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Open AccessArticle

# The Fifth Coefficient of Bazilevič Functions

by
1
Department of Mathematics, Kyungsung University, Busan 48434, Korea
2
Department of Mathematics, Swansea University, Bay Campus, Swansea SA1 8EN, UK
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(8), 1228; https://doi.org/10.3390/sym12081228
Received: 12 July 2020 / Revised: 22 July 2020 / Accepted: 25 July 2020 / Published: 26 July 2020
(This article belongs to the Special Issue Symmetry in Geometric Functions and Mathematical Analysis)

## Abstract

Let $f∈A$, the class of normalized analytic functions defined in the unit disk $D$, and be given by $f(z)=z+∑n=2∞anzn$ for $z∈D$. This paper presents a new approach to finding bounds for $|an|$. As an application, we find the sharp bound for $|a5|$ for the class $B1(α)$ of Bazilevič functions when $α≥1$.

## 1. Introduction

Let $A$ denote the class of analytic functions f in the unit disk $D = { z ∈ C : | z | < 1 }$ normalized by $f ( 0 ) = 0 = f ′ ( 0 ) − 1$. Then for $z ∈ D$, $f ∈ A$ has the following representation
$f ( z ) = z + ∑ n = 2 ∞ a n z n .$
Denote by $S$, the subset of $A$ consisting of univalent functions in $D$.
We remark at the outset that in a great number of the more familiar subclasses of $S$, sharp bounds have been found for the coefficients $| a n |$, when $2 ≤ n ≤ 4$, but bounds when $n = 5$ and beyond are much more difficult to obtain. (See, e.g., [1]).
Denote by $S ∗$, the class of starlike functions defined as follows.
Definition 1.
Let $f ∈ A .$ Then $f ∈ S ∗$ if, and only if, for $z ∈ D ,$
$Re z f ′ ( z ) f ( z ) > 0 .$
An application of the method introduced in this paper to estimate the fifth coefficient of functions in $A$, concerns the $B 1 ( α )$ Bazilevič functions defined as follows.
Definition 2.
Let $f ∈ A .$ Then $f ∈ B 1 ( α )$ if, and only if, for $α ≥ 0$, and $z ∈ D ,$
$Re f ′ ( z ) f ( z ) z α − 1 > 0 .$
We note that $B 1 ( 0 ) = S ∗$, and each of the above classes are necessarily subclasses of $S$. Apart from $α = 0$, where $| a n | ≤ n$ for $n ≥ 2 ,$ we also note that sharp bounds for $| a n |$ are known for $f ∈ B 1 ( α )$ when $α ≥ 0$ only when $2 ≤ n ≤ 4$, [2], and only partial solutions are known for $| a n |$ when $n ≥ 5$ [3,4] for $α ≥ 1 .$
It was conjectured in [4], that when $α ≥ 1$, the sharp bound for $| a n |$ when $n ≥ 2$ is given by
$| a n | ≤ 2 n − 1 + α ,$
and a partial solution to this problem in the case $n = 5$ was given in [3].
In this paper, we illustrate our method by giving a complete solution to finding the sharp bound for $| a 5 |$ when $f ∈ B 1 ( α )$ for $α ≥ 1 .$

