Argument and Coefficient Estimates for Certain Analytic Functions

Abstract

The aim of the present paper is to introduce a new class $\mathcal{G}\left(\alpha ,\delta \right)$ of analytic functions in the open unit disk and to study some properties associated with strong starlikeness and close-to-convexity for the class $\mathcal{G}\left(\alpha ,\delta \right)$. We also consider sharp bounds of logarithmic coefficients and Fekete-Szegö functionals belonging to the class $\mathcal{G}\left(\alpha ,\delta \right)$. Moreover, we provide some topics related to the results reported here that are relevant to outcomes presented in earlier research.

1. Introduction and Preliminaries

Let $\mathbb{U}$ denote the open unit dick in the complex plane $\mathbb{C}$. A function $\omega :\mathbb{U}\to \mathbb{C}$ is called a Schwarz function if $\omega$ is a analytic function in $\mathbb{U}$ with $\omega (0)=0$ and $|\omega (z)|<1$ for all $z\in \mathbb{U}$. Clearly, a Schwarz function $\omega$ is the form

$$\omega (z)=w_{1}z+w_2{z}^2+\cdots .$$

We denote by $\mathsf{\Omega }$ the set of all Schwarz functions on $\mathbb{U}$.

Let $\mathcal{A}$ be consisting of all analytic functions of the following normalized form:

$$\begin{array}{c}\hfill f(z)=z+\sum _{n=2}^{\infty }{a}_n{z}^n,\end{array}$$

(1)

in the open unit disk $\mathbb{U}$. An analytic function f is said to be univalent in a domain if it provides a one-to-one mapping onto its image: $f(z_1)=f(z_2)\Rightarrow z_1=z_2.$ Geometrically, this means that different points in the domain will be mapped into different points on the image domain. Also, let $\mathcal{S}$ be the class of functions $f\in \mathcal{A}$ which are univalent in $\mathbb{U}$. A domain D in the complex plane $\mathbb{C}$ is called starlike with respect to a point $w_0\in D$, if the line segment joining $w_0$ to every other point $w\in D$ lies in the interior of D. In other words, for any $w\in D$ and $0\le t\le 1,t{w}_0+(1-t)w\in D$. A function $f\in \mathcal{A}$ is starlike if the image $f(D)$ is starlike with respect to the origin.

For two analytic functions f and F in $\mathbb{U}$, we say that the function f is subordinate to the function F in $\mathbb{U}$ and we write $f(z)\prec F(z)$, if there exists a Schwarz function $\omega$ such that $f(z)=F\left(\omega (z)\right)$ for all $z\in \mathbb{U}$. Specifically, if the function F is univalent in $\mathbb{U}$, then we have the next equivalence:

$$f(z)\prec F(z)⟺f(0)=F(0)\mathrm{and}f(\mathbb{U})\subset F(\mathbb{U}).$$

The logarithmic coefficients ${\gamma }_n$ of $f\in \mathcal{S}$ are defined with the following series expansion:

$$\begin{array}{c}\hfill log\left({\displaystyle \frac{f(z)}{z}}\right)=2\sum _{n=1}^{\infty }{\gamma }_n(f)z^n,z\in \mathbb{U}.\end{array}$$

(2)

These coefficients are an important factor in studying diverse estimates in the theory of univalent functions. Note that we use ${\gamma }_n$ instead of ${\gamma }_n(f)$. The concept of logarithmic coefficients inspired Kayumov [1] to solve Brennan’s conjecture for conformal mappings. The importance of the logarithmic coefficients follows from Lebedev-Milin inequalities [2] (Chapter 2), see also [3,4], where estimates of the logarithmic coefficients were used to find bounds on the coefficients of f. Milin [2] conjectured the inequality

$$\sum _{m=1}^n\sum _{k=1}^m\left(k|{\gamma }_k{|}^2-\frac{1}{k}\right)\le 0(n=1,2,3,\cdots ),$$

which implies Robertson’s conjecture [5], and hence, Bieberbach’s conjecture [6]. This is the famous coefficient problem in univalent function theory. L. de Branges [7] established Bieberbach’s conjecture by proving Milin’s conjecture.

Definition 1.

Let $q,n\in \mathbb{N}$. The $q^{th}$ Hankel determinant is denote by $H_q(n)$ and defined by

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

(3)

where $a_k(k=1,2,\dots )$ are the coefficients of the Taylor series expansion of a function f of the form (1). Note that $a_1=1$.

The Hankel determinant $H_q(n)$ was defined by Pommerenke [8,9] and for fixed $q,n$ the bounds of $|H_q(n)|$ have been studied for several subfamilies of univalent functions. Different properties of these determinants can be observed in [10] (Chapter 4). The Hankel determinants $H_2(1)=a_3-a_2^2$ and $H_2(2)=a_2{a}_4-a_3^2$, are well-known as Fekete-Szegö and second Hankel determinant functionals, respectively. In addition, Fekete and Szegö [11] introduced the generalized functional $a_3-\lambda a_2^2$, where $\lambda$ is a real number. Recently, Hankel determinants and other problems for various classes of bi-univalent functions have been studied, see [12,13,14,15,16].

