Hankel Determinants and Coefficient Estimates for Starlike Functions Related to Symmetric Booth Lemniscate

Abstract

In this paper, we find Hankel determinants and coefficient bounds for a subclass of starlike functions related to Booth lemniscate. In particular, we obtain the first four sharp coefficient bounds, Hankel determinants of order two and three, and Zalcman conjecture for this class of functions.

1. Introduction

Let $\mathcal{A}$ denote the class of functions f of the form

$$f\left(\zeta \right)=\zeta +\sum _{n=2}^{\infty }{b}_n{\zeta }^n,$$

(1)

which are analytic in the open unit disk $\mathbb{E}=\left\{\zeta \in \mathbb{C}:\left|\zeta \right|<1\right\}.$ Let $\mathcal{S}$ be a subclass of $\mathcal{A}$, which contains univalent functions in $\mathbb{E}$. A function f is in class ${\mathcal{S}}^{\ast }$ of starlike functions if it satisfies $Re\left(\zeta f^{\prime }\left(\zeta \right)/f\left(\zeta \right)\right)>0$ in $\mathbb{E}$. Denote by $\mathcal{P}$, the class of functions h of the form

$$h\left(\zeta \right)=1+\sum _{n=1}^{\infty }{t}_n{\zeta }^n$$

(2)

satisfying $Re\left\{h\left(\zeta \right)\right\}>0$ in $\mathbb{E}$. A function f is said to be subordinate to a function g written as $f\prec g$, if there is a Schwarz function w with $w\left(0\right)=0$ and $\left|w\left(\zeta \right)\right|<1$ such that $f\left(\zeta \right)=g\left(w\left(\zeta \right)\right)$. In particular, if g is univalent in $\mathbb{E}$ and $f\left(0\right)=g\left(0\right)$, then $f\left(\mathbb{E}\right)\subset g\left(\mathbb{E}\right)$.

Ma and Minda [1] gave a unified presentation of various subclasses of starlike and convex functions by using subordination, where they introduced the classes

$${\mathcal{S}}^{\ast }\left(\psi \right):=\left\{f\in \mathcal{A}:\zeta f^{\prime }\left(\zeta \right)/f\left(\zeta \right)\prec \psi \left(\zeta \right)\right\}$$

and

$$\mathcal{C}\left(\psi \right):=\left\{f\in \mathcal{A}:1+\zeta f^{″}\left(\zeta \right)/f^{\prime }\left(\zeta \right)\prec \psi \left(\zeta \right)\right\}$$

of starlike and convex functions, respectively. Here, the function $\psi$ is analytic and univalent in $\mathbb{E}$, such that $\psi \left(\mathbb{E}\right)$ is convex with $\psi \left(0\right)=1$ and $Re\left\{{\psi }^{\prime }\left(\zeta \right)\right\}>0,$$z\in \mathbb{E}.$ For particular choices of function $\psi ,$ we obtain several classes of analytic and univalent functions.

Several authors have studied the subclasses of ${\mathcal{S}}^{\ast }$ of starlike and $\mathcal{C}$ of convex functions by choosing particular function $\psi$. The classes ${\mathcal{S}}^{\ast }[A,B]:={\mathcal{S}}^{\ast }((1+A\zeta )/(1+B\zeta ))$ and $\mathcal{C}[A,B]:=\mathcal{C}\left(\right(1+A\zeta )/(1+B\zeta \left)\right)$ for $-1\le B<A\le 1$, denote the classes of Janowski starlike and convex functions [2], respectively. Further, the classes of starlike and convex functions of order $\alpha \in \left[0,1\right)$ are defined by ${\mathcal{S}}^{\ast }\left(\alpha \right):={\mathcal{S}}^{\ast }[1-2\alpha ,-1]$ and $\mathcal{C}\left(\alpha \right):=\mathcal{C}[1-2\alpha ,-1]$, respectively. By choosing $\alpha =0,$ we obtain the well-known classes of starlike and convex functions, which are represented as ${\mathcal{S}}^{\ast }$ and $\mathcal{C}$, respectively. The class of strongly starlike functions of order $\beta \in \left(0,1\right]$ is given as $\mathcal{S}{\mathcal{S}}_{\beta }^{\ast }:={\mathcal{S}}^{\ast }{\left[(1+\zeta )/(1-\zeta )\right]}^{\beta }.$ The class ${\mathcal{SL}}^{\ast }:={\mathcal{S}}^{\ast }\left(\sqrt{1+\zeta }\right)$ related to lemniscate of Bernouli was introduced by Sokół and Stankiewicz [3]. The classes ${\mathcal{S}}_{\mathcal{RL}}^{\ast }:=$${\mathcal{S}}^{\ast }\left(\sqrt{2}-\left(\sqrt{2}-1\right){\left(\left(1-\zeta \right)/\left(1+2\left(\sqrt{2}-1\right)\zeta \right)\right)}^{1/2}\right)$ and ${\mathcal{S}}_e^{\ast }:={\mathcal{S}}^{\ast }\left(e^{\zeta }\right)$ were introduced by Mendiratta et al. [4,5]. The class ${\mathcal{S}}_{\mathcal{C}}^{\ast }:={\mathcal{S}}^{\ast }(1+4\zeta /3+$$2{\zeta }^2/3)$ was introduced and studied by Sharma et al. [6]. The class ${\mathcal{S}}_{{\Delta }^{\ast }}:={\mathcal{S}}^{\ast }\left(\zeta +\sqrt{1+{\zeta }^2}\right)$ was introduced by Raina et al. [7] while the class ${\mathcal{S}}_s^{\ast }:={\mathcal{S}}^{\ast }\left(1+sin\left(\zeta \right)\right)$ was studied by Cho et al. [8]. The class ${\mathcal{S}}_{cos}^{\ast }:={\mathcal{S}}^{\ast }\left(cos\left(\zeta \right)\right)$ was introduced by Bano and Raza [9]. For some recent work in this direction, we refer to [10,11,12,13,14,15,16,17,18] and references therein, which include the study of analytic functions associated with certain functions and domains such as sigmoid function, Pascal snail function, cardioid domain, petal-type domain, limacon domain, and nephroid domain. Furthermore, factional and q-fractional derivatives are applied to the classes defined by using the above domains to study various generalizations of the classes of univalent functions. For some details on the applications of fractional operators, we refer to [19,20,21,22,23,24].

