Computational Methods and Function Theory

, Volume 19, Issue 4, pp 671–685

# On the Difference of Coefficients of Bazilevič Functions

• Nak Eun Cho
• Young Jae Sim
• Derek K. Thomas
Open Access
Article

## Abstract

Let f be analytic in the unit disk $${\mathbb {D}}=\{z\in {\mathbb {C}}:|z|<1 \}$$, and $${\mathcal {S}}$$ be the subclass of normalized univalent functions given by $$f(z)=z+\sum _{n=2}^{\infty }a_n z^n$$ for $$z\in {\mathbb {D}}$$. We give bounds for $$| |a_3|-|a_2| |$$ for the subclass $${\mathcal B}(\alpha ,i \beta )$$ of generalized Bazilevič functions when $$\alpha \ge 0$$, and $$\beta$$ is real.

## Keywords

Univalent function Close-to-convex function Bazilevič function Difference of coefficients

## Mathematics Subject Classification

30C45 30C50 30C55

## 1 Introduction

Let $${{\mathcal {A}}}$$ denote the class of analytic functions f in the unit disk $${\mathbb {D}}=\{ z\in {\mathbb {C}}: |z|<1 \}$$ normalized by $$f(0)=0=f'(0)-1$$. Then for $$z\in {\mathbb {D}}$$, $$f\in {{\mathcal {A}}}$$ has the following representation
\begin{aligned} f(z) = z+ \sum _{n=2}^{\infty }a_n z^n. \end{aligned}
(1.1)
Let $${\mathcal {S}}$$ denote the subclass of all univalent (i.e., one-to-one) functions in $${{\mathcal {A}}}$$.
In 1985, de Branges [2] solved the famous Bieberbach conjecture by showing that if $$f\in {\mathcal {S}}$$, then $$|a_n| \le n$$ for $$n \ge 2$$, with equality when $$f(z)=k(z):=z/(1-z)^2$$, or a rotation. It was therefore natural to ask if for $$f\in {\mathcal {S}}$$, the inequality $$||a_{n+1}|-|a_{n}|| \le 1$$ is true when $$n \ge 2$$. This was shown not to be the case even when $$n=2$$ [4], and that the following sharp bounds hold.
\begin{aligned} -1 \le |a_3| - |a_2| \le \frac{3}{4} + e^{-\lambda _0}(2e^{-\lambda _0}-1) = 1.029\dots , \end{aligned}
where $$\lambda _0$$ is the unique value of $$\lambda$$ in $$0< \lambda <1$$, satisfying the equation $$4\lambda = e^{\lambda }$$.

Hayman [6] showed that if $$f \in {\mathcal {S}}$$, then $$| |a_{n+1}| - |a_n| | \le C$$, where C is an absolute constant. The exact value of C is unknown, best estimate to date being $$C=3.61\dots$$ [5], which because of the sharp estimate above when $$n=2$$, cannot be reduced to 1.

Denote by $${\mathcal {S}}^*$$ the subclass of $${\mathcal {S}}$$ consisting of starlike functions, i.e. functions f which map $${\mathbb {D}}$$ onto a set which is star-shaped with respect to the origin. Then it is well-known that a function $$f \in {\mathcal {S}}^*$$ if, and only if, for $$z\in {\mathbb {D}}$$
\begin{aligned} \mathrm{Re}\left\{ \frac{zf'(z)}{f(z)} \right\} >0. \end{aligned}
It was shown in [8], that when $$f \in {\mathcal {S}}^*$$, then $$| |a_{n+1}| - |a_n| | \le 1$$ is true when $$n \ge 2$$.
Next denote by $${\mathcal {K}}$$ the subclass of $${\mathcal {S}}$$ consisting of functions which are close-to-convex, i.e. functions f which map $${\mathbb {D}}$$ onto a close-to-convex domain. Then again it is well-known that a function $$f\in {\mathcal {K}}$$ if, and only if, there exists $$g \in {\mathcal {S}}^*$$ such that for $$z\in {\mathbb {D}}$$
\begin{aligned} \mathrm{Re}\left\{ \frac{zf'(z)}{g(z)} \right\} >0. \end{aligned}
(1.2)
Koepf [7] showed that if $$f\in {\mathcal {K}}$$, then $$| |a_{n+1}| - |a_n| | \le 1$$, when $$n=2$$, but establishing this inequality when $$n \ge 3$$ remains an open problem.
In 1955, Bazilevič [1] extended the notion of starlike and close-to-convex functions by showing that if $$f\in {\mathcal {A}}$$, and is given by (1.1), then if $$\alpha >0$$ and $$\beta \in {\mathbb {R}}$$, f given by
\begin{aligned} f(z) = \left( (\alpha +i\beta ) \int _{0}^{z} g^{\alpha }(t) p(t) t^{i\beta -1} dt \right) ^{1/(\alpha +i\beta )}, \end{aligned}
(1.3)
where $$g \in {\mathcal {S}}^*$$, and $$p \in {\mathcal {P}}$$, the class of functions with positive real part in $${\mathbb {D}}$$, then functions defined by (1.3) form a subset of $${\mathcal {S}}$$. Such functions are known as Bazilevič functions.
We note that in the original definition of Bazilevič functions [1], Bazilevič assumed that $$\alpha >0$$, however Sheil-Small [10], subsequently showed that when $$\alpha =0$$, such functions also belong to $${\mathcal {S}}$$, and satisfy
\begin{aligned} \frac{zf'(z)}{f(z)} \left( \frac{f(z)}{z} \right) ^{i\beta } = p(z), \end{aligned}
(1.4)
where $$p \in {\mathcal {P}}$$.

