Sandwich Theorems for a New Class of Complete Homogeneous Symmetric Functions by Using Cyclic Operator

Abstract

In this paper, we discuss and introduce a new study on the connection between geometric function theory, especially sandwich theorems, and Viete’s theorem in elementary algebra. We obtain some conclusions for differential subordination and superordination for a new formula of complete homogeneous symmetric functions class involving an ordered cyclic operator. In addition, certain sandwich theorems are found.

1. Introduction

Let $H\left(\mathbb{U}\right)$ denote the analytic function class in the open unit disk $\mathbb{U}=\{\left|z\right|<1;z \in ℂ\}$. and let $H\left[a, p\right]$ denotes the subclass of functions $f\in H\left(\mathbb{U}\right)$ as:

$$H\left[a,p\right]=\left\{f\in H:f\left(z\right)=a+a_p{z}^p+a_{p+1}{z}^{p+1}+\cdots \right\};a \in ℂ, p \in N=\left\{1, 2,\dots \right\}.$$

The resolvent $det {\left(I — A\right)}^{-1}$ of a complex matrix A is naturally an analytic function of eigenvalues $\lambda \in ℂ,$ and these eigenvalues are isolated singularities. In general, any matrix has finite eigenvalues. The resolvent set of A is defined as follows:

$$\varrho \left(A\right)=\left\{\lambda \in ℂ:\lambda I-A \mathrm{is} \mathrm{invertible}\right\}$$

The spectrum of A is expressed by $\vartheta (A)=ℂ/\varrho (A)$.

For distinct eigenvalues ${\lambda }_1,{\lambda }_2 ,\dots ,{\lambda }_m$, the polynomial

$$H\left(z\right)=h_0+h_{1}z+h_2{z}^2+\cdots +h_{\mathfrak{n}}{z}^{\mathfrak{n}}+\cdots$$

are given in terms of the eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m$ by

$$H\left(z\right)={\displaystyle \sum }_{\mathfrak{n}=0}^{\infty }{h}_{\mathfrak{n}} z^n,h_{\mathfrak{n}}={\displaystyle \sum }_{1\le {𝒾}_1,\dots ,{𝒾}_m\le 𝓂}\frac{{𝓂}_1!{𝓂}_2!\dots {𝓂}_{𝓂}!}{\mathfrak{n}!}{\lambda }_{{𝒾}_1}{\lambda }_{{𝒾}_2} \dots {\lambda }_{{𝒾}_{𝓂}} ;z \in ℂ.$$

The above formula can also be written in terms of the distinct powers of traces of the matrix as follows:

$$H\left(z\right)={\displaystyle \sum }_{\mathfrak{n}=0}^{\infty }{h}_{\mathfrak{n}} z^{\mathfrak{n}};h_{\mathfrak{n}}={\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2+\cdots +{\mathfrak{i}}_{\mathfrak{n}}=k\\ {\mathfrak{i}}_1+2{\mathfrak{i}}_2+3{\mathfrak{i}}_3+\cdots +\mathfrak{n}{\mathfrak{i}}_{\mathfrak{n}}=\mathfrak{n}\end{array}}\frac{{\left(-1\right)}^{n-k} S_1^{{\mathfrak{i}}_1}{S}_2^{{\mathfrak{i}}_2}\dots S_{\mathfrak{n}}^{{\mathfrak{i}}_{\mathfrak{n}}}}{{\mathfrak{i}}_1! {\mathfrak{i}}_2!\dots {\mathfrak{i}}_n! 1^{{\mathfrak{i}}_1} 2^{{\mathfrak{i}}_2}\dots {\mathfrak{n}}^{{\mathfrak{i}}_{\mathfrak{n}}} }$$

(1)

This class represents the subclasses of analytical functions $H\left[a,n\right]$ and denoted by $ℍ$ such that $H\left[a,1\right]=ℍ$ and has coefficients of the form (1), i.e., when the value of $n$ is equal to one and can be reduced a class $ℍ$ to “the class $H$ of normalized univalent analytical functions and composed of functions of the following form:

$$\frac{H\left(z\right)-h_0}{h_1}=z+{\displaystyle \sum }_{\mathfrak{n}=2}^{\infty }{a}_{\mathfrak{n}}{z}^{\mathfrak{n}};a_n=\frac{h_{\mathfrak{i}}}{h_1}, \left(z \in \mathbb{U}\right)$$

To each analytic function φ in open unit disk $\mathbb{U}$ into itself, we associated the composition operator $C_{\phi }$, defined by:

$$C_{\phi }f=f\circ \phi \mathrm{for} \mathrm{all} f\in H^2$$

Then, we define the ordered cyclic operator ${\mathbb{T}}^{𝕚}$ of $f$ as follows:

$$\left(C_{{\mathcal{T}}^{𝕚}}f\right)\left(z\right)=\left({\mathcal{T}}^{𝕚}\circ f\right)\left(z\right)={\mathcal{T}}^{𝕚}\left(f\left(z\right)\right)={\mathcal{T}}^{𝕚}\left(h_0+{\displaystyle \sum }_{n=1}^{\infty }{h}_n{z}^n\right)=h_{𝕚}+{\displaystyle \sum }_{n=1}^{\infty }{h}_{n+𝕚}{z}^n.$$

(2)

Let $f_1$ and $f_2$ are $analytic$ in $\mathbb{U}$, we say that the function $f_1$ is $subordinate$ to $f_2$ $or$, $f_2$ is said to be superordinate to $f_1$ if there exists a $\mathrm{Schwarz} \mathrm{function}$ $\mathbb{W}$ in $\mathbb{U}$ $with$ $\mathbb{W}\left(0\right)=0,$ and $\left|\mathbb{W}\left(z\right)\right|<1 \left(z\in \mathbb{U}\right),$ where $f_1\left(z\right)=f_2\left(\mathbb{W}\left(z\right)\right)$. In such a case, we write $f{ }_1\prec f{ }_2$ or $f{ }_1\left(z\right)\prec f{ }_2\left(z\right)$ $\left(z\in \mathbb{U}\right).$

Particularly, $f the function f_2 is univalent in$ $\mathbb{U}$, then $f_1\prec f_2$ if and only if $f_1\left(0\right)=f_2\left(0\right)$ and $f_1\left(\mathbb{U}\right)\subset f_2\left(\mathbb{U}\right)$ ([1,2]).

The set of all functions q that are analytic and injective $on \overline{\mathbb{U}}\setminus E(𝕢)$, denote $by$ $Q,$ where $\overline{\mathbb{U}}=\mathbb{U}\cup \left\{z\in \partial \mathbb{U}\right\},$ and $E(𝕢)=\left\{\xi \in \partial \mathbb{U}:𝕢\left(z\right)=\infty \right\},$ such that ${𝕢}^{\prime }\left(\xi \right)\ne 0 f or \xi \in \partial \mathbb{U}\setminus E(𝕢)$.

Definition 1[2].

Let$𝕙$and$\mathbb{k}$are two$analytic functions$in$\mathbb{U}$and$\varphi \left(r, s, t; z\right):{ℂ}^3\times \mathbb{U}\to ℂ$. If$𝕙$and$\varphi \left(𝕙\left(z\right), z𝕙\prime \left(z\right), z^2{𝕙}^{″}\left(z\right); z\right)$are$univalent functions$in$\mathbb{U}$and if$𝕙$satisfies the second-order superordination

$$\mathbb{k}\left(z\right)\prec \varphi \left(𝕙\left(z\right), z{𝕙}^{\prime }\left(z\right), z^2{𝕙}^{″}\left(z\right); z\right)$$

(3)

then $𝕙$ is called a solution of the differential superordination (3). A function $𝕢\in H\left(\mathbb{U}\right)$ is called a subordinant of (3), if $𝕙\left(z\right)\prec 𝕢\left(z\right)$ for all the functions $𝕙$ satisfying (3).

A $univalent subordinant$ $\widehat{𝕢}$ that satisfies $𝕢\left(z\right)\prec \widehat{𝕢}\left(z\right)$ for all the $subordinants$ $𝕢$ of (3), is said to be the $best$ subordinant.

Definition 2[1].

Let$\varphi :{ℂ}^3\times \mathbb{U}\to ℂ$and$\mathbb{k}\left(z\right)$be univalent in$\mathbb{U}$. If$𝕙\left(z\right)$is analytic in$\mathbb{U}$and satisfies the second-order differential subοrdination:

$$\varphi \left(𝕙\left(z\right), z{𝕙}^{\prime }\left(z\right), z^2{𝕙}^{″}\left(z\right); z\right)\prec \mathbb{k}\left(z\right)$$

(4)

Then, $𝕙$ is called a “solution of the differential subοrdination (4), and the univalent function $𝕢\left(z\right)$ is called a dominant of the solution of the differential subοrdination (4), or more simply dominant if $𝕙\left(z\right)\prec 𝕢\left(z\right)$ for all $𝕡\left(z\right)$ satisfying (4). A univalent dominant $\widehat{𝕢}\left(z\right)$ that satisfies $\widehat{𝕢}\left(z\right)\prec 𝕢\left(z\right)$ for all dominant $𝕢\left(z\right)$ of (4) is said to be the $\mathrm{best}$ dominant and is unique up to a relation of $\mathbb{U}$.

