Coefficient Inequalities of Functions Associated with Petal Type Domains

Abstract

In the theory of analytic and univalent functions, coefficients of functions’ Taylor series representation and their related functional inequalities are of major interest and how they estimate functions’ growth in their specified domains. One of the important and useful functional inequalities is the Fekete-Szegö inequality. In this work, we aim to analyze the Fekete-Szegö functional and to find its upper bound for certain analytic functions which give parabolic and petal type regions as image domains. Coefficient inequalities and the Fekete-Szegö inequality of inverse functions to these certain analytic functions are also established in this work.

1. Introduction and Preliminaries

Let $\mathcal{A}$ be the class of functions f of the form

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

(1)

which are analytic in the open unit disk $\mathcal{U}=\left\{z:\left|z\right|<1\right\}$ and $\mathcal{S}$ be the class of functions from $\mathcal{A}$ which are univalent in $\mathcal{U}$. One of the classical results regarding univalent functions related to coefficients $a_n$ of a function’s Taylor series, named as the Fekete-Szegö problem, introduced by Fekete and Szegö [1], is defined as follows:

If $f\in \mathcal{S}$ and is of the form (1), then

$$\left|a_3-\lambda a_2^2\right|\le \left\{\begin{array}{ccc}3-4\lambda ,& \mathrm{if}& \lambda \le 0,\\ 1+2exp\left(\frac{2\lambda }{\lambda -1}\right),& \mathrm{if}& 0\le \lambda \le 1,\\ 4\lambda -3,& \mathrm{if}& \lambda \ge 1.\end{array}\right.$$

This result is sharp. The Fekete-Szegö problem has a rich history in literature. Several results dealing with maximizing the non-linear functional $\left|a_3-\lambda a_2^2\right|$ for various classes and subclasses of univalent functions have been proved. The functional has been examined for $\lambda$ to be both a real and complex number. Several authors used certain classified techniques to maximize the Fekete-Szegö functional $\left|a_3-\lambda a_2^2\right|$ for different types of functions having interesting geometric characteristics of image domains. For more details and results, we refer to [1,2,3,4,5,6,7,8,9,10,11]. The function f is said to be subordinate to the function $g,$ written symbolically as $f\prec g,$ if there exists a schwarz function w such that

$$f\left(z\right)=g\left(w\left(z\right)\right),z\in \mathcal{U},$$

(2)

where $w\left(0\right)=0,\left|w\left(z\right)\right|<1$ for $z\in \mathcal{U}.$ Let P denote the class of analytic functions p such that $p\left(0\right)=1$ and $p\prec \frac{1+z}{1-z},z\in \mathcal{U}.$ For details, see [12].

In 1991, Goodman [13] initiated the concept of a conic domain by introducing generalized convex functions which generated the first parabolic region as an image domain of analytic functions. He introduced and defined the class $UCV$ of uniformly convex functions as follows:

$$UCV=\left\{f\in \mathcal{A}:\Re \left(1+\left(z-\zeta \right)\frac{f^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)>0,z,\zeta \in \mathcal{U}\right\}.$$

Later on, Rønning [14], and Ma and Minda [7] independently gave the most suitable one variable characterization of the class $UCV$ and defined it as follows:

$$UCV=\left\{f\in \mathcal{A}:\Re \left(1+\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right)>\left|\frac{z{f}^{\prime \prime }\left(z\right)}{f^{\prime }\left(z\right)}\right|,z\in \mathcal{U}\right\}.$$

This characterization gave birth to the first conic (parabolic) domain

$$\mathsf{\Omega }=\left\{w:\Re w>\left|w-1\right|\right\}.$$

This domain was then generalized by Kanas and Wiśniowska [15,16] who introduced the domain

$${\mathsf{\Omega }}_k=\left\{w:\Re w>k\left|w-1\right|,k\ge 0\right\}.$$

The conic domain ${\mathsf{\Omega }}_k$ represents the right half plane for $k=0,$ hyperbolic regions when $0<k<1,$ parabolic region for $k=1$ and elliptic regions when $k>1.$ For more details, we refer [15,16]. This conic domain ${\mathsf{\Omega }}_k$ has been extensively studied in [17,18,19]. The domain $\mathsf{\Omega }$ was also generalized by Noor and Malik [20] by introducing the domain

$$\begin{array}{c}\mathsf{\Omega }\left[A,B\right]=\left\{u+iv:{\left[\left(B^2-1\right)\left(u^2+v^2\right)-2\left(AB-1\right)u+\left(A^2-1\right)\right]}^2\right.\hfill \\ \hfill \left.>{\left(-2\left(B+1\right)\left(u^2+v^2\right)+2\left(A+B+2\right)u-2\left(A+1\right)\right)}^2+4{\left(A-B\right)}^2{v}^2\right\}.\end{array}$$

The domain $\mathsf{\Omega }\left[A,B\right]$ represents the petal type region, for more details, we refer to [20]. Now, we consider the following class of functions which take all values from the domain $\mathsf{\Omega }\left[A,B\right],-1\le B<A\le 1.$

Definition 1.

A function $p\left(z\right)$ is said to be in the class $UP\left[A,B\right],$ if and only if

$$p\left(z\right)\prec \frac{\left(A+1\right)\tilde{p}\left(z\right)-\left(A-1\right)}{\left(B+1\right)\tilde{p}\left(z\right)-\left(B-1\right)},-1\le B<A\le 1,$$

(3)

where $\tilde{p}\left(z\right)=1+\frac{2}{{\pi }^2}{\left(log\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^2$, $z\in \mathcal{U}.$

It can be seen that $\mathsf{\Omega }\left[1,-1\right]={\mathsf{\Omega }}_1=\mathsf{\Omega }$. This fact leads us to the following implications of different well-known classes of analytic functions.

