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Mathematics 2020, 8(1), 88; https://doi.org/10.3390/math8010088

Article
Argument and Coefficient Estimates for Certain Analytic Functions
by Davood Alimohammadi 1, Nak Eun Cho 2,* , Ebrahim Analouei Adegani 3 and Ahmad Motamednezhad 3
1
Department of Mathematics, Faculty of Science, Arak University, Arak 38156-8-8349, Iran
2
Department of Applied Mathematics, College of Natural Sciences, Pukyong National University, Busan 608-737, Korea
3
Faculty of Mathematical Sciences, Shahrood University of Technology, Shahrood P.O. Box 316-36155, Iran
*
Author to whom correspondence should be addressed.
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 $G α , δ$ of analytic functions in the open unit disk and to study some properties associated with strong starlikeness and close-to-convexity for the class $G α , δ$. We also consider sharp bounds of logarithmic coefficients and Fekete-Szegö functionals belonging to the class $G α , δ$. 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
$ω ( z ) = w 1 z + w 2 z 2 + ⋯ .$
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:
$f ( z ) = z + ∑ n = 2 ∞ a n z n ,$
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: $f ( z 1 ) = f ( z 2 ) ⇒ z 1 = z 2 .$ 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 segment joining $w 0$ to every other point $w ∈ D$ lies in the interior of D. In other words, for any $w ∈ D$ and $0 ≤ t ≤ 1 , t w 0 + ( 1 − t ) w ∈ D$. 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 $f ( z ) ≺ F ( z )$, if there exists a Schwarz function $ω$ such that $f ( z ) = F ω ( z )$ for all $z ∈ U$. Specifically, if the function F is univalent in $U$, then we have the next equivalence:
$f ( z ) ≺ F ( z ) ⟺ f ( 0 ) = F ( 0 ) and f ( U ) ⊂ F ( U ) .$
The logarithmic coefficients $γ n$ of $f ∈ S$ are defined with the following series expansion:
$log f ( z ) z = 2 ∑ n = 1 ∞ γ n ( f ) z n , z ∈ U .$
These coefficients are an important factor in studying diverse estimates in the theory of univalent functions. Note that we use $γ n$ instead of $γ n ( f )$. The concept of logarithmic coefficients inspired Kayumov  to solve Brennan’s conjecture for conformal mappings. The importance of the logarithmic coefficients follows from Lebedev-Milin inequalities  (Chapter 2), see also [3,4], where estimates of the logarithmic coefficients were used to find bounds on the coefficients of f. Milin  conjectured the inequality
$∑ m = 1 n ∑ k = 1 m k | γ k | 2 − 1 k ≤ 0 ( n = 1 , 2 , 3 , ⋯ ) ,$
which implies Robertson’s conjecture , and hence, Bieberbach’s conjecture . This is the famous coefficient problem in univalent function theory. L. de Branges  established Bieberbach’s conjecture by proving Milin’s conjecture.
Definition 1.
Let $q , n ∈ N$. The $q t h$ Hankel determinant is denote by $H q ( n )$ and defined by
$H q ( n ) = a n a n + 1 ⋯ a n + q − 1 a n + 1 a n + 2 ⋯ a n + q ⋮ ⋮ ⋱ ⋮ a n + q − 1 a n + q ⋯ a n + 2 q − 2 ,$
where $a k ( k = 1 , 2 , … )$ are the coefficients of the Taylor series expansion of a function f of the form (1). Note that $a 1 = 1$.
