}}{2}z + \left\{ \frac{p\_{1}c\_{2}}{2} + \left( \frac{p\_{2}}{4} - \frac{p\_{1}}{4} + \left( \frac{(q - 1)\,p\_{1}^{2}}{8} \right) \right) c\_{1}^{2} \right\} z^{2} \\ &\quad + \left\{ \frac{p\_{1}c\_{3}}{2} + \left( \frac{p\_{2}}{2} - \frac{p\_{1}}{2} + \left( \frac{(q - 1)\,p\_{1}^{2}}{4} \right) \right) c\_{1}c\_{2} \\ &\quad + \left( \frac{p\_{1}}{8} - \frac{p\_{2}}{4} - \frac{(q - 1)\,p\_{1}^{2}}{8} + \frac{p\_{3}}{8} - \frac{(q - 1)\,p\_{1}p\_{2}}{8} + \frac{(q - 1)^{2}}{32} \overline{p\_{1}^{3}} \right) c\_{1}^{3} \right] z^{3} \right] + \cdots \cdots \tag{17} \end{split}$$

Now, upon expanding the left-hand side of (15), we have

$$\begin{split} \frac{dz \left(D\_{q} \mathcal{Z}\_{q}^{\lambda} f\right)(z)}{f(z)} &= 1 + q \psi\_{1} a\_{2} z + \left\{ \left(q + q^{2}\right) \psi\_{2} a\_{3} - q \psi\_{1}^{2} a\_{2}^{2} \right\} z^{2} \\ &+ \left\{ \left(q + q^{2} + q^{3}\right) \psi\_{3} a\_{4} - \left(2q + q^{2}\right) \psi\_{1} \psi\_{2} a\_{2} a\_{3} + q \psi\_{1}^{3} a\_{2}^{3} \right\} z^{3} + \dotsb \end{split} \tag{18}$$

Finally, by comparing the corresponding coefficients in (17) and (18) along with Lemma 3, we obtain the result asserted by Theorem 1.

**Theorem 2.** *If the function f* (*z*) *given by* (1) *belongs to the class* ST (*k*, *λ*, *q*), *then*

$$\begin{split} \mathcal{T}\_{3}(2) &\leq \left[ \left( \frac{1+q}{2q\psi\_{1}} \right) p\_{1}^{2} + \left( \frac{1+q}{4\left(q+q^{2}+q^{3}\right)\psi\_{3}} \right) \left[\Omega\_{1} + \Omega\_{2} \right] \right] \\ &\cdot \left[ 4\left( \frac{(1+q)^{2}}{16q^{2}\psi\_{1}^{2}} \right) p\_{1}^{2} + 16\left|\Omega\_{3}\right| + \frac{p\_{1}^{2}}{4q^{2}\psi\_{2}^{2}} + 2\Omega\_{5} p\_{1}^{2} \left| 2 - \frac{\Omega\_{4}}{\Omega\_{5}p\_{1}^{2}} \right| \right], \end{split}$$

*where*

$$\begin{aligned} \Omega\_1 &= 2p\_1 + 4\left|p\_2 - p\_1 + \frac{(2+q^2)}{4q}p\_1^2\right|, \\ \Omega\_2 &= \left|2p\_3 + 2p\_1 - 4p\_2 - \left(2\left(1+q^2\right) - q\right)p\_1^2\right| \\ &+ \left(\frac{4q^2 - 3q + 2}{q}\right)p\_1p\_2 + \left(\frac{q^2 + q + 1}{2q^2}p\_1^3\right)\right|, \\ \Omega\_3 &= \frac{1}{2q^2\psi\_2^2} \left(\frac{p\_2}{4} - \frac{p\_1}{4} + \frac{\left(q^2 + 1\right)p\_1^2}{8q}\right)^2 - \Omega\_5 \cdot \left[\frac{p\_3}{4} + \frac{p\_1}{4} - \frac{p\_2}{2}\right.\end{aligned}$$

$$\begin{aligned} &-\frac{\left[2\left(1+q^2\right) - q\right]}{8q}p\_1^2 + \frac{4q^2 - 3q + 2}{8q}p\_1p\_2 + \left(\frac{q^2 + 2q - 1}{16q^2}\right)p\_1^3\right], \\ \Omega\_4 &= \frac{p\_1}{2q^2\psi\_2^2}\left(\frac{p\_2}{4} - \frac{p\_1}{4} + \frac{\left(q^2 + 1\right)p\_1^2}{8q}\right) - \Omega\_5 p\_1\left(p\_2 - p\_1 + \frac{\left(2 + q^2\right)p\_1^2}{4q}\right), \\ \Omega\_5 &= \frac{(1+q)^2}{16q^2\left(1+q + q^2\right)}\Psi\_{10}\eta\_3 \end{aligned}$$

