††footnotetext:  File: main.tex, printed: 28-8-2024, 5.44

# On the elliptic harmonic mappings and sense-preserving harmonic mappings

Ming-Sheng Liu${}^{~{}\mathbf{*}}$ M.-S. Liu, School of Mathematical Sciences,
South China Normal University, Guangzhou, Guangdong 510631, China.
<EMAIL_ADDRESS>and  Hao XU  H. XU, School of Mathematical Sciences, South
China Normal University, Guangzhou, Guangdong 510631, China<EMAIL_ADDRESS>
###### Abstract.

In this paper, we first establish two versions of Landau-Bloch type theorem
for $(K,K^{\prime})$-elliptic harmonic mappings with a bounded minimum
distortion. Next, we provide several coefficient estimates and a conjecture
for $(K,K^{\prime})$-elliptic harmonic mappings. Then, we establish three new
versions of Landau-Bloch type theorem for sense-preserving harmonic mappings.
Finally, we establish two sharp versions of Landau-Bloch type theorem for
certain harmonic mappings. These results are sharp in some given cases and
improve the related results of different authors.

###### Key words and phrases:

harmonic mappings, sense-preserving harmonic mappings, $K$-quasiregular
harmonic mappings, $(K,K^{{}^{\prime}})$-elliptic harmonic mappings, Landau-
Bloch type theorems
∗ Corresponding author

###### 2010 Mathematics Subject Classification:

Primary 30C50, 31A05; Secondary 32A18, 30C62, 33E05.

##  1\. Introduction and Preliminaries

Suppose that $f(z)=u(z)+iv(z)$ is a twice continuously differentiable function
in the unit disk $\mathbb{D}=\\{z\in\mathbb{C}:|z|<1\\}$. Then $f$ is a
harmonic mapping in $\mathbb{D}$ if and only if $f$ satisfies $\Delta
f=4f_{z\bar{z}}=\frac{\partial^{2}f}{\partial
x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}=0$ for all $z=x+iy\in\mathbb{D}$
(cf. [13]), where we use the common notations for its formal derivatives:

$f_{z}=\frac{1}{2}\left(f_{x}-if_{y}\right),\quad
f_{\bar{z}}=\frac{1}{2}\left(f_{x}+if_{y}\right).$

For such function $f$, we define the maximum and minimum distortions [3] as
follows:

$\Lambda_{f}=\max_{0\leqslant\theta\leqslant
2\pi}\left|f_{z}+e^{-2i\theta}f_{\bar{z}}\right|=\left|f_{z}\right|+\left|f_{\bar{z}}\right|,$

and

$\lambda_{f}=\min_{0\leqslant\theta\leqslant
2\pi}\left|f_{z}+e^{-2i\theta}f_{\bar{z}}\right|=\|f_{z}|-|f_{\bar{z}}||.$

Since $\mathbb{D}$ is simply connected, $f(z)$ can be written as $f=h+\bar{g}$
with $f(0)=h(0)$, $g(z)$ and $h(z)$ are analytic in $\mathbb{D}$ (cf. [13]).
Thus the Jacobian of $f$ is given by

$J_{f}=\left|f_{z}\right|^{2}-\left|f_{z}\right|^{2}=\left|h^{\prime}\right|^{2}-\left|g^{\prime}\right|^{2}.$

Notice that $\left|J_{f}\right|=\lambda_{f}\cdot\Lambda_{f}$. It is known [18]
that a harmonic mapping is locally univalent if and only if $J_{f}\neq 0$. A
harmonic mappings $f$ is said to be sense-preserving if $J_{f}>0$. Therefore,
a sense-preserving harmonic mapping $f$ is said to be $K$-quasiregular
harmonic $(K\geq 1)$ on $\mathbb{D}$ if $\Lambda_{f}(z)\leq K\lambda_{f}(z)$
for all $z\in\mathbb{D}$. There is currently great interest in harmonic
mappings due to their relevance for fluid flows (please refer to the
discussion in [1, 10]).

Definition 1.1. ([24]) Let $\Omega\subset\mathbb{C}$ be a domain. A mapping
$f:\Omega\rightarrow\mathbb{C}$ is said to be absolutely continuous on lines,
$ACL$ in brief, in a domain $\Omega$ if for every closed rectangle
$R\subset\Omega$ with sides parallel to the axes $x$ and $y,f$ is absolutely
continuous on almost every horizontal line and almost every vertical line in
$R$. Such a mapping $f$ has partial derivatives $f_{x}$ and $f_{y}$ a.e. in
$\Omega$. Moreover, we say $f\in ACL^{2}$ if $f\in ACL$ and its partial
derivatives are locally $L^{2}$ integrable in $\Omega$.

Definition 1.2. ([14]) A sense-preserving and continuous mapping $f$ of
$\mathbb{D}$ onto $\mathbb{C}$ is said to be
$\left(K,K^{\prime}\right)$-elliptic mapping if

(1) $f$ is $ACL^{2}$ in $\mathbb{D},J_{f}\neq 0$ a.e. in $\mathbb{D}$;

(2) $\exists K\geq 1$ and $K^{\prime}\geq 0$ such that
$\left\|D_{f}\right\|^{2}\leq KJ_{f}+K^{\prime}\text{ a.e. in }\mathbb{D}$.

It is easy to see that $\left\|D_{f}(z)\right\|=\Lambda_{f}(z)$. Thus, a
$\left(K,0\right)$-elliptic mapping becomes $K$-quasiregular mapping. For more
details on elliptic mappings, please refer to ([8, 9, 14, 23]).

The classical Landau theorem asserts that if $f$ is holomorphic in
$\mathbb{D}$ such that $f^{\prime}(0)=1$ and $|f(z)|<M$ for $z\in\mathbb{D}$,
then $f$ is univalent in the disk
$\mathbb{D}_{\rho_{0}}=\\{z\in\mathbb{C}:|z|<\rho_{0}\\}$, and
$f(\mathbb{D}_{\rho_{0}})$ contains a disk $\mathbb{D}_{R_{0}}$, where

(1.1)
$\displaystyle\rho_{0}=\frac{1}{M+\sqrt{M^{2}-1}}=M-\sqrt{M^{2}-1}~{}\mbox{
and }~{}R_{0}=M\rho_{0}^{2}.$

This result is sharp, with the extremal function

(1.2) $\displaystyle f_{0}(z)=Mz\left(\frac{1-Mz}{M-z}\right).$

The Bloch theorem asserts the existence of a positive constant number b such
that if $f$ is an analytic function on the unit disk $\mathbb{D}$ with
$f^{\prime}(0)=1$, then $f(\mathbb{D})$ contains a disk of radius $b$, that
is, a disk of radius $b$ which is the univalent image of some region in
$\mathbb{D}$. Such a disk is called “schlicht disk” for $f$. The supremum of
all such constants $b$ is called the Bloch constant (see [3, 15]).

In 2000, H.H. Chen et al. [3] obtained two versions of Laudau-type theorems
for bounded harmonic mappings in the unit disk $\mathbb{D}$. However, their
results are not sharp. Dorff and Nowak [11], Grigoryan [12], S.L. Chen et al.
[5, 6, 7], Huang et al. [16, 25], Liu et al. [19, 20, 22] and H.H. Chen et al.
[4] improved their results respectively. In particular, the following sharp
version of Landau-type theorem for harmonic mappings with bounded maximum
distortion was obtained in [16, 20, 22].

Theorem A. ([16, 20, 22]) Let $f=h+\overline{g}$ be a harmonic mapping in the
unit disk $\mathbb{D}$ such that $f(0)=0$ and $\lambda_{f}(0)=1$.

  1. (1)

If $\varLambda_{f}(z)\leq 1$ for all $z\in{\mathbb{D}}$, then $f$ is univalent
in $\mathbb{D}$ and $f(\mathbb{D})$ contains a schlicht disk $\mathbb{D}$, and
the result is sharp.

  2. (2)

If $\varLambda_{f}(z)<\varLambda$ for all $z\in{\mathbb{D}}$, then
$\varLambda>1$ and $f$ is univalent in the disk $\mathbb{D}_{\rho}$ with
$\rho=\frac{1}{\Lambda}$, and $f(\mathbb{D}_{\rho})$ contains a schlicht disk
$\mathbb{D}_{R}$, where

$R=\varLambda+(\varLambda^{3}-\varLambda)\ln\Big{(}1-\frac{1}{\varLambda^{2}}\Big{)}.$

The result is sharp.

The sharp form of Landau-type theorem for harmonic mappings with
$f(0)=0,\lambda_{f}(0)=1$ and $|f(z)|\leq 1$ in the unit disk was also
obtained in [19]. In 2020, Liu et al. proved the sharp result of Landau-Bloch
type theorem for strongly-bounded harmonic mappings when $M>1$ in [21], and
obtained several new sharp versions of Landau-Bloch type theorems of harmonic
mappings. We recall one of these results as follows.

Theorem B. ([21, Theorem 3.5]) Suppose that $M>1$. Let $f(z)$ be a harmonic
mapping in the unit disk $\mathbb{D}$ with $f(0)=\lambda_{f}(0)-1=0$, and

$f(z)=\sum_{n=1}^{\infty}a_{n}z^{n}+\sum_{n=1}^{\infty}\overline{b_{n}}\overline{z^{n}}$

satisfying the following inequality

$\displaystyle\sum\limits_{n=2}^{\infty}n(|a_{n}|+|b_{n}|)r^{n-1}\leq\frac{(M^{2}-1)(2Mr-r^{2})}{(M-r)^{2}},\quad
0\leq r\leq\rho_{0}=M-\sqrt{M^{2}-1}.$

Then $f(z)$ is univalent in the disk $\mathbb{D}_{\rho_{0}}$ and
$f(\mathbb{D}_{\rho_{0}})$ contains a schlicht disk $\mathbb{D}_{R_{0}}$,
where $R_{0}=M\rho_{0}^{2}$. This result is sharp, with
$f_{0}(z)=Mz\frac{1-Mz}{M-z}$ being an extremal mapping.