For $$\alpha \ge 0$$ and $$\beta \in {\mathbb {R}}$$, we denote functions defined as in (1.3) and (1.4) by $${\mathcal {B}}(\alpha ,i \beta )$$, and note that the class $${\mathcal {B}}(\alpha ,0) \equiv {\mathcal {B}}(\alpha )$$ has been extensively studied, and that $${\mathcal {B}}(0,0) \equiv {\mathcal {S}}^*$$ and $${\mathcal {B}}(1,0) \equiv {\mathcal {K}}$$.

Another well studied subclass of $${\mathcal {B}}(\alpha ,i \beta )$$ is the class $${\mathcal {B}}_1(\alpha ,i\beta )$$, where $$\beta =0$$ and the starlike function $$g(z) \equiv {z}$$, (see e.g. [11]). This class is usually denoted by $${\mathcal {B}}_1(\alpha )$$. Although much is known about the initial coefficients of functions in $${\mathcal {B}}_1(\alpha )$$, there appears to be no published information concerning the difference of coefficients. We also note that $${\mathcal {B}}_1(1,0)$$ reduces to the class of functions in $${\mathcal {R}}$$ such that their derivatives have positive real part for $$z\in {\mathbb {D}}$$, and that the class $${\mathcal {B}}_1(1,i\beta )$$ has been little studied.

In this paper we present some inequalities for $$| |a_{3}| - |a_2||$$ when $$f \in {\mathcal {B}}(\alpha ,i \beta )$$, obtaining sharp bounds when $$f \in {\mathcal {B}}(\alpha )$$, and $$f \in {\mathcal {B}}_1(\alpha ,i \beta )$$ when $$\alpha \ge 0$$ and $$\beta \in {\mathbb {R}}$$. We also give the sharp bounds for $$| |a_{3}| - |a_2||$$, when $$f\in {\mathcal {B}}(0,i \beta )$$.

## 2 Preliminary Lemmas

Denote by $${\mathcal {P}}$$, the class of analytic functions p with positive real part on $${\mathbb {D}}$$ given by
\begin{aligned} p(z)=1+\sum _{n=1}^{\infty }p_n z^n. \end{aligned}
(2.1)
We will use the following properties for the coefficients of functions $${\mathcal {P}}$$, given by (2.1).

### Lemma 2.1

[9] For $$p \in {\mathcal {P}}$$ and $$\nu \in {\mathbb {C}}$$,
\begin{aligned} \left| p_2 - \frac{\nu }{2} p_1^2 \right| \le 2 \max \left\{ |\nu -1|;1 \right\} , \end{aligned}
and
\begin{aligned} \left| p_2 - \frac{1}{2}p_1^2 \right| \le 2 - \frac{1}{2} |p_1|^2. \end{aligned}
Both inequalities are sharp.

### Lemma 2.2

[3] If $$p \in {\mathcal {P}}$$, then
\begin{aligned} p_1 = 2\zeta _1 \end{aligned}
(2.2)
and
\begin{aligned} p_2 = 2\zeta _1^2 + 2(1-|\zeta _1|^2)\zeta _2 \end{aligned}
(2.3)
for some $$\zeta _i \in \overline{{\mathbb {D}}}$$, $$i \in \{ 1,2 \}$$. For $$\zeta _1 \in {\mathbb {T}}$$, the boundary of $${\mathbb {D}}$$, there is a unique function $$p \in {\mathcal {P}}$$ with $$p_1$$ as in (2.2), namely,
\begin{aligned} p(z) = \frac{1+\zeta _1 z}{1-\zeta _1 z} \quad (z\in {\mathbb {D}}). \end{aligned}
For $$\zeta _1\in {\mathbb {D}}$$ and $$\zeta _2 \in {\mathbb {T}}$$, there is a unique function $$p \in {\mathcal {P}}$$ with $$p_1$$ and $$p_2$$ as in (2.2) and (2.3), namely,
\begin{aligned} p(z) = \frac{1+( {\overline{\zeta }}_1 \zeta _2 +\zeta _1 )z + \zeta _2 z^2}{1+( {\overline{\zeta }}_1 \zeta _2 -\zeta _1 )z - \zeta _2 z^2} \quad (z\in {\mathbb {D}}). \end{aligned}

We will also need the following well-known result.

### Lemma 2.3

[7, Lem. 3] Let $$g \in {\mathcal {S}}^*$$ and be given by $$g(z)=z+\sum _{n=2}^{\infty } b_n z^n$$. Then for any $$\lambda \in {\mathbb {C}}$$,
\begin{aligned} \left| b_3 - \lambda b_2^2 \right| \le \max \left\{ 1; |3-4\lambda | \right\} . \end{aligned}
The inequality is sharp when $$g(z) = k(z)$$ if $$|3-4\lambda | \ge 1$$, and when $$g(z)=(k(z^2))^{1/2}$$ if $$|3-4\lambda | < 1$$.

