Third-Order Hankel and Toeplitz Determinants for Starlike Functions Connected with the Sine Function

Abstract

Let ${\mathcal{S}}_s^{*}$ be the class of normalized functions f defined in the open unit disk $\mathbb{D}=\{z:|z|<1\}$ such that the quantity $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ lies in an eight-shaped region in the right-half plane and satisfying the condition $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec 1+sinz(z\in \mathbb{D}).$ In this paper, we aim to investigate the third-order Hankel determinant $H_3\left(1\right)$ and Toeplitz determinant $T_3\left(2\right)$ for this function class ${\mathcal{S}}_s^{*}$ associated with sine function and obtain the upper bounds of the determinants $H_3\left(1\right)$ and $T_3\left(2\right)$.

1. Introduction

Let $\mathcal{A}$ denote the class of functions f which are analytic in the open unit disk $\mathbb{D}=\{z:|z|<1\}$ of the form

$$f\left(z\right)=z+a_2{z}^2+a_3{z}^3+\cdots (z\in \mathbb{D})$$

(1)

and let $\mathcal{S}$ denote the subclass of $\mathcal{A}$ consisting of univalent functions.

Suppose that $\mathcal{P}$ denotes the class of analytic functions p normalized by

$$p\left(z\right)=1+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots$$

and satisfying the condition

$$\Re \left(p\right(z\left)\right)>0(z\in \mathbb{D}).$$

We easily see that, if $p\left(z\right)\in \mathcal{P}$, then a Schwarz function $\omega \left(z\right)$ exists with $\omega \left(0\right)=0$ and $\left|\omega \right(z\left)\right|<1$, such that (see [1])

$$p\left(z\right)=\frac{1+w\left(z\right)}{1-w\left(z\right)}(z\in \mathbb{D}).$$

Very recently, Cho et al. [2] introduced the following function class ${\mathcal{S}}_s^{*}$, which are associated with sine function:

$$S_s^{*}:=\left\{f\in \mathcal{A}:\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec 1+sinz(z\in \mathbb{D})\right\},$$

(2)

where “≺” stands for the subordination symbol (for details, see [3]) and also implies that the quantity $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ lies in an eight-shaped region in the right-half plane.

The $q^{th}$ Hankel determinant for $q\ge 1$ and $n\ge 1$ of functions f was stated by Noonan and Thomas [4] as

$$H_q\left(n\right)=\left|\begin{array}{cccc}{a}_n& a_{n+1}& \cdots & a_{n+q-1}\\ a_{n+1}& a_{n+2}& \cdots & a_{n+q}\\ ⋮& ⋮& & ⋮& \\ a_{n+q-1}& a_{n+q}& \cdots & a_{n+2q-2}\end{array}\right|(a_1=1).$$

This determinant has been considered by several authors, for example, Noor [5] determined the rate of growth of $H_q\left(n\right)$ as $n\to \infty$ for functions $f\left(z\right)$ given by Equation (1) with bounded boundary and Ehrenborg [6] studied the Hankel determinant of exponential polynomials.

In particular, we have

$$H_3\left(1\right)=\left|\begin{array}{c}{a}_1{a}_2{a}_3\hfill \\ a_2{a}_3{a}_4\hfill \\ a_3{a}_4{a}_5\hfill \end{array}\right|(n=1,q=3).$$

Since $f\in \mathcal{S},a_1=1,$

$$H_3\left(1\right)=a_3(a_2{a}_4-a_3^2)-a_4(a_4-a_2{a}_3)+a_5(a_3-a_2^2).$$

We note that $|H_2\left(1\right)|=|a_3-a_2^2|$ is the well-known Fekete-Szego functional (see, for example, [7,8,9]).

