††footnotetext:  File: Ahamed-Vasu-Himadri-p1_-12-12-2020.tex, printed:
2024-8-27, 22.35

# Bohr radius for certain close-to-convex harmonic mappings

Molla Basir Ahamed  Molla Basir Ahamed, School of Basic Science, Indian
Institute of Technology Bhubaneswar, Bhubaneswar-752050, Odisha, India.
<EMAIL_ADDRESS>,  Vasudevarao Allu  Vasudevarao Allu, School of Basic
Science, Indian Institute of Technology Bhubaneswar, Bhubaneswar-752050,
Odisha, India<EMAIL_ADDRESS>and  Himadri Halder  Himadri Halder, School
of Basic Science, Indian Institute of Technology Bhubaneswar,
Bhubaneswar-752050, Odisha, India<EMAIL_ADDRESS>
###### Abstract.

Let $\mathcal{H}$ be the class of harmonic functions $f=h+\bar{g}$ in the unit
disk $\mathbb{D}:=\\{z\in\mathbb{C}:|z|<1\\}$, where $h$ and $g$ are analytic
in $\mathbb{D}$. Let

$\mathcal{P}_{\mathcal{H}}^{0}(\alpha)=\\{f=h+\overline{g}\in\mathcal{H}:(h^{\prime}(z)-\alpha)>|g^{\prime}(z)|\;\mbox{with}\;0\leq\alpha<1,\;g^{\prime}(0)=0,\;z\in\mathbb{D}\\}$

be the class of close-to-convex mappings defined by Li and Ponnusamy [34]. In
this paper, we obtain the sharp Bohr-Rogosinski radius, improved Bohr radius
and refined Bohr radius for the class $\mathcal{P}_{\mathcal{H}}^{0}(\alpha)$.

###### Key words and phrases:

Analytic, univalent, harmonic functions; starlike, convex, close-to-convex
functions; coefficient estimate, growth theorem, Bohr radius.

###### AMS Subject Classification: Mathematics Subject Classification:

Primary 30C45, 30C50, 30C80

##  1\. Introduction

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

(1.1) $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$

such that $|f(z)|<1$ in $\mathbb{D}$, then the majorant series $M_{f}(r)$
associated with $f$ satisfies the following inequality

(1.2) $M_{f}(r):=\sum_{n=0}^{\infty}|a_{n}|r^{n}\leq
1\quad\mbox{for}\quad|z|=r\leq 1/3,$

and the constant $1/3$, known as Bohr radius, cannot be improved. In 1914, H.
Bohr [17] obtained the inequality (1.2) for $r\leq 1/6$ and subsequently
improved by $1/3$ later, Weiner, Riesz and Schur independently obtained the
constant $1/3$. An observation shows that the quantity $1-|a_{0}|$ is equal to
$d(f(0),\partial f(\mathbb{D}))$. Therefore, the inequality (1.2) is called
Bohr inequality, can be written in the following form

(1.3) $\sum_{n=1}^{\infty}|a_{n}z^{n}|\leq 1-|a_{0}|=d(f(0),\partial
f(\mathbb{D}))$

for $|z|=r\leq 1/3$, where $d$ is the Euclidean distance. 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 $|z|=r\leq 1/3$, are sometimes said to
satisfy the classical Bohr phenomenon. The notion of the Bohr phenomenon can
be generalized to the class $\mathcal{F}$ consisting of analytic functions $f$
from $\mathbb{D}$ to a given domain $\varOmega\subseteq\mathbb{C}$ such that
$f(\mathbb{D})\subseteq\varOmega$ and the class $\mathcal{F}$ is said to
satisfy the Bohr phenomenon if there exists largest radius
$r_{\varOmega}\in(0,1)$ such that the inequality (1.3) holds for $|z|=r\leq
r_{\varOmega}$ and for all functions $f\in\mathcal{F}$. We say the largest
radius $r_{\varOmega}$ is the Bohr radius for the class $\mathcal{F}$. The
Bohr radius has been obtained for the class $\mathcal{F}$ when $\varOmega$ is
convex domain [8], simply connected domain [1], the exterior of the closed
unit disk, the punctured unit disk, and concave wedge domain (see [9]). In
$1997$, Boas and Khavinson [15] generalized the Bohr inequality in several
complex variables by finding multidimensional Bohr radius. In $2020$, Liu and
Ponnusamy [36] 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 [33] obtained
several interesting improved versions of Bohr inequality. For more results on
this, we refer the reader to glance through the articles (see [22, 23, 27, 32,
33, 35, 38, 40]). In 2017, Kayumov and Ponnusamy [27] introduced Bohr-
Rogosinski radius motivated by Rogosinski radius for bounded analytic
functions in $\mathbb{D}$. Rogosinski radius is defined as follows: Let
$f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be analytic in $\mathbb{D}$ and its
corresponding partial sum of $f$ is defined by
$S_{N}(z):=\sum_{n=0}^{N-1}a_{n}z^{n}$. Then, for every $N\geq 1$, we have
$|\sum_{n=0}^{N-1}a_{n}z^{n}|<1$ 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

(1.4) $R_{N}^{f}(z):=|f(z)|+\sum_{n=N}^{\infty}|a_{n}||z|^{n}.$

It is worth to point out that
$|S_{N}(z)|=\big{|}f(z)-\sum_{n=N}^{\infty}a_{n}z^{n}\big{|}\leq|R_{N}^{f}(z)|$.
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 [27].

######  Theorem 1.5.

[27] Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be analytic in $\mathbb{D}$ and
$|f(z)|\leq 1$. Then

(1.6) $|f(z)|+\sum_{n=N}^{\infty}|a_{n}||z|^{n}\leq 1$

for $|z|=r\leq R_{N}$, 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,

(1.7) $|f(z)|^{2}+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)|z|^{n}\leq 1$

for $R^{\prime}_{N}$, where $R^{\prime}_{N}$ is the positive root of the
equation $(1+r)r^{N}-(1-r)^{2}$. The radius $R^{\prime}_{N}$ is the best
possible.

Recently, Kayumov and Ponnusamy [27] have proved the following improved
version of Bohr’s inequality.

######  Theorem 1.8.

[27] Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be analytic in $\mathbb{D}$,
$|f(z)|\leq 1$ and $S_{r}$ denote the image of the subdisk $|z|<r$ under
mapping $f$. Then

(1.9)
$B_{1}(r):=\sum_{n=0}^{\infty}|a_{n}|r^{n}+\frac{16}{9}\left(\frac{S_{r}}{\pi}\right)\leq
1\quad\mbox{for}\quad r\leq\frac{1}{3}$

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

(1.10)
$B_{2}(r):=|a_{0}|^{2}+\sum_{n=1}^{\infty}|a_{n}|r^{n}+\frac{9}{8}\left(\frac{S_{r}}{\pi}\right)\leq
1\quad\mbox{for}\quad r\leq\frac{1}{2}$

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

Bohr’s phenomenon for the complex-valued harmonic mappings have been studied
extensivel by many authors (see [1, 3, 13, 12]). Improved Bohr inequality for
locally univalent harmonic mappings have been discussed by Evdoridis et al.
[21].

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
$J_{f}(z):=|h^{\prime}(z)|^{2}-|g^{\prime}(z)|^{2}$, 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 $h(0)=h^{\prime}(0)-1=0$ and $g(0)=0$. A function
$f\in\mathcal{H}$ is said to be in $\mathcal{H}_{0}$ if $g^{\prime}(0)=0$.
Thus, every $f=h+\overline{g}\in\mathcal{H}_{0}$ has the following form

(1.11)
$f(z)=h(z)+\overline{g(z)}=z+\sum_{n=2}^{\infty}a_{n}z^{n}+\overline{\sum_{n=2}^{\infty}b_{n}z^{n}}.$

In 2013, Ponnusamy et al. [39] considered the following classes

$\mathcal{P}_{\mathcal{H}}^{0}=\\{f=h+\bar{g}\in\mathcal{H}:h^{\prime}(z)>|g^{\prime}(z)|\quad\mbox{with}\quad
g^{\prime}(0)=0\quad\mbox{for}\quad z\in\mathbb{D}\\}.$

Motivated by the class $\mathcal{P}_{\mathcal{H}}^{0}$, Li and Ponnusamy [34]
have studied the following class $\mathcal{P}_{\mathcal{H}}^{0}(\alpha)$
defined by

$\mathcal{P}_{\mathcal{H}}^{0}(\alpha)=\\{f=h+\overline{g}\in\mathcal{H}:(h^{\prime}(z)-\alpha)>|g^{\prime}(z)|\;\mbox{with}\;0\leq\alpha<1,\;g^{\prime}(0)=0\;\mbox{for}\;z\in\mathbb{D}\\}.$

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

######  Lemma 1.12.

[34] Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ and be given by (1.11).
Then for any $n\geq 2$,

  1. (i)

$\displaystyle|a_{n}|+|b_{n}|\leq\frac{2(1-\alpha)}{n};$

  2. (ii)

$\displaystyle||a_{n}|-|b_{n}||\leq\frac{2(1-\alpha)}{n};$

  3. (iii)

$\displaystyle|a_{n}|\leq\frac{2(1-\alpha)}{n}.$

All the inequalities are sharp, with extremal function
$f(z)=(1-\alpha)(-z-2\,\;\log(1-z))+\alpha z$.

######  Lemma 1.13.

[13] Let $f=h+\overline{g}\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ with
$0\leq\alpha<1$. Then

(1.14)
$|z|+\sum\limits_{n=2}^{\infty}\dfrac{2(1-\alpha)(-1)^{n-1}}{n}|z|^{n}\leq|f(z)|\leq|z|+\sum\limits_{n=2}^{\infty}\dfrac{2(1-\alpha)}{n}|z|^{n}.$

Both inequalities are sharp.