For $\alpha \in [0,1)$, we denote by ${\mathcal{S}}^{*}(\alpha )$ the subclass of $\mathcal{A}$ including of all $f\in \mathcal{A}$ for which f is a starlike function of order $\alpha$ in $\mathbb{U}$, with

$$Re{\displaystyle \frac{z{f}^{\prime }(z)}{f(z)}}>\alpha (z\in \mathbb{U}).$$

Also, for $\alpha \in (0,1]$, we denote by ${\tilde{\mathcal{S}}}^{*}(\alpha )$ the subclass of $\mathcal{A}$ consisting of all $f\in \mathcal{A}$ for which f is a strongly starlike function of order $\alpha$ in $\mathbb{U}$, with

$$\left|Arg\left({\displaystyle \frac{z{f}^{\prime }(z)}{f(z)}}\right)\right|<{\displaystyle \frac{\alpha \pi }{2}}(z\in \mathbb{U}).$$

Note that ${\tilde{\mathcal{S}}}^{*}(1)={\mathcal{S}}^{*}(0)={\mathcal{S}}^{*}$, the class of starlike functions in $\mathbb{U}$.

For $\alpha \in (0,1]$, we denote by $\tilde{\mathcal{C}}(\alpha )$ the subclass of $\mathcal{A}$ including all of $f\in \mathcal{A}$ for which

$$\left|Arg\left(f^{\prime }(z)\right)\right|<{\displaystyle \frac{\alpha \pi }{2}}(z\in \mathbb{U}).$$

Note that $\tilde{\mathcal{C}}(1)=\mathcal{C}$, the subclass of close-to-convex functions in $\mathbb{U}$. Here we understand that $Argw$ is a number in $(-\pi ,\pi ].$

For $\alpha \in (0,1]$, Nunokawa and Saitoh in [17] defined the more general class $\mathcal{G}(\alpha )$ consisting of all $f\in \mathcal{A}$ satisfying

$$Re\left(1+{\displaystyle \frac{z{f}^{″}(z)}{f^{\prime }(z)}}\right)<1+\frac{\alpha }{2}(z\in \mathbb{U}).$$

They proved that $\mathcal{G}(\alpha )$ is a subclass of ${\mathcal{S}}^{*}$. Ozaki in [18] showed that every function $\mathcal{G}(1)$ is univalent in the unit disk $\mathbb{U}$. In the following, Umezawa [19], Sakaguchi [20] and Singh and Singh [21] obtained some geometric properties of $\mathcal{G}(1)$ including, convex in one direction, close-to-convex and starlike, respectively. Obradović et al. in [22] proved the sharp coefficient bounds for the moduli of the Taylor coefficients $a_n$ of $f\in \mathcal{G}(\alpha )$ and determined the sharp bound for the Fekete-Szegö functional for functions in $\mathcal{G}(\alpha )$ with complex parameter $\lambda$. Also, Ponnusamy et al. [22,23] studied bounds for the logarithmic coefficients for functions in $\mathcal{G}(\alpha )$.

Here, we introduce a class as follows:

Definition 2.

For $\alpha ,\delta \in (0,1]$, we define the subclass $\mathcal{G}\left(\alpha ,\delta \right)$ of $\mathcal{A}$ as the following:

$$\mathcal{G}\left(\alpha ,\delta \right):=\left\{f\in \mathcal{A}:|Arg\left(\frac{2+\alpha }{\alpha }-\frac{2}{\alpha }\left(1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right)\right)|<{\displaystyle \frac{\delta \pi }{2}}(z\in \mathbb{U})\right\}.$$

It is clear that $\mathcal{G}\left(\alpha ,1\right)=\mathcal{G}(\alpha )$ for $\alpha \in (0,1]$. Let $\alpha ,\delta \in (0,1]$, identity function on $\mathbb{U}$ belongs to $\mathcal{G}\left(\alpha ,\delta \right)$ which implies that $\mathcal{G}\left(\alpha ,\delta \right)\ne \varnothing$. By means of the principle of subordination between analytic functions, we deduce

$$\mathcal{G}\left(\alpha ,\delta \right):=\left\{f\in \mathcal{A}:1+{\displaystyle \frac{z{f}^{″}(z)}{f^{\prime }(z)}}\prec -\frac{\alpha }{2}{\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\delta }+\frac{2+\alpha }{2}:=\varphi (z)(z\in \mathbb{U})\right\}.$$

(4)

Since the function f defined by

$${\displaystyle f(z)={\int }_0^{z}exp\left({\int }_0^x\frac{-\frac{\alpha }{2}{\left(\frac{1+t}{1-t}\right)}^{\delta }+\frac{\alpha }{2}}{t}\mathrm{d}t\right)\mathrm{d}x(z\in \mathbb{U})}$$

satisfies

$$1+{\displaystyle \frac{z{f}^{″}(z)}{f^{\prime }(z)}}=\varphi (z)\prec \varphi (z),$$

we deduce $f\in \mathcal{G}\left(\alpha ,\delta \right)$.

The aim of the present paper is to study some geometric properties for the class $\mathcal{G}\left(\alpha ,\delta \right)$ such as strongly starlikeness and close-to-convexity. Also we investigate sharp bounds on logarithmic coefficients and Fekete-Szegö functionals for functions belonging to the class $\mathcal{G}\left(\alpha ,\delta \right)$, which incorporate some known results as the special cases.

2. Some Properties of the Class $\mathcal{G}\left(\alpha ,\delta \right)$

We denote by Q the class of all complex-valued functions q for which q is univalent at each $\overline{\mathbb{U}}\backslash \mathrm{E}(q)$ and $q^{\prime }(\xi )\ne 0$ for all $\xi \in \partial \mathbb{U}\backslash \mathrm{E}(q)$ where

$$\mathrm{E}(q)=\left\{\xi \in \partial \mathbb{U}:\underset{z\to \xi }{lim}q(z)=\infty \right\}.$$

The following lemmas will be required to establish our main results.