*and pj* (*j* = 1, 2) *are positive and are the coefficients of the functions pk* (*z*) *defined by* (6)*.*

**Proof.** Upon comparing the corresponding coefficients in (17) and (18), we find that

$$a\_2 = \frac{(1+q)\,p\_1c\_1}{4q\psi\_1},$$

$$a\_3 = \frac{1}{2q\psi\_2} \left[ \frac{p\_1 c\_2}{2} + \left( \frac{p\_2}{4} - \frac{p\_1}{4} + \frac{\left(q^2 + 1\right)}{8q} \right) c\_1^2 \right],\tag{20}$$

$$\begin{split} a\_{4} &= \frac{(1+q)}{4\left(q+q^{2}+q^{3}\right)\psi\_{3}} \left[ p\_{1}c\_{3} + \left(p\_{2}-p\_{1} + \frac{(2+q^{2})}{4q}\right)c\_{1}c\_{2} \right. \\ &\left. + \left(\frac{p\_{3}}{4} + \frac{p\_{1}}{4} - \frac{p\_{2}}{2} - \frac{\left(2\left(1+q^{2}\right)-q\right)p\_{1}^{2}}{8q} + \frac{\left(4q^{2}-3q+2\right)}{8q}p\_{1}p\_{2} \right. \\ &\left. + \frac{\left(q^{2}+2q-1\right)}{16q^{2}}p\_{1}^{3}\right]c\_{1}^{3} \right]. \end{split} \tag{21}$$

By a simple computation, T<sup>3</sup> (2) can be written as follows:

$$\mathcal{T}\_3(2) = (a\_2 - a\_4) \left( a\_2^2 - 2a\_3^2 + a\_2 a\_4 \right).$$

Now, if *f* ∈ ST (*k*, *λ*, *q*), then it is clearly seen that

$$\begin{aligned} \left| a\_2 - a\_4 \right| &\le \left| a\_2 \right| + \left| a\_4 \right| \\ &\le \left( \frac{1+q}{2q\psi\_1} \right) p\_1^2 + \left( \frac{1+q}{4\left( q + q^2 + q^3 \right) \psi\_3} \right) \left( \Omega\_1 + \Omega\_2 \right) .\end{aligned}$$

We need to maximize *a*2 <sup>2</sup> <sup>−</sup> <sup>2</sup>*a*<sup>2</sup> <sup>3</sup> + *a*2*a*<sup>4</sup> for a function *f* ∈ ST (*k*, *λ*, *q*). So, by writing *a*2, *a*3, and *a*<sup>4</sup> in terms of *c*1, *c*2, and *c*3, with the help of (19)–(21), we get

$$\begin{aligned} & \left| a\_2^2 - 2a\_3^2 + a\_2 a\_4 \right| \\ &= \left| \left( \frac{(1+q)^2}{16q^2 \psi\_1^2} \right) p\_1^2 c\_1^2 - \Omega\_3 c\_1^4 - \Omega\_4 c\_1^2 c\_2 - \frac{p\_1^2}{8q^2 \psi\_2^2} c\_2^2 + \Omega\_5 p\_1^2 c\_1 c\_3 \right|. \end{aligned} \tag{22}$$

Finally, by applying the trigonometric inequalities, Lemmas 2 and 3 along with (22), we obtain the result asserted by Theorem 2.

As an application of Theorem 2, we first set *ψn*−<sup>1</sup> = 1 and *k* = 0 and then let *q* → 1− . We thus arrive at the following known result.