The organization of this paper is follows: In section 2 we obtain sharp Bohr-
Rogosinski radius for the class $\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ of
close-to-convex harmonic mappings. In section 3, we establish interesting
sharp improved-Bohr radius $\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$. In section
4, we prove sharp refined-Bohr radius as well as Bohr-type inequality for the
class $\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$. Section 6 is devoted for the
proofs of main the results.

##  2\. Bohr-Rogosinski Radius for the class
$\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$

We first prove the following Bohr-Rogosinski radius for the class
$\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$.

######  Theorem 2.1.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then, for
$N\geq 2$,

(2.2) $|f(z)|+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)|z|^{n}\leq
d\left(f(0),\partial f(\mathbb{D})\right)$

for $|z|=r\leq r_{{}_{N}}(\alpha)$, where $r_{{}_{N}}(\alpha)$ is the smallest
root of the equation

(2.3)
$r-1-2(1-\alpha)\left(r-1+\ln\,(2(1-r)^{2})+\sum_{n=1}^{N-1}\frac{r^{n}}{n}\right)=0\;\;\mbox{in}\;\;(0,1).$

The constant $r_{{}_{N}}(\alpha)$ is the best possible.

Figure 1. The graph of $r_{3}(\alpha)$ and $r_{10}(\alpha)$ in $(0,1)$. $\;\;\alpha$ | $\;\;0.1$ | $\;\;0.2$ | $\;\;0.3$ | $\;\;0.4$ | $\;\;0.5$ | $\;\;0.6$ | $\;\;0.7$ | $\;\;0.8$ | $\;\;0.9$
---|---|---|---|---|---|---|---|---|---
$r_{2}(\alpha)$ | $0.2771$ | $0.3115$ | $0.3477$ | $0.3866$ | $0.4296$ | $0.4785$ | $0.5367$ | $0.6109$ | $0.7187$
$r_{3}(\alpha)$ | $0.3121$ | $0.3493$ | $0.3877$ | $0.4281$ | $0.4717$ | $0.5201$ | $0.5764$ | $0.6463$ | $0.7453$
$r_{6}(\alpha)$ | $0.3248$ | $0.3653$ | $0.4070$ | $0.4508$ | $0.4978$ | $0.5493$ | $0.6080$ | $0.6786$ | $0.7736$
$r_{10}(\alpha)$ | $0.3251$ | $0.3657$ | $0.4978$ | $0.4522$ | $0.4999$ | $0.5527$ | $0.6130$ | $0.6859$ | $0.7832$
Table 1. This table shows the value of the roots $r_{{}_{N}}(\alpha)$ for
different values of $\alpha$ in $[0,1)$ and $N=2,3,6,10$.

######  Theorem 2.4.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then,
$N\geq 2$,

(2.5) $|f(z)|^{2}+\sum_{n=N}^{\infty}|a_{n}||z|^{n}\leq d\left(f(0),\partial
f(\mathbb{D})\right)$

for $|z|=r\leq r_{{}_{N}}(\alpha)$, where $r_{{}_{N}}(\alpha)\in(0,1)$ is the
smallest root of the equation

(2.6)
$\bigg{(}r-2(1-\alpha)(r+\ln\;(1-r))\bigg{)}^{2}-2(1-\alpha)\left(\ln\,(2-2r)-1+\sum_{n=2}^{N-1}\frac{r^{n}}{n}\right)-1=0.$

The constant $r_{{}_{N}}(\alpha)$ is the best possible.

Figure 2. The graph of $r_{3}(\alpha)$ and $r_{8}(\alpha)$ of the equation (2.6). $\;\;\alpha$ | $\;\;0.1$ | $\;\;0.2$ | $\;\;0.3$ | $\;\;0.4$ | $\;\;0.5$ | $\;\;0.6$ | $\;\;0.7$ | $\;\;0.8$ | $\;\;0.9$
---|---|---|---|---|---|---|---|---|---
$r_{3}(\alpha)$ | $0.4102$ | $0.4399$ | $0.4708$ | $0.5038$ | $0.5399$ | $0.5807$ | $0.6291$ | $0.6903$ | $0.7783$
$r_{8}(\alpha)$ | $0.4304$ | $0.4613$ | $0.4933$ | $0.5273$ | $0.5644$ | $0.6060$ | $0.6547$ | $0.7152$ | $0.7994$
Table 2. Values of $r_{N}(\alpha)$ for $N=3$ and $8$ when $\alpha\in[0,1)$.

######  Theorem 2.7.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then for a
positive integer $N\geq 2$,

(2.8) $|f(z^{m})|+\sum_{n=N}^{\infty}|a_{n}||z|^{n}\leq d\left(f(0),\partial
f(\mathbb{D})\right)$

for $|z|=r\leq r_{m,N}(\alpha)$, where $r_{m,N}(\alpha)\in(0,1)$ is the
smallest root of the equation

(2.9)
$r^{m}-1-2(1-\alpha)\left(r^{m}-1+\ln\,((1-r^{m})(2-2r))+\sum_{n=1}^{N-1}\frac{r^{n}}{n}\right)=0.$

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

Figure 3. The graphs $r_{2,2}(\alpha)$, $r_{3,2}(\alpha)$ and
$r_{7,2}(\alpha)$ of the equation (2.9)

Figure 4. The graphs $r_{25,2}(\alpha)$, $r_{150,2}(\alpha)$ and
$r_{5,3}(\alpha)$ of the equation (2.9).

Figure 5. The graphs $r_{15,3}(\alpha)$, $r_{25,3}(\alpha)$ and
$r_{180,3}(\alpha)$ of the equation (2.9).

Figure 6. The graphs $r_{5,5}(\alpha)$, $r_{15,5}(\alpha)$ and $r_{35,5}(\alpha)$ of the equation (2.9). $\alpha$ | $r_{2,2}(\alpha)$ | $r_{3,2}(\alpha)$ | $r_{7,2}(\alpha)$ | $r_{25,2}(\alpha)$ | $r_{50,2}(\alpha)$ | $r_{90,2}(\alpha)$ | $r_{150,2}(\alpha)$
---|---|---|---|---|---|---|---
$0.1$ | $0.2016$ | $0.2157$ | $0.2201$ | $0.2201$ | $0.2201$ | $0.2202$ | $0.2202$
$0.2$ | $0.2436$ | $0.2639$ | $0.2724$ | $0.2725$ | $0.2725$ | $0.2725$ | $0.2725$
$0.3$ | $0.2905$ | $0.3187$ | $0.3344$ | $0.3346$ | $0.3346$ | $0.3346$ | $0.3346$
$0.4$ | $0.3433$ | $0.3805$ | $0.4083$ | $0.4093$ | $0.4093$ | $0.4093$ | $0.4093$
$0.5$ | $0.4030$ | $0.4499$ | $0.4962$ | $0.5000$ | $0.5000$ | $0.5000$ | $0.5000$
$0.6$ | $0.4710$ | $0.5268$ | $0.5970$ | $0.6105$ | $0.6106$ | $0.6106$ | $0.6106$
$0.7$ | $0.5498$ | $0.6119$ | $0.7031$ | $0.7430$ | $0.7433$ | $0.7433$ | $0.7433$
$0.8$ | $0.6443$ | $0.7069$ | $0.8033$ | $0.8772$ | $0.8877$ | $0.8884$ | $0.8884$
$0.9$ | $0.7667$ | $0.8184$ | $0.8944$ | $0.9542$ | $0.9707$ | $0.9796$ | $0.9848$
Table 3. The roots of $r_{m,N}(\alpha)$ for different values of $m=2,3,7,25,50,90,150$ with $N=2$ and $\alpha\in[0,1)$. $\alpha$ | $r_{5,3}(\alpha)$ | $r_{15,3}(\alpha)$ | $r_{35,3}(\alpha)$ | $r_{85,3}(\alpha)$ | $r_{180,3}(\alpha)$ | $r_{5,5}(\alpha)$ | $r_{15,5}(\alpha)$ | $r_{35,5}(\alpha)$
---|---|---|---|---|---|---|---|---
$0.1$ | $0.6435$ | $0.6922$ | $0.6936$ | $0.6936$ | $0.6936$ | $0.7283$ | $0.8048$ | $0.8145$
$0.2$ | $0.6744$ | $0.7298$ | $0.7326$ | $0.7326$ | $0.7326$ | $0.7503$ | $0.8371$ | $0.8397$
$0.3$ | $0.7045$ | $0.7664$ | $0.7716$ | $0.7716$ | $0.7716$ | $0.7717$ | $0.8479$ | $0.8639$
$0.4$ | $0.7344$ | $0.8019$ | $0.8109$ | $0.8111$ | $0.8111$ | $0.7930$ | $0.8678$ | $0.8872$
$0.5$ | $0.7647$ | $0.8360$ | $0.8508$ | $0.8515$ | $0.8515$ | $0.8147$ | $0.8870$ | $0.9094$
$0.6$ | $0.7902$ | $0.8687$ | $0.8901$ | $0.8930$ | $0.8930$ | $0.8374$ | $0.9057$ | $0.9300$
$0.7$ | $0.8299$ | $0.8996$ | $0.9257$ | $0.9345$ | $0.9349$ | $0.8621$ | $0.9345$ | $0.9486$
$0.8$ | $0.8677$ | $0.9293$ | $0.9545$ | $0.9686$ | $0.9732$ | $0.8905$ | $0.9440$ | $0.9653$
$0.9$ | $0.9141$ | $0.9593$ | $0.9771$ | $0.9875$ | $0.9924$ | $0.9267$ | $0.9657$ | $0.9807$
Table 4. This table shows the roots $r_{m,N}(\alpha)$ for different values of
$m=5,15,35,85,180$ with $N=3$ and $m=5,15,35$ with $N=5$.

For different values of $\alpha$, $m$ and $N$, in above the corresponding
radii are represented on $x$-axis cut by the increasing curves all have
asymptotes at $x=1$ have been shown in Figures 3-6.