• $UP\left[A,B\right]\subset P\left(\frac{3-A}{3-B}\right),$ the well-known class of functions with real part greater than $\frac{3-A}{3-B},$ see [12].
• $UP\left[1,-1\right]=\mathcal{P}\left(\tilde{p}\right),$ the well-known class of functions, introduced by Kanas and Wiśniowska [4,21].

Now we consider the following classes $UCV\left[A,B\right]$ of uniformly Janowski convex functions and $ST\left[A,B\right]$ of corresponding Janowski starlike functions (see [20] ) as follows.

Definition 2.

A function $f\in \mathcal{A}$ is said to be in the class $UCV\left[A,B\right],$ $-1\le B<A\le 1,$ if and only if

$$\Re \left(\frac{\left(B-1\right)\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}-\left(A-1\right)}{\left(B+1\right)\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}-\left(A+1\right)}\right)>\left|\frac{\left(B-1\right)\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}-\left(A-1\right)}{\left(B+1\right)\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}-\left(A+1\right)}-1\right|,$$

or equivalently,

$$\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}\in UP\left[A,B\right].$$

(4)

Definition 3.

A function $f\in \mathcal{A}$ is said to be in the class $ST\left[A,B\right],$ $-1\le B<A\le 1,$ if and only if

$$\Re \left(\frac{\left(B-1\right)\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\left(A-1\right)}{\left(B+1\right)\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\left(A+1\right)}\right)>\left|\frac{\left(B-1\right)\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\left(A-1\right)}{\left(B+1\right)\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}-\left(A+1\right)}-1\right|,$$

or equivalently,

$$\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\in UP\left[A,B\right].$$

(5)

It can easily be seen that $f\in UCV\left[A,B\right]⟺z{f}^{\prime }\in ST\left[A,B\right].$ It is clear that $UCV\left[1,-1\right]=UCV$ and $ST\left[1,-1\right]=ST,$ the well-known classes of uniformly convex and corresponding starlike functions respectively, introduced by Goodman [13] and Rønning [22].

In 1994, Ma and Minda [7] found the maximum bound of Fekete-Szegö functional $\left|a_3-\lambda a_2^2\right|$ for uniformly convex functions of class $UCV$ and then Kanas [21] investigated the same for the functions of class $\mathcal{P}\left(\tilde{p}\right).$ Our aim is to solve this classical Fekete-Szegö problem for the functions of classes $UP\left[A,B\right],UCV\left[A,B\right]$ and $ST\left[A,B\right].$ We need the following lemmas (see [7]) to prove our results.

Lemma 1.

If $p\left(z\right)=1+p_{1}z+p_2{z}^2+\cdots$ is a function with positive real part in $\mathcal{U},$ then, for any complex number μ,

$$\left|p_2-\mu p_1^2\right|\le 2max\left\{1,\left|2\mu -1\right|\right\}$$

and the result is sharp for the functions

$$p_0\left(z\right)=\frac{1+z}{1-z}\mathrm{or}{p}_{*}\left(z\right)=\frac{1+z^2}{1-z^2},\left(z\in \mathcal{U}\right).$$

Lemma 2.

If $p\left(z\right)=1+p_{1}z+p_2{z}^2+\cdots$ is a function with positive real part in $\mathcal{U},$ then, for any real number v,

$$\left|p_2-v{p}_1^2\right|\le \left\{\begin{array}{c}-4v+2,v\le 0,\\ 2,0\le v\le 1,\\ 4v-2,v\ge 1.\end{array}\right.$$

When $v<0$ or $v>1$, the equality holds if and only if $p\left(z\right)$ is $\frac{1+z}{1-z}$ or one of its rotations. If $0<v<1$, then, the equality holds if and only if $p\left(z\right)=\frac{1+z^2}{1-z^2}$ or one of its rotations. If $v=0$, the equality holds if and only if,

$$p\left(z\right)=\left(\frac{1+\eta }{2}\right)\frac{1+z}{1-z}+\left(\frac{1-\eta }{2}\right)\frac{1-z}{1+z}\left(0\le \eta \le 1\right),$$

or one of its rotations. If $v=1,$ then, the equality holds if and only if $p\left(z\right)$ is reciprocal of one of the function such that equality holds in the case of $v=0$. Although the above upper bound is sharp, when $0<v<1$, it can be improved as follows:

$$\left|p_2-v{p}_1^2\right|+{\left|p_1\right|}^2\le 2\left(0<v\le \frac{1}{2}\right)$$

and

$$\left|p_2-v{p}_1^2\right|+\left(1-v\right){\left|p_1\right|}^2\le 2\left(\frac{1}{2}<v\le 1\right).$$

2. Main Results

Theorem 1.