Piejko and Sokół [25] introduced a one-parameter family of functions given by

$${\mathcal{J}}_{\alpha }\left(\zeta \right)=\frac{\zeta }{1-\alpha {\zeta }^2}=\sum _{n=1}^{\infty }{\alpha }^{n-1}{\zeta }^{2n-1},\zeta \in \mathbb{E},0\le \alpha \le 1.$$

The function ${\mathcal{J}}_{\alpha }$ is starlike univalent when $0\le \alpha <1$ and convex for $0\le \alpha \le 3-2\sqrt{2}.$ It is observed that ${\mathcal{J}}_{\alpha }\left(\mathbb{E}\right)=\mathcal{D}\left(\alpha \right),$ where

$$\mathcal{D}\left(\alpha \right)=\{x+iy\in \mathbb{C}:{(x^2+y^2)}^2-\frac{x^2}{{(1-\alpha )}^2}-\frac{y^2}{{(1+\alpha )}^2}<0,0\le \alpha <1\}$$

and

$$\mathcal{D}\left(1\right)=\{x+iy\in \mathbb{C}:\forall s\in (-\infty ,\frac{-i}{2}]\cup [\frac{i}{2},\infty ),[x+iy\ne is]\}.$$

It is clear that the curve

$${(x^2+y^2)}^2-\frac{x^2}{{(1-\alpha )}^2}-\frac{y^2}{{(1+\alpha )}^2}=0,(x,y)\ne (0,0),$$

is the Booth lemniscate of elliptic type. Karger et al. [26] have introduced the class $\mathcal{BS}\left(\alpha \right)$ of starlike univalent functions related to Booth lemniscate. It is defined as:

$$\mathcal{BS}\left(\alpha \right)=\left\{f\in \mathcal{A}:\frac{\zeta f^{\prime }\left(\zeta \right)}{f\left(\zeta \right)}-1\prec \frac{\zeta }{1-\alpha {\zeta }^2},0\le \alpha \le 1,\zeta \in \mathbb{E}\right\}.$$

(3)

The authors also found some non-sharp coefficient bounds for the class $\mathcal{BS}\left(\alpha \right).$ Cho et al. [27] studied the differential subordination and radius results for the class $\mathcal{BS}\left(\alpha \right).$ The class $\mathcal{BS}\left(\alpha \right)$ is further studied in [28,29].

Pommerenke [30] introduced the q-th Hankel determinant for analytic function, and it is stated as:

$$H_{q,n}\left(f\right):=\left|\begin{array}{cccc}{b}_n\hfill & b_{n+1}\hfill & \dots \hfill & b_{n+q-1}\hfill \\ b_{n+1}\hfill & b_{n+2}\hfill & \dots \hfill & b_{n+q}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ b_{n+q-1}\hfill & b_{n+q}\hfill & \dots \hfill & b_{n+2q-2}\hfill \end{array}\right|,$$

where $n\ge 1$ and $q\ge 1.$ It is easy to see that $H_{2,2}\left(f\right)=b_2{b}_4-b_3^2$ and

$$H_{3,1}\left(f\right)=2b_2{b}_3{b}_4-b_3^3-b_4^2+b_3{b}_5-b_2^2{b}_5.$$

(4)

These Hankel determinants for different subclasses of analytic and univalent functions have been investigated by many authors. Recently, sharp bounds for $|H_3\left(1\right)\left(f\right)|$ were obtained using a result from [31]; see [32,33,34,35,36,37] for some detailed work on Hankel determinants. A new form for the fourth Hankel determinant is given in [38], which is studied for a new subclass of analytic functions introduced, and the upper bound of the fourth Hankel determinant for this class is obtained. A new class of analytic functions associated with exponential functions is introduced in [39] and the upper bound of the third Hankel determinant is found. Sine function is used in [40] to introduce a new class of analytic functions, for which the second Hankel inequality is discussed.

In the 1960s, L. Zalcman conjectured that if $f\in \mathcal{S}$, then

$$\left|b_n^2-b_{2n-1}\right|\le {(n-1)}^2,n\ge 2,$$

(5)

which would be sharp for the Koebe function. The Zalcman conjecture implies the famous Bieberbach conjecture $\left|b_n\right|\le n$ for $n\ge 2$; see [41,42]. As example of research, in [43] a class of starlike functions with respect to symmetric points is defined involving the sine function, and the third Hankel determinant and Zalcman functional are investigated.

In the present research, we determine the upper bound of Hankel determinants of order two and three for functions in the class $\mathcal{BS}\left(\alpha \right).$ We also find the first four sharp coefficient bounds and Zalcman conjecture for the class $\mathcal{BS}\left(\alpha \right).$

We need the following results of class $\mathcal{P}$ to prove our theorems.

Lemma 1

([44]). If $h\in \mathcal{P}$ is defined in (2), $0\le B\le 1$, $B(2B-1)\le D\le B$, then

$$\left|t_3-2B{t}_1{t}_2+D{t}_1^3\right|\le 2.$$

Lemma 2

([45]). If $h\in \mathcal{P}$ is defined in (2), $0<a<1$, $0<b<1$ and if

$$8a(1-a)\{{(b\beta -2\lambda )}^2+{(b\left(a+b\right)-\beta )}^2\}+b(1-b){(\beta -2ab)}^2\le 4b^{2}a{(1-b)}^2(1-a).$$

Then

$$\left|\lambda t_1^4+a{t}_2^2+2b{t}_1{t}_3-\frac{3}{2}\beta t_1^2{t}_2-t_4\right|\le 2.$$

Lemma 3

([31,46]). If $h\in \mathcal{P}$ is defined in (2) and $t_1>0,$ then

$$\begin{array}{ccc}\hfill 2t_2& =& t_1^2+\delta (4-t_1^2),\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill 4t_3& =& t_1^3+2(4-t_1^2)t_1\delta -(4-t_1^2)t_1{\delta }^2+2(4-t_1^2){(1-|\delta |}^2)\eta ,\hfill \end{array}$$

(7)

$$\begin{array}{ccc}\hfill 8t_4& =& t_1^4+(4-t_1^2)\delta (t_1^2({\delta }^2-3\delta +3)+4\delta )-4(4-t_1^2){(1-|\delta |}^2\left)\right(t_1(\delta -1)\eta \hfill \\ & & +\overline{\delta }{\eta }^2{-(1-|\eta |}^2\left)\rho \right),\hfill \end{array}$$

(8)

for some ρ, δ and η such that $\left|\rho \right|\le 1$, $\left|\delta \right|\le 1$ and $\left|\eta \right|\le 1$.

