*Article* **Argument and Coefficient Estimates for Certain Analytic Functions**

#### **Davood Alimohammadi 1, Nak Eun Cho 2,\*, Ebrahim Analouei Adegani 3 and Ahmad Motamednezhad 3**


Received: 3 December 2019; Accepted: 2 January 2020; Published: 5 January 2020

**Abstract:** The aim of the present paper is to introduce a new class G (*<sup>α</sup>*, *δ*) of analytic functions in the open unit disk and to study some properties associated with strong starlikeness and close-to-convexity for the class G (*<sup>α</sup>*, *<sup>δ</sup>*). We also consider sharp bounds of logarithmic coefficients and Fekete-Szegö functionals belonging to the class G (*<sup>α</sup>*, *<sup>δ</sup>*). Moreover, we provide some topics related to the results reported here that are relevant to outcomes presented in earlier research.

**Keywords:** starlike function; subordinate; univalent function

**MSC:** Primary 30C45; Secondary 30C80

#### **1. Introduction and Preliminaries**

Let U denote the open unit dick in the complex plane C. A function *ω* : U → C is called a *Schwarz function* if *ω* is a analytic function in U with *ω*(0) = 0 and |*ω*(*z*)| < 1 for all *z* ∈ U. Clearly, a Schwarz function *ω* is the form

$$
\omega(z) = w\_1 z + w\_2 z^2 + \dotsb \dotsb
$$

We denote by Ω the set of all Schwarz functions on U.

Let A be consisting of all analytic functions of the following normalized form:

$$f(z) = z + \sum\_{n=2}^{\infty} a\_n z^n,\tag{1}$$

in the open unit disk 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*(*<sup>z</sup>*1) = *f*(*<sup>z</sup>*2) ⇒ *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 S be the class of functions *f* ∈ A which are univalent in U. A domain *D* in the complex plane C is called *starlike* with respect to a point *w*0 ∈ *D*, if the line segmen<sup>t</sup> joining *w*0 to every other point *w* ∈ *D* lies in the interior of *D*. In other words, for any *w* ∈ *D* and 0 ≤ *t* ≤ 1, *tw*0 + (1 − *<sup>t</sup>*)*w* ∈ *D*. A function *f* ∈ A is starlike if the image *f*(*D*) is starlike with respect to the origin.

For two analytic functions *f* and *F* in U, we say that the function *f is subordinate to* the function *F* in U and we write *f*(*z*) ≺ *<sup>F</sup>*(*z*), if there exists a Schwarz function *ω* such that *f*(*z*) = *Fω*(*z*) for all *z* ∈ U. Specifically, if the function *F* is univalent in U, then we have the next equivalence:

$$f(z) \prec F(z) \iff f(0) = F(0) \quad \text{and} \quad f(\mathbb{U}) \subset F(\mathbb{U}).$$

The logarithmic coefficients *γn* of *f* ∈ S are defined with the following series expansion:

$$2\log\left(\frac{f(z)}{z}\right) = 2\sum\_{n=1}^{\infty} \gamma\_n(f)z^n, \; z \in \mathbb{U}.\tag{2}$$

These coefficients are an important factor in studying diverse estimates in the theory of univalent functions. Note that we use *γn* instead of *<sup>γ</sup>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

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

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* ∈ N*. The qth Hankel determinant is denote by Hq*(*n*) *and defined by*

$$H\_{\emptyset}(n) = \begin{vmatrix} a\_{\emptyset} & a\_{n+1} & \dots & a\_{n+q-1} \\ a\_{n+1} & a\_{n+2} & \dots & a\_{n+q} \\ \vdots & \vdots & \ddots & \vdots \\ a\_{n+q-1} & a\_{n+q} & \dots & a\_{n+2q-2} \end{vmatrix} \tag{3}$$

*where ak* (*k* = 1, 2, ...) *are the coefficients of the Taylor series expansion of a function f of the form* (1)*. Note that a*1 = 1*.*

The Hankel determinant *Hq*(*n*) was defined by Pommerenke [8,9] and for fixed *q*, *n* the bounds of |*Hq*(*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 *<sup>H</sup>*2(1) = *a*3 − *a*22 and *<sup>H</sup>*2(2) = *a*2*a*4 − *a*23, are well-known as *Fekete-Szegö* and *second Hankel determinant functionals*, respectively. In addition, Fekete and Szegö [11] introduced the generalized functional *a*3 − *<sup>λ</sup>a*22, where *λ* is a real number. Recently, Hankel determinants and other problems for various classes of bi-univalent functions have been studied, see [12–16].

For *α* ∈ [0, <sup>1</sup>), we denote by S∗(*α*) the subclass of A including of all *f* ∈ A for which *f* is a *starlike function of order α* in U, with

$$\operatorname{Re}\frac{zf'(z)}{f(z)} > \alpha \quad (z \in \mathbb{U}).$$

Also, for *α* ∈ (0, 1], we denote by S <sup>∗</sup>(*α*) the subclass of A consisting of all *f* ∈ A for which *f* is a *strongly starlike function of order α* in U, with

$$\left| \text{Arg} \left( \frac{zf'(z)}{f(z)} \right) \right| < \frac{\alpha \pi}{2} \quad (z \in \mathbb{U}).$$

Note that S ∗(1) = S∗(0) = S∗, the class of *starlike functions* in U.

