Geometric Studies and the Bohr Radius for Certain Normalized Harmonic Mappings

Abstract

Let $\mathcal {H}$ be the class of harmonic functions $f=h+\overline{g}$ in the unit disk $\mathbb {D}:=\{z\in \mathbb {C}:|z|<1\}$, where h and g are analytic in $\mathbb {D}$. In 2020, N. Ghosh and V. Allu introduced the class $\mathcal {P}_{\mathcal {H}}^0(M)$ of normalized harmonic mappings defined by $\mathcal {P}_{\mathcal {H}}^0(M)=\{f=h+\overline{g}\in \mathcal {H}: \text {Re}(zh''(z))>-M+|zg''(z)|\;\text {with}\;M>0, g'(0)=0, z\in \mathbb {D}\}$. In this paper, we investigate various geometric properties such as starlikeness, convexity, convex combination and convolution for functions in the class $\mathcal {P}_{\mathcal {H}}^0(M)$. Furthermore, we determine the sharp Bohr–Rogosinski radius, improved Bohr radius and refined Bohr radius for the class $\mathcal {P}_{\mathcal {H}}^0(M)$.

1 Introduction and Preliminaries

The classical inequality of Bohr says that if f is an analytic function in the unit disk $\mathbb {D}:=\{z\in \mathbb {C}:|z|<1\}$ with the following Taylor series expansion

$$\begin{aligned} f(z)=\sum _{n=0}^{\infty } a_nz^n \end{aligned}$$

(1.1)

such that $|f(z)|<1$ in $\mathbb {D}$. Then

$$\begin{aligned} \sum _{n=0}^{\infty }|a_n|r^n\le 1\;\text {for}\;|z|=r\le \frac{1}{3}. \end{aligned}$$

(1.2)

Here 1/3 is known as Bohr radius and it can’t be improved, while the inequality (1.2) is known as Bohr inequality. In 1914, H. Bohr [10] obtained the inequality (1.2) for $r\le 1/6$ but later Weiner, Riesz and Schur [12] independently improved it to 1/3. An observation shows that the quantity $1-|a_0|$ is equal to $d\left( f(0), \partial \mathbb {D}\right)$, where d is the Euclidean distance and $\partial \mathbb {D}$ is the boundary of $\mathbb {D}$. Therefore, the inequality (1.2) also can be written as

$$\begin{aligned} \sum _{n=1}^{\infty }|a_nz^n|\le 1-|a_0|=d\left( f(0), \partial \mathbb {D}\right) \;\text {for}\;|z|=r\le \frac{1}{3}.\end{aligned}$$

(1.3)

It is important to note that the constant 1/3 is independent of the coefficients of the Taylor series (1.1). This fact can be elucidated by saying that Bohr inequality occurs in the class $\mathcal {B}$ of analytic self maps of the unit disk $\mathbb {D}$. Analytic functions $f\in \mathcal {B}$ of the form (1.1) satisfying the inequality (1.2) for $|z|=r\le 1/3$, are said to satisfy the classical Bohr phenomenon. The concept of Bohr phenomenon can be generalized to the class $\mathcal {A}$ consisting of analytic functions of the form (1.1) which map from $\mathbb {D}$ into a given domain $\Theta \subseteq \mathbb {C}$ such that $f(\mathbb {D})\subseteq \Theta$. The class $\mathcal {A}$ is said to satisfy the Bohr phenomenon if $\exists$ largest radius $r_{\Theta }\in (0, 1)$ such that (1.3) holds for $|z|=r \le r_{\Theta }$. Here $r_{\Theta }$ is known as Bohr radius for the class $\mathcal {A}$. The Bohr radius has been obtained for the class $\mathcal {A}$ when $\Theta$ is convex domain [4], simply connected domain [1], the exterior of the closed unit disk, the punctured unit disk, and concave wedge domain (see [5]). In 1997, Boas and Khavinson [9] generalized the Bohr inequality in several complex variables by finding multidimensional Bohr radius. In 2021, Liu and Ponnusamy [22] obtained multidimensional analogues of refined Bohr inequality.

There are many improved versions of Bohr’s inequality (1.2) in various forms obtained by several authors. In 2020, Kayumov and Ponnusamy [20] obtained several interesting improved versions of Bohr inequality. For more results on this, we refer the reader to glance through the articles (see [16,17,18,19,20,21, 23, 24]). In 2017, Kayumov and Ponnusamy [18] introduced Bohr–Rogosinski radius motivated by Rogosinski radius for bounded analytic functions in $\mathbb {D}$. Rogosinski radius is defined as follows: Let $f(z)=\sum _{n=0}^{\infty }a_nz^n$ be analytic in $\mathbb {D}$ and its corresponding partial sum of f is defined by $S_N(z):=\sum _{n=0}^{N-1}a_nz^n$. Then, for every $N\ge 1$, we have $|\sum _{n=0}^{N-1}a_nz^n|<1$ in the disk $|z|<1/2$ and the radius 1/2 is sharp. Motivated by Rogosinski radius, Kayumov and Ponnusamy have considered the Bohr–Rogosinski sum $R_N^f(z)$ which is defined by

$$\begin{aligned} R_N^f(z):=|f(z)|+\sum _{n=N}^{\infty }|a_n||z|^n. \end{aligned}$$

(1.4)

It is worth to point out that $|S_N(z)|=\left| f(z)-\sum _{n=N}^{\infty }a_nz^n\right| \le R_N^f(z)$. Therefore, it is easy to see that the validity of Bohr-type radius for $R_N^f(z)$, which is related to the classical Bohr sum (Majorant series) in which f(0) is replaced by f(z), gives Rogosinski radius in the case of bounded analytic functions in $\mathbb {D}$. We have the following interesting results by Kayumov and Ponnusamy [18].

Theorem A

[18] Let $f(z)=\sum _{n=0}^{\infty } a_nz^n$ be analytic in $\mathbb {D}$ and $|f(z)|\le 1$. Then

$$\begin{aligned} |f(z)|+\sum _{n=N}^{\infty }|a_n||z|^n\le 1\end{aligned}$$

(1.5)

for $|z|=r\le R_N$, where $R_N$ is the positive root of the equation $\psi _N(r)=0$, $\psi _n(r)=2(1+r)r^N-(1-r)^2$. The radius $R_N$ is the best possible. Moreover,

$$\begin{aligned} |f(z)|^2+\sum _{n=N}^{\infty }|a_n||z|^n\le 1\end{aligned}$$

(1.6)

for $R_N'$, where $R_N'$ is the positive root of the equation $(1+r)r^N-(1-r)^2=0$. The radius $R_N'$ is the best possible.

In 2020, Kayumov and Ponnusamy [20] have proved the following several improved versions of Bohr’s inequality for analytic functions.

Theorem B

[20] Let $f(z)=\sum _{n=0}^{\infty } a_nz^n$ be analytic in $\mathbb {D}$, $|f(z)|\le 1$ and $S_r$ denotes the area of the image of the subdisk $|z|<r$ under mapping f. Then

$$\begin{aligned} B_1(r):=\sum _{n=0}^{\infty } |a_n|r^n+\frac{16}{9}\left( \frac{S_r}{\pi }\right) \le 1\;\text {for}\;r\le \frac{1}{3} \end{aligned}$$

(1.7)

and the numbers 1/3, 16/9 cannot be improved. Moreover,

$$\begin{aligned} B_2(r):=|a_0|^2+\sum _{n=1}^{\infty } a_nr^n+\frac{9}{8}\left( \frac{S_r}{\pi }\right) \le 1\;\text {for}\;r\le \frac{1}{2} \end{aligned}$$

(1.8)

and the numbers 1/2, 9/8 cannot be improved.

In 2020, Ponnusamy et al. [23] established the following refined Bohr inequality by applying a refined version of the coefficient inequalities.

Theorem C

[23] Let $f(z)=\sum _{n=0}^{\infty } a_nz^n$ be analytic in $\mathbb {D}$ and $|f(z)|\le 1$. Then

$$\begin{aligned} \sum _{n=0}^\infty |a_n|r^n+\left( \frac{1}{1+|a_0|}+\frac{r}{1-r}\right) \sum _{n=1}^\infty |a_n|^2r^{2n}\le 1\end{aligned}$$

(1.9)

for $r\le 1/(2+|a_0|)$ and the numbers $1/(1+|a_0|)$ and $1/(2+|a_0|)$ cannot be improved. Moreover,

$$\begin{aligned} |a_0|^2+\sum _{n=1}^\infty |a_n|r^n+\left( \frac{1}{1+|a_0|}+\frac{r}{1-r}\right) \sum _{n=1}^\infty |a_n|^2r^{2n}\le 1\end{aligned}$$

for $r\le 1/2$ and the numbers $1/(1+|a_0|)$ and 1/2 cannot be improved.

Bohr’s phenomenon for the complex-valued harmonic mappings have been studied extensively by many authors (see [1, 2, 6, 7]). Improved Bohr inequality for locally univalent harmonic mappings have been discussed by Evdoridis et al. [13].

A complex-valued function $f=u+iv$ is harmonic if u and v are real-harmonic in $\mathbb {D}$. Every harmonic function f has the canonical representation $f=h+\overline{g}$, where h and g are analytic in $\mathbb {D}$ known respectively as the analytic and co-analytic parts of f. A locally univalent harmonic function f is said to be sense-preserving if the Jacobian of f, defined by $J_f(z):=|h'(z)|^2-|g'(z)|^2$, is positive in $\mathbb {D}$ and sense-reversing if $J_f(z)$ is negative in $\mathbb {D}$. Let $\mathcal {H}$ be the class of all complex- valued harmonic functions $f=h+\overline{g}$ defined in $\mathbb {D}$, where h and g are analytic in $\mathbb {D}$ such that $h(0)=h'(0)-1=0$ and $g(0)=0$. If the co-analytic part $g(z) \equiv 0$ in $\mathbb {D}$, then the class $\mathcal {H}$ reduces to the class $\mathcal {A}$ of analytic functions in $\mathbb {D}$ with $f(0)=0$ and $f'(0)=1$. A function $f\in \mathcal {H}$ is said to be in $\mathcal {H}_0$ if $g'(0)=0$. Thus, every $f=h+\overline{g}\in \mathcal {H}_0$ has the following form

$$\begin{aligned} f(z)=h(z)+\overline{g}(z)=z+\sum _{n=2}^\infty a_nz^n+\overline{\sum _{n=2}^\infty b_nz^n}. \end{aligned}$$

(1.10)

A domain $\Omega$ is called starlike with respect to a point $z_0\in \Omega$ if the line segment joining $z_0$ to any point in $\Omega$ lies in $\Omega$. In particular, if $z_0=0$, then $\Omega$ is simply called starlike. A complex-valued harmonic mapping $f\in \mathcal {H}$ is said to be starlike if $f(\mathbb {D})$ is starlike. We denote the class of harmonic starlike functions in $\mathbb {D}$ by $\mathcal {S}_{\mathcal {H}}^*$. A domain $\Omega$ is called convex if it is starlike with respect to every point in $\Omega$. A function $f\in \mathcal {H}$ is said to be convex if $f(\mathbb {D})$ is convex. We denote $\mathcal {K}_{\mathcal {H}}$ by the class of harmonic convex mappings in $\mathbb {D}$.

