Improved Upper Bounds of the Third-Order Hankel Determinant for Ozaki Close-to-Convex Functions

Abstract

$\mathcal{L}etN$ be the class of functions that convex in one direction and $\mathcal{M}$ denote the class of functions $z{f}^{\prime }\left(z\right)$, where $f\in \mathcal{N}$. In the paper, the third-order Hankel determinants for these classes are estimated. The estimates of ${\mathcal{H}}_{3,1}\left(f\right)$ obtained in the paper are improved.

1. Introduction and Definitions

Let $\mathcal{H}\left(\mathcal{U}\right)$ denote the class of functions analytic in the open unit disk $\mathcal{U}=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ and $\mathcal{A}$ be the class of functions $f\in \mathcal{H}\left(\mathcal{U}\right)$, with the form

$$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n$$

(1)

with the standard normalization $f\left(0\right)=0,f^{\prime }\left(0\right)=1.$ Let $\mathcal{S}$ denote the subclass of $\mathcal{A}$ consisting of functions that are univalent in $\mathcal{U}.$

Using ${\mathcal{H}}_{k,n}\left(f\right)$ where $k,n\in \mathbb{N}=\{1,2,···\},$ we denote the Hankel determinant of functions $f\in \mathcal{A}$ of the form (1), which is defined by

$${\mathcal{H}}_{k,n}\left(f\right):=\left|\begin{array}{cccc}{a}_n\hfill & a_{n+1}\hfill & \dots \hfill & a_{n+k-1}\hfill \\ a_{n+1}\hfill & a_{n+2}\hfill & \dots \hfill & a_{n+k}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ a_{n+k-1}\hfill & a_{n+k}\hfill & \dots \hfill & a_{n+2k-2}\hfill \end{array}\right|(a_1=1).$$

Several researchers, including Pommerenke [1,2], Hayman [3], Noonan and Thomas [4], and Ehrenborg [5], have studied the Hankel determinant ${\mathcal{H}}_{k,n}\left(f\right)$ and presented some remarkable results, which are useful, for example, in showing that a function of bounded characteristic in $\mathcal{U}$. In particular, many results [6,7,8] are known concerning the second Hankel determinant ${\mathcal{H}}_{2,1}$ and ${\mathcal{H}}_{2,2}$ when $f\in \mathcal{S}$. In many recent papers [9,10,11,12,13,14,15,16,17,18], the third Hankel determinant

$$H_{3,1}\left(f\right)=a_3{a}_5+2a_2{a}_3{a}_4-a_3^3-a_4^2-a_2^2{a}_5$$

(2)

has been studied, which is quite complicated. Recently, there have been many results for the subclasses of $\mathcal{S}$. For starlike and convex functions the sharp bounds for the third Hankel determinant are $\frac{4}{9}$ ([19]) and $\frac{4}{135}$ ([20]). For the class $\mathcal{F}\in \mathcal{A}$ of functions satisfying $\mathcal{R}e\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\}>-\frac{1}{2},z\in \mathcal{U}$, Kowalczyk et al. [11] obtained the sharp estimate $|{\mathcal{H}}_{3,1}|\le \frac{1}{16}$. Other related work is published in [21,22,23,24].

In the paper, we study the upper bounds of the third-order Hankel determinant ${\mathcal{H}}_{3,1}\left(f\right)$ for the following classes

$$\mathcal{M}=\left\{f\in \mathcal{A}:\mathcal{R}e\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}<\frac{3}{2}\right\},$$

$$\mathcal{N}=\left\{f\in \mathcal{A}:\mathcal{R}e\{1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\}<\frac{3}{2}\right\}.$$

This problem was studied by Prajapat et al. [9] (see [18]).

In 1941, Ozaki [25] introduced and studied the class $\mathcal{N}$. Later, Sakaguchi [26] and R. Singh and S. Singh [27] showed, respectively, that functions in $\mathcal{N}$ are close to convex and starlike. In 2013, Obradović [28] derived the sharp bound of $|a_n|\le \frac{1}{n(n-1)}$ in the class $\mathcal{N}$. Ponnusamy [29] obtained the bounds of initial logarithmic coefficients for $f\in \mathcal{N}$.