## 2. Auxiliary Results

Denote by $P$, the class of analytic functions p with positive real part on $D$ given by
$p ( z ) = 1 + ∑ n = 1 ∞ p n z n .$
Lemma 1
([5]). If the functions
$1 + ∑ n = 1 ∞ b n z n and 1 + ∑ n = 1 ∞ c n z n$
belong to $P$, then the same is true of the function
$1 + 1 2 ∑ n = 1 ∞ b n c n z n .$
Lemma 2
([6]). Let $h ( z ) = 1 + β 1 z + β 2 z 2 + ⋯$ and $1 + G ( z ) = 1 + d 1 z + d 2 z 2 + ⋯$ be functions in $P$, $ϵ 0 = 1$, and
$ϵ n = 1 2 n 1 + 1 2 ∑ k = 1 n n k β k , n ∈ N .$
If $A n$$( n ∈ N )$ is defined by
$∑ n = 1 ∞ ( − 1 ) n + 1 ϵ n − 1 G n ( z ) = ∑ n = 1 ∞ A n z n ,$
then $| A n | ≤ 2$.
We first outline the method of proof.
Let $p ∈ P$ be in the form (3), and
$Ψ = p 4 + B 1 p 1 4 + B 2 p 1 2 p 2 + B 3 p 1 p 3 + B 4 p 2 2$
with $B i ∈ C$, $i ∈ { 1 , 2 , 3 , 4 }$. Assume that there exists $q ∈ P$ of the form $q ( z ) = 1 + ∑ n = 1 ∞ b n z n$. Then by Lemma 1 the function
$1 + 1 2 ( p ( z ) − 1 ) ∗ ( q ( z ) − 1 ) = 1 + ∑ n = 1 ∞ 1 2 b n p n z n$
also belongs to $P$. Let
$1 + G ( z ) = 1 + 1 2 ( p ( z ) − 1 ) ∗ ( q ( z ) − 1 ) = 1 + ∑ n = 1 ∞ v n z n .$
Then $1 + G ( z ) ∈ P ,$ and $v n = b n p n / 2$, $n ∈ N$.
Now assume that $h ( z ) = 1 + ∑ n = 1 ∞ u n z n ∈ P$. Then by Lemma 2 we obtain $| A 4 | ≤ 2$, where
$A 4 = 1 2 ϵ 0 b 4 p 4 − 1 4 ϵ 1 b 2 2 p 2 2 − 1 2 ϵ 1 b 1 b 3 p 1 p 3 + 3 8 ϵ 2 b 1 2 b 2 p 1 2 p 2 − 1 16 ϵ 3 b 1 4 p 1 4 .$
Here, $ϵ i$, $i ∈ { 1 , 2 , 3 , 4 }$, are given by
$ϵ 0 = 1 ,$
$ϵ 1 = 1 2 1 + 1 2 u 1 ,$
$ϵ 2 = 1 4 1 + u 1 + 1 2 u 2 ,$
$ϵ 3 = 1 8 1 + 3 2 u 1 + 3 2 u 2 + 1 2 u 3 .$
Hence we have the following.
(A) Let $p ∈ P$ be in the form (3). If there exist q, $h ∈ P$ such that q and h are represented by
$q ( z ) = 1 + ∑ n = 1 ∞ b n z n , h ( z ) = 1 + ∑ n = 1 ∞ u n z n ,$
respectively, with
$b 4 = 2 B 1 = − 1 128 1 + 3 2 u 1 + 3 2 u 2 + 1 2 u 3 b 1 4 B 2 = 3 32 1 + u 1 + 1 2 u 2 b 1 2 b 2 B 3 = − 1 4 1 + 1 2 u 1 b 1 b 3 B 4 = − 1 8 1 + 1 2 u 1 b 2 2 ,$
then $| Ψ | = | A 4 | ≤ 2$, where Ψ and $A 4$ are given by (5) and (6), respectively.