The Hankel determinant $H q ( n )$ 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  (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  introduced the generalized functional $a 3 − λ a 2 2$, where $λ$ 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 $α ∈ [ 0 , 1 )$, 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
$Re z f ′ ( z ) f ( z ) > α ( z ∈ U ) .$
Also, for $α ∈ ( 0 , 1 ]$, we denote by $S ˜ * ( α )$ the subclass of $A$ consisting of all $f ∈ A$ for which f is a strongly starlike function of order $α$ in $U$, with
$Arg z f ′ ( z ) f ( z ) < α π 2 ( z ∈ U ) .$
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
$Arg f ′ ( z ) < α π 2 ( z ∈ U ) .$
Note that $C ˜ ( 1 ) = C$, the subclass of close-to-convex functions in $U$. Here we understand that $Arg w$ is a number in $( − π , π ] .$
For $α ∈ ( 0 , 1 ]$, Nunokawa and Saitoh in  defined the more general class $G ( α )$ consisting of all $f ∈ A$ satisfying
$Re 1 + z f ″ ( z ) f ′ ( z ) < 1 + α 2 ( z ∈ U ) .$
They proved that $G ( α )$ is a subclass of $S *$. Ozaki in  showed that every function $G ( 1 )$ is univalent in the unit disk $U$. In the following, Umezawa , Sakaguchi  and Singh and Singh  obtained some geometric properties of $G ( 1 )$ including, convex in one direction, close-to-convex and starlike, respectively. Obradović et al. in  proved the sharp coefficient bounds for the moduli of the Taylor coefficients $a n$ 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 $G α , δ$ of $A$ as the following:
$G α , δ : = f ∈ A : | Arg 2 + α α − 2 α 1 + z f ″ ( z ) f ′ ( z ) | < δ π 2 ( z ∈ U ) .$
It is clear that $G α , 1 = G ( α )$ for $α ∈ ( 0 , 1 ]$. Let $α , δ ∈ ( 0 , 1 ]$, identity function on $U$ belongs to $G α , δ$ which implies that $G α , δ ≠ ∅$. By means of the principle of subordination between analytic functions, we deduce
$G α , δ : = f ∈ A : 1 + z f ″ ( z ) f ′ ( z ) ≺ − α 2 1 + z 1 − z δ + 2 + α 2 : = ϕ ( z ) ( z ∈ U ) .$
Since the function f defined by
$f ( z ) = ∫ 0 z exp ∫ 0 x − α 2 1 + t 1 − t δ + α 2 t d t d x ( z ∈ U )$
satisfies
$1 + z f ″ ( z ) f ′ ( z ) = ϕ ( z ) ≺ ϕ ( z ) ,$
we deduce $f ∈ G α , δ$.
The aim of the present paper is to study some geometric properties for the class $G α , δ$ 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 $G α , δ$, which incorporate some known results as the special cases.

## 2. Some Properties of the Class $G α , δ$

We denote by Q the class of all complex-valued functions q for which q is univalent at each $U ¯ \ E ( q )$ and $q ′ ( ξ ) ≠ 0$ for all $ξ ∈ ∂ U \ E ( q )$ where
$E ( q ) = ξ ∈ ∂ U : lim z → ξ q ( z ) = ∞ .$
The following lemmas will be required to establish our main results.
Lemma 1
( (Lemma 2.2d (i))). Let $q ∈ Q$ with $q ( 0 ) = a$ and let $p ( z ) = a + p n z n + …$ 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 $ξ 0 ∈ ∂ U \ E ( q )$ such that $p z : z ∈ U , | z | < | z 0 | ⊂ q ( U )$,
$p ( z 0 ) = q ( ξ 0 ) .$
Lemma 2.
(see [25,26]) Let the function p given by
$p ( z ) = 1 + ∑ n = 1 ∞ c n z n$
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
$| arg p ( z ) | < β π 2 ( | z | < | z 0 | )$
and
$| arg p ( z 0 ) | = β π 2 ,$
for some $β > 0 ,$ then
$z 0 p ′ ( z 0 ) p ( z 0 ) = i k β ( i = − 1 ) ,$
where
$k ≥ a + a − 1 2 ≥ 1 when arg p ( z 0 ) = β π 2$
and
$k ≤ − a + a − 1 2 ≤ − 1 when arg p ( z 0 ) = − β π 2 ,$
where
$p ( z 0 ) 1 / β = ± i a and a > 0 .$
Theorem 1.