On the other hand, Thomas and Halim [10] defined the symmetric Toeplitz determinant $T_q\left(n\right)$ as follows:

$$T_q\left(n\right)=\left|\begin{array}{cccc}{a}_n& a_{n+1}& \cdots & a_{n+q-1}\\ a_{n+1}& a_n& \cdots & a_{n+q}\\ ⋮& ⋮& & ⋮\\ a_{n+q-1}& a_{n+q}& \cdots & a_n\end{array}\right|(n\ge 1,q\ge 1).$$

The Toeplitz determinants are closely related to Hankel determinants. Hankel matrices have constant entries along the reverse diagonal, whereas Toeplitz matrices have constant entries along the diagonal. For a good summary of the applications of Toeplitz matrices to the wide range of areas of pure and applied mathematics, we can refer to [11].

As a special case, when $n=2$ and $q=3$, we have

$$T_3\left(2\right)=\left|\begin{array}{c}{a}_2{a}_3{a}_4\hfill \\ a_3{a}_2{a}_3\hfill \\ a_4{a}_3{a}_2\hfill \end{array}\right|.$$

In recent years, many authors studied the second-order Hankel determinant $H_2\left(2\right)$ and the third-order Hankel determinant $H_3\left(1\right)$ for various classes of functions (the interested readers can see, for instance, [12,13,14,15,16,17,18,19,20,21,22,23,24,25]). However, apart from the work in [10,21,26,27], there appears to be little literature dealing with Toeplitz determinants. Inspired by the aforementioned works, in this paper, we aim to investigate the third-order Hankel determinant $H_3\left(1\right)$ and Toeplitz determinant $T_3\left(2\right)$ for the above function class ${\mathcal{S}}_s^{*}$ associated with sine function, and obtain the upper bounds of the above determinants.

2. Main Results

To prove our desired results, we need the following lemmas.

Lemma 1.

If $p\left(z\right)\in \mathcal{P}$, then exists some $x,z$ with $\left|x\right|\le 1$ (see [28]), $\left|z\right|\le 1$, such that

$$2c_2=c_1^2+x(4-c_1^2),$$

$$4c_3=c_1^3+2c_{1}x(4-c_1^2)-(4-c_1^2)c_1{x}^2+2(4-c_1^2){(1-|x|}^2)z.$$

Lemma 2.

Let $p\left(z\right)\in \mathcal{P}$ (see [29]), then

$$|c_n|\le 2,n=1,2,\cdots .$$

We now state and prove the main results of our present investigation.

Theorem 1.

If the function $f\left(z\right)\in {\mathcal{S}}_s^{*}$ and of the form Equation (1), then

$$|a_2|\le 1,|a_3|\le \frac{1}{2},|a_4|\le \frac{5}{9},\left|a_5\right|\le \frac{47}{72}.$$

(3)

Proof. 

Since $f\left(z\right)\in {\mathcal{S}}_s^{*}$, according to subordination relationship, so there exists a Schwarz function $\omega \left(z\right)$ with $\omega \left(0\right)=0$ and $\left|\omega \right(z\left)\right|<1$, such that

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}=1+sin\left(\omega \left(z\right)\right).$$

Now,

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

$$=(1+\sum _{n=2}^{\infty }n{a}_n{z}^{n-1})[1-a_{2}z+(a_2^2-a_3)z^2-(a_2^3-2a_2{a}_3+a_4)z^3$$

$$+(a_2^4-3a_2^2{a}_3+2a_2{a}_4+a_3^2-a_5)z^4+\cdots ]$$

$$=1+a_{2}z+(2a_3-a_2^2)z^2+(a_2^3-3a_2{a}_3+3a_4)z^3$$

$$+(4a_5-a_2^4+4a_2^2{a}_3-4a_2{a}_4-2a_3^2)z^4+\cdots .$$

(4)

Define a function

$$p\left(z\right)=\frac{1+\omega \left(z\right)}{1-\omega \left(z\right)}=1+c_{1}z+c_2{z}^2+\cdots .$$

Clearly, we have $p\left(z\right)\in \mathcal{P}$ and

$$\omega \left(z\right)=\frac{p\left(z\right)-1}{1+p\left(z\right)}=\frac{c_{1}z+c_2{z}^2+c_3{z}^3+\cdots }{2+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots }.$$

(5)