For *α* ∈ (0, 1], we denote by C (*α*) the subclass of A including all of *f* ∈ A for which

$$\left| \text{Arg} \left( f'(z) \right) \right| < \frac{\alpha \pi}{2} \qquad (z \in \mathbb{U}).$$

Note that C (1) = C, the subclass of *close-to-convex functions* in U. Here we understand that Arg *w* is a number in (−*π*, *<sup>π</sup>*].

For *α* ∈ (0, 1], Nunokawa and Saitoh in [17] defined the more general class G(*α*) consisting of all *f* ∈ A satisfying

$$\operatorname{Re}\left(1+\frac{zf''(z)}{f'(z)}\right) < 1+\frac{\mathfrak{a}}{2} \qquad (z\in \mathbb{U}).$$

They proved that G(*α*) is a subclass of S∗. Ozaki in [18] showed that every function G(1) is univalent in the unit disk U. In the following, Umezawa [19], Sakaguchi [20] and Singh and Singh [21] obtained some geometric properties of G(1) including, convex in one direction, close-to-convex and starlike, respectively. Obradovi´c et al. in [22] proved the sharp coefficient bounds for the moduli of the Taylor coefficients *an* of *f* ∈ G(*α*) and determined the sharp bound for the Fekete-Szegö functional for functions in G(*α*) with complex parameter *λ*. Also, Ponnusamy et al. [22,23] studied bounds for the logarithmic coefficients for functions in G(*α*).

Here, we introduce a class as follows:

**Definition 2.** *For α*, *δ* ∈ (0, 1]*, we define the subclass* G (*<sup>α</sup>*, *δ*) *of* A *as the following:*

$$\mathcal{G}\left(\mathfrak{a},\delta\right) := \left\{ f \in \mathcal{A} : \left| \operatorname{Arg} \left( \frac{2+a}{a} - \frac{2}{a} \left( 1 + \frac{zf''(z)}{f'(z)} \right) \right) \right| < \frac{\delta \pi}{2} \quad (z \in \mathbb{U} \ ) \right\}.$$

It is clear that G (*<sup>α</sup>*, 1) = G(*α*) for *α* ∈ (0, 1]. Let *α*, *δ* ∈ (0, 1], identity function on U belongs to G (*<sup>α</sup>*, *δ*) which implies that G (*<sup>α</sup>*, *δ*) = ∅. By means of the principle of subordination between analytic functions, we deduce

$$\mathcal{G}\left(a,\delta\right) := \left\{ f \in \mathcal{A} : 1 + \frac{zf''(z)}{f'(z)} \prec -\frac{a}{2} \left( \frac{1+z}{1-z} \right)^{\delta} + \frac{2+a}{2} := \phi(z) \quad (z \in \mathbb{U}) \right\}.\tag{4}$$

Since the function *f* defined by

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

satisfies

$$1 + \frac{z f''(z)}{f'(z)} = \phi(z) \prec \phi(z).$$

we deduce *f* ∈ G (*<sup>α</sup>*, *<sup>δ</sup>*).

The aim of the present paper is to study some geometric properties for the class G (*<sup>α</sup>*, *δ*) 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 G (*<sup>α</sup>*, *<sup>δ</sup>*), which incorporate some known results as the special cases.

#### **2. Some Properties of the Class** G (*<sup>α</sup>*, *δ*)

We denote by *Q* the class of all complex-valued functions *q* for which *q* is univalent at each U \ <sup>E</sup>(*q*) and *q*(*ξ*) = 0 for all *ξ* ∈ *∂*U \ <sup>E</sup>(*q*) where

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

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

**Lemma 1** ([24] (Lemma 2.2d (i)))**.** *Let q* ∈ *Q with q*(0) = *a and let p*(*z*) = *a* + *pnz<sup>n</sup>* + ... *be analytic in* U *with p*(*z*) ≡ 1 *and n* ≥ 1*. If p is not subordinate to q in* U *then there exist z*0 ∈ U *and ξ*0 ∈ *∂*U \ *<sup>E</sup>*(*q*) *such that p* (*z*) : *z* ∈ U, |*z*| < |*<sup>z</sup>*0| ⊂ *q*(U)*,*

$$p(z\_0) = q(\mathfrak{z}\_0).$$

**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* U *with p*(0) = 1 *and p*(*z*) = 0 *for all z* ∈ U. *If there exists a point z*0 ∈ U *with*

$$|\arg\left(p(z)\right)| < \frac{\beta \pi}{2} \qquad \left(|z| < |z\_0|\right)$$

*and*

$$|\arg\left(p(z\_0)\right)| = \frac{\beta \pi}{2}\gamma$$

*for some β* > 0, *then*

$$\frac{z\_0 p'(z\_0)}{p(z\_0)} = ik\beta \qquad (i = \sqrt{-1})\nu$$

*where*

$$k \ge \frac{a + a^{-1}}{2} \ge 1 \quad \text{when} \quad \arg \left( p(z\_0) \right) = \frac{\beta \pi}{2} \tag{5}$$