Let $p\in UP\left[A,B\right]$, $-1\le B<A\le 1$ and of the form $p\left(z\right)=1+{\sum }_{n=1}^{\infty }{p}_n{z}^n.$ Then, for a complex number $\mu ,$ we have

$$\left|p_2-\mu p_1^2\right|\le \frac{4}{{\pi }^2}\left(A-B\right).max\left(1,\left|\frac{4}{{\pi }^2}\left(B+1\right)-\frac{2}{3}+4\mu \left(\frac{A-B}{{\pi }^2}\right)\right|\right)$$

(6)

and for a real number $\mu ,$ we have

$$\left|p_2-\mu p_1^2\right|\le \frac{2\left(A-B\right)}{{\pi }^2}\left\{\begin{array}{cc}\begin{array}{c}\frac{4}{3}-\frac{8}{{\pi }^2}\left(B+1\right)-\frac{8}{{\pi }^2}\left(A-B\right)\mu ,\end{array}& \begin{array}{c}\mu \le -\frac{{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B},\\ -----------\end{array}\\ \begin{array}{c}2,\end{array}& \begin{array}{c}-\frac{{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B}\le \mu \\ \le \frac{5{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B},\\ -----------\end{array}\\ -\frac{4}{3}+\frac{8}{{\pi }^2}\left(B+1\right)+\frac{8}{{\pi }^2}\left(A-B\right)\mu ,& \mu \ge \frac{5{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B}.\end{array}\right.$$

(7)

These results are sharp and the equality in (6) holds for the functions

$$p_1\left(z\right)=\frac{\frac{2(A+1)}{{\pi }^2}{\left(log\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^2+2}{\frac{2(B+1)}{{\pi }^2}{\left(log\frac{1+\sqrt{z}}{1-\sqrt{z}}\right)}^2+2}$$

(8)

or

$$p_2\left(z\right)=\frac{\frac{2(A+1)}{{\pi }^2}{\left(log\frac{1+z}{1-z}\right)}^2+2}{\frac{2(B+1)}{{\pi }^2}{\left(log\frac{1+z}{1-z}\right)}^2+2}.$$

(9)

When $\mu <-\frac{{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B}$ or $\mu >\frac{5{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B},$ the equality in (7) holds for the function $p_1\left(z\right)$ or one of its rotations. If $-\frac{{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B}<\mu <\frac{5{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B}$, then, the equality in (7) holds for the function $p_2\left(z\right)$ or one of its rotations. If $\mu =-\frac{{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B}$, the equality in (7) holds for the function

$$p_3\left(z\right)=\left(\frac{1+\eta }{2}\right)p_1\left(z\right)+\left(\frac{1-\eta }{2}\right)p_1\left(-z\right),\left(0\le \eta \le 1\right),$$

(10)

or one of its rotations. If $\mu =\frac{5{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B},$ then, the equality in (7) holds for the functions $p\left(z\right)$ which is reciprocal of one of the function such that equality holds in the case for $\mu =-\frac{{\pi }^2}{12\left(A-B\right)}-\frac{B+1}{A-B}$.

Proof. 

For $h\in P$ and of the form $h\left(z\right)=1+{\sum }_{n=1}^{\infty }{c}_n{z}^n,$ we consider

$$h\left(z\right)=\frac{1+w\left(z\right)}{1-w\left(z\right)},$$

where $w\left(z\right)$ is such that $w\left(0\right)=0$ and $\left|w\left(z\right)\right|<1.$ It follows easily that

$$\begin{array}{ccc}\hfill w\left(z\right)& =& {\displaystyle \frac{h\left(z\right)-1}{h\left(z\right)+1}}\hfill \\ & =& {\displaystyle \frac{\left(1+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots \right)-1}{(1+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots )+1}}\hfill \\ & =& {\displaystyle \frac{1}{2}}{c}_{1}z+\left({\displaystyle \frac{1}{2}}{c}_2-{\displaystyle \frac{1}{4}}{c}_1^2\right)z^2+\left({\displaystyle \frac{1}{2}}{c}_3-{\displaystyle \frac{1}{2}}{c}_2{c}_1+{\displaystyle \frac{1}{8}}{c}_1^3\right)z^3+\cdots .\hfill \end{array}$$

(11)

Now, if $\tilde{p}\left(z\right)=1+R_{1}z+R_2{z}^2+\cdots ,$ then from (11), one may have

$$\begin{array}{ccc}\hfill \tilde{p}\left(w\left(z\right)\right)& =& 1+R_{1}w\left(z\right)+R_2{\left(w\left(z\right)\right)}^2+R_3{\left(w\left(z\right)\right)}^3+\cdots \hfill \\ & =& 1+R_1\left(\frac{1}{2}{c}_{1}z+\left(\frac{1}{2}{c}_2-\frac{1}{4}{c}_1^2\right)z^2+\left(\frac{1}{2}{c}_3-\frac{1}{2}{c}_2{c}_1+\frac{1}{8}{c}_1^3\right)z^3+\cdots \right)\hfill \\ & & +R_2{\left(\frac{1}{2}{c}_{1}z+\left(\frac{1}{2}{c}_2-\frac{1}{4}{c}_1^2\right)z^2+\left(\frac{1}{2}{c}_3-\frac{1}{2}{c}_2{c}_1+\frac{1}{8}{c}_1^3\right)z^3+\cdots \right)}^2\hfill \\ & & +R_3{\left(\frac{1}{2}{c}_{1}z+\left(\frac{1}{2}{c}_2-\frac{1}{4}{c}_1^2\right)z^2+\left(\frac{1}{2}{c}_3-\frac{1}{2}{c}_2{c}_1+\frac{1}{8}{c}_1^3\right)z^3+\cdots \right)}^3+\cdots ,\hfill \end{array}$$

where $R_1=\frac{8}{{\pi }^2},R_2=\frac{16}{3{\pi }^2}$ and $R_3=\frac{184}{45{\pi }^2}$, see [21]. Using these, the above series reduces to

$$\tilde{p}\left(w\left(z\right)\right)=1+\frac{4}{{\pi }^2}{c}_{1}z+\frac{4}{{\pi }^2}\left(c_2-\frac{1}{6}{c}_1^2\right)z^2+\frac{4}{{\pi }^2}\left(c_3-\frac{1}{3}{c}_2{c}_1+\frac{2}{45}{c}_1^3\right)z^3+\cdots .$$