## 3 The class $${\mathcal {B}}(\alpha ,i\beta )$$

We begin by proving the following inequalities for $$f \in {\mathcal {B}}(\alpha ,i\beta )$$.

### Theorem 3.1

Let $$f \in {\mathcal {B}}(\alpha ,i\beta )$$ and be given by (1.1). If $$0 \le \alpha \le (\sqrt{17}-1)/2$$ and $$\beta \in {\mathbb {R}}$$, then
\begin{aligned} -1 \le |a_3|-|a_2| \le \frac{2+\alpha }{ |2+\alpha +i\beta | }. \end{aligned}
(3.1)

### Proof

Recall that $$|a_2|-|a_3| \le 1$$ for all $$f\in {\mathcal {S}}$$ [4, Thm. 3.11]. So, since $${\mathcal {B}}(\alpha ,i\beta ) \subset {\mathcal {S}}$$ for all $$\alpha \ge 0$$ and $$\beta \in {\mathbb {R}}$$, it is sufficient to prove the upper bound in (3.1).

Let $$f \in {\mathcal {B}}(\alpha ,i\beta )$$ be of the form (1.1). Then from (1.3) we have
\begin{aligned} \left( \frac{zf'(z)}{f(z)} \right) \left( \frac{f(z)}{g(z)} \right) ^{\alpha } \left( \frac{f(z)}{z} \right) ^{i\beta } = p(z), \end{aligned}
for some $$g\in {\mathcal {S}}^*$$ and $$p\in {\mathcal {P}}$$. Writing
\begin{aligned} g(z)=z+\sum _{n=2}^{\infty }b_n z^n \quad \text {and}\quad p(z)=1+\sum _{n=1}^{\infty }p_n z^n \end{aligned}
and equating the coefficients, we obtain
\begin{aligned} a_2 = \frac{\alpha b_2 + p_1}{ 1+\alpha +i\beta } \end{aligned}
(3.2)
and
\begin{aligned} \begin{aligned} a_3 =&\frac{p_2}{2+\alpha +i\beta } - \frac{ (-1+\alpha +i\beta ) p_1^2 }{ 2(1+\alpha +i\beta )^2 } + \frac{ \alpha (3+\alpha +i\beta )b_2 p_1 }{ (1+\alpha +i\beta )^2 (2+\alpha +i\beta ) } \\&+ \frac{ \alpha b_3 }{ 2+\alpha +i\beta } + \frac{ \alpha (-1+\alpha -2i\beta -i\alpha \beta +\beta ^2) b_2^2}{ 2(2+\alpha +i\beta )(1+\alpha +i\beta )^2 }. \end{aligned} \end{aligned}
(3.3)
Let $$\mu _1 = (3+\alpha +i\beta )/(2(2+\alpha +i\beta ))$$, and suppose that $$|a_2| \le 1/|\mu _1|$$.
Then by Lemmas 2.1 and 2.3 we have
\begin{aligned} \begin{aligned} |a_3 - \mu _1 a_2^2|&= \left| \frac{1}{2+\alpha +i\beta } \left( p_2 - \frac{1}{2}p_1^2 + \alpha \left( b_3 - \frac{1}{2} b_2^2 \right) \right) \right| \\&\le \frac{2+\alpha }{|2+\alpha +i\beta |}. \end{aligned} \end{aligned}
(3.4)
Thus from (3.4) we obtain
\begin{aligned} |a_3| - |a_2| \le |a_3| - |\mu _1| |a_2|^2 \le |a_3 - \mu _1 a_2^2 | \le \frac{2+\alpha }{|2+\alpha +i\beta |}. \end{aligned}
Now assume that $$1/|\mu _1| \le |a_2| \le 2$$, and let $$\mu _2 =1/(2+\alpha +i\beta )$$. Then
\begin{aligned} a_3 - \mu _2 a_2^2 = \Psi _1 + \frac{1}{2+\alpha +i\beta }\Psi _2, \end{aligned}
(3.5)
where
\begin{aligned} \Psi _1 = \frac{\alpha b_3}{2+\alpha +i\beta } - \frac{ \alpha (1+i\beta )b_2^2 }{ 2(1+\alpha +i\beta )(2+\alpha +i\beta ) }, \end{aligned}
and
\begin{aligned} \Psi _2 = \frac{ \alpha b_2 p_1 }{ (1+\alpha +i\beta ) } - \frac{ (\alpha +i\beta ) p_1^2 }{ 2(1+\alpha +i\beta ) } + p_2. \end{aligned}
Put $$\mu = (1+i\beta )/(2(1+\alpha +i\beta ))$$. Then it is easily seen that $$|3-4\mu | = |1+3\alpha +i\beta |/|1+\alpha +i\beta | \ge 1$$. Thus Lemma 2.3 gives
\begin{aligned} |\Psi _1| \le \frac{\alpha }{ |2+\alpha +i\beta | }|3-4\mu | = \frac{\alpha |1+3\alpha +i\beta |}{ |2+\alpha +i\beta ||1+\alpha +i\beta | }. \end{aligned}
(3.6)
Next use (2.2) and (2.3) in Lemma 2.2 to obtain
\begin{aligned} \Psi _2 = \frac{2\alpha b_2 \zeta _1}{1+\alpha +i\beta } + \frac{2\zeta _1^2}{1+\alpha +i\beta } + 2\left( 1-|\zeta _1|^2\right) \zeta _2, \end{aligned}
where $$\zeta _i \in \overline{{\mathbb {D}}}$$ ($$i=1$$, 2). The triangle inequality and $$|b_2|\le 2$$ then gives
\begin{aligned} |\Psi _2| \le \psi ( |\zeta _1| ), \end{aligned}
(3.7)
where
\begin{aligned} \psi (x) = 2+ \frac{4\alpha }{|1+\alpha +i\beta |}x + 2\left( \frac{1-|1+\alpha +i\beta |}{|1+\alpha +i\beta |} \right) x^2 \end{aligned}
with $$x\in [0,1]$$.
Let $$x_0 = \alpha /( |1+\alpha +i\beta |-1 )$$, so that $$x_0 \in [0,1]$$, and $$\psi$$ has a unique critical point at $$x=x_0$$. Since $$\psi$$ has a negative leading coefficient, it follows from (3.7) that for all $$x\in [0,1]$$,