Recently, Miller and Mocanu [1] obtained sufficient conditions on the functions $\mathbb{k}, 𝕢,$ and $\varphi$ for which the following implication holds:

$$\mathbb{k}\left(z\right)\prec \varphi \left(𝕙\left(z\right), z{𝕙}^{\prime }\left(z\right), z^2{𝕙}^{″}\left(z\right); z\right)\to 𝕢\left(z\right)\prec 𝕙\left(z\right)$$

Using these results, Bulboaca [3] considered some classes of first-order differential superordinations, as well as integral operators preserving superordination [4]. Ali et al. [5,6], Atshan and Hadi [7], Atshan and Ali [8,9], and (see [10,11,12,13,14,15,16,17]) obtained results of subordination and superordination to analytic functions in $\mathbb{U}$. Lately, Al-Ameedee et al. [18,19], Atshan et al. [20,21,22], Bulboaca [23], Selvaraj and Karthikeyan [24] and (see [25,26,27,28,29,30,31,32,33,34]) got sandwich results to some classes of analytic functions. Further differential subοrdination results can be found in [35,36] for different orders.

Lemma 1.

Let$H\in ℍ and ,h_n\in ℂ,$we define the ordered cyclic operator of degree 1

$${\mathcal{T}}^1:ℍ\to ℍ, by$$

$${\mathcal{T}}^1\left(H\left(z\right)\right)={\displaystyle \sum }_{k=0}^n h_{n+1}{z}^n, z\in \mathbb{U}$$

and

$${\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)={\displaystyle \sum }_{k=0}^n h_{n+𝕚}{z}^n, z\in \mathbb{U}$$

where

$${\mathcal{T}}^0\left(H\left(z\right)\right)=H\left(z\right).$$

Proof 1.

Let $H\left(z\right)\in ℍ,$ then

$${\mathcal{T}}^1\left(H\left(z\right)\right)=\left(C_{{\mathcal{T}}^1}H\right)\left(z\right)=\left(H\circ {\mathcal{T}}^1\right)\left(z\right)={\mathcal{T}}^1\left({\displaystyle {\displaystyle \sum }_{k=1}^{\infty }}{h}_k{z}^k\right)={\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{h}_{n+1}{z}^n,$$

and

$$\begin{array}{c}{\mathcal{T}}^2\left(H\left(z\right)\right)=\left(C_{{\mathcal{T}}^2}f\right)\left(z\right)=\left(H\circ {\mathcal{T}}^2\right)\left(z\right)=\left(H\circ \mathcal{T}\circ \mathcal{T}\right)\left(z\right)=\mathcal{T}\left(\mathcal{T}\left({\displaystyle {\displaystyle \sum }_{k=1}^{\infty }}{h}_k{z}^k\right)\right)\\ =\mathcal{T}\left({\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{h}_{n+1}{z}^n\right)={\displaystyle {\displaystyle \sum }_{n=1}^{\infty }}{h}_{n+2}{z}^n,\end{array}$$

and so on

$${\mathcal{T}}^{𝕚}\left(f\left(z\right)\right)=\left(C_{{\mathcal{T}}^{𝕚}}H\right)\left(z\right)=\left(f\circ {\mathcal{T}}^{𝕚}\right)\left(z\right)=\left(H\circ \underset{𝕚-times}{\underbrace{\mathcal{T}\dots \circ \mathcal{T}}}\right)\left(z\right)=\mathcal{T}\left(\mathcal{T} \left(\dots \mathcal{T}\left({\displaystyle \sum }_{n=0}^{\infty }{h}_n{z}^n\right)\dots \right)\right)={\displaystyle \sum }_{n=0}^{\infty }{h}_{n+𝕚}{z}^n.$$

(5)

□

This completes the proof.

By simple calculation and using Newton’s identities, we obtain

$$z{\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)}^{\prime }=-{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\mathbb{S}\left(z\right).$$

Also

$$\frac{z{\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)}^{\prime }}{{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)}=-\mathbb{S}\left(z\right)=\frac{z{\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }}{{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)} .$$

(6)

2. Preliminaries

In order to demonstrate our results of differential $\mathrm{sub}\mathsf{o}\mathrm{rdination}$ and $\mathrm{superordination}$, the following definitions and known results are used.

Definition 3.[37].

A polynomial$p(x_1,x_2 ,\dots ,x_{\mathfrak{n}})$is called a$symmetric polynomial$if it satisfies:

$$p(x_{\phi \left(1\right)},x_{\phi \left(2\right)} ,\dots ,x_{\phi \left(\mathfrak{n}\right)})=p(x_1,x_2 ,\dots ,x_{\mathfrak{n}})$$

for all$permutations$ $\phi$of{1, …, n} such that${\Lambda }_{\mathfrak{n}}$denoted to the space of all$symmetric$polynomials in$x_1,x_2 ,\dots ,x_{\mathfrak{n}}$.

Definition 4.[37].

Suppose$x_1,x_2 ,\dots ,x_{\mathfrak{n}}$are the$n$roots of a polynomial

$$x^n+a_1{x}^{n-1}+\cdots +a_{n-1}x+a_n$$

then

$$\begin{array}{c}{e}_0=1,e_1(x_1,x_2 ,\dots ,x_{\mathfrak{n}})={\displaystyle \sum }_{i=1}^{\mathfrak{n}}{x}_i=-a_1,e_2(x_1,x_2 ,\dots ,x_{\mathfrak{n}})={\displaystyle \sum }_{1\le i_1<i_2\le n}^n{x}_{i_1}{x}_{i_2}=a_2,\\ \dots ,e_m(x_1,x_2 ,\dots ,x_{\mathfrak{n}})={\displaystyle \sum }_{1\le i_1<\dots <i_m\le \mathfrak{n} }^n{x}_{i_1}\dots x_{i_m}={\left(-1\right)}^m{a}_m,\dots ,e_n(x_1,x_2 ,\dots ,x_{\mathfrak{n}})=x_1{x}_2 \dots x_{\mathfrak{n}} \end{array}$$

The polynomial$e_m(x_1,x_2 ,\dots ,x_{\mathfrak{n}})$is called them-th symmetric polynomial in$x_1,x_2 ,\dots ,x_{\mathfrak{n} }$.

Definition 5.[37].

For each$k\ge 0$, the complete symmetric polynomial is the sum of all monomials of k-degree as follows:

$$h_k(x_1,x_2 ,\dots ,x_{\mathfrak{n}})={{\displaystyle \sum }}_{d_1+\dots +d_{\mathfrak{n}}=k}{x}_1{}^{d_1}\dots x_{\mathfrak{n}}{}^{d_{\mathfrak{n}}}$$

Particularly $h_0(x_1,x_2 ,\dots ,x_{\mathfrak{n}})=1$. It is not hard to see that

$$h_k(x_1,x_2 ,\dots ,x_{\mathfrak{n}})={{\displaystyle \sum }}_{\lambda \in \mathsf{{\rm P}}\left(\mathrm{k},\mathrm{n}\right)}{m}_{\lambda }(x_1,x_2 ,\dots ,x_n)$$

such that $m_{\lambda }$ is the partition of k.

Thus, for each $k\in {\mathsf{{\rm Z}}}^{+}$, there exists exactly one complete homogeneous symmetric polynomial of k-degree in $\mathfrak{n}$ variables.

Lemma 2.[37].

The symmetry between$h$and$e$suggests the introduction of the following map$\omega :{\Lambda }_{\mathfrak{n}}\to {\Lambda }_{\mathfrak{n}}$such that

$$\omega ({{\displaystyle \sum }}_{m_1, \dots ,m_{\mathfrak{n}}}{a}_{m_1, \dots ,m_{\mathfrak{n}}}{e}_1^{m_1}+\cdots +e_{\mathfrak{n}}^{m_{\mathfrak{n}}})={{\displaystyle \sum }}_{m_1, \dots ,m_{\mathfrak{n}}}{a}_{m_1, \dots ,m_{\mathfrak{n}}}{h}_1^{m_1}+\cdots +h_{\mathfrak{n}}^{m_{\mathfrak{n}}},$$

and

$$\omega ({{\displaystyle \sum }}_{m_1, \dots ,m_{\mathfrak{n}}}{a}_{m_1, \dots ,m_{\mathfrak{n}}}{e}_1^{m_1}+\cdots +e_{\mathfrak{n}}^{m_{\mathfrak{n}}})={{\displaystyle \sum }}_{m_1, \dots ,m_{\mathfrak{n}}}{a}_{m_1, \dots ,m_{\mathfrak{n}}}{e}_1^{m_1}+\cdots +e_{\mathfrak{n}}^{m_{\mathfrak{n}}}.$$

It has the following properties:

(1)

$\omega$ is a ring isomorphism, i.e.,

$$\omega \left(\rho +q\right)=\omega \left(\rho \right)+\omega \left(q\right), \omega \left( \rho · q\right)=\omega \left(\rho \right)·\omega \left(q\right),$$

for $\rho , q\in \mathfrak{n}$.