Lemma 1

([24] (Lemma 2.2d (i))). Let $q\in Q$ with $q(0)=a$ and let $p(z)=a+p_n{z}^n+\dots$ be analytic in $\mathbb{U}$ with $p(z)\equiv 1$ and $n\ge 1$. If p is not subordinate to q in $\mathbb{U}$ then there exist $z_0\in \mathbb{U}$ and ${\xi }_0\in \partial \mathbb{U}\backslash E(q)$ such that $\left\{p\left(z\right):z\in \mathbb{U},|z|<|z_0|\right\}\subset q(\mathbb{U})$,

$$p(z_0)=q({\xi }_0).$$

Lemma 2.

(see [25,26]) Let the function p given by

$$p(z)=1+\sum _{n=1}^{\infty }{c}_n{z}^n$$

be analytic in $\mathbb{U}$ with $p(0)=1$ and $p(z)\ne 0$ for all $z\in \mathbb{U}.$ If there exists a point $z_0\in \mathbb{U}$ with

$$|arg\left(p(z)\right)|<{\displaystyle \frac{\beta \pi }{2}}(|z|<|z_0|)$$

and

$$|arg\left(p(z_0)\right)|={\displaystyle \frac{\beta \pi }{2}},$$

for some $\beta >0,$ then

$${\displaystyle \frac{z_0{p}^{\prime }(z_0)}{p(z_0)}}=ik\beta (i=\sqrt{-1}),$$

where

$$k\ge {\displaystyle \frac{a+a^{-1}}{2}}\ge 1\mathrm{when}arg\left(p(z_0)\right)={\displaystyle \frac{\beta \pi }{2}}$$

(5)

and

$$k\le -{\displaystyle \frac{a+a^{-1}}{2}}\le -1\mathrm{when}arg\left(p(z_0)\right)=-{\displaystyle \frac{\beta \pi }{2}},$$

(6)

where

$${\left[p(z_0)\right]}^{1/\beta }=\pm ia\mathrm{and}a>0.$$

Theorem 1.

Let $\alpha ,\beta \in (0,1]$. If $f\in \mathcal{A}$ satisfies the condition

$$|Arg\left(\frac{2+\alpha }{\alpha }-\frac{2}{\alpha }\left(1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right)\right)|<Arctan\left(\frac{4\beta }{2+\alpha }\right),$$

(7)

then

$$|Arg\left({\displaystyle \frac{z{f}^{\prime }(z)}{f(z)}}\right)|<{\displaystyle \frac{\beta \pi }{2}}(z\in \mathbb{U}).$$

Proof. 

Let $f\in \mathcal{A}$ and define the function $p:\mathbb{U}\to \mathbb{C}$ by

$$p(z)={\displaystyle \frac{z{f}^{\prime }(z)}{f(z)}}=1+\sum _{n=1}^{\infty }{c}_n{z}^n(z\in \mathbb{U}).$$

Then it follows that p is analytic in $\mathbb{U}$, $p(0)=1$,

$$1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}=p(z)+\frac{z{p}^{\prime }(z)}{p(z)}(z\in \mathbb{U})$$

and $p(z)\ne 0$ for all $z\in \mathbb{U}$. In fact, if p has a zero $z_0\in \mathbb{U}$ of order m, then we may write

$$p(z)={(z-z_0)}^m{p}_1(z)(m\in \mathbb{N}=1,2,3,\cdots ),$$

where $p_1$ is analytic in $\mathbb{U}$ with $p_1(z_0)\ne 0.$ Then

$$\frac{2+\alpha }{\alpha }-\frac{2}{\alpha }\left(p(z)+\frac{z{p}^{\prime }(z)}{p(z)}\right)=\frac{2+\alpha }{\alpha }-\frac{2}{\alpha }\left(p(z)+\frac{z{p}_1^{\prime }(z)}{p_1(z)}+\frac{mz}{z-z_0}\right).$$

Thus, choosing $z\to z_0$, suitably the argument of the right-hand of the above equality can take any value between $-\pi$ and $\pi$, which contradicts (7).

Define the function $q:\overline{\mathbb{U}}\backslash \left\{1\right\}\to \mathbb{C}$ by

$$q(z)={\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\beta }(z\in \overline{\mathbb{U}}\backslash \left\{1\right\}).$$

Then $q\in Q$, $q(0)=1$ and $\mathrm{E}(q)=\left\{1\right\}$. It is clear that $|Arg\left(p(z)\right)|<{\displaystyle \frac{\beta \pi }{2}}$ for all $z\in \mathbb{U}$ if and only if $p\prec q$ on $\mathbb{U}$. Let $|Arg\left(p(z_1)\right)|\ge {\displaystyle \frac{\beta \pi }{2}}$ for some $z_1\in \mathbb{U}$. Then p is not subordinate to q. By Lemma 1 there exists $z_0\in \mathbb{U}$ and ${\xi }_0\in \partial \mathbb{U}\backslash \left\{1\right\}$ such that $\left\{p\left(z\right):z\in \mathbb{U},|z|<|z_0|\right\}\subset q(\mathbb{U})$ and $p(z_0)=q({\xi }_0)$. Therefore,

$$|Arg\left(p(z)\right)|<{\displaystyle \frac{\beta \pi }{2}},$$

for all $z\in \mathbb{U}$ with $|z|<|z_0|$ and

$$|Arg\left(p(z_0)\right)|={\displaystyle \frac{\beta \pi }{2}}.$$

Then, Lemma 2, gives us that

$${\displaystyle \frac{z_0{p}^{\prime }(z_0)}{p(z_0)}}=ik\beta ,$$

where ${\left[p(z_0)\right]}^{\frac{1}{\beta }}=\pm ia(a>0)$ and k is given by (5) or (6).