\begin{aligned} |\Psi _2 |\le \psi (x_0) = 2+\frac{2\alpha ^2}{ |1+\alpha +i\beta | (|1+\alpha +i\beta |-1) } \quad (x\in [0,1]). \end{aligned}
(3.8)
Therefore from (3.5), (3.6) and (3.10) we obtain
\begin{aligned} \begin{aligned} |a_3 - \mu _2 a_2^2|&\le \frac{1}{|2+\alpha +i\beta |} \left( 2 + \frac{\alpha |1+3\alpha +i\beta |}{ |1+\alpha +i\beta | } + \frac{2\alpha ^2}{ |1+\alpha +i\beta | ( |1+\alpha +i\beta | -1) } \right) \\&=: \Psi (\alpha ,\beta ). \end{aligned} \end{aligned}
Next write $$y:=|a_2|$$, and assume that $$y\in [1/|\mu _1|,{\tilde{x}}]$$, where
\begin{aligned} {\tilde{x}}= \frac{2\alpha +2}{ |1+\alpha +i\beta | }, \end{aligned}
(3.9)
so that
\begin{aligned} |a_3| - |a_2| \le |a_3 - \mu _2 a_2^2| + |\mu _2||a_2|^2 - |a_2| \le \Psi (\alpha ,\beta ) + \varphi (y), \end{aligned}
(3.10)
where $$\varphi$$ is defined by
\begin{aligned} \varphi (y) = \frac{1}{|2+\alpha +i\beta |}y^2 - y\quad (y\in [1/| \mu _1|, \tilde{x}]). \end{aligned}
Since $$\varphi$$ is convex on $$[1/|\mu _1|,{\tilde{x}}]$$,
\begin{aligned} \varphi (y) \le \max \{ \varphi (1/|\mu _1|); \varphi ({\tilde{x}}) \} \end{aligned}
(3.11)
for all $$y\in [1/|\mu _1|,{\tilde{x}}]$$.
Thus in order to establish the upper bound in (3.1), we use (3.10) and (3.11), and need to show that
\begin{aligned} \Psi (\alpha ,\beta ) + \varphi \left( \frac{1}{|\mu _1|}\right) \le \frac{2+\alpha }{ |2+\alpha +i\beta | } \end{aligned}
(3.12)
and
\begin{aligned} \Psi (\alpha ,\beta ) + \varphi ({\tilde{x}}) \le \frac{2+\alpha }{ |2+\alpha +i\beta | }. \end{aligned}
(3.13)
We first obtain (3.12).
Since
\begin{aligned} \frac{4}{ |3+\alpha +i\beta | } -2 <0 \quad \text {and}\quad \frac{ |2+\alpha +i\beta | }{ |3+\alpha +i\beta | } \ge \frac{2+\alpha }{3+\alpha }, \end{aligned}
(3.12) holds provided
\begin{aligned} \begin{aligned} A_1&:= \frac{ \alpha |1+3\alpha +i\beta | }{ |1+\alpha +i\beta | } + \frac{ 2\alpha ^2 }{ |1+\alpha +i\beta | ( |1+\alpha +i\beta |-1) }\\&+ \frac{ 4(2+\alpha ) |2+\alpha +i\beta | }{ (3+\alpha )|3+\alpha +i\beta | } -\alpha \\&\le \frac{ 2(2+\alpha ) |2+\alpha +i\beta | }{ 3+\alpha } =: A_2. \end{aligned} \end{aligned}
Clearly $$A_1 \le A_2$$ is true when $$\alpha =0$$. For $$\alpha >0$$, using the inequalities
\begin{aligned} \frac{ |1+3\alpha +i\beta | }{ |1+\alpha +i\beta | } \le \frac{1+3\alpha }{1+\alpha }, \quad \frac{1}{ |1+\alpha +i\beta | } \le \frac{1}{1+\alpha } \end{aligned}
and
\begin{aligned} \frac{1}{ |1+\alpha +i\beta |-1 } \le \frac{1}{\alpha }, \end{aligned}
it follows that
\begin{aligned} \frac{1}{2}(A_1-A_2) \le |2+\alpha +i\beta | \left( \frac{\alpha }{ |2+\alpha +i\beta | } + \frac{2(2+\alpha )}{ (3+\alpha ) |3+\alpha +i\beta | } - \frac{2+\alpha }{3+\alpha } \right) .\nonumber \\ \end{aligned}
(3.14)
We next note that the following is valid provided $$\alpha \in [0,(\sqrt{17}-1)/2]$$.
\begin{aligned} \frac{\alpha }{ |2+\alpha +i\beta | } + \frac{2(2+\alpha )}{ (3+\alpha )|3+\alpha +i\beta | } \le \frac{\alpha }{2+\alpha } + \frac{ 2(2+\alpha ) }{ (3+\alpha )^2 } \le \frac{2+\alpha }{3+\alpha }.\quad \end{aligned}
(3.15)
Thus from (3.15) and (3.14), $$A_1 \le A_2$$ and (3.12) is established, providing $$\alpha \in [0,(\sqrt{17}-1)/2]$$.
Next we prove (3.13), which is satisfied if $$B_1 \le B_2$$, where
\begin{aligned} B_1:= \alpha ( |1+3\alpha +i\beta | - |1+\alpha +i\beta | ) + \frac{ 2\alpha ^2 }{ |1+\alpha +i\beta |-1 } + \frac{ (2\alpha +2)^2 }{ |1+\alpha +i\beta | } \end{aligned}
and
\begin{aligned} B_2:= 2(1+\alpha ) |2+\alpha +i\beta |. \end{aligned}
A similar process to the above gives
\begin{aligned} B_1 \le 2\alpha ^2 + 2\alpha + \frac{ (2\alpha +2)^2 }{ 1+\alpha } = 2(1+a)(2+a) \le B_2, \end{aligned}
which proves inequality (3.13), and so the proof of Theorem 3.1 is complete. $$\square$$