(12)

Since $p\in UP\left[A,B\right]$, so from relations (2), (3) and (12), one may have

$$\begin{array}{ccc}\hfill p\left(z\right)& =& \frac{\left(A+1\right)\tilde{p}\left(w\left(z\right)\right)-\left(A-1\right)}{\left(B+1\right)\tilde{p}\left(w\left(z\right)\right)-\left(B-1\right)}\hfill \\ & =& \frac{2+\left(A+1\right)\frac{4}{{\pi }^2}{c}_{1}z+\left(A+1\right)\frac{4}{{\pi }^2}\left(c_2-\frac{1}{6}{c}_1^2\right)z^2+\cdots }{2+\left(B+1\right)\frac{4}{{\pi }^2}{c}_{1}z+\left(B+1\right)\frac{4}{{\pi }^2}\left(c_2-\frac{1}{6}{c}_1^2\right)z^2+\cdots }.\hfill \end{array}$$

This implies that

$$\begin{array}{ccc}\hfill p\left(z\right)& =& 1+\left(A-B\right)\frac{2}{{\pi }^2}{c}_{1}z+\left(A-B\right)\frac{2}{{\pi }^2}\left(c_2-\frac{1}{6}{c}_1^2-\frac{2}{{\pi }^2}\left(B+1\right)c_1^2\right)z^2+\hfill \\ & & \left(A-B\right)\frac{8}{{\pi }^2}\left(\left(\frac{{\left(B+1\right)}^2}{{\pi }^4}+\frac{B+1}{6{\pi }^2}+\frac{1}{90}\right)c_1^3-\left(\frac{B+1}{{\pi }^2}+\frac{1}{12}\right)c_2{c}_1+\frac{1}{4}{c}_3\right)z^3+\cdots .\hfill \end{array}$$

(13)

If $p\left(z\right)=1+{\sum }_{n=1}^{\infty }{p}_n{z}^n,$ then equating coefficients of z and $z^2,$ one may have

$$\begin{array}{ccc}\hfill p_1& =& \frac{2}{{\pi }^2}\left(A-B\right)c_1,\hfill \\ \hfill p_2& =& \frac{2}{{\pi }^2}\left(A-B\right)\left(c_2-\frac{1}{6}{c}_1^2-\frac{2}{{\pi }^2}\left(B+1\right)c_1^2\right).\hfill \end{array}$$

Now for a complex number $\mu ,$ consider

$$p_2-\mu p_1^2=\frac{2}{{\pi }^2}\left(A-B\right)\left[c_2-c_1^2\left(\frac{1}{6}+\frac{2}{{\pi }^2}\left(B+1\right)+\mu \frac{2}{{\pi }^2}\left(A-B\right)\right)\right].$$

This implies that

$$\left|p_2-\mu p_1^2\right|=\frac{2}{{\pi }^2}\left(A-B\right)\left|c_2-\left(\frac{1}{6}+\frac{2}{{\pi }^2}\left(B+1\right)+\mu \frac{2}{{\pi }^2}\left(A-B\right)\right)c_1^2\right|.$$

(14)

Using Lemma 1, one may have

$$\left|p_2-\mu p_1^2\right|\le \frac{2}{{\pi }^2}\left(A-B\right).2max\left(1,\left|2v-1\right|\right),$$

where

$$v=\frac{1}{6}+\frac{2}{{\pi }^2}\left(B+1\right)+\mu \frac{2}{{\pi }^2}\left(A-B\right).$$

This leads us to the required inequality (6) and applying Lemma 2 to the expression (14) for real number $\mu$, we get the required inequality (7). Sharpness follows from the functions $p_i\left(z\right);i=1,2,3$, defined by (8)–(10), and the following series form.

$$\begin{array}{ccc}\hfill p_1\left(z\right)& =& 1+\frac{4\left(A-B\right)}{{\pi }^2}z+\frac{8\left(A-B\right)}{{\pi }^2}\left(\frac{1}{3}-\frac{2\left(B+1\right)}{{\pi }^2}\right)z^2+\hfill \\ & & \frac{16\left(A-B\right)}{{\pi }^2}\left(4\left(\frac{{\left(B+1\right)}^2}{{\pi }^4}+\frac{B+1}{6{\pi }^2}+\frac{1}{90}\right)-2\left(\frac{B+1}{{\pi }^2}+\frac{1}{12}\right)+\frac{1}{4}\right)z^3+\cdots ,\hfill \end{array}$$

$$\begin{array}{ccc}\hfill p_2\left(z\right)& =& 1+\frac{4\left(A-B\right)}{{\pi }^2}{z}^2+\frac{8\left(A-B\right)}{{\pi }^2}\left(\frac{1}{3}-\frac{2\left(B+1\right)}{{\pi }^2}\right)z^4+\hfill \\ & & \frac{16\left(A-B\right)}{{\pi }^2}\left(4\left(\frac{{\left(B+1\right)}^2}{{\pi }^4}+\frac{B+1}{6{\pi }^2}+\frac{1}{90}\right)-2\left(\frac{B+1}{{\pi }^2}+\frac{1}{12}\right)+\frac{1}{4}\right)z^6+\cdots .\hfill \end{array}$$

 □

Corollary 1.