**Corollary 1** (see [25])**.** *If the function f* (*z*) *given by* (1) *belongs to the class* S∗, *then*

$$\mathcal{T}\_3\left(2\right) \le 84.$$

**Theorem 3.** *If the function f* (*z*) *given by* (1) *belongs to the class* ST (*k*, *λ*, *q*), *then*

$$\left| a\_2 a\_4 - a\_3^2 \right| \le \frac{1}{4\eta^2 \psi\_2^2} \left| p\_{1'}^2 \right. \tag{23}$$

*where k* ∈ [0, 1] *and pj* (*j* = 1, 2, 3) *are positive and are the coefficients of the functions pk* (*z*) *defined by* (6)*.*

**Proof.** Making use of (19)–(21), we find that

*<sup>a</sup>*2*a*<sup>4</sup> <sup>−</sup> *<sup>a</sup>*<sup>2</sup> <sup>3</sup> <sup>=</sup> *<sup>A</sup>* (*q*) 16*q*2*ψ*1*ψ*<sup>3</sup> *p*2 <sup>1</sup>*c*1*c*<sup>3</sup> + *A* (*q*) *ψ*<sup>2</sup> <sup>2</sup> − *ψ*1*ψ*<sup>3</sup> 16*q*2*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *<sup>p</sup>*<sup>1</sup> *<sup>p</sup>*<sup>2</sup> <sup>−</sup> *<sup>A</sup>* (*q*) *<sup>ψ</sup>*<sup>2</sup> <sup>2</sup> − *ψ*1*ψ*<sup>3</sup> 16*q*2*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*2 1 + *A* (*q*) 2 + *q*<sup>2</sup> *ψ*2 <sup>2</sup> − 2 1 + *q*<sup>2</sup> *ψ*1*ψ*<sup>3</sup> 64*q*2*ψ*1*ψ*<sup>3</sup> *p*3 1 *c*2 <sup>1</sup>*c*<sup>2</sup> + 1 16*q*2*ψ*<sup>2</sup> 2 *p*2 1*c*2 2 + *A* (*q*) 64*q*2*ψ*1*ψ*<sup>3</sup> *p*<sup>1</sup> *p*<sup>3</sup> + *A* (*q*) *ψ*<sup>2</sup> <sup>2</sup> − *ψ*1*ψ*<sup>3</sup> 64*q*2*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*2 <sup>1</sup> + *<sup>ψ</sup>*1*ψ*<sup>3</sup> <sup>−</sup> *<sup>A</sup>* (*q*) *<sup>ψ</sup>*<sup>2</sup> 2 32*q*2*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*<sup>1</sup> *p*<sup>2</sup> + 2 1 + *q*<sup>2</sup> *<sup>ψ</sup>*1*ψ*<sup>3</sup> <sup>−</sup> 2 1 + *q*<sup>2</sup> − *q A* (*q*) *ψ*<sup>2</sup> 2 128*q*3*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*3 1 + *<sup>A</sup>* (*q*) <sup>4</sup>*q*<sup>2</sup> <sup>−</sup> <sup>3</sup>*<sup>q</sup>* <sup>+</sup> <sup>2</sup> *ψ*2 <sup>2</sup> − 2 1 + *q*<sup>2</sup> *ψ*1*ψ*<sup>3</sup> 128*q*3*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*2 <sup>1</sup> *p*<sup>2</sup> + *<sup>A</sup>* (*q*) *<sup>q</sup>*<sup>2</sup> <sup>+</sup> <sup>2</sup>*<sup>q</sup>* <sup>−</sup> <sup>1</sup> *ψ*2 <sup>2</sup> <sup>−</sup> 1 + *q*<sup>2</sup> <sup>2</sup> *<sup>ψ</sup>*1*ψ*<sup>3</sup> 256*q*4*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*4 <sup>1</sup> <sup>−</sup> <sup>1</sup> 64*q*2*ψ*<sup>2</sup> 2 *p*2 2 % *c*4 1, (24)

where

$$A\left(q\right) = \frac{\left(1+q\right)^2}{1+q+q^2}.$$

We substitute the values of *<sup>c</sup>*<sup>2</sup> and *<sup>c</sup>*<sup>3</sup> from the above Lemma and, for simplicity, take *<sup>Y</sup>* <sup>=</sup> <sup>4</sup> <sup>−</sup> *<sup>c</sup>*<sup>2</sup> 1 and *Z* = (1 − |*x*| <sup>2</sup>)*z*. Without loss of generality, we assume that *c* = *c*<sup>1</sup> (0 *c* -2), so that