When $$\beta =0$$, we deduce the following, noting that when $$\alpha =1$$, we obtain the inequality $$||a_3|-|a_2|| \le 1$$ for $$f\in {\mathcal {K}}$$ obtained in [7].

### Corollary 3.1

Let $$f\in {\mathcal {B}}(\alpha )$$. Then $$||a_3|-|a_2|| \le 1$$ provided $$0\le \alpha \le (\sqrt{17}-1)/2]=1.56\dots$$.

The inequality is sharp when both the functions f and g are the Koebe function.

We end this section by noting from the definition, since $${\mathcal {B}}_1(0,i \beta ) \equiv {\mathcal {B}}(0,i \beta )$$, the following is an immediate consequence of Theorem 4.1 below.

### Theorem 3.2

Let $$f \in {\mathcal {B}}(0,i \beta )$$, and be given by (1.1) with $$\beta \in {\mathbb {R}}$$. Then
\begin{aligned} - \frac{2}{ \sqrt{ |1+i\beta |^2 + |3+i\beta | } } \le |a_3| - |a_2| \le \frac{2}{ |2+i\beta | }. \end{aligned}
(3.16)
Both inequalities are sharp.

## 4 The class $${\mathcal {B}}_{1}(\alpha ,i\beta )$$,

We next consider the class $${\mathcal {B}}_{1}(\alpha ,i\beta )$$, recalling that $$f\in {\mathcal {B}}_{1}(\alpha ,i\beta )$$ if, and only if, for $$\alpha \ge 0$$ and $$\beta \in {\mathbb {R}}$$,
\begin{aligned} \mathrm{Re}\left\{ \frac{zf'(z)}{f(z)} \left( \frac{f(z)}{z} \right) ^{\alpha +i\beta } \right\} >0 \quad (z\in {\mathbb {D}}). \end{aligned}
We find the sharp upper and lower bounds of $$|a_3|-|a_2|$$ over the class $${\mathcal {B}}_{1}(\alpha ,i \beta )$$.

### Theorem 4.1

Let $$f \in {\mathcal {B}}_{1}(\alpha ,i \beta )$$ for $$\alpha \ge 0$$ and $$\beta \in {\mathbb {R}}$$, and be given by (1.1). Then
\begin{aligned} - \frac{2}{ \sqrt{ |1+\alpha +i\beta |^2 + |3+\alpha +i\beta | } } \le |a_3| - |a_2| \le \frac{2}{ |2+\alpha +i\beta | }. \end{aligned}
(4.1)
Both inequalities are sharp.

