1 Introduction

Let \(H({\mathbb {D}})\) denote the class of holomorphic functions in the unit disk \({\mathbb {D}}=\{z:\, |z|<1\}\). Given a function \(f\in H({\mathbb {D}})\) define \(B_f\) to be the least upper bound of all numbers \(b>0\) such that there exists a number \(z_0\in {\mathbb {C}}\) and a region \(\Omega \subseteq {\mathbb {D}}\) which is univalently mapped onto \(\{z\in {\mathbb {C}}: |z-z_0|<b\}\) by \(f\). The Bloch’s constant \(B\) is defined as (cf. [2])

$$\begin{aligned} B:=\inf \{B_f: f\in H({\mathbb {D}})\ \text{ and }\ f'(0)=1\}. \end{aligned}$$

The best known bounds for \(B\) at present are (cf. [1, 3])

$$\begin{aligned} \frac{\sqrt{3}}{4}+2\times 10^{-4}\le B \le \frac{\sqrt{3}-1}{2}\cdot \frac{\Gamma (1/3)\Gamma (11/12)}{\Gamma (1/4)}, \end{aligned}$$

where \(\Gamma \) is the Gamma function. However, the exact value of \(B\) is still unknown.

Let \(H_M({\mathbb {D}})\) denote the class of functions \(f\in H({\mathbb {D}})\) with \(|f(z)|<M\) for \(z\in {\mathbb {D}}\). The classical Landau theorem states that if \(f\in H_M({\mathbb {D}})\) with \(f(0)=f'(0)-1=0\), then \(f\) is univalent in the disk \(|z|<\rho _0\) with \(\rho _0=\frac{1}{M+\sqrt{M^2-1}}\) and \(f(|z|<\rho _0)\) contains a disk \(|w|<\sigma _0\) with \(\sigma _0=M\rho _0^2\). This result is sharp (cf. [8]).

Define complex derivatives of \(w(z)\) as follows:

$$\begin{aligned} w_z:=\frac{1}{2}\left( w_x-iw_y\right) \ \ ~\text{ and }~\ \ w_{\bar{z}}:=\frac{1}{2}\left( w_x+iw_y\right) , \end{aligned}$$
(1.1)

where \(z=x+iy\). A complex-valued function \(w\) on \({\mathbb {D}}\) is harmonic if it is twice continuously differentiable and satisfies Laplace’s equation:

$$\begin{aligned} \Delta w=4w_{z\bar{z}}=\frac{\partial ^2 w}{\partial x^2}+\frac{\partial ^2 w}{\partial y^2}=0\ \ \text{ for }\ \ z\in {\mathbb {D}}. \end{aligned}$$

We refer to the book of Duren [5, 6] for good setting of harmonic mappings. Let

$$\begin{aligned} \Lambda _w(z)=\max \limits _{0\le \alpha \le \pi }|e^{i\alpha }w_z(z)+e^{-i\alpha }w_{\bar{z}}(z)|=|w_z(z)|+|w_{\bar{z}}(z)| \end{aligned}$$

and

$$\begin{aligned} \lambda _w(z)=\min \limits _{0\le \alpha \le \pi }|e^{i\alpha }w_z(z)+e^{-i\alpha }w_{\bar{z}}(z)|=||w_z(z)|-|w_{\bar{z}}(z)||. \end{aligned}$$

It is well known that \(w(z)\) is locally univalent and sense-preserving in \({\mathbb {D}}\) if and only if its Jacobian satisfies the following condition.

$$\begin{aligned} J_w(z)=|w_z(z)|^2-|w_{\bar{z}}(z)|^2>0\ \quad \text{ for }\ z \in {\mathbb {D}}. \end{aligned}$$

We know from Poisson formula that every bounded harmonic mapping \(w\) defined in \({\mathbb {D}}\) has the representation

$$\begin{aligned} w(z)=P[F](z)=\int \limits _0^{2\pi } P(r,t-\varphi )F(e^{it})\,dt, \quad z=re^{i\varphi }\in {\mathbb {D}}, \end{aligned}$$
(1.2)

where \(F\) is the boundary function defined on the unit circle \({\mathbb {T}}:=\{z:\, |z|=1\}\) and

$$\begin{aligned} P(r,t-\varphi )=\frac{1}{2\pi }\frac{1-r^2}{1-2r\cos (t-\varphi )+r^2}, \end{aligned}$$

denote the Poisson kernel. In what follows we write \(F(t)\) instead of \(F(e^{it})\) for the boundary function (cf. [9, 10]).