Definition A

The polylogarithm $Li_s(z)$, also known as Jonquière’s function, is a special function of order s and argument z

$$\begin{aligned} Li_s(z)=\sum _{n=1}^\infty \frac{z^n}{n^s}=z+\frac{z^2}{2^s}+\frac{z^3}{3^s}+\cdots \end{aligned}$$

defined in the complex plane over the unit disk. The special case $s=1$ involves the ordinary natural logarithm, $Li_1(z)=-\ln (1-z)$, while the special cases $s=2$ and $s=3$ are called the dilogarithm (also known as Spence’s function) and trilogarithm respectively.

In 2020, Ghosh and Allu [14] considered the following class for $M>0$,

$$\begin{aligned} \mathcal {P}_{\mathcal {H}}^0(M)=\{f=h+\overline{g}\in \mathcal {H}: \text {Re}\left( zh''(z)\right) >-M+|zg''(z)|\;\text {with}\; g'(0)=0\;\text {for}\; z\in \mathbb {D}\}. \end{aligned}$$

The organization of this paper is: In Sect. 3, we discuss some geometric properties for functions in the class $\mathcal {P}_{\mathcal {H}}^0(M)$. In Sect. 4, we obtain sharp Bohr–Rogosinski radius for the class $\mathcal {P}_{\mathcal {H}}^0(M)$. In Sect. 5, we establish interesting sharp improved Bohr radius $\mathcal {P}_{\mathcal {H}}^0(M)$. In Sect. 6, we prove sharp refined Bohr radius for the class $\mathcal {P}_{\mathcal {H}}^0(M)$. The rest sections are for lemmas and proofs of the main results.

2 Some Lemmas

We have the following lemmas related to coefficient bounds and growth estimates for the class $\mathcal {P}_{\mathcal {H}}^0(M)$.

Lemma 2.1

[14] The harmonic map $f=h+\overline{g}$ belongs to $\mathcal {P}_{\mathcal {H}}^0(M)$ if and only if the function $F_\epsilon =h +\epsilon g$ belongs to $\mathcal {P}(M$) for $|\epsilon |=1$, where $\mathcal {P}(M)$ is defined by

$$\begin{aligned} \mathcal {P}(M):= \left\{ \phi \in \mathcal {A}: \text {Re}\left( z\phi ''(z)\right)>-M\;\;\text {for}\; M > 0\;\text {and}\;z\in \mathbb {D}\right\} . \end{aligned}$$

(2.1)

Lemma 2.2

[14] Let $f \in \mathcal {P}_{\mathcal {H}}^0(M)$ for $M>0$ be given by (1.10). Then for each $n\ge 2$, we have $|b_n|\le \frac{2M}{n(n-1)}$. The result is sharp with $f(z)=z-\frac{M}{n(n-1)}\overline{z^n}$ being extremal.

Remark 2.1

We have found some typographical error in Lemma 2.2 of [14] and the correct statement is given below:

If $f \in \mathcal {P}_{\mathcal {H}}^0(M)$ for $M>0$ is given by (1.10). Then for each $n\ge 2$, we have $|b_n|\le \frac{M}{n(n-1)}$. The result is sharp with $f(z)=z-\frac{M}{n(n-1)}\overline{z^n}$ being extremal.

Lemma 2.3

[14] Let $f \in \mathcal {P}_{\mathcal {H}}^0(M)$ for $M>0$ be given by (1.10). Then for any $n\ge 2$, we have (i) $|a_n|+|b_n|\le \frac{2M}{n(n-1)}$; (ii) $\left| |a_n|-|b_n|\right| \le \frac{2M}{n(n-1)}$; (iii) $|a_n|\le \frac{2M}{n(n-1)}$.

The result is sharp for the function f given by $f'(z)=1-2M\ln (1-z)$.

Lemma 2.4

[7, 14] Let $f=h+\overline{g}\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $M>0$ be given by (1.10). Then

$$\begin{aligned} |z|+2M\sum _{n=2}^\infty \frac{(-1)^{n-1}}{n(n-1)}|z|^n\le |f(z)|\le |z|+2M\sum _{n=2}^\infty \frac{1}{n(n-1)}|z|^n.\end{aligned}$$

Both the inequalities are sharp for the function $f_M=z+2M\sum _{n=2}^\infty \frac{1}{n(n-1)}z^n$.

To prove our convolution results, we need the following definitions and lemmas.

Definition 2.1

[11, 25] Let $\psi _1$ and $\psi _2$ be two analytic functions in $\mathbb {D}$ given by $\psi _1(z)=\sum _{n=0}^{\infty }a_nz^n$ and $\psi _2(z)=\sum _{n=0}^{\infty }b_nz^n$. The convolution (or, Hadamard product) is defined by

$$\begin{aligned} \left( \psi _1*\psi _2\right) (z)=\sum _{n=0}^{\infty }a_nb_nz^n\;\;\text {for}\; z\in \mathbb {D}.\end{aligned}$$

Definition 2.2

[15] For harmonic functions $f_1=h_1+\overline{g_1}$ and $f_2=h_2+\overline{g_2}$ in $\mathcal {H}$, the convolution is defined as $f_1*f_2=h_1*h_2+\overline{g_1*g_2}$.

Definition 2.3

[25] A sequence $\{a_n\}$ of non-negative numbers is said to be a convex null sequence if $a_n\rightarrow 0$ as $n\rightarrow \infty$ and $a_0-a_1\ge a_1-a_2\ge \cdots \ge a_{n-1}-a_n\ge \cdots \ge 0$.

Lemma 2.5

[25] Let $\{a_n\}$ be a convex null sequence. Then the function p given by $p(z)=\frac{a_0}{2}+\sum \limits _{n=1}^{\infty } a_nz^n$ is analytic in $\mathbb {D}$ and $\text {Re}(p(z))>0, z\in \mathbb {D}$.

Lemma 2.6

[25] Let the function p be analytic in $\mathbb {D}$ with $p(0)=1$ and $\text {Re}\;p(z)>\frac{1}{2}$ in $\mathbb {D}$. Then for any analytic function f in $\mathbb {D}$, the function $p*f$ takes values in the convex hull of the image of $\mathbb {D}$ under f.

Lemma 2.7

Let $\mathcal {P}(M)$ be the subclass of $\mathcal {A}$ defined in (2.1). If $F\in \mathcal {P}(M)$ with $0<M\le \frac{3}{5}$. Then $\text {Re}\left( \frac{F(z)}{z}\right) >\frac{1}{2}$.

Proof

Let $F\in \mathcal {P}(M)$ be given by $F(z)=z+\sum \limits _{n=2}^\infty A_nz^n$. Then, we have

$$\begin{aligned} \text {Re}\left( zF''(z)\right)>-M\Rightarrow & {} \text {Re}\left( M+\sum _{n=2}^\infty n(n-1)A_nz^{n-1}\right)>0\\\Rightarrow & {} \text {Re}\left( 1+\frac{1}{2M}\sum _{n=2}^\infty n(n-1)A_nz^{n-1}\right) >\frac{1}{2}\;\;\text {for}\;\;z\in \mathbb {D}. \end{aligned}$$

Let $p(z)=1+\frac{1}{2M}\sum \nolimits _{n=2}^\infty n(n-1)A_nz^{n-1}$. Then $p(0)=1$ and $\text {Re}(p(z))>\frac{1}{2}$ in $\mathbb {D}$. Now, we consider a sequence $\{c_n\}$ defined by $c_0=1$ and $c_{n-1}=\frac{2M}{n(n-1)}$ for $n\ge 2$. It is clear that $c_n\rightarrow 0$ as $n\rightarrow \infty$. Note that $c_0-c_1=1-M$ and $c_1-c_2=2\,M/3$. So $c_0-c_1\ge c_1-c_2\ge \cdots \ge c_{n-1}-c_n\ge \cdots \ge 0$ is possible only when $0<M\le 3/5$. Thus $\{c_n\}$ is a convex null sequence. In view of Lemma 2.5, the function

$$\begin{aligned} q(z)=\frac{1}{2}+\sum _{n=2}^\infty \frac{2M}{n(n-1)}z^{n-1} \end{aligned}$$

is analytic in $\mathbb {D}$ with $\text {Re} (q(z))>0$. Now,

$$\begin{aligned} \frac{F(z)}{z}=1+\sum \limits _{n=2}^\infty A_nz^{n-1}= & {} p(z)*\left( 1+\sum _{n=2}^\infty \frac{2M}{n(n-1)}z^{n-1}\right) \nonumber \\= & {} p(z)*\left( q(z)+\frac{1}{2}\right) . \end{aligned}$$

(2.2)

In view of Lemma 2.6 and (2.2), we have $\text {Re}\left( \frac{F(z)}{z}\right) >\frac{1}{2}$ for $z\in \mathbb {D}$. This completes the proof. $\square$

Lemma 2.8

Let $F_1,F_2\in \mathcal {P}(M)$ with $0<M\le \frac{3}{5}$, where $\mathcal {P}(M)$ is defined in (2.1). Then $F_1*F_2\in \mathcal {P}(M)$.

Proof

Let $F_1(z)=z+\sum \nolimits _{n=2}^\infty A_nz^n$ and $F_2(z)=z+\sum \nolimits _{n=2}^\infty B_nz^n$. Then the convolution of $F_1$ and $F_2$ is given by

$$\begin{aligned} F(z)=F_1(z)*F_2(z)=z+\sum \limits _{n=2}^\infty A_nB_nz^n. \end{aligned}$$

Now,

$$\begin{aligned} zF''(z)=\sum \limits _{n=2}^\infty n(n-1)A_nB_nz^{n-1}=\left( \frac{F_2(z)}{z}\right) *\left( zF_1''(z)\right) .\end{aligned}$$

(2.3)

Since $F_1, F_2\in \mathcal {P}(M)$, so $\text {Re}(zF_1''(z))>-M$ and in view of Lemma 2.7, we have $\text {Re}\left( \frac{F_2(z)}{z}\right) >\frac{1}{2}$. In view of Lemma 2.6 and (2.3), we have $\text {Re}(zF''(z))>-M$ in $\mathbb {D}$. Therefore $F=F_1*F_2\in \mathcal {P}(M)$. This completes the proof. $\square$

Lemma 2.9

[8] Let $f=h+\overline{g}$ be given by (1.10).

• (i) If $\sum \nolimits _{n=2}^{\infty } n\left( |a_n|+|b_n|\right) \le 1$, then f is starlike in $\mathbb {D}$;

• (ii) If $\sum \nolimits _{n=2}^{\infty } n^2\left( |a_n|+|b_n|\right) \le 1$, then f is convex in $\mathbb {D}$.

3 Convex Combinations and Convolutions

In this section, we will show that $\mathcal {P}_{\mathcal {H}}^0(M)$ is closed under convex combinations and convolutions.

Theorem 3.1

The class $\mathcal {P}_{\mathcal {H}}^0(M)$ is closed under convex combinations.