### Proof

From (3.2), (3.3) (with $$b_2=b_3=0$$), and Lemma 2.2, we obtain
\begin{aligned} a_2 = \frac{2\zeta _1}{1+\alpha +i\beta } \end{aligned}
and
\begin{aligned} a_3 = \left( \frac{2}{2+\alpha +i\beta } - \frac{2(-1+\alpha +i\beta )}{ (1+\alpha +i\beta )^2 } \right) \zeta _1^2 + \frac{2}{2+\alpha +i\beta } \left( 1- |\zeta _1|^2\right) \zeta _2 \end{aligned}
for some $$\zeta _i \in \overline{{\mathbb {D}}}$$ ($$i=1,2$$). The triangle inequality gives
\begin{aligned} |a_3|-|a_2| \le \psi ( |\zeta _1| ), \end{aligned}
(4.2)
where
\begin{aligned} \psi (x) = \kappa _2 x^2+ \kappa _1 x+ \kappa _0 \quad (x\in [0,1]) \end{aligned}
with
\begin{aligned} \kappa _2= & {} \left| \frac{2}{2+\alpha +i\beta } - \frac{2(-1+\alpha +i\beta )}{ (1+\alpha +i\beta )^2 } \right| - \frac{2}{|2+\alpha +i\beta |}, \\ \kappa _1= & {} - \frac{2}{|1+\alpha +i\beta |}, \quad \text {and} \quad \kappa _0= \frac{2}{|2+\alpha +i\beta |}. \end{aligned}
We first prove the upper bound in (4.1).
If $$\kappa _2 \le 0$$, then since $$\kappa _1<0$$, we have $$\psi '(x) = 2\kappa _2x+\kappa _1<0$$ for all $$x\in [0,1]$$. Thus
\begin{aligned} \psi (x) \le \psi (0) = \kappa _0 \quad (x\in [0,1]). \end{aligned}
(4.3)
Suppose next that $$\kappa _2>0$$. We now note that $$\kappa _2+\kappa _1 \le 0$$, since
\begin{aligned} \begin{aligned} \frac{1}{2}(\kappa _2+\kappa _1)&\le \frac{|-1+\alpha +i\beta |}{|1+\alpha +i\beta |^2} - \frac{1}{|1+\alpha +i\beta |} \\&= \frac{1}{|1+\alpha +i\beta |} \left( \frac{|-1+\alpha +i\beta |}{|1+\alpha +i\beta |} -1 \right) \end{aligned} \end{aligned}
and $$|1+\alpha +i\beta | \ge |-1+\alpha +i\beta |.$$
Since $$\kappa _2>0$$, $$\psi$$ is a quadratic function with positive leading coefficient, and $$\psi (1) = \kappa _2+\kappa _1+\kappa _0 \le \kappa _0=\psi (0)$$, it follows that
\begin{aligned} \psi (x) \le \max \{ \psi (0); \psi (1) \} = \psi (0) = \kappa _0 \quad (x\in [0,1]). \end{aligned}
(4.4)
Thus from (4.2), (4.3) and (4.5) we obtain
\begin{aligned} |a_3| - |a_2| \le \kappa _0 = \frac{2}{|2+\alpha +i\beta |}. \end{aligned}
We next prove the lower bound in (4.1).
Write
\begin{aligned} |a_3|-|a_2| = \frac{2}{|2+\alpha +i\beta |} \Psi , \end{aligned}
(4.5)
where
\begin{aligned} \Psi = \left| R_1 e^{i\theta } \zeta _1^2 + (1-\zeta _1^2) \zeta _2 \right| - R_2 \zeta _1 \end{aligned}
with
\begin{aligned} R_1 = \left| \frac{3+\alpha +i\beta }{(1+\alpha +i\beta )^2} \right| , \quad \theta = \arg \left( \frac{3+\alpha +i\beta }{(1+\alpha +i\beta )^2} \right) \end{aligned}
and
\begin{aligned} R_2 = \left| \frac{2+\alpha +i\beta }{1+\alpha +i\beta } \right| , \end{aligned}
so that we need to show that
\begin{aligned} \Psi \ge \frac{ -R_2 }{ \sqrt{R_1+1} }. \end{aligned}
Since both $${\mathcal {B}}_1(\alpha ,i \beta )$$ and $${\mathcal {P}}$$ are rotationally invariant, we may assume that $$\zeta _1 \in [0,1]$$.
Now write $$\zeta _2 = s e^{i\varphi }$$ with $$s\in [0,1]$$ and $$\varphi \in {\mathbb {R}}$$, so that
\begin{aligned} \Psi = \left| R_1 e^{i(\theta -\varphi )} \zeta _1^2 + (1-\zeta _1^2)s \right| - R_2 \zeta _1. \end{aligned}
Then
\begin{aligned} \begin{aligned} \Psi&= \sqrt{ R_1^2 \zeta _1^4 +2R_1\zeta _1^2(1-\zeta _1^2)s\cos (\theta -\varphi ) + (1-\zeta _1^2)^2s^2 } - R_2\zeta _1 \\&\ge \left| R_1 \zeta _1^2 - (1-\zeta _1^2)s \right| - R_2 \zeta _1, \end{aligned} \end{aligned}