*and*

$$k \le -\frac{a+a^{-1}}{2} \le -1 \quad \text{when} \quad \arg\left(p(z\_0)\right) = -\frac{\beta \pi}{2},\tag{6}$$

*where*

$$\left[p(z\_0)\right]^{1/\beta} = \pm ia \quad \text{and} \quad a > 0.$$

**Theorem 1.** *Let α*, *β* ∈ (0, 1]*. If f* ∈ A *satisfies the condition*

$$\left| \left| \operatorname{Arg} \left( \frac{2+\kappa}{\alpha} - \frac{2}{\alpha} \left( 1 + \frac{zf''(z)}{f'(z)} \right) \right) \right| \right| < \operatorname{Arg} \tan \left( \frac{4\beta}{2+\alpha} \right), \tag{7}$$

*then*

$$|\operatorname{Arg}\left(\frac{zf'(z)}{f(z)}\right)| < \frac{\beta\pi}{2} \quad (z \in \mathbb{U}).$$

**Proof.** Let *f* ∈ A and define the function *p* : U → C by

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

Then it follows that *p* is analytic in U, *p*(0) = 1,

$$1 + \frac{z f''(z)}{f'(z)} = p(z) + \frac{z p'(z)}{p(z)} \quad (z \in \mathbb{U})$$

and *p*(*z*) = 0 for all *z* ∈ U. In fact, if *p* has a zero *z*0 ∈ U of order *m*, then we may write

$$p(z) = (z - z\_0)^m p\_1(z) \quad (m \in \mathbb{N} = 1, 2, 3, \dots),$$

where *p*1 is analytic in U with *p*1(*<sup>z</sup>*0) = 0. Then

$$\frac{2+a}{a} - \frac{2}{a} \left( p(z) + \frac{z p'(z)}{p(z)} \right) = \frac{2+a}{a} - \frac{2}{a} \left( p(z) + \frac{z p\_1'(z)}{p\_1(z)} + \frac{mz}{z - z\_0} \right).$$

Thus, choosing *z* → *z*0, suitably the argumen<sup>t</sup> of the right-hand of the above equality can take any value between −*π* and *π*, which contradicts (7).

Define the function *q* : U \ {1} → C by

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

Then *q* ∈ *Q*, *q*(0) = 1 and <sup>E</sup>(*q*) = {1}. It is clear that Arg *p*(*z*) <sup>&</sup>lt; *βπ*2 for all *z* ∈ U if and only if *p* ≺ *q* on U. Let Arg *<sup>p</sup>*(*<sup>z</sup>*1) <sup>≥</sup> *βπ*2 for some *z*1 ∈ U. Then *p* is not subordinate to *q*. By Lemma 1 there exists *z*0 ∈ U and *ξ*0 ∈ *∂*U \ {1} such that *p* (*z*) : *z* ∈ U, |*z*| < |*<sup>z</sup>*0| ⊂ *q*(U) and *p*(*<sup>z</sup>*0) = *q*(*ξ*0). Therefore,

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

for all *z* ∈ U with |*z*| < |*<sup>z</sup>*0| and

$$|\operatorname{Arg}\left(p(z\_0)\right)| = \frac{\beta \pi}{2}.$$

Then, Lemma 2, gives us that

$$\frac{z\_0 p'(z\_0)}{p(z\_0)} = ik\beta\_\prime$$

where [*p*(*<sup>z</sup>*0)] 1 *β* = ±*ia* (*a* > 0) and *k* is given by (5) or (6).

Define the function *g* : (0, *a*) → R by

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

Then *g* is a differentiable function on (0, *a*) and *g*(*t*) > 0 for all *t* ∈ (0, *<sup>a</sup>*). This implies that the function *h* : (0, *a*) → R defined by

$$h(t) = \operatorname{Arctan}\left(\mathcal{g}(t)\right) \quad t \in (0, a)\_r$$

is a non-decreasing function on (0, *<sup>a</sup>*). Thus

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

*Mathematics* **2020**, *8*, 88

> Therefore, we have

$$\arctan\left(\frac{\frac{2}{2+\alpha}\left(a^{\beta}\sin\frac{\beta\pi}{2}+\beta\right)}{1-\frac{2}{2+\alpha}a^{\beta}\cos\frac{\beta\pi}{2}}\right) \ge \arctan\left(\frac{2\beta}{2+\alpha}\right).\tag{8}$$

Now we consider six cases for estimation of Arg *<sup>p</sup>*(*<sup>z</sup>*0) as follows:

Case 1. Arg *<sup>p</sup>*(*<sup>z</sup>*0) = *βπ*2 and 1 − 2 2+*α aβ* cos *βπ*2 > 0. In this case we have [*p*(*<sup>z</sup>*0)] 1*β* = *ia* (*a* > <sup>0</sup>), and *k* ≥ 1. Therefore,

$$\begin{split} \text{Arg}\left(\frac{2+\kappa}{a}\left(1-\frac{2}{2+a}\left(p(z\_0)+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right)\right) &= \text{Arg}\left(1-\frac{2}{2+a}a^{\theta}\cos\frac{\beta\pi}{2}-i\frac{2}{2+a}\left(a^{\theta}\sin\frac{\beta\pi}{2}+k\beta\right)\right) \\ &= \text{Arg}\left(\frac{-\frac{2}{2+a}\left(a^{\theta}\sin\frac{\beta\pi}{2}+k\beta\right)}{1-\frac{2}{2+a}a^{\theta}\cos\frac{\beta\pi}{2}}\right) \\ &\leq \text{Arg}\left(\frac{-\frac{2}{2+a}\left(a^{\theta}\sin\frac{\beta\pi}{2}+\beta\right)}{1-\frac{2}{2+a}a^{\theta}\cos\frac{\beta\pi}{2}}\right) \\ &= -\text{Area}\left(\frac{\frac{2}{2+a}\left(a^{\theta}\sin\frac{\beta\pi}{2}+\beta\right)}{1-\frac{2}{2+a}a^{\theta}\cos\frac{\beta\pi}{2}}\right) \\ &= -h(a) \\ &\leq -\text{Area}\left(\frac{2\beta}{2+a}\right). \end{split}$$

Now applying (8) and (9) we ge<sup>t</sup>

$$\begin{split} \mathrm{Arg}\left(\frac{2+\kappa}{\alpha}\left(1-\frac{2}{2+\kappa}\left(p(z\_{0})+\frac{z\_{0}p'(z\_{0})}{p(z\_{0})}\right)\right)\right) &= \mathrm{Arg}\left(1-\frac{2}{2+\kappa}\left(p(z\_{0})+\frac{z\_{0}p'(z\_{0})}{p(z\_{0})}\right)\right) \\ &= \mathrm{Arg}\left(1-\frac{2}{2+\kappa}\left(1+\frac{z\_{0}f''(z\_{0})}{f'(z\_{0})}\right)\right) \\ &\leq -\mathrm{Arg}\left(\frac{\frac{2}{2+\kappa}\left(a^{6}\sin\frac{\beta\pi}{2}+\beta\right)}{1-\frac{2}{2+\kappa}a^{6}\cos\frac{\beta\pi}{2}}\right) \\ &\leq -\mathrm{Arg}\left(\frac{2\beta}{2+\kappa}\right), \end{split}$$

which contradicts (7).

Case 2. Arg *<sup>p</sup>*(*<sup>z</sup>*0) = *βπ*2 and 1 − 2 2+*α aβ* cos *βπ*2 = 0. In this case, we have *p*(*<sup>z</sup>*0) = *aβ*(cos *βπ*2 + *i* sin *βπ*2 ) and *k* ≥ 1. Thus − 2 2+*α aβ* sin *βπ*2 + *kβ* < 0 and so

$$\begin{aligned} \operatorname{Arg}\left(\frac{2+a}{a}\left(1-\frac{2}{2+a}\left(p(z\_0)+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right)\right) &= \operatorname{Arg}\left(-i\frac{2}{2+a}\left(a^\beta \sin\frac{\beta\pi}{2}+k\beta\right)\right) \\ &= -\frac{\pi}{2} < -\operatorname{Arg}\left(\frac{2\beta}{2+a}\right), \end{aligned}$$

which contradicts (7).

Case 3. Arg *<sup>p</sup>*(*<sup>z</sup>*0) = *βπ*2 and 1 − 2 2+*α aβ* cos *βπ*2 < 0. In this case, we have *p*(*<sup>z</sup>*0) = *aβ*(cos *βπ*2 + *i* sin *βπ*2 ) and *k* ≥ 1. Thus

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

Therefore,

$$\begin{split} \text{Arg}\left(\frac{2+a}{a}\left(1-\frac{2}{2+a}\left(p(z\_0)+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right)\right) &= \text{Arg}\left(1-\frac{2}{2+a}a^{\beta}\cos\frac{\beta\pi}{2}-i\frac{2}{2+a}\left(a^{\beta}\sin\frac{\beta\pi}{2}+k\beta\right)\right) \\ &= -\pi+\text{Area}\left(\frac{-\frac{2}{2+a}\left(a^{\beta}\sin\frac{\beta\pi}{2}+k\beta\right)}{1-\frac{2}{2+a}a^{\beta}\cos\frac{\beta\pi}{2}}\right) \\ &< -\pi+\frac{\pi}{2} \\ &= -\frac{\pi}{2} \\ &< -\text{Area}\left(\frac{2\beta}{2+a}\right), \end{split}$$

which contradicts (7).

Case 4. Arg *<sup>p</sup>*(*<sup>z</sup>*0) = − *βπ*2 and 1 − 2 2+*α aβ* cos *βπ*2 > 0. In this case we have *p*(*<sup>z</sup>*0) = *aβ*(cos *βπ*2 − *i* sin *βπ*2 ) and *k* ≤ −1. Thus − 2 2+*α* −*<sup>a</sup><sup>β</sup>* sin *βπ*2 + *kβ* < 0. Now , applying (8) we ge<sup>t</sup>