Let $p\in UP\left[1,-1\right]=\mathcal{P}\left(p_1\right)=\mathcal{P}\left(\tilde{p}\right)$ and of the form $p\left(z\right)=1+{\sum }_{n=1}^{\infty }{p}_n{z}^n.$ Then, for a complex number $\mu ,$ we have

$$\left|p_2-\mu p_1^2\right|\le \frac{8}{{\pi }^2}.max\left(1,\left|\frac{8\mu }{{\pi }^2}-\frac{2}{3}\right|\right)$$

(15)

and for real number $\mu ,$ we have

$$\left|p_2-\mu p_1^2\right|\le \frac{4}{{\pi }^2}\left\{\begin{array}{cc}\frac{4}{3}-\frac{16}{{\pi }^2}\mu ,& \mu \le -\frac{{\pi }^2}{24},\\ 2,& -\frac{{\pi }^2}{24}\le \mu \le \frac{5{\pi }^2}{24},\\ -\frac{4}{3}+\frac{16}{{\pi }^2}\mu ,& \mu \ge \frac{5{\pi }^2}{24}.\end{array}\right.$$

(16)

These inequalities are sharp.

In [4,21], Kanas studied the class $\mathcal{P}\left(p_k\right)$ which consists of functions who take all values from the conic domain ${\mathsf{\Omega }}_k$. Kanas [21] found the bound of Fekete-Szegö functional for the class $\mathcal{P}\left(p_k\right)$ whose particular case for $k=1$ is as follows:

Let $p\left(z\right)=1+b_{1}z+b_2{z}^2+b_3{z}^3+\cdots \in \mathcal{P}\left(p_1\right).$ Then, for real number $\mu ,$ we have

$$\left|b_2-\mu b_1^2\right|\le \frac{8}{{\pi }^2}\left\{\begin{array}{cc}1-\frac{8}{{\pi }^2}\mu ,& \mu \le 0,\\ 1,& \mu \in \left(0,1\right],\\ 1+\frac{8}{{\pi }^2}\left(\mu -1\right),& \mu \ge 1.\end{array}\right.$$

(17)

We observe that Corollary 1 improves the bounds of the Fekete-Szegö functional $\left|p_2-\mu p_1^2\right|$ for the functions of class $\mathcal{P}\left(p_1\right)$.

Theorem 2.

Let $f\in UCV\left[A,B\right]$, $-1\le B<A\le 1$ and of the form (1). Then, for a real number $\mu ,$ we have

$$\left|a_3-\mu a_2^2\right|\le \frac{A-B}{3{\pi }^2}\left\{\begin{array}{cc}\begin{array}{c}\frac{4}{3}-\frac{8}{{\pi }^2}\left(B+1\right)+\frac{4}{{\pi }^2}\left(A-B\right)\left(2-3\mu \right),\end{array}& \begin{array}{c}\mu \le \frac{2}{3}-\frac{{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)},\\ ------------\end{array}\\ \begin{array}{c}2,\end{array}& \begin{array}{c}\frac{2}{3}-\frac{{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)}\le \mu \\ \le \frac{2}{3}+\frac{5{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)},\\ ------------\end{array}\\ -\frac{4}{3}+\frac{8}{{\pi }^2}\left(B+1\right)-\frac{4}{{\pi }^2}\left(A-B\right)\left(2-3\mu \right),& \mu \ge \frac{2}{3}+\frac{5{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)}.\end{array}\right.$$

(18)

This result is sharp.

Proof. 

If $f\in UCV\left[A,B\right]$, $-1\le B<A\le 1,$ then it follows from relations (2)–(4),

$$\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}=\frac{\left(A+1\right)\tilde{p}\left(w\left(z\right)\right)-\left(A-1\right)}{\left(B+1\right)\tilde{p}\left(w\left(z\right)\right)-\left(B-1\right)},$$

where $w\left(z\right)$ is such that $w\left(0\right)=0$ and $\left|w\left(z\right)\right|<1.$ The right hand side of above expression gets its series form from (13) and reduces to

$$\begin{array}{ccc}\hfill {\displaystyle \frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}}& =& 1+\left(A-B\right){\displaystyle \frac{2}{{\pi }^2}}{c}_{1}z+\left(A-B\right){\displaystyle \frac{2}{{\pi }^2}}\left(c_2-{\displaystyle \frac{1}{6}}{c}_1^2-{\displaystyle \frac{2}{{\pi }^2}}\left(B+1\right)c_1^2\right)z^2+\hfill \\ & & \left(A-B\right){\displaystyle \frac{8}{{\pi }^2}}\left(\left({\displaystyle \frac{{\left(B+1\right)}^2}{{\pi }^4}}+{\displaystyle \frac{B+1}{6{\pi }^2}}+{\displaystyle \frac{1}{90}}\right)c_1^3-\left({\displaystyle \frac{B+1}{{\pi }^2}}+{\displaystyle \frac{1}{12}}\right)c_2{c}_1+{\displaystyle \frac{1}{4}}{c}_3\right)z^3+\cdots .\hfill \end{array}$$

(19)