(2)

$\omega \left(e_i\right)=h_i and \omega \left(h_i\right)=e_i$.

(3)

${\omega }^2=id$.

$\mathrm{We} \mathrm{see} \mathrm{that}$ ${\mathsf{\Lambda }}_{\mathfrak{n}}=ℂ[h_1,\dots ,h_{\mathfrak{n}}$]. In $\mathrm{other} \mathrm{words}$, if we define for ${\lambda }^{\prime }=\left({\lambda }_1^{\prime },\dots ,{\lambda }_k^{\prime }\right)\in {\rho }^{\prime }\left(k,\mathfrak{n}\right).$

$$h_{{\lambda }^{\prime }}(x_1,x_2 ,\dots ,x_{\mathfrak{n}})=h_{{\lambda }_1^{\prime }}(x_1,x_2 ,\dots ,x_{\mathfrak{n}})\dots h_{{\lambda }_k^{\prime }}(x_1,x_2 ,\dots ,x_{\mathfrak{n}}),$$

${\left\{h_{{\lambda }^{\prime }}\right\}}_{{\lambda }^{\prime }\in {\rho }^{\prime }\left(k,\mathfrak{n}\right)}$ is a basis of ${\mathsf{\Lambda }}_k^{\mathfrak{n}}$.

Theorem 1.[37].

For$r \ge 1$, the$r-th$Newton$polynomial$(power sum) in$x_1,x_2 ,\dots ,x_{\mathfrak{n}}$is${\rho }_r(x_1,x_2 ,\dots ,x_{\mathfrak{n}})=x_1^r+\cdots +x_{\mathfrak{n}}^r .$

The generating function for them is

$$P_{\mathfrak{n}}={\displaystyle \sum }_{r\ge 1} {\rho }_r(x_1,x_2 ,\dots ,x_{\mathfrak{n}})t^{r-1}={\displaystyle \sum }_{i=1}^{\mathfrak{n}}{\displaystyle \sum }_{r\ge 1}{x}_i^r{t}^{r-1}={\displaystyle \sum }_{i=1}^{\mathfrak{n}}\frac{x_i}{1-x_{i}t}=\frac{d}{dt}log\frac{1}{{{\displaystyle \prod }}_{i=1}^{\mathfrak{n}}1-x_{i}t}.$$

By comparing this formula, we get:

$$P_{\mathfrak{n}}\left(t\right)=\frac{{\mathsf{H}}_{\mathfrak{n}}^{\prime }\left(t\right)}{{\mathsf{H}}_{\mathfrak{n}}\left(t\right)}=\frac{{\mathsf{{\rm E}}}_{\mathfrak{n}}^{\prime }\left(-t\right)}{{\mathsf{{\rm E}}}_{\mathfrak{n}}\left(-t\right)} ,$$

and by applying $\omega$, yields to:

• $\omega ({\rho }_{\mathfrak{n}})\left(t\right)=h_{\mathfrak{n}}\left(-t\right)$ or equivalently,
• $\omega ({\rho }_r)={\left(-1\right)}^{r-1}{h}_r,$ one also has $H_{\mathfrak{n}}^{\prime }\left(t\right)={\rho }_{\mathfrak{n}}\left(t\right)H_{\mathfrak{n}}\left(t\right) ,{\mathsf{{\rm E}}}_{\mathfrak{n}}^{\prime }\left(t\right)={\rho }_{\mathfrak{n}}\left(t\right){\mathsf{{\rm E}}}_{\mathfrak{n}}\left(-t\right)$.

Equivalently

$$k h_k={\displaystyle \sum }_{r=1}^k{\rho }_r{h}_{k-r} ,also k e_k={\displaystyle \sum }_{r=1}^k{\left(-1\right)}^{r-1}{\rho }_r{e}_{k-r}$$

These are $\mathrm{called} \mathrm{the} \mathrm{Newton} \mathrm{formulas}$.

Definition 6.[38].

A permutation matrix is a square matrix that has inputs$0, 1$derived from the identity matrix of the same size by a permutation of rows. There are$n!$permutation matrices of size$n$.

Definition 7.[39].

A bounded$linear operator T$on a Hilbert space$H$is called$cyclic operator$if there exists a vector$x\in H$and the set span $\left\{T^{\mathfrak{n}}x: \mathfrak{n}= 0, 1, \dots \right\}$is dense in H. The vector x is called a$cyclic vector$for the operator T.

Theorem 2.[39].

Let$S, T, X$b bounded operators on a Hilbert space$H$satisfying a conjugate relation$SX=XT$, if T is$cyclic operator$and$X$has a dense range, then$S$is cyclic operator too.

Proposition 1.[40].

Let$T$be “an operator on a Hilbert space H that has diagonal matrix” A= diag$\left({\lambda }_1,{\lambda }_2,,\dots \right)$with respect to some orthonormal basis {$e_n$}. Then$T$is cyclic if and only if the diagonal entries$\left\{{\lambda }_j\right\}$are distinct.

Definition 8.[35,41].

The set of all functions$f\left(z\right)$that are$analytic$and$injective$ $on \overline{\mathbb{U}}\setminus E\left(f\right)$, denote$by$ $Q,$ $where$ $\overline{\mathbb{U}}=\mathbb{U}\cup \left\{z\in \partial \mathbb{U}\right\}$and

$$E\left(f\right)=\left\{\xi \in \partial \mathbb{U}:f\left(z\right)=\infty \right\},$$

(7)

such that$f^{\prime }\left(\xi \right)\ne 0for\xi \in \partial \mathbb{U}\setminus E\left(f\right)$.

Lemma 3.[1].

Let the function$𝕢\left(z\right)$be univalent in the open unit disc$\mathbb{U}$and let$\theta$and$\phi$be analytic in a domain$D$containing$𝕢\left(\mathbb{U}\right)$with$\phi \left(w\right)\ne 0$when$w\in 𝕢\left(\mathbb{U}\right)$. Put$Q\left(z\right)=z{𝕢}^{\prime }\left(z\right)\phi \left(𝕢\left(z\right)\right) and 𝕙\left(z\right)=\theta \left(𝕢\left(z\right)+Q\left(z\right)\right)$.

Suppose that

(1)

$Q\left(z\right)$ is starlike univalent in $\mathbb{U}$,

(2)

$Re\left(\frac{z{𝕙}^{\prime }\left(z\right)}{Q\left(z\right)}\right)>0, z\in \mathbb{U}.$

If $𝕙$ is analytic in $\mathbb{U}$ $with$ $𝕙\left(0\right)=𝕢\left(0\right), 𝕙\left(\mathbb{U}\right)\subseteq D$ and

$$\theta \left(𝕙\left(z\right)\right)+z{𝕙}^{\prime }\left(z\right)\phi \left(𝕙\left(z\right)\right)\prec \theta \left(𝕢\left(z\right)\right)+z{𝕢}^{\prime }\left(z\right)\phi \left(𝕢\left(z\right)\right),$$

(8)

then $𝕙\left(z\right)\prec 𝕢\left(z\right), and 𝕢\left(z\right)$ is the $\mathrm{best}$ dominant.

Lemma 4.[35].

Let$𝕢\left(z\right)$be$convex$ $univalent$function$in$open unit disk$\mathbb{U}$, let$\psi \in ℂ, \gamma \in {ℂ}^{*}=ℂ\setminus \left\{0\right\}$with

$$Re\left(1+\frac{z{𝕢}^{″}\left(z\right)}{𝕢\prime \left(z\right)}\right)>max\left\{0,-Re\left(\frac{\psi }{\gamma }\right)\right\}.$$

(9)

$\mathrm{If}$ $𝕙\left(z\right)$ is analytic in $\mathbb{U}$ with $𝕙\left(0\right)$ =$𝕢\left(0\right)$ and

$$\psi 𝕙\left(z\right)+\gamma z{𝕙}^{\prime }\left(z\right)\prec \psi 𝕢\left(z\right)+\gamma z{𝕢}^{\prime }\left(z\right),$$

(10)

then $𝕙\left(z\right)$ $\prec 𝕢\left(z\right);z\in \mathbb{U}$ and $𝕢$ is the $best$ dominant.

Lemma 5.[23].