2. Main Results

In the following first theorem proved, sharp bounds of the first four coefficients are obtained for functions in class $\mathcal{BS}\left(\alpha \right)$ defined by (3)$.$

Theorem 1.

Let $f\in \mathcal{BS}\left(\alpha \right)$ be of the form (1)$.$ Then

$$\begin{array}{ccc}\hfill \left|b_n\right|& \le & \frac{1}{n-1},0\le \alpha \le 1,n=2,3,4,\hfill \\ \hfill \left|b_5\right|& \le & \frac{1}{4},0\le \alpha \le {\alpha }^{\ast },\hfill \end{array}$$

where ${\alpha }^{\ast }\simeq 403549870$ is the solution of $73224{\alpha }^2-1008\alpha -1301=0.$ Results are sharp for the function

$$f_n\left(\zeta \right)=\zeta exp\left(\frac{{tan}^{-1}\left(\sqrt{\alpha }{\zeta }^n\right)}{n\sqrt{\alpha }}\right),n=1,2,3,4.$$

Proof. 

Let $f\in \mathcal{BS}\left(\alpha \right).$ Then

$$\frac{\zeta f^{\prime }\left(\zeta \right)}{f\left(\zeta \right)}\prec 1+\frac{\zeta }{1-\alpha {\zeta }^2},0\le \alpha \le 1.$$

By using the definition of subordination, we have

$$\frac{\zeta f^{\prime }\left(\zeta \right)}{f\left(\zeta \right)}=1+\frac{w\left(\zeta \right)}{1-\alpha w^2\left(\zeta \right)},$$

(9)

where w is analytic and maps origin onto the origin and $\left|w\left(\zeta \right)\right|<1$ for $\zeta \in \mathbb{E}$. Now for $h\in \mathcal{P}$, we have

$$w\left(\zeta \right)=\frac{h\left(\zeta \right)-1}{h\left(\zeta \right)+1}.$$

Let h be of the form (2)$.$ Then

$$\begin{array}{ccc}\hfill 1+\frac{w\left(\zeta \right)}{1-\alpha w^2\left(\zeta \right)}& =& 1+\frac{1}{2}{t}_1\zeta +\left(\frac{1}{2}{t}_2-\frac{1}{4}{t}_1^2\right){\zeta }^2+\left(\frac{1}{8}(1+\alpha )t_1^3-\frac{1}{2}{t}_1{t}_2+\frac{1}{2}{t}_3\right){\zeta }^3\hfill \\ & & +\left(\frac{1}{16}(-1-3\alpha )t_1^4+\frac{3}{8}(1+\alpha )t_1^2{t}_2-\frac{1}{2}{t}_1{t}_3+\frac{1}{2}{t}_4-\frac{1}{4}{t}_2^2\right){\zeta }^4+\cdots .\hfill \end{array}$$

We also have

$$\begin{array}{ccc}\hfill \frac{\zeta f^{\prime }\left(\zeta \right)}{f\left(\zeta \right)}& =& 1+b_2\zeta +\left[2b_3-b_2^2\right]{\zeta }^2+\left[3b_4-3b_2{b}_3+b_2^3\right]{\zeta }^3\hfill \\ & & +\left(4b_5-4b_2{b}_4-2b_3^2+4b_3{b}_2^2-b_2^4\right){\zeta }^4+\cdots .\hfill \end{array}$$

Putting these values in (9) and comparing the coefficients, we obtain

$$\begin{array}{ccc}\hfill b_2& =& \frac{1}{2}{t}_1,\hfill \end{array}$$

(10)

$$\begin{array}{ccc}\hfill b_3& =& \frac{1}{4}{t}_2,\hfill \end{array}$$

(11)

$$\begin{array}{ccc}\hfill b_4& =& -\frac{1}{24}{t}_1{t}_2+\frac{1}{24}{t}_1^3\alpha +\frac{1}{6}{t}_3,\hfill \end{array}$$

(12)

$$\begin{array}{ccc}\hfill b_5& =& -\frac{5}{192}\alpha t_1^4-\frac{1}{24}{t}_1{t}_3-\frac{1}{32}{t}_2^2+\frac{\left(1+9\alpha \right)}{96}{t}_1^2{t}_2+\frac{1}{8}{t}_4.\hfill \end{array}$$

(13)

Results for $n=2,3$ is a simple application of the coefficient bounds for class $\mathcal{P}$. For $n=4,$ consider

$$\left|b_4\right|=\frac{1}{6}\left|t_3-\frac{1}{4}{t}_1{t}_2+\frac{1}{4}{t}_1^3\alpha \right|=\frac{1}{6}\left|t_3-2B{t}_1{t}_2+D{t}_1^3\right|.$$

Here, $B=\frac{1}{8}$ and $D=\frac{1}{4}\alpha .$ Now $0<B<1$ and $D\le B$ for $0\le \alpha \le 1.$ Additionally, $B\left(2B-1\right)=\frac{-3}{32}<D.$ Using Lemma 1, we obtain the required result. Now,

$$\begin{array}{ccc}\hfill \left|b_5\right|& =& \frac{1}{8}\left|\frac{-5\alpha }{24}{t}_1^4+\frac{1}{4}{t}_2^2+\frac{1}{3}{t}_1{t}_3-\frac{\left(1+9\alpha \right)}{12}{t}_1^2{t}_2-t_4\right|\hfill \\ & =& \frac{1}{8}\left|\lambda t_1^4+a{t}_2^2+2b{t}_1{t}_3-\frac{3}{2}\beta t_1^2{t}_2-t_4\right|,\hfill \end{array}$$

where $\lambda =\frac{-5\alpha }{24},a=\frac{1}{4},b=\frac{1}{6}$ and $\beta =\frac{1+9\alpha }{18}.$ Consider

$$\begin{array}{ccc}& & 8a(1-a)\{{(b\beta -2\lambda )}^2+{(b\left(a+b\right)-\beta )}^2\}+b(1-b){(\beta -2ab)}^2-4b^{2}a{(1-b)}^2(1-a)\hfill \\ & =& \frac{1}{\mathrm{93,312}}\left\{\mathrm{73,224}{\alpha }^2-1008\alpha -1301\right\}.\hfill \end{array}$$