Recently, Allu and Kumar [2] established two new versions of Laudau-Bloch type
theorem for $(K,K^{{}^{\prime}})$-elliptic harmonic mappings and
K-quasiregular mappings as follows.

Theorem C. ([2, Theorem 2.5]) Let $f=h+\overline{g}$ be a
$(K,K^{{}^{\prime}})$-elliptic harmonic mapping in $\mathbb{D}$ such that
$f(0)=0,\lambda_{f}(0)=1$ and $\lambda_{f}(z)\leq\Lambda$ for
$z\in\mathbb{D}$. Then $f$ is univalent in the disk $\mathbb{D}_{\rho_{1}}$
and $f(\mathbb{D}_{\rho_{1}})$ contains a schlicht disk $\mathbb{D}_{R_{1}}$,
where

$\rho_{1}=\frac{1}{1+K\Lambda+\sqrt{K^{\prime}}},\quad
R_{1}=1+(K\Lambda+\sqrt{K^{\prime}})\ln\Big{(}1-\frac{1}{1+K\Lambda+\sqrt{K^{\prime}}}\Big{)}.$

Theorem D. ([2, Theorem 2.7]) Let $f=h+\overline{g}$ be a K-quasiregular
mapping in $\mathbb{D}$ such that $f(0)=0,J_{f}(0)=1$ and
$\lambda_{f}(z)\leq\Lambda$ for $z\in\mathbb{D}$. Then $f$ is univalent in the
disk $\mathbb{D}_{\rho_{2}}$ and $f(\mathbb{D}_{\rho_{2}})$ contains a
schlicht disk $\mathbb{D}_{R_{2}}$, where

$\rho_{2}=\frac{1}{1+K^{3/2}\Lambda},\quad
R_{2}=\frac{1}{\sqrt{K}}+K\Lambda\ln\Big{(}1-\frac{1}{1+K^{3/2}\Lambda}\Big{)}.$

However, Theorems C and D are not sharp. It is then natural to raise the
following problems:

######  Problem 1.

Can we improve the results in Theorems C and D? Can we establish several sharp
results?

The paper is organized as follows. In Section 2, we present statements of our
theorems which improve Theorems C and D, and several new versions of Landau-
Bloch type theorem and coefficient estimates for sense-preserving harmonic
mappings and normalized $(K,K^{{}^{\prime}})$-elliptic harmonic mappings,
$K$-quasiregular mappings are also provided. In particular, these results are
sharp in some given cases. In Section 3, we state a couple of lemmas which are
needed for the proofs of our main results. In Section 4, we present the proofs
of the main results. In particular, Theorem 1, Corollary 2, Theorems 5, 6, 8
and 9 provide an affirmative answer to Problem 1.

##  2\. Statement of Main Results

We first improve Theorems C and D as follows:

######  Theorem 1.

Let $f=h+\overline{g}$ be a $(K,K^{{}^{\prime}})$-elliptic harmonic mapping
defined in the unit disk $\mathbb{D}$ such that $f(0)=0,\lambda_{f}(0)=1$ and
$\lambda_{f}(z)<\Lambda$ for all $z\in\mathbb{D}$. Then $\Lambda>1$, $f$ is
univalent in the disk $\mathbb{D}_{r_{1}}$ with

$r_{1}=\frac{2}{K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}}}\geq\frac{1}{K\Lambda+\sqrt{K^{\prime}}}>\rho_{1},$

and $f(\mathbb{D}_{r_{1}})$ contains a schlicht disk with radius

$\sigma_{1}=\frac{1}{r_{1}}+\Big{(}\frac{1}{r_{1}^{3}}-\frac{1}{r_{1}}\Big{)}\ln\Big{(}1-r_{1}^{2}\Big{)}>R_{1}.$

When $K=1,\,K^{\prime}=0$, the radii $r_{1}=\frac{1}{\Lambda}$ and
$\sigma_{1}=\Lambda+(\Lambda^{3}-\Lambda)\ln\Big{(}1-\frac{1}{\Lambda^{2}}\Big{)}$
are sharp.

######  Remark 1.

We remark that Theorem 1 improves Theorem C. In order to be more explicit, we
compute the approximate values obtained from Theorem 1 and the corresponding
values obtained from Theorem C for various choices of $K$, $K^{\prime}$ and
$\Lambda$. From Table 1, we can see that, for the same $K$, $K^{\prime}$ and
$\Lambda$, $r_{1}>\rho_{1}$, $\sigma_{1}>R_{1}$.

Table 1. Values of $r_{1}$, $\rho_{1}$; $\sigma_{1}$, $R_{1}$ in Theorem 1, Theorem C ($K$, $K^{\prime}$, $\Lambda$) | (1.0,1.1,1.2) | (1.2,1.3,1.4) | (1.4,1.5,1.6) | (1.6,1.7,1.8) | (1.8,1.9,2.0) | (2.0,2.1,2.2)
---|---|---|---|---|---|---
$r_{1}$ | 0.5530 | 0.4432 | 0.3598 | 0.2956 | 0.2459 | 0.2069
$\rho_{1}$ | 0.3078 | 0.2618 | 0.2240 | 0.1929 | 0.1673 | 0.1460
$\sigma_{1}$ | 0.3100 | 0.2377 | 0.1882 | 0.1523 | 0.1255 | 0.1049
$R_{1}$ | 0.1727 | 0.1441 | 0.1214 | 0.1003 | 0.0887 | 0.0768

Setting $K^{\prime}=0$ in Theorem 1, we get the following improved version of
[2, Corollary 2.6].

######  Corollary 1.

Let $f=h+\overline{g}$ be a $K$-quasiregular mapping defined in the unit disk
$\mathbb{D}$ such that $f(0)=0,\lambda_{f}(0)=1$ and $\lambda_{f}(z)<\Lambda$
for all $z\in\mathbb{D}$. Then $f$ is univalent in $\mathbb{D}_{r_{2}}$ and
$f(\mathbb{D}_{r_{2}})$ contains a schlicht disk $\mathbb{D}_{\sigma_{2}}$,
where

$r_{2}=r_{2}(K,\Lambda)=\frac{1}{K\Lambda},\quad\sigma_{2}=K\Lambda+(K^{3}\Lambda^{3}-K\Lambda)\ln\Big{(}1-\frac{1}{K^{2}\Lambda^{2}}\Big{)}.$

When $K=1$, the result is sharp with an extremal function being
$f_{1}(z)=\Lambda\int_{0}^{z}{\frac{\Lambda z-1}{\Lambda-z}dz}$.

######  Theorem 2.

Let $f=h+\overline{g}$ be a $(K,K^{{}^{\prime}})$-elliptic harmonic mapping in
$\mathbb{D}$ such that $f(0)=0,\,J_{f}(0)=1$ and $\lambda_{f}(z)<\Lambda$ for
all $z\in\mathbb{D}$. Then $\Lambda>\frac{1}{\sqrt{K+K^{\prime}}}$, $f$ is
univalent in the disk $\mathbb{D}_{r_{3}}$ and $f(\mathbb{D}_{r_{3}})$
contains a schlicht disk $\mathbb{D}_{\sigma_{3}}$, where

$r_{3}=r_{3}(K,K^{\prime},\Lambda)=\frac{2}{\sqrt{K+K^{\prime}}\,(K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}})},$
$\sigma_{3}=\frac{1}{\sqrt{K+K^{\prime}}}\bigg{\\{}\frac{1}{r_{3}}+\Big{(}\frac{1}{r_{3}^{3}}-\frac{1}{r_{3}}\Big{)}\ln\Big{(}1-r_{3}^{2}\Big{)}\bigg{\\}}.$

When $K=1,\,K^{\prime}=0$, the radii $r_{3}=\frac{1}{\Lambda}$ and
$\sigma_{3}=\Lambda+(\Lambda^{3}-\Lambda)\ln\Big{(}1-\frac{1}{\Lambda^{2}}\Big{)}$
are sharp.

Setting $K^{\prime}=0$ in Theorem 2, we get the following result, which
improves Theorem D.

######  Corollary 2.

Let $f=h+\overline{g}$ be a K-quasiregular mapping defined on unit disk
$\mathbb{D}$ such that $f(0)=0,J_{f}(0)=1$ and $\lambda_{f}(z)<\Lambda$ for
all $z\in\mathbb{D}$. Then $\Lambda>\frac{1}{\sqrt{K}}$, $f$ is univalent in
the disk $\mathbb{D}_{r_{3}^{\prime}}$ and $f(\mathbb{D}_{r_{3}^{\prime}})$
contains a schlicht disk $\mathbb{D}_{\sigma_{3}^{\prime}}$, where

$r_{3}^{\prime}=\frac{1}{K^{3/2}\Lambda}>\rho_{2},\quad\sigma_{3}^{\prime}=K\Lambda+(K^{4}\Lambda^{3}-K\Lambda)\ln\Big{(}1-\frac{1}{K^{3}\Lambda^{2}}\Big{)}>R_{2}.$

When $K=1$, the radii $r_{3}^{\prime}=\frac{1}{\Lambda}$ and
$\sigma_{3}^{\prime}=\Lambda+(\Lambda^{3}-\Lambda)\ln\Big{(}1-\frac{1}{\Lambda^{2}}\Big{)}$
are sharp with an extremal function being
$f_{1}(z)=\Lambda\int_{0}^{z}{\frac{\Lambda z-1}{\Lambda-z}dz}$.

######  Remark 2.

From Table 2, we can see that, for the same $K$ and $\Lambda$,
$r_{3}^{\prime}>\rho_{2}$, $\sigma_{3}^{\prime}>R_{2}$.