In 2022, Obradović and Tuneski [10] improved Bansal’s inequality for $f\in \mathcal{N}$ and showed that $|{\mathcal{H}}_{3,1}\left(f\right)|\le \frac{17}{1080}$. Later, Zaprawa [18] proved Obradović’s conjecture $|{\mathcal{H}}_{3,1}\left(f\right)|\le \frac{19}{2160}$ for $f\in \mathcal{N}$. We will significantly improve the estimate of the third Hankel determinant for the class $\mathcal{M}$ and $\mathcal{N}$.

In this paper, we use a method based on the estimates of the coefficients of the Schwartz function. This method is different from the commonly used method, which is the main reason for the improvement in the estimate for the class mentioned above.

To obtain the main results, we will need the following, almost forgotten, result of Carleson ([30]).

Lemma 1.

Let $\omega =c_{1}z+c_2{z}^2+c_3{z}^3+···$ be a Schwarz function. Then

$$\left|c_2\right|\le 1-{\left|c_1\right|}^2,|c_3|\le 1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|},|c_4|\le 1-|c_1{|}^2-{|c_2|}^2.$$

2. Main Results

We begin with improvements in the upper bound of the third Hankel determinant for the class $\mathcal{M}$.

Theorem 1.

Let $(x_0,y_0)\approx (0.458445,0.631054)$ be the solution of the system of equations

$$\left\{\begin{array}{c}-30x^7+(12y-110)x^6+(-3y-132)x^5+(-28y^2-90y-34)x^4+(-70y^2-119y+40)x^3\hfill \\ +(-42y^2-27y+24)x^2+(-13y^3+8y^2+30y+2)x-8y^4-13y^3+8y^2+13y=0,\hfill \\ 12x^6-28x^5+(-112y-110)x^4+(-112y-36)x^3+(-129y^2+104)x^2\hfill \\ +(-102y^2-64y+88)x+64y^3+27y^2-64y+18=0.\hfill \end{array}\right.$$

If $f\in \mathcal{M}$, then

$$\begin{array}{ccc}\hfill \left|H_{3,1}\left(f\right)\right|& \le & \frac{1}{144}[-24x_0^5+(12y_0-20)x_0^4+(-52y_0+24)x_0^3+(-32y_0^2-18y_0+4)x_0^2\hfill \\ & & +(32y_0^2+52y_0)x_0+9y_0^3-32y_0^2+18y_0+16+\frac{16y_0^4}{{(1+x_0)}^2}-\frac{52x_0{y}_0^3}{1+x_0}-\frac{24x_0^3{y}_0^2}{1+x_0}]\hfill \\ & & \approx \frac{28.1080}{144}=0.195194\dots .\hfill \end{array}$$

Proof. 

For a function $f\in \mathcal{M}$, there exists a Schwarz function $\omega \left(z\right)$, such that

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=\frac{3}{2}-\frac{1+\omega \left(z\right)}{2(1-\omega (z\left)\right)},$$

i.e.,

$$z{f}^{\prime }\left(z\right)(1-\omega \left(z\right))=f\left(z\right)(1-2\omega \left(z\right)).$$

By comparing the coefficients in the above expression, we receive

$$a_2=-c_1,a_3=-\frac{c_2}{2},a_4=-\frac{2c_3+c_1{c}_2}{6},a_5=-\frac{2c_1^2{c}_2+4c_1{c}_3+3c_2^2+6c_4}{24}.$$

(3)

From (2) and (3), we achieve

$$\left|H_{3,1}\left(f\right)\right|=\frac{1}{144}\left|18c_2{c}_4-16c_3^2+36c_1^2{c}_4+12c_1^4{c}_2-4c_1^2{c}_2^2+27c_2^3+24c_1^3{c}_3-52c_1{c}_2{c}_3\right|.$$

(4)