######  Theorem 2.10.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then

(2.11) $\displaystyle
r+|h(r)|^{p}+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}\leq
d\left(f(0),\partial f(\mathbb{D})\right),\;\text{for}\;\;r\leq
r_{p}(\alpha),$

where $r_{p}(\alpha)$ is the smallest root of the equation

(2.12)
$r^{p}+r-1-2(1-\alpha)\left(r-1+\ln\,(2-2r)\right)=0\;\;\mbox{in}\;\;(0,1).$

The radius $r_{p}(\alpha)$ is the best possible.

Figure 7. The roots $r_{{}_{{}_{7}}}(\alpha)$ and $r_{{}_{35}}(\alpha)$ of the equation $r^{p}+r-1-2(1-\alpha)\left(r-1+\ln\,(2-2r)\right)=0$ in $(0,1)$. $\alpha$ | $0.1$ | $0.2$ | $0.3$ | $0.4$ | $0.5$ | $0.6$ | $0.7$ | $0.8$ | $0.9$
---|---|---|---|---|---|---|---|---|---
$r_{7}(\alpha)$ | $0.3249$ | $0.3653$ | $0.4069$ | $0.4503$ | $0.4963$ | $0.5456$ | $0.5992$ | $0.6579$ | $0.7231$
$r_{35}(\alpha)$ | $0.3251$ | $0.3657$ | $0.4078$ | $0.4522$ | $0.5000$ | $0.5529$ | $0.6136$ | $0.6872$ | $0.7867$
Table 5. The roots $r_{{}_{7}}(\alpha)$ and $r_{{}_{35}}(\alpha)$ of the
equation $r^{p}+r-1-2(1-\alpha)\left(r-1+\ln\,(2-2r)\right)=0$ for
$\alpha\in[0,1)$.

##  3\. Improved Bohr Radius for the class
$\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$

In $2020$, Kayumov and Ponnusamy [33] have obtained several improved versions
of Bohr inequality for analytic functions.

######  Theorem 3.1.

[33] Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be analytic in $\mathbb{D}$,
$|f(z)|\leq 1$ and $S_{r}$ denotes the image of the subdisk $|z|<r$ under
mapping $f$. Then

(3.2)
$B_{1}(r):=\sum_{n=0}^{\infty}|a_{n}|r^{n}+\frac{16}{9}\left(\frac{S_{r}}{\pi}\right)\leq
1\quad\mbox{for}\quad r\leq\frac{1}{3}$

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

(3.3)
$B_{2}(r):=|a_{0}|^{2}+\sum_{n=1}^{\infty}|a_{n}|r^{n}+\frac{9}{8}\left(\frac{S_{r}}{\pi}\right)\leq
1\quad\mbox{for}\quad r\leq\frac{1}{2}$

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

######  Theorem 3.4.

[33] Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be analytic in $\mathbb{D}$ and
$|f(z)|\leq 1$. Then

(3.5)
$|a_{0}|+\sum_{n=1}^{\infty}\bigg{(}|a_{n}|+\frac{1}{2}|a_{n}|^{2}\bigg{)}r^{n}\leq
1\quad\mbox{for}\quad r\leq\frac{1}{3}$

and the numbers $1/3$ and $1/2$ cannot be improved.

The primary object of this section is to generalize the harmonic versions of
Theorem 3.1 and Theorem 3.4 for the class
$\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$. It is interesting to investigate
Theorem 3.1, when ${S_{r}}/{\pi}$ has certain power. Therefore in order to
generalize Theorem 3.1, we consider a ${N^{\rm th}}$ degree polynomial in
$S_{r}/\pi$ of the form

$P\left(\frac{S_{r}}{\pi}\right)=\left(\frac{S_{r}}{\pi}\right)^{N}+\left(\frac{S_{r}}{\pi}\right)^{N-1}+\cdots+\frac{S_{r}}{\pi}.$

######  Theorem 3.6.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then

(3.7)
$r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+P\left(\frac{S_{r}}{\pi}\right)\leq
d\left(f(0),\partial f(\mathbb{D})\right)$

for $r\leq r_{{}_{N}}(\alpha)$, where $P(w)=w^{N}+w^{N-1}+\cdots+w$, a
polynomial in $w$ of degree $N-1$, and $r_{{}_{N}}(\alpha)\in(0,1)$ is the
smallest root of the equation

(3.8)
$r-1-2(1-\alpha)\left(r-1+\ln\,(2-2r)\right)+P\left(r^{2}-4(1-\alpha)^{2}(r^{2}+\ln\;(1-r^{2}))\right)=0.$

Th constant $r_{{}_{N}}(\alpha)$ is the best possible.

$\alpha$ | $0.1$ | $0.2$ | $0.3$ | $0.4$ | $0.5$ | $0.6$ | $0.7$ | $0.8$ | $0.9$
---|---|---|---|---|---|---|---|---|---
$r_{2}(\alpha)$ | $0.2734$ | $0.3027$ | $0.3320$ | $0.3618$ | $0.3923$ | $0.4241$ | $0.4574$ | $0.4927$ | $0.5303$
$r_{3}(\alpha)$ | $0.2732$ | $0.3023$ | $0.3314$ | $0.3607$ | $0.3907$ | $0.4217$ | $0.4540$ | $0.4878$ | $0.5230$
$r_{4}(\alpha)$ | $0.2732$ | $0.3023$ | $0.3313$ | $0.3606$ | $0.3905$ | $0.4213$ | $0.4533$ | $0.4867$ | $0.5212$
$r_{5}(\alpha)$ | $0.2732$ | $0.3023$ | $0.3313$ | $0.3606$ | $0.3904$ | $0.4213$ | $0.4532$ | $0.4864$ | $0.5208$
Table 6. The roots $r_{{}_{N}}(\alpha)$ of equation (3.8) when $N=2,3,4,5$ and
$\alpha\in[0,1)$.

As a consequence of Theorem 3.6, we obtain the following interesting
corollary.

######  Corollary 3.9.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then

(3.10)
$r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+\left(\frac{S_{r}}{\pi}\right)\leq
d\left(f(0),\partial f(\mathbb{D})\right)$

for $r\leq r_{\alpha}$, where $r_{\alpha}\in(0,1)$ is the smallest root of the
equation

(3.11)
$r^{2}+r-1-2(1-\alpha)(3-2\alpha)\left(r+\ln\,(1-r)\right)-2(1-\alpha)(\ln
2-1)=0.$

The radius $r_{\alpha}$ is the best possible.

Figure 8. The figure demonstrates the roots $r(\alpha)$ when $\alpha\in[0,1)$. $\alpha$ | $0.1$ | $0.2$ | $0.3$ | $0.4$ | $0.5$ | $0.6$ | $0.7$ | $0.8$ | $0.9$
---|---|---|---|---|---|---|---|---|---
$r(\alpha)$ | $0.2322$ | $0.2635$ | $0.2967$ | $0.3323$ | $0.3707$ | $0.4125$ | $0.4579$ | $0.5074$ | $0.5610$
Table 7. The roots $r(\alpha)$ of the equation (3.11) when $\alpha\in[0,1)$.

The polylogarithm function is defined by a power series in $z$, which is also
a Dirichlet series in $s$. That is

$Li_{s}(z)=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{s}}=z+\frac{z^{2}}{2^{s}}+\frac{z^{3}}{3^{s}}+\cdots,$

is valid for arbitrary complex order $s$ and for all complex arguments $z$
with $|z|<1.$ Therefore, the dilogarithm function is denoted by $Li_{2}(z)$,
is a particular case of the polylogarithm. The following theorem is the
generalization of the harmonic version of Theorem 3.4 by considering the right
hand side $d\left(f(0),\partial f(\mathbb{D})\right)$ instead of $1$.

######  Theorem 3.12.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then

(3.13)
$r+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|+(|a_{n}|+|b_{n}|)^{2}\right)r^{n}\leq
d\left(f(0),\partial f(\mathbb{D})\right),\;\text{for}\;\;r\leq r_{\alpha},$

where $r_{\alpha}\in(0,1)$ is the smallest root of the equation

(3.14)
$r-1-2(1-\alpha)\left(r-1+\ln\,(2-2r)\right)+4(1-\alpha^{2})(Li_{2}(r)-r)=0,$

where $Li_{2}(z)$ is a dilogarithm. The constant $r_{\alpha}$ is best
possible.

##  4\. Refined Bohr Radius for the class
$\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$

In $2020$, Ponnusamy et al. [38] established the following refined Bohr
inequality by applying a refined version of the coefficient inequalities.

######  Theorem 4.1.

[38] Let $f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}$ be analytic in $\mathbb{D}$ and
$|f(z)|\leq 1$. Then

$\sum_{n=0}^{\infty}|a_{n}|r^{n}+\left(\frac{1}{1+|a_{0}|}+\frac{r}{1-r}\right)\sum_{n=1}^{\infty}|a_{n}|^{2}r^{2n}\leq
1$

for $r\leq 1/(2+|a_{0}|)$ and the numbers $1/(1+|a_{0}|)$ and $1/(2+|a_{0}|)$
cannot be improved. Moreover,

$|a_{0}|^{2}+\sum_{n=1}^{\infty}|a_{n}|r^{n}+\left(\frac{1}{1+|a_{0}|}+\frac{r}{1-r}\right)\sum_{n=1}^{\infty}|a_{n}|^{2}r^{2n}\leq
1$

for $r\leq 1/2$ and the numbers $1/(1+|a_{0}|)$ and $1/2$ cannot be improved.

Next we prove the harmonic analogue of Theorem 4.1.

######  Theorem 4.2.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then

(4.3) $\displaystyle
r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+\frac{1}{1-r^{N}}\sum_{n=2}^{\infty}n(|a_{n}|+|b_{n}|)^{2}r^{2n}$
$\displaystyle\leq d\left(f(0),\partial
f(\mathbb{D})\right)\;\text{for}\;\;r\leq r_{{}_{N}}(\alpha),$

where $r_{{}_{N}}(\alpha)\in(0,1)$ is the smallest root of the equation

(4.4)
$r-1-2(1-\alpha)\left(r-1+\ln\,(2-2r)+\frac{2(1-\alpha)}{1-r^{N}}(r^{2}+\ln\,(1-r^{2}))\right)=0.$

Here $r_{{}_{N}}(\alpha)$ is the best possible.