Table 2. Values of $r_{3}^{\prime}$, $\rho_{3}$; $\sigma_{3}^{\prime}$, $R_{3}$ in Corollary 2, Theorem D ($K$, $\Lambda$) | (1.0, 1.2) | (1.2, 1.4) | (1.4, 1.6) | (1.6, 1.8) | (1.8, 2.0) | (2.0, 2.2) | (2.2, 2.4)
---|---|---|---|---|---|---|---
$r_{3}^{\prime}$ | 0.8333 | 0.5434 | 0.3773 | 0.2745 | 0.2070 | 0.1607 | 0.1277
$\rho_{2}$ | 0.4545 | 0.3521 | 0.2739 | 0.2154 | 0.1715 | 0.1385 | 0.1132
$\sigma_{3}^{\prime}$ | 0.5740 | 0.2768 | 0.1676 | 0.1113 | 0.0783 | 0.0573 | 0.0433
$R_{2}$ | 0.2726 | 0.1838 | 0.1281 | 0.0920 | 0.0679 | 0.0514 | 0.0397

Next, we provide several results and a conjecture of coefficient estimates for
normalized $(K,K^{{}^{\prime}})$-elliptic harmonic mappings.

######  Theorem 3.

Let $f=h+\overline{g}$ be a $(K,K^{{}^{\prime}})$-elliptic harmonic mapping in
$\mathbb{D}$ such that $f(0)=0,\,J_{f}(0)=1$ and $\lambda_{f}(z)\leq\Lambda$
for all $z\in\mathbb{D}$, and

(2.1) $\displaystyle h(z)=\sum\limits_{n=1}^{\infty}{a_{n}z^{n}},\quad
g(z)=\sum\limits_{n=1}^{\infty}{b_{n}z^{n}}.$

Then, we have that

$\displaystyle\frac{1}{\sqrt{K+K^{\prime}}}\leq|a_{1}|+|b_{1}|\leq\sqrt{K+K^{\prime}},$

and

$\left|a_{n}\right|+\left|b_{n}\right|\leq\frac{(K+K^{\prime})(K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}})^{2}-4}{2n(K+K^{\prime})(K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}})},\,n=2,3,\ldots.$

When $K=1,K^{\prime}=0$, the above estimates are sharp for all $n=2,3,\ldots$,
with the extremal functions

(2.2) $\displaystyle
f_{n}(z)=\Lambda^{2}z-\int_{0}^{z}{\frac{\Lambda^{3}-\Lambda}{\Lambda+z^{n-1}}dz}.$

Applying a similar method as for that used in Theorem 3, we have the following
result.

######  Corollary 3.

Let $f=h+\overline{g}$ be a $(K,K^{{}^{\prime}})$-elliptic harmonic mapping in
$\mathbb{D}$ such that $f(0)=0,\,\lambda_{f}(0)=1$ and
$\lambda_{f}(z)\leq\Lambda$ for all $z\in\mathbb{D}$, and let $h,g$ be given
by (2.1). Then, we have that

$\displaystyle 1\leq|a_{1}|+|b_{1}|\leq\frac{K+\sqrt{K^{2}+4K^{\prime}}}{2},$

and

$\left|a_{n}\right|+\left|b_{n}\right|\leq\frac{(K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}})^{2}-4}{2n(K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}})},\,n=2,3,\ldots.$

When $K=1,K^{\prime}=0$, the above estimates are sharp for all $n=2,3,\ldots$,
with the extremal functions given by (2.2).

Setting $K^{\prime}=0$ in Theorem 3, we get the following result.

######  Corollary 4.

Let $f=h+\overline{g}$ be a K-quasiregular mapping in $\mathbb{D}$ such that
$f(0)=0,J_{f}(0)=1$ and $\lambda_{f}(z)\leq\Lambda$ for all $z\in\mathbb{D}$,
and let $h,g$ be given by (2.1). Then

$\frac{1}{\sqrt{K}}\leq|a_{1}|+|b_{1}|\leq\sqrt{K},\quad\left|a_{n}\right|+\left|b_{n}\right|\leq\frac{K^{3}\Lambda^{2}-1}{n\Lambda
K^{2}},\,n=2,3,\ldots.$

When $K=1$, the above estimates are sharp for all $n=2,3,\ldots$, with the
extremal functions given by (2.2).

Setting $K^{\prime}=0$ in Corollary 3, we get the following result.

######  Corollary 5.

Let $f=h+\overline{g}$ be a K-quasiregular mapping in $\mathbb{D}$ such that
$f(0)=0,\lambda_{f}(0)=1$ and $\lambda_{f}(z)\leq\Lambda$ for all
$z\in\mathbb{D}$, and let $h,g$ be given by (2.1). Then,

$1\leq|a_{1}|+|b_{1}|\leq
K,\quad\left|a_{n}\right|+\left|b_{n}\right|\leq\frac{K^{2}\Lambda^{2}-1}{n\Lambda
K},n=2,3,\ldots.$

When $K=1$, the above estimates are sharp for all $n=2,3,\ldots$, with the
extremal functions $f_{n}(z)$ given by (2.2).

Note that

$\frac{K^{2}\Lambda^{2}-1}{n\Lambda
K}=K\frac{\Lambda^{2}-\frac{1}{K^{2}}}{n\Lambda},$

we propose the following conjecture.

######  Conjecture 1.

Let $F=h+\overline{g}$ be a $K$-quasiregular mapping in unit disk $\mathbb{D}$
such that $F(0)=0,\lambda_{F}(0)=1$ and $\lambda_{F}(z)\leq\Lambda$ for all
$z\in\mathbb{D}$, and let $h,g$ be given by (2.1). Then,

$\left|a_{n}\right|+\left|b_{n}\right|\leq
K\frac{\Lambda^{2}-1}{n\Lambda},n=2,3,\ldots,$

and the above estimates are sharp for all $n=2,3\ldots$, with the extremal
functions:

$F_{n}(z)=\frac{K+1}{2}\bigg{(}\Lambda^{2}z-\int_{0}^{z}{\frac{\Lambda^{3}-\Lambda}{\Lambda+z^{n-1}}}dz\bigg{)}+\overline{\frac{K-1}{2}\bigg{(}\Lambda^{2}z-\int_{0}^{z}{\frac{\Lambda^{3}-\Lambda}{\Lambda+z^{n-1}}dz}\bigg{)}}.$

Now, we establish three versions of Landau-Bloch type theorem for sense-
preserving harmonic mappings and a new version of Landau-Bloch type theorem
for K-quasiregular mapping as follows:

######  Theorem 4.

Let $f=h+\overline{g}$ be a sense-preserving harmonic mapping in the unit disk
$\mathbb{D}$ such that $f(0)=0,h^{\prime}(0)-1=g^{\prime}(0)=0$ and
$\lambda_{f}(z)\leq\lambda$ for all $z\in\mathbb{D}$. Then $\lambda\geq 1$,
$f$ is univalent in the disk $\mathbb{D}_{r_{4}}$ with
$r_{4}=r_{4}(\lambda)=\frac{1}{2(\sqrt{2}+1)\lambda}$, and
$f(\mathbb{D}_{r_{4}})$ contains a schlicht disk with radius

$\sigma_{4}=\frac{(\sqrt{2}+1)\lambda}{2}\bigg{\\{}1+\Big{(}(\sqrt{2}+1)^{2}\lambda^{2}-1\Big{)}\ln\Big{(}1-\frac{(\sqrt{2}-1)^{2}}{\lambda^{2}}\Big{)}\bigg{\\}}.$

######  Theorem 5.

Suppose that $M>1$. Let $f=h+\overline{g}$ be a sense-preserving harmonic
mapping in the unit disk $\mathbb{D}$ such that
$f(0)=f_{z}(0)-1=f_{\overline{z}}(0)=0$ and $|h(z)|\leq M$ for all
$z\in\mathbb{D}$. Then $f(z)$ is univalent in $\mathbb{D}_{\rho_{0}}$ with
$\rho_{0}=\frac{1}{M+\sqrt{M^{2}-1}}$. The radius $\rho_{0}$ is sharp with the
extremal function $f_{0}(z)=Mz\frac{1-Mz}{M-z}$.

######  Theorem 6.

Suppose that $\Lambda>1$. Let $f=h+\overline{g}$ be a sense-preserving
harmonic mapping in the unit disk $\mathbb{D}$ such that
$f(0)=f_{z}(0)-1=f_{\overline{z}}(0)=0$ and $|h^{\prime}(z)|<\Lambda$ for all
$z\in\mathbb{D}$. Then $f(z)$ is univalent in $\mathbb{D}_{r_{6}}$ with
$r_{6}=\frac{1}{\Lambda}$. The radius $r_{6}$ is sharp with the extremal
function $f_{1}(z)=\int_{0}^{z}{\Lambda\frac{1-\Lambda z}{\Lambda-z}dz}$.

######  Theorem 7.

Let $f=h+\overline{g}$ be a K-quasiregular mapping defined on unit disk
$\mathbb{D}$ such that $f(0)=0,\lambda_{f}(0)=1$ and $|h(z)|\leq M$ for all
$z\in\mathbb{D}$. Then $f$ is univalent in the disk $\mathbb{D}_{r_{7}}$ and
$f(\mathbb{D}_{r_{7}})$ contains a schlicht disk $\mathbb{D}_{R_{7}}$, where

$r_{7}=\frac{K+1}{8KM},\quad\sigma_{7}=\frac{2KM}{K+1}\bigg{\\{}1+\Big{(}\Big{(}\frac{4KM}{K+1}\Big{)}^{2}-1\Big{)}\ln\bigg{(}1-\frac{(K+1)^{2}}{16K^{2}M^{2}}\bigg{)}\bigg{\\}}.$

Finally, we establish two sharp versions of Landau-Bloch type theorems for
certain harmonic mappings.