On the other hand,

$$1+sin\left(\omega \left(z\right)\right)=1+\frac{1}{2}{c}_{1}z+(\frac{c_2}{2}-\frac{c_1^2}{4})z^2+(\frac{5c_1^3}{48}+\frac{c_3-c_1{c}_2}{2})z^3$$

$$+(\frac{c_4}{2}+\frac{5c_1^2{c}_2}{16}-\frac{c_2^2}{4}-\frac{c_1{c}_3}{2}-\frac{c_1^4}{32})z^4+\cdots .$$

(6)

Comparing the coefficients of $z,z^2,z^3,z^4$ between Equations (4) and (6), we obtain

$$a_2=\frac{c_1}{2},a_3=\frac{c_2}{4},a_4=\frac{c_3}{6}-\frac{c_1{c}_2}{24}-\frac{c_1^3}{144},a_5=\frac{c_4}{8}-\frac{c_1{c}_3}{24}+\frac{5c_1^4}{1152}-\frac{c_1^2{c}_2}{192}-\frac{c_2^2}{32}.$$

(7)

By using Lemma 2, we thus know that

$$|a_2|\le 1,|a_3|\le \frac{1}{2},|a_4|\le \frac{5}{9},\left|a_5\right|\le \frac{47}{72}.$$

The proof of Theorem 1 is completed.  □

Theorem 2.

If the function $f\left(z\right)\in {\mathcal{S}}_s^{*}$ and of the form in Equation (1), then we have

$$|a_3-a_2^2|\le \frac{1}{2}.$$

(8)

Proof. 

According to Equation (7), we have

$$|a_3-a_2^2|=\left|\frac{c_2}{4}-\frac{c_1^2}{4}\right|.$$

By applying Lemma 1, we get

$$|a_3-a_2^2|=\left|\frac{x(4-c_1^2)}{8}-\frac{c_1^2}{8}\right|.$$

Let $\left|x\right|=t,t\in [0,1],c_1=c,c\in [0,2].$ Then, using the triangle inequality, we obtain

$$|a_3-a_2^2|\le \frac{t(4-c^2)}{8}+\frac{c^2}{8}.$$

Suppose that

$$F(c,t)=\frac{t(4-c^2)}{8}+\frac{c^2}{8},$$

then $\forall t\in (0,1),\forall c\in (0,2),$

$$\frac{\partial F}{\partial t}=\frac{4-c^2}{8}>0,$$

which shows that $F(c,t)$ is an increasing function on the closed interval [0,1] about t. Therefore, the function $F(c,t)$ can get the maximum value at $t=1$, that is, that

$$maxF(c,t)=F(c,1)=\frac{(4-c^2)}{8}+\frac{c^2}{8}=\frac{1}{2}.$$

Thus, obviously,

$$|a_3-a_2^2|\le \frac{1}{2}.$$

The proof of Theorem 2 is thus completed.  □

Theorem 3.

If the function $f\left(z\right)\in {\mathcal{S}}_s^{*}$ and of the form in Equation (1), then we have

$$|a_2{a}_3-a_4|\le \frac{1}{3}.$$

(9)

Proof. 

From Equation (7), we have

$$\begin{array}{c}|a_2{a}_3-a_4|=|\frac{c_1{c}_2}{8}+\frac{c_1^3}{144}-\frac{c_3}{6}+\frac{c_1{c}_2}{24}|\\ =|\frac{c_1{c}_2}{6}-\frac{c_3}{6}+\frac{c_1^3}{144}|.\end{array}$$

Now, in view of Lemma 1, we get

$$|a_2{a}_3-a_4|=\left|\frac{7c_1^3}{144}+\frac{(4-c_1^2)c_1{x}^2}{24}-\frac{(4-c_1^2){(1-|x|}^2)z}{12}\right|.$$

Let $\left|x\right|=t,t\in [0,1],c_1=c,c\in [0,2].$ Then, using the triangle inequality, we deduce that

$$|a_2{a}_3-a_4|\le \frac{7c^3}{144}+\frac{(4-c^2)c{t}^2}{24}+\frac{(4-c^2)(1-t^2)}{12}.$$