Figure 9. The roots $r_{2}(\alpha)$ and $r_{25}(\alpha)$ when $\alpha\in[0,1)$ have been shown in Figure 9. $\alpha$ | $0.1$ | $0.2$ | $0.3$ | $0.4$ | $0.5$ | $0.6$ | $0.7$ | $0.8$ | $0.9$
---|---|---|---|---|---|---|---|---|---
$r_{2}(\alpha)$ | $0.3148$ | $0.3527$ | $0.3920$ | $0.4338$ | $0.4793$ | $0.5304$ | $0.5904$ | $0.6651$ | $0.7693$
$r_{25}(\alpha)$ | $0.3158$ | $0.3542$ | $0.3942$ | $0.4368$ | $0.4835$ | $0.5361$ | $0.5977$ | $0.6741$ | $0.7792$
Table 8. In this table, we obtained the roots of $r_{{}_{N}}(\alpha)$ for
$N=2$ and $25$ when $\alpha\in[0,1)$.

In the following, we prove two interesting results which are harmonic analogue
of refined Bohr inequality.

######  Theorem 4.5.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then

(4.6) $\displaystyle
r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+\left(\frac{1}{1+|a_{2}|+|b_{2}|}+\frac{r^{m}}{1-r^{m}}\right)\sum_{n=3}^{\infty}n^{m-1}(|a_{n}|+|b_{n}|)^{m}r^{mn}$
$\displaystyle\leq d\left(f(0),\partial
f(\mathbb{D})\right),\;\text{for}\;\;r\leq r_{{m}}(\alpha),$

where $r_{{m}}(\alpha)\in(0,1)$ is the smallest root of the equation

(4.7) $\displaystyle r-1-2(1-\alpha)(r-1+\ln\,(2-2r))$
$\displaystyle-2^{m}(1-\alpha)^{m}\left(\frac{1}{1+|a_{2}|+|b_{2}|}+\frac{r^{m}}{1-r^{m}}\right)\left(r^{m}+\frac{r^{2m}}{2}+\ln\,(1-r^{m})\right)=0.$

The constant $r_{{m}}(\alpha)$ is the best possible.

######  Theorem 4.8.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then

(4.9) $\displaystyle
r+\frac{\left(1-(1+|a_{2}|+|b_{2}|-(|a_{2}|+|b_{2}|)^{2})\right)r}{1-(|a_{2}|+|b_{2}|)r}+\sum_{n=3}^{\infty}(|a_{n}|+|b_{n}|)r^{n}$
$\displaystyle\leq d\left(f(0),\partial
f(\mathbb{D})\right)\;\text{for}\;\;r\leq r_{\alpha},$

where $r_{\alpha}\in(0,1)$ is the smallest root of the equation

(4.10)
$r-\frac{\left(1-(|a_{2}|+|b_{2}|)^{2}\right)r}{1-(|a_{2}|+|b_{2}|)r}-2(1-\alpha)\left(r+\frac{r^{2}}{2}-1+\ln\,(2-2r)\right)=0.$

The constant $r_{\alpha}$ is the best possible.

##  5\. Bohr-Type Inequality for the class
$\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$

We now prove the following Bohr-type inequality for the class of functions
$\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$.

######  Theorem 5.1.

Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given by (1.11). Then

(5.2)
$\displaystyle|f(z)|+\sqrt{|{J}_{f}(z)|}r+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)r^{n}\leq
d\left(f(0),\partial f(\mathbb{D})\right)\;\text{for}\;\;r\leq
r_{{}_{N}}(\alpha),$

where $r_{{}_{N}}(\alpha)\in(0,1)$ is the smallest root of the equation

(5.3) $\displaystyle
r-1-2(1-\alpha)\left(2r-1+\frac{r^{2}}{2}+\cdots+\frac{r^{N-1}}{N-1}+\ln
2+2\ln(1-r)\right)$
$\displaystyle\quad\quad+\left(\alpha+(1-\alpha)\left(\frac{1+r}{1-r}\right)\right)r=0.$

The radius $r_{{}_{N}}(\alpha)$ is the best possible.

##  6\. Proof of the main results

Proof of Theorem 2.1. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given
by (1.11). Then from Theorem 2.1, we have

(6.1)
$|f(z)|\geq|z|+(1-\alpha)\sum\limits_{n=2}^{\infty}\dfrac{2(-1)^{n-1}}{n}|z|^{n}\quad\mbox{for
}\quad|z|<1.$

The Euclidean distance between $f(0)$ and the boundary of $f(\mathbb{D})$ is
given by

(6.2) $d(f(0),\partial f(\mathbb{D}))=\liminf\limits_{|z|\rightarrow
1}|f(z)-f(0)|.$

Since $f(0)=0$, from (1.14) and (6.2) we obtain

(6.3) $d(f(0),\partial f(\mathbb{D}))\geq
1+\sum\limits_{n=2}^{\infty}2(1-\alpha)\dfrac{(-1)^{n-1}}{n}.$

Using Lemmas 1.12 and 1.13, for $|z|=r_{{}_{N}}(\alpha)$, we obtain

(6.4) $\displaystyle|f(z)|+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)r^{n}$
$\displaystyle\leq
r+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}+\sum_{n=N}^{\infty}\frac{2(1-\alpha)r^{n}}{n}$
$\displaystyle=r-2(1-\alpha)(r+\ln(1-r))-2(1-\alpha)\left(\ln(1-r)+\sum_{n=1}^{N-1}\frac{r^{n}}{n}\right)$
$\displaystyle=r-2(1-\alpha)\left(r+\ln((1-r)^{2})+\sum_{n=1}^{N-1}\frac{r^{n}}{n}\right).$

It is easy to see that

(6.5)
$r-2(1-\alpha)\left(r+\ln((1-r)^{2})+\sum_{n=1}^{N-1}\frac{r^{n}}{n}\right)\leq
1+2(1-\alpha)(\ln 2-1)$

for $r\leq r_{{}_{N}}(\alpha)$, where $r_{{}_{N}}(\alpha)$ is the smallest
root of

$r-1-2(1-\alpha)\left(r-1+\ln(2(1-r)^{2})+\sum_{n=1}^{N-1}\frac{r^{n}}{n}\right)=0$

in $(0,1)$. Let $H_{1}:[0,1)\rightarrow\mathbb{R}$ be defined by

$H_{1}(r):=r-1-2(1-\alpha)\left(r-1+\ln(2(1-r)^{2})+\sum_{n=1}^{N-1}\frac{r^{n}}{n}\right).$

The existence of a root $r_{N}(\alpha)$ is ensured by the following fact that
$H_{1}$ is a continuous function with the properties
$H_{1}(0)=-1-2(1-\alpha)(\ln 2-1)<0$ and $\displaystyle\lim_{r\rightarrow
1}H_{1}(r)=+\infty.$ Let $r_{N}(\alpha)$ to be the smallest root of
$H_{1}(r)=0$ in $(0,1)$. Therefore, we have $H_{1}(r_{N}(\alpha))=0$. That is

(6.6) $\displaystyle
r_{N}(\alpha)-1-2(1-\alpha)\left(r_{N}(\alpha)-1+\ln(2(1-r_{N}(\alpha))^{2})+\sum_{n=1}^{N-1}\frac{r^{n}_{N}(\alpha)}{n}\right)=0.$

In view of (6.3), (6.4) and (6.5) for $|z|=r\leq r_{N}(\alpha)$, it follows
that

$|f(z)|+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)r^{n}\leq d(f(0),\partial
f(\mathbb{D})).$

In order to show that the constant $r_{{}_{N}}(\alpha)$ is the best possible
constant, we consider the following function $f=f_{\alpha}$ by

(6.7) $f_{\alpha}(z)=z+\sum_{n=2}^{\infty}\frac{2(1-\alpha)z^{n}}{n}.$

It is easy to show that $f_{\alpha}\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$.
For $f=f_{\alpha}$, it can be seen that

(6.8) $d(f(0),\partial f(\mathbb{D}))=1+2(1-\alpha)(\ln 2-1).$

For $f=f_{\alpha}$ and $|z|=r_{N}(\alpha)$, a simple computation using (6.6)
and (6.8) shows that

$\displaystyle|f(z)|+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)r^{n}$
$\displaystyle=r_{{}_{N}}(\alpha)+\sum_{n=2}^{\infty}\frac{2(1-\alpha)(r_{{}_{N}}(\alpha))^{n}}{n}+\sum_{n=N}^{\infty}\frac{2(1-\alpha)(r_{{}_{N}}(\alpha))^{n}}{n}$
$\displaystyle=r_{{}_{N}}(\alpha)-2(1-\alpha)\left(r_{{}_{N}}(\alpha)+\ln((1-r_{{}_{N}}(\alpha))^{2})+\sum_{n=1}^{N-1}\frac{r^{n}_{{}_{N}}(\alpha)}{n}\right)$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)$ $\displaystyle=d(f(0),\partial
f(\mathbb{D})).$

Therefore, the radius $r_{{}_{N}}(\alpha)$ is the best possible. This
completes the proof. ∎

Proof of Theorem 2.4. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given
by (1.11). Then in view of Lemmas 1.12 and 1.13, for $|z|=r$, we obtain

(6.9) $\displaystyle|f(z)|^{2}+\sum_{n=N}^{\infty}|a_{n}||z|^{n}$
$\displaystyle\leq\left(r+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}\right)^{2}+\sum_{n=N}^{\infty}\frac{2(1-\alpha)r^{n}}{n}.$