We now recall a recent result of Cho et al. [7], where they obtained the following parametric formulas for the initial coefficients of Carathéodory functions (see also [8]). We recall the Möbius transformation $ψ ζ : D → D$, $ζ ∈ D$, defined by
$ψ ζ ( z ) = z − ζ 1 − ζ ¯ z$
and let
$L ( z ) = 1 + z 1 − z , z ∈ D .$
Lemma 3
([7]). If $p ∈ P$ is of the form (3), then
$p 1 = 2 ζ 1 ,$
$p 2 = 2 ζ 1 2 + 2 ( 1 − | ζ 1 | 2 ) ζ 2$
and
$p 3 = 2 ζ 1 3 + 4 ( 1 − | ζ 1 | 2 ) ζ 1 ζ 2 − 2 ( 1 − | ζ 1 | 2 ) ζ ¯ 1 ζ 2 2 + 2 ( 1 − | ζ 1 | 2 ) ( 1 − | ζ 2 | 2 ) ζ 3$
for some $ζ i ∈ D ¯$, $i ∈ { 1 , 2 , 3 }$. For $ζ 1 ∈ T$, there is a unique function $p ∈ P$ with $p 1$ as in (9), namely,
$p ( z ) = 1 + ζ 1 z 1 − ζ 1 z , z ∈ D .$
For $ζ 1 ∈ D$ and $ζ 2 ∈ T$, there is a unique function $p = L ∘ ω ∈ P$ with $p 1$ and $p 2$ as in (9) and (10), where
$ω ( z ) = z ψ − ζ 1 ( ζ 2 z ) , z ∈ D ,$
i.e.,
$p ( z ) = 1 + ( ζ ¯ 1 ζ 2 + ζ 1 ) z + ζ 2 z 2 1 + ( ζ ¯ 1 ζ 2 − ζ 1 ) z − ζ 2 z 2 , z ∈ D .$
For $ζ 1$, $ζ 2 ∈ D$ and $ζ 3 ∈ T$, there is a unique function $p = L ∘ ω ∈ P$ with $p 1$, $p 2$ and $p 3$ as in (9)–(11), where
$ω ( z ) = z ψ − ζ 1 ( z ψ − ζ 2 ( ζ 3 z ) ) , z ∈ D ,$
i.e.,
$p ( z ) = 1 + ( ζ ¯ 2 ζ 3 + ζ ¯ 1 ζ 2 + ζ 1 ) z + ( ζ ¯ 1 ζ 3 + ζ 1 ζ ¯ 2 ζ 3 + ζ 2 ) z 2 + ζ 3 z 3 1 + ( ζ ¯ 2 ζ 3 + ζ ¯ 1 ζ 2 − ζ 1 ) z + ( ζ ¯ 1 ζ 3 − ζ 1 ζ ¯ 2 ζ 3 − ζ 2 ) z 2 − ζ 3 z 3 , z ∈ D .$
Conversely, if $ζ 1$, $ζ 2 ∈ D$ and $ζ 3 ∈ D ¯$ are given, then we can construct a (unique) function $p ∈ P$ of the form (3) so that $p i$, $i ∈ { 1 , 2 , 3 }$, satisfying the identities in (9)–(11). For this, we define
$ω ( z ) = ω ζ 1 , ζ 2 , ζ 3 ( z ) = z ψ − ζ 1 ( z ψ − ζ 2 ( ζ 3 z ) ) , z ∈ D ,$
where $ψ ζ$ is the function defined as in (7). Then $ω ∈ B$. Moreover, if we define $p ( z ) = ( 1 + ω ( z ) ) / ( 1 − ω ( z ) )$, $z ∈ D$, then p is represented by (3), where $p 1$, $p 2$ and $p 3$ satisfy the identities in (9)–(11) (see the proof of ([7], Lemma 2.4)).
Assume that the function $q ( z ) = 1 + ∑ n = 1 ∞ b n z n ∈ P$ is constructed by $ξ 1$, $ξ 2 ∈ D$, $ξ 3 ∈ D ¯ ,$ and the function $h ( z ) = 1 + ∑ n = 1 ∞ u n z n ∈ P$ is constructed by $ζ 1$, $ζ 2 ∈ D$, $ζ 3 ∈ D ¯$. Namely, $q = L ∘ ω 1$ and $h = L ∘ ω 2$, where L is the function defined by (8), $ω 1 ( z ) = z ψ − ξ 1 ( z ψ − ξ 2 ( ξ 3 z ) )$ and $ω 2 ( z ) = z ψ − ζ 1 ( z ψ − ζ 2 ( ζ 3 z ) )$. Then by combining the above argument, we conclude (B) below.
