On the Difference of Coefficients of Bazilevič Functions

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.

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. $\square$