The above relation has negative value when $\alpha \le {\alpha }^{\ast }\simeq 0.1403549870,$ where ${\alpha }^{\ast }$ is the root of the equation $\mathrm{73,224}{\alpha }^2-1008\alpha -1301=0.$ Using Lemma 2, we obtain the required result. Consider the function $f_n:\mathbb{E}\to \mathbb{C}$ defined as $f_n\left(\zeta \right)=\zeta exp(\frac{{tan}^{-1}\left(\sqrt{\alpha }{\zeta }^{n-1}\right)}{\left(n-1\right)\sqrt{\alpha }}.$ Then

$$\frac{\zeta f_n^{\prime }\left(\zeta \right)}{f_n}=1+\frac{{\zeta }^n}{1-\alpha {\zeta }^{2n}}.$$

Hence, it is clear that $f_n\in \mathcal{BS}\left(\alpha \right).$ Now

$$\begin{array}{ccc}\hfill f_1\left(\zeta \right)& =& \zeta exp\left(\frac{{tan}^{-1}\left(\sqrt{\alpha }\zeta \right)}{\sqrt{\alpha }}\right)=\zeta +{\zeta }^2+\frac{1}{2}{\zeta }^3+\frac{1}{6}(1+2\alpha ){\zeta }^4+\cdots ,\hfill \end{array}$$

(14)

$$\begin{array}{ccc}\hfill f_2\left(\zeta \right)& =& \zeta exp\left(\frac{{tan}^{-1}\left(\sqrt{\alpha }{\zeta }^2\right)}{2\sqrt{\alpha }}\right)=\zeta +\frac{1}{2}{\zeta }^3+\frac{1}{8}{\zeta }^5+\cdots ,\hfill \end{array}$$

(15)

$$\begin{array}{ccc}\hfill f_3\left(\zeta \right)& =& \zeta exp\left(\frac{{tan}^{-1}\left(\sqrt{\alpha }{\zeta }^3\right)}{3\sqrt{\alpha }}\right)=\zeta +\frac{1}{3}{\zeta }^4+\frac{1}{18}{\zeta }^7+\cdots ,\hfill \end{array}$$

(16)

$$\begin{array}{ccc}\hfill f_4\left(\zeta \right)& =& \zeta exp\left(\frac{{tan}^{-1}\left(\sqrt{\alpha }{\zeta }^4\right)}{4\sqrt{\alpha }}\right)=\zeta +\frac{1}{4}{\zeta }^5+\cdots .\hfill \end{array}$$

(17)

Hence, the result is sharp for the function $f_n\left(\zeta \right)=\zeta exp\left(\frac{{tan}^{-1}\left(\sqrt{\alpha }{\zeta }^n\right)}{\left(n\right)\sqrt{\alpha }}\right),n=1,2,3,4.$ □

The following result investigates the sharp upper bound of $\left|b_2{b}_3-b_4\right|$ for the functions of class $\mathcal{BS}\left(\alpha \right)$ defined by (3)$.$

Theorem 2.

Let $f\in \mathcal{BS}\left(\alpha \right)$ be of the form (1)$.$ Then

$$\left|b_2{b}_3-b_4\right|\le \frac{1}{3}.$$

Result is sharp.

Proof. 

Consider

$$\left|b_2{b}_3-b_4\right|=\frac{1}{6}\left|t_1{t}_2-\frac{1}{4}{t}_1^3\alpha -t_3\right|.$$

(18)

Putting the values of $t_2$ and $t_3$ from (6) and (7) in (18)$,$ we can write

$$\begin{array}{ccc}\hfill \left|b_2{b}_3-b_4\right|& =& \frac{1}{24}\left|4t_1{t}_2-t_1^3\alpha -4t_3\right|\hfill \\ & =& \frac{1}{24}\left|\left(1-\alpha \right)t_1^3+(4-t_1^2){\delta }^2{t}_1-2(4-t_1^2)(1-{\left|\delta \right|}^2)\eta \right|\hfill \\ & \le & \frac{1}{24}\left[\left(1-\alpha \right)t^3+(4-t^2)u^{2}t+2(4-t^2)(1-u^2)\right],\hfill \end{array}$$

where $t_1=t\in \left[0,2\right]$ and $\left|\delta \right|=u\in \left[0,1\right]$. Let

$$\phi \left(t,u\right)=\left(1-\alpha \right)t^3+(4-t^2)u^{2}t+2(4-t^2)(1-u^2).$$

Differentiating $\phi$ with respect to $u,$ then

$$\frac{\partial \phi \left(t,u\right)}{\partial u}=2u(4-t^2)\left(t-2\right).$$

Now, $\frac{\partial \phi \left(t,u\right)}{\partial u}\le 0$ for $u\in [0,1]$ and $t\in [0,2].$ Hence, $\phi \left(t,u\right)$ is decreasing. This implies that $max\phi (t,u)=\phi (t,0).$ Let

$$\phi (t,0)={\phi }_1\left(t\right)=\left(1-\alpha \right)t^3+2(4-t^2).$$

Now ${\phi }_1^{\prime }\left(t\right)=3\left(1-\alpha \right)t^2-4t$ and ${\phi }_1^{″}\left(0\right)=-4<0.$ Hence $max{\phi }_1\left(t\right)={\phi }_1\left(0\right)=8.$ This implies that

$$\left|b_2{b}_3-b_4\right|\le \frac{1}{3}.$$

The result is sharp for the function $f_3$ given in (16)$.$ □

Now, we compute the sharp upper bound of the second Hankel determinant for the class $\mathcal{BS}\left(\alpha \right)$ defined by (3)$.$

Theorem 3.

Let $f\in \mathcal{BS}\left(\alpha \right)$ be of the form (1)$.$ Then

$$\left|H_{2,2}\left(f\right)\right|\le \frac{1}{4}.$$

Result is sharp.

Proof. 