(B) Let $p ∈ P$ be in the form (3). If there exist $ζ 1$, $ζ 2$, $ξ 1$, $ξ 2 ∈ D$, $ζ 3$, $ξ 3 ∈ D ¯$ satisfying the following conditions
$b 1 = 2 ξ 1 b 2 = 2 ξ 1 2 + 2 ( 1 − | ξ 1 | 2 ) ξ 2 b 3 = 2 ξ 1 3 + 4 ( 1 − | ξ 1 | 2 ) ξ 1 ξ 2 − 2 ( 1 − | ξ 1 | 2 ) ξ ¯ 1 ξ 2 2 + 2 ( 1 − | ξ 1 | 2 ) ( 1 − | ξ 2 | 2 ) ξ 3 u 1 = 2 ζ 1 u 2 = 2 ζ 1 2 + 2 ( 1 − | ζ 1 | 2 ) ζ 2 u 3 = 2 ζ 1 3 + 4 ( 1 − | ζ 1 | 2 ) ζ 1 ζ 2 − 2 ( 1 − | ζ 1 | 2 ) ζ ¯ 1 ζ 2 2 + 2 ( 1 − | ζ 1 | 2 ) ( 1 − | ζ 2 | 2 ) ζ 3 b 4 = 2 B 1 = − 1 128 1 + 3 2 u 1 + 3 2 u 2 + 1 2 u 3 b 1 4 B 2 = 3 32 1 + u 1 + 1 2 u 2 b 1 2 b 2 B 3 = − 1 4 1 + 1 2 u 1 b 1 b 3 B 4 = − 1 8 1 + 1 2 u 1 b 2 2 ,$
then $| Ψ | = | A 4 | ≤ 2$, where Ψ and $A 4$ are given by (5) and (6), respectively.
Since the system of equations in (B) has many solutions, we now place some restrictions on the parameters sufficient for our purpose.
We fix
$q ( z ) = q τ ( z ) = 1 + 2 τ z + z 2 1 − z 2 .$
Then if $τ ∈ [ − 1 , 1 ]$, $q ∈ P$ and is given by $q ( z ) = 1 + ∑ n = 1 ∞ b n z n ,$ with
$b 1 = b 3 = 2 τ , and b 2 = b 4 = 2 .$
We also assume that $ζ i$, $i ∈ { 1 , 2 , 3 }$ take real values. Then the identities for $u i$, $i ∈ { 1 , 2 , 3 }$ become
$u 1 = 2 ζ 1 u 2 = 2 ζ 1 2 + 2 ( 1 − ζ 1 2 ) ζ 2 u 3 = 2 ζ 1 3 + 4 ( 1 − ζ 1 2 ) ζ 1 ζ 2 − 2 ( 1 − | ζ 1 | 2 ) ζ 1 ζ 2 2 + 2 ( 1 − ζ 1 2 ) ( 1 − ζ 2 2 ) ζ 3 .$
Thus we are able to conclude the following.
(C) Let $p ∈ P$ be in the form (3). If there exist $ζ 1$, $ζ 2 ∈ ( − 1 , 1 )$ and $ζ 3$, $τ ∈ [ − 1 , 1 ]$ such that
$u 1 = 2 ζ 1 u 2 = 2 ζ 1 2 + 2 ( 1 − ζ 1 2 ) ζ 2 u 3 = 2 ζ 1 3 + 4 ( 1 − ζ 1 2 ) ζ 1 ζ 2 − 2 ( 1 − | ζ 1 | 2 ) ζ 1 ζ 2 2 + 2 ( 1 − ζ 1 2 ) ( 1 − ζ 2 2 ) ζ 3 b 4 = 2 B 1 = − 1 8 1 + 3 2 u 1 + 3 2 u 2 + 1 2 u 3 τ 4 B 2 = 3 4 1 + u 1 + 1 2 u 2 τ 2 B 3 = − 1 + 1 2 u 1 τ 2 B 4 = − 1 2 1 + 1 2 u 1 ,$
then $| Ψ | = | A 4 | ≤ 2$, where Ψ and $A 4$ are given by (5) and (6), respectively.