Proof

Let $f_i=h_i+\overline{g_i}\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $1\le i\le n$ and $\sum \nolimits _{i=1}^n t_i=1$, where $0\le t_i\le 1$ for each i. Then, we have

$$\begin{aligned} \text {Re}\left( zh_i''(z)\right) >-M+|zg_i''(z)| \;\text {with}\; h_i(0)=g_i(0)=h_i'(0)-1=g_i'(0)=0\;\text {for}\;1\le i\le n. \end{aligned}$$

The convex combination of the $f_i$’s can be written as

$$\begin{aligned} f(z)=\sum _{i=1}^n t_if_i(z)=h(z)+\overline{g(z)},\end{aligned}$$

where $h(z)=\sum \nolimits _{i=1}^n t_i h_i(z)$ and $g(z)=\sum \nolimits _{i=1}^n t_ig_i(z)$. Then both h and g are analytic in $\mathbb {D}$ with $h(0)=g(0)=h'(0)-1=g'(0)=0$. Now,

$$\begin{aligned} \text {Re}\left( zh''(z)\right)= & {} \text {Re}\left( z\sum _{i=1}^nt_ih_i''\right) =\sum _{i=1}^n t_i\text {Re}\left( zh_i''\right) \\> & {} \sum _{i=1}^nt_i \left( -M+|zg_i''(z)|\right) =-M+\sum _{i=1}^nt_i\left| zg_i''(z)\right| \\\ge & {} -M+\left| z\left( \sum _{i=1}^nt_i g_i''(z)\right) \right| = -M+|zg''(z)|. \end{aligned}$$

This shows that $f\in \mathcal {P}_{\mathcal {H}}^0(M)$. This completes the proof.$\square$

Theorem 3.2

Let $F_1,F_2\in \mathcal {P}_{\mathcal {H}}^0(M)$ with $0<M\le \frac{3}{5}$. Then $F_1*F_2\in \mathcal {P}_{\mathcal {H}}^0(M)$.

Proof

Let $F_1=h_1+\overline{g_1}$ and $F_2=h_2+\overline{g_2}$ be two functions in $\mathcal {P}_{\mathcal {H}}^0(M)$. Then the convolution of $F_1$ and $F_2$ is given by $F_1*F_2=h_1*h_2+\overline{g_1*g_2}$. To show that $F_1*F_2\in \mathcal {P}_{\mathcal {H}}^0(M)$, it is sufficient to show that $F=h_1*h_2+\epsilon \left( g_1*g_2\right) \in \mathcal {P}(M)$ for each $\epsilon$ with $|\epsilon |=1$. By Lemma 2.1, we have $h_1+\epsilon g_1, h_2+\epsilon g_2\in \mathcal {P}(M)$ for each $\epsilon$ with $|\epsilon |=1$. Thus, we deduce that

$$\begin{aligned} F=h_1*h_2+\epsilon \left( g_1*g_2\right) =\frac{1}{2}\left( (h_1-g_1)*(h_2-\epsilon g_2)\right) +\frac{1}{2}\left( (h_1+g_1)*(h_2+\epsilon g_2)\right) . \end{aligned}$$

In view of Lemma 2.8, we have $(h_1-g_1)*(h_2-\epsilon g_2), (h_1+g_1)*(h_2+\epsilon g_2)\in \mathcal {P}(M)$. Then in view of Theorem 3.1, we get $F\in \mathcal {P}(M)$. Hence $\mathcal {P}_{\mathcal {H}}^0(M)$ is closed under convolution. This completes the proof. $\square$

In 2002, Goodloe [15] considered the Hadamard product of a harmonic function with an analytic function as follows.

$$\begin{aligned} f\tilde{*}\phi =h*\phi +\overline{g*\phi },\end{aligned}$$

where $f=h+\overline{g}$ is harmonic and $\phi$ is an analytic function in $\mathbb {D}$.

Theorem 3.3

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ and $\phi \in \mathcal {A}$ be such that $\text {Re}\left( \frac{\phi (z)}{z}\right) >\frac{1}{2}$ for $z\in \mathbb {D}$. Then $f\tilde{*}\phi \in \mathcal {P}_{\mathcal {H}}^0(M)$.

Proof

Let $f=h+\overline{g}\in \mathcal {P}_{\mathcal {H}}^0(M)$. In view of Lemma 2.1, we have $f_1=h+\epsilon g\in \mathcal {P}(M)$ for each $\epsilon$ with $|\epsilon |=1$. To prove that $f\tilde{*}\phi =h*\phi +\overline{g*\phi }\in \mathcal {P}_{\mathcal {H}}^0(M)$, it is sufficient to show that $F(z)=h*\phi +\epsilon (g*\phi )\in \mathcal {P}(M)$ for each $\epsilon (|\epsilon |=1)$. Since $f_1\in \mathcal {P}(M)$ and $\phi \in \mathcal {A}$, so we assume that $f_1(z)=z+\sum _{n=2}^\infty A_nz^n$ and $\phi (z)=z+\sum _{n=2}^\infty B_nz^n$. Then, we deduce that

$$\begin{aligned}{} & {} F=f_1*\phi =z+\sum \limits _{n=2}^\infty A_nB_nz^n\nonumber \\ \text {and}{} & {} zF''=\sum \limits _{n=2}^\infty n(n-1)A_nB_nz^{n-1}=\left( \frac{\phi (z)}{z}\right) *(zf_1''(z)). \end{aligned}$$

(3.1)

Since $\text {Re}\left( \frac{\phi (z)}{z}\right) >\frac{1}{2}$ and $f_1\in \mathcal {P}(M)$, $\text {Re}(zf_1''(z))>-M$, so in view of Lemma 2.6 and (3.1), we have $\text {Re}(zF''(z))>-M$ in $\mathbb {D}$. Hence $F\in \mathcal {P}(M)$. This completes the proof. $\square$

Corollary 3.1

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ and $\phi \in \mathcal {K}$, where $\mathcal {K}$ denotes the family of all convex functions in $\mathbb {D}$. Then $f\tilde{*}\phi \in \mathcal {P}_{\mathcal {H}}^0(M)$.

Proof

Since $\phi \in \mathcal {K}$, so $\text {Re}\left( \frac{\phi (z)}{z}\right) >\frac{1}{2}$ for $z\in \mathbb {D}$. The result immediately follows from Theorem 3.3.$\square$

Theorem 3.4

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ and $\phi \in \mathcal {A}$ be such that $\text {Re}\left( \frac{\phi (z)}{z}\right) >\frac{1}{2}$ for $z\in \mathbb {D}$. Then $f*\left( \phi +\beta \overline{\phi }\right) \in \mathcal {P}_{\mathcal {H}}^0(M)$, where $|\beta |=1$.

Proof

Let $f=h+\overline{g}\in \mathcal {P}_{\mathcal {H}}^0(M)$. Then $h(z)=z+\sum _{n=2}^\infty a_nz^n$, $g(z)=\sum _{n=2}^\infty b_nz^n$. Now,

$$\begin{aligned} f*\left( \phi +\beta \overline{\phi }\right) =\left( h+\overline{g}\right) *\left( \phi +\overline{\overline{\beta }\phi }\right) =h*\phi +\overline{\overline{\beta }(g*\phi )}.\end{aligned}$$

To prove that $f*\left( \phi +\beta \overline{\phi }\right) \in \mathcal {P}_{\mathcal {H}}^0(M)$, it is sufficient to show that $f_{\epsilon }=h*\phi +\epsilon \overline{\beta }(g*\phi )\in \mathcal {P}(M)$ for each $\epsilon (|\epsilon |=1)$. Let $\phi (z)=z+\sum \nolimits _{n=2}^\infty C_nz^n$. For each $|\epsilon |=1$, we have

$$\begin{aligned} zf_{\epsilon }''(z)=\left( \frac{\phi (z)}{z}\right) *\left( z\left( h(z)+\epsilon \overline{\beta }g(z)\right) ''\right) .\end{aligned}$$

(3.2)

Since $f=h+\overline{g}\in \mathcal {P}_{\mathcal {H}}^0(M)$, so in view of Lemma 2.1, we have $h+\epsilon \overline{\beta } g\in \mathcal {P}(M)$ for $\epsilon$, $\beta$ with $|\epsilon \overline{\beta }|=1$, i.e., $|\beta |=1$. Thus, we have

$$\begin{aligned} \text {Re}\left( z\left( h(z)+\epsilon \overline{\beta }g(z)\right) ''\right) >-M\;\text {for}\;z\in \mathbb {D}. \end{aligned}$$

Since $\text {Re}\left( \frac{\phi (z)}{z}\right) >\frac{1}{2}$ for $z\in \mathbb {D}$, so in view of Lemma 2.6 and (3.2), we have

$$\begin{aligned} \text {Re}\left( zf_{\epsilon }''(z)\right) >-M\;\text {for}\;z\in \mathbb {D}.\end{aligned}$$

Hence $f_{\epsilon }\in \mathcal {P}(M)$. This completes the proof. $\square$

Corollary 3.2

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ and $\phi \in \mathcal {K}$, where $\mathcal {K}$ denotes the family of all convex functions in $\mathbb {D}$. Then $f*\left( \phi +\beta \overline{\phi }\right) \in \mathcal {P}_{\mathcal {H}}^0(M)$, where $|\beta |=1$.

Proof

Since $\phi \in \mathcal {K}$, so $\text {Re}\left( \frac{\phi (z)}{z}\right) >\frac{1}{2}$ for $z\in \mathbb {D}$. The result immediately follows from Theorem 3.4.$\square$

By Lemmas 2.3 and 2.9, it is possible to show that each $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ is convex (resp. starlike) in some disk D, i.e., f(D) is a convex domain (resp. f(D) is a domain starlike with respect to the origin).

Theorem 3.5

Let $f=h+\overline{g}\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). Then f is starlike in $|z|<1-e^{-\frac{1}{2M}}=r^*$ and convex in $|z|<r_c$, where $r_c\in (0,1)$ is the smallest root of the equation $\frac{r}{1-r}-\ln (1-r)-\frac{1}{2M}=0$.

Proof

Let $0<r<1$ and $f_r(z)=\frac{1}{r}f(rz)=z+\sum _{n=2}^{\infty } a_nr^{n-1}z^n+\overline{\sum _{n=2}^{\infty } b_nr^{n-1}z^n}$ for $z\in \mathbb {D}$. For convenience, we let $S_1=\sum _{n=2}^{\infty } n\left( |a_n|+|b_n|\right) r^{n-1}$ and $S_2=\sum _{n=2}^{\infty } n^2\left( |a_n|+|b_n|\right) r^{n-1}$. According to Lemma 2.9, it suffices to show that $S_1\le 1$ for $|z|=r<r^*$ and $S_2\le 1$ for $|z|=r<r_c$. In view of Lemma 2.3, we have

$$\begin{aligned} S_1\le 2M\sum _{n=1}^{\infty }\frac{r^n}{n}=-2M\ln (1-r).\end{aligned}$$

Thus, $S_1\le 1$ if $r< 1-e^{-\frac{1}{2M}}=r^*$. Again

$$\begin{aligned} S_2\le 2M\sum _{n=1}^{\infty }\left( r^n+\frac{r^n}{n}\right) =2M\left( \frac{r}{1-r}-\ln (1-r)\right) .\end{aligned}$$

Thus, $S_2\le 1$ if $r<r_c$, where $r_c\in (0,1)$ is the smallest root of the equation $\frac{r}{1-r}-\ln (1-r)-\frac{1}{2M}=0$. This completes the proof. $\square$

4 Bohr–Rogosinski Radius for the Class $\mathcal {P}_{\mathcal {H}}^0(M)$

In 2023, Ahamed et al. [3] obtained the following results regarding Bohr–Rogosinski radius for the class $\mathcal {P}_{\mathcal {H}}^0(M)$.