$$\begin{split} \text{Arg}\left(\frac{2+a}{\pi}\left(1-\frac{a}{2+a}\left(p(z\_0)+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right)\right) &= \text{Arg}\left(1-\frac{2}{2+a}\left(a^{\beta}\frac{-i\delta\pi}{2}+ik\beta\right)\right) \\ &= \text{Arctan}\left(\frac{-\frac{2}{2+a}\left(-a^{\beta}\sin\frac{\beta\pi}{2}+k\beta\right)}{1-\frac{2}{2+a}a^{\beta}\cos\frac{\beta\pi}{2}}\right) \\ &\geq \text{Arctan}\left(\frac{-\frac{2}{2+a}\left(-a^{\beta}\sin\frac{\beta\pi}{2}-\beta\right)}{1-\frac{2}{2+a}a^{\beta}\cos\frac{\beta\pi}{2}}\right) \\ &= \text{Arctan}\left(\frac{\frac{2}{2+a}\left(a^{\beta}\sin\frac{\beta\pi}{2}+\beta\right)}{1-\frac{2}{2+a}a^{\beta}\cos\frac{\beta\pi}{2}}\right) \\ &\geq \text{Arctan}\left(\frac{2\beta}{2+a}\right), \end{split}$$

which contradicts (7).

> For other cases applying the same method in Case 2. and Case 3. with *k* ≤ −1 we obtain

$$\operatorname{Arg}\left(\frac{2+a}{a}\left(1-\frac{2}{2+a}\left(p(z\_0)+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right)\right) \ge \arctan\left(\frac{2\beta}{2+a}\right),$$

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

**Corollary 1.** *Let α*, *β* ∈ (0, 1] *and δ* = 2*π* Arctan 2*β* <sup>2</sup>+*α* . *If f* ∈ G (*<sup>α</sup>*, *<sup>δ</sup>*)*, then f* ∈ <sup>S</sup>∗(*β*)*.*

**Theorem 2.** *Let α*, *β* ∈ (0, 1]*. If f* ∈ A *and*

$$\left| \left| \operatorname{Arg} \left( \frac{2 + \kappa}{\alpha} - \frac{2}{\alpha} \left( 1 + \frac{z f''(z)}{f'(z)} \right) \right) \right| < \arctan \left( \frac{2\beta}{\alpha} \right) \right| \tag{10}$$

*then*

$$\left| \text{Arg}\left( f'(z) \right) \right| < \frac{\beta \pi}{2} \quad (z \in \mathbb{U}).$$

*Mathematics* **2020**, *8*, 88

**Proof.** Define the function *p* : U → C by

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

Then *p* is analytic in U, *p*(0) = 1,

$$1 + \frac{zf''(z)}{f'(z)} = 1 + \frac{zp'(z)}{p(z)}.$$

and *p*(*z*) = 0 for all *z* ∈ U. If there exists a point *z*0 ∈ U such that

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

for all *z* ∈ U with |*z*| < |*<sup>z</sup>*0| and

$$\left| \operatorname{Arg} \left( p(z\_0) \right) \right| = \frac{\beta \pi}{2}.$$

Then, Lemma 2, gives us that

$$\frac{z\_0 p'(z\_0)}{p(z\_0)} = ik\beta\_r$$

where [*p*(*<sup>z</sup>*0)] 1 *β* = ±*ia* (*a* > 0) and *k* is given by (5) or (6). For the case Arg *<sup>p</sup>*(*<sup>z</sup>*0) = *απ*2 when

$$[p(z\_0)]^{\frac{1}{\tilde{p}}} = ia \qquad (a > 0)$$

and *k* ≥ 1, we have

$$\begin{split} \mathrm{Arg}\left(\frac{2+\alpha}{\alpha}\left(1-\frac{2}{2+\alpha}\left(1+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right)\right) &= \mathrm{Arg}\left(1-\frac{2}{2+\alpha}\left(1+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right) \\ &= \mathrm{Arg}\left(1-\frac{2}{2+\alpha}\left(1+ik\beta\right)\right) \\ &= \mathrm{Arg}\tan\left(\frac{-2k\beta}{\alpha}\right) \\ &\leq -\mathrm{Arg}\tan\left(\frac{2\beta}{\alpha}\right), \end{split}$$

which contradicts (10).

Next, for the case Arg *<sup>p</sup>*(*<sup>z</sup>*0) = −*απ*2 when

$$p(z\_0) = -ia \qquad (a > 0)$$

and *k* ≤ −1, using the same method as before, we can obtain

$$\begin{split} \mathrm{Arg}\left(\frac{2+\alpha}{\alpha}\left(1-\frac{2}{2+\alpha}\left(1+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right)\right) &= \mathrm{Arg}\left(1-\frac{2}{2+\alpha}\left(1+\frac{z\_0 p'(z\_0)}{p(z\_0)}\right)\right) \\ &= \mathrm{Arg}\left(1-\frac{2}{2+\alpha}\left(1+ik\mathcal{G}\right)\right) \\ &= \mathrm{Arg}\tan\left(\frac{-2k\beta}{\alpha}\right) \\ &\ge \mathrm{Arg}\tan\left(\frac{2\beta}{\alpha}\right), \end{split}$$

which is a contradicts (10).

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

$$|\operatorname{Arg}\left(f'(z)\right)| < \frac{\beta\pi}{2} \quad (z \in \mathbb{U}).$$

and hence the proof is completed.