## 3. The Fifth Coefficient of $B 1 ( α )$ Bazilevič Functions

Lemma 4.
([9] Cohn’s rule) Let $t ( z ) = a 0 + a 1 z + ⋯ + a n z n$ be a polynomial of degree n and
$t ∗ ( z ) = z n t ( 1 / z ¯ ) ¯ = a ¯ n + a ¯ n − 1 z + ⋯ + a ¯ 0 z n .$
Let r and s be the number of zeros of t inside and on the unit circle $| z | = 1$, respectively. If $| a 0 | < | a n |$, then
$t 1 ( z ) = a ¯ n t ( n ) − a 0 t ∗ ( z ) z$
is a polynomial of degree $n − 1$ and has $r 1 = r − 1$ and $s 1 = s$ number of zeros inside and on the unit circle $| z | = 1$, respectively.
We now use the above method to find the sharp bound for $| a 5 |$ when $α ≥ 1 .$
Theorem 1.
Let $f ∈ B 1 ( α )$, and be given by (1), then $| a 5 | ≤ 2 / ( 4 + α )$, provided $α ≥ 1$. The inequality is sharp, with extreme function $f ∈ A$ defined by
$f ′ ( z ) f ( z ) z α − 1 = 1 + z 5 1 − z 5 .$
Proof.
From (2) we can write
$f ′ ( z ) f ( z ) z α − 1 = p ( z ) ,$
for some function $p ∈ P$ of the form (3). Putting the series (1) and (3) into (14) by equating the coefficients we get
$p 1 = ( 1 + α ) a 2 , p 2 = 1 2 ( 2 + α ) [ ( − 1 + α ) a 2 2 + 2 a 3 ] ,$
$p 3 = 1 6 ( 3 + α ) [ ( 2 − 3 α + α 2 ) a 2 3 + 6 ( − 1 + α ) a 2 a 3 + 6 a 4 ]$
and
$p 4 = 1 24 ( 4 + α ) [ ( − 6 + 11 α − 6 α 2 + α 3 ) a 2 4 + 12 ( 3 − 3 α + α 2 ) a 2 2 a 3 + 24 ( − 1 + α ) a 2 a 4 + 12 ( ( − 1 + α ) a 3 2 + 2 a 5 ) ] .$
From the equalities (15)–(17), $a 5$ can be written as
$a 5 = 1 4 + α · Ψ ,$
where
$Ψ = p 4 + ( 4 + α ) ( 1 − α ) ( μ 1 p 1 p 3 + μ 2 p 2 2 + μ 3 p 1 2 p 2 + μ 4 p 1 4 ) ,$
with
$μ 1 = 1 ( 1 + α ) ( 3 + α ) , μ 2 = 1 2 ( 2 + α ) 2 ,$
$μ 3 = 1 − 2 α 2 ( 1 + α ) 2 ( 2 + α ) , and μ 4 = ( 3 α − 1 ) ( 2 α − 1 ) 24 ( 1 + α ) 4 .$
Thus it is enough to show that $| Ψ | ≤ 2$.
When $α = 1$, it is clear that $| Ψ | ≤ 2$ holds trivially, and so we can assume that $α > 1$.
Put
$ζ 1 = − 25 + 28 α + 10 α 2 + α 3 ( 3 + 4 α + α 2 ) 2 ,$
$ζ 2 = 12 + 42 α + 32 α 2 + 9 α 3 + α 4 3 ( 34 + 52 α + 32 α 2 + 9 α 3 + α 4 ) ,$
and
$ζ 3 = − 256 + 684 α + 770 α 2 + 438 α 3 + 134 α 4 + 21 α 5 + α 6 4 ( 5 + 3 α + α 2 ) ( 57 + 156 α + 163 α 2 + 82 α 3 + 20 α 4 + 2 α 5 ) ,$
then $| ζ i | < 1$ ($i = 1 , 2 , 3$) when $α > 1$.
Now let $ω ∈ B$ be defined by (12). Thus, the function k defined by $k = L ∘ ω$, where L is given by (8), belongs to $P$, with $k ( 0 ) = 1$.
Setting
$k ( z ) = 1 + ∑ n = 1 ∞ β n z n ,$
we obtain
$β 1 = − 2 ( 25 + 28 α + 10 α 2 + α 3 ) ( 3 + 4 α + α 2 ) 2 ,$
$β 2 = 2 ( 187 + 262 α + 173 α 2 + 94 α 3 + 41 α 4 + 10 α 5 + α 6 ) 3 ( 1 + α ) 4 ( 3 + α ) 2 ,$
and
$β 3 = − 2 ( 427 + 696 α + 688 α 2 + 641 α 3 + 427 α 4 + 161 α 5 + 30 α 6 + 2 α 7 ) 3 ( 1 + α ) 6 ( 3 + α ) 2 .$
