Sufficiency Conditions for a Class of Convex Functions Connected with Tangent Functions Associated with the Combination of Babalola Operators and Binomial Series

Abstract

In this paper, we create a new subclass of convex functions given with tangent functions applying the combination of Babalola operators and Binomial series. Moreover, we obtain several important geometric results, including sharp coefficient bounds, sharp Fekete–Szego inequality, Kruskal inequality, and growth and distortion estimates. Furthermore, for functions with logarithmic coefficients, we compute sharp Fekete–Szego inequality and sharp coefficient bounds.

1. Introduction

Let $\mathcal{A}$ represent the family of analytic functions as follows:

$$\mathcal{G}\left(\zeta \right)=\zeta +\sum _{j=2}^{\infty }{a}_j{\zeta }^j,(\Omega =\{\zeta :\left|\zeta \right|<1,\zeta \in \mathbb{C}\},$$

(1)

and $\mathcal{S}$ be a class of all functions belonging to $\mathcal{A}$ which are univalent functions.

If $\mathcal{G}$ and Υ are analytic functions in $\mathcal{A}$, then $\mathcal{G}$ is subordinate to Υ, written $\mathcal{G}\prec \mathrm{{\rm Y}}$ if there exists an analytic Schwarz function $\varpi$ in $\mathcal{A}$ with $\varpi \left(0\right)=0$ and $\left|\varpi \left(\zeta \right)\right|<1$ for all $\zeta \in \Omega ,$ such that $\mathcal{G}\left(\zeta \right)=\mathrm{{\rm Y}}\left(\varpi \right(\zeta \left)\right).$ Furthermore, if the function $\mathrm{{\rm Y}}$ is univalent in $\mathcal{A},$ then we have

$$\mathcal{G}\left(\zeta \right)\prec \mathrm{{\rm Y}}\left(\zeta \right)\iff \mathcal{G}\left(0\right)=\mathrm{{\rm Y}}\left(0\right)\mathrm{and}\mathcal{G}(\Omega )\subset \mathrm{{\rm Y}}(\Omega ).$$

Furthermore, let the function $\mathcal{H}\in \mathcal{A}$ be given by

$$\mathcal{H}\left(\zeta \right)=\zeta +\sum _{j=2}^{\infty }{c}_j{\zeta }^j(\zeta \in \Omega ).$$

The Hadamard product or convolution of $\mathcal{F}$ and $\mathcal{H}$ is given by

$$\left(\mathcal{G}\ast \mathcal{H}\right)\left(\zeta \right)=\zeta +\sum _{j=2}^{\infty }{a}_j{c}_j{\zeta }^j,\left(\zeta \in \Omega \right).$$

(2)

The function $\mathcal{G}$ is a convex function of order $\beta ,$ $0\le \beta <1$, such that

$$\mathcal{K}:=\left\{\mathcal{G}\in \mathcal{A}:Re\frac{{\left(\zeta {\mathcal{G}}^{\prime }\left(\zeta \right)\right)}^{\prime }}{{\mathcal{G}}^{\prime }\left(\zeta \right)}>\beta ,0\le \beta <1,\zeta \in \Omega \right\}.$$

(3)

Babalola [1] defined the operator ${\mathcal{L}}_{\upsilon }^{\sigma }:\mathcal{A}\to \mathcal{A}$ as

$${\mathcal{L}}_{\upsilon }^{\sigma }\mathcal{G}\left(\zeta \right)=\left({\rho }_{\sigma }\ast {\rho }_{\sigma ,\upsilon }^{-1}\ast \mathcal{G}\right)\left(\zeta \right),$$

(4)

where

$${\rho }_{\sigma ,\upsilon }\left(\zeta \right)=\frac{\zeta }{{\left(1-\zeta \right)}^{\sigma -\upsilon +1}},\sigma -\upsilon +1>0,{\rho }_{\sigma }={\rho }_{\sigma ,0},$$

and ${\rho }_{\sigma ,\upsilon }^{-1}$ is

$${\rho }_{\sigma ,\upsilon }^{-1}\left(\zeta \right)=\zeta +\sum _{j=2}^{\infty }\frac{(\sigma -\upsilon )!}{(\sigma +j-\upsilon -1)!}{\zeta }^j\left(\sigma ,\upsilon \in \mathbb{N}=\{1,2,3,\dots \}\right).$$

For $\zeta \in \Omega ,$ then (4) is equivalent to

$${\mathcal{L}}_{\upsilon }^{\sigma }\mathcal{G}\left(\zeta \right)=\zeta +\sum _{j=2}^{\infty }\left[\frac{\Gamma (\sigma +j)}{\Gamma (\sigma +1)}·\frac{(\sigma -\upsilon )!}{(\sigma +j-\upsilon -1)!}\right]a_j{\zeta }^j.$$

Making use of the binomial series,

$${\left(1-\delta \right)}^n=\sum _{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right){\left(-1\right)}^i{\delta }^i\left(n\in \mathbb{N}\right).$$