Assume that

$$F(c,t)=\frac{7c^3}{144}+\frac{(4-c^2)c{t}^2}{24}+\frac{(4-c^2)(1-t^2)}{12}.$$

Therefore, we have, $\forall t\in (0,1),\forall c\in (0,2)$

$$\frac{\partial F}{\partial t}=\frac{(4-c^2)t(c-2)}{12}<0,$$

namely, $F(c,t)$ is an decreasing function on the closed interval [0,1] about t. This implies that the maximum value of $F(c,t)$ occurs at $t=0,$ which is

$$maxF(c,t)=F(c,0)=\frac{(4-c^2)}{12}+\frac{7c^3}{144}.$$

Define

$$G\left(c\right)=\frac{(4-c^2)}{12}+\frac{7c^3}{144},$$

clearly, the function $G\left(c\right)$ has a maximum value attained at $c=0,$ also which is

$$|a_2{a}_3-a_4|\le G\left(0\right)=\frac{1}{3}.$$

The proof of Theorem 3 is completed.  □

Theorem 4.

If the function $f\left(z\right)\in {\mathcal{S}}_s^{*}$ and of the form in Equation (1), then we have

$$|a_2{a}_4-a_3^2|\le \frac{1}{4}.$$

(10)

Proof. 

Suppose that $f\left(z\right)\in {\mathcal{S}}_s^{*}$, then from Equation (7), we have

$$\begin{array}{c}|a_2{a}_4-a_3^2|=\left|\frac{c_1{c}_3}{12}-\frac{c_1^2{c}_2}{48}+\frac{c_1^4}{48}-\frac{c_2^2}{16}\right|.\hfill \end{array}$$

Now, in terms of Lemma 1, we obtain

$$\begin{array}{c}|a_2{a}_4-a_3^2|=\left|\frac{c_1{c}_3}{12}-\frac{c_1^2{c}_2}{48}-\frac{c_1^4}{288}-\frac{c_2^2}{16}\right|\hfill \\ =\left|-\frac{5c_1^4}{576}-\frac{x^2{c}_1^2(4-c_1^2)}{48}-\frac{x^2{(4-c_1^2)}^2}{64}+\frac{c_1(4-c_1^2){(1-|x|}^2)z}{24}\right|.\hfill \end{array}$$

Let $\left|x\right|=t,t\in [0,1],c_1=c,c\in [0,2].$ Then, using the triangle inequality, we get

$$|a_2{a}_4-a_3^2|\le \frac{t^2{c}^2(4-c^2)}{48}+\frac{(1-t^2)c(4-c^2)}{24}+\frac{t^2{(4-c^2)}^2}{64}+\frac{5c^4}{576}.$$

Putting

$$F(c,t)=\frac{t^2{c}^2(4-c^2)}{48}+\frac{(1-t^2)c(4-c^2)}{24}+\frac{t^2{(4-c^2)}^2}{64}+\frac{5c^4}{576},$$

then, $\forall t\in (0,1),\forall c\in (0,2),$ we have

$$\frac{\partial F}{\partial t}=\frac{t(c^2-8c+12)(4-c^2)}{96}>0,$$

which implies that $F(c,t)$ increases on the closed interval [0,1] about t. That is, that $F(c,t)$ have a maximum value at $t=1,$ which is

$$maxF(c,t)=F(c,1)=\frac{c^2(4-c^2)}{48}+\frac{{(4-c^2)}^2}{64}+\frac{5c^4}{576}.$$

Setting

$$G\left(c\right)=\frac{c^2(4-c^2)}{48}+\frac{{(4-c^2)}^2}{64}+\frac{5c^4}{576},$$

then we have

$$G^{\prime }\left(c\right)=\frac{c(4-c^2)}{24}-\frac{c^3}{24}-\frac{c(4-c^2)}{16}+\frac{5c^3}{144}.$$

If $G^{\prime }\left(c\right)=0,$ then the root is $c=0.$ In addition, since $G^{″}\left(0\right)=-\frac{1}{12}<0,$ so the function $G\left(c\right)$ can take the maximum value at $c=0,$ which is

$$|a_2{a}_4-a_3^2|\le G\left(0\right)=\frac{1}{4}.$$

The proof of Theorem 4 is completed.  □

Theorem 5.