Consider

$$\left|H_{2,2}\left(f\right)\right|=\left|b_3^2-b_2{b}_4\right|=\left|-\frac{1}{48}\alpha t_1^4+\frac{1}{48}{t}_1^2{t}_2-\frac{1}{12}{t}_3{t}_1+\frac{1}{16}{t}_2^2\right|.$$

(19)

Putting the values of $t_2$ and $t_3$ from (6) and (7) in (19), we can write

$$\begin{array}{ccc}\hfill \left|H_{2,2}\left(f\right)\right|& =& \frac{1}{192}\left|\left(1-4\alpha \right)t_1^4+3{(4-t_1^2)}^2{\delta }^2+4(4-t_1^2){\delta }^2{t}_1^2-8(4-t_1^2)(1-{\left|\delta \right|}^2)\eta t_1\right|\hfill \\ & \le & \frac{1}{192}\left\{\left|1-4\alpha \right|t^4+3{(4-t^2)}^2{u}^2+4(4-t^2)u^2{t}^2+8(4-t^2)(1-u^2)t\right\},\hfill \end{array}$$

where $t_1=t\in \left[0,2\right]$ and $\left|\delta \right|=u\in \left[0,1\right]$. Suppose

$$\phi \left(t,u\right)=\left|1-4\alpha \right|t^4+3{(4-t^2)}^2{u}^2+4(4-t_1^2)u^2{t}^2+8(4-t^2)(1-u^2)t.$$

Differentiating $\phi$ with respect to $u,$ we obtain

$$\frac{\partial \phi \left(t,u\right)}{\partial u}=2(4-t^2)\left(6-t\right)\left(2-t\right)u.$$

Now, $\frac{\partial \psi \left(t,u\right)}{\partial u}>0$ for $u\in [0,1]$ and $t\in [0,2].$ Hence, $\phi$ is increasing function of $\phi$. This implies that $max\phi \left(t,u\right)=\phi \left(t,1\right).$ Suppose that

$$\phi \left(t,1\right)={\phi }_1\left(t\right)=\left|1-4\alpha \right|t^4+3{(4-t^2)}^2+4(4-t^2)t^2.$$

Differentiating ${\phi }_1$ with respect to $t,$ then

$${\phi }_1^{\prime }\left(t\right)=\left\{\begin{array}{c}-16t\left(\alpha t^2+1\right),0\le \alpha \le 1/4,\\ 4t\left\{(4\alpha -2)t^2-4\right\},1/4\le \alpha \le 1.\end{array}\right.$$

For the case $0\le \alpha \le 1/4,$ the function ${\phi }_1^{\prime }\left(t\right)<0$. This implies that $max{\phi }_1\left(t\right)={\phi }_1\left(0\right)=48.$ For the case $1/4\le \alpha \le 1,$ the equation

$$4t\left\{(4\alpha -2)t^2-4\right\}=0$$

has three roots, namely $t=0,t=\pm \frac{2}{\sqrt{4\alpha -2}}.$ Now ${\phi }_1^{\prime \prime }\left(t\right)=12\left(4\alpha -2\right)t^2-16.$ It is easy to see that ${\phi }_1^{\prime \prime }\left(\pm \frac{2}{\sqrt{4\alpha -2}}\right)=32>0$ and ${\phi }_1^{\prime \prime }\left(0\right)=-16.$ This shows that $max{\phi }_1\left(t\right)={\phi }_1\left(0\right)=48.$ Hence, we have the required result. Inequality is sharp for the function $f_2$ given in (15)$.$ □

The next result establishes the sharp inequality for upper bound of $\left|b_3^2-b_5\right|$ for the functions of class $\mathcal{BS}\left(\alpha \right)$ defined by (3)$.$

Theorem 4.

Let $f\in \mathcal{BS}\left(\alpha \right)$ be of the form (1)$.$ Then,

$$\left|b_3^2-b_5\right|\le \frac{1}{4},0\le \alpha \le {\alpha }_1,$$

where ${\alpha }_1\simeq 0.28494$ is the solution of $53784{\alpha }^2-16992\alpha +475=0.$ The result is sharp for the function $f_4$ given in (17)$.$

Proof. 

Now

$$\begin{array}{ccc}\hfill \left|b_3^2-b_5\right|& =& \frac{1}{8}\left|\frac{5}{24}\alpha t_1^4+\frac{3}{4}{t}_2^2+\frac{1}{3}{t}_1{t}_3-\frac{1+9\alpha }{12}{t}_1^2{t}_2-t_4\right|\hfill \\ & =& \frac{1}{8}\left|\lambda t_1^4+a{t}_2^2+2b{t}_1{t}_3-\frac{3}{2}\beta t_1^2{t}_2-t_4\right|,\hfill \end{array}$$

where $\lambda =\frac{5\alpha }{24},a=\frac{3}{4},b=\frac{1}{6}$ and $\beta =\frac{1+9\alpha }{18}.$ Consider

$$\begin{array}{ccc}& & 8a(1-a)\{{(b\beta -2\lambda )}^2+{(b\left(a+b\right)-\beta )}^2\}+b(1-b){(\beta -2ab)}^2-4b^{2}a{(1-b)}^2(1-a)\hfill \\ & =& \frac{1}{93312}\left\{53784{\alpha }^2-3168\alpha -1301\right\}.\hfill \end{array}$$

(20)

Equation (20) has negative value when $\alpha \le {\alpha }_1=118/747+\sqrt{143934}/2988\simeq 0.28494,$ where ${\alpha }_1$ is the root of the equation $53784{\alpha }^2-16992\alpha +475=0.$ Using Lemma 2, we obtain the required result. □

Finally, we compute the upper bound of third Hankel determinant for the class $\mathcal{BS}\left(\alpha \right)$ defined by (3)$.$

Theorem 5.

Let $f\in \mathcal{BS}\left(\alpha \right)$ be of the form (1)$.$ Then

$$|H_{3,1}\left(f\right)|\le \left\{\begin{array}{c}\frac{5}{16},\alpha \in \left[0,\frac{23}{27}\right),\\ \frac{-\mathrm{41,796}{\alpha }^4+\mathrm{95,363}{\alpha }^3+648{\alpha }^{3}A-1729{\alpha }^{2}A-\mathrm{33,336}{\alpha }^2-\mathrm{12,105}\alpha +528A\alpha +129A-621}{-3888{(4{\alpha }^2-4\alpha -1)}^2},\alpha \in \left[\frac{23}{27},1\right)\end{array}\right.$$

where $A=\sqrt{-129-528\alpha +1729{\alpha }^2-648{\alpha }^3}$.

Proof. 