(4.6)
with equality when $$\cos (\theta -\varphi ) = -1$$.
Suppose that $$R_1 \zeta _1^2 - (1-\zeta _1^2)s \le 0$$, then $$\zeta _1 \le \sqrt{s/(R_1+s)} =: \eta _1$$, and so by (4.6) it follows that
\begin{aligned} \begin{aligned} \Psi&\ge -(R_1+s)\zeta _1^2 - R_2\zeta _1 +s \\&\ge -(R_1+s)\eta _1^2 - R_2\eta _1 +s \\&= -R_2 \sqrt{ \frac{s}{R_1+s} } \\&\ge \frac{ -R_2 }{ \sqrt{R_1+1} }, \end{aligned} \end{aligned}
since $$s \le 1$$.
If $$R_1 \zeta _1^2 - (1-\zeta _1^2)s \ge 0$$, then $$\zeta _1 \ge \eta _1$$, and define $$\phi$$ by
\begin{aligned} \phi (x) = (R_1+s)x^2 - R_2 x -s, \end{aligned}
and let
\begin{aligned} \eta _2 = \frac{R_2}{2(R_1+s)} \end{aligned}
be the unique critical point of $$\phi$$. Then by (4.6) we have
\begin{aligned} \Psi \ge \phi (\zeta _1). \end{aligned}
(4.7)
The condition $$\eta _2 \ge \eta _1$$ is equivalent to the inequality $$4s^2 + 4R_1 s- R^2_2 \ge 0$$, which holds for $$0 \le s \le \lambda$$, where
\begin{aligned} \lambda = \lambda _{\alpha ,\beta } := \frac{1}{2}\left( -R_1 + \sqrt{R_1^2 + R_2^2} \right) . \end{aligned}
It is easily seen that $$\lambda <1$$ since
\begin{aligned} R_2^2 = \frac{ (2+\alpha )^2 + \beta ^2 }{ (1+\alpha )^2 + \beta ^2 } \le \left( \frac{2+\alpha }{1+\alpha } \right) ^2 \le 4 < 4+ R_1, \end{aligned}
for $$\alpha \ge 0$$, and $$\beta \in {\mathbb {R}}$$.
We also note that $$R_2-2R_1 < 2$$, since
\begin{aligned} R_2 - 2R_1 < R_2 \le \frac{2+\alpha }{1+\alpha } \le 2. \end{aligned}
We consider next the case $$R_2 \le 2R_1$$, where $$\eta _1 \le 1$$ for all $$s\in [0,1]$$, and distinguish two sub-cases, $$\eta _2 \le \eta _1$$, and $$\eta _2 \ge \eta _1$$.
When $$s\in [\lambda ,1]$$, we have $$\eta _2 \le \eta _1$$, and so from (4.7) we obtain
\begin{aligned} \Psi \ge \phi (\eta _1) = -R_2 \sqrt{ \frac{s}{R_1+s} } \ge \frac{-R_2}{\sqrt{R_1+1}} \end{aligned}
(4.8)
since $$s\in [0,1]$$. When $$s\in [0,\lambda ]$$, we have $$\eta _2 \ge \eta _1$$. This, and (4.7), implies that
\begin{aligned} \Psi \ge \phi (\eta _2) = - \left( s+\frac{R_2^2}{4(R_1+s)} \right) = -\frac{1}{4}h(s), \end{aligned}
(4.9)
where h is defined by
\begin{aligned} h(x) = 4x + \frac{R_2^2}{R_1+x}. \end{aligned}
(4.10)
Differentiating h gives
\begin{aligned} (R_1+x)^2 h'(x) = 4x^2 + 8R_1x+4R_1^2 - R_2^2. \end{aligned}
Since $$4R_1^2 - R_2^2 = (2R_1+R_2)(2R_1-R_2) \ge 0$$, h is increasing on the interval $$[0,\lambda ]$$, and so from (4.9) we have
\begin{aligned} \Psi \ge - \frac{1}{4}h(\lambda ) = - \left( \lambda + \frac{ R_2^2 }{ 4(R_1 +\lambda ) } \right) . \end{aligned}
(4.11)
Next note that
\begin{aligned} \frac{ R_2 }{ \sqrt{R_1+1} } \ge \lambda + \frac{ R_2^2 }{ 4(R_1 +\lambda ) }, \end{aligned}
(4.12)
since
\begin{aligned} \lambda + \frac{ R_2^2 }{ 4(R_1 +\lambda ) } \le \frac{ R_2 \sqrt{\lambda } }{ \sqrt{R_1 +\lambda } }, \end{aligned}
provided $$\sqrt{ \lambda (R_1+1) } \le \sqrt{ R_1+\lambda }$$ which is valid for all $$\alpha \ge 0$$ and $$\beta \in {\mathbb {R}}$$ since $$\lambda < 1$$.
Thus it follows from (4.8), (4.11) and (4.12) that
\begin{aligned} \Psi \ge \frac{-R_2}{ \sqrt{R_1+1} } \end{aligned}
is true provided $$R_2 \le 2R_1$$.