For harmonic mappings in \({\mathbb {D}}\), under suitable restriction we can obtain its Bloch and Landau theorems. For \(r>0\), let \({\mathbb {D}}_r=\{z\in {\mathbb {C}}:\ |z|<r\}\). In [4], Chen et al. proved the Landau theorem for harmonic mappings as follows.

Theorem A

Let \(w\) be a harmonic mapping of \({\mathbb {D}}\) satisfies \(w(0)=w_{\bar{z}}(0)=w_z(0)-1=0\) and \(|w(z)|<M\) for \(z\in {\mathbb {D}}\). Then \(w\) is univalent in the disk \({\mathbb {D}}_{r_1}\), where

$$\begin{aligned} r_1=\frac{\pi ^2}{16mM}\approx \frac{1}{11.105M} \end{aligned}$$
(1.3)

and \(w({\mathbb {D}}_{r_1})\) contains a schlicht disk \({\mathbb {D}}_{R_1}\), where

$$\begin{aligned} R_1=\frac{r_1}{2}\approx \frac{1}{22.21M}. \end{aligned}$$
(1.4)

Here \(m\approx 6.85\) is the minimum of the function \((3-r^2)/(r(1-r^2))\) for \(0<r<1\).

By using sharp coefficients estimate, the authors in [7] improved a version of Landau theorem for the class of bounded harmonic mappings.

Theorem B

Let \(w\) be a harmonic mapping of \({\mathbb {D}}\) satisfies \(w(0)=w_{\bar{z}}(0)=w_z(0)-1=0\) and \(|w(z)|<M\) for \(z\in {\mathbb {D}}\). Then \(w\) is close-to-convex (univalent) in the disk \({\mathbb {D}}_{r_2}\), where

$$\begin{aligned} r_2=1-\sqrt{\frac{4M}{4M+\pi }} \end{aligned}$$
(1.5)

and \(w({\mathbb {D}}_{r_2})\) contains a schlicht disk \({\mathbb {D}}_{R_2}\), where

$$\begin{aligned} R_2=r_2-\frac{4M}{\pi }\frac{r_2^2}{1-r_2}. \end{aligned}$$
(1.6)

By using Poisson formula, we obtain better estimates on Bloch constants for planar harmonic mappings.

2 Auxiliary Results

Lemma 2.1

Let \(z=\rho e^{i\theta }\in {\mathbb {D}}\). Then

$$\begin{aligned} \frac{1}{\pi }\int \limits _0^{2\pi } \frac{|z||2-e^{-it}z|}{|e^{it}-z|^2}dt =\frac{2\rho \sqrt{2-\rho ^2}}{1-\rho ^2}, \end{aligned}$$
(2.1)

where \(0\le \rho <1\) and \(0\le \theta \le 2\pi \).

Proof

Let \(\zeta =e^{it}\in {\mathbb {T}}\). According to the residue theorem, we see that

$$\begin{aligned} \frac{1}{\pi }\int \limits _0^{2\pi } \frac{|z||2-e^{-it}z|}{|e^{it}-z|^2}dt= & {} \frac{1}{\pi }\int \limits _0^{2\pi } \frac{\rho \sqrt{4-4\rho \cos (\theta -t)+\rho ^2}}{1-2\rho \cos (\theta -t)+\rho ^2}dt\\= & {} \frac{1}{\pi i}\oint \limits _{|\zeta |=1} \frac{\rho \sqrt{(4+\rho ^2)-2\rho e^{-i\theta }(e^{2i\theta }+\zeta ^2)/\zeta }}{(1+\rho ^2)\zeta -\rho e^{-i\theta }(e^{2i\theta }+\zeta ^2)}d\zeta \\= & {} \frac{1}{\pi i}\oint \limits _{|\zeta |=1} \frac{\rho \sqrt{(4+\rho ^2)-2\rho e^{-i\theta }(e^{2i\theta }+\zeta ^2)/\zeta }}{-\rho e^{-i\theta }[\zeta -\rho e^{i\theta }][\zeta -(1/\rho )e^{i\theta }]}d\zeta \\= & {} \frac{2\rho \sqrt{2-\rho ^2}}{1-\rho ^2}. \end{aligned}$$