If the function $f\left(z\right)\in {\mathcal{S}}_s^{*}$ and of the form in Equation (1), then we have

$$|a_2^2-a_3^2|\le \frac{5}{4}.$$

(11)

Proof. 

Suppose that $f\left(z\right)\in {\mathcal{S}}_s^{*}$, then, by using Equation (7), we have

$$\begin{array}{c}|a_2^2-a_3^2|=|\frac{c_1^2}{4}-\frac{c_2^2}{16}|.\hfill \end{array}$$

Next, according to Lemma 1, we obtain

$$\begin{array}{c}|a_2^2-a_3^2|=\left|\frac{c_1^2}{4}-\frac{c_2^2}{16}\right|\hfill \\ =\left|\frac{c_1^2}{4}-\frac{c_1^4}{64}-\frac{x{c}_1^2(4-c_1^2)}{32}-\frac{x^2{(4-c_1^2)}^2}{64}\right|.\hfill \end{array}$$

Let $\left|x\right|=t,t\in [0,1],c_1=c,c\in [0,2].$ Then, by applying the triangle inequality, we get

$$|a_2^2-a_3^2|\le \frac{c^2}{4}+\frac{c^4}{64}+\frac{t{c}^2(4-c^2)}{32}+\frac{t^2{(4-c^2)}^2}{64}.$$

Taking

$$F(c,t)=\frac{c^2}{4}+\frac{c^4}{64}+\frac{t{c}^2(4-c^2)}{32}+\frac{t^2{(4-c^2)}^2}{64}.$$

Then, $\forall t\in (0,1),\forall c\in (0,2),$ we have

$$\frac{\partial F}{\partial t}=\frac{c^2(4-c^2)}{32}+\frac{t{(4-c^2)}^2}{32}>0,$$

which implies that $F(c,t)$ increases on the closed interval [0,1] about t. Namely, the maximum value of $F(c,t)$ attains at $t=1,$ which is

$$maxF(c,t)=F(c,1)=\frac{c^2}{4}+\frac{c^4}{64}+\frac{c^2(4-c^2)}{32}+\frac{{(4-c^2)}^2}{64}.$$

Let

$$G\left(c\right)=\frac{c^2}{4}+\frac{c^4}{64}+\frac{c^2(4-c^2)}{32}+\frac{{(4-c^2)}^2}{64},$$

then

$$\begin{array}{c}{G}^{\prime }\left(c\right)=\frac{c}{2}>0,\forall c\in (0,2).\hfill \end{array}$$

Therefore, the function $G\left(c\right)$ is an increasing function on the closed interval [0,2] about c, and thus $G\left(c\right)$ has a maximum value attained at $c=2,$ which is

$$|a_2^2-a_3^2|\le G\left(2\right)=\frac{5}{4}.$$

The proof of Theorem 5 is completed.  □

Theorem 6.

If the function $f\left(z\right)\in {\mathcal{S}}_s^{*}$ and of the form in Equation (1), then we have

$$|a_2{a}_3-a_3{a}_4|\le \frac{13}{12}.$$

(12)

Proof. 

Assume that $f\left(z\right)\in {\mathcal{S}}_s^{*}$, then from Equation (7), we obtain

$$\begin{array}{c}|a_2{a}_3-a_3{a}_4|=|\frac{c_1{c}_2}{8}+\frac{c_1^3{c}_2}{576}-\frac{c_2{c}_3}{24}+\frac{c_1{c}_2^2}{96}|.\hfill \end{array}$$