*<sup>a</sup>*2*a*<sup>4</sup> <sup>−</sup> *<sup>a</sup>*<sup>2</sup> <sup>3</sup> = *<sup>q</sup>* (<sup>1</sup> <sup>−</sup> *<sup>q</sup>*) *<sup>A</sup>* (*q*) *<sup>ψ</sup>*<sup>2</sup> 2 128*q*2*ψ*1*ψ*<sup>3</sup> *p*3 <sup>1</sup> + *A* (*q*) 64*q*2*ψ*1*ψ*<sup>3</sup> *p*<sup>1</sup> *p*<sup>3</sup> + *<sup>A</sup>* (*q*) <sup>4</sup>*q*<sup>2</sup> <sup>−</sup> <sup>3</sup>*<sup>q</sup>* <sup>+</sup> <sup>2</sup> *ψ*2 <sup>2</sup> − 2 1 + *q*<sup>2</sup> *ψ*1*ψ*<sup>3</sup> 128*q*3*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*2 <sup>1</sup> *p*<sup>2</sup> + *<sup>A</sup>* (*q*) *<sup>q</sup>*<sup>2</sup> <sup>+</sup> <sup>2</sup>*<sup>q</sup>* <sup>−</sup> <sup>1</sup> *ψ*2 <sup>2</sup> <sup>−</sup> 1 + *q*<sup>2</sup> <sup>2</sup> *<sup>ψ</sup>*1*ψ*<sup>3</sup> 256*q*4*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*4 <sup>1</sup> <sup>−</sup> <sup>1</sup> 64*q*2*ψ*<sup>2</sup> 2 *p*2 2 % *c*4 + *A* (*q*) *ψ*<sup>2</sup> <sup>2</sup> − *ψ*1*ψ*<sup>3</sup> 32*q*2*ψ*1*ψ*<sup>2</sup> <sup>2</sup>*ψ*<sup>3</sup> *p*<sup>1</sup> *p*<sup>2</sup> + *A* (*q*) 2 + *q*<sup>2</sup> *ψ*2 <sup>2</sup> − 2 1 + *q*<sup>2</sup> *ψ*1*ψ*<sup>3</sup> 128*q*2*ψ*1*ψ*<sup>3</sup> *p*3 1 % *c*2*xY* · <sup>−</sup> *<sup>A</sup>* (*q*) 64*q*2*ψ*1*ψ*<sup>3</sup> *p*2 <sup>1</sup>*c*2*Yx*<sup>2</sup> <sup>−</sup> <sup>1</sup> 64*q*2*ψ*<sup>2</sup> 2 *p*2 <sup>1</sup>*x*2*Y*<sup>2</sup> + *A* (*q*) 32*q*2*ψ*1*ψ*<sup>3</sup> *p*2 1*cYZ*% . (25)

Upon setting *Z* = (1 − |*x*| <sup>2</sup>)*z* and taking the moduli in (25) and using trigonometric inequality, we find that

$$\begin{split} \left| a\_2 a\_4 - a\_3^2 \right| &\le \left| \lambda\_1 \right| c^4 + \left| \lambda\_2 \right| \left| \mathbf{x} \right| \mathbf{y}^2 c^2 + \frac{A \left( q \right)}{64 q^2 \psi\_1 \psi\_3} \left| p\_1^2 \mathbf{y} \right| \mathbf{x} \big|^2 c^2 \\ &+ \frac{1}{64 q^2 \psi\_2^2} \left| p\_1^2 \left| \mathbf{x} \right|^2 \mathbf{y}^2 + \frac{A \left( q \right)}{32 q^2 \psi\_1 \psi\_3} p\_1^2 c^2 \mathbf{y} \left( 1 - \left| \mathbf{x} \right|^2 \right) \\ &= \Lambda \left( c, \left| \mathbf{x} \right| \right), \end{split} \tag{26}$$