Define the function $g:(0,a)\to \mathbb{R}$ by

$$g(t)=\frac{\frac{2}{2+\alpha }\left(t^{\beta }sin(\frac{\beta \pi }{2})+\beta \right)}{1-\frac{2}{2+\alpha }{t}^{\beta }cos\frac{\beta \pi }{2}}t\in (0,a).$$

Then g is a differentiable function on $(0,a)$ and $g^{\prime }(t)>0$ for all $t\in (0,a)$. This implies that the function $h:(0,a)\to \mathbb{R}$ defined by

$$h(t)=Arctan\left(g(t)\right)t\in (0,a),$$

is a non-decreasing function on $(0,a)$. Thus

$$h(a)\ge \underset{t\to 0^{+}}{lim}h(t)=Arctan\left(\frac{2\beta }{2+\alpha }\right).$$

Therefore, we have

$$Arctan\left(\frac{\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\ge Arctan\left(\frac{2\beta }{2+\alpha }\right).$$

(8)

Now we consider six cases for estimation of $Arg\left(p(z_0)\right)$ as follows:

Case 1. $Arg\left(p(z_0)\right)={\displaystyle \frac{\beta \pi }{2}}$ and $1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}>0$. In this case we have ${\left[p(z_0)\right]}^{\frac{1}{\beta }}=ia(a>0),$ and $k\ge 1$. Therefore,

$$\begin{array}{cc}\hfill Arg\left(\frac{2+\alpha }{\alpha }\left(1-\frac{2}{2+\alpha }\left(p(z_0)+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\right)& =Arg\left(1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}-i\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)\right)\hfill \\ \hfill & =Arctan\left(\frac{-\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\hfill \\ \hfill & \le Arctan\left(\frac{-\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\hfill \\ \hfill & =-Arctan\left(\frac{\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\hfill \\ \hfill & =-h(a)\hfill \\ \hfill & \le -Arctan\left(\frac{2\beta }{2+\alpha }\right).\hfill \end{array}$$

(9)

Now applying (8) and (9) we get

$$\begin{array}{cc}\hfill Arg\left(\frac{2+\alpha }{\alpha }\left(1-\frac{2}{2+\alpha }\left(p(z_0)+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\right)& =Arg\left(1-\frac{2}{2+\alpha }\left(p(z_0)+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\hfill \\ \hfill & =Arg\left(1-\frac{2}{2+\alpha }\left(1+\frac{z_0{f}^{\prime \prime }(z_0)}{f^{\prime }(z_0)}\right)\right)\hfill \\ \hfill & \le -Arctan\left(\frac{\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\hfill \\ \hfill & \le -Arctan\left(\frac{2\beta }{2+\alpha }\right),\hfill \end{array}$$

which contradicts (7).

Case 2. $Arg\left(p(z_0)\right)={\displaystyle \frac{\beta \pi }{2}}$ and $1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}=0$. In this case, we have $p(z_0)=a^{\beta }(cos\frac{\beta \pi }{2}+isin\frac{\beta \pi }{2})$ and $k\ge 1$. Thus $-\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)<0$ and so

$$\begin{array}{cc}\hfill Arg\left(\frac{2+\alpha }{\alpha }\left(1-\frac{2}{2+\alpha }\left(p(z_0)+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\right)& =Arg\left(-i\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)\right)\hfill \\ \hfill & =-\frac{\pi }{2}<-Arctan\left(\frac{2\beta }{2+\alpha }\right),\hfill \end{array}$$

which contradicts (7).

Case 3. $Arg\left(p(z_0)\right)={\displaystyle \frac{\beta \pi }{2}}$ and $1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}<0$. In this case, we have $p(z_0)=a^{\beta }(cos\frac{\beta \pi }{2}+isin\frac{\beta \pi }{2})$ and $k\ge 1$. Thus

$$\frac{-\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}>0.$$

Therefore,

$$\begin{array}{cc}\hfill Arg\left(\frac{2+\alpha }{\alpha }\left(1-\frac{2}{2+\alpha }\left(p(z_0)+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\right)& =Arg\left(1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}-i\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)\right)\hfill \\ \hfill & =-\pi +Arctan\left(\frac{-\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\hfill \\ \hfill & <-\pi +\frac{\pi }{2}\hfill \\ \hfill & =-\frac{\pi }{2}\hfill \\ \hfill & <-Arctan\left(\frac{2\beta }{2+\alpha }\right),\hfill \end{array}$$

which contradicts (7).