This completes the proof. \(\square \)

Lemma 2.2

Let \(0\le \alpha \le 2\pi \) and \(w=P[F]\) be a harmonic mapping in \({\mathbb {D}}\) with the boundary function \(F\). Then

$$\begin{aligned}&e^{i\alpha }w_z(z)+e^{-i\alpha }w_{\bar{z}}(z)\nonumber \\&\quad =\,e^{i\alpha }w_z(0)+e^{-i\alpha }w_{\bar{z}}(0)+ \frac{1}{\pi }\int \limits _0^{2\pi }\mathrm{Re\,}\left\{ e^{i\alpha }\frac{2z-e^{-it}z^2}{(e^{it}-z)^2}\right\} F(t)dt, \end{aligned}$$
(2.2)

where \(z=re^{i\theta }\in {\mathbb {D}}\).

Proof

For each \(z=re^{i\theta }\in {\mathbb {D}}\), it follows from (1.2) that

$$\begin{aligned} w(z)=P[F](z) = \frac{1}{2\pi }\int \limits _0^{2\pi } \frac{e^{it}F(t)}{e^{it}-z}\,dt+ \frac{1}{2\pi }\overline{\int \limits _0^{2\pi }\frac{z \overline{F(t)}}{e^{it}-z}\,dt}. \end{aligned}$$
(2.3)

Then \(w_z(z)=\frac{1}{2\pi }\int \limits _0^{2\pi } \frac{e^{it}F(t)}{(e^{it}-z)^2}dt\) and \(w_{\bar{z}}(z)=\frac{1}{2\pi }\int \limits _0^{2\pi } \frac{e^{-it}F(t)}{(e^{-it}-\bar{z})^2}dt\). Let \(\alpha \in [0, 2\pi ]\) be an arbitrary constant. Then

$$\begin{aligned} e^{i\alpha }w_z(z)= & {} e^{i\alpha }w_z(0) +\frac{e^{i\alpha }}{2\pi }\int \limits _0^{2\pi } \left[ \frac{e^{it}}{(e^{it}-z)^2}-\frac{1}{e^{it}}\right] F(t)dt \\= & {} e^{i\alpha }w_z(0)+\frac{e^{i\alpha }}{2\pi }\int \limits _0^{2\pi } \frac{2z-z^2e^{-it}}{(e^{it}-z)^2}F(t)dt. \end{aligned}$$

Similarly, we can obtain

$$\begin{aligned} e^{-i\alpha }w_{\bar{z}}(z)=e^{-i\alpha }w_{\bar{z}}(0) +\frac{e^{-i\alpha }}{2\pi }\int \limits _0^{2\pi } \frac{2\bar{z}-\bar{z}^2e^{it}}{(e^{-it}-\bar{z})^2}F(t)dt. \end{aligned}$$

Equality (2.2) holds directly from the above two equalities. \(\square \)

3 Main Results

Theorem 3.1

Let \(w=P[F]\) be a harmonic mapping of \({\mathbb {D}}\) satisfies \(w(0)=w_{\bar{z}}(0)=w_z(0)-1=0\) and \(|w(z)|< M\) for \(z\in {\mathbb {D}}\), where \(M>1\) and \(F\) is the boundary function. Then \(w\) is univalent in the disk \({\mathbb {D}}_{r_0}\), where

$$\begin{aligned} r_0=\sqrt{1-\frac{2M}{\sqrt{1+4M^2}}} \end{aligned}$$
(3.1)

is the root of the equation \(1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}=0\) and \(w({\mathbb {D}}_{r_0})\) contains a schlicht disk \({\mathbb {D}}_{R_0}\), where

$$\begin{aligned} R_0 =r_0-\log \frac{\sqrt{2-r_0^2}+1}{\sqrt{2-r_0^2}-1} +2\sqrt{2-r_0^2}+2\left[ \log (\sqrt{2}+1)-\sqrt{2}\right] . \end{aligned}$$
(3.2)