In 2022, El-Deeb and Lupas [2] introduced the linear differential operator as follows:

$${\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,0}\mathcal{G}\left(\zeta \right)=\mathcal{G}\left(\zeta \right),$$

$$\begin{array}{ccc}\hfill {\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,1}\mathcal{G}\left(\zeta \right)& =& {\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma }\mathcal{G}\left(\zeta \right)={\left(1-\delta \right)}^n{\mathcal{L}}_{\upsilon }^{\sigma }\mathcal{G}\left(\zeta \right)+\left[1-{\left(1-\delta \right)}^n\right]\zeta {\left({\mathcal{L}}_{\upsilon }^{\sigma }\mathcal{G}\right)}^{\prime }\left(\zeta \right)\hfill \\ & =& \zeta +\sum _{j=2}^{\infty }\left[1+\left(j-1\right)c^n\left(\delta \right)\right]\left[\frac{\Gamma (\sigma +j)}{\Gamma (\sigma +1)}·\frac{(\sigma -\upsilon )!}{(\sigma +j-\upsilon -1)!}\right]a_j{\zeta }^j\hfill \\ & & .\hfill \\ & & .\hfill \\ & & .\hfill \\ \hfill {\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)& =& {\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma }\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m-1}\mathcal{G}\left(\zeta \right)\right)\hfill \\ & =& {\left(1-\delta \right)}^n{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m-1}\mathcal{G}\left(\zeta \right)+\left[1-{\left(1-\delta \right)}^n\right]\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m-1}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\hfill \\ & =& \zeta +\sum _{j=2}^{\infty }{\left[1+\left(j-1\right)c^n\left(\delta \right)\right]}^m\left[\frac{\Gamma (\sigma +j)}{\Gamma (\sigma +1)}·\frac{(\sigma -\upsilon )!}{(\sigma +j-\upsilon -1)!}\right]a_j{\zeta }^j\hfill \\ & =& \zeta +\sum _{j=2}^{\infty }{\psi }_j{a}_j{\zeta }^j,\hfill \\ & & \left(\delta >0;n,\sigma ,\upsilon \in \mathbb{N};m\in {\mathbb{N}}_0=\mathbb{N}\cup \left\{0\right\}\right),\hfill \end{array}$$

(5)

where

$${\psi }_j={\left[1+\left(j-1\right)c^n\left(\delta \right)\right]}^m\left[\frac{\Gamma (\sigma +j)}{\Gamma (\sigma +1)}·\frac{(\sigma -\upsilon )!}{(\sigma +j-\upsilon -1)!}\right],$$

(6)

and

$$c^n\left(\delta \right)=\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right){\left(-1\right)}^{i+1}{\delta }^i\left(n\in \mathbb{N}\right).$$

From (5), we obtain that

$$c^n\left(\delta \right)\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }={\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m+1}\mathcal{G}\left(\zeta \right)-\left[1-c^n\left(\delta \right)\right]{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right).$$

(7)

The integrated families of starlike and convex functions were developed in 1985 by Padmanabhan et al. [3], utilizing the theory of convolution along with the function $\frac{\zeta }{{\left(1-\zeta \right)}^a}$, where $a\in \mathbb{R}$. By taking a regular function $\varphi \left(\zeta \right)$ with $\varphi \left(0\right)=1$ and $h\left(\zeta \right)\in \mathcal{A},$ Shanmugam [4] expanded on the concept of [3] and introduced the functions class ${\mathcal{S}}_h^{\ast }\left(\varphi \right)$ as

$${\mathcal{S}}_h^{\ast }\left(\varphi \right)=\left\{\mathcal{G}\in \mathcal{A}:\frac{\zeta {(\mathcal{G}\ast h)}^{\prime }}{(\mathcal{G}\ast h)}\prec \varphi \left(\zeta \right),\zeta \in \Omega \right\}.$$

(8)

By taking $h\left(\zeta \right)=\frac{\zeta }{1-\zeta }$ or $\frac{\zeta }{{\left(1-\zeta \right)}^2}$, we derive the famous classes ${\mathcal{S}}^{\ast }\left(\varphi \right)$ and $\mathcal{C}\left(\varphi \right)$ of Ma- and Minda-type starlike and convex functions defined by [5]. Further, by choosing $\varphi \left(\zeta \right)=\frac{1+\zeta }{1-\zeta }$, these classes reduce to ${\mathcal{S}}^{\ast }$ and $\mathcal{C}$.

By choice $\varphi \left(\zeta \right)$ in the generic form of ${\mathcal{S}}^{\ast }\left(\varphi \right)$ and $\mathcal{C}\left(\varphi \right)$, numerous scholars have defined and investigated a variety of intriguing subclasses of analytic and univalent functions in the recent past. Here are a few of them highlighted:

Let $\varphi \left(\zeta \right)=\frac{1+F\zeta }{1+G\zeta },$$-1\le G<F\le 1$. Then ${\mathcal{S}}^{\ast }[F,G]={\mathcal{S}}^{\ast }\left(\frac{1+F\zeta }{1+G\zeta }\right)$ is the class of Janowski starlike functions; see [6]. For $\varphi \left(\zeta \right)=cos\zeta ,$ the class ${\mathcal{S}}_{cos\zeta }^{\ast }$ was studied by Bano et al. [7], whereas for $\varphi \left(\zeta \right)=cosh\zeta ,$ the function class ${\mathcal{S}}_{cosh\zeta }^{\ast }$ was introduced and studied by Alotaibi et al. [8]. For $\varphi \left(\zeta \right)=e^{\zeta },$ the class ${\mathcal{S}}_e^{\ast }$ was defined and studied by Mendiratta [9]; see [10]. For $\varphi \left(\zeta \right)=1+sin\zeta ,$ the class ${\mathcal{S}}^{\ast }\left(\varphi \right)$ reduces to ${\mathcal{S}}_{sin}^{\ast }$; it was presented and examined by Cho et al. [11]. For $\varphi \left(\zeta \right)=1+{sinh}^{-1}\left(\zeta \right)$, the family ${\mathcal{S}}^{\ast }\left(\varphi \right)$ was established and studied by Kumar et al. [12]; for more details, see [13]. For $\varphi \left(\zeta \right)=\frac{2}{1+e^{-\zeta }},$ the class ${\mathcal{S}}^{\ast }\left(\varphi \right)$ reduces to ${\mathcal{S}}_{sig}^{\ast }$; see [14,15,16]. The class ${\mathcal{S}}_{tanh\zeta }^{\ast }={\mathcal{S}}^{\ast }\left(\varphi \left(\zeta \right)\right),$ for $\varphi \left(\zeta \right)=1+tanh\zeta ,$ was established by Khalil et al. [17]; see [18].