Case 4. $Arg\left(p(z_0)\right)=-{\displaystyle \frac{\beta \pi }{2}}$ and $1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}>0$. In this case we have $p(z_0)=a^{\beta }(cos\frac{\beta \pi }{2}-isin\frac{\beta \pi }{2})$ and $k\le -1$. Thus $-\frac{2}{2+\alpha }\left(-a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)<0$. Now, applying (8) we get

$$\begin{array}{cc}\hfill Arg\left(\frac{2+\alpha }{\alpha }\left(1-\frac{\alpha }{2+\alpha }\left(p(z_0)+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\right)& =Arg\left(1-\frac{2}{2+\alpha }\left(a^{\beta }{e}^{\frac{-i\beta \pi }{2}}+ik\beta \right)\right)\hfill \\ \hfill & =Arctan\left(\frac{-\frac{2}{2+\alpha }\left(-a^{\beta }sin\frac{\beta \pi }{2}+k\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\hfill \\ \hfill & \ge Arctan\left(\frac{-\frac{2}{2+\alpha }\left(-a^{\beta }sin\frac{\beta \pi }{2}-\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\hfill \\ \hfill & =Arctan\left(\frac{\frac{2}{2+\alpha }\left(a^{\beta }sin\frac{\beta \pi }{2}+\beta \right)}{1-\frac{2}{2+\alpha }{a}^{\beta }cos\frac{\beta \pi }{2}}\right)\hfill \\ \hfill & \ge Arctan\left(\frac{2\beta }{2+\alpha }\right),\hfill \end{array}$$

which contradicts (7).

For other cases applying the same method in Case 2. and Case 3. with $k\le -1$ we obtain

$$\begin{array}{c}\hfill Arg\left(\frac{2+\alpha }{\alpha }\left(1-\frac{2}{2+\alpha }\left(p(z_0)+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\right)\ge Arctan\left(\frac{2\beta }{2+\alpha }\right),\end{array}$$

which contradicts (7). Hence the proof is completed. □

Corollary 1.

Let $\alpha ,\beta \in (0,1]$ and $\delta =\frac{2}{\pi }Arctan\left(\frac{2\beta }{2+\alpha }\right).$ If $f\in \mathcal{G}\left(\alpha ,\delta \right)$, then $f\in {\tilde{\mathcal{S}}}^{*}(\beta )$.

Theorem 2.

Let $\alpha ,\beta \in (0,1]$. If $f\in \mathcal{A}$ and

$$|Arg\left(\frac{2+\alpha }{\alpha }-\frac{2}{\alpha }\left(1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}\right)\right)|<Arctan\left(\frac{2\beta }{\alpha }\right),$$

(10)

then

$$|Arg\left(f^{\prime }(z)\right)|<{\displaystyle \frac{\beta \pi }{2}}(z\in \mathbb{U}).$$

Proof. 

Define the function $p:\mathbb{U}\to \mathbb{C}$ by

$$p(z)=f^{\prime }(z)=1+\sum _{n=1}^{\infty }{c}_n{z}^n(z\in \mathbb{U}).$$

Then p is analytic in $\mathbb{U}$, $p(0)=1$,

$$1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}=1+\frac{z{p}^{\prime }(z)}{p(z)}.$$

and $p(z)\ne 0$ for all $z\in \mathbb{U}$. If there exists a point $z_0\in \mathbb{U}$ such that

$$|Arg\left(p(z)\right)|<{\displaystyle \frac{\beta \pi }{2}},$$

for all $z\in \mathbb{U}$ with $|z|<|z_0|$ and

$$|Arg\left(p(z_0)\right)|={\displaystyle \frac{\beta \pi }{2}}.$$

Then, Lemma 2, gives us that

$${\displaystyle \frac{z_0{p}^{\prime }(z_0)}{p(z_0)}}=ik\beta ,$$

where ${\left[p(z_0)\right]}^{\frac{1}{\beta }}=\pm ia(a>0)$ and k is given by (5) or (6).

For the case $Arg\left(p(z_0)\right)={\displaystyle \frac{\alpha \pi }{2}}$ when

$$p(z_0){]}^{\frac{1}{\beta }}=ia(a>0)$$

and $k\ge 1,$ we have

$$\begin{array}{cc}\hfill Arg\left(\frac{2+\alpha }{\alpha }\left(1-\frac{2}{2+\alpha }\left(1+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\right)& =Arg\left(1-\frac{2}{2+\alpha }\left(1+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\hfill \\ \hfill & =Arg\left(1-\frac{2}{2+\alpha }\left(1+ik\beta \right)\right)\hfill \\ \hfill & =Arctan\left(\frac{-2k\beta }{\alpha }\right)\hfill \\ \hfill & \le -Arctan\left(\frac{2\beta }{\alpha }\right),\hfill \end{array}$$

which contradicts (10).

Next, for the case $Arg\left(p(z_0)\right)=-{\displaystyle \frac{\alpha \pi }{2}}$ when

$$p(z_0)=-ia(a>0)$$

and $k\le -1,$ using the same method as before, we can obtain

$$\begin{array}{cc}\hfill Arg\left(\frac{2+\alpha }{\alpha }\left(1-\frac{2}{2+\alpha }\left(1+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\right)& =Arg\left(1-\frac{2}{2+\alpha }\left(1+\frac{z_0{p}^{\prime }(z_0)}{p(z_0)}\right)\right)\hfill \\ \hfill & =Arg\left(1-\frac{2}{2+\alpha }\left(1+ik\beta \right)\right)\hfill \\ \hfill & =Arctan\left(\frac{-2k\beta }{\alpha }\right)\hfill \\ \hfill & \ge Arctan\left(\frac{2\beta }{\alpha }\right),\hfill \end{array}$$

which is a contradicts (10).

Consequently, from the two above-discussed contradictions, it follows that

$$|Arg\left(f^{\prime }(z)\right)|<{\displaystyle \frac{\beta \pi }{2}}(z\in \mathbb{U}).$$

and hence the proof is completed. □

Corollary 2.