By using triangle inequality and Lemma in (4), we come across

$$\begin{array}{ccc}\hfill 144\left|H_{3,1}\left(f\right)\right|& & \le 18|c_2||c_4|+16|c_3{|}^2+36|c_1{|}^2|c_4|+12|c_1{|}^4|c_2|+4|c_1{|}^2|c_2{|}^2+27{|c_2|}^3\hfill \\ & & +24|c_1{|}^3|c_3|+52|c_1||c_2||c_3|\hfill \\ & & \le 18|c_2|\left(1-|c_1{|}^2-{|c_2|}^2\right)+16{\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)}^2+36{|c_1|}^2\hfill \\ & & \left(1-|c_1{|}^2-{|c_2|}^2\right)+12|c_1{|}^4|c_2|+4|c_1{|}^2|c_2{|}^2+27{|c_2|}^3\hfill \\ & & +24|c_1{|}^3\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+51|c_1||c_2|\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)\hfill \\ & & =-24|c_1{|}^5+\left(12|c_2|-20\right)|c_1{|}^4+\left(-52|c_2|+24\right){|c_1|}^3+\left(-32|c_2{|}^2\right.\hfill \\ & & \left.-18|c_2|+4\right)|c_1{|}^2+\left(32|c_2{|}^2+52|c_2|\right)|c_1|+9|c_2{|}^3-32|c_2{|}^2+18|c_2|\hfill \\ & & +16+\frac{16|c_2{|}^4}{(1+|c_1{|)}^2}-\frac{52|c_1||c_2{|}^3}{1+|c_1|}-\frac{24|c_1{|}^3{|c_2|}^2}{1+|c_1|}.\hfill \end{array}$$

By putting $x=|c_1|$ and $y=|c_2|$ in above expression, we obtain

$$\begin{array}{ccc}\hfill 144|H_{3,1}\left(f\right)|& \le & -24x^5+(12y-20)x^4+(-52y+24)x^3+(-32y^2-18y+4)x^2\hfill \\ & & +(32y^2+52y)x+9y^3-32y^2+18y+16+\frac{16y^4}{{(1+x)}^2}-\frac{52x{y}^3}{1+x}-\frac{24x^3{y}^2}{1+x}\hfill \\ & & =F(x,y)\hfill \end{array}$$

We continue by finding the maximum of the function F on the region $\Omega =\{(x,y):0\le x\le 1,0\le y\le 1-x^2\}$. Differentiating F partially with respect to x and y, we obtain

$$\begin{array}{cc}\hfill \frac{\partial F}{\partial x}& =-120x^4+(48y-80)x^3+(-156y+72)x^2+(-64y^2-36y+8)x+32y^2+52y\hfill \\ & -\frac{32y^4}{{(1+x)}^3}-\frac{52y^3}{{(1+x)}^2}-\frac{24(3x^2+2x^3)y^2}{{(1+x)}^2}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \frac{\partial F}{\partial y}& =12x^4-52x^3+(-64y-18)x^2+(64y+52)x+27y^2-64y+18+\frac{64y^3}{{(1+x)}^2}\hfill \\ & -\frac{156x{y}^2}{1+x}-\frac{48x^{3}y}{1+x}.\hfill \end{array}$$

By putting $\frac{\partial F}{\partial x}=0,\frac{\partial F}{\partial y}=0$ and simplifying, we receive

$$\left\{\begin{array}{c}-30x^7+(12y-110)x^6+(-3y-132)x^5+(-28y^2-90y-34)x^4+(-70y^2-119y+40)x^3\hfill \\ +(-42y^2-27y+24)x^2+(-13y^3+8y^2+30y+2)x-8y^4-13y^3+8y^2+13y=0,\hfill \\ 12x^6-28x^5+(-112y-110)x^4+(-112y-36)x^3+(-129y^2+104)x^2\hfill \\ +(-102y^2-64y+88)x+64y^3+27y^2-64y+18=0.\hfill \end{array}\right.$$