Now, by using Lemma 1, we see that

$$\begin{array}{c}|a_2{a}_3-a_3{a}_4|=\left|\frac{c_1{c}_2}{8}+\frac{c_1^3{c}_2}{576}-\frac{c_2{c}_3}{24}+\frac{c_1{c}_2^2}{96}\right|\hfill \\ =\left|\frac{c_1^3}{16}-\frac{c_1^5}{576}-\frac{11x{c}_1^3(4-c_1^2)}{1152}+\frac{x{c}_1(4-c_1^2)}{16}+\frac{x^2{c}_1(4-c_1^2)[c_1^2+x(4-c_1^2)]}{192}+\frac{c_1{x}^2{(4-c_1^2)}^2}{128}+\frac{{(1-|x|}^2)z(4-c_1^2)[x(4-c_1^2)+c_1^2]}{96}\right|.\hfill \end{array}$$

If we let $\left|x\right|=t,t\in [0,1],c_1=c,c\in [0,2],$ then, using the triangle inequality, we have

$$|a_2{a}_3-a_3{a}_4|\le \frac{c^3}{16}+\frac{c^5}{576}+\frac{11t{c}^3(4-c^2)}{1152}+\frac{t(4-c^2)}{8}+\frac{t^2[c^2+t(4-c^2)](4-c^2)}{96}+\frac{t^2{(4-c^2)}^2}{64}+\frac{(4-c^2)[t(4-c^2)+c^2]}{96}.$$

Setting

$$F(c,t)=\frac{c^3}{16}+\frac{c^5}{576}+\frac{11t{c}^3(4-c^2)}{1152}+\frac{t(4-c^2)}{8}+\frac{t^2[c^2+t(4-c^2)](4-c^2)}{96}+\frac{t^2{(4-c^2)}^2}{64}+\frac{(4-c^2)[t(4-c^2)+c^2]}{96}.$$

Then, we easily see that, $\forall t\in (0,1),\forall c\in (0,2),$

$$\frac{\partial F}{\partial t}=\frac{11c^3(4-c^2)}{1152}+\frac{(4-c^2)}{8}+\frac{t[c^2+t(4-c^2)](4-c^2)}{48}+\frac{t^2{(4-c^2)}^2}{96}+\frac{t{(4-c^2)}^2}{32}+\frac{{(4-c^2)}^2}{96}>0,$$

which implies that $F(c,t)$ is an increasing function on the closed interval [0,1] about t. That is, that the maximum value of $F(c,t)$ occurs at $t=1,$ which is

$$maxF(c,t)=F(c,1)=\frac{c^3}{16}+\frac{c^5}{576}+\frac{11c^3(4-c^2)}{1152}+\frac{(4-c^2)}{8}+\frac{(4-c^2)}{24}+\frac{{(4-c^2)}^2}{64}+\frac{(4-c^2)}{24}.$$

Taking

$$G\left(c\right)={\displaystyle \frac{c^3}{16}}+{\displaystyle \frac{c^5}{576}}+{\displaystyle \frac{11c^3(4-c^2)}{1152}}+{\displaystyle \frac{(4-c^2)}{8}}+{\displaystyle \frac{(4-c^2)}{24}}+{\displaystyle \frac{{(4-c^2)}^2}{64}}+{\displaystyle \frac{(4-c^2)}{24}},$$

then

$$\begin{array}{c}{G}^{\prime }\left(c\right)={\displaystyle \frac{3c^2}{16}}+{\displaystyle \frac{5c^4}{576}}+{\displaystyle \frac{11c^2(4-c^2)}{384}}-{\displaystyle \frac{11c^4}{576}}-{\displaystyle \frac{c(4-c^2)}{16}}-{\displaystyle \frac{c}{12}},\hfill \\ G^{″}\left(c\right)={\displaystyle \frac{3c}{8}}+{\displaystyle \frac{5c^3}{144}}+{\displaystyle \frac{11c(4-2c^2)}{192}}-{\displaystyle \frac{11c^3}{144}}-{\displaystyle \frac{(4-c^2)}{16}}+{\displaystyle \frac{c^2}{8}}-{\displaystyle \frac{1}{12}}.\hfill \end{array}$$

We easily find that $c=0$ is the root of the function $G^{\prime }\left(c\right)=0$, since $G^{″}\left(0\right)<0$, which implies that the function $G\left(c\right)$ can reach the maximum value at $c=0,$ also which is

$$|a_2{a}_3-a_3{a}_4|\le G\left(0\right)=\frac{13}{12}.$$

The proof of Theorem 6 is completed.  □

Theorem 7.