######  Theorem 8.

Suppose that $M>1$. Let $f=h+\overline{g}$ be a harmonic mapping in the unit
disk $\mathbb{D}$ such that $f(0)=0,\,\lambda_{f}(0)=1$, and let $h(z),g(z)$
satisfy

$h(z)=\sum\limits_{n=1}^{\infty}{a_{n}z^{n}},\quad
g(z)=\sum\limits_{n=1}^{\infty}{b_{n}z^{n}},z\in\mathbb{D},$

and for $|z|<\rho_{0}=\frac{1}{M+\sqrt{M^{2}-1}}$, we have

$|h^{\prime}(z)-h^{\prime}(0)|+|g^{\prime}(z)-g^{\prime}(0)|\leq\frac{(M^{2}-1)(2M|z|-|z|^{2})}{(M-|z|)^{2}}.$

Then $f(z)$ is univalent in $\mathbb{D}_{\rho_{0}}$ and
$f(\mathbb{D}_{\rho_{0}})$ contains a schlicht disk $\mathbb{D}_{R_{0}}$,
where $R_{0}=M\rho_{0}^{2}$. This result is sharp, with an extremal function
being $f_{0}(z)=Mz\frac{1-Mz}{M-z}$.

######  Remark 3.

Theorem 8 improves Theorem B, since for $|z|=r<1$, we have

$|h^{\prime}(z)-h^{\prime}(0)|+|g^{\prime}(z)-g^{\prime}(0)|\leq\sum\limits_{n=2}^{\infty}n(|a_{n}|+|b_{n}|)r^{n-1}.$

######  Theorem 9.

Let $f(z)=h(z)+\overline{g(z)}$ be a harmonic mapping in $\mathbb{D}$ with
$h(z)$ $g(z)$ given by (2.1), such that $f(0)=0,\lambda_{f}(0)=1$ and
$|zf_{z}(z)+\overline{z}f_{\overline{z}}(z)|<\Lambda$ for all
$z\in\mathbb{D}$, and $a_{n}b_{n}=0$ for $n=2,3,\ldots$. Then $\Lambda\geq 1$
and $f(z)$ is univalent in $\mathbb{D}_{r_{9}}$, and $f(\mathbb{D}_{r_{9}})$
contains a schlicht disk $\mathbb{D}_{\sigma_{9}}$, where
$r_{9}=\frac{1}{\Lambda}$, $\sigma_{9}=\Lambda-\sqrt{\Lambda^{2}-1}$. The
radius $r_{9}$ is sharp, with the extremal functions
$f_{1}(z)=\Lambda\int_{0}^{z}\frac{1-\Lambda z}{\Lambda-z}dz$. Moreover, when
$\Lambda=1$, $\sigma_{9}=1$ is also sharp.

##  3\. Key lemmas

In order to establish our main results, we need the following lemmas.

Lemma E. (see, for example, p.35 in [15]) Let
$f(z)=a_{0}+a_{1}z+\ldots+a_{n}z^{n}+\ldots$ be analytic in $\mathbb{D}$ and
let $|f(z)|\leq 1$ for all $z\in\mathbb{D}$. Then $|a_{n}|\leq 1-|a_{0}|^{2}$
for $n=1,2,\ldots$.

Lemma F. ([21, Lemma 2.2]) Suppose that $\rho>0,\,\sigma>0$, $h(z)$ and $g(z)$
are holomorphic in $\mathbb{D}$ with $h(0)=g(0)=0$. Then, for every complex
number $\lambda$ with $\left|\lambda\right|=1$, $f_{\lambda}(z)=h(z)+\lambda
g(z)$ is univalent in $\mathbb{D}_{\rho}$ and
$f_{\lambda}(\mathbb{D}_{\rho})\supseteq\mathbb{D}_{\sigma}$ if and only if
for every complex number $\lambda$ with $\left|\lambda\right|=1$,
$F_{\lambda}(z)=h(z)+\lambda\overline{g(z)}$ is univalent on
$\mathbb{D}_{\rho}$ and
$F_{\lambda}(\mathbb{D}_{\rho})\supseteq\mathbb{D}_{\sigma}$.

Lemma G. (see, for example, p.297 in [17]) If $0<x<1$. Then $\ln
x>\frac{1}{2}(x-\frac{1}{x})$.

Next, we establish a new lemma, which play a key role in the proof of our main
results.

Lemma H. Suppose that $x\geq 1$. Then
$1+x\ln\frac{x}{1+x}<\frac{1}{2(x+0.5)}$.

Proof: Let $f(x)=1+x\ln\frac{x}{1+x}-\frac{1}{2x+1}$. Then, a direct
computation yields

$f^{\prime}(x)=\ln\frac{x}{1+x}+\frac{1}{1+x}+\frac{1}{2(x+0.5)^{2}},$

$f^{\prime}(1)=\frac{1}{2}+\frac{2}{9}-\ln 2>0$ and

$\lim_{x\to+\infty}f^{\prime}(x)=0.$

Also, for $x\geq 1$, we have

$\displaystyle f^{\prime\prime}(x)$ $\displaystyle=$
$\displaystyle\frac{1}{x(1+x)}-\frac{1}{(1+x)^{2}}-\frac{1}{(x+0.5)^{3}}=\frac{1}{x(1+x)^{2}}-\frac{1}{(x+0.5)^{3}}$
$\displaystyle=$
$\displaystyle-\frac{\frac{1}{2}x^{2}+\frac{1}{8}(2x-1)}{x(1+x)^{2}(x+0.5)^{3}}<0,$

showing that $f^{\prime}(x)$ is strictly decreasing on $[1,\infty)$, and thus,
we have that $f^{\prime}(x)>0$ in $[1,\infty)$. This implies that $f(x)$ is
strictly increasing in $[1,\infty)$. We note that $f(1)=1-\ln 2-\frac{1}{3}<0$
and

$\lim_{x\to+\infty}f(x)=0,$

and therefore, we obtain that $f(x)<0$ in $[1,\infty)$, and the proof is
complete. $\Box$

##  4\. Proofs of the main results

###  4.1. Proof of Theorem 1

Since $f=h+\overline{g}$ is a $(K,K^{{}^{\prime}})$-elliptic harmonic mapping
and $\lambda_{f}(z)<\Lambda$ for all $z\in\mathbb{D}$, we have that
$\Lambda>\lambda_{f}(0)=1$, and for each $z\in\mathbb{D}$,

$\Lambda_{f}^{2}(z)\leq KJ_{f}(z)+K^{\prime}\leq
K\Lambda_{f}(z)\lambda_{f}(z)+K^{\prime}<K\Lambda\Lambda_{f}(z)+K^{\prime},$

that is,

$\Lambda_{f}^{2}(z)-K\Lambda\Lambda_{f}(z)-K^{\prime}<0.$

Hence, for $z\in\mathbb{D}$, we have that

(4.1)
$\displaystyle\Lambda_{f}(z)<\frac{K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}}}{2}.$

Note that $f=h+\overline{g}$ is a harmonic mapping in $\mathbb{D}$ with
$f(0)=0,\lambda_{f}(0)=1$, it follows from Theorem A that $f(z)$ is univalent
in the disk $\mathbb{D}_{r_{1}}$ with

$r_{1}=\frac{2}{K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}}}\geq\frac{1}{K\Lambda+\sqrt{K^{\prime}}}>\rho_{1},$

and $f(\mathbb{D}_{r_{1}})$ contains a schlicht disk with radius

$\sigma_{1}=\frac{1}{r_{1}}+\Big{(}\frac{1}{r_{1}^{3}}-\frac{1}{r_{1}}\Big{)}\ln\Big{(}1-r_{1}^{2}\Big{)}.$

When $K=1,\,K^{\prime}=0$, we have that $\Lambda(z)=\lambda(z)<\Lambda$ for
$z\in\mathbb{D}$, it follows from Theorem A that the radii
$r_{1}=\frac{1}{\Lambda}$ and
$\sigma_{1}=\Lambda+(\Lambda^{3}-\Lambda)\ln\Big{(}1-\frac{1}{\Lambda^{2}}\Big{)}$
are sharp.

Finally, we verify that $\sigma_{1}>R_{1}$. In fact, it follows from Lemmas G
and H that

$\displaystyle\sigma_{1}$ $\displaystyle\geq$ $\displaystyle
K\Lambda+\sqrt{K^{\prime}}+\Big{(}(K\Lambda+\sqrt{K^{\prime}})^{3}-K\Lambda-\sqrt{K^{\prime}}\Big{)}\ln\Big{(}1-\frac{1}{(K\Lambda+\sqrt{K^{\prime}})^{2}}\Big{)}$
$\displaystyle>$ $\displaystyle
K\Lambda+\sqrt{K^{\prime}}+\Big{(}(K\Lambda+\sqrt{K^{\prime}})^{3}-K\Lambda-\sqrt{K^{\prime}}\Big{)}\cdot$
$\displaystyle\cdot\frac{1}{2}\Big{(}\frac{(K\Lambda+\sqrt{K^{\prime}})^{2}-1}{(K\Lambda+\sqrt{K^{\prime}})^{2}}-\frac{(K\Lambda+\sqrt{K^{\prime}})^{2}}{(K\Lambda+\sqrt{K^{\prime}})^{2}-1}\Big{)}=\frac{1}{2(K\Lambda+\sqrt{K^{\prime}})}$
$\displaystyle>$ $\displaystyle\frac{1}{2(K\Lambda+\sqrt{K^{\prime}}+0.5)}$
$\displaystyle>$ $\displaystyle
1+(K\Lambda+\sqrt{K^{\prime}})\ln\Big{(}1-\frac{1}{1+K\Lambda+\sqrt{K^{\prime}}}\Big{)}=R_{1}.\hskip
142.26378pt\Box$