A simple computation shows that

(6.10)
$\left(r+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}\right)^{2}+\sum_{n=N}^{\infty}\frac{2(1-\alpha)r^{n}}{n}\leq
1+2(1-\alpha)(\ln 2-1)$

for $r\leq r_{N}(\alpha)$, where $r_{N}(\alpha)$ is the smallest root of
$H_{2}(r)=0$ in $(0,1)$, where $H_{2}:[0,1)\rightarrow\mathbb{R}$ is defined
by

$H_{2}(r)=\bigg{(}r-2(1-\alpha)(r+\ln(1-r))\bigg{)}^{2}-2(1-\alpha)\left(\ln(2-2r)-1+\sum_{n=1}^{N-1}\frac{r^{n}}{n}\right)-1.$

Then $H_{2}$ is a continuous function with $H_{2}(0)=-1-2(1-\alpha)(\ln
2-1)<0$ and $\displaystyle\lim_{r\rightarrow 1}H_{2}(r)=+\infty.$ Therefore,
$H_{2}(r)=0$ has a root in $(0,1)$ and we choose the smallest root to be
$r_{{}_{N}}(\alpha)$. Therefore, we have $H_{2}(r_{{}_{N}}(\alpha))=0$. That
is

(6.11)
$\displaystyle\bigg{(}r_{N}(\alpha)-2(1-\alpha)(r_{N}(\alpha)+\ln(1-r_{N}(\alpha)))\bigg{)}^{2}$
$\displaystyle-2(1-\alpha)\left(\ln(2-2r_{N}(\alpha))-1+\sum_{n=1}^{N-1}\frac{r^{n}_{N}(\alpha)}{n}\right)-1=0.$

Using (6.3), (6.9) and (6.10) for $|z|=r\leq r_{{}_{N}}(\alpha)$, we obtain

$|f(z)|^{2}+\sum_{n=N}^{\infty}|a_{n}|r^{n}\leq d(f(0),\partial
f(\mathbb{D})).$

In order to show that $r_{{}_{N}}(\alpha)$ is the best possible, we consider
the function $f=f_{\alpha}$ defined by (6.7). For $f=f_{\alpha}$ and
$|z|=r_{{}_{N}}(\alpha)$, a simple computation using (6.8) and (6.11) shows
that

$\displaystyle|f(z)|^{2}+\sum_{n=N}^{\infty}|a_{n}|r^{n}$
$\displaystyle=\left(r_{{}_{N}}(\alpha)+\sum_{n=2}^{\infty}\frac{2(1-\alpha)(r_{{}_{N}}(\alpha))^{n}}{n}\right)^{2}+\sum_{n=N}^{\infty}\frac{2(1-\alpha)(r_{{}_{N}}(\alpha))^{n}}{n}$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)$ $\displaystyle=d(f(0),\partial
f(\mathbb{D})).$

Therefore, the radius $r_{{}_{N}}(\alpha)$ is the best possible. ∎

Proof of Theorem 2.7. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given
by (1.11). Using Lemmas 1.12 and 1.13 for $|z|=r$, we obtain

(6.12) $\displaystyle|f(z^{m})|+\sum_{n=N}^{\infty}|a_{n}||z|^{n}$
$\displaystyle\leq
r^{m}+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{mn}}{n}+\sum_{n=N}^{\infty}\frac{2(1-\alpha)r^{n}}{n}.$

A simple computation shows that

(6.13) $\displaystyle
r^{m}+\sum_{n=2}^{\infty}\frac{2(1-\alpha)(r^{m})^{n}}{n}+\sum_{n=N}^{\infty}\frac{2(1-\alpha)r^{n}}{n}$
$\displaystyle=r^{m}-2(1-\alpha)(r^{m}+\ln(1-r^{m}))-2(1-\alpha)\left(r+\ln(1-r)+\sum_{n=2}^{N-1}\frac{r^{n}}{n}\right)$
$\displaystyle=r^{m}-2(1-\alpha)\left(r^{m}+r+\ln(1-r)(1-r^{m})+\sum_{n=2}^{N-1}\frac{r^{n}}{n}\right)$
$\displaystyle\leq 1+2(1-\alpha)(\ln 2-1)$

for $r\leq r{{}_{m,N}}(\alpha)$, where $r_{{}_{m,N}}(\alpha)$ is the smallest
root of $H_{3}(r)=0$ in $(0,1)$, where $H_{3}:[0,1)\rightarrow\mathbb{R}$ is
defined by

$H_{3}(r):=r^{m}-1-2(1-\alpha)\left(r^{m}+r-1+\ln(2-2r)(1-r^{m})+\sum_{n=2}^{N-1}\frac{r^{n}}{n}\right).$

In view of the same line of argument as in the proof of Theorem 2.1, we can
show that $H_{3}(r)=0$ has a root in $(0,1)$ and we choose
$r_{{}_{m,N}}(\alpha)$ to be the smallest root of $H_{3}(r)$. Therefore, we
have $H_{3}(r_{{}_{m,N}}(\alpha))=0$. That is

(6.14) $\displaystyle r_{{}_{m,N}}(\alpha)-1-2(1-\alpha)G_{m,N}(r)=0,$

where

$G_{m,N}=\left(r_{{}_{m,N}}(\alpha)+r_{{}_{N}}(\alpha)-1+\ln\left((2-2r_{N}(\alpha))(1-r_{{}_{m,N}}(\alpha))\right)+\sum_{n=2}^{N-1}\frac{r^{n}_{{}_{m,N}}(\alpha)}{n}\right).$

In view of (6.3), (6.12) and (6.13) for $|z|=r\leq r_{{}_{m,N}}(\alpha)$, we
obtain

$|f(z^{m})|+\sum_{n=N}^{\infty}|a_{n}|r^{n}\leq d(f(0),\partial
f(\mathbb{D})).$

To show that the radius $r_{{}_{m,N}}(\alpha)$ is the best possible, we
consider the function $f=f(\alpha)$ defined by (6.7). For $f=f_{\alpha}$ and
$|z|=r_{{}_{m,N}}(\alpha)$, a simple calculation using (6.8) and (6.14) shows
that

$\displaystyle|f(z^{m})|+\sum_{n=N}^{\infty}|a_{n}|r^{n}$
$\displaystyle=r^{m}_{{}_{m,N}}(\alpha)-2(1-\alpha)\left(r^{m}_{{}_{m,N}}(\alpha)+r+\ln\bigg{(}(1-r_{{}_{m,N}}(\alpha))(1-r^{m}_{{}_{m,N}}(\alpha))\bigg{)}+\sum_{n=2}^{N-1}\frac{r^{n}_{{}_{m,N}}(\alpha)}{n}\right)$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)=d(f(0),\partial f(\mathbb{D})).$

Hence, the radius $r_{{}_{m,N}}(\alpha)$ is the best possible. This completes
the proof. ∎

Proof of Theorem 2.10. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be
given by (1.11). Applying Lemmas 1.12 and 1.13 for $|z|=r$, we obtain

(6.15) $\displaystyle r+|h(r)|^{p}+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}$
$\displaystyle\leq
r+|h(r)|^{p}+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}.$

It is not difficult to show that

(6.16) $\displaystyle
r+|h(r)|^{p}+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}$
$\displaystyle=r^{p}+r-2(1-\alpha)(r+\ln(1-r))$ $\displaystyle\leq
1+2(1-\alpha)(\ln 2-1)$

for $r\leq r_{p}(\alpha)$, where $r_{p}(\alpha)$ is the smallest root of
$H_{4}(r)=0$ in $(0,1)$ and $H_{4}:[0,1)\rightarrow\mathbb{R}$ is defined by

$H_{4}(r):=r^{p}+r-1-2(1-\alpha)(r-1+\ln(2-2r)).$

By the same argument used in the proof of Theorem 2.1, we can show that
$H_{4}(r)$ has a root in $(0,1)$ and we choose $r_{p}(\alpha)$ to be the
smallest root of $H_{4}(r)$. Therefore, $H_{4}(r_{p}(\alpha))=0$. That is

(6.17) $\displaystyle
r^{p}_{p}(\alpha)+r_{p}(\alpha)-1-2(1-\alpha)(r_{p}(\alpha)-1+\ln(2-2r_{p}(\alpha)))=0,$

Using (6.3), (6.15) and (6.16) for $|z|=r\leq r_{p}(\alpha)$, we obtain

$r+|h(r)|^{p}+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}\leq d(f(0),\partial
f(\mathbb{D})).$

In order to show that $r_{p}(\alpha)$ is the best possible, we consider the
function $f=f_{\alpha}$ defined by (6.7). For $f=f_{\alpha}$ and
$|z|=r_{p}(\alpha)$, a simple calculation using (6.8) and (6.17) shows that

$\displaystyle
r_{p}(\alpha)+|h(r_{p}(\alpha))|^{p}+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}$
$\displaystyle=r^{p}_{p}(\alpha)+r_{p}(\alpha)-2(1-\alpha)(r_{p}(\alpha)+\ln(1-r_{p}(\alpha)))$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)$ $\displaystyle=d(f(0),\partial
f(\mathbb{D})).$

Therefore, the radius $r_{p}(\alpha)$ is the best possible. This completes the
proof. ∎

Proof of Theorem 3.6. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given
by (1.11). For the analytic functions $h$ and $g$, the area of the disk
$|z|<r$ under the harmonic map $f$ is $S_{r}$ is given by

(6.18) $\displaystyle S_{r}=\iint\limits_{D_{r}}$
$\displaystyle\left(|h^{\prime}(z)|^{2}-|g^{\prime}(z)|^{2}\right)dxdy,$
(6.19) $\displaystyle\frac{1}{\pi}\iint\limits_{D_{r}}$
$\displaystyle|h^{\prime}(z)|^{2}dxdy=\sum_{n=1}^{\infty}n|a_{n}|^{2}r^{2n},$
(6.20) $\displaystyle\frac{1}{\pi}\iint\limits_{D_{r}}$
$\displaystyle|g^{\prime}(z)|^{2}dxdy=\sum_{n=2}^{\infty}n|b_{n}|^{2}r^{2n}.$