Let$𝕢\left(z\right)$be$convex$univalent in the unit disk$\mathbb{U}$and$let$ $\theta and \phi$be$analytic$in a domain$D$containing$𝕢\left(\mathbb{U}\right).$Suppose that

(1)

$Re\left\{\frac{{\theta }^{\prime }\left(𝕢\left(z\right)\right)}{\phi \left(𝕢\left(z\right)\right)}\right\}>0 for z\in \mathbb{U}$,

(2)

$z𝕢\prime \left(z\right)\phi \left(𝕢\left(z\right)\right)$is$starlike$ univalent in $z\in \mathbb{U}.$

If $𝕙\left(z\right)\in H\left[𝕢\left(0\right),1\right]\cap Q,$ $with$ $𝕙\left(\mathbb{U}\right)\subseteq D,$ $and$ $\theta (𝕙\left(z\right)+z𝕙\prime \left(z\right)\phi \left(𝕙\left(z\right)\right)$ is univalent in $\mathbb{U}$, and

$$\theta \left(𝕢\left(z\right)\right)+z{𝕢}^{\prime }\left(z\right)\phi \left(𝕢\left(z\right)\right)\prec \theta \left(𝕙\left(z\right)\right)+z{𝕙}^{\prime }\left(z\right)\phi \left(𝕙\left(z\right)\right),$$

(11)

then $𝕢\left(z\right)\prec 𝕙\left(z\right),and 𝕢\left(z\right);z\in \mathbb{U}$ is the $best$ subordinant.

Lemma 6. [2].

Let$𝕢\left(z\right)$be convex univalent in$\mathbb{U} and 𝕢\left(0\right)=1.$ $Let$ $\gamma \in ℂ,that$ $Re\left\{\gamma \right\}>0$. I$f$ $𝕙\left(z\right)\in H\left[𝕢\left(0\right),1\right]\cap Q$ $and$ $𝕙\left(z\right)+\gamma z𝕙\prime \left(z\right)$is univalent in$\mathbb{U}$, then

$$𝕢\left(z\right)+\gamma z{𝕢}^{\prime }\left(z\right)\prec 𝕙\left(z\right)+\gamma z{𝕙}^{\prime }\left(z\right),$$

(12)

which implies that$𝕢\left(\mathit{z}\right)\prec 𝕙\left(\mathit{z}\right) and 𝕢\left(\mathit{z}\right)$is the $best$subordinant.

3. Derivation of the Formula for $h_k$s in Terms of $s_k$s

If $A={\left(aij\right)}_{m\times m}, a_{ij }\in ℂ$, be a diagonal complex matrix, the rational polynomial $\frac{1}{det \left(I — A\right) }$ which factors into $\frac{1}{{{\displaystyle \prod }}_{l=1}^m\left(1-z{\lambda }_l\right)}$ where ${\lambda }_l$ is the eigenvalues of the matrix. The coefficients $h_1, h_2, .··, h_n ,\dots$ of this polynomial

$$H\left(z\right)=h_0+h_{1}z+h_2{z}^2+\dots +h_k{z}^k+\dots ;h_0=1,z\in \mathbb{U}$$

are given in terms of the eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_{𝓂}$ by

$$h_k({\lambda }_1,{\lambda }_2 ,\dots ,{\lambda }_{𝓂})={\displaystyle \sum }_{i_1+\cdots +i_m=k}^n{\lambda }_1{}^{i_1}\dots {\lambda }_{𝓂}{}^{i_{𝓂}} ,$$

$$h_k({\lambda }_1,{\lambda }_2 ,\dots ,{\lambda }_{𝓂})={\displaystyle \sum }_{1\le {\mathfrak{i}}_1,\dots ,{\mathfrak{i}}_k\le 𝓂}\frac{{𝓂}_1!{𝓂}_2!\dots {𝓂}_m!}{k!}{\lambda }_{{\mathfrak{i}}_1}{\lambda }_{{\mathfrak{i}}_2} \dots {\lambda }_{{\mathfrak{i}}_{𝓂}}$$

The symmetrical powers of eigenvalue.${\lambda }_i$ are defined by $S_{\mathfrak{n}}=trace \left(A^{\mathfrak{n}}\right)={\displaystyle \sum }_{l=1}^m{\lambda }_l^{\mathfrak{n}}$.The summation here is over all the eigenvalues of $A$.

Consider the formal power series $\mathbb{S}\left(z\right)={\displaystyle \sum }_{n=0}^{\infty }{S}_{\mathfrak{n}} z^n$ and $H\left(z\right)={\displaystyle \sum }_{\mathfrak{n}=0}^{\infty }{\left(-1\right)}^{n-1}{h}_{\mathfrak{n}} z^{\mathfrak{n}}$. It is a convenience to take $S_0=0 \mathrm{and} h_0=1$.

By using identities similar to Newton’s identities and applying $\omega$, one gets:

$\omega (h_{\mathfrak{n}})\left(t\right)=a_{\mathfrak{n}}\left(-t\right)$ or equivalently, $\omega (h_{\mathfrak{n}})={\left(-1\right)}^{n-1}{a}_{\mathfrak{n}}$, we obtain identities equivalent to the formal differential equation:

$$\mathbb{S}\left(z\right)H\left(z\right)+z{H}^{\prime }\left(z\right)=0.$$

This may be resolved by separating the variables:

$$\mathbb{S}\left(z\right)=-\frac{z{\mathsf{H}}^{\prime }\left(z\right)}{\mathsf{H}\left(z\right)},$$

and

$$\int \mathbb{S}\left(z\right)dz=-\int \frac{z{\mathsf{H}}^{\prime }\left(z\right)}{\mathsf{H}\left(z\right)}dz=-\ln \mathsf{H}\left(z\right)+c.$$

We can include the term on the left side for each term to be obtained

$$\int {\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{S}_{\mathfrak{n}} z^{n-1}dz={\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{S}_{\mathfrak{n}}\frac{z^{\mathfrak{n}}}{\mathfrak{n}}=-\ln \mathsf{H}\left(z\right)+c.$$

When $z=0,$ the left side is 0, and the right side is c. Therefore, $c=0,$ and we have two sets of power whose coefficients imply $h_{\mathfrak{n}}$ and $S_{\mathfrak{n}}$.

Since $\ln H\left(z\right)=-{\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{S}_{\mathfrak{n}}\frac{z^{\mathfrak{n}}}{\mathfrak{n}},$ and that yields $H\left(z\right)=e^{-{\displaystyle \sum }_{n=1}^{\infty }{S}_{\mathfrak{n}}\frac{z^{\mathfrak{n}}}{\mathfrak{n}}}$.

Expansion using the power series to the exponential function,

$$H\left(z\right)=1-\frac{1}{1!}\left({\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{S}_{\mathfrak{n}}\frac{z^{\mathfrak{n}}}{\mathfrak{n}}\right)+\frac{1}{2!}{\left({\displaystyle \sum }_{n=1}^{\infty }{S}_{\mathfrak{n}}\frac{z^{\mathfrak{n}}}{\mathfrak{n}}\right)}^2-\frac{1}{3!}{\left({\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{S}_{\mathfrak{n}}\frac{z^{\mathfrak{n}}}{\mathfrak{n}}\right)}^3+\dots .$$

Hence, collect coefficients of $z^n$ in this series, as above,

$$\begin{gathered} {\left(-1\right)}^{n-1}{h}_{\mathfrak{n}}=\frac{1}{1!}\frac{S_{\mathfrak{n}}}{\mathfrak{n}}+{\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2=2\\ {\mathfrak{i}}_1,{\mathfrak{i}}_2\ge 1\end{array}}\frac{1}{2!}\frac{S_{{\mathfrak{i}}_1} S_{{\mathfrak{i}}_2}}{{\mathfrak{i}}_1 {\mathfrak{i}}_2}-{\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2+{\mathfrak{i}}_3=3\\ {\mathfrak{i}}_1,{\mathfrak{i}}_2,{\mathfrak{i}}_3\ge 1\end{array}}\frac{1}{3!}\frac{S_{{\mathfrak{i}}_1} S_{{\mathfrak{i}}_2 }{S}_{{\mathfrak{i}}_3}}{{\mathfrak{i}}_1 {\mathfrak{i}}_2 {\mathfrak{i}}_3}+\dots \\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}={\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2+\cdots +{\mathfrak{i}}_{\mathfrak{n}}=k\\ {\mathfrak{i}}_1+2{\mathfrak{i}}_2+3{\mathfrak{i}}_3+\cdots +\mathfrak{n}{\mathfrak{i}}_{\mathfrak{n}}=\mathfrak{n}\end{array}}\frac{{\left(-1\right)}^{n-k} S_1^{{\mathfrak{i}}_1}{S}_2^{{\mathfrak{i}}_2}\dots S_{\mathfrak{n}}^{{\mathfrak{i}}_{\mathfrak{n}}}}{{\mathfrak{i}}_1! {\mathfrak{i}}_2!\dots {\mathfrak{i}}_n! 1^{{\mathfrak{i}}_1} 2^{{\mathfrak{i}}_2}\dots {\mathfrak{n}}^{{\mathfrak{i}}_{\mathfrak{n}}} }. \end{gathered}$$