###  4.2. Proof of Theorem 2

Since $f=h+\overline{g}$ is a $(K,K^{{}^{\prime}})$-elliptic harmonic mapping,
$J_{f}(0)=1$ and $\lambda_{f}(z)<\Lambda$ for all $z\in\mathbb{D}$, we have
that $\Lambda>\lambda_{f}(0)$, and

$\Lambda_{f}^{2}(0)\leq KJ_{f}(0)+K^{\prime}=K+K^{\prime},$

that is,

$\Lambda_{f}(0)\leq\sqrt{K+K^{\prime}}.$

Hence,

(4.2)
$\displaystyle\Lambda>\lambda_{f}(0)=\frac{J_{f}(0)}{\Lambda_{f}(0)}\geq\frac{1}{\sqrt{K+K^{\prime}}}.$

Let $F(z)=\frac{f(z)}{\lambda_{f}(0)}$, then $F(z)$ is a harmonic mapping in
$\mathbb{D}$, $\lambda_{F}(0)=1$, and by (4.1), we have

$\Lambda_{F}(z)=\frac{\Lambda_{f}(z)}{\lambda_{f}(0)}\leq\sqrt{K+K^{\prime}}\cdot\frac{K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}}}{2}$

for $z\in\mathbb{D}$. So, it follows from Theorem A that $F(z)$ is univalent
in the disk $\mathbb{D}_{r_{3}}$ with

$r_{3}=\frac{2}{\sqrt{K+K^{\prime}}\,(K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}})},$

and $F(\mathbb{D}_{r_{3}})$ contains a schlicht disk with radius

$\sigma_{3}^{\prime\prime}=\frac{1}{r_{3}}+\Big{(}\frac{1}{r_{3}^{3}}-\frac{1}{r_{3}}\Big{)}\ln\Big{(}1-r^{2}_{3}\Big{)}.$

Note that

$\displaystyle\lambda_{f}(0)\sigma_{3}^{\prime\prime}\geq\frac{1}{\sqrt{K+K^{\prime}}}\bigg{\\{}\frac{1}{r_{3}}+\Big{(}\frac{1}{r_{3}^{3}}-\frac{1}{r_{3}}\Big{)}\ln\Big{(}1-r_{3}^{2}\Big{)}\bigg{\\}}=\sigma_{3}.$

Hence, $f(z)$ is univalent in the disk $\mathbb{D}_{r_{3}}$ and
$f(\mathbb{D}_{r_{3}})$ contains a schlicht disk $\mathbb{D}_{\sigma_{3}}$.

When $K=1,\,K^{\prime}=0$, we have that $\Lambda(z)=\lambda(z)<\Lambda$ for
$z\in\mathbb{D}$, it follows from Theorem A that the radii
$r_{3}=\frac{1}{\Lambda}$ and
$\sigma_{3}=\Lambda+(\Lambda^{3}-\Lambda)\ln\Big{(}1-\frac{1}{\Lambda^{2}}\Big{)}$
are sharp. $\Box$

###  4.3. Proof of Corollary 2

By Theorem 2, we only need to verify that $\sigma_{3}^{\prime}>R_{3}$. In
fact, it follows from Lemmas G and H that

$\displaystyle\hskip 51.21504pt\sigma_{3}^{\prime}$ $\displaystyle>$
$\displaystyle
K\Lambda+(K^{4}\Lambda^{3}-K\Lambda)\cdot\frac{1}{2}\Big{(}\frac{K^{3}\Lambda^{2}-1}{K^{3}\Lambda^{2}}-\frac{K^{3}\Lambda^{2}}{K^{3}\Lambda^{2}-1}\Big{)}=\frac{1}{2K^{2}\Lambda}$
$\displaystyle>$
$\displaystyle\frac{1}{\sqrt{K}}\cdot\frac{1}{2(K^{3/2}\Lambda+0.5)}$
$\displaystyle>$
$\displaystyle\frac{1}{\sqrt{K}}\bigg{(}1+K^{3/2}\Lambda\ln\Big{(}1-\frac{1}{1+K^{3/2}\Lambda}\Big{)}\bigg{)}=R_{3}.\hskip
102.43008pt\Box$

###  4.4. Proof of Theorem 3

Since $f=h+\overline{g}$ is a $(K,K^{{}^{\prime}})$-elliptic harmonic mapping,
$J_{f}(0)=1$ and $\lambda_{f}(z)<\Lambda$ for all $z\in\mathbb{D}$, it follows
from (4.1) and (4.2) that

$\displaystyle||a_{1}|-|b_{1}||$ $\displaystyle=$
$\displaystyle\lambda_{f}(0)\geq\frac{1}{\sqrt{K+K^{\prime}}},$
$\displaystyle\Lambda_{f}(z)$ $\displaystyle<$
$\displaystyle\Lambda_{1}:=\frac{K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}}}{2},\,z\in\mathbb{D}.$

Hence

$\displaystyle\frac{1}{\sqrt{K+K^{\prime}}}\leq\lambda_{f}(0)\leq|a_{1}|+|b_{1}|=\frac{J_{f}(0)}{\lambda_{f}(0)}=\frac{1}{\lambda_{f}(0)}\leq\sqrt{K+K^{\prime}}.$

Let $H(z)=h(z)+e^{i\alpha}g(z),\,0\leq\alpha\leq 2\pi$, then $H$ is
holomorphic in $\mathbb{D}$ and for $z\in\mathbb{D}$,

$|H^{\prime}(z)|=\bigg{|}\sum\limits_{n=1}^{\infty}{n(a_{n}+e^{i\alpha}b_{n})z^{n-1}}\bigg{|}\leq\Lambda_{f}(z)<\Lambda_{1}.$

By Lemma E, we have that

$\Big{|}\frac{n(a_{n}+e^{i\alpha}b_{n})}{\Lambda_{1}}\Big{|}\leq
1-\Big{|}\frac{a_{1}+e^{i\alpha}b_{1}}{\Lambda_{1}}\Big{|}^{2}\leq
1-\Big{|}\frac{|a_{1}|-|b_{1}|}{\Lambda_{1}}\Big{|}^{2}\leq
1-\Big{|}\frac{1}{\sqrt{K+K^{\prime}}\Lambda_{1}}\Big{|}^{2}.$

Then, it follows from the arbitrary of $\alpha$ that

$|a_{n}|+|b_{n}|\leq\frac{(K+K^{\prime})\Lambda_{1}^{2}-1}{n(K+K^{\prime})\Lambda_{1}}=\frac{(K+K^{\prime})(K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}})^{2}-4}{2n(K+K^{\prime})(K\Lambda+\sqrt{K^{2}\Lambda^{2}+4K^{\prime}})},\quad
n=2,3,\ldots.$

When $K=1,\,K^{\prime}=0$, the above estimates are sharp for all
$n=2,3,\ldots$, with the extremal functions

$None$
$f_{n}(z)=\Lambda^{2}z-\int_{0}^{z}{\frac{\Lambda^{3}-\Lambda}{\Lambda+z^{n-1}}dz}=z+\frac{\Lambda^{2}-1}{n\Lambda}z^{n}+\sum\limits_{k=2}^{\infty}\frac{(-1)^{k+1}(\Lambda^{2}-1)}{(kn-k+1)\Lambda^{k}}z^{kn-k+1}.$

###  4.5. Proof of Theorem 4

Since $f=h+\overline{g}$ is a sense-preserving harmonic mapping defined on
unit disk $\mathbb{D}$ such that $f(0)=0,h^{\prime}(0)-1=g^{\prime}(0)=0$, we
have that $|h^{\prime}(z)|>|g^{\prime}(z)|$ for $z\in\mathbb{D}$, this implies
that $w(z)=\frac{g^{\prime}(z)}{h^{\prime}(z)}$ is holomorphic in
$\mathbb{D}$, $w(0)=0$ and $|w(z)|<1$ for $z\in\mathbb{D}$.

By Schwarz’s lemma, we have that $|w(z)|\leq|z|$, i.e.,
$|g^{\prime}(z)|\leq|z||h^{\prime}(z)|$.

Since $\lambda_{f}(z)=|h^{\prime}(z)|-|g^{\prime}(z)|\leq\lambda$, we have
that

$|h^{\prime}(z)|\leq\frac{\lambda}{1-|z|}\mbox{ and
}|g^{\prime}(z)|\leq\frac{\lambda|z|}{1-|z|}\mbox{ for }z\in\mathbb{D}.$

So,

$\Lambda_{f}(z)=|h^{\prime}(z)|+|g^{\prime}(z)|\leq\frac{\lambda(1+|z|)}{1-|z|}.$

Let $P(z)=(\sqrt{2}+1)f(\frac{z}{\sqrt{2}+1}),\,z\in\mathbb{D}$, a simple
computation yields that $\lambda_{P}(0)=\lambda_{f}(0)=1$, and

$\Lambda_{P}(z)=\Lambda_{f}(\frac{z}{(\sqrt{2}+1)})\leq\frac{\lambda(1+|z/(\sqrt{2}+1)|)}{1-|z/(\sqrt{2}+1)|}\leq(\sqrt{2}+1)\lambda.$

Therefore, by Theorem A, we obtain that $P(z)$ is univalent in the disk
$\mathbb{D}_{r_{4}^{\prime}}$ and $F(\mathbb{D}_{r_{4}^{\prime}})$ contains a
schlicht disk $\mathbb{D}_{\sigma_{4}^{\prime}}$, where