Taking motivation and inspiration from the work mentioned above, we present the following subfamily of holomorphic functions.

Definition 1.

Let $\mathcal{G}$ be given in (1). Then

$$\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }\iff \mathcal{G}\in \mathcal{A}\mathit{and}\frac{{\left(\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\right)}^{\prime }}{{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }}\prec 1+\frac{tan\zeta }{2},\zeta \in \Omega .$$

(9)

Geometrically, the family ${\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ includes all the functions $\mathcal{G}$ that the image domain of $1+\frac{tan\zeta }{2},$ for a specified radius.

2. Set of Lemmas

We use the following lemmas in our results.

Let $\mathcal{P}$ stand for the family of all holomorphic functions p that have a positive real portion and be represented by the following series:

$$p\left(\zeta \right)=1+\sum _{j=1}^{\infty }{\alpha }_j{\zeta }^j,\zeta \in \Omega .$$

(10)

Lemma 1

([19]). If $p\in \mathcal{P}$, then the following estimations hold:

$$\left|c_j\right|\le 2,j\ge 1,$$

(11)

and

$$\left|c_{j+n}-\mu c_j{c}_n\right|<2,0<\mu \le 1,$$

(12)

and for $\eta \in \mathbb{C}$, we have

$$\left|p_2-\eta p_1^2\right|<2max\left\{1,\left|2\eta -1\right|\right\}.$$

(13)

Lemma 2

([20]). If $p\in \mathcal{P}$ and has the form (10), then

$$|{\alpha }_1{p}_1^3-{\alpha }_2{p}_1{p}_2+{\alpha }_3{p}_3|\le 2|{\alpha }_1|+2|{\alpha }_2-2{\alpha }_1|+2|{\alpha }_1-{\alpha }_2+{\alpha }_3|,$$

(14)

where ${\alpha }_1,{\alpha }_2$, and ${\alpha }_3$ are real numbers.

Lemma 3

([21]). Let ${\chi }_1,{\sigma }_1,{\psi }_1$, and ${\varrho }_1$ satisfy the inequalities ${\chi }_1,{\varrho }_1\in \left(0,1\right)$ and

$$\begin{array}{cc}\hfill & 8{\varrho }_1\left(1-{\varrho }_1\right)\left[{\left({\chi }_1{\sigma }_1-2{\psi }_1\right)}^2+{\left({\chi }_1\left({\varrho }_1+{\chi }_1\right)-{\sigma }_1\right)}^2\right]\hfill \\ \hfill & +{\chi }_1\left(1-{\chi }_1\right){\left({\sigma }_1-2{\varrho }_1{\chi }_1\right)}^2\hfill \\ \hfill & \le 4{\chi }_1^2{\left(1-{\chi }_1\right)}^2{\varrho }_1\left(1-{\varrho }_1\right).\hfill \end{array}$$

If $h\in \mathcal{P}$ and is of the form (10), then

$$\left|{\psi }_1{p}_1^4+{\varrho }_1{p}_2^2+2{\chi }_1{p}_1{p}_3-\frac{3}{2}{\sigma }_1{p}_1^2{p}_2-p_4\right|\le 2.$$

Lemma 4

([22]). Let $p\in \mathcal{P}$ and x and ζ belong to Ω; then, we have

$$\begin{array}{ccc}\hfill 2c_2& =& c_1^2+x\left(4-c_1^2\right),\hfill \\ \hfill 4c_3& =& 2x\left(4-c_1^2\right)c_1-x^2\left(4-c_1^2\right)c_1+2\zeta \left(1-{\left|x\right|}^2\right)\left(4-c_1^2\right)+c_1^3.\hfill \end{array}$$

where $c_2$ and $c_3$ are studied in [21].

The current study attempts to determine the necessary and sufficient conditions, the radius of convexity, growth and distortion estimates, the Kruskal inequality, sharp coefficient bounds, sharp Fekete–Szego inequality, and logarithmic coefficient estimates for the subclass ${\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ of class $\mathcal{A}$ associated with tangent functions.

3. Main Results

Theorem 1.

Let $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ given in (1). Then,

$$\frac{1}{\zeta }\left[{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \left(\frac{\zeta -\mathcal{M}{\zeta }^2}{{\left(1-\zeta \right)}^3}\right)\right]\ne 0,$$

(15)

where

$$\mathcal{M}=\frac{4+tan\left(e^{i\theta }\right)}{2}.$$

(16)

Proof. 