If the function $f\left(z\right)\in {\mathcal{S}}_s^{*}$ and of the form in Equation (1), then we have

$$|H_3\left(1\right)|\le \frac{275}{432}\approx 0.637.$$

(13)

Proof. 

Since

$$H_3\left(1\right)=a_3(a_2{a}_4-a_3^2)-a_4(a_4-a_2{a}_3)+a_5(a_3-a_2^2),$$

by applying the triangle inequality, we get

$$|H_3\left(1\right)|\le |a_3\left|\right|a_2{a}_4-a_3^2|+|a_4\left|\right|a_4-a_2{a}_3|+|a_5\left|\right|a_3-a_2^2|.$$

(14)

Now, substituting Equations (3), (8), (9) and (10) into Equation (14), we easily obtain the desired assertion (Equation (13)).  □

Theorem 8.

If the function $f\left(z\right)\in {\mathcal{S}}_s^{*}$ and of the form in Equation (1), then we have

$$|T_3\left(2\right)|\le \frac{139}{72}\approx 1.931.$$

(15)

Proof. 

Because

$$T_3\left(2\right)=a_2(a_2^2-a_3^2)-a_3(a_2{a}_3-a_3{a}_4)+a_4(a_3^2-a_2{a}_4),$$

by using the triangle inequality, we obtain

$$|T_3\left(2\right)|\le |a_2\left|\right|a_2^2-a_3^2|+|a_3\left|\right|a_2{a}_3-a_3{a}_4|+|a_4\left|\right|a_3^2-a_2{a}_4|.$$

(16)

Next, from Equations (3), (10), (11) and (12), we immediately get the desired assertion (Equation (15)).  □

Finally, we give two examples to illustrate our results obtained.

Example 1.

If we take the function $f\left(z\right)=e^z-1=z+{\sum }_{n=2}^{\infty }\frac{z^n}{n!}\in {\mathcal{S}}_s^{*}$, then we obtain

$$\begin{array}{cc}\hfill |H_3\left(1\right)|\le & |a_3\left|\right|a_2{a}_4-a_3^2|+|a_4\left|\right|a_4-a_2{a}_3|+|a_5\left|\right|a_3-a_2^2|\hfill \\ \hfill =\frac{1}{3!}\times & |\frac{1}{2!}\times \frac{1}{4!}-\frac{1}{3!}\times \frac{1}{3!}|+\frac{1}{4!}\times |\frac{1}{4!}-\frac{1}{2!}\times \frac{1}{3!}|+\frac{1}{5!}\times |\frac{1}{3!}-\frac{1}{2!}\times \frac{1}{2!}|\hfill \\ & \approx 0.004<0.637.\hfill \end{array}$$

Example 2.

If we set the function $f\left(z\right)=-log(1-z)=z+{\sum }_{n=2}^{\infty }\frac{z^n}{n}\in {\mathcal{S}}_s^{*}$, then we get

$$\begin{array}{cc}\hfill |T_3\left(2\right)|\le & |a_2\left|\right|a_2^2-a_3^2|+|a_3\left|\right|a_2{a}_3-a_3{a}_4|+|a_4\left|\right|a_3^2-a_2{a}_4|\hfill \\ \hfill =\frac{1}{2}\times & |\frac{1}{2}\times \frac{1}{2}-\frac{1}{3}\times \frac{1}{3}|+\frac{1}{3}\times |\frac{1}{2}\times \frac{1}{3}-\frac{1}{3}\times \frac{1}{4}|+\frac{1}{4}\times |\frac{1}{3}\times \frac{1}{3}-\frac{1}{2}\times \frac{1}{4}|\hfill \\ & \approx 0.107<1.931.\hfill \end{array}$$