$r_{4}^{\prime}=\frac{1}{(\sqrt{2}+1)\lambda},\,\sigma_{4}^{\prime}=(\sqrt{2}+1)\lambda+\Big{(}(\sqrt{2}+1)^{3}\lambda^{3}-(\sqrt{2}+1)\lambda\Big{)}\ln\Big{(}1-\frac{1}{(\sqrt{2}+1)^{2}\lambda^{2}}\Big{)}.$

Hence, $f$ is univalent in $\mathbb{D}_{r_{4}}$ and $f(\mathbb{D}_{r_{4}})$
contains a schlicht disk $\mathbb{D}_{\sigma_{4}}$, where

$\displaystyle r_{4}$ $\displaystyle=$
$\displaystyle\frac{r_{4}^{\prime}}{2}=\frac{1}{2(\sqrt{2}+1)\lambda},\quad$
$\displaystyle\sigma_{4}$ $\displaystyle=$
$\displaystyle\frac{\sigma_{4}^{\prime}}{2}=\frac{(\sqrt{2}+1)\lambda}{2}\bigg{\\{}1+\Big{(}(\sqrt{2}+1)^{2}\lambda^{2}-1\Big{)}\ln\Big{(}1-\frac{(\sqrt{2}-1)^{2}}{\lambda^{2}}\Big{)}\bigg{\\}}.\hskip
56.9055pt\Box$

###  4.6. Proof of Theorem 5

We adopt a similar method as for that used in [4, Theorem 2]. Let
$F(z)=\int_{0}^{z}{f_{z}(z)dz}$ for $z\in\mathbb{D}$, then it follows from
$f_{z}(z)$ is holomorphic in $\mathbb{D}$ that $F(z)$ is holomorphic in
$\mathbb{D}$. Since $f(0)=f_{z}(0)-1=f_{\overline{z}}(0)=0$ and $|h(z)|\leq M$
for $z\in\mathbb{D}$, we have that

$F(0)=0,\quad
F^{\prime}(0)=f_{z}(0)=1,\quad|F(z)|=\bigg{|}\int_{0}^{z}{f_{z}(z)dz}\bigg{|}=\bigg{|}\int_{0}^{z}{h^{\prime}(z)dz}\bigg{|}=|h(z)|\leq
M.$

By the classical Landau theorem, we have that $F(z)$ is univalent in
$\mathbb{D}_{\rho_{0}}$ and $F(\mathbb{D}_{\rho_{0}})$ contains a disk
$\mathbb{D}_{R_{0}}$, where $\rho_{0}$ is given by (1.1) and
$R_{0}=M\rho_{0}^{2}$.

Since $f(z)$ is a sense-preserving harmonic mapping in $\mathbb{D}$, we have
that $|f_{z}(z)|>|f_{\overline{z}}(z)|$ for all $z\in\mathbb{D}$, this implies
that $f_{z}(z)\neq 0$ for all $z\in\mathbb{D}$. Thus, the function
$\frac{\overline{f_{\overline{z}}}}{f_{z}}$ is holomorphic and has its modulus
less than 1 on $\mathbb{D}$.

Let $\Delta=F(\mathbb{D}_{\rho_{0}})$, then $F(z)$ is biholomorphic from
$\mathbb{D}_{\rho_{0}}$ onto $\Delta$. By applying Schwarz’s lemma to $F(z)$,
it is easy to verify that $\Delta\subset\mathbb{D}$, therefore, the
composition mapping $g(\zeta)=f\circ F^{-1}(\zeta)$ is a harmonic mapping in
$\Delta\,(\subset\mathbb{D})$, and

$g_{\zeta}=\frac{f_{z}(z)}{F^{\prime}(z)}=1,\quad|g_{\overline{\zeta}}|=\bigg{|}\frac{f_{\overline{z}}}{\overline{f_{z}}}\bigg{|}<1$

for $\zeta=F(z)\in\Delta$. Thus, for any $\zeta_{1},\zeta_{2}\in\Delta$ with
$\zeta_{1}\neq\zeta_{2}$, we have

$|g(\zeta_{1})-g(\zeta_{2})|\geq\bigg{|}\int_{\overline{\zeta_{1},\zeta_{2}}}{g_{\zeta}d\zeta}\bigg{|}-\bigg{|}\int_{\overline{\zeta_{1},\zeta_{2}}}{g_{\overline{\zeta}}d\overline{\zeta}}\bigg{|}>0.$

This shows that $g$ is univalent on $\Delta$ and, consequently, $f(z)$ is
univalent on $\mathbb{D}_{\rho_{0}}$.

Finally, we prove the sharpness of $\rho_{0}$. To this end, we consider the
harmonic mapping

$f(z)=h(z)=Mz\frac{1-Mz}{M-z}.$

Note that $f(0)=0,f_{\overline{z}}(z)\equiv 0$ and for $z\in\mathbb{D}$,

$|h(z)|=|f(z)|=\Big{|}Mz\frac{1-Mz}{M-z}\Big{|}<M,\quad
f^{\prime}(z)=M^{2}\frac{(z-M-\sqrt{M^{2}-1})(z-\rho_{0})}{(M-z)^{2}},$

we obtain that $f_{z}(0)-1=f_{\overline{z}}(0)=0$,
$|f_{z}(z)|>|f_{\overline{z}}(z)|\equiv 0$ for
$z\in\mathbb{D}\backslash\\{\rho_{0}\\}$, and $f^{\prime}(\rho_{0})=0$, which
proves the sharpness of $\rho_{0}$. $\Box$

###  4.7. Proof of Theorem 6

As in the proof of Theorem 5, the function $F(z):=\int_{0}^{z}{f_{z}(z)dz}$ is
holomorphic in $\mathbb{D}$. Since $f(0)=f_{z}(0)-1=f_{\overline{z}}(0)=0$ and
$|h^{\prime}(z)|<\Lambda$ for $z\in\mathbb{D}$, we have that

$F(0)=0,\quad
F^{\prime}(0)=f_{z}(0)=1,\quad|F^{\prime}(z)|=|f_{z}(z)|=|h^{\prime}(z)|<\Lambda\quad\mbox{
for }z\in\mathbb{D}.$

By Theorem A, we obtain that $F(z)$ is univalent in the disk
$\mathbb{D}_{r_{6}}$ with $r_{6}=\frac{1}{\Lambda}$.

Note that

$|F(z)|=\Big{|}\int_{0}^{z}h^{\prime}(t)dt\Big{|}\leq\int_{0}^{z}|h^{\prime}(t)||dt|\leq\Lambda|z|<\Lambda\quad\mbox{
for }z\in\mathbb{D}.$

By using an analogous proof of Theorem 5 and the extremal function
$f_{1}(z)=h(z)=\int_{0}^{z}{\Lambda\frac{1-\Lambda z}{\Lambda-z}dz}$, we may
prove that $f(z)$ is univalent on $\mathbb{D}_{r_{6}}$, and the radius $r_{6}$
is sharp. $\Box$

###  4.8. Proof of Theorem 7

Since $f$ is a $K$-quasiregular mapping in $\mathbb{D}$, we have that

$|f_{z}|+|f_{\overline{z}}|\leq K(|f_{z}|-|f_{\overline{z}}|),$

that is,

$|f_{\overline{z}}|\leq\frac{K-1}{K+1}|f_{z}|.$

Because $h(0)=0$ and $|h(z)|<M$ for $z\in\mathbb{D}$, by Cauchy’s estimates
theorem, we have

$|f_{z}(z)|=|h^{\prime}(z)|\leq\frac{M}{1-|z|}$

for $z\in\mathbb{D}$. Then

$\Lambda_{f}(z)=|f_{z}|+|f_{\overline{z}}|\leq\frac{2KM}{(K+1)(1-|z|)}.$

Let $P(z)=2f(\frac{z}{2}),\,z\in\mathbb{D}$. Then
$\lambda_{P}(0)=\lambda_{f}(0)=1$, and

$\Lambda_{P}(z)=\Lambda_{f}(\frac{z}{2})\leq\frac{2KM}{(K+1)(1-|z/2|)}\leq\frac{4KM}{K+1}.$

Therefore, by Theorem A, we obtain that $P(z)$ is univalent in the disk
$\mathbb{D}_{r_{7}^{\prime}}$ and $F(\mathbb{D}_{r_{7}^{\prime}})$ contains a
schlicht disk $\mathbb{D}_{\sigma_{7}^{\prime}}$, where

$r_{7}^{\prime}=\frac{K+1}{4KM},\quad\sigma_{7}^{\prime}=\frac{4KM}{K+1}+\Big{(}\Big{(}\frac{4KM}{K+1}\Big{)}^{3}-\frac{4KM}{K+1}\Big{)}\ln\bigg{(}1-\frac{(K+1)^{2}}{16K^{2}M^{2}}\bigg{)}.$

Hence, $f$ is univalent in $\mathbb{D}_{r_{7}}$ and $f(\mathbb{D}_{r_{7}})$
contains a schlicht disk $\mathbb{D}_{\sigma_{7}}$, where

$\displaystyle r_{7}$ $\displaystyle=$
$\displaystyle\frac{r_{7}^{\prime}}{2}=\frac{K+1}{8KM},\quad$
$\displaystyle\sigma_{7}$ $\displaystyle=$
$\displaystyle\frac{\sigma_{7}^{\prime}}{2}=\frac{2KM}{K+1}\bigg{\\{}1+\Big{(}\Big{(}\frac{4KM}{K+1}\Big{)}^{2}-1\Big{)}\ln\bigg{(}1-\frac{(K+1)^{2}}{16K^{2}M^{2}}\bigg{)}\bigg{\\}}.\hskip
56.9055pt\Box$

###  4.9. Proof of Theorem 8

Since $\lambda_{f}(0)=1$, we have that $||a_{1}|-|b_{1}||=\lambda_{f}(0)=1$.