From (4), we have

$$H_{3,1}\left(f\right)=2b_2{b}_3{b}_4-b_2^2{b}_5-b_3^3+b_3{b}_5-b_4^2.$$

(21)

By putting the values from (10)–(13) in (21), we have

$$H_{3,1}\left(f\right)=\frac{1}{2304}\left[\begin{array}{c}(-4{\alpha }^2+15\alpha )t_1^6+(-37\alpha -6)t_2{t}_1^4+(-32\alpha +24)t_3{t}_1^3\hfill \\ +((54\alpha -4)t_2^2-72t_4)t_1^2+104t_2{t}_1{t}_3-54t_2^3+72t_2{t}_4-64t_3^2\hfill \end{array}\right],$$

where $0<\alpha <1$. By using the invariance of class $\mathcal{P}$ under the rotation, we take $t_1\in [0,2]$. Let $t:=t_1.$ Then, by using the Equalities (6)–(8) and upon some simplification, we are able to obtain

$$H_{3,1}\left(f\right)=\frac{1}{2304}\left({{\rm Y}}_1(t,\delta )+{{\rm Y}}_2(t,\delta )\eta +{{\rm Y}}_3(t,\delta ){\eta }^2+\Psi (t,\delta ,\eta )\rho \right),$$

where $\rho ,\eta ,\delta \in \overline{\mathbb{E}},$

$$\begin{array}{ccc}\hfill {{\rm Y}}_1(t,\delta )& =& t^6\left(2\alpha -\frac{1}{4}-4{\alpha }^2\right)+(4-t^2)\left(-\frac{15}{2}{t}^4\alpha \delta +8t^4\alpha {\delta }^2-18t^2{\delta }^2+\frac{3}{4}{t}^4\delta \right.\hfill \\ & & \left.+\frac{5}{2}{t}^4{\delta }^2-\frac{9}{2}{t}^4{\delta }^3\right)+{(4-t^2)}^2\left(\frac{27}{2}{t}^2\alpha {\delta }^2-\frac{21}{2}{\delta }^3{t}^2+\frac{1}{2}{\delta }^4{t}^2+\frac{9}{4}{t}^2{\delta }^2\right.\hfill \\ & & \left.+18{\delta }^3\right)-\frac{27}{4}{\delta }^3{(4-t^2)}^3,\hfill \\ \hfill {{\rm Y}}_2(t,\delta )& =& -2t(4-t^2){(1-|\delta |}^2)\left(t^2(-2-9\delta +8\alpha )+(4-t^2)({\delta }^2-6\delta )\right),\hfill \\ \hfill {{\rm Y}}_3(t,\delta )& =& -2(4-t^2){(1-|\delta |}^2)\left((4-t^2){(8+|\delta |}^2)-9t^2\overline{\delta }\right),\hfill \\ \hfill \Psi (t,\delta ,\eta )& =& 18(4-t^2){(1-|\delta |}^2\left)\right(1-{\left|\eta \right|}^2)\left((4-t^2)\delta -t^2\right).\hfill \end{array}$$

By taking $u:=\left|\delta \right|$, $v:=\left|\eta \right|$ and using $\left|\rho \right|\le 1$, we have

$$\begin{array}{cc}\hfill |H_{3,1}\left(f\right)|& \le \frac{1}{2304}\left(|{{\rm Y}}_1(t,\delta )|+|{{\rm Y}}_2(t,\delta )|v+|{{\rm Y}}_3(t,\delta )|v^2+|\Psi (t,\delta ,\eta )|\right)\hfill \\ \hfill & =:G(t,u,v),\hfill \end{array}$$

where

$$G(t,u,v):=\frac{1}{2304}\left(g_1(t,u)+g_2(t,u)v+g_3(t,u)v^2+g_4(t,u)(1-v^2)\right),$$

(22)

with

$$\begin{array}{ccc}\hfill g_1(t,u):=& & t^6{\left(2\alpha -\frac{1}{2}\right)}^2+(4-t^2)\left(\frac{15}{2}{t}^{4}u\alpha +8t^4{u}^2\alpha +18t^2{u}^2+\frac{3}{4}{t}^{4}u\right.\hfill \\ & & \left.+\frac{5}{2}{t}^4{u}^2+\frac{9}{2}{t}^4{u}^3\right)+{(4-t^2)}^2\left(\frac{27}{2}{t}^2{u}^2\alpha +\frac{21}{2}{u}^3{t}^2+\frac{1}{2}{u}^4{t}^2+\frac{9}{4}{t}^2{u}^2\right.\hfill \\ & & \left.+18u^3\right)+\frac{27}{4}{u}^3{(4-t^2)}^3,\hfill \\ \hfill g_2(t,u):=& & 2t(4-t^2)(1-u^2)\left(t^2(2+9u+8\alpha )+(4-t^2)(u^2+6u)\right),\hfill \\ \hfill g_3(t,u):=& & 2(4-t^2)(1-u^2)\left((4-t^2)(8+u^2)+9t^{2}u\right),\hfill \\ \hfill g_4(t,u):=& & 18(4-t^2)(1-u^2)\left((4-t^2)u+t^2\right).\hfill \end{array}$$

Thus, we need to maximize $G(t,u,v)$ over the closed cuboid $S:[0,2]\times [0,1]\times [0,1]$. To do this, we find the maximum values in the interior of the six faces, on the twelve edges and in the interior of S.

I. 

Interior points of cuboid:

Let $(t,u,v)\in (0,2)\times (0,1)\times (0,1)$. Differentiating $G(t,u,v)$ with respect to v, we obtain (after some simplification)

$$\begin{array}{cc}\hfill \frac{\partial G}{\partial v}& =\frac{1}{1152}(4-t^2)(1-u^2)[2v(u-1)((4-t^2)(u-8)+9t^2)\hfill \\ \hfill & +t(u(4-t^2)(u+6)+t^2(9u+8\alpha +2))].\hfill \end{array}$$

So that $\frac{\partial G}{\partial v}=0$ when

$$v=\frac{t(u(4-t^2)(u+6)+t^2(9u+8\alpha +2))}{2(1-u)((4-t^2)(u-8)+9t^2)}:=v_0.$$

If $v_0$ is a critical point inside S, then $v_0\in (0,1)$, which is possible only if

$$t(u(4-t^2)(u+6)+t^2(9u+8\alpha +2))<2(1-u)((4-t^2)(u-8)+9t^2)$$

(23)

and

$$t^2>\frac{4(u-8)}{u-17}.$$

(24)

Let $g\left(u\right):=4(u-8)/(u-17).$ Since $g^{\prime }\left(u\right)<0$ for $(0,1)$, $g\left(u\right)$ is decreasing in $(0,1)$, and so $t^2>\frac{7}{4}$. A simple exercise shows that (23) does not hold in this case for all values of $u\in \left[\frac{7}{18},1\right)$; thus, there are no critical points of G in $(0,2)\times \left[\frac{7}{18},1\right)\times (0,1)$.