Applying Newton’s methods to the above equations in Maple Software, we obtain

$$\begin{array}{c}\left\{\begin{array}{c}{x}_0\approx 0.458445,\hfill \\ y_0\approx 0.631054,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_1\approx -1,\hfill \\ y_1\approx 0,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_2\approx 0.358816,\hfill \\ y_2\approx 0.886184,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_3\approx 0.672790,\hfill \\ y_3\approx -1.280037,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_4\approx 40.019196,\hfill \\ y_4\approx 142.738372,\hfill \end{array}\right.\hfill \\ \left\{\begin{array}{c}{x}_5\approx -0.217838,\hfill \\ y_5\approx -1.300191.\hfill \end{array}\right.\left\{\begin{array}{c}{x}_6\approx -0.865959,\hfill \\ y_6\approx -0.198154.\hfill \end{array}\right.\left\{\begin{array}{c}{x}_7\approx -1.00406,\hfill \\ y_7\approx -0.051781.\hfill \end{array}\right.\left\{\begin{array}{c}{x}_8\approx -1.109198,\hfill \\ y_8\approx 0.031831.\hfill \end{array}\right.\hfill \end{array}$$

Thus, in $\Omega$, there is a critical point $\left(x_0,y_0\right)\approx \left(0.458445,0.631054\right)$ satisfying $y\le 1-x^2$, for which

$$F\left(x_0,y_0\right)\approx 28.1080\dots$$

Therefore, we continue studying F on the edges of $\Omega .$

For $x=0,$

$$F(0,y)=16y^4+9y^3-32y^2+18y+16\le F(0,1)=27.$$

For $y=0,$

$$F(x,0)=-24x^5-20x^4+24x^3+4x^2+16\le F(0.5759\dots ,0)\approx 18.1903\dots$$

On the edge $y=1-x^2$, $F(x,y)$ becomes

$$F(x,0)=-9x^6-27x^4+9x^2+27\le F(\frac{\sqrt{-9+6\sqrt{3}}}{3},0)=16\sqrt{3}\approx 27.7182.$$

Thus, we get

$$\begin{array}{cc}\hfill |H_{3,1}\left(f\right)|& \le \frac{28.1080}{144}\approx 0.195194\dots \hfill \end{array}$$

We complete the proof of Theorem 1. □

Remark 1.

The estimates of the third Hankel determinant $\left|H_{3,1}\left(f\right)\right|$ of Theorem 1 are more accurate than that obtained in ([18], Theorem 1).

Theorem 2.

Let $(x_0,y_0)\approx (0.417110,0.514879)$ is the approximate root of the system of equations

$$\left\{\begin{array}{c}-60x^7+(24y-212)x^6+(6y-240)x^5+(-59y^2-150y-52)x^4+(-145y^2-224y+64)x^3\hfill \\ +(-81y^2-72y+24)x^2+(-22y^3+25y^2+42y-4)x-20y^4-22y^3+20y^2+22y=0,\hfill \\ 6x^6-10x^5+(-59y-50)x^4+(-54y-24)x^3+(-64y^2+5y+44)x^2+(-62y^2-40y+46)x\hfill \\ +40y^3+2y^2-40y+12=0.\hfill \end{array}\right.$$

If $f\in \mathcal{N}$, then

$$\begin{array}{ccc}\hfill \left|H_{3,1}\left(f\right)\right|& \le & \frac{1}{8640}[-72x_0^5+(36y_0-48)x_0^4+(-132y_0+72)x_0^3+(-105y_0^2-72y_0-12)x_0^2\hfill \\ & & +(120y_0^2+132y_0)x_0+4y_0^3-120y_0^2+72y_0+60+\frac{60y_0^4}{{(1+x_0)}^2}-\frac{132x_0{y}_0^3}{1+x_0}-\frac{72x_0^3{y}_0^2}{1+x_0}]\hfill \\ & & \approx \frac{88.353}{8640}=0.01202\dots .\hfill \end{array}$$

Proof. 