If $f\left(z\right)=z+{\sum }_{n=2}^{\infty }{a}_n{z}^n,$ then one may have

$$\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}=1+2a_{2}z+\left(6a_3-4a_2^2\right)z^2+\left(12a_4-18a_2{a}_3+8a_2^3\right)z^3+\cdots .$$

(20)

From (19) and (20), comparison of coefficients of z and $z^2$ gives

$$a_2=\frac{1}{{\pi }^2}\left(A-B\right)c_1$$

(21)

and

$$6a_3-4a_2^2=\left(A-B\right)\frac{2}{{\pi }^2}\left(c_2-\frac{1}{6}{c}_1^2-\frac{2}{{\pi }^2}\left(B+1\right)c_1^2\right).$$

This implies, by using (21), that

$$a_3=\frac{1}{3{\pi }^2}\left(A-B\right)\left(c_2-\frac{1}{6}{c}_1^2-\frac{2}{{\pi }^2}\left(B+1\right)c_1^2+\frac{2}{{\pi }^2}\left(A-B\right)c_1^2\right).$$

(22)

Now, for a real number $\mu ,$ consider

$$\begin{array}{ccc}\hfill \left|a_3-\mu a_2^2\right|& =& \left|\left(A-B\right)\frac{1}{3{\pi }^2}\left(c_2-\frac{1}{6}{c}_1^2-\frac{2}{{\pi }^2}\left(B+1\right)c_1^2\right)+\frac{2}{3{\pi }^4}{\left(A-B\right)}^2{c}_1^2-\mu \frac{1}{{\pi }^4}{\left(A-B\right)}^2{c}_1^2\right|\hfill \\ & =& \frac{A-B}{3{\pi }^2}\left|c_2-c_1^2\left(\frac{1}{6}+\frac{2}{{\pi }^2}\left(B+1\right)-\frac{2}{{\pi }^2}\left(A-B\right)+\frac{3\mu }{{\pi }^2}\left(A-B\right)\right)\right|\hfill \\ & =& \frac{A-B}{3{\pi }^2}\left|c_2-v{c}_1^2\right|,\hfill \end{array}$$

where

$$v=\frac{1}{6}+\frac{2}{{\pi }^2}\left(B+1\right)-\frac{1}{{\pi }^2}\left(A-B\right)\left(2-3\mu \right).$$

Applying Lemma 2 leads us to the required result. The inequality (18) is sharp and equality holds for $\mu <\frac{2}{3}-\frac{{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)}$ or $\mu >\frac{2}{3}+\frac{5{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)}$ when $f\left(z\right)$ is $f_1\left(z\right)$ or one of its rotations, where $f_1\left(z\right)$ is defined such that $\frac{{\left(z{f}_1^{\prime }\left(z\right)\right)}^{\prime }}{f_1^{\prime }\left(z\right)}=p_1\left(z\right).$ If $\frac{2}{3}-\frac{{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)}<\mu <\frac{2}{3}+\frac{5{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)}$, then, the equality holds for the function $f_2\left(z\right)$ or one of its rotations, where $f_2\left(z\right)$ is defined such that $\frac{{\left(z{f}_2^{\prime }\left(z\right)\right)}^{\prime }}{f_2^{\prime }\left(z\right)}=p_2\left(z\right).$ If $\mu =\frac{2}{3}-\frac{{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)}$, the equality holds for the function $f_3\left(z\right)$ or one of its rotations, where $f_3\left(z\right)$ is defined such that $\frac{{\left(z{f}_3^{\prime }\left(z\right)\right)}^{\prime }}{f_3^{\prime }\left(z\right)}=p_3\left(z\right).$ If $\mu =\frac{2}{3}+\frac{5{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)},$ then, the equality holds for $f\left(z\right)$, which is such that $\frac{{\left(z{f}^{\prime }\left(z\right)\right)}^{\prime }}{f^{\prime }\left(z\right)}$ is reciprocal of one of the function such that equality holds in the case of $\mu =\frac{2}{3}-\frac{{\pi }^2}{18\left(A-B\right)}-\frac{2\left(B+1\right)}{3\left(A-B\right)}$. □

For $A=1,B=-1,$ the above result takes the following form which is proved by Ma and Minda [8].

Corollary 2.

Let $f\in UCV\left[1,-1\right]=UCV$ and of the form (1). Then, for a real number $\mu ,$

$$\left|a_3-\mu a_2^2\right|\le \frac{2}{3{\pi }^2}\left\{\begin{array}{cc}\frac{4}{3}+\frac{8}{{\pi }^2}\left(2-3\mu \right),& \mu \le \frac{2}{3}-\frac{{\pi }^2}{36},\\ 2,& \frac{2}{3}-\frac{{\pi }^2}{36}\le \mu \le \frac{2}{3}+\frac{5{\pi }^2}{36},\\ -\frac{4}{3}-\frac{8}{{\pi }^2}\left(2-3\mu \right),& \mu \ge \frac{2}{3}+\frac{5{\pi }^2}{36}.\end{array}\right.$$

This result is sharp.

Theorem 3.