Theorem D

[3] Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). Then

$$\begin{aligned} |f(z)|+\sum _{n=N}^\infty \left( |a_n|+|b_n|\right) |z|^n\le d\left( f(0), \partial f(\mathbb {D})\right) \end{aligned}$$

(4.1)

for $|z|=r\le r_N(M)$ with $N\ge 2$, where $r_N(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned} r-1+2M\left( 2r-1+2(1-r)\ln (1-r)-\sum _{n=2}^{N-1}\frac{r^n}{n(n-1)}+\ln 4\right) =0. \end{aligned}$$

The constant $r_N(M)$ is the best possible.

Theorem E

[3] Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). Then

$$\begin{aligned} |f(z)|^2+\sum _{n=N}^\infty \left( |a_n|+|b_n|\right) |z|^n\le d\left( f(0), \partial f(\mathbb {D})\right) \end{aligned}$$

(4.2)

for $|z|=r\le r_N(M)$ with $N\ge 2$, where $r_N(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned}{} & {} \left( r+2M\left( r+(1-r)\ln (1-r)\right) \right) ^2\\{} & {} \quad +2M\left( r-1+(1-r)\ln (1-r) -\sum _{n=2}^{N-1}\frac{r^n}{n(n-1)}+\ln 4\right) =1.\end{aligned}$$

The constant $r_N(M)$ is the best possible.

Theorem F

[3] Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). Then

$$\begin{aligned} |f(z^m)|+\sum _{n=N}^\infty |a_n||z|^n\le d\left( f(0), \partial f(\mathbb {D})\right) \end{aligned}$$

(4.3)

for $|z|=r\le r_{m,N}(M)$ with $N\ge 2$, where $r_{m,N}(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned}{} & {} r^m-2M\\{} & {} \quad \left( r^m+r-1+\left( 1-r^m\right) \ln \left( 1-r^m\right) +(1-r)\ln (1-r)+\sum _{n=2}^{N-1}\frac{r^n}{n(n-1)}+\ln 4\right) =1.\end{aligned}$$

The constant $r_{m,N}(M)$ is the best possible.

Note that, $0<M<\frac{1}{2(\ln 4-1)}$ in Theorems D-F. Now we focus on the following question.

Question 4.1

Can we further reduce the Bohr–Rogosinski radius for the class $\mathcal {P}_{\mathcal {H}}^0(M)$ in Theorems D and E?

Corresponding to the question above, we first prove the following Bohr–Rogosinski radius for the class $\mathcal {P}_{\mathcal {H}}^0(M)$.

Theorem 4.1

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $0<M<\frac{1}{2(\ln 4-1)}$ be given by (1.10). Then

$$\begin{aligned} |f(z)|+\sum _{n=N_1}^\infty |a_n||z|^n+\sum _{n=N_2}^\infty |b_n||z|^n\le d\left( f(0), \partial f(\mathbb {D})\right) \end{aligned}$$

(4.4)

for $|z|=r\le r_{N_1,N_2}(M)$ with $N_1\ge N_2\ge 2$, where $r_{N_1,N_2}(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned} r-1+2M\left( 2r-1+2\ln 2+2(1-r)\ln (1-r)-\sum _{n=2}^{N_1-1}\frac{r^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) =0. \end{aligned}$$

The constant $r_{N_1,N_2}(M)$ is the best possible (Figs. 1 and 2).

Fig. 1

The graphs of $r_{3,2}(M)$ and $r_{5,3}(M)$ in (0, 1)

Fig. 2

The graphs of $r_ {5, 4}(M)$ and $r_ {7, 3}(M)$ in (0, 1)

Table 1 This table shows the values of the roots $r_{N_1,N_2}(M)$ for different values of M in $\left( 0,\frac{1}{2(\ln 4-1)}\right)$ with $N_1=3,5,7$ and $N_2=2,3,4$

Remark 4.1

Clearly Theorem 4.1 holds for the small Bohr–Rogosinski radius than the radius in Theorem D. It can be checked from the Table 1, e.g., when $N=3$, then $r_3(0.4)=0.527$ [3] and $r_{3,2}(0.4)=0.497629$.

Theorem 4.2

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $0<M<\frac{1}{2(\ln 4-1)}$ be given by (1.10). Then

$$\begin{aligned} |f(z^m)|^l+\sum _{n=N_1}^\infty |a_n||z^n+\sum _{n=N_2}^\infty |b_n||z|^n\le d\left( f(0), \partial f(\mathbb {D})\right) \end{aligned}$$

(4.5)

for $|z|=r\le r_{l,m,N_1,N_2}(M)$ with $N_1\ge N_2\ge 2$ and $l,m\in \mathbb {N}$, where $r_{l,m,N_1,N_2}(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned}{} & {} \left[ r^m+2M\left\{ r^m+(1-r^m)\ln (1-r^m)\right\} \right] ^l+2M\left( (1-r)\ln (1-r)+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}\right. \\ {}{} & {} \left. -1+2\ln 2+\frac{r^{N_1-1}}{N_1-1}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) -1=0.\end{aligned}$$

The constant $r_{l,m,N_1,N_2}(M)$ is the best possible (Figs. 3 and 4).

Fig. 3

The graphs of $r_ {2, 1, 3, 2} (M)$ and $r_ {2, 1, 5, 3} (M)$ in (0, 1)

Fig. 4

The graphs of $r_ {2, 2, 3, 2}(M)$ and $r_ {3, 2, 5, 3} (M)$ in (0, 1)

Table 2 This table shows the values of the roots $r_{l,m,N_1,N_2}(M)$ for different values of M in $\left( 0,\frac{1}{2(\ln 4-1)}\right)$ with $l=2,3$, $m=1,2,3$, $N_1=3,5,7$ and $N_2=2,3$

Remark 4.2

Clearly Theorem 4.2 holds for the small Bohr–Rogosinski radius than the radius in Theorem E. It can be checked from the Table 2, e.g., when $N=3$, then $r_3(1.29)=0.053$ [3], and $r_{2,1,3,2}(1.29)=0.0434376$.

Theorem 4.3

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $0<M<\frac{1}{2(\ln 4-1)}$ be given by (1.10). Then

$$\begin{aligned} |f(z^m)|^l+\sum _{n=N}^\infty |a_nb_n||z|^n\le d\left( f(0), \partial f(\mathbb {D})\right) \end{aligned}$$

(4.6)

for $|z|=r\le r_{l,m,N}(M)$ with $N\ge 2$ and $l,m\in \mathbb {N}$, where $r_{l,m,N}(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned}{} & {} \left[ r^m+2M\left\{ r^m+(1-r^m)\ln (1-r^m)\right\} \right] ^l-1+8M^2(r-1) \left( \ln (1-r)+\sum _{n=1}^{N-2}\frac{r^n}{n}\right) \\{} & {} \quad +4M^2(1+r)\left( Li_2(r)-\sum _{n=1}^{N-2}\frac{r^n}{n^2}\right) -8M^2\frac{r^{N-1}}{N-1}-4M^2\frac{r^{N-1}}{(N-1)^2}-2M(1-2\ln 2)=0,\end{aligned}$$

where $Li_2(r)$ is the dilogarithm function. The constant $r_{l,m,N}(M)$ is the best possible (Figs. 5 and 6; Table 3).

Fig. 5

The graphs of $r_ {1, 1, 3} (M)$ and $r_ {1, 2, 3} (M)$ in (0, 1)

Fig. 6

The graphs of $r_ {1, 2, 4} (M)$ and $r_ {2, 2, 4} (M)$ in (0, 1)

Table 3 This table shows the values of the roots $r_{l,m,N}(M)$ for different values of M in $\left( 0,\frac{1}{2(\ln 4-1)}\right)$ with $N=3,4,5$, $m=1,2$ and $l=1,2$

Theorem 4.4

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $0<M<\frac{1}{2(\ln 4-1)}$ be given by (1.10). Then

$$\begin{aligned} r+k_1|h(r)|^p+k_2|g(r)|^q+\sum _{n=N_1}^\infty |a_n|r^n+\sum _{n=N_2}^\infty |b_n|r^n\le d\left( f(0), \partial f(\mathbb {D})\right) \end{aligned}$$

(4.7)

for $r\le r_{p,k_1,q,k_2,N_1,N_2}(M)$ with $N_1\ge N_2\ge 2$, $p,q\in \mathbb {N}$ and $k_1,k_2\in \mathbb {R}$, where $r_{p,k_1,q,k_2,N_1,N_2}(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned}{} & {} r+k_1 r^p+k_2 r^q-1+2M\left( (1-r)\ln (1-r)-1+2\ln 2+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}+\frac{r^{N_1-1}}{N_1-1}\right. \\ {}{} & {} \left. +\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) =0. \end{aligned}$$

The radius $r_{p,k_1,q,k_2,N_1,N_2}(M)$ is the best possible (Figs. 7; Table 4).

Fig. 7

The graphs of $r_ {25, 100, 8, 1000, 3, 2}(M)$ and $r_ {7, 200, 9, 200, 5, 2}(M)$ in (0, 1)

Table 4 This table shows the values of the roots $r_{p,k_1,q,k_2,N_1,N_2}(M)$ for different values of M in $\left( 0,\frac{1}{2(\ln 4-1)}\right)$ with $p=7,25,70$, $k_1=100,200,2000$, $q=8,9,90$, $k_2=200,1000,2000$, $N_1=3,5$ and $N_2=2,3$

5 Proof of the Theorems 4.1–4.4

Proof of Theorem 4.1

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be as (1.10). Using Lemma 2.4, we get

$$\begin{aligned} |z| +2M\sum _{n=2}^{\infty }\frac{(-1)^n}{n(n-1)}|z|^n\le |f(z)|, \;\;\text {where}\;\;|z|<1.\end{aligned}$$

(5.1)

Since $f(0)=0$, so the Euclidean distance between f(0) and the boundary of $f(\mathbb {D})$ is $d\left( f(0),\partial f(\mathbb {D})\right) :=\liminf \limits _{|z|\rightarrow 1}|f(z)-f(0)|=\liminf \limits _{|z|\rightarrow 1}|f(z)|$. Thus from (5.1), we get

$$\begin{aligned} 1+2M(1-2\ln 2)\le d\left( f(0),\partial f(\mathbb {D})\right) .\end{aligned}$$

(5.2)