One also has

$$h_{\mathfrak{n}}={\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2+\cdots +{\mathfrak{i}}_{\mathfrak{n}}=k\\ {\mathfrak{i}}_1+2{\mathfrak{i}}_2+3{\mathfrak{i}}_3+\cdots +\mathfrak{n}{\mathfrak{i}}_{\mathfrak{n}}=\mathfrak{n}\end{array}}\frac{{\left(-1\right)}^{n-k} S_1^{{\mathfrak{i}}_1}{S}_2^{{\mathfrak{i}}_2}\dots S_{\mathfrak{n}}^{{\mathfrak{i}}_{\mathfrak{n}}}}{{\mathfrak{i}}_1! {\mathfrak{i}}_2!\dots {\mathfrak{i}}_n! 1^{{\mathfrak{i}}_1} 2^{{\mathfrak{i}}_2}\dots {\mathfrak{n}}^{{\mathfrak{i}}_{\mathfrak{n}}} }$$

and, indeed, the extent of the coefficient.

$$\frac{n!}{{\mathfrak{i}}_1!1^{{\mathfrak{i}}_1} {\mathfrak{i}}_2!2^{{\mathfrak{i}}_2}\dots {\mathfrak{i}}_n! n^{{\mathfrak{i}}_n}} .$$

This corresponds to the number of permutations of $n$ symbols consisting of ${\mathfrak{i}}_{j }$- cycles of length $j=1,2,\dots \mathfrak{n} ,$

$$h_{\mathfrak{n}}={\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2+\cdots +{\mathfrak{i}}_{\mathfrak{n}}=k\\ {\mathfrak{i}}_1+2{\mathfrak{i}}_2+3{\mathfrak{i}}_3+\cdots +\mathfrak{n}{\mathfrak{i}}_{\mathfrak{n}}=\mathfrak{n}\end{array}}\frac{\mathfrak{n}!}{\mathfrak{n}!}.\frac{{\left(-1\right)}^{n-k} S_1^{{\mathfrak{i}}_1}{S}_2^{{\mathfrak{i}}_2}\dots S_{\mathfrak{n}}^{{\mathfrak{i}}_{\mathfrak{n}}}}{{\mathfrak{i}}_1! {\mathfrak{i}}_2!\dots {\mathfrak{i}}_n! 1^{{\mathfrak{i}}_1} 2^{{\mathfrak{i}}_2}\dots {\mathfrak{n}}^{{\mathfrak{i}}_{\mathfrak{n}}} } .$$

It also provides the computations, viz

$${\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2+\cdots +{\mathfrak{i}}_{\mathfrak{n}}=k\\ {\mathfrak{i}}_1+2{\mathfrak{i}}_2+3{\mathfrak{i}}_3+\cdots +\mathfrak{n}{\mathfrak{i}}_{\mathfrak{n}}=\mathfrak{n}\end{array}}\frac{\mathfrak{n}!}{{𝒾}_1!1^{{\mathfrak{i}}_1} {\mathfrak{i}}_2!2^{{\mathfrak{i}}_2}\dots {\mathfrak{i}}_{\mathfrak{n}}! {\mathfrak{n}}^{{\mathfrak{i}}_n}} ,$$

it equals to $\left|s\left(\mathfrak{n},k\right)\right|={(-1)}^{n-k}s\left(\mathfrak{n},k\right),$ where $s\left(\mathfrak{n},k\right)$ is the well-known Sterling numbers of the first kind.

$$h_n={\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2+\cdots +{\mathfrak{i}}_{\mathfrak{n}}=k\\ {\mathfrak{i}}_1+2{\mathfrak{i}}_2+3{\mathfrak{i}}_3+\cdots +\mathfrak{n}{\mathfrak{i}}_{\mathfrak{n}}=\mathfrak{n}\end{array}}\frac{ s\left(\mathfrak{n},k\right) S_1^{{\mathfrak{i}}_1}{S}_2^{{\mathfrak{i}}_2}\dots S_{\mathfrak{n}}^{\mathfrak{n}}}{\mathfrak{n}! } .$$

4. Derivation of an Ordered Cyclic Operator

Let $C_m$ be an ordered cyclic subgroup of symmetric group $S_m$ and ${\mathcal{M}}_m$ be an ordered cyclic matrix for symmetric matrices of size m as follows:

$${𝓂}_m^0=I_m=\left[\begin{array}{cccccc}1& 0& 0& 0& \dots & 0\\ 0& 1& 0& 0& \dots & 0\\ 0& 0& 1& 0& \dots & 0\\ 0& 0& 0& 1& \dots & 0\\ ⋮& ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& 0& 0& 0& \dots & 1\end{array}\right],{𝓂}_m^1=\left[\begin{array}{cccccc}0& 1& 0& \dots & 0& 0\\ 0& 0& 1& \dots & 0& 0\\ 0& 0& 0& \ddots & 0& 0\\ 0& 0& 0& \dots & 1& 0\\ 0& 0& 0& \dots & 0& 1\\ 1& 0& 0& \dots & 0& 0\end{array}\right]\dots ,$$

$${𝓂}_m^{m-1}=\left[\begin{array}{cccccc}0& 0& 0& \dots & 0& 1\\ 1& 0& 0& \dots & 0& 0\\ 0& 1& 0& \dots & 0& 0\\ 0& 0& 1& \dots & 0& 0\\ 0& 0& 0& \ddots & 0& 0\\ 0& 0& 0& \dots & 1& 0\end{array}\right] .$$

We define the isomorphism $\mathcal{T}:(C_m,\circ )\to ({\mathcal{M}}_m,.)$ such that $\mathcal{T}\left(c_m^{𝕚}\right)={𝓂}_m^{𝕚}$ for all $𝕚$.

Also these matrices act on the sequence $\left\{{𝒶}_n\right\}$ as follows:

$$\begin{array}{c}{𝓂}_m^{𝕚}\left[\begin{array}{c}{𝒶}_k\\ {𝒶}_{k+1}\\ ⋮\\ {𝒶}_{k+n-1}\end{array}\right]=\left[\begin{array}{c}{𝒶}_{k+𝕚}\\ {𝒶}_{k+𝕚+1}\\ ⋮\\ {𝒶}_{k+𝕚+n-1}\end{array}\right], \mathrm{it} \mathrm{follows} \mathrm{from} \mathrm{that}:\\ {𝓂}_m^{𝕚}\left(C^T\left(p\right)\left[\begin{array}{c}{𝒶}_k\\ {𝒶}_{k+1}\\ ⋮\\ {𝒶}_{k+n-1}\end{array}\right]\right)={𝓂}_m^{𝕚}\left(\left[\begin{array}{c}{𝒶}_{k+1}\\ {𝒶}_{k+2}\\ ⋮\\ {𝒶}_{k+n}\end{array}\right]\right)=\left[\begin{array}{c}{𝒶}_{k+𝕚+1}\\ {𝒶}_{k+𝕚+2}\\ ⋮\\ {𝒶}_{k+𝕚+n}\end{array}\right].\end{array}$$

It’s satisfies for each powers of companion matrix, thus the action of ordered cyclic operator of elementary symmetric polynomials and so on by taking $tr\left({𝓂}_m^{𝕚}{\left(C^T\left(p\right)\right)}^k\right);k=1,2,\dots ,n$, that yields to:

$$\begin{array}{c}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}tr\left({𝓂}_m^{𝕚}{\left(C^T\left(p\right)\right)}^1\right)=-a_{1+𝕚}=S_1 ,\\ tr\left({𝓂}_m^{𝕚}{\left(C^T\left(p\right)\right)}^2\right)=a_{1+𝕚}^2-2a_{2+𝕚}=S_2;a_{2+𝕚}=\frac{1}{2}\left({\left( S_1\right)}^2-S_2\right),\\ tr\left({𝓂}_m^{𝕚}{\left(C^T\left(p\right)\right)}^3\right)=-a_{1+𝕚}^3-a_{3+𝕚}+3a_{1+𝕚}{a}_{2+i}{a}_{3+𝕚}=S_3;a_{3+𝕚}=\frac{1}{3}\left(-{\left( S_1\right)}^3+3S_1 S_2-S_3\right) \dots .\end{array}$$

It follows from acting of ordered cyclic operator on characteristic polynomial ${\mathcal{T}}^{𝕚}\left(p\left(z\right)\right)={\mathcal{T}}^{𝕚}\left(a_0{z}^n+a_1{z}^{n-1}+\dots +a_{n-1}z+a_n\right)=a_{𝕚}{z}^n+a_{1+𝕚}{z}^{n-1}+\dots +a_{n-1+𝕚}z+a_{n+𝕚};n mod 𝕚.$

By applying $\omega$, one gets:

$\omega (a_n)\left(t\right)=h_n\left(-t\right)$ or equivalently,

$\omega (a_n)={\left(-1\right)}^{n-1}{h}_n$. One also has

$${\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)={\mathcal{T}}^{𝕚}\left(h_0+h_{1}z+h_2{z}^2+\dots +h_k{z}^k+\dots \right)=h_{𝕚}+h_{1+𝕚}z+h_{2+𝕚}{z}^2+\dots +h_{k+𝕚}{z}^k+\dots .$$

Consider the formal power series $\mathbb{S}\left(z\right)={\displaystyle \sum }_{n=0}^{\infty }{S}_n z^n$ and ${\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)={\displaystyle \sum }_{n=0}^{\infty }{\left(-1\right)}^{n-1}{h}_{n+𝕚} z^n$. It is convenient to take $S_0=0 and a_0=1$.