Therefore, in view of (6.18), (6.19) and (6.20) and Lemma 1.12, we obtain

$\displaystyle\frac{S_{r}}{\pi}$
$\displaystyle=\frac{1}{\pi}\iint\limits_{D_{r}}\left(|h^{\prime}(z)|^{2}-|g^{\prime}(z)|^{2}\right)dxdy$
$\displaystyle=r^{2}+\sum_{n=2}^{\infty}|a_{n}|^{2}r^{2n}-\sum_{n=2}^{\infty}n|b_{n}|^{2}r^{2n}$
$\displaystyle=r^{2}+\sum_{n=2}^{\infty}n\left(|a_{n}|+|b_{n}|\right)\left(|a_{n}|-|b_{n}|\right)r^{2n}$
$\displaystyle\leq
r^{2}+\sum_{n=2}^{\infty}\frac{4(1-\alpha)^{2}r^{2n}}{n^{2}}$
$\displaystyle=r^{2}-4(1-\alpha)^{2}\left(r^{2}+\ln(1-r^{2})\right).$

Using Lemmas 1.12 and 1.13 for $|z|=r$, we obtain

(6.21) $\displaystyle
r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+P\left(\frac{S_{r}}{\pi}\right)$
$\displaystyle\leq
r+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}+P(r^{2}-4(1-\alpha)^{2}\left(r^{2}+\ln(1-r^{2})\right))$
$\displaystyle=r-2(1-\alpha)(r+\ln(1-r))+P(r^{2}-4(1-\alpha)^{2}\left(r^{2}+\ln(1-r^{2})\right))$
$\displaystyle\leq 1+2(1-\alpha)(\ln 2-1)$

for $r\leq r(\alpha)$, where $r(\alpha)$ is the smallest root of $H_{5}(r)=0$
in $(0,1)$ where $H_{5}:[0,1)\rightarrow\mathbb{R}$ be defined by

$H_{5}(r):=r-1-2(1-\alpha)(r-1+\ln(2-2r))+P(r^{2}-4(1-\alpha)^{2}\left(r^{2}+\ln(1-r^{2})\right)).$

Clearly, $H_{5}(r(\alpha))=0$. That is

(6.22) $\displaystyle
r(\alpha)-1-2(1-\alpha)(r(\alpha)-1+\ln(2-2r(\alpha)))+P(G(r,\alpha))=0,$

where

$G(r,\alpha)=r(\alpha)^{2}-4(1-\alpha)^{2}\left(r(\alpha)^{2}+\ln(1-r(\alpha)^{2})\right).$

From (6.3), (6.21) and (6.22), $|z|=r\leq r(\alpha)$, we obtain

$r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+P\left(\frac{S_{r}}{\pi}\right)\leq
d(f(0),\partial f(\mathbb{D})).$

To show that the radius $r(\alpha)$ is the best possible, we consider the
function defined by (6.7). For $f=f_{\alpha}$ and $|z|=r(\alpha)$, a simple
computation using (6.8) and (6.22) shows that

$\displaystyle
r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+P\left(\frac{S_{r}}{\pi}\right)$
$\displaystyle=r-2(1-\alpha)(r+\ln(1-r))+P(G_{(}r,\alpha))$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)=d(f(0),\partial f(\mathbb{D})).$

Thus, the radius $r(\alpha)$ is the best possible. This completes the proof. ∎

Proof of Theorem 3.12. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be
given by (1.11). Using Lemmas 1.12 and 1.13, for $|z|=r$, we obtain

(6.23)
$\displaystyle\;\;\;\;r+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|+(|a_{n}|+|b_{n}|)^{2}\right)r^{n}$
$\displaystyle\leq
r+\sum_{n=2}^{\infty}\left(\frac{2(1-\alpha)}{n}+\frac{4(1-\alpha)^{2}}{n^{2}}\right)r^{n}.$

An easy computation shows that

(6.24) $\displaystyle
r+\sum_{n=2}^{\infty}\left(\frac{2(1-\alpha)}{n}+\frac{4(1-\alpha)^{2}}{n^{2}}\right)r^{n}$
$\displaystyle=r-2(1-\alpha)(r+\ln(1-r))+4(1-\alpha)^{2}(Li_{2}(r)-r)$
$\displaystyle\leq 1+2(1-\alpha)(\ln 2-1)$

for $r\leq r(\alpha)$, where $r(\alpha)$ is the smallest root of $H_{6}(r)=0$
in $(0,1)$ and $H_{6}:[0,1)\rightarrow\mathbb{R}$ be defined by

$H_{6}(r):=r-2(1-\alpha)(r-1+\ln(2-2r))+4(1-\alpha)^{2}(Li_{2}(r)-r)-1.$

Thus, we have $H_{6}(r(\alpha))=0$. That is

(6.25) $\displaystyle
r(\alpha)-2(1-\alpha)(r(\alpha)-1+\ln(2-2r(\alpha)))+4(1-\alpha)^{2}(Li_{2}(r(\alpha))-r(\alpha))-1=0.$

Using (6.3), (6.23) and (6.24) for $|z|=r\leq r(\alpha)$, we obtain

$r+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|+(|a_{n}|+|b_{n}|)^{2}\right)r^{n}\leq
d(f(0),\partial f(\mathbb{D})).$

In order to show that $r(\alpha)$ is the best possible, we consider the
function defined by (6.7). For $f=f_{\alpha}$ and $|z|=r(\alpha)$, a simple
computation using (6.8) and (6.25) shows that

$\displaystyle
r(\alpha)+\sum_{n=2}^{\infty}\left(|a_{n}|+|b_{n}|+(|a_{n}|+|b_{n}|)^{2}\right)r^{n}(\alpha)$
$\displaystyle=r(\alpha)-2(1-\alpha)(r(\alpha)+\ln(1-r(\alpha))+4(1-\alpha)^{2}(Li_{2}(r(\alpha))-r(\alpha))$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)$ $\displaystyle=d(f(0),\partial
f(\mathbb{D})).$

Therefore, the radius $r(\alpha)$ is the best possible. This completes the
proof. ∎

Proof of Theorem 4.2. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given
by (1.11). In view of Lemmas 1.12 and 1.13 for $|z|=r$, we obtain

(6.26) $\displaystyle
r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+\frac{1}{1-r^{N}}\sum_{n=2}^{\infty}n(|a_{n}|+|b_{n}|)^{2}r^{2n}$
$\displaystyle\leq
r+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}+\frac{1}{1-r^{N}}\sum_{n=2}^{\infty}\frac{4(1-\alpha)^{2}r^{2n}}{n}.$

A simple computation shows that

(6.27) $\displaystyle
r+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}+\frac{1}{1-r^{N}}\sum_{n=2}^{\infty}\frac{4(1-\alpha)^{2}r^{2n}}{n}$
$\displaystyle=r-2(1-\alpha)(r+\ln(1-r))-\frac{4(1-\alpha)^{2}}{1-r^{N}}(r^{2}+\ln(1-r^{2}))$
$\displaystyle\leq 1+2(1-\alpha)(\ln 2-1)$

for $r\leq r(\alpha)$, where $r(\alpha)$ is the smallest root of $H_{7}(r)=0$
in $(0,1)$ and $H_{7}:[0,1)\rightarrow\mathbb{R}$ be defined by

$H_{7}(r):=r-2(1-\alpha)(r-1+\ln(2-2r))-\frac{4(1-\alpha)^{2}}{1-r^{N}}(r^{2}+\ln(1-r^{2}))-1.$

Thus, we have $H_{7}(r(\alpha))=0$. That is

(6.28) $\displaystyle r(\alpha)-2(1-\alpha)(r-1+\ln(2-2(r(\alpha)))$
$\displaystyle-\frac{4(1-\alpha)^{2}}{1-r^{N}(\alpha))}\bigg{(}r^{2}(\alpha)+\ln(1-r^{2}(\alpha))\bigg{)}-1=0.$

Using (6.3), (6.26) and (6.27) for $|z|=r\leq r(\alpha)$, we obtain

$r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+\frac{1}{1-r^{N}}\sum_{n=2}^{\infty}n(|a_{n}|+|b_{n}|)^{2}r^{2n}\leq
d(f(0),\partial f(\mathbb{D})).$

In order to show that $r(\alpha)$ is the best possible, we consider the
function $f=f_{\alpha}$ defined by (6.7). For $f=f_{\alpha}$ and
$|z|=r(\alpha)$, a simple calculation using (6.8) and (6.28) shows that

$\displaystyle
r(\alpha)+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}(\alpha)+\frac{1}{1-r^{N}(\alpha)}\sum_{n=2}^{\infty}n(|a_{n}|+|b_{n}|)^{2}r^{2n}(\alpha)$
$\displaystyle=r(\alpha)-2(1-\alpha)(r(\alpha)+\ln(1-r(\alpha)))-\frac{4(1-\alpha)^{2}}{1-r^{N}(\alpha)}(r^{2}(\alpha)+\ln(1-r^{2}(\alpha)))$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)$ $\displaystyle=d(f(0),\partial
f(\mathbb{D})).$

Hence the radius $r(\alpha)$ is the best possible. This completes the proof. ∎

Proof of Theorem 4.5. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given
by (1.11). Using Lemmas 1.12 and 1.13 for $|z|=r$, we obtain

(6.29) $\displaystyle
r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+\left(\frac{1}{1+|a_{2}|+|b_{2}|}+\frac{r^{m}}{1-r^{m}}\right)\sum_{n=3}^{\infty}n^{m-1}(|a_{n}|+|b_{n}|)^{m}r^{mn}$
$\displaystyle\leq
r+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}+\left(\frac{1}{1+|a_{2}|+|b_{2}|}+\frac{r^{m}}{1-r^{m}}\right)\sum_{n=3}^{\infty}\frac{2^{m}(1-\alpha)^{m}(r^{m})^{n}}{n}.$