As $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ is analytic in $\mathcal{A},$ so $\frac{1}{\zeta }\mathcal{G}\left(\zeta \right)\ne 0$ for all $\zeta$ in $\Omega ,$ then by using the definition of subordination and (9), we have

$$\frac{{\left(\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\right)}^{\prime }}{{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }}=1+tan\omega \left(\zeta \right),$$

(17)

where $\omega \left(\zeta \right)$ is a Schwarz function. Let $\omega \left(\zeta \right)=e^{i\theta },$$-\pi \le \theta \le \pi .$ Then, (17) becomes

$$\frac{\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{″}}{{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }}\ne \frac{tan\left(e^{i\theta }\right)}{2},$$

which implies

$${\zeta }^2{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{″}-\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\frac{tan\left(e^{i\theta }\right)}{2}\ne 0.$$

(18)

It can be easily seen that

$${\zeta }^2{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{″}-\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }={\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \frac{\zeta \left(1+\zeta \right)}{{\left(1-\zeta \right)}^3},$$

(19)

and

$$\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }={\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \frac{\zeta }{{\left(1-\zeta \right)}^2}.$$

Using (19) and some simple calculations, (18) becomes

$${\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \left(\frac{\zeta -\mathcal{M}{\zeta }^2}{{\left(1-\zeta \right)}^3}\right)\ne 0.$$

(20)

From (20), we will obtain (15), where $\mathcal{M}$ is given in (16). □

Theorem 2.

Let $\mathcal{G}\in \mathcal{A}$. Then, $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ if

$$\sum _{j=2}^{\infty }\left[\frac{2j\left(2+tan\left(e^{i\theta }\right)\right)-4j^2}{tan\left(e^{i\theta }\right)}\right]{\psi }_j{a}_j{\zeta }^{j-1}-1\ne 0.$$

(21)

Proof. 

If $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$, then from Theorem 1, we have

$$\frac{1}{\zeta }\left[{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \left(\frac{\zeta -\mathcal{M}{\zeta }^2}{{\left(1-\zeta \right)}^3}\right)\right]\ne 0,$$

where $\mathcal{M}$ is given in (16). The above relation implies that

$$\frac{1}{\zeta }\left[\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \frac{\zeta }{{\left(1-\zeta \right)}^3}\right)-\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \frac{\mathcal{M}{\zeta }^2}{{\left(1-\zeta \right)}^3}\right)\right]\ne 0.$$

Since ${\zeta }^2=\zeta \left(1+\zeta \right)-\zeta ,$ we have

$$\frac{1}{\zeta }\left[\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \frac{\zeta }{{\left(1-\zeta \right)}^3}\right)-\mathcal{M}\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \frac{\zeta \left(1+\zeta \right)}{{\left(1-\zeta \right)}^3}-{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\ast \frac{\zeta }{{\left(1-\zeta \right)}^3}\right)\right]\ne 0.$$

(22)

Now, applying (19) and some properties of convolution, (22) reduces to

$$\frac{1}{\zeta }\left[\left(\frac{1}{2}{\zeta }^2{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{″}+\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\right)-\mathcal{M}\left({\zeta }^2{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{″}\right)\right]\ne 0.$$

Using (1) and after some simplification, we obtain (21). □

Theorem 3.

Let $\mathcal{G}\in \mathcal{A}$ be given in (1). Then, $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ if

$$\sum _{j=2}^{\infty }\left(\left|\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}\right|\right)\left|{\psi }_j\right|\left|a_j\right|<1.$$

(23)

Proof. 

To demonstrate the necessary outcome, we employ relation (21) as

$$\begin{array}{ccc}& & \left|1-\sum _{j=2}^{\infty }\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}{\psi }_j{a}_j{\zeta }^{j-1}\right|\hfill \\ & >& 1-\sum _{j=2}^{\infty }\left|\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}\right|\left|{\psi }_j\right|\left|a_j\right|{\left|\zeta \right|}^{j-1}.\hfill \end{array}$$

(24)

From (23), we have

$$1-\sum _{j=2}^{\infty }\left|\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}\right|\left|{\psi }_j\right|\left|a_j\right|>0.$$

(25)

From (24) and (25), we obtain the intended outcome by applying Theorem 2. □

Theorem 4.

Let $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }.$ Then, $\mathcal{G}$ is convex of order $\beta ,$ $0\le \beta <1$ and $\left|\zeta \right|<r_1$, where

$$r_1=\underset{j\ge 2}{inf}{\left(\left|\frac{\left(1-\beta \right)\left(4j-2\left(2+tan\left(e^{i\theta }\right)\right)\right){\psi }_j}{\left(j-\beta \right)tan\left(e^{i\theta }\right)}\right|\right)}^{\frac{1}{j-1}}.$$

(26)

Proof. 

It is sufficient to show that

$$\left|\frac{{\left(\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\right)}^{\prime }}{{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }}-1\right|\le 1-\beta .$$

(27)

From (1), we have

$$\left|\frac{\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{″}}{{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }}\right|\le \frac{{\displaystyle \sum _{j=2}^{\infty }}j\left(j-1\right){\psi }_j{a}_j{\left|\zeta \right|}^{j-1}}{1-\sum _{j=2}^{\infty }j{\psi }_j{a}_j{\left|\zeta \right|}^{j-1}}.$$

(28)

The inequality (28) is bounded above by $1-\beta ,$ if

$$\sum _{j=2}^{\infty }\left[\frac{j\left(j-1\right)+j\left(1-\beta \right)}{1-\beta }\right]{\psi }_j\left|a_j\right|{\left|\zeta \right|}^{j-1}\le 1.$$

(29)

However, by Theorem 3, the above inequality is true if

$$\sum _{j=2}^{\infty }\left|\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}{\psi }_j\right|\left|a_j\right|<1.$$

(30)

Then, the inequality (29) becomes

$$\left[\frac{j\left(j-\beta \right)}{1-\beta }\right]{\left|\zeta \right|}^{j-1}\le \left|\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}{\psi }_j\right|.$$

Simple math provides

$$r_1=\underset{j\ge 2}{inf}{\left(\left|\frac{\left(1-\beta \right)\left(4j-2\left(2+tan\left(e^{i\theta }\right)\right)\right){\psi }_j}{\left(j-\beta \right)tan\left(e^{i\theta }\right)}\right|\right)}^{\frac{1}{j-1}}.$$

The desired outcome is demonstrated. □

4. Growth and Distortion Estimates

Theorem 5.