Then, by using identities similar to Newton’s identities and applying $\omega$, one gets:

$\omega (h_n)\left(t\right)=a_n\left(-t\right)$ or equivalently, $\omega (h_n)={\left(-1\right)}^{n-1}{a}_n,$ we obtain identities equivalent to the formal differential equation:

$$\mathbb{S}\left(\mathrm{z}\right){\mathcal{T}}^{𝕚}\left(\mathrm{H}\left(\mathrm{z}\right)\right)+ \mathrm{z}{\mathcal{T}}^{𝕚}\left( {\mathrm{H}}^{\prime }\left(\mathrm{z}\right)\right)=0,$$

$$\mathbb{S}\left(\mathrm{z}\right){\mathcal{T}}^{𝕚}\left(\mathrm{H}\left(\mathrm{z}\right)\right)+ \mathrm{z}{\mathcal{T}}^{𝕚}\left(\acute{\mathrm{H}}\left(\mathrm{z}\right)\right)=\left({{\displaystyle \sum }}_{\mathrm{n}=0}^{\infty }{\mathrm{S}}_{\mathrm{n}}{ \mathrm{z}}^{\mathrm{n}}\right)\left({{\displaystyle \sum }}_{\mathrm{n}=0}^{\infty }{\left(-1\right)}^{\mathrm{n}-1}{\mathrm{h}}_{\mathrm{n}+𝕚}{ \mathrm{z}}^{\mathrm{n}}\right)+{{\displaystyle \sum }}_{\mathrm{n}=0}^{\infty }{\left(-1\right)}^{\mathrm{n}-1}{\mathrm{h}}_{\mathrm{n}+𝕚}{ \mathrm{z}}^{\mathrm{n}}.$$

This can be solved by separating the variables:

$$\mathbb{S}\left(\mathrm{z}\right)=-\frac{\mathrm{z}{\mathcal{T}}^{𝕚}\left({\mathrm{H}}^{\prime }\left(\mathrm{z}\right)\right)}{{\mathcal{T}}^{𝕚}\left(\mathrm{H}\left(\mathrm{z}\right)\right)};𝕚=0,1,2,\dots ,$$

and

$$\int \mathbb{S}\left(\mathrm{z}\right)\mathrm{dz}=-\int \frac{\mathrm{z}{\mathcal{T}}^{𝕚}\left({\mathrm{H}}^{\prime }\left(\mathrm{z}\right)\right)}{{\mathcal{T}}^{𝕚}\left(\mathrm{H}\left(\mathrm{z}\right)\right)}\mathrm{dz}=-\ln {\mathcal{T}}^{𝕚}\left(\mathrm{H}\left(\mathrm{z}\right)\right)+c.$$

We can include the term on the left side for each term to be obtained

$$\int {\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{\mathrm{S}}_{\mathfrak{n}}{ \mathrm{z}}^{\mathfrak{n}-1}\mathrm{dz}={\displaystyle \sum }_{\mathrm{n}=1}^{\infty }{\mathrm{S}}_{\mathfrak{n}}\frac{{\mathrm{z}}^{\mathfrak{n}}}{\mathfrak{n}}=-\ln {\mathcal{T}}^{𝕚}\left(\mathrm{H}\left(\mathrm{z}\right)\right)+c.$$

When $\mathrm{z}=0,$ the left side is 0 and the right side is c. Therefore, $c=0$ and we have two sets of power whose coefficients imply ${\mathrm{h}}_{\mathrm{n}+𝕚}$ and ${\mathrm{S}}_{\mathrm{n}}$.

Since $\ln {\mathcal{T}}^{𝕚}\left(\mathrm{H}\left(\mathrm{z}\right)\right)=-{\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{\mathrm{S}}_{\mathfrak{n}}\frac{{\mathrm{z}}^{\mathfrak{n}}}{\mathfrak{n}},$ and that yields ${\mathcal{T}}^{𝕚}\left(\mathrm{H}\left(\mathrm{z}\right)\right)={\mathrm{e}}^{-{\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{\mathrm{S}}_{\mathrm{n}}\frac{{\mathrm{z}}^{\mathfrak{n}}}{\mathrm{n}}}$.

Expansion using the power series to the exponential function,

$${\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)=1-\frac{1}{1!}\left({\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{S}_{\mathfrak{n}}\frac{z^{\mathfrak{n}}}{\mathfrak{n}}\right)+\frac{1}{2!}{\left({\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{S}_n\frac{z^{\mathfrak{n}}}{\mathfrak{n}}\right)}^2-\frac{1}{3!}{\left({\displaystyle \sum }_{\mathfrak{n}=1}^{\infty }{S}_{\mathfrak{n}}\frac{z^{\mathfrak{n}}}{\mathfrak{n}}\right)}^3+\dots .$$

Hence, collect coefficients of $z^n$ in this series, as above,

$$h_{n+𝕚}={\displaystyle \sum }_{\begin{array}{c}{\mathfrak{i}}_1+{\mathfrak{i}}_2+\cdots +{\mathfrak{i}}_{\left(n+𝕚\right)}=k\\ {\mathfrak{i}}_1+2{\mathfrak{i}}_2+3{\mathfrak{i}}_3+\cdots +\left(n+𝕚\right)𝒾{\mathfrak{i}}_{\left(n+𝕚\right)}=\left(n+𝕚\right)\end{array}}\frac{{\left(-1\right)}^{n-k} S_1^{{\mathfrak{i}}_1}{S}_2^{{\mathfrak{i}}_2}\dots S_{\left(n+𝕚\right)}^{{\mathfrak{i}}_{\left(n+𝕚\right)}}}{{\mathfrak{i}}_1! {\mathfrak{i}}_2!\dots {\mathfrak{i}}_n! 1^{{\mathfrak{i}}_1} 2^{{\mathfrak{i}}_2}\dots {\left(n+𝕚\right)}^{{𝒾}_{\left(n+𝕚\right)}} }.$$

5. Differential Subοrdination Results

Theorem 3.

Let$𝕢\left(z\right)be$convex univalent in$with$ $𝕢\left(0\right)=1,$ $\lambda >0, and \eta \in ℂ\setminus \left\{0\right\}$. Assume that

$$Re\left\{1+\frac{z{𝕢}^{″}\left(z\right)}{{𝕢}^{\prime }\left(z\right)}\right\}>max\left\{0,-Re\left(\frac{1}{\eta }\right)\right\},$$

(13)

if

$$e_1\left(z\right)=\left(1+\eta \right)\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)-\eta \left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)\left(\frac{z{\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }}{{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)}\right),$$

(14)

$$e_1\left(z\right)\prec 𝕢\left(z\right)+\eta z{𝕢}^{\prime }\left(z\right),$$

(15)

then

$${\mathcal{T}}^{𝕚}H\left(z\right)\prec 𝕢\left(z\right),$$

(16)

and$𝕢\left(z\right)$is the$best$dominant.

Proof 2.

If we consider the analytic function

$$𝕡\left(z\right)={\mathcal{T}}^{𝕚}H\left(z\right); 𝕚=0,1,2,\dots , z\in \mathbb{U},$$

(17)

differentiating (17) with respect to $z ,$ we have

$${𝕡}^{\prime }\left(z\right)={\left({\mathcal{T}}^{𝕚}H\left(z\right)\right)}^{\prime },\mathrm{z}{𝕡}^{\prime }\left(z\right)=-𝕡\left(z\right)\mathbb{S}\left(\mathrm{z}\right),\frac{\mathrm{z}{𝕡}^{\prime }\left(z\right)}{𝕡\left(z\right)}=-\mathbb{S}\left(\mathrm{z}\right).$$

Now, using the identity (6), we obtain

$$\frac{z𝕡\prime \left(z\right)}{𝕡\left(z\right)}=\frac{z{\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }}{{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)},$$

Therefore,

$$z{𝕡}^{\prime }\left(z\right)=\frac{z {\mathcal{T}}^{𝕚}\left(H\left(z\right)\right){\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }}{{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)} .$$

(18)

Since $e_1\left(z\right)=\left(1+\eta \right)\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)-\eta \left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)\left(\frac{z{\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }}{{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)}\right),$

$$e_1\left(z\right)\prec 𝕢\left(z\right)+\eta z{𝕢}^{\prime }\left(z\right).$$

The subοrdination (15) is equivalent to

$$𝕡\left(z\right)+\eta z{𝕡}^{\prime }\left(z\right)\prec 𝕢\left(z\right)+\eta \lambda z{𝕢}^{\prime }\left(z\right).$$

(19)

Application of Lemma 3 with $\beta =\eta , \alpha =1,$ we obtain (16). □

$Taking 𝕢\left(z\right)=\frac{1+Az}{1+\mathrm{B}z} \left(-1\le \mathrm{B}<\mathrm{A}\le 1\right),$ in Theorem 3, we $\mathrm{get} \mathrm{the} \mathrm{following} \mathrm{result}$.