Let $α , β ∈ ( 0 , 1 ]$. If $f ∈ A$ satisfies the condition
$| Arg 2 + α α − 2 α 1 + z f ″ ( z ) f ′ ( z ) | < Arctan 4 β 2 + α ,$
then
$| Arg z f ′ ( z ) f ( z ) | < β π 2 ( z ∈ U ) .$
Proof.
Let $f ∈ A$ and define the function $p : U → C$ by
$p ( z ) = z f ′ ( z ) f ( z ) = 1 + ∑ n = 1 ∞ c n z n ( z ∈ U ) .$
Then it follows that p is analytic in $U$, $p ( 0 ) = 1$,
$1 + z f ″ ( z ) f ′ ( z ) = p ( z ) + z p ′ ( z ) p ( z ) ( z ∈ U )$
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
$p ( z ) = ( z − z 0 ) m p 1 ( z ) ( m ∈ N = 1 , 2 , 3 , ⋯ ) ,$
where $p 1$ is analytic in $U$ with $p 1 ( z 0 ) ≠ 0 .$ Then
$2 + α α − 2 α p ( z ) + z p ′ ( z ) p ( z ) = 2 + α α − 2 α p ( z ) + z p 1 ′ ( z ) p 1 ( z ) + m z z − z 0 .$
Thus, choosing $z → z 0$, suitably the argument 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
$q ( z ) = 1 + z 1 − z β ( z ∈ U ¯ \ { 1 } ) .$
Then $q ∈ Q$, $q ( 0 ) = 1$ and $E ( q ) = { 1 }$. It is clear that $| Arg p ( z ) | < β π 2$ for all $z ∈ U$ if and only if $p ≺ q$ on $U$. Let $| Arg p ( z 1 ) | ≥ β π 2$ 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 $p z : z ∈ U , | z | < | z 0 | ⊂ q ( U )$ and $p ( z 0 ) = q ( ξ 0 )$. Therefore,
$| Arg p ( z ) | < β π 2 ,$
for all $z ∈ U$ with $| z | < | z 0 |$ and
$| Arg p ( z 0 ) | = β π 2 .$
Then, Lemma 2, gives us that
$z 0 p ′ ( z 0 ) p ( z 0 ) = i k β ,$
where $[ p ( z 0 ) ] 1 β = ± i a ( a > 0 )$ and k is given by (5) or (6).
Define the function $g : ( 0 , a ) → R$ by
$g ( t ) = 2 2 + α t β sin ( β π 2 ) + β 1 − 2 2 + α t β cos β π 2 t ∈ ( 0 , a ) .$
Then g is a differentiable function on $( 0 , a )$ and $g ′ ( t ) > 0$ for all $t ∈ ( 0 , a )$. This implies that the function $h : ( 0 , a ) → R$ defined by
$h ( t ) = Arctan g ( t ) t ∈ ( 0 , a ) ,$
is a non-decreasing function on $( 0 , a )$. Thus
$h ( a ) ≥ lim t → 0 + h ( t ) = Arctan 2 β 2 + α .$
Therefore, we have
$Arctan 2 2 + α a β sin β π 2 + β 1 − 2 2 + α a β cos β π 2 ≥ Arctan 2 β 2 + α .$
Now we consider six cases for estimation of $Arg p ( z 0 )$ as follows:
Case 1. $Arg p ( z 0 ) = β π 2$ and $1 − 2 2 + α a β cos β π 2 > 0$. In this case we have $[ p ( z 0 ) ] 1 β = i a ( a > 0 ) ,$ and $k ≥ 1$. Therefore,
$Arg 2 + α α 1 − 2 2 + α p ( z 0 ) + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α a β cos β π 2 − i 2 2 + α a β sin β π 2 + k β = Arctan − 2 2 + α a β sin β π 2 + k β 1 − 2 2 + α a β cos β π 2 ≤ Arctan − 2 2 + α a β sin β π 2 + β 1 − 2 2 + α a β cos β π 2 = − Arctan 2 2 + α a β sin β π 2 + β 1 − 2 2 + α a β cos β π 2 = − h ( a ) ≤ − Arctan 2 β 2 + α .$
Now applying (8) and (9) we get