In view of Lemmas 2.2, 2.3 and 2.4, we now deduce for $N_1\ge N_2\ge 2$ that

$$\begin{aligned}{} & {} |f(z)|+\sum _{n=N_1}^{\infty }|a_n|r^n+\sum _{n=N_2}^{\infty }|b_n|r^n\nonumber \\{} & {} \quad =|f(z)|+\sum _{n=N_1}^{\infty }|a_n|r^n+\sum _{n=N_1}^{\infty }|b_n|r^n+\sum _{n=N_2}^{N_1-1}|b_n|r^n\nonumber \\{} & {} \quad \le r+2M\left( \sum _{n=2}^{\infty }\frac{r^n}{n(n-1)}+\sum _{n=N_1}^{\infty }\frac{r^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) \nonumber \\{} & {} \quad = r+2M\left( 2r+2(1-r)\ln (1-r)-\sum _{n=2}^{N_1-1}\frac{r^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) .\nonumber \\ \end{aligned}$$

(5.3)

Now, we deduce that

$$\begin{aligned}{} & {} r+2M\left( 2r+2(1-r)\ln (1-r)-\sum _{n=2}^{N_1-1}\frac{r^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) \nonumber \\{} & {} \quad \le 1+2M(1-2\ln 2) \end{aligned}$$

(5.4)

for $0<r\le r_{N_1,N_2}(M)<1$, where $r_{N_1,N_2}(M)$ is the smallest root of the equation

$$\begin{aligned} r-1+2M\left( 2r-1+2\ln 2+2(1-r)\ln (1-r)-\sum _{n=2}^{N_1-1}\frac{r^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) =0.\end{aligned}$$

To ensure about the existence of a root $r_{N_1,N_2}(M)$, we construct the function $\mathcal {G}_1:[0,1)\rightarrow \mathbb {R}$ such that

$$\begin{aligned}{} & {} \mathcal {G}_1(r):=r-1+2M\\{} & {} \quad \left( 2r-1+2\ln 2+2(1-r)\ln (1-r)-\sum _{n=2}^{N_1-1}\frac{r^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) .\end{aligned}$$

It is clear that (i) $\mathcal {G}_1$ is continuous on [0, 1), (ii) $\mathcal {G}_1(0)=-1+2M(2\ln 2-1)<0$ and (iii) $\lim \limits _{r\rightarrow 1}\mathcal {G}_1(r)>0$, since $\lim \limits _{r\rightarrow 1}(1-r)\ln (1-r)=0$. Thus the claim follows from Intermediate value theorem. Thus, we have

$$\begin{aligned}{} & {} r_{N_1,N_2}(M)-1+2M\left( 2r_{N_1,N_2}(M)-1+2\ln 2+2(1-r_{N_1,N_2}(M))\ln (1-r_{N_1,N_2}(M))\right. \nonumber \\{} & {} \left. -\sum _{n=2}^{N_1-1}\frac{r_{N_1,N_2}^n(M)}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r_{N_1,N_2}^n(M)}{2n(n-1)}\right) =0. \end{aligned}$$

(5.5)

Combining (5.2), (5.3) and (5.4) for $|z|=r\le r_{N_1,N_2}(M)$, we deduce that

$$\begin{aligned} |f(z)|+\sum _{n=N_1}^{\infty }|a_n|r^n+\sum _{n=N_2}^{\infty }|b_n|r^n\le d\left( f(0),\partial f(\mathbb {D})\right) .\end{aligned}$$

Now we show that the radius $r_{N_1,N_2}(M)$ is the best possible. We set

$$\begin{aligned} f:=f_{M}(z)=z+2M\sum _{n=2}^{\infty }\frac{z^n}{n(n-1)}.\end{aligned}$$

(5.6)

Note that $f_{M}(0)=0$, $f_{M}\in \mathcal {P}_{\mathcal {H}}^0(M)$. For $z=r$, we have

$$\begin{aligned}{} & {} |f_{M}(r)-f_{M}(0)|=|f_{M}(r)|=\left| r+2M\sum _{n=2}^{\infty }\frac{(r)^n}{n(n-1)}\right| =r+2M\sum _{n=2}^{\infty }\frac{r^n}{n(n-1)}\nonumber \\\text {and}{} & {} \liminf _{r\rightarrow 1^{-}}|f_{M}(r)|=1+2M. \end{aligned}$$

(5.7)

and for $z=-r$, we have

$$\begin{aligned}{} & {} |f_{M}(-r)-f_{M}(0)|=\left| -r+2M\sum _{n=2}^{\infty }\frac{(-r)^n}{n(n-1)}\right| =r+2M\sum _{n=2}^{\infty }\frac{(-1)^{n-1}r^n}{n(n-1)}\nonumber \\\text {and}{} & {} \liminf _{r\rightarrow 1^{-}}|f_{M}(-r)|=1+2M(1-2\ln 2).\end{aligned}$$

(5.8)

From (5.7) and (5.8), we have

$$\begin{aligned}{} & {} d\left( f_{M}(0),\partial f_{M}(\mathbb {D})\right) =\liminf _{|z|=r\rightarrow 1^{-}}|f_{M}(z)-f_{M}(0)|=1+2M(1-2\ln 2). \end{aligned}$$

(5.9)

For $|z|=r_{N_1,N_2}(M)$, it follows from (5.5) and (5.9) that

$$\begin{aligned}{} & {} |f(z)|+\sum _{n=N_1}^{\infty }|a_n|r_{N_{1,2}}^n(M)+\sum _{n=N_2}^{\infty }|b_n|r_{N_{1,2}}^n(M)\\{} & {} =r_{N_1,N_2}(M)+2M\left( 2r_{N_1,N_2}(M)+2\left( 1-r_{N_1,N_2}(M)\right) \ln \left( 1-r_{N_1,N_2}(M)\right) \right. \nonumber \\ {}{} & {} \left. -\sum _{n=2}^{N_1-1}\frac{r_{N_1,N_2}^n(M)}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r_{N_1,N_2}^n(M)}{2n(n-1)}\right) =1+2M(1-2\ln 2)=d\left( f_{M}(0),\partial f_{M}(\mathbb {D})\right) . \end{aligned}$$

Thus the result follows. $\square$

Proof of Theorem 4.2

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). Note that $N_1\ge N_2\ge 2$ and $l,m\in \mathbb {N}$. In view of Lemmas 2.2, 2.3 and 2.4, we get

$$\begin{aligned}{} & {} |f(z^m)|^l+\sum _{n=N_1}^\infty |a_n||z|^n+\sum _{n=N_2}^\infty |b_n||z|^n\nonumber \\{} & {} \le \left( r^m+2M\sum _{n=2}^\infty \frac{r^{mn}}{n(n-1)}\right) ^l+2M\left( \sum _{n=N_1}^{\infty }\frac{r^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) \nonumber \\{} & {} =\left[ r^m+2M\left\{ r^m+(1-r^m)\ln (1-r^m)\right\} \right] ^l+2M\left( (1-r)\ln (1-r)+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}\right. \nonumber \\ {}{} & {} \left. +\frac{r^{N_1-1}}{N_1-1}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) \end{aligned}$$

(5.10)

Similarly as the proof of Theorem 4.1, we get (5.2) and

$$\begin{aligned}{} & {} \left[ r^m+2M\left\{ r^m+(1-r^m)\ln (1-r^m)\right\} \right] ^l+2M\left( (1-r)\ln (1-r)+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}\right. \nonumber \\ {}{} & {} \left. +\frac{r^{N_1-1}}{N_1-1}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) \le 1+2M(1-2\ln 2)\end{aligned}$$

(5.11)

for $0<r\le r_{l,m,N_1,N_2}(M)$, where $r_{l,m,N_1,N_2}(M)$ is the smallest root of $\mathcal {G}_2(r)=0$ in (0, 1), where $\mathcal {G}_2:[0,1)\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned}{} & {} \mathcal {G}_2(r)=\left[ r^m+2M\left\{ r^m+(1-r^m)\ln (1-r^m)\right\} \right] ^l\\{} & {} \quad +2M\left( (1-r)\ln (1-r)+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}\right. \\{} & {} \quad \left. -1+2\ln 2+\frac{r^{N_1-1}}{N_1-1}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) -1.\end{aligned}$$

Similarly as the proof of Theorem 4.1, we have

$\mathcal {G}_2(r_{l,m,N_1,N_2}(M))=0$, i.e.,

$$\begin{aligned}{} & {} \left[ r_{l,m,N_1,N_2}^m(M)\!+\!2M\left\{ r_{l,m,N_1,N_2}^m(M)+(1\!-\!r_{l,m,N_1,N_2}^m(M)) \ln (1\!-\!r_{l,m,\!\!N_1,\!N_2}^m(M))\right\} \right] ^l\nonumber \\{} & {} \quad +2M\left( (1-r_{l,m,N_1,N_2}(M))\ln (1-r_{l,m,N_1,N_2}(M))+(1-r_{l,m,N_1,N_2}(M))\right. \nonumber \\{} & {} \left. \sum _{n=1}^{N_1-2}\frac{r_{l,m,N_1,N_2}^n(M)}{n}\right. \nonumber \\{} & {} \quad \left. -1+2\ln 2+\frac{r_{l,m,N_1,N_2}^{N_1-1}(M)}{N_1-1}+\sum _{n=N_2}^{N_1-1} \frac{r_{l,m,N_1,N_2}^n(M)}{2n(n-1)}\right) -1=0.\end{aligned}$$

(5.12)

Combining (5.2), (5.10) and (5.11), we obtain for $|z|=r\le r_{l,m,N_1,N_2}(M)$

$$\begin{aligned} |f(z^m)|^l+\sum _{n=N_1}^\infty |a_n||z^n+\sum _{n=N_2}^\infty |b_n||z|^n\le d\left( f(0),\partial f(\mathbb {D})\right) .\end{aligned}$$

In order to show that $r_{l,m,N_1,N_2}(M)$ is the best possible, we consider the function $f=f_{M}$ defined by (5.6) and we again get (5.9). For $|z|=r_{l,m,N_1,N_2}(M)$, it follows from (5.9) and (5.12) that

$$\begin{aligned}{} & {} |f(z^m)|^l+\sum _{n=N_1}^\infty |a_n||z^n+\sum _{n=N_2}^\infty |b_n||z|^n \\{} & {} \quad =\left[ r_{l,m,N_1,N_2}^m(M)+2M\left\{ r_{l,m,N_1,N_2}^m(M)+(1-r_{l,m,N_1,N_2}^m(M))\ln (1-r_{l,m,N_1,N_2}^m(M))\right\} \right] ^l\nonumber \\{} & {} \quad +2M(1-r_{l,m,N_1,N_2}(M))\ln (1-r_{l,m,N_1,N_2}(M))\nonumber \\ {}{} & {} \quad +2M\left( (1-r_{l,m,N_1,N_2}(M))\sum _{n=1}^{N_1-2}\frac{r_{l,m,N_1,N_2}^n(M)}{n}+\frac{r_{l,m,N_1,N_2}^{N_1-1}(M)}{N_1-1}+\sum _{n=N_2}^{N_1-1}\frac{r_{l,m,N_1,N_2}^n(M)}{2n(n-1)}\right) \\{} & {} \quad =1+2M(1-2\ln 2)=d\left( f_{M}(0),\partial f_{M}(\mathbb {D})\right) .\end{aligned}$$

Therefore, the radius $r_{l,m,N_1,N_2}(M)$ is the best possible. Thus the result follows. $\square$