Let $f\in ST\left[A,B\right]$, $-1\le B<A\le 1$ and of the form (1). Then, for a real number $\mu ,$

$$\left|a_3-\mu a_2^2\right|\le \frac{A-B}{{\pi }^2}\left\{\begin{array}{cc}\begin{array}{c}\frac{4}{3}-\frac{8}{{\pi }^2}\left(B+1\right)+\frac{8}{{\pi }^2}\left(A-B\right)\left(1-2\mu \right),\end{array}& \begin{array}{c}\mu \le \frac{1}{2}-\frac{{\pi }^2}{24\left(A-B\right)}-\frac{B+1}{2\left(A-B\right)},\\ ------------\end{array}\\ \begin{array}{c}2,\end{array}& \begin{array}{c}\frac{1}{2}-\frac{{\pi }^2}{24\left(A-B\right)}-\frac{B+1}{2\left(A-B\right)}\le \mu \\ \le \frac{1}{2}+\frac{5{\pi }^2}{24\left(A-B\right)}-\frac{B+1}{2\left(A-B\right)},\\ ------------\end{array}\\ -\frac{4}{3}+\frac{8}{{\pi }^2}\left(B+1\right)-\frac{8}{{\pi }^2}\left(A-B\right)\left(1-2\mu \right),& \mu \ge \frac{1}{2}+\frac{5{\pi }^2}{24\left(A-B\right)}-\frac{B+1}{2\left(A-B\right)}.\end{array}\right.$$

(23)

This result is sharp.

Proof. 

The proof follows similarly as in Theorem 2. □

For $A=1,B=-1,$ the above result reduces to the following form.

Corollary 3.

Let $f\in ST\left[1,-1\right]$ and of the form (1). Then, for a real number $\mu ,$

$$\left|a_3-\mu a_2^2\right|\le \frac{2}{{\pi }^2}\left\{\begin{array}{cc}\frac{4}{3}+\frac{16}{{\pi }^2}\left(1-2\mu \right),& \mu \le \frac{1}{2}-\frac{{\pi }^2}{48},\\ 2,& \frac{1}{2}-\frac{{\pi }^2}{48}\le \mu \le \frac{1}{2}+\frac{5{\pi }^2}{48},\\ -\frac{4}{3}-\frac{16}{{\pi }^2}\left(1-2\mu \right),& \mu \ge \frac{1}{2}+\frac{5{\pi }^2}{48}.\end{array}\right.$$

Now we consider the inverse function $\mathcal{F}$ which maps petal type regions to the open unit disk $\mathcal{U}$, defined as $\mathcal{F}\left(w\right)=\mathcal{F}\left(f\left(z\right)\right)=z,$ $z\in \mathcal{U}$ and we find the following coefficient bound for inverse functions. As the classes $UCV\left[A,B\right]$ and $ST\left[A,B\right]$ are the subclasses of $\mathcal{S}.$ Thus the existence of such inverse functions to the functions from $UCV\left[A,B\right]$ and $ST\left[A,B\right]$ is assured.

Theorem 4.

Let $w=f\left(z\right)\in UCV\left[A,B\right],$$-1\le B<A\le 1$ and $\mathcal{F}\left(w\right)=f^{-1}\left(w\right)=w+{\sum }_{n=2}^{\infty }{d}_n{w}^n.$ Then,

$$\left|d_n\right|\le \frac{4\left(A-B\right)}{n\left(n-1\right){\pi }^2}\left(n=2,3,4\right).$$

Proof. 

Since $\mathcal{F}\left(w\right)=\mathcal{F}\left(f\left(z\right)\right)=z,$ so it is easy to see that

$$d_2=-a_2,d_3=2a_2^2-a_3,d_4=-a_4+5a_2{a}_3-5a_2^3.$$

By using (21) and (22), one can have

$$d_2=\frac{-1}{{\pi }^2}\left(A-B\right)c_1$$

(24)

and

$$d_3=\frac{A-B}{3{\pi }^2}\left[\left(\frac{1}{6}+\frac{2}{{\pi }^2}\left(B+1\right)+\frac{4}{{\pi }^2}\left(A-B\right)\right)c_1^2-c_2\right].$$

(25)

From (19) and (20), comparison of $z^3$ gives

$$\begin{array}{ccc}\hfill a_4& =& \frac{A-B}{3{\pi }^2}\left[\left(\frac{1}{45}+\frac{1}{{\pi }^2}\left(\frac{1}{3}\left(B+1\right)-\frac{1}{4}\left(A-B\right)\right)+\frac{1}{{\pi }^4}\left(2{\left(B+1\right)}^2-3\left(A-B\right)\left(B+1\right)+{\left(A-B\right)}^2\right)\right)c_1^3\right.\hfill \\ & & \left.-\left(\frac{1}{6}+\frac{1}{{\pi }^2}\left(2\left(B+1\right)-\frac{3}{2}\left(A-B\right)\right)\right)c_2{c}_1+\frac{1}{2}{c}_3\right].\hfill \end{array}$$

Using the values of $a_n;n=2,3,4,$ we get

$$\begin{array}{c}\hfill \begin{array}{ccc}\hfill d_4& =& -{\displaystyle \frac{A-B}{3{\pi }^2}}\left[\left({\displaystyle \frac{1}{45}}+{\displaystyle \frac{1}{3{\pi }^2}}\left(B+1+{\displaystyle \frac{7}{4}}\left(A-B\right)\right)+{\displaystyle \frac{1}{{\pi }^4}}\left(2{\left(B+1\right)}^2+7\left(A-B\right)\left(B+1\right)+6{\left(A-B\right)}^2\right)\right)c_1^3\right.\hfill \\ & & \left.-\left({\displaystyle \frac{1}{6}}+{\displaystyle \frac{2}{{\pi }^2}}\left(B+1+{\displaystyle \frac{7}{4}}\left(A-B\right)\right)\right)c_2{c}_1+{\displaystyle \frac{1}{2}}{c}_3\right].\hfill \end{array}\end{array}$$