where

$$\begin{split} \lambda\_{1} &= \frac{q\left(1-q\right)A\left(q\right)\psi\_{2}^{2}}{128q^{2}\psi\_{1}\psi\_{3}}p\_{1}^{3} + \frac{A\left(q\right)}{64q^{2}\psi\_{1}\psi\_{3}}p\_{1}p\_{3} \\ &+ \left(\frac{A\left(q\right)\left(4q^{2}-3q+2\right)\psi\_{2}^{2}-2\left(1+q^{2}\right)\psi\_{1}\psi\_{3}}{128q^{3}\psi\_{1}\psi\_{2}^{2}\psi\_{3}}\right)p\_{1}^{2}p\_{2} \\ &+ \left(\frac{A\left(q\right)\left(q^{2}+2q-1\right)\psi\_{2}^{2}-\left(1+q^{2}\right)^{2}\psi\_{1}\psi\_{3}}{256q^{4}\psi\_{1}\psi\_{2}^{2}\psi\_{3}}\right)p\_{1}^{4} - \frac{1}{64q^{2}\psi\_{2}^{2}}p\_{2}^{2} \\ \lambda\_{2} &= \frac{A\left(q\right)\psi\_{2}^{2}-\psi\_{1}\psi\_{3}}{32q^{2}\psi\_{1}\psi\_{2}^{2}\psi\_{3}}p\_{1}p\_{2} + \frac{A\left(q\right)\left(2+q^{2}\right)\psi\_{2}^{2}-2\left(1+q^{2}\right)\psi\_{1}\psi\_{3}}{128q^{2}\psi\_{1}\psi\_{3}}p\_{1}^{3}. \end{split}$$

Now, trivially, we have

Λ (|*x*|) > 0

on [0, 1], and so

$$
\Lambda\left(|x|\right) \le \Lambda\left(1\right).
$$

Hence, by puting *<sup>Y</sup>* <sup>=</sup> <sup>4</sup> <sup>−</sup> *<sup>c</sup>*<sup>2</sup> <sup>1</sup> and after some simplification, we have

$$\begin{split} \left| a\_2 a\_4 - a\_3^2 \right| &= \left( |\lambda\_1| - |\lambda\_2| + \frac{\psi\_1 \psi\_3 - A \left( q \right) \psi\_2^2}{64 q^2 \psi\_1 \psi\_3} p\_1^2 \right) c^4 \\ &+ \left( 4 \left| \lambda\_2 \right| + \left( \frac{A \left( q \right) \psi\_2^2 - \psi\_1 \psi\_3}{16 q^2 \psi\_1 \psi\_3} p\_1^2 \right) \right) c^2 + \frac{1}{4 q^2 \psi\_2^2} p\_1^2 \\ &= G \left( c \right) . \end{split} \tag{27}$$

For optimum value of *G* (*c*), we consider *G* (*c*) = 0, which implies that *c* = 0. So *G* (*c*) has a maximum value at *c* = 0. We therefore conclude that the maximum value of *G* (*c*) is given by

$$\frac{1}{4q^2\psi\_2^2}p\_{1\prime}^2$$

which occurs at *c* = 0 or

$$\mathcal{L}^2 = -\frac{128\left|\lambda\_2\right|q^2\psi\_1\psi\_3 + 4A\left(q\right)\psi\_2^2 - 2\psi\_1\psi\_3p\_1^2}{\left(64q^2\left(\left|\lambda\_1\right| - \left|\lambda\_2\right|\right)\Psi\_1\psi\_3 + \psi\_1\psi\_3 - A\left(q\right)\psi\_2^2p\_1^2\right)}.$$

This completes the proof of Theorem 3.

If we put *ψn*−<sup>1</sup> = 1 and let *q* → 1− in Theorem 3, we have the following known result.

**Corollary 2** (see [26])**.** *If the function f* (*z*) *given by* (1) *belongs to the class k-*ST , *where k* ∈ [0, 1] , *then*

$$\left| a\_2 a\_4 - a\_3^2 \right| \le \frac{p\_1^2}{4}.$$

If we put

$$p\_1 = 2 \quad \text{and} \quad \psi\_{n-1} = 1,$$

by letting *q* → 1− in Theorem 3, we have the following known result.

**Corollary 3** (see [18])**.** *If f* ∈ S∗, *then*

$$\left| a\_2 a\_4 - a\_3^2 \right| \le 1.$$

By letting *k* = 1, *ψn*−<sup>1</sup> = 1, *q* → 1− and

$$p\_1 = \frac{8}{\pi^2}, \quad p\_2 = \frac{16}{3\pi^2} \quad \text{and} \quad p\_3 = \frac{184}{45\pi^2}$$

in Theorem 3, we have the following known result.

**Corollary 4** (see [27])**.** *If the function f* (*z*) *given by* (1) *belong to the class* SP, *then*

$$\left| a\_2 a\_4 - a\_3^2 \right| \le \frac{16}{\pi^4}.$$