Suppose that there is a critical point $(\tilde{t},\tilde{u},\tilde{v})$ of G existing in the interior of cuboid S. Clearly, it must satisfy that $\tilde{u}<\frac{7}{18}$. From the above discussion, it can be also known that $\tilde{t}>\frac{548}{299}$ and when $\tilde{u}\to 0,$ then $\tilde{t}\to 1.433783077$. Thus, we conclude that a possible solution exists in $\left[1.433783077,2\right)\times \left(0,\frac{7}{18}\right),$ for inequality (23). A computation shows

$$\frac{\partial G}{\partial t}{|}_{y=y_0}\ne 0,$$

in this interval. Therefore, no critical point exists in the interior of S.

II. 

Interior of all the six faces of the cuboid:

On the face $t=0$, $G(t,u,v)$ reduces to

$$h_1(u,v):=G(0,u,v)=\frac{2(1-u^2)(u-1)(u-8)v^2+9u(3u^2+2)}{144},x,y\in (0,1).$$

As $h_1$ has no critical point in $(0,1)\times (0,1)$ since

$$\frac{\partial h_1}{\partial y}=\frac{(1-u^2)(u-1)(u-8)v}{36}\ne 0,x,y\in (0,1).$$

On the face $t=2$, $G(t,u,v)$ reduces to

$$G(2,u,v)=\frac{{(4\alpha -1)}^2}{144}\le \frac{1}{16},u,v\in (0,1).$$

On $u=0$, $G(t,u,v)$ takes the form $G(t,0,v)$ which is given by

$$h_2(t,v):=\frac{8v^2(4-t^2)(32-17t^2)+16t^3(4-t^2)(4\alpha +1)v+{(4\alpha -1)}^2{t}^6-72t^4+288t^2}{9216},$$

where $t\in (0,2)$ and $v\in (0,1)$. We solve $\frac{\partial h_2}{\partial v}=0$ and $\frac{\partial h_2}{\partial t}=0$ to obtain the critical points. On solving $\frac{\partial h_2}{\partial v}=0$, we obtain

$$v=-\frac{(4\alpha +1)t^3}{32-17t^2}=:v_1.$$

(25)

For the given bound of v, $v_1$ must belong to $(0,1)$, which implies that $t>t_0$, $t_0\approx 1.371988681$. From $\frac{\partial h_2}{\partial t}=0$, we have

$$(272t^2-800)v^2+8t(12-5t^2)(4\alpha +1)v+3{(4\alpha -1)}^2{t}^4-144t^2+288=0.$$

(26)

By substituting (25) in (26) and simplifying, we obtain

$$\begin{array}{c}98304+(2448{\alpha }^2-3400\alpha +153)t^8+(14336\alpha -14256-6144{\alpha }^2)t^6\\ -(16384\alpha -79968)t^4-153600t^2=0.\end{array}$$

(27)

After some computations, we see that there are different solutions of (27) for different values of $\alpha \in (0,1)$, which does not satisfy the condition $t>t_0$. So $h_2$ has no maxima in $(0,2)\times (0,1)$.

On $u=1$, $G(t,u,v)$ can be written as

$$h_3(t,v):=G(t,1,v)=\frac{720+(4{\alpha }^2-4\alpha -1)t^6+2(3-23\alpha )t^4+(216\alpha -184)t^2}{2304},t\in (0,2).$$

The equation $\frac{\partial h_3}{\partial t}=0$ has five roots that depend on the value of $\alpha$, from which two roots make two cases for $\alpha$.

Case 1.

For all the values of $\alpha$, the root $t=:t_0=0$ is satisfied. Since $h_3$ has minimum value at this root, we neglect it.

Case 2.

For $\alpha \in \left[\frac{23}{27},1\right)$, the root

$$t=:t_1=\frac{\sqrt{-(24{\alpha }^2-24\alpha -6)(-23\alpha +3+A)}}{-3(4{\alpha }^2-4\alpha -1)},t\in [0,2],$$

where $A=\sqrt{-129-528\alpha +1729{\alpha }^2-648{\alpha }^3}$, is satisfied. Since $h_3$ achieves its maxima at $t_1$, hence we may write that

$$\begin{array}{cc}\hfill h_3(t,v)& =\frac{\begin{array}{c}-41796{\alpha }^4+95363{\alpha }^3+648{\alpha }^{3}A-1729{\alpha }^{2}A-33336{\alpha }^2\\ -12105\alpha +528A\alpha +129A-621\end{array}}{-3888{(4{\alpha }^2-4\alpha -1)}^2}\hfill \\ \hfill & \le \frac{53\sqrt{424}}{486}-\frac{7505}{3888}\approx 0.315250606,t\in [0,2],v\in [0,1],\hfill \end{array}$$

for $\alpha \in \left[\frac{23}{27},1\right)$.

On $v=0$, $G(t,u,v)$ reduces to

$$h_4(t,u):=G(t,u,0)=\frac{1}{9216}\left(\begin{array}{c}{(4\alpha -1)}^2{t}^6+(4-t^2)\left(u(4-t^2)\right(15u^2{t}^2+2u^3{t}^2\hfill \\ +9t^{2}u+108u^2+72+54t^{2}u\alpha )+72t^2+3t^{4}u\hfill \\ +32t^4{u}^2\alpha +30t^{4}u\alpha +10t^4{u}^2+18t^4{u}^3)\hfill \end{array}\right).$$

After some computations, we see that the system of equations $\frac{\partial h_4}{\partial u}=0$ and $\frac{\partial h_4}{\partial t}=0$ has roots depending upon $\alpha .$ In particular, we have solution $(1.999999994,0.06954329322)$ for $\alpha =0.9,$ and $(1.999999994,0.2109628665)$ for $\alpha =1$ in $(0,2)\times (0,1)$. We also see that the maximum of $h_4(t,u)$ for these points is attained at $(1.999999994,0.2109628665).$ Thus, we conclude that $h_4(t,u)\le 0.06249999999$.