Next assume that $$R_2 \ge 2R_1$$. In this case there exists $$s \in [0,1]$$, such that $$\eta _2 \ge 1$$.

Setting $${\hat{\lambda }} = (R_2 - 2R_1)/2$$ it follows that $$0< {\hat{\lambda }}< \lambda <1$$.

When $$s\in [\lambda ,1]$$, we have $$\eta _2 \le \eta _1$$, and a similar method to that used in the case $$R_2 \le 2R_1$$ gives
\begin{aligned} \Psi \ge \frac{-R_2}{ \sqrt{R_1+1} }. \end{aligned}
When $$s\in [{\hat{\lambda }},\lambda ]$$, we have $$\eta _2 \ge \eta _1$$, and so the function h, defined by (4.10), is increasing on $$[{\hat{\lambda }},\lambda ]$$ since
\begin{aligned} \begin{aligned} (R_1+x)^2 h'(x)&= 4x^2 + 8R_1x+4R_1^2 - R_2^2 \\&\ge 4{\hat{\lambda }}^2 + 8R_1{\hat{\lambda }}+4R_1^2 - R_2^2 =0 \quad (x\in [{\hat{\lambda }},\lambda ]). \end{aligned} \end{aligned}
Thus from (4.11) and (4.12), we have
\begin{aligned} \Psi \ge -\frac{1}{4}h(\lambda ) \ge \frac{-R_2}{ \sqrt{R_1+1} }. \end{aligned}
When $$s\in [0,{\hat{\lambda }}]$$, we have $$\eta _2 \ge 1$$, which implies
\begin{aligned} \Psi \ge \phi (1) = R_1 - R_2. \end{aligned}
(4.13)
Finally from (4.13), in order to establish the left hand inequality in (4.1), it is enough to show that
\begin{aligned} \frac{R_2}{ \sqrt{R_1+1} } \ge R_2 - R_1. \end{aligned}
(4.14)
Since
\begin{aligned} R_1 - R_2 + \frac{R_2}{ \sqrt{R_1 +1} } = R_1 R_2 \left( \frac{1}{R_2} - \frac{1}{ R_1 +1 + \sqrt{R_1+1} } \right) , \end{aligned}
and since $$R_1>0$$ and $$R_2>0$$, (4.14) is satisfied, if for $$\alpha \ge 0$$ and $$\beta \in {\mathbb {R}}$$
\begin{aligned} \sqrt{R_1+1} > R_2 - R_1 -1. \end{aligned}
(4.15)
Since
\begin{aligned} R_2-R_1-1 = \frac{1}{|1+\alpha +i\beta |} \left( |2+\alpha +i\beta | - |1+\alpha +i\beta | - \frac{|3+\alpha +i\beta |}{|1+\alpha +i\beta |} \right) \end{aligned}
and
\begin{aligned} |2+\alpha +i\beta | \le |1+\alpha +i\beta | + 1 < |1+\alpha +i\beta | + \frac{|3+\alpha +i\beta |}{|1+\alpha +i\beta |}, \end{aligned}
it follows that $$R_2-R_1-1<0<\sqrt{R_1+1}$$, which establishes (4.15), and hence (4.14).

Thus the proof of the inequalities for $$|a_3|-|a_2|$$ is complete.

In order to show that the inequalities are sharp, first let the function $$f_1$$ be defined by (1.3) with $$g(z)=z$$ and $$p(z)=(1+z^2)/(1-z^2)$$. Then $$f_1 \in {\mathcal {B}}_1(\alpha ,i \beta )$$ with
\begin{aligned} f_1(z) = z +\frac{2}{2+\alpha +i\beta }z^3+ \cdots . \end{aligned}
Thus the upper bound in (4.1) is sharp.
Next put $$\zeta _1=1/\sqrt{R_1+1}$$, and $$\zeta _2=se^{i\varphi }$$ with $$s=1$$ and $$\varphi =\theta -\pi$$. Then
\begin{aligned} \Psi = \left| R_1e^{i(\theta -\varphi )} \zeta _1^2 + (1-\zeta _1^2)s\right| - R_2 \zeta _1 = - \frac{R_2}{\sqrt{R_1+1}}. \end{aligned}
(4.16)
Since $$\zeta _1 \in {\mathbb {D}}$$ and $$\zeta _2 \in {\mathbb {T}}$$, it follows from Lemma 2.2 that the function $${\hat{p}}$$ defined by
\begin{aligned} \begin{aligned} {\hat{p}}(z)&= \frac{1 + (\zeta _1 \zeta _2 + \zeta _1)z + \zeta _2 z^2}{1 + (\zeta _1 \zeta _2 - \zeta _1)z - \zeta _2 z^2} \\&= \frac{ \sqrt{R_1+1} + (e^{i\varphi }+1)z + \sqrt{R_1+1}e^{i\varphi }z^2}{ \sqrt{R_1+1} + (e^{i\varphi }-1)z - \sqrt{R_1+1}e^{i\varphi }z^2} \end{aligned} \end{aligned}
belongs to $${\mathcal {P}}$$. Now let the function $$f_2$$ be defined by (1.3) with $$g(z)=z$$ and $$p={\hat{p}}$$. Then $$f_2 \in {\mathcal {B}}_1(\alpha ,i\beta )$$. From (4.5) and (4.16), we obtain
\begin{aligned} |a_3|-|a_2| = \frac{2}{|2+\alpha +i\beta |} \Psi = - \frac{2}{ \sqrt{ |1+\alpha +i\beta |^2 + |3+\alpha +i\beta | } }, \end{aligned}
which shows that the left hand equality in (4.1) is sharp.

This completes the proof of Theorem 4.1. $$\square$$

## Notes

### Acknowledgements

The first 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. 2019R1I1A3A0105086). The second author was 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).

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## Authors and Affiliations

• Nak Eun Cho
• 1
• Young Jae Sim
• 2
• Derek K. Thomas
• 3
Email author
1. 1.Department of Applied MathematicsPukyong National UniversityBusanKorea
2. 2.Department of MathematicsKyungsung UniversityBusanKorea
3. 3.Department of MathematicsSwansea UniversitySwanseaUK