Proof of Theorem 4.3

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). Note that $N\ge 2$ and $l,m\in \mathbb {N}$. In view of Lemmas 2.2, 2.3 and 2.4, we obtain

$$\begin{aligned}{} & {} |f(z^m)|^l+\sum _{n=N}^\infty |a_nb_n||z|^n\nonumber \\{} & {} \quad \le \left( r^m+2M\sum _{n=2}^\infty \frac{r^{mn}}{n(n-1)}\right) ^l+4M^2\sum _{n=N}^\infty \frac{r^{n}}{n^2(n-1)^2}.\end{aligned}$$

(5.13)

Similarly as the proof of Theorem 4.1, we get (5.2) and

$$\begin{aligned}{} & {} \left( r^m+2M\sum _{n=2}^\infty \frac{r^{mn}}{n(n-1)}\right) ^l+4M^2\sum _{n=N}^\infty \frac{r^{n}}{n^2(n-1)^2}\nonumber \\{} & {} \quad \le \left[ r^m\!+\!2M\left\{ r^m\!+\!(1\!-\!r^m)\ln (1\!-\!r^m)\right\} \right] ^l\!+\!8M^2(r\!-\!1) \left( \ln (1\!-\!r)\!+\!\sum _{n\!=\!1}^{N-2}\frac{r^n}{n}\right) \nonumber \\{} & {} \quad \quad +4M^2(1+r)\left( Li_2(r)-\sum _{n=1}^{N-2}\frac{r^n}{n^2}\right) -8M^2\frac{r^{N-1}}{N-1}-4M^2\frac{r^{N-1}}{(N-1)^2}\nonumber \\{} & {} \quad \le 1+2M(1-2\ln 2) \end{aligned}$$

(5.14)

for $0< r\le r_{l,m,N}(M)$, where $r_{l,m,N}(M)$ is the smallest root of $\mathcal {G}_3(r)=0$ in (0, 1) and $\mathcal {G}_3:[0,1)\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned}{} & {} \mathcal {G}_3(r):= \left[ r^m+2M\left\{ r^m+(1-r^m)\ln (1-r^m)\right\} \right] ^l-1+8M^2(r-1)\\{} & {} \quad \left( \ln (1-r)+\sum _{n=1}^{N-2}\frac{r^n}{n}\right) +4M^2(1+r)\left( Li_2(r)-\sum _{n=1}^{N-2}\frac{r^n}{n^2}\right) -8M^2\frac{r^{N-1}}{N-1}-4M^2\frac{r^{N-1}}{(N-1)^2}-2M(1-2\ln 2).\end{aligned}$$

Similarly as the proof of Theorem 4.1, we have $\mathcal {G}_3\left( r_{l,m,N}(M)\right) =0$, i.e.,

$$\begin{aligned}{} & {} \left[ r_{l,m,N}^m(M)+2M\left\{ r_{l,m,N}^m(M)+(1-r_{l,m,N}^m(M))\ln (1-r_{l,m,N}^m(M))\right\} \right] ^l\nonumber \\{} & {} \quad -1-2M(1-2\ln 2)\nonumber \\{} & {} \quad +8M^2(r_{l,m,N}(M)-1) \left( \ln (1-r_{l,m,N}(M))+\sum _{n=1}^{N-2}\frac{r_{l,m,N}^n(M)}{n}\right) -8M^2\frac{r_{l,m,N}^{N-1}(M)}{N-1}\nonumber \\{} & {} \quad +4M^2(1+r_{l,m,N}(M))\left( Li_2(r_{l,m,N}(M))-\sum _{n=1}^{N-2}\frac{r_{l,m,N}^n(M)}{n^2}\right) -4M^2\frac{r_{l,m,N}^{N-1}(M)}{(N-1)^2}=0.\nonumber \\ \end{aligned}$$

(5.15)

Combining (5.2), (5.13) and (5.14), we obtain for $|z|=r\le r_{l,m,N}(M)$

$$\begin{aligned} |f(z^m)|^l+\sum _{n=N}^\infty |a_nb_n||z|^n\le d\left( f(0),\partial f(\mathbb {D})\right) .\end{aligned}$$

In order to show that $r_{l,m,N}(M)$ is the best possible, we consider the function $f=f_ M$ defined by (5.6) and we again get (5.9). For $|z|=r_{l,m,N}(M)$, it follows from (5.9) and (5.15) that

$$\begin{aligned}{} & {} |f(z^m)|^l+\sum _{n=N}^\infty |a_nb_n|r_{l,m,N}^n(M)\\{} & {} \quad =\left[ r_{l,m,N}^m(M)+2M\left\{ r_{l,m,N}^m(M)+(1-r_{l,m,N}^m(M))\ln (1-r_{l,m,N}^m(M))\right\} \right] ^l\\{} & {} \quad \quad +8M^2(r_{l,m,N}(M)-1) \left( \ln (1-r_{l,m,N}(M))+\sum _{n=1}^{N-2}\frac{r_{l,m,N}^n(M)}{n}\right) -8M^2\frac{r_{l,m,N}^{N-1}(M)}{N-1}\\{} & {} \quad \quad +4M^2(1+r_{l,m,N}(M))\left( Li_2(r_{l,m,N}(M))-\sum _{n=1}^{N-2}\frac{r_{l,m,N}^n(M)}{n^2}\right) -4M^2\frac{r_{l,m,N}^{N-1}(M)}{(N-1)^2}\\{} & {} \quad =1+2M(1-2\ln 2)=d\left( f_M(0),\partial f_M(\mathbb {D})\right) .\end{aligned}$$

Therefore, the radius $r_{l,m,N}(M)$ is the best possible. Thus the results follows. $\square$

Proof of Theorem 4.4

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). We know that, if $\phi$ is analytic in $\mathbb {D}$ with $\phi (0)=0$ and $|\phi (z)|<1$, $\forall z\in \mathbb {D}$, then by Schwarz Lemma, we have $|\phi (z)|\le |z|$. Note that $N_1\ge N_2\ge 2$, $p,q\in \mathbb {N}$ and $k_1,k_2\in \mathbb {R}$. In view of Lemmas 2.2, 2.3 and 2.4, we obtain

$$\begin{aligned}{} & {} r+k_1|h(r)|^p+k_2|g(r)|^q+\sum _{n=N_1}^\infty |a_n|r^n+\sum _{n=N_2}^\infty |b_n|r^n\nonumber \\{} & {} \quad \le r+k_1 r^p+k_2 r^q+\sum _{n=N_1}^\infty \frac{2Mr^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{Mr^n}{n(n-1)}.\end{aligned}$$

(5.16)

Now, we deduce that

$$\begin{aligned}{} & {} r+k_1 r^p+k_2 r^q+\sum _{n=N_1}^\infty \frac{2Mr^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{Mr^n}{n(n-1)}\nonumber \\{} & {} \quad = r+k_1 r^p+k_2 r^q+2M\left( \sum _{n=N_1}^\infty \frac{r^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) \nonumber \\{} & {} \quad =r+k_1 r^p+k_2 r^q+2M\left( (1-r)\ln (1-r)+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}+\frac{r^{N_1-1}}{N_1-1}\right. \nonumber \\{} & {} \left. \quad \quad +\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) \le 1+2M(1-2\ln 2)\end{aligned}$$

(5.17)

for $0<r\le r_{p,k_1,q,k_2,N_1,N_2}(M)$, where $r_{p,k_1,q,k_2,N_1,N_2}(M)$ is the smallest root of $\mathcal {G}_4(r)=0$ in (0, 1) and $\mathcal {G}_4:[0,1)\rightarrow \mathbb {R}$ is defined by

$$\begin{aligned}{} & {} \mathcal {G}_4(r):=r+k_1 r^p+k_2 r^q-1+2M\\{} & {} \left( (1-r)\ln (1-r)-1+2\ln 2+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}\right. \\ {}{} & {} \left. +\frac{r^{N_1-1}}{N_1-1}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) .\end{aligned}$$

Similarly as the proof of Theorem 4.1, we have $\mathcal {G}_4\left( r_{p,k_1,q,k_2,N_1,N_2}(M)\right) =0$, i.e.,

$$\begin{aligned}{} & {} r_{p,k_1,q,k_2,N_1,N_2}(M)+k_1 r_{p,k_1,q,k_2,N_1,N_2}^p(M)+k_2 r_{p,k_1,q,k_2,N_1,N_2}^q(M)-1\nonumber \\{} & {} \quad \quad +2M\left( (1-r_{p,k_1,q,k_2,N_1,N_2}(M))\ln (1-r_{p,k_1,q,k_2,N_1,N_2}(M)) -1+2\ln 2\right. \nonumber \\{} & {} \quad \quad +\left. \frac{r_{p,k_1,q,k_2,N_1,N_2}^{N_1-1}(M)}{N_1-1}\right. \nonumber \\{} & {} \quad \quad \left. +(1-r_{p,k_1,q,k_2,N_1,N_2}(M))\sum _{n=1}^{N_1-2}\frac{r_{p,k_1,q,k_2,N_1,N_2}^n(M)}{n}+\sum _{n=N_2}^{N_1-1}\frac{2r_{p,k_1,q,k_2,N_1,N_2}^n(M)}{2n(n-1)}\right) =0.\nonumber \\ \end{aligned}$$

(5.18)

Combining (5.2), (5.16) and (5.17), we obtain for $|z|=r\le r_{p,k_1,q,k_2,N_1,N_2}(M)$

$$\begin{aligned} r+k_1|h(r)|^p+k_2|g(r)|^q+\sum _{n=N_1}^\infty |a_n|r^n+\sum _{n=N_2}^\infty |b_n|r^n\le d\left( f(0),\partial f(\mathbb {D})\right) .\end{aligned}$$

In order to show that $r_{p,k_1,q,k_2,N_1,N_2}(M)$ is the best possible, we consider the function $f=f_M$ defined by (5.6) and we again get (5.9). For $|z|=r_{p,k_1,q,k_2,N_1,N_2}(M)$, it follows from (5.9) and (5.18) that

$$\begin{aligned}{} & {} r_{p,k_1,q,k_2,N_1,N_2}(M)+k_1|h(r_{p,k_1,q,k_2,N_1,N_2}(M))|^p+k_2|g(r_{p,k_1,q,k_2,N_1,N_2}(M))|^q\\{} & {} \quad \quad +\sum _{n=N_1}^\infty |a_n|r_{p,k_1,q,k_2,N_1,N_2}^n(M)+\sum _{n=N_2}^\infty |b_n|r_{p,k_1,q,k_2,N_1,N_2}^n(M)\\{} & {} \quad =r_{p,k_1,q,k_2,N_1,N_2}(M)+k_1 r_{p,k_1,q,k_2,N_1,N_2}^p(M)+k_2 r_{p,k_1,q,k_2,N_1,N_2}^q(M)+2M\times \\{} & {} \left( (1-r_{p,k_1,q,k_2,N_1,N_2}(M))\ln (1-r_{p,k_1,q,k_2,N_1,N_2}(M))+\frac{r_{p,k_1,q,k_2,N_1,N_2}^{N_1-1}(M)}{N_1-1}\right. \nonumber \\{} & {} \quad \quad \left. +(1\!-\!r_{p,k_1,q,\!k_2,N_1,N_2}(M))\sum _{n\!=\!1}^{N_1\!-\!2}\frac{r_{p,k_1,q,k_2,N_1,N_2}^n(M)}{n}\!+\!\sum _{n\!=\!N_2}^{N_1-1}\frac{r_{p,k_1,q,k_2,N_1,N_2}^n(M)}{2n(n\!-\!1)}\right) \\{} & {} \quad =1+2M(1-2\ln 2)=d\left( f_M(0),\partial f_M(\mathbb {D})\right) .\end{aligned}$$

Therefore, the radius $r_{p,k_1,q,k_2,N_1,N_2}(M)$ is the best possible. Thus the result follows. $\square$

6 Improved Bohr Radius for the Class $\mathcal {P}_{\mathcal {H}}^0(M)$

In 2023, Ahamed et al. [3] generalized the harmonic versions of Theorem B for the class $\mathcal {P}_{\mathcal {H}}^0(M)$ and obtained the following result.