For every complex number $\lambda$ with $|\lambda|=1$, let

$f_{\lambda}(z)=h(z)+\lambda g(z)=\sum\limits_{n=1}^{\infty}{(a_{n}+\lambda
b_{n})z^{n}}.$

Since
$|h^{\prime}(z)-h^{\prime}(0)|+|g^{\prime}(z)-g^{\prime}(0)|\leq\frac{(M^{2}-1)(2M|z|-|z|^{2})}{(M-|z|)^{2}}$
for $|z|<\rho_{0}$, we have

$\displaystyle\Big{|}\sum\limits_{n=2}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}\Big{|}$ $\displaystyle\leq$
$\displaystyle|h^{\prime}(z)-h^{\prime}(0)|+|g^{\prime}(z)-g^{\prime}(0)|$
$\displaystyle\leq$
$\displaystyle\frac{(M^{2}-1)(2M|z|-|z|^{2})}{(M-|z|)^{2}}\vskip 6.0pt plus
2.0pt minus 2.0pt$ $\displaystyle<$
$\displaystyle\frac{(M^{2}-1)(2M\rho_{0}-\rho_{0}^{2})}{(M-\rho_{0})^{2}}$

for $|z|<\rho_{0}$.

We first prove $f_{\lambda}(z)$ is univalent in $\mathbb{D}_{\rho_{0}}$. To
this end, for any $z_{1}\neq z_{2}\in\mathbb{D}_{\rho_{0}}$, we have

$\displaystyle|f_{\lambda}(z_{1})-f_{\lambda}(z_{2})|$ $\displaystyle=$
$\displaystyle\bigg{|}\int_{[z_{1},z_{2}]}{\sum\limits_{n=1}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}}dz\bigg{|}\vskip 6.0pt plus 2.0pt minus 2.0pt$
$\displaystyle\geq$ $\displaystyle\bigg{|}\int_{[z_{1},z_{2}]}{(a_{1}+\lambda
b_{1})dz}\bigg{|}-\bigg{|}\int_{[z_{1},z_{2}]}{\sum\limits_{n=2}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}dz}\bigg{|}\vskip 6.0pt plus 2.0pt minus 2.0pt$
$\displaystyle\geq$
$\displaystyle|z_{1}-z_{2}|||a_{1}|-|b_{1}||-\int_{[z_{1},z_{2}]}{\bigg{|}\sum\limits_{n=2}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}\bigg{|}|dz|}\vskip 6.0pt plus 2.0pt minus 2.0pt$
$\displaystyle>$
$\displaystyle|z_{1}-z_{2}|-\int_{[z_{1},z_{2}]}{\frac{(M^{2}-1)(2M\rho_{0}-\rho_{0}^{2})}{(M-\rho_{0})^{2}}|dz|}\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle\geq$
$\displaystyle|z_{1}-z_{2}|\bigg{(}1-\frac{(M^{2}-1)(2M\rho_{0}-\rho_{0}^{2})}{(M-\rho_{0})^{2}}\bigg{)}=0.$

Thus, we have that $f_{\lambda}(z_{1})\neq f_{\lambda}(z_{2})$, which proves
the univalence of $f_{\lambda}(z)$ in $\mathbb{D}_{\rho_{0}}$.

Next, noticing that $f(0)=0$, for any
$z^{\prime}\in\partial\mathbb{D}_{\rho_{0}}$, we have

$\displaystyle|f_{\lambda}(z^{\prime})|$ $\displaystyle=$
$\displaystyle|\Sigma_{n=1}^{\infty}{(a_{n}+\lambda b_{n})z^{\prime
n}}|\geq\lambda_{f}(0)\rho_{0}-|\Sigma_{n=2}^{\infty}{(a_{n}+\lambda
b_{n})z^{\prime n}}|\vskip 6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle\geq$
$\displaystyle\rho_{0}-\int_{[0,z^{\prime}]}{|\Sigma_{n=2}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}|}}|dz|\vskip 6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle\geq$
$\displaystyle\rho_{0}-\int_{[0,z^{\prime}]}{\frac{(M^{2}-1)(2M|z|-|z|^{2})}{(M-|z|)^{2}}|dz|}\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle=$
$\displaystyle\rho_{0}-\int_{0}^{1}{\frac{(M^{2}-1)(2Mt\rho_{0}-t^{2}\rho_{0}^{2})}{(M-t\rho_{0})^{2}}d(t\rho_{0})}\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle=$
$\displaystyle\rho_{0}-\frac{(M^{2}-1)\rho_{0}^{2}}{M-\rho_{0}}=M\rho_{0}\frac{1-M\rho_{0}}{M-\rho_{0}}=M\rho_{0}^{2}=R_{0}.$

Hence, $f_{\lambda}(\mathbb{D}_{\rho_{0}})$ contains a schlicht disk
$\mathbb{D}_{R_{0}}$. By Lemma F, we obtain that $f(z)$ is univalent on
$\mathbb{D}_{\rho_{0}}$ and $f(\mathbb{D}_{\rho_{0}})$ contains a schlicht
disk $\mathbb{D}_{R_{0}}$.

Finally, we prove the sharpness of $\rho_{0}$ and $R_{0}$. We consider the
harmonic mapping $f_{0}(z)=Mz\frac{1-Mz}{M-z}$. Note that $f_{0}(0)=0$,
$\lambda_{f_{0}}(0)=1$ and

$f_{0}(z)=Mz\frac{1-Mz}{M-z}=z-\sum\limits_{n=2}^{\infty}{\frac{M^{2}-1}{M^{n-1}}z^{n}},z\in\mathbb{D},$

direct computation yields

$\bigg{|}\sum\limits_{n=2}^{\infty}{n\frac{M^{2}-1}{M^{n-1}}z^{n-1}}\bigg{|}\leq\sum\limits_{n=2}^{\infty}{n\frac{M^{2}-1}{M^{n-1}}|z|^{n-1}}=\frac{(M^{2}-1)(2M|z|-|z|^{2})}{(M-|z|)^{2}},\quad
0\leq r<1,$

thus it follows from the above results that $f_{0}(z)$ is univalent in
$\mathbb{D}_{\rho_{0}}$ and $f(\mathbb{D}_{\rho_{0}})$ contains a schlicht
disk $\mathbb{D}_{R_{0}}$.

Since

$f^{\prime}_{0}(z)=M^{2}\frac{(z-M-\sqrt{M^{2}-1})(z-\rho_{0})}{(M-z)^{2}},$

we obtain that $f^{\prime}_{0}(\rho_{0})=0$, which proves the sharpness of
$\rho_{0}$.

Noticing that $f_{0}(0)=0$,
$z^{\prime}=\rho_{0}\in\partial\mathbb{D}_{\rho_{0}}$, we have

$|f_{0}(z^{\prime})-f_{0}(0)|=|f_{0}(z^{\prime})|=M\rho_{0}\frac{1-M\rho_{0}}{M-\rho_{0}}=M\rho_{0}^{2}=R_{0}.$

Hence, the radius $R_{0}$ is also sharp. This completes the proof. $\Box$

###  4.10. Proof of Theorem 9

Let $z=re^{i\theta},\,r\in[0,1)$, $\theta\in[0,2\pi)$, then

$f(re^{i\theta})=\sum\limits_{n=1}^{\infty}{a_{n}r^{n}e^{in\theta}}+\sum\limits_{n=1}^{\infty}{\overline{b_{n}}r^{n}e^{-in\theta}}.$

Since $|zf_{z}(z)+\overline{z}f_{\overline{z}}(z)|<\Lambda$ for all
$z\in\mathbb{D}$, by Parseval’s identity, we have that

$\sum\limits_{n=1}^{\infty}{n^{2}(|a_{n}|^{2}+|b_{n}|^{2})r^{2n}}=\frac{1}{2\pi}\int_{0}^{2\pi}{|re^{i\theta}f_{z}(re^{i\theta})+\overline{re^{i\theta}}f_{\overline{z}}(re^{i\theta})|^{2}d\theta}\leq\Lambda^{2}.$

Notice that $\lambda_{f}(0)=||a_{1}|-|b_{1}||=1$, let $r\rightarrow 1^{-}$, we
have that $\Lambda\geq 1$, and

$\sum\limits_{n=2}^{\infty}{n^{2}(|a_{n}|^{2}+|b_{n}|^{2})}=\sum\limits_{n=1}^{\infty}{n^{2}(|a_{n}|^{2}+|b_{n}|^{2})}-(|a_{1}|^{2}+|b_{1}|^{2})\leq\Lambda^{2}-1.$

Since $a_{n}b_{n}=0$ for $n=2,3,\ldots$, we have that

$\displaystyle\sum\limits_{n=2}^{\infty}{n^{2}(|a_{n}|+|b_{n}|)^{2}}=\sum\limits_{n=2}^{\infty}{n^{2}(|a_{n}|^{2}+|b_{n}|^{2})}\leq\Lambda^{2}-1.$

For any complex number $\lambda$ with $|\lambda|=1$, let

$f_{\lambda}(z)=h(z)+\lambda
g(z)=\int_{0}^{z}{\sum\limits_{n=1}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}dz}.$

We first prove $f_{\lambda}(z)$ is univalent in the $\mathbb{D}_{r_{9}}$ with
$r_{9}=\frac{1}{\Lambda}$. To this end, for any $z_{1},z_{2}\in\mathbb{D}_{r}$
with $z_{1}\neq z_{2}$ and $0<r<r_{9}$, we have