Let

$\displaystyle Q_{m}(r):=\frac{1}{1+|a_{2}|+|b_{2}|}+\frac{r^{m}}{1-r^{m}}.$

A computation using $Q_{m}(r)$ shows that

(6.30) $\displaystyle
r+\sum_{n=2}^{\infty}\frac{2(1-\alpha)r^{n}}{n}+Q_{m}(r)\sum_{n=3}^{\infty}\frac{2^{m}(1-\alpha)^{m}(r^{m})^{n}}{n}$
$\displaystyle=r-2(1-\alpha)(r+\ln(1-r))-2^{m}(1-\alpha)^{m}Q_{m}(r)\left(r^{m}+\frac{r^{2m}}{2}+\ln(1-r^{m})\right)$
$\displaystyle\leq 1+2(1-\alpha)(\ln 2-1)$

for $r\leq r_{m}(\alpha)$, where $r_{m}(\alpha)$ is the smallest root of of
$H_{8}(r)$ in $(0,1)$ and

$\displaystyle H_{8}(r):=$ $\displaystyle
r-1-2(1-\alpha)\left(r-1+\ln(2-2r)\right)-2^{m}(1-\alpha)^{m}Q_{m}(r)\left(r^{m}+\frac{r^{2m}}{2}+\ln(1-r^{m})\right).$

Then we have $H_{8}(r_{m}(\alpha))=0$. That is

(6.31) $\displaystyle
r_{m}(\alpha)-1-2(1-\alpha)\left(r_{m}(\alpha)-1+\ln(2-2r_{m}(\alpha))\right)$
$\displaystyle-2^{m}(1-\alpha)^{m}Q_{m}(r)\left(r^{m}_{m}(\alpha)+\frac{r^{2m}_{m}(\alpha)}{2}+\ln(1-r^{m}_{m}(\alpha))\right)=0.$

With the help of (6.3), (6.29) and (6.30) for $|z|=r\leq r_{m}(\alpha)$, we
obtain

$\displaystyle
r+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}+\left(\frac{1}{1+|a_{2}|+|b_{2}|}+\frac{r^{m}}{1-r^{m}}\right)\sum_{n=3}^{\infty}n^{m-1}(|a_{n}|+|b_{n}|)^{m}r^{mn}$
$\displaystyle\leq d(f(0),\partial f(\mathbb{D})).$

To show that $r_{m}(\alpha)$ is the best possible, we consider the function
$f=f_{\alpha}$ defined by (6.7). For the function $f=f_{\alpha}$ and
$|z|=r_{m}(\alpha)$, an simple computation using (6.8) and (6.31) shows that

$\displaystyle
r_{m}(\alpha)+\sum_{n=2}^{\infty}(|a_{n}|+|b_{n}|)r^{n}_{m}(\alpha)+\left(\frac{1}{1+|a_{2}|+|b_{2}|}+\frac{r^{m}}{1-r^{m}_{m}(\alpha)}\right)$
$\displaystyle\quad\quad\times\sum_{n=3}^{\infty}n^{m-1}(|a_{n}|+|b_{n}|)^{m}r^{mn}_{m}(\alpha)$
$\displaystyle=r_{m}(\alpha)-2(1-\alpha)(r_{m}(\alpha)+\ln(1-r_{m}(\alpha)))$
$\displaystyle\quad\quad-2^{m}(1-\alpha)^{m}Q_{m}(r_{m}(\alpha))\left(r^{m}_{m}(\alpha)+\frac{r^{2m}_{m}(\alpha)}{2}+\ln(1-r^{m}_{m}(\alpha))\right)$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)$ $\displaystyle=d(f(0),\partial
f(\mathbb{D})).$

Therefore, the radius $r_{m}(\alpha)$ is the best possible. This completes the
proof. ∎

Proof of Theorem 4.8. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given
by (1.11). By using Lemmas 1.12 and 1.13, for $|z|=r$, we obtain

(6.32) $\displaystyle
r+\frac{[1-(1+|a_{2}|+|b_{2}|-(|a_{2}|+|b_{2}|)^{2})]r}{1-(|a_{2}|+|b_{2}|)r}+\sum_{n=3}^{\infty}(|a_{n}|+|b_{n}|)r^{n}$
$\displaystyle\leq
r+\frac{[1-(1+|a_{2}|+|b_{2}|-(|a_{2}|+|b_{2}|)^{2})]r}{1-(|a_{2}|+|b_{2}|)r}+\sum_{n=3}^{\infty}\frac{2(1-\alpha)r^{n}}{n}.$

Let ${R}(r)$ be defined

${R}(r)=\frac{[1-(1+|a_{2}|+|b_{2}|-(|a_{2}|+|b_{2}|)^{2})]r}{1-(|a_{2}|+|b_{2}|)r}.$

A simple computation shows that

(6.33)
$\displaystyle\;\;\;r+{R}(r)+\sum_{n=3}^{\infty}\frac{2(1-\alpha)r^{n}}{n}$
$\displaystyle=r+{R}(r)-2(1-\alpha)\left(r+\frac{r^{2}}{2}+\ln(1-r)\right)$
$\displaystyle\leq 1+2(1-\alpha)(\ln 2-1)$

for $r\leq r(\alpha)$, where $r(\alpha)$ is the smallest root of $H_{9}(r)$ in
$(0,1)$ and $H_{9}:[0,1)\rightarrow\mathbb{R}$ be defined by

$H_{9}(r):=r+{R}(r)-2(1-\alpha)\left(r+\frac{r^{2}}{2}-1+\ln(2-2r)\right)-1.$

Then we have $H_{9}(r(\alpha))=0$. That is

(6.34) $\displaystyle
r(\alpha)+{R}(r(\alpha))-2(1-\alpha)\left(r(\alpha)+\frac{r^{2}(\alpha)}{2}-1+\ln(2-2r(\alpha))\right)-1=0.$

From (6.3), (6.32) and (6.33) for $|z|=r\leq r(\alpha)$, we obtain

$r+\frac{[1-(1+|a_{2}|+|b_{2}|-(|a_{2}|+|b_{2}|)^{2})]r}{1-(|a_{2}|+|b_{2}|)r}+\sum_{n=3}^{\infty}(|a_{n}|+|b_{n}|)r^{n}\leq
d(f(0),\partial f(\mathbb{D})).$

To show that the radius $r(\alpha)$ is the best possible, we consider the
function $f=f_{\alpha}$ defined in (6.7). For $f=f_{\alpha}$ and
$|z|=r(\alpha)$, a simple calculation using (6.8) and (6.34) shows that

$\displaystyle
r(\alpha)+\frac{\left(1-(1+|a_{2}|+|b_{2}|-(|a_{2}|+|b_{2}|)^{2})\right)r(\alpha)}{1-(|a_{2}|+|b_{2}|)r(\alpha)}+\sum_{n=3}^{\infty}(|a_{n}|+|b_{n}|)r^{n}(\alpha)$
$\displaystyle=r(\alpha)+{R}(r(\alpha))-2(1-\alpha)\left(r(\alpha)+\frac{r^{2}(\alpha)}{2}+\ln(1-r(\alpha))\right)$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)$ $\displaystyle=d(f(0),\partial
f(\mathbb{D})).$

Hence the radius $r(\alpha)$ is the best possible. This comples the proof. ∎

Proof of Theorem 5.1. Let $f\in\mathcal{P}^{0}_{\mathcal{H}}(\alpha)$ be given
by (1.11). It is not difficult to show that

$\displaystyle|h^{\prime}(z)|\leq\alpha+(1-\alpha)\left(\frac{1+r}{1-r}\right).$

Therefore, we have

$\displaystyle|{J}_{f}(z)|=|h^{\prime}(z)|^{2}-|g^{\prime}(z)|^{2}\leq|h^{\prime}(z)|^{2}.$

Using Lemmas 1.12 and 1.13, for $|z|=r_{N}(\alpha)$, we obtain

(6.35)
$\displaystyle|f(z)|+\sqrt{|{J}_{f}(z)|}r+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)r^{n}$
$\displaystyle\leq
r+2(1-\alpha)(-r-\ln(1-r))+\left(\alpha+(1-\alpha)\left(\frac{1+r}{1-r}\right)\right)r$
$\displaystyle\quad\quad+2(1-\alpha)\left(-r-\frac{r^{2}}{2}-\cdots-\frac{r^{N-1}}{N-1}-\ln(1-r)\right)$
$\displaystyle=r-2(1-\alpha)\left(2r+\frac{r^{2}}{2}+\cdots+\frac{r^{N-1}}{N-1}+2\ln(1-r)\right)+\left(\alpha+(1-\alpha)\left(\frac{1+r}{1-r}\right)\right)r.$

A simple calculations shows that

(6.36) $\displaystyle
r-2(1-\alpha)\left(2r+\frac{r^{2}}{2}+\cdots+\frac{r^{N-1}}{N-1}+2\ln(1-r)\right)+\left(\alpha+(1-\alpha)\left(\frac{1+r}{1-r}\right)\right)r$
$\displaystyle\leq 1+2(1-\alpha)(\ln 2-1)$

for $r\leq r(\alpha)$, where $r(\alpha)$ is the smallest root of $H_{10}(r)$
in $(0,1)$. Here $H_{10}:[0,1)\rightarrow\mathbb{R}$ defined by

$\displaystyle H_{10}(r):$
$\displaystyle=r-1-2(1-\alpha)\left(2r-1+\frac{r^{2}}{2}+\cdots+\frac{r^{N-1}}{N-1}+\ln
2+2\ln(1-r)\right)$
$\displaystyle\quad\quad+\left(\alpha+(1-\alpha)\left(\frac{1+r}{1-r}\right)\right)r.$

Therefore, we have $H_{10}(r_{{}_{N}}(\alpha))=0$, which shows that

(6.37) $\displaystyle
r_{N}(\alpha)-1-2(1-\alpha)\left(2r_{N}(\alpha)-1+\frac{r^{2}_{N}(\alpha)}{2}+\cdots+\frac{r^{N-1}_{N}(\alpha)}{N-1}+\ln
2+2\ln(1-r_{N}(\alpha))\right)$
$\displaystyle\quad\quad+\left(\alpha+(1-\alpha)\left(\frac{1+r_{N}(\alpha)}{1-r_{N}(\alpha)}\right)\right)^{2}=0,$