Theorem G

[3] Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). Then

$$\begin{aligned} r+\sum _{n=2}^\infty \left( |a_n|+|b_n|\right) r^n+P\left( \frac{S_r}{\pi }\right) \le d\left( f(0),\partial f(\mathbb {D})\right) \end{aligned}$$

(6.1)

for $r\le r_N(M)$, where $P\left( \omega \right) =\omega ^N+\omega ^{N-1}+\cdots +\omega$ and $r_N(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned} r-1-2M(r-1+(1-r)\ln (1-r)+2\ln 2)+P\left( r^2+4M^2G(r)\right) =0,\nonumber \\\end{aligned}$$

(6.2)

where G(r) is defined by $G(r):=r^2\left( Li_2(r^2)-1\right) +(1-r^2)\ln (1-r^2)$. The constant $r_N(M)$ is the best possible.

In order to generalize Theorem G, we consider a N-th degree polynomial of the form

$$\begin{aligned} P\left( \omega \right) =c_{N}\left( \omega \right) ^N+c_{N-1}\left( \omega \right) ^{N-1}+c_{N-2}\left( \omega \right) ^{N-2}+\cdots +c_{1}\omega , \end{aligned}$$

(6.3)

where $c_i\in \mathbb {R}$ $(1\le i\le N)$ with $c_N\not =0$. Concerning improved Bohr radius for the class $\mathcal {P}_{\mathcal {H}}^0(M)$, we have obtain the following results.

Theorem 6.1

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $0<M<\frac{1}{2(\ln 4-1)}$ be given by (1.10). Then

$$\begin{aligned} r+\sum _{n=N_1}^\infty |a_n||z|^n+\sum _{n=N_2}^\infty |b_n||z|^n+P\left( \frac{S_r}{\pi }\right) \le d\left( f(0),\partial f(\mathbb {D})\right) \end{aligned}$$

(6.4)

for $r\le r_{N_1,N_2}(M)$ with $N_1\ge N_2\ge 2$, where $P\left( \omega \right)$ is defined in (6.3) and $r_{N_1,N_2}(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned}{} & {} r-1+2M\left( (1-r)\ln (1-r)-1+2\ln 2+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}+\frac{r^{N_1-1}}{N_1-1}\right. \\{} & {} \quad \left. +\sum _{n=N_2}^{N_1-1 }\frac{r^n}{2n(n-1)}\right) \\ {}{} & {} +P\left( r^2+4M^2\left( (r^2-1)\ln (1-r^2)+r^2(Li_2(r^2)-1)\right) \right) =0.\end{aligned}$$

The constant $r_{N_1,N_2}(M)$ is the best possible (Fig. 8 and 9; Tables 5 and 6).

Fig. 8

The graphs of $r_{3,2}(M)$ and $r_{10,5}(M)$ in (0, 1) when $P(r)=r^3+6r^2+3r$

Fig. 9

The graphs of $r_{3,2}(M)$ and $r_{10,4}(M)$ in (0, 1) when $P(r)=r^2+2r$

Table 5 This table shows the values of the roots $r_{N_1,N_2}(M)$ for different values of M in $\left( 0,\frac{1}{2(\ln 4-1)}\right)$ with $N_1=3,5,10$, $N_2=2,3,5$ and $P(r)=r^3+6r^2+3r$

Table 6 This table shows the values of the roots $r_{N_1,N_2}(M)$ for different values of M in $\left( 0,\frac{1}{2(\ln 4-1)}\right)$ with $N_1=3,10$, $N_2=2,4$ and $P(r)=r^2+2r$

As a consequence of Theorem 6.1, we obtain the following corollary.

Corollary 6.1

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $0<M<\frac{1}{2(\ln 4-1)}$ be given by (1.10). Then,

$$\begin{aligned} r+\sum _{n=N_1}^\infty |a_n|r^n+\sum _{n=N_2}^\infty |b_n|r^n+\frac{S_r}{\pi }\le d\left( f(0),\partial f(\mathbb {D})\right) \end{aligned}$$

(6.5)

for $r\le r_{N_1,N_2}(M)$, where $r_{N_1,N_2}(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned}{} & {} r-1+2M\left( (1-r)\ln (1-r)-1+2\ln 2+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}+\frac{r^{N_1-1}}{N_1-1}\right. \\{} & {} \left. +\sum _{n=N_2}^{N_1-1} \frac{r^n}{2n(n-1)}\right) +r^2+4M^2\left( (r^2-1)\ln (1-r^2)+r^2(Li_2(r^2)-1)\right) =0.\end{aligned}$$

The constant $r_{N_1,N_2}(M)$ is the best possible.

7 Proof of the Theorem 6.1

Proof of Theorem 6.1

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). For the analytic functions h and g, the area $S_r$ of the disk $|z|<r$ under the harmonic map f is given by

$$\begin{aligned}{} & {} S_r=\iint \limits _{\begin{array}{c} |z|<r \end{array}}\left( |h'(z)|^2-|g'(z)|^2\right) dx dy, \end{aligned}$$

(7.1)

$$\begin{aligned} \text {where}{} & {} \frac{1}{\pi }\iint \limits _{\begin{array}{c} |z|<r \end{array}}|h'(z)|^2dx dy=\sum _{n=1}^\infty n|a_n|^2r^{2n} \end{aligned}$$

(7.2)

$$\begin{aligned} \text {and}{} & {} \frac{1}{\pi }\iint \limits _{\begin{array}{c} |z|<r \end{array}}|g'(z)|^2dx dy=\sum _{n=2}^\infty n|b_n|^2r^{2n}. \end{aligned}$$

(7.3)

Combining (7.1), (7.2), (7.3) and Lemma 2.3, we obtain

$$\begin{aligned} \frac{S_r}{\pi }= & {} \frac{1}{\pi }\iint \limits _{\begin{array}{c} |z|<r \end{array}}\left( |h'(z)|^2-|g'(z)|^2\right) dx dy \nonumber \\= & {} r^2+\sum _{n=2}^\infty n|a_n|^2r^{2n}-\sum _{n=2}^\infty n|b_n|^2r^{2n}\nonumber \\= & {} r^2+\sum _{n=2}^\infty n\left( |a_n|+|b_n|\right) \left( |a_n|-|b_n|\right) r^{2n}\nonumber \\\le & {} r^2+\sum _{n=2}^\infty \frac{4M^2r^{2n}}{n(n-1)^2}\nonumber \\= & {} r^2+4M^2\left( (r^2-1)\ln (1-r^2)+r^2(Li_2(r^2)-1)\right) .\end{aligned}$$

(7.4)

Note that, $N_1\ge N_2\ge 2$. In view of Lemmas 2.2, 2.3 and 2.4, we obtain

$$\begin{aligned}{} & {} r+\sum _{n=N_1}^\infty |a_n|r^n+\sum _{n=N_2}^\infty |b_n|r^n+P\left( \frac{S_r}{\pi }\right) \nonumber \\{} & {} =r+\sum _{n=N_1}^\infty \left( |a_n|+|b_n|\right) r^n+\sum _{n=N_2}^{N_1-1}|b_n|r^n+P\left( \frac{S_r}{\pi }\right) \nonumber \\{} & {} \le r+\sum _{n=N_1}^\infty \frac{2Mr^n}{n(n-1)}+\sum _{n=N_2}^{N_1-1}\frac{Mr^n}{n(n-1)}+P\left( r^2+4M^2\left( (r^2-1)\ln (1-r^2)\right. \right. \nonumber \\ {}{} & {} \left. \left. +r^2(Li_2(r^2)-1)\right) \right) \nonumber \\{} & {} =r+2M\left( (1-r)\ln (1-r)+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}+\frac{r^{N_1-1}}{N_1-1}+\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) \nonumber \\{} & {} +P\left( r^2+4M^2\left( (r^2-1)\ln (1-r^2)+r^2(Li_2(r^2)-1)\right) \right) \nonumber \\ {}{} & {} \le 1+2M(1-2\ln 2)\end{aligned}$$

(7.5)

for $0<r\le r_{N_1,N_2}(M)$, where $r_{N_1,N_2}(M)$ is the smallest root of $\mathcal {G}_5(r)=0$ in (0, 1) and $\mathcal {G}_5:[0,1)\rightarrow \mathbb {R}$ be defined by

$$\begin{aligned}{} & {} \mathcal {G}_5(r):=r-1+2M\left( (1-r)\ln (1-r)-1+2\ln 2+(1-r)\sum _{n=1}^{N_1-2}\frac{r^n}{n}+\frac{r^{N_1-1}}{N_1-1}\right. \\ {}{} & {} \left. +\sum _{n=N_2}^{N_1-1}\frac{r^n}{2n(n-1)}\right) +P\left( r^2+4M^2\left( (r^2-1)\ln (1-r^2)+r^2(Li_2(r^2)-1)\right) \right) .\end{aligned}$$

Similarly as the proof of Theorem 4.1, we have $\mathcal {G}_5\left( r_{N_1,N_2}(M)\right) =0$, i.e.,

$$\begin{aligned}{} & {} r_{N_1,N_2}(M)-1+2M\left( (1-r_{N_1,N_2}(M))\ln (1-r_{N_1,N_2}(M))-1+2\ln 2\right. \nonumber \\{} & {} \left. +\sum _{n=N_2}^{N_1-1}\frac{r_{N_1,N_2}^n(M)}{2n(n-1)}\right. \nonumber \\{} & {} \left. +(1-r_{N_1,N_2}(M))\sum _{n=1}^{N_1-2}\frac{r_{N_1,N_2}^n(M)}{n} +\frac{r_{N_1,N_2}^{N_1-1}(M)}{N_1-1}\right) +P\left( r_{N_1,N_2}^2(M)+4M^2\times \right. \nonumber \\{} & {} \left. \left( \left( r_{N_1,N_2}^2(M)-1\right) \ln \left( 1-r_{N_1,N_2}^2(M)\right) +r_{N_1,N_2}^2(M)\left( Li_2(r_{N_1,N_2}^2(M))-1\right) \right) \right) =0.\nonumber \\ \end{aligned}$$

(7.6)

Combining (5.2), (7.5) and (7.6), we obtain for $|z|=r\le r_{N_1,N_2}(M)$

$$\begin{aligned} r+\sum _{n=N_1}^\infty |a_n|r^n+\sum _{n=N_2}^\infty |b_n|r^n+P\left( \frac{S_r}{\pi }\right) \le d\left( f(0),\partial f(\mathbb {D})\right) .\end{aligned}$$