With $ϵ n$ ($n ∈ N$) as in (4), we obtain
$ϵ 1 = ( 2 + α ) 2 ( − 4 + 3 α + α 2 ) 2 ( 3 + 4 α + α 2 ) 2 ,$
$ϵ 2 = ( 2 + α ) 3 ( 4 − 11 α + 5 α 2 + 2 α 3 ) 6 ( 1 + α ) 4 ( 3 + α ) 2 ,$
and
$ϵ 3 = ( 2 + α ) 4 ( − 4 + 23 α − 38 α 2 + 13 α 3 + 6 α 4 ) 24 ( 1 + α ) 6 ( 3 + α ) 2 .$
Now define q by
$q ( z ) = 1 + 2 τ z + 2 τ 2 z 2 + 2 τ z 3 + z 4 1 − z 4 = 1 + ∑ n = 1 ∞ c n z n$
with $τ : = 3 + 4 α + α 2 / ( 2 + α )$. Then $0 < τ < 1$ for $α > 1$, and
$c 1 = c 3 = 2 τ , c 2 = 2 τ 2 and c 4 = 2 .$
We shall show that q belongs $P$. Let
$ω ( z ) = q ( z ) − 1 q ( z ) + 1 = z A ( z ) B ( z ) ,$
where A and B are polynomials of degree 3, and defined by
$A ( z ) = τ + τ 2 z + τ z 2 + z 3 and B ( z ) = 1 + τ z + τ 2 z 2 + τ z 3 ,$
respectively. Then it holds that
$B ( z ) = z 3 A ( 1 / z ¯ ) ¯ .$
Define a function $A ˜$ by $A ˜ ( z ) = ( A ( z ) − τ B ( z ) ) / z$. Then a computation gives
$A ˜ ( z ) = ( 1 − τ 2 ) ( τ + z ) z = 0 .$
Therefore $A ˜$ has two zeros in $D$. By Lemma 4, the function A has exactly three zeros, say $w 1$, $w 2$ and $w 3$, in $D$. Hence, from (23) and (24), we have
$ω ( z ) = z ∏ i = 1 3 z − w i 1 − w ¯ i z ,$
which implies that $ω$ belongs to $B$. It follows from (23) that $q ∈ P$ as we asserted.
Since $p ∈ P ,$ by Lemma 1 the function
$h ( z ) : = 1 + 1 2 ∑ n = 1 ∞ c n p n z n$
also belongs to $P$. Therefore from Lemma 2 with $d n = c n p n / 2$ we have $| A 4 | ≤ 2$, where
$A 4 = 1 2 c 4 p 4 − 1 4 ϵ 1 ( c 2 2 p 2 2 + 2 c 1 c 3 p 1 p 3 ) + 3 8 ϵ 2 c 1 2 c 2 p 1 2 p 2 − 1 16 ϵ 3 c 1 4 p 1 4 .$
Finally, (19)–(22) shows that $A 4 = Ψ$, and so $| Ψ | ≤ 2$. Thus it follows from (18) that the inequality $| a 5 | ≤ 2 / ( 4 + α )$ holds.
Let $f ∈ B 1 ( α )$ be the function defined by (13). Then, by equating coefficients in (13), we get $a 2 = a 3 = a 4 = 0$ and $a 5 = 2 / ( 4 + α )$, which shows that this result is sharp. This completes the proof of Theorem 1. □

## 4. Conclusions

There are several instances in the literature where only partial solutions are known for the bounds for $| a 5 |$ for functions in subclasses of $S$ (again, see [1]). Applying the method introduced in this paper may well provide improved, or complete solutions to some of these.

## Author Contributions

All authors have equal contributions. All authors have read and agreed to the published version of the manuscript.

## Funding

This work was partially supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIP; Ministry of Science, ICT & Future Planning) (No. NRF-2017R1C1B5076778).

## Conflicts of Interest

The authors declare no conflict of interest.

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