Let $\alpha ,\beta \in (0,1]$ and $\delta =\frac{2}{\pi }Arctan\left(\frac{2\beta }{\alpha }\right)$. If $f\in \mathcal{G}\left(\alpha ,\delta \right)$, then $f\in \tilde{\mathcal{C}}(\beta )$. In other words, if $f\in \mathcal{G}\left(\alpha ,\delta \right)$, then $f(z)$ is close-to-convex (univalent) in $\mathbb{U}$.

3. Coefficient Bounds

In this section, we give a the general problem of coefficients in the class $\mathcal{G}\left(\alpha ,\delta \right)$ like the estimates of coefficients for membership of this, bounds of logarithmic coefficients and the Fekete-Szegö problem with sharp inequalities. In order to achieve our aim we need to establish some knowledge.

Lemma 3

([27] (p. 172)). Let $\omega \in \mathsf{\Omega }$ with $\omega (z)={\displaystyle \sum _{n=1}^{\infty }}{w}_n{z}^n$ for all $z\in \mathbb{U}$. Then $|w_1|\le 1$ and

$$|w_n|\le 1-|w_1{|}^2\mathrm{for}\mathrm{all}n\in \mathbb{N}\mathrm{with}n\ge 2.$$

Lemma 4

([28] (Inequality 7, p. 10)). Let $\omega \in \mathsf{\Omega }$ with $\omega (z)={\displaystyle \sum _{n=1}^{\infty }}{w}_n{z}^n$ for all $z\in \mathbb{U}$. Then

$$|w_2-t{w}_1^2|\le max\{1,|t|\}\mathrm{for}\mathrm{all}t\in \mathbb{C}.$$

The inequality is sharp for the functions $\omega (z)=z^2$ or $\omega (z)=z$.

Lemma 5

([29]). If $\omega \in \mathsf{\Omega }$ with $\omega (z)={\displaystyle \sum _{n=1}^{\infty }}{w}_n{z}^n(z\in \mathbb{U})$, then for any real numbers $q_1$ and $q_2$, we have the following sharp estimate:

$$|p_3+q_1{w}_1{w}_2+q_2{w}_1^3|\le H(q_1;q_2),$$

where

$$\begin{array}{c}\hfill H(q_1;q_2)=\left\{\begin{array}{ccc}1\hfill & \mathrm{if}\hfill & (q_1,q_2)\in D_1\cup D_2\cup \left\{(2,1)\right\},\hfill \\ |q_2|\hfill & \mathrm{if}\hfill & (q_1,q_2)\in {\cup }_{k=3}^7{D}_k,\hfill \\ \frac{2}{3}(|q_1|+1){\left(\frac{|q_1|+1}{3(|q_1|+1+q_2)}\right)}^{\frac{1}{2}}\hfill & \mathrm{if}\hfill & (q_1,q_2)\in D_8\cup D_9,\hfill \\ \frac{q_2}{3}\left(\frac{q_1^2-4}{q_1^2-4q_2}\right){\left(\frac{q_1^2-4}{3(q_2-1)}\right)}^{\frac{1}{2}}\hfill & \mathrm{if}\hfill & (q_1,q_2)\in D_{10}\cup D_{11}\backslash \left\{(2,1)\right\},\hfill \\ \frac{2}{3}(|q_1|-1){\left(\frac{|q_1|-1}{3(|q_1|-1-q_2)}\right)}^{\frac{1}{2}}\hfill & \mathrm{if}\hfill & (q_1,q_2)\in D_{12},\hfill \end{array}\right.\end{array}$$

and the sets $D_k,k=1,2,\dots ,12$ are stated as given below:

$$\begin{array}{cc}\hfill & D_1=\left\{(q_1,q_2):|q_1|\le \frac{1}{2},|q_2|\le 1\right\},\hfill \\ \hfill & D_2=\left\{(q_1,q_2):\frac{1}{2}\le |q_1|\le 2,\frac{4}{27}\left((|q_1{|+1)}^3\right)-(|q_1|+1)\le q_2\le 1\right\},\hfill \\ \hfill & D_3=\left\{(q_1,q_2):|q_1|\le \frac{1}{2},q_2\le -1\right\},\hfill \end{array}$$

$$\begin{array}{cc}\hfill & D_4=\left\{(q_1,q_2):|q_1|\ge \frac{1}{2},|q_2|\le -\frac{2}{3}(|q_1|+1)\right\},\hfill \\ \hfill & D_5=\left\{(q_1,q_2):|q_1|\le 2,q_2\ge 1\right\},\hfill \\ \hfill & D_6=\left\{(q_1,q_2):2\le |q_1|\le 4,q_2\ge \frac{1}{12}(q_1^2+8)\right\},\hfill \\ \hfill & D_7=\left\{(q_1,q_2):|q_1|\ge 4,q_2\ge \frac{2}{3}(|q_1|-1)\right\},\hfill \\ \hfill & D_8=\left\{(q_1,q_2):\frac{1}{2}\le |q_1|\le 2,-\frac{2}{3}(|q_1|+1)\le q_2\le \frac{4}{27}\left((|q_1{|+1)}^3\right)-(|q_1|+1)\right\},\hfill \\ \hfill & D_9=\left\{(q_1,q_2):|q_1|\ge 2,-\frac{2}{3}(|q_1|+1)\le q_2\le \frac{2|q_1|(|q_1|+1)}{q_1^2+2|q_1|+4}\right\},\hfill \\ \hfill & D_{10}=\left\{(q_1,q_2):2\le |q_1|\le 4,\frac{2|q_1|(|q_1|+1)}{q_1^2+2|q_1|+4}\le q_2\le \frac{1}{12}(q_1^2+8)\right\},\hfill \\ \hfill & D_{11}=\left\{(q_1,q_2):|q_1|\ge 4,\frac{2|q_1|(|q_1|+1)}{q_1^2+2|q_1|+4}\le q_2\le \frac{2|q_1|(|q_1|-1)}{q_1^2-2|q_1|+4}\right\},\hfill \\ \hfill & D_{12}=\left\{(q_1,q_2):|q_1|\ge 4,\frac{2|q_1|(|q_1|-1)}{q_1^2-2|q_1|+4}\le q_2\le \frac{2}{3}(|q_1|-1)\right\}.\hfill \end{array}$$

