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

Let S∗ s be the class of normalized functions f defined in the open unit disk D = {z : |z| < 1} such that the quantity z f ′(z) f (z) lies in an eight-shaped region in the right-half plane and satisfying the condition z f ′(z) f (z) ≺ 1 + sin z (z ∈ D). In this paper, we aim to investigate the third-order Hankel determinant H3(1) and Toeplitz determinant T3(2) for this function class S∗ s associated with sine function and obtain the upper bounds of the determinants H3(1) and T3(2).


Introduction
Let A denote the class of functions f which are analytic in the open unit disk D = {z : |z| < 1} of the form and let S denote the subclass of A consisting of univalent functions.Suppose that P denotes the class of analytic functions p normalized by and satisfying the condition (p(z)) > 0 (z ∈ D).
We easily see that, if p(z) ∈ P, then a Schwarz function ω(z) exists with ω(0) = 0 and |ω(z)| < 1, such that (see [1]) Very recently, Cho et al. [2] introduced the following function class S * s , which are associated with sine function: where "≺" stands for the subordination symbol (for details, see [3]) and also implies that the quantity z f (z) f (z) lies in an eight-shaped region in the right-half plane.
Mathematics 2019, 7, 404; doi:10.3390/math7050404www.mdpi.com/journal/mathematics The q th Hankel determinant for q ≥ 1 and n ≥ 1 of functions f was stated by Noonan and Thomas [4] as This determinant has been considered by several authors, for example, Noor [5] determined the rate of growth of H q (n) as n → ∞ for functions f (z) given by Equation (1) with bounded boundary and Ehrenborg [6] studied the Hankel determinant of exponential polynomials.
On the other hand, Thomas and Halim [10] defined the symmetric Toeplitz determinant T q (n) as follows: 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].

Main Results
To prove our desired results, we need the following lemmas.Lemma 1.If p(z) ∈ P, then exists some x, z with |x| ≤ 1(see [28]), |z| ≤ 1, such that , Lemma 2. Let p(z) ∈ P (see [29]), then We now state and prove the main results of our present investigation.
Theorem 1.If the function f (z) ∈ S * s and of the form Equation ( 1), then Proof.Since f (z) ∈ S * s , according to subordination relationship, so there exists a Schwarz function Define a function Clearly, we have p(z) ∈ P and On the other hand, Comparing the coefficients of z, z 2 , z 3 , z 4 between Equations ( 4) and ( 6), we obtain By using Lemma 2, we thus know that The proof of Theorem 1 is completed.
Theorem 2. If the function f (z) ∈ S * s and of the form in Equation ( 1), then we have Proof.According to Equation ( 7), we have By applying Lemma 1, we get Then, using the triangle inequality, we obtain Suppose that then ∀t ∈ (0, 1), ∀c ∈ (0, 2), ∂F ∂t 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 Thus, obviously, The proof of Theorem 2 is thus completed.
Theorem 3. If the function f (z) ∈ S * s and of the form in Equation ( 1), then we have Proof.From Equation ( 7), we have Now, in view of Lemma 1, we get Then, using the triangle inequality, we deduce that Assume that Therefore, we ∀t ∈ (0, 1), ∀c ∈ (0, 2) ) 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 clearly, the function G(c) has a maximum value attained at c = 0, also which is The proof of Theorem 3 is completed.
Theorem 4. If the function f (z) ∈ S * s and of the form in Equation ( 1), then we have Proof.Suppose that f (z) ∈ S * s , then from Equation ( 7), we have Now, in terms of Lemma 1, we obtain Then, the triangle inequality, we get Putting then, ∀t ∈ (0, 1), ∀c ∈ (0, 2), we have Setting then we have If G (c) = 0, then the root is c = 0.In addition, since G (0) = − 1 12 < 0, so the function G(c) can take the maximum value at c = 0, which is The proof of Theorem 4 is completed.
Theorem 5.If the function f (z) ∈ S * s and of the form in Equation ( 1), then we have Proof.Suppose that f (z) ∈ S * s , then, by using Equation ( 7), we have Next, according to Lemma 1, we obtain . Then, by applying the triangle inequality, we get Taking Then, ∀t ∈ (0, 1), ∀c ∈ (0, 2), we have which implies that F(c, t) increases on the closed interval [0,1] about t.Namely, the maximum value of F(c, t) Let . Therefore, the function G(c) is an increasing function on the closed interval [0,2] about c, and thus G(c) has a maximum value attained at c = 2, which is The proof of Theorem 5 is completed.
Proof.Assume that f (z) ∈ S * s , then from Equation (7), we obtain Now, by using Lemma 1, we see that , then, using the triangle inequality, we have
Then, we easily see that, ∀t ∈ (0, 1), ∀c ∈ (0, 2), 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 Taking We easily find that c = 0 is the root of the function G (c) = 0, sinceG (0) < 0, which implies that the function G(c) can reach the maximum value at c = 0, also which is The proof of Theorem 6 is completed.
Finally, we give two examples to illustrate our results obtained.

Theorem 6 .
If the function f (z) ∈ S * s and of the form in Equation (1), then we have |a 2 a 3 − a 3 a 4 | ≤ 13 12 .