$\displaystyle|f_{\lambda}(z_{1})-f_{\lambda}(z_{2})|$ $\displaystyle=$
$\displaystyle\bigg{|}\int_{[z_{1},z_{2}]}{\sum\limits_{n=1}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}}dz\bigg{|}\vskip 6.0pt plus 2.0pt minus 2.0pt$
$\displaystyle\geq$ $\displaystyle\bigg{|}\int_{[z_{1},z_{2}]}{(a_{1}+\lambda
b_{1})dz}\bigg{|}-\bigg{|}\int_{[z_{1},z_{2}]}{\sum\limits_{n=2}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}dz}\bigg{|}\vskip 6.0pt plus 2.0pt minus 2.0pt$
$\displaystyle\geq$
$\displaystyle|z_{1}-z_{2}|||a_{1}|-|b_{1}||-\int_{[z_{1},z_{2}]}{\bigg{|}\sum\limits_{n=2}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}\bigg{|}|dz|}$ $\displaystyle\geq$
$\displaystyle|z_{1}-z_{2}|\bigg{(}1-\sum\limits_{n=2}^{\infty}{n(|a_{n}|+|b_{n}|)r^{n-1}}\bigg{)}\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle\geq$
$\displaystyle|z_{1}-z_{2}|\bigg{(}1-\sqrt{\sum\limits_{n=2}^{\infty}{n^{2}(|a_{n}|+|b_{n}|)^{2}}\sum\limits_{n=2}^{\infty}{r^{2n-2}}}\bigg{)}\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle\geq$
$\displaystyle|z_{1}-z_{2}|\bigg{(}1-\sqrt{(\Lambda^{2}-1)\sum\limits_{n=2}^{\infty}{r^{2n-2}}}\bigg{)}\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle=$
$\displaystyle|z_{1}-z_{2}|\bigg{(}1-\frac{r\sqrt{\Lambda^{2}-1}}{\sqrt{1-r^{2}}}\bigg{)}>0,$

which implies that $f_{\lambda}(z_{1})\neq f_{\lambda}(z_{2})$, this proves
the univalence of $f_{\lambda}(z)$ in the $\mathbb{D}_{r_{9}}$.

Next, noticing that $f_{\lambda}(0)=0$, for any
$z^{\prime}\in\partial\mathbb{D}_{r_{9}}$, we have

$\displaystyle|f_{\lambda}(z^{{}^{\prime}})|$ $\displaystyle=$
$\displaystyle\bigg{|}\int_{[0,z^{\prime}]}{\sum\limits_{n=1}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}}dz\bigg{|}\vskip 6.0pt plus 2.0pt minus 2.0pt$
$\displaystyle\geq$
$\displaystyle|z^{\prime}|||a_{1}|-|b_{1}||-\int_{[0,z^{\prime}]}{\bigg{|}\sum\limits_{n=2}^{\infty}{n(a_{n}+\lambda
b_{n})z^{n-1}}\bigg{|}|dz|}\vskip 6.0pt plus 2.0pt minus 2.0pt$
$\displaystyle\geq$ $\displaystyle
r_{9}-\int_{0}^{r_{9}}{\Sigma_{n=2}^{\infty}{n(|a_{n}|+|b_{n}|)r^{n-1}dr}}\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle\geq$ $\displaystyle
r_{9}-\int_{0}^{r_{9}}{\sqrt{\sum\limits_{n=2}^{\infty}{n^{2}(|a_{n}|+|b_{n}|)^{2}}\sum\limits_{n=2}^{\infty}{r^{2n-2}}}\,dr}\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle\geq$ $\displaystyle
r_{9}-\sqrt{\Lambda^{2}-1}\int_{0}^{r_{9}}\frac{r}{\sqrt{1-r^{2}}}\,dr\vskip
6.0pt plus 2.0pt minus 2.0pt$ $\displaystyle=$ $\displaystyle
r_{9}+\sqrt{\Lambda^{2}-1}(\sqrt{1-r_{9}^{2}}-1)=\Lambda-\sqrt{\Lambda^{2}-1}=\sigma_{9}.$

Hence, $f_{\lambda}(\mathbb{D}_{r_{9}})$ contains a schlicht disk
$\mathbb{D}_{\sigma_{9}}$. By Lemma F, we have that $f(z)$ is univalent in
$\mathbb{D}_{r_{9}}$ and $f(\mathbb{D}_{r_{9}})$ contains a schlicht disk
$\mathbb{D}_{\sigma_{9}}$.

Now, we prove the sharpness of $r_{9}$. We consider the harmonic mapping

$f_{1}(z)=\int_{0}^{z}{\Lambda\frac{1-\Lambda z}{\Lambda-z}dz}.$

It is easy to verify that $f_{1}(z)$ satisfy all of the hypothesis of Theorem
9. Then it follows from the above results that the harmonic mapping $f_{1}(z)$
is univalent in $\mathbb{D}_{r_{9}}$ and $f(\mathbb{D}_{r_{9}})$ contains a
schlicht disk $\mathbb{D}_{\sigma_{9}}$. By $f_{1}^{\prime}(r_{9})=0$, we get
that the radius $r_{9}$ is sharp.

Finally, when $\Lambda=1$, it is obvious that $\sigma_{9}=r_{9}=1$ are sharp
with the extremal mapping $f_{2}(z)=z$. $\Box$

### Acknowledgments

This research is supported by Guangdong Natural Science Foundations (Grant No.
2021A1515010058).

### Conflict of Interests

The authors declare that they have no conflict of interest, regarding the
publication of this paper.

### Data Availability Statement

The authors declare that this research is purely theoretical and does not
associate with any data.

## References

  * [1] A. Aleman and A. Constantin, Harmonic maps and ideal fluid flows, Arch. Ration. Mech. Anal., 204 (2012), 479–513;
  * [2] V. Allu and R. Kumar, Landau-Bloch type theorem for elliptic and quasiregular harmonic mappings, _J. Math. Anal. Appl._ (2024), 128215, doi: https://doi.org/10.1016/j.jmaa.2024.128215.
  * [3] H.H. Chen, P.M.Gauthier and W. Hengartner, Bloch constants for planar harmonic mappings, _Proc. Amer. Math. Soc._ , 128, 3231–3240 (2000)
  * [4] H.H. Chen, P.M.Gauthier, The Landau theorem and Bloch theorem for planar harmonic and pluriharmonic mappings, _Proc Amer Math Soc_ , 139(2), 583–595 (2011)
  * [5] S. Chen, S. Ponnusamy and A. Rasila, Coefficient estimates, Landau’s theorem and Lipschitz-type spaces on planar harmonic mappings, _J Aust Math Soc_ , 96(2), 198–215 (2014)
  * [6] S. Chen, S. Ponnusamy and X. Wang, Properties of some classes of planar harmonic and planar biharmonic mappings, _Complex Anal Oper Theory_ , 2011, 5, 901–916 (2011)
  * [7] S. Chen, S. Ponnusamy and X. Wang, Coefficient estimates and Landau-Bloch’s constant for planar harmonic mappings, _Bulletin of the Malaysian Mathematical Sciences Society, Second Series_ , 34(2), 255–265 (2011)
  * [8] S. Chen and S. Ponnusamy, On certain quasiconformal and elliptic mappings, _J. Math. Anal. Appl._ , 486, 1–16 (2020)
  * [9] S. Chen, S. Ponnusamy and X. Wang, Remarks on norm estimates of the partial derivabtives for harmonic mappings and harmonic quasiregular mappings, _J. Goem. Anal._ , 31, 11051–11060 (2021)
  * [10] O. Constantin and M. J. Martin, A harmonic maps approach to fluid flows, Math. Ann., 369 (2017), 1–16.
  * [11] M. Dorff and M. Nowak, Landau’s theorem for planar harmonic mappings, _Comput Meth Funct Theory_ , 4(1), 151–158 (2000)
  * [12] A. Grigoryan, Landau and Bloch theorems for harmonic mappings, _Complex Variable Theory Appl_ , 51(1): 81–87 (2006)
  * [13] P.L. Duren, Harmonic mapping in the plan, _Cambridge University Press_ , (2004)
  * [14] R. Finn and J. Serrin, On the Holder continuity of quasiconformal and elliptic mappings, _Trans. Amer. Math. Soc._ , 89, 1–15 (1958)
  * [15] I. Graham and G. Kohr, Geometric Function Theory in One and Higher Dimensions, _Marcel Dekker Inc, New York._ , (2003)
  * [16] X.-Z. Huang, Sharp estimate on univalent radius for planar harmonic mappings with bounded Fréchet derivative (in Chinese), _Sci Sin Math_ , 44(6), 685–692 (2014)
  * [17] J. Kuang, Applied Inequalities (3nd ed.), _Shandong Science and Technology Press, Jinan_ , (2004).
  * [18] H. Lewy, On the non-vanishing of the Jacobian in certain one-to-one mappings, _Bull. Amer. Math. Soc._ , 42, 689–692 (1936)
  * [19] M.-S. Liu, Landau’s theorem for planar harmonic mappings, _Comput. Math. with Appl._ , 57(7), 1142–1146 (2009)
  * [20] M.-S. Liu, Estimates on Bloch constants for planar harmonic mappings, _Sci. China Ser. A-Math._ , 52(1), 87–93 (2009)
  * [21] M.-S. Liu, L.F. Luo and X. Luo, Landau-Bloch type theorems for strongly bounded harmonic mappings, _Monatsh Math_ , 191(1), 175–185 (2020)
  * [22] M.-S. Liu and H.H. Chen, The Landau-Bloch type theorems for planar harmonic mappings with bounded dilation, _J. Math. Anal. Appl._ , 468(2), 1066–1081 (2018)
  * [23] L. Nirenberg, On nonlinear elliptic partial differential equations and Holder continuity, _Commun. Pure. Appl. Math._ , 6, 103–156 (1953)
  * [24] S. Rickman, Quasiregular Mappings, _Springer-Verlag, Berlin_ , (1993)
  * [25] X.Q. Xia, X.Z. Huang, Estimates on Bloch constants for planar bounded harmonic mappings (in Chinese), _Chin. Ann. Math. Ser._ , 31A(6), 769–776 (2010)