$Arg 2 + α α 1 − 2 2 + α p ( z 0 ) + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α p ( z 0 ) + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α 1 + z 0 f ′ ′ ( z 0 ) f ′ ( z 0 ) ≤ − Arctan 2 2 + α a β sin β π 2 + β 1 − 2 2 + α a β cos β π 2 ≤ − Arctan 2 β 2 + α ,$
Case 2. $Arg p ( z 0 ) = β π 2$ and $1 − 2 2 + α a β cos β π 2 = 0$. In this case, we have $p ( z 0 ) = a β ( cos β π 2 + i sin β π 2 )$ and $k ≥ 1$. Thus $− 2 2 + α a β sin β π 2 + k β < 0$ and so
$Arg 2 + α α 1 − 2 2 + α p ( z 0 ) + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg − i 2 2 + α a β sin β π 2 + k β = − π 2 < − Arctan 2 β 2 + α ,$
Case 3. $Arg p ( z 0 ) = β π 2$ and $1 − 2 2 + α a β cos β π 2 < 0$. In this case, we have $p ( z 0 ) = a β ( cos β π 2 + i sin β π 2 )$ and $k ≥ 1$. Thus
$− 2 2 + α a β sin β π 2 + k β 1 − 2 2 + α a β cos β π 2 > 0 .$
Therefore,
$Arg 2 + α α 1 − 2 2 + α p ( z 0 ) + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α a β cos β π 2 − i 2 2 + α a β sin β π 2 + k β = − π + Arctan − 2 2 + α a β sin β π 2 + k β 1 − 2 2 + α a β cos β π 2 < − π + π 2 = − π 2 < − Arctan 2 β 2 + α ,$
Case 4. $Arg p ( z 0 ) = − β π 2$ and $1 − 2 2 + α a β cos β π 2 > 0$. In this case we have $p ( z 0 ) = a β ( cos β π 2 − i sin β π 2 )$ and $k ≤ − 1$. Thus $− 2 2 + α − a β sin β π 2 + k β < 0$. Now, applying (8) we get
$Arg 2 + α α 1 − α 2 + α p ( z 0 ) + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α a β e − i β π 2 + i k β = Arctan − 2 2 + α − a β sin β π 2 + k β 1 − 2 2 + α a β cos β π 2 ≥ Arctan − 2 2 + α − a β sin β π 2 − β 1 − 2 2 + α a β cos β π 2 = Arctan 2 2 + α a β sin β π 2 + β 1 − 2 2 + α a β cos β π 2 ≥ Arctan 2 β 2 + α ,$
For other cases applying the same method in Case 2. and Case 3. with $k ≤ − 1$ we obtain
$Arg 2 + α α 1 − 2 2 + α p ( z 0 ) + z 0 p ′ ( z 0 ) p ( z 0 ) ≥ Arctan 2 β 2 + α ,$
which contradicts (7). Hence the proof is completed. □
Corollary 1.
Let $α , β ∈ ( 0 , 1 ]$ and $δ = 2 π Arctan 2 β 2 + α .$ If $f ∈ G α , δ$, then $f ∈ S ˜ * ( β )$.
Theorem 2.
Let $α , β ∈ ( 0 , 1 ]$. If $f ∈ A$ and
$| Arg 2 + α α − 2 α 1 + z f ″ ( z ) f ′ ( z ) | < Arctan 2 β α ,$
then
$| Arg f ′ ( z ) | < β π 2 ( z ∈ U ) .$
Proof.
Define the function $p : U → C$ by
$p ( z ) = f ′ ( z ) = 1 + ∑ n = 1 ∞ c n z n ( z ∈ U ) .$
Then p is analytic in $U$, $p ( 0 ) = 1$,
$1 + z f ″ ( z ) f ′ ( z ) = 1 + z p ′ ( z ) p ( z ) .$
and $p ( z ) ≠ 0$ for all $z ∈ U$. If there exists a point $z 0 ∈ U$ such that
$| Arg p ( z ) | < β π 2 ,$
for all $z ∈ U$ with $| z | < | z 0 |$ and
$| Arg p ( z 0 ) | = β π 2 .$
Then, Lemma 2, gives us that
$z 0 p ′ ( z 0 ) p ( z 0 ) = i k β ,$
where $[ p ( z 0 ) ] 1 β = ± i a ( a > 0 )$ and k is given by (5) or (6).