Let $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ and $\left|\zeta \right|=r$. Then,

$$r-\left|\frac{tan\left(e^{i\theta }\right)}{\left(8-4tan\left(e^{i\theta }\right)\right){\psi }_2}\right|r^2\le \left|\mathcal{G}\left(\zeta \right)\right|\le r+\left|\frac{tan\left(e^{i\theta }\right)}{\left(8-4tan\left(e^{i\theta }\right)\right){\psi }_2}\right|r^2.$$

(31)

Proof. 

Consider

$$\begin{array}{ccc}\hfill \left|\mathcal{G}\left(\zeta \right)\right|& =& \left|\zeta +\sum _{j=2}^{\infty }{a}_j{\zeta }^j\right|\hfill \\ & \le & r+\sum _{j=2}^{\infty }\left|a_j\right|r^j.\hfill \end{array}$$

Since $r^j\le r^2$ for $j\ge 2$ and $r<1,$ we have

$$\left|\mathcal{G}\left(\zeta \right)\right|\le r+r^2\sum _{j=2}^{\infty }\left|a_j\right|.$$

(32)

Similarly,

$$\left|\mathcal{G}\left(\zeta \right)\right|\ge r-r^2\sum _{j=2}^{\infty }\left|a_j\right|.$$

(33)

Now, from (23) it is implied

$$\sum _{j=2}^{\infty }\left(\left|\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}\right|\right)\left|{\psi }_j\right|\left|a_j\right|<1.$$

Since

$$\left|\frac{16-4\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}{\psi }_2\right|\sum _{j=2}^{\infty }\left|a_j\right|\le \sum _{j=2}^{\infty }\left(\left|\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}\right|\right)\left|{\psi }_j\right|\left|a_j\right|,$$

we obtain

$$\left|\frac{8-4tan\left(e^{i\theta }\right)}{tan\left(e^{i\theta }\right)}{\psi }_2\right|\sum _{j=2}^{\infty }\left|a_j\right|<1,$$

one can easily write this as

$$\sum _{j=2}^{\infty }\left|a_j\right|<\left|\frac{tan\left(e^{i\theta }\right)}{16-4\left(2+tan\left(e^{i\theta }\right)\right){\psi }_2}\right|,$$

and placing this value in (32) and (33), the necessary inequality is obtained. □

Theorem 6.

Let $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ and $\left|\zeta \right|=r$. Then,

$$1-2\left|\frac{tan\left(e^{i\theta }\right)}{\left(8-4tan\left(e^{i\theta }\right)\right){\psi }_2}\right|r\le \left|{\mathcal{G}}^{\prime }\left(\zeta \right)\right|\le 1+2\left|\frac{tan\left(e^{i\theta }\right)}{\left(8-4tan\left(e^{i\theta }\right)\right){\psi }_2}\right|r.$$

Proof. 

Consider

$$\begin{array}{ccc}\hfill \left|{\mathcal{G}}^{\prime }\left(\zeta \right)\right|& =& \left|1+\sum _{j=2}^{\infty }j{a}_j{\zeta }^{j-1}\right|\hfill \\ & \le & 1+\sum _{j=2}^{\infty }j\left|a_j\right|r^{j-1}.\hfill \end{array}$$

Since $r^{j-1}\le r^2$ for $j\ge 2$ and $r<1,$ we have

$$\left|{\mathcal{G}}^{\prime }\left(\zeta \right)\right|\le 1+2r\sum _{j=2}^{\infty }\left|a_j\right|.$$

(34)

Similarly,

$$\left|{\mathcal{G}}^{\prime }\left(\zeta \right)\right|\ge 1-2r\sum _{j=2}^{\infty }\left|a_j\right|.$$

(35)

Now, from (23) it is implied that

$$\sum _{j=2}^{\infty }\left|\frac{4n^2-2n\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}\right|\left|a_j\right|<1.$$

Since

$$\left|\frac{16-4\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}{\psi }_2\right|\sum _{j=2}^{\infty }\left|a_j\right|\le \sum _{j=2}^{\infty }\left(\left|\frac{4j^2-2j\left(2+tan\left(e^{i\theta }\right)\right)}{tan\left(e^{i\theta }\right)}\right|\right)\left|{\psi }_j\right|\left|a_j\right|,$$

we obtain

$$\left|\frac{8-4tan\left(e^{i\theta }\right)}{tan\left(e^{i\theta }\right)}{\psi }_2\right|\sum _{j=2}^{\infty }\left|a_j\right|<1,$$

we have

$$\sum _{j=2}^{\infty }\left|a_j\right|<\left|\frac{tan\left(e^{i\theta }\right)}{\left(8-4tan\left(e^{i\theta }\right)\right){\psi }_2}\right|,$$

setting this value in (34) and (35), we thus accomplish what is needed. □

Theorem 7.