**Corollary 2.** *Let α*, *β* ∈ (0, 1] *and δ* = 2 *π* Arctan 2*β α . If f* ∈ G (*<sup>α</sup>*, *<sup>δ</sup>*)*, then f* ∈ C (*β*)*. In other words, if f* ∈ G (*<sup>α</sup>*, *<sup>δ</sup>*)*, then f*(*z*) *is close-to-convex (univalent) in* U*.*

#### **3. Coefficient Bounds**

In this section, we give a the general problem of coefficients in the class G (*<sup>α</sup>*, *δ*) 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 \Omega\$ with \$\omega(z) = \sum\_{n=1}^{\infty} w\_n z^n\$ for all \$z \in \mathbb{U}\$. Then \$|w\_1| \le 1\$ and 
$$|\omega\_1| < 1 \quad |\omega\_1|^2 \quad \text{for all } \omega \in \mathbb{U}\$. \newline \text{then } \omega > \omega\_1$$


**Lemma 4** ([28] (Inequality 7, p. 10))**.** *Let ω* ∈ Ω *with ω*(*z*) = ∞ ∑ *<sup>n</sup>*=1 *wnz<sup>n</sup> for all z* ∈ U*. Then*

$$|w\_2 - tw\_1^2| \le \max\{1, |t|\} \quad \text{for all } t \in \mathbb{C}.$$

*The inequality is sharp for the functions ω*(*z*) = *z*2 *or ω*(*z*) = *z.*

**Lemma 5** ([29])**.** *If ω* ∈ Ω *with ω*(*z*) = ∞ ∑ *<sup>n</sup>*=1 *wnz<sup>n</sup>* (*z* ∈ U)*, then for any real numbers q*1 *and q*<sup>2</sup>*, we have the following sharp estimate:*

$$|p\_3 + q\_1 w\_1 w\_2 + q\_2 w\_1^3| \le H(q\_1; q\_2)\_\lambda$$

*where*

$$H(q\_1;q\_2) = \begin{cases} 1 & \text{if } \quad (q\_1,q\_2) \in D\_1 \cup D\_2 \cup \{(2,1)\},\\ |q\_2| & \text{if } \quad (q\_1,q\_2) \in \mathbb{J}\_{k-3}^\top D\_{k},\\ \frac{2}{3}(|q\_1|+1) \left(\frac{|q\_1|+1}{3(|q\_1|+1+q\_2)}\right)^{\frac{1}{2}} & \text{if } \quad (q\_1,q\_2) \in D\_8 \cup D\_{8'}\\ \frac{q\_2}{3} \left(\frac{q\_1^2-4}{q\_1^2-4q\_2}\right) \left(\frac{q\_1^2-4}{3(q\_2-1)}\right)^{\frac{1}{2}} & \text{if } \quad (q\_1,q\_2) \in D\_{10} \cup D\_{11} \backslash \{(2,1)\},\\ \frac{2}{3}(|q\_1|-1) \left(\frac{|q\_1|-1}{3(|q\_1|-1-q\_2)}\right)^{\frac{1}{2}} & \text{if } \quad (q\_1,q\_2) \in D\_{12} \end{cases}$$

*and the sets Dk*, *k* = 1, 2, . . . , 12 *are stated as given below:*

$$\begin{aligned} D\_1 &= \left\{ (q\_1, q\_2) : |q\_1| \le \frac{1}{2}, \ |q\_2| \le 1 \right\}, \\ D\_2 &= \left\{ (q\_1, q\_2) : \frac{1}{2} \le |q\_1| \le 2, \ \frac{4}{27} \left( (|q\_1| + 1)^3 \right) - (|q\_1| + 1) \le q\_2 \le 1 \right\}, \\ D\_3 &= \left\{ (q\_1, q\_2) : |q\_1| \le \frac{1}{2}, \ q\_2 \le -1 \right\}, \end{aligned}$$

*Mathematics* **2020**, *8*, 88

*D*4 = (*q*1, *q*2) : |*q*1| ≥ 12 , |*q*2|≤− 23 (|*q*1| + 1) , *D*5 = {(*q*1, *q*2) : |*q*1| ≤ 2, *q*2 ≥ 1} , *D*6 = (*q*1, *q*2) : 2 ≤ |*q*1| ≤ 4, *q*2 ≥ 112 (*q*21 + 8) , *D*7 = (*q*1, *q*2) : |*q*1| ≥ 4, *q*2 ≥ 23 (|*q*1| − 1) , *D*8 = (*q*1, *q*2) : 12 ≤ |*q*1| ≤ 2, − 23 (|*q*1| + 1) ≤ *q*2 ≤ 427 (|*q*1<sup>|</sup> + 1)3 − (|*q*1| + 1) , *D*9 = (*q*1, *q*2) : |*q*1| ≥ 2, − 23 (|*q*1| + 1) ≤ *q*2 ≤ <sup>2</sup>|*q*1|(|*q*1| + 1) *q*21 + <sup>2</sup>|*q*1| + 4 , *D*10 = (*q*1, *q*2) : 2 ≤ |*q*1| ≤ 4, <sup>2</sup>|*q*1|(|*q*1| + 1) *q*21 + <sup>2</sup>|*q*1| + 4 ≤ *q*2 ≤ 112 (*q*21 + 8) , *D*11 = (*q*1, *q*2) : |*q*1| ≥ 4, <sup>2</sup>|*q*1|(|*q*1| + 1) *q*21 + <sup>2</sup>|*q*1| + 4 ≤ *q*2 ≤ <sup>2</sup>|*q*1|(|*q*1| − 1) *q*21 − <sup>2</sup>|*q*1| + 4 , *D*12 = (*q*1, *q*2) : |*q*1| ≥ 4, <sup>2</sup>|*q*1|(|*q*1| − 1) *q*21 − <sup>2</sup>|*q*1| + 4 ≤ *q*2 ≤ 23 (|*q*1| − 1) .