For the case $Arg p ( z 0 ) = α π 2$ when
$p ( z 0 ) ] 1 β = i a ( a > 0 )$
and $k ≥ 1 ,$ we have
$Arg 2 + α α 1 − 2 2 + α 1 + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α 1 + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α 1 + i k β = Arctan − 2 k β α ≤ − Arctan 2 β α ,$
Next, for the case $Arg p ( z 0 ) = − α π 2$ when
$p ( z 0 ) = − i a ( a > 0 )$
and $k ≤ − 1 ,$ using the same method as before, we can obtain
$Arg 2 + α α 1 − 2 2 + α 1 + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α 1 + z 0 p ′ ( z 0 ) p ( z 0 ) = Arg 1 − 2 2 + α 1 + i k β = Arctan − 2 k β α ≥ Arctan 2 β α ,$
Consequently, from the two above-discussed contradictions, it follows that
$| Arg f ′ ( z ) | < β π 2 ( z ∈ U ) .$
and hence the proof is completed. □
Corollary 2.
Let $α , β ∈ ( 0 , 1 ]$ and $δ = 2 π Arctan 2 β α$. If $f ∈ G α , δ$, then $f ∈ C ˜ ( β )$. In other words, if $f ∈ G α , δ$, 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 $G α , δ$ 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
( (p. 172)). Let $ω ∈ Ω$ with $ω ( z ) = ∑ n = 1 ∞ w n z n$ for all $z ∈ U$. Then $| w 1 | ≤ 1$ and
Lemma 4
( (Inequality 7, p. 10)). Let $ω ∈ Ω$ with $ω ( z ) = ∑ n = 1 ∞ w n z n$ for all $z ∈ U$. Then
The inequality is sharp for the functions $ω ( z ) = z 2$ or $ω ( z ) = z$.
Lemma 5
(). If $ω ∈ Ω$ with $ω ( z ) = ∑ n = 1 ∞ w n z n ( z ∈ U )$, then for any real numbers $q 1$ and $q 2$, we have the following sharp estimate:
$| p 3 + q 1 w 1 w 2 + q 2 w 1 3 | ≤ H ( q 1 ; q 2 ) ,$
where
$H ( q 1 ; q 2 ) = 1 if ( q 1 , q 2 ) ∈ D 1 ∪ D 2 ∪ { ( 2 , 1 ) } , | q 2 | if ( q 1 , q 2 ) ∈ ∪ k = 3 7 D k , 2 3 ( | q 1 | + 1 ) | q 1 | + 1 3 ( | q 1 | + 1 + q 2 ) 1 2 if ( q 1 , q 2 ) ∈ D 8 ∪ D 9 , q 2 3 q 1 2 − 4 q 1 2 − 4 q 2 q 1 2 − 4 3 ( q 2 − 1 ) 1 2 if ( q 1 , q 2 ) ∈ D 10 ∪ D 11 \ { ( 2 , 1 ) } , 2 3 ( | q 1 | − 1 ) | q 1 | − 1 3 ( | q 1 | − 1 − q 2 ) 1 2 if ( q 1 , q 2 ) ∈ D 12 ,$
and the sets $D k , k = 1 , 2 , … , 12$ are stated as given below:
$D 1 = ( q 1 , q 2 ) : | q 1 | ≤ 1 2 , | q 2 | ≤ 1 , D 2 = ( q 1 , q 2 ) : 1 2 ≤ | q 1 | ≤ 2 , 4 27 ( | q 1 | + 1 ) 3 − ( | q 1 | + 1 ) ≤ q 2 ≤ 1 , D 3 = ( q 1 , q 2 ) : | q 1 | ≤ 1 2 , q 2 ≤ − 1 ,$
$D 4 = ( q 1 , q 2 ) : | q 1 | ≥ 1 2 , | q 2 | ≤ − 2 3 ( | q 1 | + 1 ) , D 5 = ( q 1 , q 2 ) : | q 1 | ≤ 2 , q 2 ≥ 1 , D 6 = ( q 1 , q 2 ) : 2 ≤ | q 1 | ≤ 4 , q 2 ≥ 1 12 ( q 1 2 + 8 ) , D 7 = ( q 1 , q 2 ) : | q 1 | ≥ 4 , q 2 ≥ 2 3 ( | q 1 | − 1 ) , D 8 = ( q 1 , q 2 ) : 1 2 ≤ | q 1 | ≤ 2 , − 2 3 ( | q 1 | + 1 ) ≤ q 2 ≤ 4 27 ( | q 1 | + 1 ) 3 − ( | q 1 | + 1 ) , D 9 = ( q 1 , q 2 ) : | q 1 | ≥ 2 , − 2 3 ( | q 1 | + 1 ) ≤ q 2 ≤ 2 | q 1 | ( | q 1 | + 1 ) q 1 2 + 2 | q 1 | + 4 , D 10 = ( q 1 , q 2 ) : 2 ≤ | q 1 | ≤ 4 , 2 | q 1 | ( | q 1 | + 1 ) q 1 2 + 2 | q 1 | + 4 ≤ q 2 ≤ 1 12 ( q 1 2 + 8 ) , D 11 = ( q 1 , q 2 ) : | q 1 | ≥ 4 , 2 | q 1 | ( | q 1 | + 1 ) q 1 2 + 2 | q 1 | + 4 ≤ q 2 ≤ 2 | q 1 | ( | q 1 | − 1 ) q 1 2 − 2 | q 1 | + 4 , D 12 = ( q 1 , q 2 ) : | q 1 | ≥ 4 , 2 | q 1 | ( | q 1 | − 1 ) q 1 2 − 2 | q 1 | + 4 ≤ q 2 ≤ 2 3 ( | q 1 | − 1 ) .$
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
$φ ( z ) = 1 + B 1 z + B 2 z 2 + B 3 z 3 + … , z ∈ U , with B 1 ≠ 0 .$
Lemma 6
( (Theorem 2)). Let the function $f ∈ K ( φ )$. Then the logarithmic coefficients of f satisfy the inequalities
$| γ 1 | ≤ | B 1 | 4 ,$
$| γ 2 | ≤ | B 1 | 12 if | 4 B 2 + B 1 2 | ≤ 4 | B 1 | , | 4 B 2 + B 1 2 | 48 if | 4 B 2 + B 1 2 | > 4 | B 1 | ,$
and if $B 1 , B 2$, and $B 3$ are real values,
$| γ 3 | ≤ | B 1 | 24 H ( q 1 ; q 2 ) ,$
where $H ( q 1 ; q 2 )$ is given by Lemma 5, $q 1 = B 1 + 4 B 2 B 1 2$ and $q 2 = B 2 + 2 B 3 B 1 2$. The bounds (12) and (13) are sharp.
Theorem 3.
Let $f ∈ G α , δ$. Then
$| a 2 | ≤ α δ 2 , | a 3 | ≤ α δ 6 , | a 4 | ≤ α δ 12 H ( q 1 ; q 2 ) ,$
where $H ( q 1 ; q 2 )$ is given by Lemma 5,
$q 1 = − 3 α δ 2 + 2 δ and q 2 = δ 2 − 3 α 2 + α 2 2 + 2 3 + 1 3 .$
The first two bounds are sharp.
Proof.