Using (6.3), (6.35) and (6.36) for $|z|=r\leq r_{{}_{N}}(\alpha)$, we obtain

$|f(z)|+|{J}_{f}(z)|+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)r^{n}\leq
d(f(0),\partial f(\mathbb{D})).$

In order to show that $r_{{}_{N}}(\alpha)$ is sharp, we consider the function
$f=f_{\alpha}$ defined by (6.7). For $f=f_{\alpha}$ and
$|z|=r_{{}_{N}}(\alpha)$, a simple calculation using (6.8) and (6.37) shows
that

$\displaystyle|f(z)|+\sqrt{|{J}_{f}(z)|}r+\sum_{n=N}^{\infty}(|a_{n}|+|b_{n}|)r^{n}_{{}_{N}}(\alpha)$
$\displaystyle=r_{{}_{N}}(\alpha)-2(1-\alpha)\left(2r_{{}_{N}}(\alpha)+\frac{r^{2}_{{}_{N}}(\alpha)}{2}+\cdots+\frac{r^{N-1}_{{}_{N}}(\alpha)}{N-1}+2\ln(1-r_{{}_{N}}(\alpha))\right)$
$\displaystyle\quad\quad+\left(\alpha+(1-\alpha)\left(\frac{1+r_{{}_{N}}(\alpha)}{1-r_{{}_{N}}(\alpha)}\right)\right)r$
$\displaystyle=1+2(1-\alpha)(\ln 2-1)$ $\displaystyle=d(f(0),\partial
f(\mathbb{D})).$

Therefore, the radius $r_{{}_{N}}(\alpha)$ is the best possible. This
completes the proof. ∎

Acknowledgment: The first author is supported by the Institute Post Doctoral
Fellowship of IIT Bhubaneswar, India, the second author is supported by SERB-
MATRICS, and third author is supported by CSIR, India.

## References

  * [1] Y. Abu-Muhanna, Bohr’s phenomenon in subordination and bounded harmonic classes, Complex Var. Elliptic Equ. 55 (2010), 1071–1078.
  * [2] Y. Abu-Muhanna and R. M. Ali, Bohr’s phenomenon for analytic functions into the exterior of a compact convex body, J. Math. Anal. Appl. 379 (2011), 512–517.
  * [3] Y.Abu Muhanna, R. M. Ali, Z. C. Ng and S. F. M Hasni, Bohr radius for subordinating families of analytic functions and bounded harmonic mappings, J. Math. Anal.Appl. 420 (2014), 124–136.
  * [4] L. Aizenberg, A. Aytuna and P. Djakov, Generalization of theorem on Bohr for bases in spaces of holomorphic functions of several complex variables, J. Math. Anal.Appl. 258 (2001), 429–447.
  * [5] M. B. Ahamed, Vasudevarao Allu and Himadri Halder, The Bohr phenomenon for analytic functions on simply connected domains, https://arxiv.org/pdf/2011.13890.pdf.
  * [6] L. Aizenberg, A. Aytuna and P. Djakov, An abstract approach to Bohr’s phenomenon, Proc. Amer. Math. Soc. 128 (9) (2000), 2611-–2619.
  * [7] L. Aizenberg and N. Tarkhanov, A Bohr phenomenon for elliptic equations, Proc. Lond. Math. Soc. 82(2) (2001), 385–401.
  * [8] L.Aizenberg, Generalization of results about the Bohr radius for power series, Stud. Math. 180 (2007), 161–168.
  * [9] R. M. Ali, R.W. Barnard and A.Yu. Solynin, A note on Bohr’s phenomenon for power series, J. Math. Anal.Appl. 449 (2017), 154-167.
  * [10] R. M . Ali, R.W. Barnard and A.Yu. Solynin, A note on Bohr’s phenomenon for power series, J. Math. Anal.Appl. 449 (2017), 154-167.
  * [11] S. A. Alkhaleefah, I.R. Kayumov and S. Ponnusamy, On the Bohr inequality with a fixed zero coefficient, Proc. Amer. Math. Soc. 147 (2019) 5263-5274.
  * [12] Vasudevarao Allu and Himadri Halder, Bohr inequality for certain harmonic mappings, see https://arxiv.org/pdf/2009.08683.pdf.
  * [13] Vasudevarao Allu and Himadri Halder, Bhor phenomenon for certain subclasses of Harmonic Mappings, see https://arxiv.org/pdf/2006.11622.pdf.
  * [14] C. B${\rm\acute{E}}$n${\rm\acute{E}}$teau, A. Dahlner and D. Khavinson, Remarks on the Bohr phenomenon, Comput. Methods Funct. Theory, 4 (1) (2004), 1-–19.
  * [15] H.P. Boas and D. Khavinson, Bohr’s power series theorem in several variables, Proc. Amer. Math. Soc. 125 (10) (1997), 2975-–2979.
  * [16] B. Bhowmik and N. Das, Bohr phenomenon for subordinating families of certain univalent functions, J. Math. Anal. Appl. 462 (2018), 1087-1098.
  * [17] H. Bohr, A theorem concerning power series, Proc. Lond. Math. Soc. s2-13 (1914), 1–5.
  * [18] A. Defant and L. Frerick, A logarithmic lower bound for multi-dimensional bohr radii, Israel J. Math. 152 (1) (2006), 17–-28.
  * [19] P. L. Duren, Univalent Functions (Grundlehren der mathematischen Wisseenschaften 259, New York, Berlin, Heidelberg, Tokyo) Springer-Verlag, 1983.
  * [20] P.L. Duren, Harmonic mapping in the plan, Cambridge University Press, (2004).
  * [21] S. Evdoridis, S. Ponnusamy and A. Rasila, Improved Bohr’s inequality for locally univalent harmonic mappings, Indag. Math. (N.S.) 30 (1) (2019), 201–-213.
  * [22] Y. Huang, M.-S. Liu and S. Ponnusamy, Refined Bohr-type inequalities with area measure for bounded analytic functions, Anal. Math. Phys. 10, 50 (2020), 21 pages; https://doi.org/10.1007/s13324-020-000393-0.
  * [23] A. Ismagilov, I. R. Kayumov and S. Ponnusamy, Sharp Bohr type inequality, J. Math. Anal. Appl. 489 (2020), 124–147.
  * [24] Nirupam Ghosh and A. Vasudevarao, Some basic properties of certain subclass of harmonic univalent functions, Complex Var. Elliptic Equ. 63 (2018), 1687–1703.
  * [25] Nirupam Ghosh and A.Vasudevarao, On a subclass of harmonic close-to-convex mappings, Monatsh. Math. 188 (2019), 247-267.
  * [26] Nirupam Ghosh and A. Vasudevarao, On some subclasses of harmonic mappings, Bull. Aust. Math. Soc. 101 (2020), 130–140.
  * [27] I.R. Kayumov and S. Ponnusamy, Bohr-Rogosinski radius for analytic functions, preprint, see https://arxiv.org/abs/1708.05585.
  * [28] I.R. Kayumov and S. Ponnusamy, Bohr inequality for odd analytic functions, Comput. Methods Funct. Theory, 17 (2017) 679–688.
  * [29] I.R. Kayumov, S. Ponnusamy and N. Shakirov, Bohr radius for localy unovalent harmonic mappings, Math. Nachr. 291 (2018), 1757-1768.
  * [30] A. Kayumova, I. R. Kayumov and S. Ponnusamy, Bohr’s inequality for harmonic mappings and beyond, Mathematics and computing, 245–256, Commun. Comput. Inf. Sci., 834, Springer, Singapore, 2018.
  * [31] I. R. Kayumov and S. Ponnusamy, Bohr’s inequalities for the analytic functions with lacunary series and harmonic functions, J. Math. Anal. Appl. 465(2018), 857–871.
  * [32] I.R. Kayumov and S. Ponnusamy, On a powered Bohr inequality, Ann. Acad. Sci. Fenn. Math. 44 (2019) 301-310.
  * [33] I. R. Kayumov and S. Ponnusamy, Improved version of Bohr’s inequalities, C. R. Math. Acad. Sci. Paris 358 (5) (2020), 615–-620.
  * [34] L. Li and S. Ponnusamy, Injectivity of sections of univalent harmonic mappings, Nonlinear Analysis 89 (2013), 276–283.
  * [35] G. Liu, Z. Liu and S. Ponnusamy, Refined Bohr inequality for bounded analytic functions, priprint, see https://arxiv.org/pdf/2006.08930.
  * [36] M. S. Liu and S. Ponnusamy, Multidimensional analogues of refined Bohr’s inequality, Proc. Amer. Math. Soc., (to appear).
  * [37] Vern I. Paulsen, Gelu Popescu and Dinesh Singh, On Bohr’s inequality, Proc. Lond. Math. Soc. s3-85 (2002), 493–512.
  * [38] S. Ponnusamy, R. Viajayakumar and K-J Wirths, New inequalities for the coefficients of unimodular bounded functions, Results Math (2020) 75 : 107.
  * [39] S. Ponnusamy, H. Yamamoto and H. Yanagihara, Variability regions for certain families of harmonic univalent mappings, Complex Var. Elliptic Equ. 58 (1) (2013), 23–34.
  * [40] S. Ponnusamy and K-J Wirths, Bohr Type Inequalities for Functions with a Multiple Zero at the Origin, Compt. Method Funct. Theory, 20(2020), 550–570.
  * [41] G. Popescu, Multivariable Bohr inequalities, Trans. Amer. Math. Soc. 359(11)(2007), 5283-–5317.
  * [42] Derek K. Thomas, Nikola Tuneski and Allu Vasudevarao, Univalent functions. A primer, De Gruyter Studies in Mathematics, 69. De Gruyter, Berlin, 2018.