Assume that $f\in \mathcal{N}$. From the definition, we know there is a Schwarz function $\omega$ such that

$$1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}=\frac{3}{2}-\frac{1+\omega \left(z\right)}{2(1-\omega (z\left)\right)},$$

i.e.,

$${\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }(1-\omega \left(z\right))=f^{\prime }\left(z\right)(1-2\omega \left(z\right)).$$

Using some easy computation, comparing the coefficients in the above expression, we receive

$$a_2=-\frac{c_1}{2},a_3=-\frac{c_2}{6},a_4=-\frac{2c_3+c_1{c}_2}{24},a_5=-\frac{2c_1^2{c}_2+4c_1{c}_3+3c_2^2+6c_4}{120}.$$

(5)

From (2) and (5), we achieve

$$\left|H_{3,1}\left(f\right)\right|=\frac{1}{8640}\left|36c_1^4{c}_2+72c_1^3{c}_3-132c_1{c}_2{c}_3+3c_1^2{c}_2^2+108c_1^2{c}_4+76c_2^3-60c_3^2+72c_2{c}_4\right|.$$

(6)

Applying the triangle inequality and Lemma in (6), we obtain

$$\begin{array}{ccc}\hfill 8640\left|H_{3,1}\left(f\right)\right|& & \le 36|c_1{|}^4|c_2|+72|c_1{|}^3|c_3|+132|c_1||c_2||c_3|+3|c_1{|}^2|c_2{|}^2+108|c_1{|}^2|c_4|+76|c_2{|}^3\hfill \\ & & +60|c_3{|}^2+72|c_2||c_4|\hfill \\ & & \le 36|c_1{|}^4|c_2|+72|c_1{|}^3\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+132|c_1||c_2|\hfill \\ & & \left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)+3|c_1{|}^2|c_2{|}^2+108|c_1{|}^2(1-|c_1{|}^2-|c_2{|}^2)\hfill \\ & & +76|c_2{|}^3+60{\left(1-|c_1{|}^2-\frac{|c_2{|}^2}{1+|c_1|}\right)}^2+72|c_2|\left(1-|c_1{|}^2-{|c_2|}^2\right)\hfill \\ & & =-72|c_1{|}^5+\left(36|c_2|-48\right)|c_1{|}^4+\left(-132|c_2|+72\right){|c_1|}^3+\left(-105|c_2{|}^2\right.\hfill \\ & & \left.-72|c_2|-12\right)|c_1{|}^2+\left(120|c_2{|}^2+132|c_2|\right)|c_1|+4|c_2{|}^3-120{|c_2|}^2\hfill \\ & & +72|c_2|+60+\frac{60|c_2{|}^4}{(1+|c_1{|)}^2}-\frac{132|c_1||c_2{|}^3}{1+|c_1|}-\frac{72|c_1{|}^3{|c_2|}^2}{1+|c_1|}.\hfill \end{array}$$

Putting $|c_1|=x$ and $|c_2|=y$ in above expression, we come across

$$\begin{array}{ccc}\hfill 9640\left|H_{3,1}\left(f\right)\right|& \le & -72x^5+(36y-48)x^4+(-132y+72)x^3+(-105y^2-72y-12)x^2\hfill \\ & & +(120y^2+132y)x+4y^3-120y^2+72y+60+\frac{60y^4}{{(1+x)}^2}-\frac{132x{y}^3}{1+x}-\frac{72x^3{y}^2}{1+x}\hfill \\ & & =\mathrm{{\rm Y}}(x,y)\hfill \end{array}$$

where $x\in [0,1]$ and $y\in [0,1-x^2].$

In order to caculate the maximum of the function $\mathrm{{\rm Y}}$ on the region $\Omega =\{(x,y):0\le x\le 1,0\le y\le 1-x^2\}$, we take the partial derivative with respect to x and y, respectively, and we receive

$$\begin{array}{cc}\hfill \frac{\partial \mathrm{{\rm Y}}}{\partial x}& =-360x^4+(144y-192)x^3+(-396y+216)x^2+(-210y^2-144y-24)x\\ & +120y^2+132y-\frac{120y^4}{{(1+x)}^3}-\frac{132y^3}{{(1+x)}^2}-\frac{72(3x^2+2x^3)y^2}{{(1+x)}^2}\end{array}$$

and

$$\begin{array}{cc}\hfill \frac{\partial \mathrm{{\rm Y}}}{\partial y}& =36x^4-132x^3+(-210y-72)x^2+(240y+132)x+12y^2-240y+72+\frac{240y^3}{{(1+x)}^2}\hfill \\ & -\frac{396x{y}^2}{1+x}-\frac{144x^{3}y}{1+x}.\hfill \end{array}$$