(26)

Now, from (24) and (25), one can have

$$\left|d_2\right|\le \frac{2}{{\pi }^2}\left(A-B\right)$$

and

$$\begin{array}{ccc}\hfill \left|d_3\right|& \le & \frac{A-B}{3{\pi }^2}\left|\frac{1}{6}+\frac{2}{{\pi }^2}\left(B+1\right)+\frac{4}{{\pi }^2}\left(A-B\right)\right|\left|c_2-c_1^2\right|\hfill \\ & & +\frac{A-B}{3{\pi }^2}\left|\frac{5}{6}-\frac{2}{{\pi }^2}\left(B+1\right)-\frac{4}{{\pi }^2}\left(A-B\right)\right|\left|c_2\right|.\hfill \end{array}$$

Application of the bounds $\left|c_2-c_1^2\right|\le 2$ and $\left|c_2\right|\le 2$ (see Lemma 2 for $v=1$ and $v=0$) gives $\left|d_3\right|\le \frac{2\left(A-B\right)}{3{\pi }^2}.$ Lastly, (26) reduces to

$$\left|d_4\right|\le \frac{A-B}{3{\pi }^2}\left[\left|{\lambda }_1\right|\left|c_3-2c_2{c}_1+c_1^3\right|+\left|{\lambda }_2\right|\left|c_3-c_2{c}_1\right|+\left|{\lambda }_3\right|\left|c_3\right|\right],$$

(27)

where

$${\lambda }_1=\frac{1}{45}+\frac{1}{3{\pi }^2}\left(B+1+\frac{7}{4}\left(A-B\right)\right)+\frac{1}{{\pi }^4}\left(2{\left(B+1\right)}^2+7\left(A-B\right)\left(B+1\right)+6{\left(A-B\right)}^2\right),$$

$${\lambda }_2=\frac{11}{90}+\frac{4}{3{\pi }^2}\left(B+1+\frac{7}{4}\left(A-B\right)\right)-\frac{2}{{\pi }^4}\left(2{\left(B+1\right)}^2+7\left(A-B\right)\left(B+1\right)+6{\left(A-B\right)}^2\right)$$

and

$${\lambda }_3=\frac{16}{45}-\frac{5}{3{\pi }^2}\left(B+1+\frac{7}{4}\left(A-B\right)\right)+\frac{1}{{\pi }^4}\left(2{\left(B+1\right)}^2+7\left(A-B\right)\left(B+1\right)+6{\left(A-B\right)}^2\right).$$

Applying the bounds $\left|c_3-2c_2{c}_1+c_1^3\right|\le 2,$ see [23], $\left|c_3-c_2{c}_1\right|\le 2$ and $\left|c_3\right|\le 2,$ see [7] to the right hand side of (27) and using the fact that ${\lambda }_i\ge 0;i=1,2,3,$ we have $\left|d_4\right|\le \frac{A-B}{3{\pi }^2}$ and this completes the proof. □

For $A=1,B=-1,$ the above result takes the following form which is proved by Ma and Minda [8].

Corollary 4.

Let $w=f\left(z\right)\in UCV$ and $\mathcal{F}\left(w\right)=f^{-1}\left(w\right)=w+{\sum }_{n=2}^{\infty }{d}_n{w}^n.$ Then,

$$\left|d_n\right|\le \frac{8}{n\left(n-1\right){\pi }^2}\left(n=2,3,4\right).$$

Theorem 5.

Let $w=f\left(z\right)\in UCV\left[A,B\right],$$-1\le B<A\le 1$ and $\mathcal{F}\left(w\right)=f^{-1}\left(w\right)=w+{\sum }_{n=2}^{\infty }{d}_n{w}^n.$ Then, for a real number $\mu ,$ we have

$$\left|d_3-\mu d_2^2\right|\le \frac{A-B}{3{\pi }^2}\left\{\begin{array}{cc}\begin{array}{c}\frac{4}{3}-\frac{8}{{\pi }^2}\left(B+1\right)-\frac{4}{{\pi }^2}\left(A-B\right)\left(4-3\mu \right),\end{array}& \begin{array}{c}\mu \ge \frac{4}{3}+\frac{{\pi }^2}{18\left(A-B\right)}+\frac{2\left(B+1\right)}{3\left(A-B\right)},\\ ------------\end{array}\\ \begin{array}{c}2,\end{array}& \begin{array}{c}\frac{4}{3}-\frac{5{\pi }^2}{18\left(A-B\right)}+\frac{2\left(B+1\right)}{3\left(A-B\right)}\le \mu \\ \le \frac{4}{3}+\frac{{\pi }^2}{18\left(A-B\right)}+\frac{2\left(B+1\right)}{3\left(A-B\right)},\\ ------------\end{array}\\ -\frac{4}{3}+\frac{8}{{\pi }^2}\left(B+1\right)+\frac{4}{{\pi }^2}\left(A-B\right)\left(4-3\mu \right),& \mu \le \frac{4}{3}-\frac{5{\pi }^2}{18\left(A-B\right)}+\frac{2\left(B+1\right)}{3\left(A-B\right)}.\end{array}\right.$$

This result is sharp.

Proof. 

The proof follows directly from (24), (25) and Lemma 2. □