We assume that *ϕ* is a univalent function in the unit disk U satisfying *ϕ*(0) = 1 such that it has the power series expansion of the following form

$$\varphi(z) = 1 + B\_1 z + B\_2 z^2 + B\_3 z^3 + \dots, \; z \in \mathbb{U}, \quad \text{with} \quad B\_1 \neq 0. \tag{11}$$

**Lemma 6** ([30] (Theorem 2))**.** *Let the function f* ∈ K(*ϕ*)*. Then the logarithmic coefficients of f satisfy the inequalities*


$$|\gamma\_2| \le \begin{cases} \frac{|B\_1|}{12} & \text{if } \quad |4B\_2 + B\_1^2| \le 4|B\_1|, \\\\ \frac{|4B\_2 + B\_1^2|}{48} & \text{if } \quad |4B\_2 + B\_1^2| > 4|B\_1|, \end{cases} \tag{13}$$

*and if B*1, *B*2*, and B*3 *are real values,*

$$|\gamma\_3| \le \frac{|B\_1|}{24} H(q\_1; q\_2),\tag{14}$$

*where <sup>H</sup>*(*q*1; *q*2) *is given by Lemma* 5*, q*1 = *B*1+ <sup>4</sup>*B*2 *B*1 2 *and q*2 = *B*2+ <sup>2</sup>*B*3 *B*1 2 *. The bounds* (12) *and* (13) *are sharp.*

**Theorem 3.** *Let f* ∈ G (*<sup>α</sup>*, *δ*)*. Then*

$$|a\_2| \le \frac{\alpha \delta}{2}, \quad |a\_3| \le \frac{\alpha \delta}{6}, \quad |a\_4| \le \frac{\alpha \delta}{12} H(q\_1; q\_2)\_\delta$$

*where <sup>H</sup>*(*q*1; *q*2) *is given by Lemma* 5*,*

$$q\_1 = \frac{-3\alpha\delta}{2} + 2\delta \quad \text{and} \quad q\_2 = \delta^2 \left(\frac{-3\alpha}{2} + \frac{\alpha^2}{2} + \frac{2}{3}\right) + \frac{1}{3}.$$

*The first two bounds are sharp.* **Proof.** Set *g*(*z*) =: *z f* (*z*), where *f* ∈ G (*<sup>α</sup>*, *δ*) and suppose that *g*(*z*) = *z* + ∞ ∑ *n*=2 *bnzn*. Hence *bn* = *nan* for *n* ≥ 1. Then from (4), it follows that

$$\begin{aligned} \frac{zg'(z)}{g(z)} &\prec -\frac{\alpha}{2} \left( \frac{1+z}{1-z} \right)^{\delta} + \frac{2+\alpha}{2} =: \phi(z) \\ &= 1 - \alpha \delta z - \alpha \delta^2 z^2 - \frac{1}{3} \alpha \delta (2\delta^2 + 1) z^3 + \dotsb \\ &:= 1 + B\_1 z + B\_2 z^2 + B\_3 z^3 + \dotsb \ . \end{aligned}$$

Now, by the definition of the subordination, there is a *ω* ∈ Ω with *ω*(*z*) = ∑∞*<sup>n</sup>*=<sup>1</sup>*wnz<sup>n</sup>* so that

$$\begin{aligned} \frac{z g'(z)}{g(z)} &= \phi(\omega(z)) \\ &= 1 + B\_1 w\_1 z + (B\_1 w\_2 + B\_2 w\_1^2) z^2 + (B\_1 w\_3 + 2 w\_1 w\_2 B\_2 + B\_3 w\_1^3) z^3 + \cdots \end{aligned}$$

From the above equality, it concludes that

$$\begin{cases} b\_2 = B\_1 w\_1 \\ 2b\_3 - b\_2^2 = B\_1 w\_2 + B\_2 w\_1^2 \\ 3b\_4 - 3b\_2 b\_3 + b\_2^3 = B\_1 w\_3 + 2w\_1 w\_2 B\_2 + B\_3 w\_1^3. \end{cases}$$

First, for *b*2, from Lemma 3 we ge<sup>t</sup> |*b*2| ≤ *αδ*, and so |*<sup>a</sup>*2| ≤ *αδ*2 . Next, utilizing Lemma 3 for *b*3 and using |*<sup>B</sup>*2 + *<sup>B</sup>*21|≤|*<sup>B</sup>*1|, we have

$$\begin{aligned} |b\_3| \leq & \frac{|B\_1|(1 - |w\_1|^2) + |B\_2 + B\_1^2| |w\_1|^2}{2} \\ = & \frac{|B\_1| + \left\lceil |B\_2 + B\_1^2| - |B\_1| \right\rceil |w\_1|^2}{2} \\ \leq & \frac{|B\_1|}{2} = \frac{\alpha \delta}{2} .\end{aligned}$$