Set $g ( z ) = : z f ′ ( z )$, where $f ∈ G α , δ$ and suppose that $g ( z ) = z + ∑ n = 2 ∞ b n z n$. Hence $b n = n a n$ for $n ≥ 1$. Then from (4), it follows that
$z g ′ ( z ) g ( z ) ≺ − α 2 1 + z 1 − z δ + 2 + α 2 = : ϕ ( z ) = 1 − α δ z − α δ 2 z 2 − 1 3 α δ ( 2 δ 2 + 1 ) z 3 + ⋯ : = 1 + B 1 z + B 2 z 2 + B 3 z 3 + ⋯ .$
Now, by the definition of the subordination, there is a $ω ∈ Ω$ with $ω ( z ) = ∑ n = 1 ∞ w n z n$ so that
$z g ′ ( z ) g ( z ) = ϕ ( ω ( z ) ) = 1 + B 1 w 1 z + ( B 1 w 2 + B 2 w 1 2 ) z 2 + ( B 1 w 3 + 2 w 1 w 2 B 2 + B 3 w 1 3 ) z 3 + ⋯ .$
From the above equality, it concludes that
$b 2 = B 1 w 1 2 b 3 − b 2 2 = B 1 w 2 + B 2 w 1 2 3 b 4 − 3 b 2 b 3 + b 2 3 = B 1 w 3 + 2 w 1 w 2 B 2 + B 3 w 1 3 .$
First, for $b 2$, from Lemma 3 we get $| b 2 | ≤ α δ$, and so $| a 2 | ≤ α δ 2$. Next, utilizing Lemma 3 for $b 3$ and using $| B 2 + B 1 2 | ≤ | B 1 |$, we have
Ultimately, utilizing Lemma 5 for $a 4$, we have
where
$q 1 = 3 2 B 1 + 2 B 2 B 1 = − 3 α δ 2 + 2 δ and q 2 = 3 2 B 2 + 1 2 B 1 2 + B 3 B 1 = δ 2 − 3 α 2 + α 2 2 + 2 3 + 1 3 .$
The extremal functions for the initial coefficients $a n ( n = 2 , 3 )$ are of the form:
$f n ( z ) = ∫ 0 z exp ∫ 0 x ϕ ( t n ) − 1 t d t d x = z − α β n ( n + 1 ) z n + 1 + α β 2 ( α / n − 1 ) 2 n ( 2 n + 1 ) z 2 n + 1 + ⋯ ,$
obtained by taking $ω ( z ) = z n$ in (4). Therefore, this completes the proof. □
Theorem 4.
Let $f ∈ G α , δ$. Then
$| γ 1 | ≤ α δ 4 , | γ 2 | ≤ α δ 12 , | γ 3 | ≤ α δ 24 H ( q 1 ; q 2 ) ,$
where $H ( q 1 ; q 2 )$ is given by Lemma 5, $q 1 = − α δ + 4 δ 2$, and $q 2 = − α δ 2 + 2 ( 2 δ 2 + 1 ) 3 2$. 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 $f n ( z )$ for $n = 1 , 2$, respectively. Therefore, this completes the proof. □
Theorem 5.
Let $f ∈ G α , δ$. Then we have sharp inequalities for complex parameter μ
Proof.
Let $f ∈ G α , δ$, then from (4), by the definition of the subordination, there is a $ω ∈ Ω$ with $ω ( z ) = ∑ n = 1 ∞ w n z n$ so that
$1 + z f ″ ( z ) f ′ ( z ) = ϕ ( ω ( z ) ) = 1 + B 1 w 1 z + ( B 1 w 2 + B 2 w 1 2 ) z 2 + ⋯ .$
Therefore, we get that
$2 a 2 = B 1 w 1 and 6 a 3 − 4 a 2 2 = B 1 w 2 + B 2 w 1 2 .$
Form the above equalities, we have
$| a 3 − μ a 2 2 | = 1 6 | B 1 | | w 2 + ν w 1 2 | .$
The results are obtained by the application of Lemma 4 with $ν = B 2 B 1 + B 1 ( 1 − 3 μ 2 )$, where $B 1 = − α δ$ and $B 2 = − α δ 2$. 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 $δ = 1$ in Theorem 3, we get the result obtained in  (Theorem 1) for $n = 2 , 3 , 4$.
(ii)
Setting $δ = 1$ in Theorem 3, we have the result obtained in  (Theorem 2.10).
(iii)
Letting $δ = 1$ in Theorem 4, we obtain a correction of the result presented in  (Theorem 2).

## 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.

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