We assume that $\phi$ is a univalent function in the unit disk $\mathbb{U}$ satisfying $\phi (0)=1$ such that it has the power series expansion of the following form

$$\phi (z)=1+B_{1}z+B_2{z}^2+B_3{z}^3+\dots ,z\in \mathbb{U},\mathrm{with}{B}_1\ne 0.$$

(11)

Lemma 6

([30] (Theorem 2)). Let the function $f\in \mathcal{K}(\phi )$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_1|\le {\displaystyle \frac{|B_1|}{4}},$$

(12)

$$\begin{array}{cc}\hfill |{\gamma }_2|\le & \left\{\begin{array}{ccc}{\displaystyle \frac{|B_1|}{12}}\hfill & \mathrm{if}\hfill & |4B_2+B_1^2|\le 4|B_1|,\hfill \\ \hfill & \hfill \\ {\displaystyle \frac{|4B_2+B_1^2|}{48}}\hfill & \mathrm{if}\hfill & |4B_2+B_1^2|>4|B_1|,\hfill \end{array}\right.\hfill \end{array}$$

(13)

and if $B_1,B_2$, and $B_3$ are real values,

$$\begin{array}{c}\hfill |{\gamma }_3|\le {\displaystyle \frac{|B_1|}{24}}H(q_1;q_2),\end{array}$$

(14)

where $H(q_1;q_2)$ is given by Lemma 5, $q_1=\frac{B_1+\frac{4B_2}{B_1}}{2}$ and $q_2=\frac{B_2+\frac{2B_3}{B_1}}{2}$. The bounds (12) and (13) are sharp.

Theorem 3.

Let $f\in \mathcal{G}\left(\alpha ,\delta \right)$. Then

$$|a_2|\le \frac{\alpha \delta }{2},|a_3|\le {\displaystyle \frac{\alpha \delta }{6}},|a_4|\le {\displaystyle \frac{\alpha \delta }{12}}H(q_1;q_2),$$

where $H(q_1;q_2)$ is given by Lemma 5,

$$q_1=\frac{-3\alpha \delta }{2}+2\delta \mathrm{and}{q}_2={\delta }^2\left(\frac{-3\alpha }{2}+\frac{{\alpha }^2}{2}+\frac{2}{3}\right)+\frac{1}{3}.$$

The first two bounds are sharp.

Proof. 

Set $g(z)=:z{f}^{\prime }(z)$, where $f\in \mathcal{G}\left(\alpha ,\delta \right)$ and suppose that $g(z)=z+{\displaystyle \sum _{n=2}^{\infty }}{b}_n{z}^n$. Hence $b_n=n{a}_n$ for $n\ge 1$. Then from (4), it follows that

$$\begin{array}{cc}\hfill {\displaystyle \frac{z{g}^{\prime }(z)}{g(z)}}& \prec -\frac{\alpha }{2}{\left({\displaystyle \frac{1+z}{1-z}}\right)}^{\delta }+\frac{2+\alpha }{2}=:\varphi (z)\hfill \\ \hfill & =1-\alpha \delta z-\alpha {\delta }^2{z}^2-\frac{1}{3}\alpha \delta (2{\delta }^2+1)z^3+\cdots \hfill \\ \hfill & :=1+B_{1}z+B_2{z}^2+B_3{z}^3+\cdots .\hfill \end{array}$$

Now, by the definition of the subordination, there is a $\omega \in \mathsf{\Omega }$ with $\omega (z)={\sum }_{n=1}^{\infty }{w}_n{z}^n$ so that

$$\begin{array}{cc}\hfill \frac{z{g}^{\prime }(z)}{g(z)}=& \varphi (\omega (z))\hfill \\ \hfill =& 1+B_1{w}_{1}z+(B_1{w}_2+B_2{w}_1^2)z^2+(B_1{w}_3+2w_1{w}_2{B}_2+B_3{w}_1^3)z^3+\cdots .\hfill \end{array}$$

From the above equality, it concludes that

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{b}_2=B_1{w}_1\hfill & \hfill \\ 2b_3-b_2^2=B_1{w}_2+B_2{w}_1^2\hfill & \hfill \\ 3b_4-3b_2{b}_3+b_2^3=B_1{w}_3+2w_1{w}_2{B}_2+B_3{w}_1^3.\hfill \end{array}\right.\end{array}$$

First, for $b_2$, from Lemma 3 we get $|b_2|\le \alpha \delta$, and so $|a_2|\le \frac{\alpha \delta }{2}$. Next, utilizing Lemma 3 for $b_3$ and using $|B_2+B_1^2|\le |B_1|$, we have

$$\begin{array}{cc}\hfill |b_3|\le & {\displaystyle \frac{|B_1|(1-|w_1{|}^2)+|B_2+B_1^2||w_1{|}^2}{2}}\hfill \\ \hfill =& {\displaystyle \frac{|B_1|+\left[|B_2+B_1^2|-|B_1|\right]{|w_1|}^2}{2}}\hfill \\ \hfill \le & {\displaystyle \frac{|B_1|}{2}}={\displaystyle \frac{\alpha \delta }{2}}.\hfill \end{array}$$