Corollary 1.

Let$\eta \in ℂ\setminus \left\{0\right\}$, and suppose that

$$Re\left(\frac{1+z}{1-z}\right)>max\left\{0,-Re\left(\frac{1}{\eta }\right)\right\}.$$

If $H\in ℍ$ satisfies the following subοrdination condition:

$$\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)+\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)\left(\frac{z{\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }}{{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)}\right)\prec \frac{1+z}{1-z}+\frac{2z}{{\left(1-z\right)}^2} ,$$

then

$$z{\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }\prec \frac{1+z}{1-z} ,$$

and $\frac{1+z}{1-z}$ is the $best dominant$.

Theorem 4.

Let$𝕢\left(z\right)$be convex univalent in unit disk$\mathbb{U}$ $with$ $𝕢\left(0\right)=1, 𝕢\left(z\right)\ne 0$and$\frac{z{𝕢}^{\prime }\left(z\right)}{𝕢\left(z\right)}$is starlike univalent$\mathbb{U}$,$\eta \in ℂ\setminus \left\{0\right\}, and a,\lambda ,\mu ,\sigma ,\varrho ,\alpha , \xi \in ℂ, H\in ℍ$, and suppose that$H and 𝕢$satisfy the next two conditions

$$tz\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)+\left(1-t\right)z\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)\ne 0, \left(z\in \mathbb{U}, 0\le t\le 1\right),$$

(20)

and

$$Re\left\{1+\frac{\alpha }{\eta }𝕢\left(z\right)+\frac{2\xi }{\eta }{\left(𝕢\left(z\right)\right)}^2-z\frac{𝕢\prime \left(z\right)}{𝕢\left(z\right)}+z\frac{{𝕢}^{″}\left(z\right)}{{𝕢}^{\prime }\left(z\right)}\right\}>0.$$

(21)

If

$$e_2\left(z\right)=a+\lambda 𝕢𝕢\left(z\right)+\mu \xi {𝕢}^2\left(z\right)+\varrho \frac{z{𝕢}^{\prime }\left(z\right)}{𝕢\left(z\right)},$$

(22)

such that

$$\begin{array}{cc}{e}_2\left(z\right)=& \sigma +\alpha \left(tz\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)+\left(1-t\right)z\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)\right)\hfill \\ & +\xi \left(tz\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)+\left(1-t\right)z\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)\right)\\ & +\eta \left[\frac{tz{\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }+\left(1-t\right)z{\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)}^{\prime }}{t{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right){\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)}\right],\end{array}$$

(23)

and

$$e_2\left(z\right)\prec \sigma +\alpha 𝕢\left(z\right)+\xi {\left(𝕢\left(z\right)\right)}^2+\eta \frac{z{𝕢}^{\prime }\left(z\right)}{𝕢\left(z\right)},$$

(24)

then

$$tz{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right)z{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\prec 𝕢\left(z\right),$$

(25)

and $𝕢\left(z\right)$ is the $\mathrm{best} \mathrm{dominant}$.

Proof 3.

Suppose that $𝕡\left(z\right)$ is an analytic function and is defined as:

$$𝕡\left(z\right)=t{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right){\mathcal{T}}^{𝕚}\left(H\left(z\right)\right).$$

(26)

Then, the function $𝕡\left(z\right)$ is analytic in $\mathbb{U}$ and $𝕡\left(0\right)=1,$ differentiating (26) with respect to $z,$ we get

$$\frac{z𝕡\prime \left(z\right)}{𝕡\left(z\right)}=\frac{tz\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)\prime +\left(1-t\right)z\left({\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\right)\prime }{t{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)f\left(z\right)+\left(1-t\right){\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)}.$$

(27)

By setting $\theta \left(\omega \right)=\sigma +\alpha \omega +\xi {\omega }^2 and \varphi \left(\omega \right)=\frac{\eta }{\omega },$ it can be easily observed that $\theta \left(\omega \right)$ is analytic in $ℂ , \varphi \left(\omega \right)$ is analytic in $ℂ\setminus \left\{0\right\} and \varphi \left(\omega \right)\ne 0 , \omega \in ℂ\setminus \left\{0\right\}.$$\mathrm{In} \mathrm{addition}, \mathrm{we} \mathrm{get}$

$$\begin{array}{cc}Q\left(z\right)=z{𝕢}^{\prime }\left(z\right)\varphi \left(z\right)& =\eta \frac{z{𝕢}^{\prime }\left(z\right)}{𝕢\left(z\right)} , \left(z\in \mathbb{U}\right) and 𝕙\left(z\right)=\theta \left(𝕢\left(z\right)\right)+Q\left(z\right)\hfill \\ & =\sigma +\alpha {𝕢}^{\prime }\left(z\right)+\xi {\left(𝕢\left(z\right)\right)}^2+\eta \frac{z𝕢\prime \left(z\right)}{𝕢\left(z\right)} , \end{array}$$

It is clear that $Q\left(z\right)$ $\mathrm{is} \mathrm{starlike} \mathrm{univalent}$ in $\mathbb{U}$, and that

$$Re\left(\frac{z{𝕙}^{\prime }\left(z\right)}{Q\left(z\right)}\right)=Re\left(1+\frac{\alpha }{\eta }𝕢\left(z\right)+\frac{2\xi }{\eta }{\left(𝕢\left(z\right)\right)}^2-z\frac{𝕢\prime \left(z\right)}{𝕢\left(z\right)}+z\frac{{𝕢}^{″}\left(z\right)}{{𝕢}^{\prime }\left(z\right)}\right)>0;\left(z\in \mathbb{U}\right).$$

By using (27), hypothesis (24) can be $\mathrm{equivalently} \mathrm{written}$ as

$$\theta \left(𝕡\left(z\right)\right)+z{𝕡}^{\prime }\left(z\right)\varphi \left(𝕡\left(z\right)\right)\prec \varphi \left(𝕢\left(z\right)\right)+z{𝕢}^{\prime }\left(z\right)\varphi \left(𝕢\left(z\right)\right),$$

Thus$, \mathrm{by} \mathrm{applying} \mathrm{Lemma}$4, the function $𝕢\left(z\right) \mathrm{is}$ the $\mathrm{best}$ dominant. □

6. Differential Superordination Results

Theorem 5.

Let$𝕢\left(z\right)$be a$convex univalent function$in$\mathbb{U}$with$𝕢\left(0\right)=1, Re\left\{\eta \right\}>0$. Let$H\in ℍ,$satisfies

$${\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\in H\left[𝕢\left(0\right),1\right]\cap Q.$$

If $\mathrm{the} \mathrm{function}$ $e_1\left(z\right)$ defined by (14) is $u\mathrm{nivalent}$ $\mathrm{in}$ $\mathbb{U}$ and

$$𝕢\left(z\right)+\eta z{𝕢}^{\prime }\left(z\right)\prec e_1\left(z\right),$$

(28)

then

$$𝕢\left(z\right)\prec {\mathcal{T}}^{𝕚}\left(H\left(z\right)\right) ,$$

(29)

and $𝕢\left(z\right)$ is the $\mathrm{best}$ subordinant.

Proof .

Suppose that $𝕡\left(z\right)$ is an analytic function and is defined $\mathrm{as}:$

$$𝕡\left(z\right)={\mathcal{T}}^{𝕚}\left(H\left(z\right)\right).$$

(30)

Differentiating (30) with respect to $z$, we have

$$\frac{z𝕡\prime \left(z\right)}{𝕡\left(z\right)}=\frac{z{\left({\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\right)}^{\prime }}{{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)}.$$

(31)

After some computation and using (6), from (31), we get

$$e_1\left(z\right)=𝕡\left(z\right)+\eta z{𝕡}^{\prime }\left(z\right),$$

and by applying Lemma 5, we get the following result.

Taking $𝕢\left(z\right)=\frac{1+Az}{1+\mathrm{B}z} , \left(-1\le \mathrm{B}<\mathrm{A}\le 1\right),$ in Theorem 5, we get the following corollary. □

Corollary 2.