For $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon },$ the coefficient bounds are

$$\begin{array}{ccc}\hfill \left|a_2\right|& \le & \frac{1}{4{\psi }_2},\hfill \end{array}$$

(36)

$$\begin{array}{ccc}\hfill \left|a_3\right|& \le & \frac{1}{12{\psi }_3},\hfill \end{array}$$

(37)

$$\begin{array}{ccc}\hfill \left|a_4\right|& \le & \frac{1}{24{\psi }_4},\hfill \end{array}$$

(38)

$$\begin{array}{ccc}\hfill \left|a_5\right|& \le & \frac{1}{40{\psi }_5},\hfill \end{array}$$

(39)

and

$$\left|a_3-\vartheta a_2^2\right|\le \frac{1}{12{\psi }_3}max\left\{1,\left|\frac{3\vartheta {\psi }_3-2{\psi }_2^2}{4{\psi }_2^2}\right|\right\}.$$

(40)

The above outcomes (36), (37), (38), and (39) are sharp for functions given below, respectively:

$${\mathcal{G}}_1\left(\zeta \right)=\underset{0}{\overset{\zeta }{\int }}exp\underset{0}{\overset{\zeta }{\int }}\frac{\frac{tanx}{2}}{x}dx=\zeta +\frac{1}{4}{\zeta }^2+\frac{1}{24}{\zeta }^3+\cdots ,$$

(41)

$$\begin{array}{ccc}\hfill {\mathcal{G}}_2\left(\zeta \right)& =& \underset{0}{\overset{\zeta }{\int }}exp\underset{0}{\overset{\zeta }{\int }}\frac{\frac{tan{x}^2}{2}}{x}dx=\zeta +\frac{{\zeta }^3}{12}+\frac{{\zeta }^5}{160}+\cdots ,\hfill \end{array}$$

(42)

$$\begin{array}{ccc}\hfill {\mathcal{G}}_3\left(\zeta \right)& =& \underset{0}{\overset{\zeta }{\int }}exp\underset{0}{\overset{\zeta }{\int }}\frac{\frac{tan{x}^3}{2}}{x}dx=\zeta +\frac{{\zeta }^4}{24}+\frac{{\zeta }^7}{504}+\cdots ,\hfill \end{array}$$

(43)

$$\begin{array}{ccc}\hfill {\mathcal{G}}_4\left(\zeta \right)& =& \underset{0}{\overset{\zeta }{\int }}exp\underset{0}{\overset{\zeta }{\int }}\frac{\frac{tan{x}^4}{2}}{x}dx=\zeta +\frac{{\zeta }^5}{40}+\frac{{\zeta }^9}{1152}+\cdots ,\hfill \end{array}$$

(44)

and the bound (40) is extreme for the function defined in (42).

Proof. 

As $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon },$ then by the definition

$$\frac{{\left(\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\right)}^{\prime }}{{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }}\prec \frac{2+tan\left(\zeta \right)}{2},$$

which can be written as

$$\frac{{\left(\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\right)}^{\prime }}{{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }}=\frac{2+tan\left(\omega \left(\zeta \right)\right)}{2},$$

where $\omega \left(\zeta \right)$ is Schwarz function with properties that

$$\zeta \left(0\right)=0\mathrm{and}\left|\omega \left(\zeta \right)\right|<1.$$

Now, let

$$\frac{{\left(\zeta {\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }\right)}^{\prime }}{{\left({\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)\right)}^{\prime }}=1+2a_2{\psi }_2\zeta +\left(6{\psi }_3{a}_3-4{\psi }_2^2{a}_2^2\right){\zeta }^2+\left(12{\psi }_4{a}_4-18{\psi }_2{\psi }_3{a}_2{a}_3+8{\psi }_2^3{a}_2^3\right){\zeta }^3+\cdots ,$$

(45)

and

$$\begin{array}{ccc}\hfill 1+\frac{tan\left(\zeta \left(\chi \right)\right)}{2}& =& 1+\frac{1}{4}{c}_1\zeta +\left(\frac{1}{4}{c}_2-\frac{1}{8}{c}_1^2\right){\zeta }^2+\left(\frac{1}{12}{c}_1^3-\frac{1}{4}{c}_2{c}_1+\frac{1}{4}{c}_3\right){\zeta }^3\hfill \\ & & +\left(-\frac{1}{16}{c}_1^4+\frac{1}{4}{c}_1^2{c}_2-\frac{1}{4}{c}_3{c}_1-\frac{1}{8}{c}_2^2+\frac{1}{4}{c}_4\right){\zeta }^4+\dots .\hfill \end{array}$$

(46)

Comparing (45) and (46), we have

$$\begin{array}{ccc}{a}_2\hfill & =& \frac{1}{8{\psi }_2}{c}_1,\hfill \end{array}$$

(47)

$$\begin{array}{ccc}{a}_3\hfill & =& \frac{1}{24{\psi }_3}{c}_2-\frac{1}{96{\psi }_3}{c}_1^2,\hfill \end{array}$$

(48)

$$\begin{array}{ccc}{a}_4\hfill & =& \frac{17}{4608{\psi }_4}{c}_1^3-\frac{5}{384{\psi }_4}{c}_2{c}_1+\frac{1}{48{\psi }_4}{c}_3,\hfill \end{array}$$

(49)

$$\begin{array}{ccc}{a}_5\hfill & =& -\frac{1}{80}\left(\frac{157}{1152{\psi }_5}{c}_1^4-\frac{29}{48{\psi }_5}{c}_1^2{c}_2+\frac{2}{3{\psi }_5}{c}_3{c}_1+\frac{3}{8{\psi }_5}{c}_2^2-c_4\right).\hfill \end{array}$$

(50)

Then, by applying (11) to (47), we have

$$\left|a_2\right|\le \frac{1}{4{\psi }_2}.$$

Furthermore, applying (12) with $n=1$ to (48), we obtain

$$\left|a_3\right|\le \frac{1}{12{\psi }_3}.$$

For Equation (49), applying Lemma 2, we obtain

$$\left|a_4\right|\le \frac{1}{24{\psi }_4},$$

and from (50), we have

$$\begin{array}{ccc}\hfill \left|a_5\right|& =& \left|-\frac{1}{80}\right|\left|\frac{157}{1152}{c}_1^4-\frac{29}{48}{c}_1^2{c}_2+\frac{2}{3}{c}_3{c}_1+\frac{3}{8}{c}_2^2-c_4\right|\hfill \\ & \le & \frac{1}{40{\psi }_5}(\mathrm{by}\mathrm{Lemma}3).\hfill \end{array}$$

Now, from (47) and (48), we have

$$\left|a_3-\vartheta a_2^2\right|=\frac{1}{24{\psi }_3}\left|c_2-\frac{3\vartheta {\psi }_3-2{\psi }_2^2}{4{\psi }_2^2}{c}_1^2\right|.$$

Furthermore, applying (13) to the above relation, we achieve our goals. □

Theorem 8.