By putting $\frac{\partial \mathrm{{\rm Y}}}{\partial x}=0,\frac{\partial \mathrm{{\rm Y}}}{\partial y}=0$ and simplifying, we come across

$$\left\{\begin{array}{c}-60x^7+(24y-212)x^6+(6y-240)x^5+(-59y^2-150y-52)x^4+(-145y^2-224y+64)x^3\hfill \\ +(-81y^2-72y+24)x^2+(-22y^3+25y^2+42y-4)x-20y^4-22y^3+20y^2+22y=0,\hfill \\ 6x^6-10x^5+(-59y-50)x^4+(-54y-24)x^3+(-64y^2+5y+44)x^2+(-62y^2-40y+46)x\hfill \\ +40y^3+2y^2-40y+12=0.\hfill \end{array}\right.$$

By applying Newton’s methods to the above equations in Maple Software, we receive

$$\begin{array}{c}\left\{\begin{array}{c}{x}_0\approx 0.417110,\hfill \\ y_0\approx 0.514879,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_1\approx -1,\hfill \\ y_1\approx 0,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_2\approx 0.026119,\hfill \\ y_2\approx -1.150345,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_3\approx 0.213779,\hfill \\ y_3\approx 0.991641,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_4\approx 0..661308,\hfill \\ y_4\approx -1.149135,\hfill \end{array}\right.\hfill \\ \left\{\begin{array}{c}{x}_5\approx 40.425660,\hfill \\ y_5\approx 142.247907,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_6\approx -0.862110,\hfill \\ y_6\approx -0.189623,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_7\approx -1.001355,\hfill \\ y_7\approx -0.027737,\hfill \end{array}\right.\left\{\begin{array}{c}{x}_8\approx -1.103190,\hfill \\ y_8\approx 0.025415.\hfill \end{array}\right.\hfill \end{array}$$

Then, there is a critical point $(x_0,y_0)$ satisfying $y\le 1-x^2$ at which $\mathrm{{\rm Y}}(x,y)$ obtains its maximum. Thus, we have

$$\mathrm{{\rm Y}}(x,y)\le \mathrm{{\rm Y}}(x_0,y_0)\approx 88.353\dots$$

Therefore, we continue studying $\mathrm{{\rm Y}}$ on the edges of $\Omega .$

For $x=0,$

$$\mathrm{{\rm Y}}(0,y)=60y^4+4y^3-120y^2+72y+60\le \mathrm{{\rm Y}}(0,1)=76.$$

For $y=0,$

$$\mathrm{{\rm Y}}(x,0)=-72x^5-48x^4+72x^3-12x^2+60\le \mathrm{{\rm Y}}(0.4583\dots ,0)\approx 60.8369\dots$$

For $y=1-x^2$, $G(x,y)$ reduce

$$\mathrm{{\rm Y}}(x,0)=-25x^6-90x^4+39x^2+76\le \mathrm{{\rm Y}}(\frac{\sqrt{5}}{5},0)=80.$$

Thus, we obtain

$$\begin{array}{cc}\hfill \left|H_{3,1}\left(f\right)\right|& \le \frac{88.353}{8640}=0.01202\dots \hfill \end{array}$$

The proof of Theorem 2 is completed. □

Remark 2.

The bound of the third Hankel determinant $|H_{3,1}\left(f\right)|$ in Theorem 2 are more accurate than that in [[18], Theorem 5].

3. Conclusions

In this paper, a new method of finding the third Hankel determinant for close- to-convex functions was proposed. The bounds of the third Hankel determinant for the classes $\mathcal{M}$ and $\mathcal{N}$, derived with the new method, are better than those obtained by Zaprawa [15].

$$|H_{3,1}\left(f\right)|\le \frac{43}{144}for f\in \mathcal{M},$$

$$|H_{3,1}\left(f\right)|\le \frac{19}{2160}for f\in \mathcal{N}.$$

The advantage of the method is the possibility of calculating the bound of these functionals when the function coefficients are real.