$Let$ $-1\le \mathrm{B}⟨A\le 1 , \eta \in ℂ\setminus \left\{0\right\} with Re\left\{\eta \right\}⟩0$, also let

$${\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\in H\left[𝕢\left(0\right),1\right]\cap Q.$$

$\mathrm{If} \mathrm{the} \mathrm{function} e_1\left(z\right)$ given by (14) is $\mathrm{univalent}$ in $\mathbb{U}$ and $H\in ℍ$ satisfies the following $\mathrm{superordination} \mathrm{condition}$

$$\frac{1+\mathrm{A}z}{1+\mathrm{B}\mathrm{B}z}+\eta \frac{\left(\mathrm{A}-\mathrm{B}\right)z}{{\left(1+\mathrm{B}z\right)}^2}\prec e_1\left(z\right),$$

then

$$\frac{1+\mathrm{A}z}{1+\mathrm{B}z}\prec {\mathcal{T}}^{𝕚}\left(H\left(z\right)\right),$$

and the function $\frac{1+Az}{1+\mathrm{B}z}$ is the $\mathrm{best}$ subordinant.

Theorem 6.

Let$𝕢\left(z\right)$be convex univalent in unit disk$\mathbb{U}$, with$𝕢\left(0\right)=1, 𝕢\left(z\right)\ne 0, and \frac{z{𝕢}^{\prime }\left(z\right)}{𝕢\left(z\right)},$is starlike in$\mathbb{U}$, let$\eta \in ℂ\setminus \left\{0\right\} and \sigma ,\alpha \in ℂ.$Further, assume that$𝕢$satisfies

$$Re\left\{\left(\alpha +2\xi 𝕢\left(z\right)\right)\frac{𝕢\left(z\right)𝕢\prime \left(z\right)}{\eta }\right\}>0, \left(z\in \mathbb{U}\right).$$

(32)

Let $H\left(z\right)\in ℍ$, and suppose that $H\left(z\right)$ satisfies the next condition

$$tz{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right)z{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\ne 0;\left(z\in \mathbb{U}\right), \left(0\le t\le 1\right)$$

(33)

and

$$tz{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right)z{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\in H\left[𝕢\left(0\right),1\right]\cap Q.$$

(34)

$\mathrm{If} \mathrm{the} \mathrm{function} e_2\left(z\right), \mathrm{given} \mathrm{by} \left(22\right) \mathrm{is} \mathrm{univalent} \mathrm{in} \mathbb{U} ,\mathrm{and}$

$$\sigma +\alpha 𝕢\left(z\right)+\xi {\left(𝕢\left(z\right)\right)}^2+\eta \frac{z{𝕢}^{\prime }\left(z\right)}{𝕢\left(z\right)}\prec e_2\left(z\right),$$

(35)

then

$$𝕢\left(z\right)\prec tz{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right)z{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right),$$

(36)

and $𝕢\left(z\right)$ is the $\mathrm{best}$ subordinant.

Proof 4.

Let the function $𝕡\left(z\right)$ defined on $\mathbb{U}$ by (24).

Then, a computation shows that

$$\frac{z{𝕡}^{\prime }\left(z\right)}{𝕡\left(z\right)}=\frac{tz{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)\prime +\left(1-t\right)z{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\prime }{t{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right)\mathbb{T}{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)},$$

(37)

by setting $\Theta \left(w\right)=\sigma +\alpha w+\xi w^2, and\varphi \left(w\right)=\frac{\eta }{w},\left(w\in ℂ\setminus \left\{0\right\}\right)$.

We see that $\Theta \left(w\right)$ is $\mathrm{analytic}$ in $ℂ, \varphi \left(w\right)$ is $\mathrm{analytic}$ in $ℂ\setminus \left\{0\right\},$ and that $\varphi \left(w\right)\ne 0 \left(w\in ℂ\setminus \left\{0\right\}\right)$. In addition, we get

$$Q\left(z\right)=z{𝕢}^{\prime }\left(z\right)\varphi \left(𝕢\left(z\right)\right)=\eta \frac{z{𝕢}^{\prime }\left(z\right)}{𝕢\left(z\right)} ;z\in \mathbb{U}.$$

It observed $that$ $Q\left(z\right)$ is starlike univalent $in$ $\mathbb{U}, \mathrm{and} \mathrm{that}$

$$Re\left(\frac{z{\Theta }^{\prime }\left(𝕢\left(z\right)\right)}{\varphi \left(𝕢\left(z\right)\right)}\right)=Re\left\{\alpha +2\xi 𝕢\left(z\right)\frac{𝕢\left(z\right)𝕢\prime \left(z\right)}{\eta }\right\}>0 .$$

By making use of (37), hypothesis (35) can be written as:

$$\Theta \left(𝕢\left(z\right)\right)+z{𝕢}^{\prime }\left(z\right)\varphi \left(𝕢\left(z\right)\right)\prec \Theta \left(𝕡\left(z\right)\right)+z{𝕡}^{\prime }\left(z\right)\varphi \left(𝕡\left(z\right)\right),$$

Thus, the proof is complete by applying Lemma 6. □

7. Sandwich Results

Combination Theorem 3 with Theorem 5 to obtain the following theorem.

Theorem 7.

Let${𝕢}_1 and {𝕢}_2$be convex univalent functions$in$ $\mathbb{U}$with${𝕢}_1\left(0\right)={𝕢}_2\left(0\right)=1 and {𝕢}_2$satisfies(13). Suppose that$Re\left\{y\right\}>0, \lambda \in ℂ\setminus \left\{0\right\}.$If$H\in ℍ,$such that

$${\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\in H\left[𝕢\left(0\right),1\right]\cap Q,$$

and the function$e_1\left(z\right)$is univalent in$\mathbb{U}$and satisfies

$${𝕢}_1\left(z\right)+\eta z{𝕢}_1{}^{\prime }\left(z\right)\prec e_1\left(z\right)\prec {𝕢}_2\left(z\right)+\eta z{𝕢}_2{}^{\prime }\left(z\right),$$

(38)

where$e_1\left(z\right)$is given by (14), then

$${𝕢}_1\left(z\right)\prec {\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\prec {𝕢}_2\left(z\right),$$

where${𝕢}_1 and {𝕢}_2,$are respectively the$best subordinant and best dominant$ of $\left(38\right)$.

Combining Theorem 4 with Theorem 6, we obtain the following $\mathrm{sandwich} \mathrm{theorem}:$

Theorem 8.

$Let {𝕢}_j , be two convex univalent functions in \mathbb{U}, such that {𝕢}_j\left(0\right)=1, {𝕢}_j\left(z\right)\ne 0, and \frac{z{𝕢}_j^{\prime }\left(z\right)}{{𝕢}_j\left(z\right)} \left(j=1,2\right) is starlike univalent in \mathbb{U}, let \lambda ,\eta \in ℂ\setminus \left\{0\right\} and \sigma ,\alpha ,\xi \in ℂ.$Suppose that${𝕢}_1 and {𝕢}_2$ satisfies (22) $and$(32), respectively.

$\mathrm{If}$ $H\in ℍ,$ and suppose that $H$ satisfies the next condition

$$tz{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right)z{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\ne 0, \left(z\in U, 0\le t\le 1\right),$$

and

$$tz{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right)z{\mathcal{T}}^{𝕚}\left(f\left(z\right)\right)\in H\left[q\left(0\right),1\right]\cap Q.$$

$\mathrm{If} \mathrm{the} \mathrm{function} e_2\left(z\right) \mathrm{given} \mathrm{by} \left(23\right) \mathrm{is} \mathrm{univalent} \mathrm{in} \mathbb{U} , \mathrm{and}$

$$\sigma +\alpha {𝕢}_1\left(z\right)+\xi {\left({𝕢}_1\left(z\right)\right)}^2+\eta \frac{z{𝕢}_1{}^{\prime }\left(z\right)}{{𝕢}_1\left(z\right)}\prec e_2\left(z\right)\prec \sigma +\alpha {𝕢}_2\left(z\right)+\xi {\left({𝕢}_2\left(z\right)\right)}^2+\eta \frac{z{𝕢}_2{}^{\prime }\left(z\right)}{{𝕢}_2\left(z\right)},$$

(39)

then

$${𝕢}_1\left(z\right)\prec tz{\mathcal{T}}^{𝕚+1}\left(H\left(z\right)\right)+\left(1-t\right)z{\mathcal{T}}^{𝕚}\left(H\left(z\right)\right)\prec {𝕢}_2\left(z\right),$$

where q₁ and q₂ are the $\mathrm{best}$ subodinant and $\mathrm{best}$ dominant respectively of (7.2).

8. Conclusions

We introduce a new study on the connection between geometric function theory, especially sandwich theorems, and Viete’s theorem in elementary algebra. We obtain some conclusions for differential subordination and superordination for a new formula of complete homogeneous symmetric functions class involving an ordered cyclic operator. In addition, certain sandwich theorems are found. These properties and results are symmetrical with differential superordination properties to form sandwich theorems. We have different results than the other authors. We have opened some windows to allow authors to generalize our new subclasses in order to obtain new results in the theory of univalent and multivalent functions using the results of the paper.