Ultimately, utilizing Lemma 5 for $a_4$, we have

$$\begin{array}{cc}\hfill |b_4|\le & {\displaystyle \frac{B_1}{3}}|c_3+\left({\displaystyle \frac{3}{2}}{B}_1+{\displaystyle \frac{2B_2}{B_1}}\right)w_1{w}_2+\left({\displaystyle \frac{3}{2}}{B}_2+{\displaystyle \frac{1}{2}}{B}_1^2+{\displaystyle \frac{B_3}{B_1}}\right)w_1^3|\hfill \\ \hfill \le & {\displaystyle \frac{B_1}{3}}H(q_1;q_2),\hfill \end{array}$$

where

$$q_1={\displaystyle \frac{3}{2}}{B}_1+{\displaystyle \frac{2B_2}{B_1}}=\frac{-3\alpha \delta }{2}+2\delta \mathrm{and}{q}_2={\displaystyle \frac{3}{2}}{B}_2+{\displaystyle \frac{1}{2}}{B}_1^2+{\displaystyle \frac{B_3}{B_1}}={\delta }^2\left(\frac{-3\alpha }{2}+\frac{{\alpha }^2}{2}+\frac{2}{3}\right)+\frac{1}{3}.$$

The extremal functions for the initial coefficients $a_n(n=2,3)$ are of the form:

$$\begin{array}{c}\hfill {\displaystyle f_n(z)={\int }_0^{z}exp\left({\int }_0^x\frac{\varphi (t^n)-1}{t}\mathrm{d}t\right)\mathrm{d}x=z-\frac{\alpha \beta }{n(n+1)}{z}^{n+1}+\frac{\alpha {\beta }^2(\alpha /n-1)}{2n(2n+1)}{z}^{2n+1}+\cdots ,}\end{array}$$

obtained by taking $\omega (z)=z^n$ in (4). Therefore, this completes the proof. □

Theorem 4.

Let $f\in \mathcal{G}\left(\alpha ,\delta \right)$. Then

$$|{\gamma }_1|\le {\displaystyle \frac{\alpha \delta }{4}},|{\gamma }_2|\le {\displaystyle \frac{\alpha \delta }{12}},|{\gamma }_3|\le {\displaystyle \frac{\alpha \delta }{24}}H(q_1;q_2),$$

where $H(q_1;q_2)$ is given by Lemma 5, $q_1=\frac{-\alpha \delta +4\delta }{2}$, and $q_2=\frac{-\alpha {\delta }^2+\frac{2(2{\delta }^2+1)}{3}}{2}$. The first two bounds are sharp.

Proof. 

The results are concluded from Theorem 6 by setting $\phi :=\varphi$. Also, two first bounds are sharp for $f_n(z)$ for $n=1,2$, respectively. Therefore, this completes the proof. □

Theorem 5.

Let $f\in \mathcal{G}\left(\alpha ,\delta \right)$. Then we have sharp inequalities for complex parameter μ

$$|a_3-\mu a_2^2| \le \left\{\begin{array}{ccc}\frac{\alpha {\delta }^2}{6}\left|1-\alpha +\frac{3\mu }{2}\alpha \right|\hfill & for& \left|\mu +\frac{2}{3\alpha }(1-\alpha )\right|\ge \frac{2}{3\alpha \delta },\hfill \\ \frac{\alpha \delta }{6}\hfill & for& \left|\mu +\frac{2}{3\alpha }(1-\alpha )\right|<\frac{2}{3\alpha \delta }.\hfill \end{array}\right.$$

Proof. 

Let $f\in \mathcal{G}\left(\alpha ,\delta \right)$, then from (4), by the definition of the subordination, there is a $\omega \in \mathsf{\Omega }$ with $\omega (z)={\sum }_{n=1}^{\infty }{w}_n{z}^n$ so that

$$\begin{array}{c}\hfill 1+\frac{z{f}^{″}(z)}{f^{\prime }(z)}=\varphi (\omega (z))=1+B_1{w}_{1}z+(B_1{w}_2+B_2{w}_1^2)z^2+\cdots .\end{array}$$

Therefore, we get that

$$2a_2=B_1{w}_1\mathrm{and}6a_3-4a_2^2=B_1{w}_2+B_2{w}_1^2.$$

Form the above equalities, we have

$$|a_3-\mu a_2^2|=\frac{1}{6}|B_1||w_2+\nu w_1^2|.$$

The results are obtained by the application of Lemma 4 with $\nu =\left[\frac{B_2}{B_1}+B_1(1-\frac{3\mu }{2})\right]$, where $B_1=-\alpha \delta$ and $B_2=-\alpha {\delta }^2$. Equality is attained in the first inequality by the function $f=f_1$ and in the second inequality for $f=f_2$. □

Remark 1.

(i) 

Taking into account $\delta =1$ in Theorem 3, we get the result obtained in [31] (Theorem 1) for $n=2,3,4$.

(ii) 

Setting $\delta =1$ in Theorem 3, we have the result obtained in [23] (Theorem 2.10).

(iii) 

Letting $\delta =1$ in Theorem 4, we obtain a correction of the result presented in [31] (Theorem 2).