In order to show that $r_{N_1,N_2}(M)$ is the best possible, we consider the function $f=f_M$ defined by (5.6) and we again get (5.9). For $|z|=r_{N_1,N_2}(M)$, it follows from (5.9) and (7.6) that

$$\begin{aligned}{} & {} r_{N_1,N_2}(M)+\sum _{n=N_1}^\infty |a_n|r_{N_1,N_2}^n(M)+\sum _{n=N_2}^\infty |b_n|r_{N_1,N_2}^n(M)+P\left( \frac{S_r}{\pi }\right) \\{} & {} =r_{N_1,N_2}(M)+2M\left( (1-r_{N_1,N_2}(M))\ln (1-r_{N_1,N_2}(M))+\sum _{n=N_2}^{N_1-1}\frac{r_{N_1,N_2}^n(M)}{2n(n-1)}\right. \nonumber \\ {}{} & {} \left. +(1-r_{N_1,N_2}(M))\sum _{n=1}^{N_1-2}\frac{r_{N_1,N_2}^n(M)}{n}+\frac{r_{N_1,N_2}^{N_1-1}(M)}{N_1-1}\right) +P\left( r_{N_1,N_2}^2(M)+4M^2\times \right. \nonumber \\ {}{} & {} \left. \left( \left( r_{N_1,N_2}^2(M)-1\right) \ln \left( 1-r_{N_1,N_2}^2(M)\right) +r_{N_1,N_2}^2(M)\left( Li_2(r_{N_1,N_2}^2(M))-1\right) \right) \right) \\{} & {} =1+2M(1-2\ln 2)=d\left( f_M(0),\partial f_M(\mathbb {D})\right) .\end{aligned}$$

Hence the radius $r_{N_1,N_2}(M)$ is the best possible. This completes the proof. $\square$

8 Refined Bohr Radius for the Class $\mathcal {P}_{\mathcal {H}}^0(M)$

In this section, we establish some sharp results on the harmonic analogue of Theorem C for the class $\mathcal {P}_{\mathcal {H}}^0(M)$ as follows.

Theorem 8.1

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ for $0<M<\frac{1}{2(\ln 4-1)}$ be given by (1.10). Then

$$\begin{aligned}{} & {} |f(z)|^p+\sum _{n=N}^\infty \left( |a_n|+|b_n|\right) r^n+\frac{r^q}{1-r^q}\sum _{n=N}^\infty (n(n-1))^{q-1}\left( |a_n|+|b_n|\right) ^{q}r^{qn}\\{} & {} \le d\left( f(0),\partial f(\mathbb {D})\right) \end{aligned}$$

for $r\le r_{p,q,N}(M)$ with $p,q(\not =1)\in \mathbb {N}$ and $N\ge 2$, where $r_{p,q,N}(M)\in (0,1)$ is the smallest root of the equation

$$\begin{aligned}{} & {} \left[ r+2M\left\{ r+(1-r)\ln (1-r)\right\} \right] ^p-1+2M\left( (1-r)\ln (1-r)+(1-r)\sum _{n=1}^{N-2}\frac{r^n}{n}\right. \\{} & {} \left. -1+2\ln 2+\frac{r^{N-1}}{N-1}\right) +\frac{(2rM)^q}{1-r^q}\left( (1-r^q)\ln (1-r^q)+(1-r^q)\sum _{n=1}^{N-2}\frac{r^{nq}}{n}+\frac{r^{q(N-1)}}{N-1}\right) =0.\nonumber \end{aligned}$$

(8.1)

The constant $r_{p,q,N}(M)$ is the best possible (Figs. 10 and 11; Table 7).

Fig. 10

The graphs of $r_ {1, 2, 3}(M)$ and $r_ {1, 2, 10}(M)$ in (0, 1)

Fig. 11

The graphs of $r_ {2, 3, 5}(M)$ and $r_ {3, 3, 5}(M)$ in (0, 1)

Table 7 This table shows the values of the roots $r_{p,q,N}(M)$ for different values of M in $\left( 0,\frac{1}{2(\ln 4-1)}\right)$ with $p=1,2,3$, $q=2,3$ and $N=3,5,10$

9 Proof of the Theorem 8.1

Proof of Theorem 8.1

Let $f\in \mathcal {P}_{\mathcal {H}}^0(M)$ be given by (1.10). Note that $p,q(\not =1)\in \mathbb {N}$ and $N\ge 2$. In view of Lemmas 2.3 and 2.4, we obtain

$$\begin{aligned}{} & {} |f(z)|^p+\sum _{n=N}^\infty \left( |a_n|+|b_n|\right) r^n+\frac{r^q}{1-r^q}\sum _{n=N}^\infty (n(n-1))^{q-1}\left( |a_n|+|b_n|\right) ^{q}r^{qn}\nonumber \\{} & {} \le \left( r+2M\sum _{n=2}^\infty \frac{r^{n}}{n(n-1)}\right) ^p+\sum _{n=N}^\infty \frac{2Mr^{n}}{n(n-1)}+\frac{r^q}{1-r^q}\sum _{n=N}^\infty \frac{2^qM^qr^{qn}}{n(n-1)}.\end{aligned}$$

(9.1)

Now,

$$\begin{aligned}{} & {} \left( r+2M\sum _{n=2}^\infty \frac{r^{n}}{n(n-1)}\right) ^p+\sum _{n=N}^\infty \frac{2Mr^{n}}{n(n-1)}+\frac{r^q}{1-r^q}\sum _{n=N}^\infty \frac{2^qM^qr^{qn}}{n(n-1)}\\{} & {} =\left[ r+2M\left\{ r+(1-r)\ln (1-r)\right\} \right] ^p\\{} & {} +2M\left( (1-r)\ln (1-r)+(1-r) \sum _{n=1}^{N-2}\frac{r^n}{n}+\frac{r^{N-1}}{N-1}\right) \\ {}{} & {} +\frac{(2rM)^q}{1-r^q}\left( (1-r^q)\ln (1-r^q)+(1-r^q)\sum _{n=1}^{N-2}\frac{r^{nq}}{n}+\frac{r^{q(N-1)}}{N-1}\right) \\{} & {} \le 1+2M(1-2\ln 2) \end{aligned}$$

for $0<r\le r_{p,q,N}(M)$, where $r_{p,q,N}(M)$ is the smallest root of $\mathcal {G}_7(r)=0$ in (0, 1) and $\mathcal {G}_7:[0,1)\rightarrow \mathbb {R}$ be defined by

$$\begin{aligned}{} & {} \mathcal {G}_7(r):=\left[ r+2M\left\{ r+(1-r)\ln (1-r)\right\} \right] ^p-1+2M\left( (1-r)\ln (1-r)+(1-r)\sum _{n=1}^{N-2}\frac{r^n}{n}\right. \\{} & {} \left. -1+2\ln 2+\frac{r^{N-1}}{N-1}\right) +\frac{(2rM)^q}{1-r^q}\left( (1-r^q)\ln (1-r^q)+(1-r^q)\sum _{n=1}^{N-2}\frac{r^{nq}}{n}+\frac{r^{q(N-1)}}{N-1}\right) .\end{aligned}$$

Similarly as the proof of Theorem 4.1, we have $\mathcal {G}_7\left( r_{p,q,N}(M)\right) =0$, i.e.,

$$\begin{aligned}{} & {} \left[ r_{p,q,N}(M)+2M\left\{ r_{p,q,N}(M)+(1-r_{p,q,N}(M))\ln (1-r_{p,q,N}(M))\right\} \right] ^p\nonumber \\{} & {} +2M\left( (1-r_{p,q,N}(M))\times \right. \nonumber \\{} & {} \left. \ln (1-r_{p,q,N}(M))+(1-r_{p,q,N}(M))\sum _{n=1}^{N-2}\frac{r_{p,q,N}^n(M)}{n}-1+2\ln 2+\frac{r_{p,q,N}^{N-1}(M)}{N-1}\right) -1\nonumber \\{} & {} +\frac{(2r_{p,q,N}(M)M)^q}{1-r_{p,q,N}^q(M)}\left( (1-r_{p,q,N}^q(M))\ln (1-r_{p,q,N}^q(M))+(1-r_{p,q,N}^q(M))\sum _{n=1}^{N-2}\frac{r_{p,q,N}^{nq}(M)}{n}\right. \nonumber \\ {}{} & {} \left. +\frac{r_{p,q,N}^{q(N-1)}(M)}{N-1}\right) =0.\end{aligned}$$

(9.2)

Combining (5.2), (9.1) and (9.2), we obtain for $|z|=r\le r_{p,q,N}(M)$

$$\begin{aligned}{} & {} |f(z)|^p+\sum _{n=N}^\infty \left( |a_n|+|b_n|\right) r^n+\frac{r^q}{1-r^q}\sum _{n=N}^\infty (n(n-1))^{q-1}\left( |a_n|+|b_n|\right) ^{q}r^{qn}\\{} & {} \le d\left( f(0),\partial f(\mathbb {D})\right) .\end{aligned}$$

In order to show that $r_{p,q,N}(M)$ is the best possible, we consider the function $f=f_M$ defined by (5.6) and we again get (5.9). For $|z|=r_{p,q,N}(M)$, it follows from (5.9) and (9.2) that

$$\begin{aligned}{} & {} |f(z)|^p+\sum _{n=N}^\infty \left( |a_n|+|b_n|\right) r_{p,q,N}^n(M)+\frac{r_{p,q,N}^q(M)}{1-r_{p,q,N}^q(M)}\sum _{n=N}^\infty (n(n-1))^{q-1}\\{} & {} \left( |a_n|+|b_n|\right) ^{q}r_{p,q,N}^{qn}(M)\\{} & {} \quad =\left[ r_{p,q,N}(M)+2M\left\{ r_{p,q,N}(M)+(1-r_{p,q,N}(M))\ln (1-r_{p,q,N}(M))\right\} \right] ^p\\{} & {} \quad \quad +2M\left( (1-r_{p,q,N}(M))\times \right. \nonumber \\{} & {} \quad \left. \ln (1-r_{p,q,N}(M))+(1-r_{p,q,N}(M))\sum _{n=1}^{N-2}\frac{r_{p,q,N}^n(M)}{n} +\frac{r_{p,q,N}^{N-1}(M)}{N-1}\right) +\frac{(2r_{p,q,N}(M)M)^q}{1-r_{p,q,N}^q(M)}\times \nonumber \\{} & {} \left( (1-r_{p,q,N}^q(M))\ln (1-r_{p,q,N}^q(M))+(1-r_{p,q,N}^q(M))\sum _{n=1}^{N-2}\frac{r_{p,q,N}^{nq}(M)}{n}+\frac{r_{p,q,N}^{q(N-1)}(M)}{N-1}\right) \\{} & {} \quad =1+2M(1-2\ln 2)=d\left( f_M(0),\partial f_M(\mathbb {D})\right) .\end{aligned}$$

Hence the radius $r_{p,q,N}(M)$ is the best possible. Thus the result follows. $\square$