Let $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon },$ then

$$\left|a_2{a}_3-a_4\right|\le \frac{1}{24{\psi }_4}.$$

The outcome is sharp for the function defined in (43).

Proof. 

From (47)–(49), we have

$$\left|a_2{a}_3-a_4\right|=\left|\left(\frac{1}{768{\psi }_2{\psi }_3}+\frac{17}{4608{\psi }_4}\right)c_1^3-\left(\frac{1}{192{\psi }_2{\psi }_3}+\frac{5}{384{\psi }_4}\right)c_2{c}_1+\frac{1}{48{\psi }_4}{c}_3\right|,$$

applying Lemma 2, we achieve the intended outcomes. □

5. Kruskal Inequality

Here, we will provide direct evidence of the inequality

$$\left|a_j^p-a_2^{p\left(j-1\right)}\right|\le 2^{p\left(j-1\right)}-j^p,$$

over the class ${\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }$ for the choice of $j=4,$$p=1$, and for $j=5,$$p=1$. For a class of univalent functions as a whole, Kruskal introduced and demonstrated this inequality in [23]. For some recent investigations on Kruskal inequality, we refer the readers to see [15,24].

Theorem 9.

If $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon },$ then

$$\left|a_4-a_2^3\right|\le \frac{1}{24{\psi }_4}.$$

The outcome is sharp for the function defined in (43).

Proof. 

From (47) and (49), we have

$$\left|a_4-a_2^3\right|=\left|\left(\frac{17}{4608{\psi }_4}-\frac{1}{512{\psi }_2^3}\right)c_1^3-\frac{5}{384{\psi }_4}{c}_2{c}_1+\frac{1}{48{\psi }_4}{c}_3\right|,$$

by applying Lemma 2, we obtain

$$\left|a_4-a_2^3\right|\le \frac{1}{24{\psi }_4}.$$

□

Theorem 10.

If $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon },$ then

$$\left|a_5-a_2^4\right|\le \frac{1}{40{\psi }_5}.$$

The outcome is sharp for the function defined in (44).

Proof. 

From (47) and (50), we have

$$\begin{array}{ccc}\hfill \left|a_5-a_2^4\right|& =& \left|-\frac{1}{80{\psi }_5}\right|\left|\left(\left(\frac{157}{1152}-\frac{{\psi }_5}{4096{\psi }_2^4}\right)c_1^4-\frac{29}{48}{c}_1^2{c}_2+\frac{2}{3}{c}_3{c}_1+\frac{3}{8}{c}_2^2-c_4\right)\right|\hfill \\ & \le & \frac{1}{40{\psi }_5}(\mathrm{by}\mathrm{Lemma}3).\hfill \end{array}$$

□

6. Logarithmic Coefficients for the Family ${\mathcal{C}}_{tan}^{\mathit{\sigma },\mathit{m},\mathit{n},\mathit{\delta },\mathit{\upsilon }}$

The logarithmic coefficients of ${\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\in \mathcal{S}$, denoted by ${\kappa }_n={\kappa }_n\left(\mathcal{G}\right),$ are defined by the following series expansion:

$$log\frac{{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}\left(\zeta \right)}{\zeta }=2\sum _{j=1}^{\infty }{\kappa }_j{\zeta }^j.$$

For the function ${\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}\mathcal{G}$, given by (5), the logarithmic coefficients are as follows:

$$\begin{array}{ccc}{\kappa }_1\hfill & =& \frac{1}{2}{\psi }_2{a}_2,\hfill \end{array}$$

(51)

$$\begin{array}{ccc}{\kappa }_2\hfill & =& \frac{1}{2}\left({\psi }_3{a}_3-\frac{1}{2}{\psi }_2^2{a}_2^2\right),\hfill \end{array}$$

(52)

$$\begin{array}{ccc}{\kappa }_3\hfill & =& \frac{1}{2}\left({\psi }_4{a}_4-{\psi }_2{\psi }_3{a}_2{a}_3+\frac{1}{3}{\psi }_2^3{a}_2^3\right),\hfill \end{array}$$

(53)

$$\begin{array}{ccc}{\kappa }_4\hfill & =& \frac{1}{2}\left({\psi }_5{a}_5-{\psi }_2{\psi }_4{a}_2{a}_4-{\psi }_2^2{\psi }_3{a}_2^2{a}_3-\frac{1}{2}{\psi }_3^2{a}_3^2-\frac{1}{4}{\psi }_2^4{a}_2^4\right).\hfill \end{array}$$

(54)

Theorem 11.

If $\mathcal{G}$ has the form (1) and belongs to ${\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon },$ then

$$\begin{array}{ccc}\hfill \left|{\kappa }_1\right|& \le & \frac{1}{8{\psi }_2},\hfill \\ \hfill \left|{\kappa }_2\right|& \le & \frac{1}{24{\psi }_3},\hfill \\ \hfill \left|{\kappa }_3\right|& \le & \frac{1}{48{\psi }_4},\hfill \\ \hfill \left|{\kappa }_4\right|& \le & \frac{1}{80{\psi }_5}.\hfill \end{array}$$

The bounds of Theorem 11 are precise and cannot be improved further.