Ultimately, utilizing Lemma 5 for *a*4, we have

$$\begin{aligned} |b\_4| &\le \frac{B\_1}{3} \left| c\_3 + \left( \frac{3}{2} B\_1 + \frac{2B\_2}{B\_1} \right) w\_1 w\_2 + \left( \frac{3}{2} B\_2 + \frac{1}{2} B\_1^2 + \frac{B\_3}{B\_1} \right) w\_1^3 \right| \\ &\le \frac{B\_1}{3} H(q\_1; q\_2), \end{aligned}$$

where

$$q\_1 = \frac{3}{2}B\_1 + \frac{2B\_2}{B\_1} = \frac{-3a\delta}{2} + 2\delta \quad \text{and} \quad q\_2 = \frac{3}{2}B\_2 + \frac{1}{2}B\_1^2 + \frac{B\_3}{B\_1} = \delta^2 \left(\frac{-3a}{2} + \frac{a^2}{2} + \frac{2}{3}\right) + \frac{1}{3}A$$

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

$$f\_n(z) = \int\_0^z \exp\left(\int\_0^x \frac{\phi(t^n) - 1}{t} dt\right) \mathrm{d}x = z - \frac{a\beta}{n(n+1)} z^{n+1} + \frac{a\beta^2 (a/n - 1)}{2n(2n+1)} z^{2n+1} + \cdots, \quad z \in \mathbb{R}$$

obtained by taking *ω*(*z*) = *z<sup>n</sup>* in (4). Therefore, this completes the proof.

**Theorem 4.** *Let f* ∈ G (*<sup>α</sup>*, *δ*)*. Then*

$$|\gamma\_1| \le \frac{\alpha \delta}{4}, \quad |\gamma\_2| \le \frac{\alpha \delta}{12}, \quad |\gamma\_3| \le \frac{\alpha \delta}{24} H(q\_1; q\_2)\_{\prime}$$

*where <sup>H</sup>*(*q*1; *q*2) *is given by Lemma* 5*, q*1 = −*αδ*+4*δ* 2 *, and q*2 = −*αδ*2<sup>+</sup> <sup>2</sup>(2*δ*2+<sup>1</sup>) 3 2 *. The first two bounds are sharp.*

**Proof.** The results are concluded from Theorem 6 by setting *ϕ* := *φ*. Also, two first bounds are sharp for *fn*(*z*) for *n* = 1, 2, respectively. Therefore, this completes the proof.

**Theorem 5.** *Let f* ∈ G (*<sup>α</sup>*, *δ*)*. Then we have sharp inequalities for complex parameter μ*

$$\left| a\_3 - \mu a\_2^2 \right| \le \begin{cases} \frac{a\delta^2}{6} \left| 1 - \alpha + \frac{3\mu}{2}\alpha \right| & \text{for} \quad \left| \mu + \frac{2}{3\alpha}(1 - \alpha) \right| \ge \frac{2}{3\alpha\delta}, \\\\ \frac{a\delta}{6} & \text{for} \quad \left| \mu + \frac{2}{3\alpha}(1 - \alpha) \right| < \frac{2}{3\alpha\delta}. \end{cases}$$

**Proof.** Let *f* ∈ G (*<sup>α</sup>*, *<sup>δ</sup>*), then from (4), by the definition of the subordination, there is a *ω* ∈ Ω with *ω*(*z*) = ∑∞*<sup>n</sup>*=<sup>1</sup> *wnz<sup>n</sup>* so that

$$1 + \frac{z f''(z)}{f'(z)} = \phi(\omega(z)) = 1 + B\_1 w\_1 z + (B\_1 w\_2 + B\_2 w\_1^2) z^2 + \dotsb \dotsb$$

Therefore, we ge<sup>t</sup> that

$$2a\_2 = B\_1 w\_1 \qquad \text{and} \qquad 6a\_3 - 4a\_2^2 = B\_1 w\_2 + B\_2 w\_1^2.$$

Form the above equalities, we have

$$\left| a\_3 - \mu a\_2^2 \right| = \frac{1}{6} |B\_1| \left| w\_2 + \nu w\_1^2 \right|.$$

The results are obtained by the application of Lemma 4 with *ν* = &*B*2 *B*1 + *<sup>B</sup>*1(<sup>1</sup> − 3*μ*2 )', where *B*1 = −*αδ* and *B*2 = −*αδ*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.**


**Author Contributions:** Investigation, D.A., N.E.C., E.A.A. and A.M.; Writing—original draft, E.A.A.; Writing—review and editing, N.E.C. The authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript.

**Funding:** The second author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2019R1I1A3A01050861).

**Acknowledgments:** The authors would like to express their gratitude to the referees for many valuable suggestions regarding a previous version of this paper.

**Conflicts of Interest:** The authors declare no conflict of interest.