On $v=1$, $G(t,u,v)$ can be written as

$$\begin{array}{cc}\hfill G(t,u,1)& =\frac{1}{9216}\left(\begin{array}{c}{(4\alpha +1)}^2{t}^6+(4-t^2)\left((4-t^2)\right(-8t{u}^4-56u^2+180u^3\hfill \\ +54t^2{u}^2\alpha -8u^4-48t{u}^3+15u^3{t}^2+8t{u}^2+9t^2{u}^2+2u^4{t}^2\hfill \\ +48tu+64)+3t^{4}u+18t^4{u}^3-16t^3{u}^2-72t^3{u}^3+72t^2{u}^2\hfill \\ +30t^{4}u\alpha +72t^{3}u+72t^{2}u+32t^4{u}^2\alpha -64u^2{t}^3\alpha \hfill \\ +16t^3+10t^4{u}^2+64t^3\alpha -72u^3{t}^2)\hfill \end{array}\right).\hfill \\ \hfill & =:h_5(t,u).\hfill \end{array}$$

We see that the system of equations $\frac{\partial h_5}{\partial u}=0$ and $\frac{\partial h_5}{\partial t}=0$ has roots depending upon $\alpha .$ In particular, $(0.1095160521,0.1289017775)$ for $\alpha =0$, $(0.1097676030,0.1276141649)$ for $\alpha =0.1$, $(0.1099305749,0.1264798400)$ for $\alpha =0.2$, $(0.1101093763,0.1252480620)$ for $\alpha =0.3$, and so on in $(0,2)\times (0,1)$. The max value is on the interval $(0.1107687201,0.1163024505)$ for $\alpha =1,$ that is $h_5(t,u)\le 0.1108016596$.

III. 

On the vertices of the cuboid:

$$\begin{array}{cc}\hfill G(0,0,0)& =0,G(0,0,1)=\frac{1}{36},G(0,1,0)=\frac{5}{16},G(0,1,1)=\frac{5}{16},\hfill \\ \hfill G(2,0,0)& =G(2,0,1)=G(2,1,0)=G(2,1,1)=\frac{1}{16}.\hfill \end{array}$$

IV. 

On the edges of the cuboid:

Finally, we find the points of maxima of $G(t,u,v)$ on the 12 edges of S.

$$\begin{array}{cc}\hfill G(t,0,0)& =\frac{{(4\alpha -1)}^2{t}^6-72t^4+288t^2}{9216}\le G(t_1,0,0)\hfill \\ \hfill & =\frac{(\sqrt{30+48\alpha -96{\alpha }^2}-6)(\sqrt{30+48\alpha -96{\alpha }^2}-4+32{\alpha }^2-16\alpha )}{-24{(4\alpha -1)}^4}\hfill \\ \hfill & \le \frac{1}{24}\approx 0.04166666667,t\in (0,2).\hfill \end{array}$$

where

$$t=:t_1=\frac{2\sqrt{-{(4\alpha -1)}^2(\sqrt{30-96{\alpha }^2+48\alpha }-6)}}{{(4{\alpha }^2-1)}^2}.$$

for $t\in [0,2]$ and $\alpha \in (0,\frac{1+\sqrt{6}}{4})\approx (0,0.8623724358)$

$$\begin{array}{cc}\hfill G(t,0,1)& =\frac{1024+{(4\alpha -1)}^2{t}^6-16(1+4\alpha )t^5+64t^4+64(1+4\alpha )t^3-512t^2}{9216}\le G(0,0,1)\hfill \\ \hfill & =\frac{1}{9}\approx 0.11111,t\in (0,2).\hfill \end{array}$$

$$\begin{array}{cc}\hfill G(t,1,0)& =\frac{720+(-4\alpha +4{\alpha }^2-1)t^6+2(3-23\alpha )t^4+(216\alpha -184)t^2}{2304}\le G(t_2,1,0)\hfill \\ \hfill & =\frac{\begin{array}{c}-41796{\alpha }^4+95363{\alpha }^3+648{\alpha }^{3}A-1729{\alpha }^{2}A-33336{\alpha }^2-12105\alpha \\ +528\alpha A+129A-621\end{array}}{-3888{(-4\alpha -1+4{\alpha }^2)}^2}\hfill \\ \hfill & \le \frac{53\sqrt{424}}{486}-\frac{7505}{3888}\approx 0.315250606,t\in (0,2),\hfill \end{array}$$

where

$$t:=t_2=\frac{\sqrt{-(24{\alpha }^2-24\alpha -6)(-23\alpha +3+\sqrt{1729{\alpha }^2-528\alpha -129-648{\alpha }^3})}}{-3(4{\alpha }^2-4\alpha -1)}\in (0,2)$$

for $\alpha \in \left(\frac{23}{27},1\right)$ and $A=\sqrt{-129-528\alpha +1729{\alpha }^2-648{\alpha }^3}$.

$$G(0,u,0)=\frac{u(2+3u^2)}{16}\le G(0,1,0)=\frac{5}{16}\approx 0.3125,u\in (0,1)$$

$$G(0,u,1)=\frac{-2u^4+45u^3-14u^2+16}{144}\le G(0,0,1)=\frac{1}{9},u\in (0,1).$$

$$\begin{array}{cc}\hfill G(2,u,0)& =\frac{1}{16},u\in (0,1).\hfill \\ \hfill G(2,u,1)& =\frac{1}{16},u\in (0,1).\hfill \\ \hfill G(0,0,v)& =\frac{1}{9}{v}^2\le \frac{1}{9},v\in (0,1).\hfill \\ \hfill G(0,1,v)& =\frac{5}{16}\approx 0.3125,v\in (0,1).\hfill \\ \hfill G(2,0,v)& =\frac{1}{16},v\in (0,1).\hfill \\ \hfill G(2,1,v)& =\frac{1}{16},v\in (0,1).\hfill \end{array}$$

Now, by viewing all the above cases, we get

$$\begin{array}{cc}\hfill |H_{3,1}\left(f\right)|& \le \frac{\begin{array}{c}-41796{\alpha }^4+95363{\alpha }^3+648{\alpha }^{3}A-1729{\alpha }^{2}A-33336{\alpha }^2\\ -12105\alpha +528\alpha A+129A-621\end{array}}{-3888{(-4\alpha -1+4{\alpha }^2)}^2}\hfill \\ \hfill & \le \frac{53\sqrt{424}}{486}-\frac{7505}{3888}\approx 0.315250606,\hfill \end{array}$$

where $A=\sqrt{-129-528b+1729b^2-648b^3}$ for $\alpha \in \left(\frac{23}{27},1\right)$ and $|H_{3,1}\left(f\right)|\le \frac{5}{16}$ for $\alpha \in \left[0,\frac{23}{27}\right)$. Hence, the proof is completed. □

3. Conclusions

In this paper, we studied the starlike functions associated with booth lemniscate, defined in the open unit disk, given in (3). Certain inequalities, such as coefficient bounds and upper bounds of Hankel determinants, were established. Based on our findings, Zalcman conjecture was proposed for said starlike functions. For future work, many fractional operators can be applied on the discussed class $\mathcal{BS}\left(\alpha \right)$. Some suitable fractional operators for the said purpose can be found in [19,20,21,22,47,48] and the references therein.