Proof. 

Now, from (51)–(54) and (47)–(50), we obtain

$$\begin{array}{cc}{\kappa }_1\hfill & =\frac{1}{16{\psi }_2}{c}_1,\hfill \end{array}$$

(55)

$$\begin{array}{cc}{\kappa }_2\hfill & =\frac{1}{48{\psi }_3}{c}_2-\frac{7}{768{\psi }_2^2}{c}_1^2,\hfill \end{array}$$

(56)

$$\begin{array}{cc}{\kappa }_3\hfill & =\frac{13}{4608{\psi }_2^3}{c}_1^3-\frac{7}{768{\psi }_2{\psi }_3}{c}_2{c}_1+\frac{1}{96{\psi }_4}{c}_3,\hfill \end{array}$$

(57)

$$\begin{array}{cc}{\kappa }_4\hfill & =-\frac{1561}{1474560{\psi }_2^4}{c}_1^4+\frac{413}{92160{\psi }_2^2{\psi }_3}{c}_1^2{c}_2-\frac{7}{1280{\psi }_2{\psi }_4}{c}_3{c}_1-\frac{1}{360{\psi }_2^2}{c}_2^2+\frac{1}{160{\psi }_5}{c}_4,\hfill \end{array}$$

(58)

$$\left|{\kappa }_1\right|\le \frac{1}{8{\psi }_2}.$$

From (56) and using (12), we obtain

$$\left|{\kappa }_2\right|\le \frac{1}{24{\psi }_3}.$$

Applying Lemma 2 to Equation (57), we obtain

$$\left|{\kappa }_3\right|\le \frac{1}{48{\psi }_4}.$$

Furthermore, using Lemma 3 to (58) we obtain

$$\left|{\kappa }_4\right|\le \frac{1}{80{\psi }_5}.$$

Proof for sharpness: since

$$\begin{array}{ccc}\hfill log\frac{{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}{\mathcal{G}}_1\left(\zeta \right)}{\zeta }& =& 2\sum _{j=2}^{\infty }\kappa \left({\mathcal{G}}_1\right){\zeta }^j=\frac{1}{4{\psi }_2}\zeta +\cdots ,\hfill \\ \hfill log\frac{{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}{\mathcal{G}}_2\left(\zeta \right)}{\zeta }& =& 2\sum _{j=2}^{\infty }\kappa \left({\mathcal{G}}_2\right){\zeta }^j=\frac{1}{12{\psi }_3}{\zeta }^2+\cdots ,\hfill \\ \hfill log\frac{{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}{\mathcal{G}}_3\left(\zeta \right)}{\zeta }& =& 2\sum _{j=2}^{\infty }\kappa \left({\mathcal{G}}_3\right){\zeta }^j=\frac{1}{24{\psi }_4}{\zeta }^3+\cdots ,\hfill \\ \hfill log\frac{{\mathcal{D}}_{n,\delta ,\upsilon }^{\sigma ,m}{\mathcal{G}}_4\left(\zeta \right)}{\zeta }& =& 2\sum _{j=2}^{\infty }\kappa \left({\mathcal{G}}_4\right){\zeta }^j=\frac{1}{40{\psi }_5}{\zeta }^4+\cdots ,\hfill \end{array}$$

it follows that these inequalities are obtained for the functions ${\mathcal{G}}_n\left(\zeta \right)$ for $n=1,2,3$, and 4 defined in (41)–(44). □

Theorem 12.

Let $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon }.$ Then, for complex number $\lambda ,$ we have

$$\left|{\kappa }_2-\lambda {\kappa }_1^2\right|\le \frac{1}{24{\psi }_3}max\left\{1,\frac{\left|3\lambda {\psi }_3-{\psi }_2^2\right|}{8{\psi }_2^2}\right\}.$$

The result is the best possible.

Proof. 

From (55) and (56), we have

$$\left|{\kappa }_2-\lambda {\kappa }_1^2\right|=\frac{1}{48{\psi }_3}\left|c_2-\frac{7{\psi }_2^2+3{\psi }_3\lambda }{16{\psi }_2^2}{c}_1^2\right|.$$

Applying (13) to the preceding equation yields the desired outcome. □

Theorem 13.

If $\mathcal{G}\in {\mathcal{C}}_{tan}^{\sigma ,m,n,\delta ,\upsilon },$ then

$$\left|{\kappa }_1{\kappa }_2-{\kappa }_3\right|\le \frac{1}{48{\psi }_4}.$$

The outcome is extremal.

Proof. 

From (55)–(57), we have

$$\left|{\kappa }_1{\kappa }_2-{\kappa }_3\right|=\left|\frac{125}{36864{\psi }_2^3}{c}_1^3-\frac{1}{96{\psi }_2{\psi }_3}{c}_2{c}_1+\frac{1}{96{\psi }_4}{c}_3\right|,$$

applying Lemma 2, we achieve the intended outcomes. □

7. Conclusions

In the present investigation, we investigated a new subclass of holomorphic convex functions associated with tangent functions. We focused our attention on deriving necessary and sufficient conditions based on the convolution operator, radius of convexity, growth and distortion approximation, sharp initial four coefficient bounds, sharp Fekete–Szegö approximation, and sharp Kruskal inequality for functions belonging to this class. Hopefully, this work will open new directions for research in GFT and related areas.