Proof

For \(0<r<1\), take \(z_1\), \(z_2\in {\mathbb {D}}_r\). Let \(\ell : z(x)=z_1+(z_2-z_1)x=\rho (x)e^{i\theta (x)}\) be the segment line of \(z_1\) and \(z_2\), where \(0\le x\le 1\). Then

$$\begin{aligned} |w(z_1)-w(z_2)|= & {} \left| \int \limits _{\ell }\left[ w_z(z(x))z'(x)+w_{\bar{z}}(z(x)) \overline{z'(x)}\right] dx\right| \\= & {} |z_1-z_2|\left| \int \limits _0^1\left[ e^{i\alpha }w_z(z(x)) +e^{-i\alpha }w_{\bar{z}}(z(x))\right] dx\right| , \end{aligned}$$

where \(\alpha =\arg (z_1-z_2)\in [0, 2\pi ]\). Since \(0\le \rho (x)<r<1\) and the function \(\frac{s\sqrt{2-s^2}}{1-s^2}\) is an increasing function for \(0\le s<1\), we see that

$$\begin{aligned} \int \limits _0^1\frac{\rho (x)\sqrt{2-\rho ^2(x)}}{1-\rho ^2(x)}dx\le \frac{r\sqrt{2-r^2}}{1-r^2}. \end{aligned}$$

Applying \(w_{\bar{z}}(0)=w_z(0)-1=0\) together with (2.2) and Lemma 2.1 we see that

$$\begin{aligned}&\left| \int \limits _0^1 \left[ e^{i\alpha }w_z(z(x)) +e^{-i\alpha }w_{\bar{z}}(z(x))\right] dx\right| \\&\quad =\, \left| e^{i\alpha } +\frac{1}{\pi }\int \limits _0^1 \left( \int \limits _0^{2\pi }\mathrm{Re\,}\left[ e^{i\alpha }\frac{2z(x)-e^{-it}z^2(x)}{(e^{it}-z(x))^2} \right] F(t)dt\right) dx\right| \\&\quad \ge 1-\frac{M}{\pi }\int \limits _0^1 \left( \int \limits _0^{2\pi } \frac{|z(x)||2-e^{-it}z(x)|}{|e^{it}-z(x)|^2}dt\right) dx\\&\quad =1-M\int \limits _0^1\frac{2\rho (x)\sqrt{2-\rho ^2(x)}}{1-\rho ^2(x)}dx\\&\quad \ge 1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}. \end{aligned}$$

Hence,

$$\begin{aligned} |w(z_1)-w(z_2)|\ge |z_1-z_2| \left( 1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}\right) . \end{aligned}$$
(3.3)

Let \(1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}=0\). Then \(r_0=\sqrt{1-\frac{2M}{\sqrt{1+4M^2}}}\). This shows that \(w(z)\) is univalent in the disk \({\mathbb {D}}_{r_0}\). Let \(z=r_0e^{i\varphi }\in \partial {\mathbb {D}}_{r_0}\). Then according to (2.2) and Lemma 2.1, we have

$$\begin{aligned} |w(r_0e^{i\varphi })|= & {} \left| \int \limits _{0}^{r_0}\left[ w_z(xe^{i\varphi })e^{i\varphi } +w_{\bar{z}}(xe^{i\varphi })e^{-i\varphi } \right] dx\right| \\\ge & {} r_0-\int \limits _0^{r_0}\frac{2x\sqrt{2-x^2}}{1-x^2}dx\\= & {} r_0-\log \frac{\sqrt{2-r_0^2}+1}{\sqrt{2-r_0^2}-1} +2\sqrt{2-r_0^2}+2 \left[ \log (\sqrt{2}+1)-\sqrt{2}\right] \\= & {} R_0. \end{aligned}$$

We see that \(w({\mathbb {D}}_{r_0})\) contains the schlicht disk \({\mathbb {D}}_{R_0}\).

The proof is completed. \(\square \)

Remark 3.2

For \(1<M\le 5.07\), we see that \(r_2<r_0\). Let

$$\begin{aligned} r_0^*=\left\{ \begin{array} {r@{\ }l} r_0 \ \ &{} \text{ if }\ \ 1<M\le 5.07\\ \\ r_2 \ \ &{} \text{ if }\ \ M>5.07. \end{array}\right. \end{aligned}$$

Then \(w(z)\) is univalent in the disk \(D_{r_0^*}\), where \(r_0^{*}\ge r_2>r_1\). This shows that our Theorem 3.1 has improved the former results.

Theorem 3.3

Let \(w=P[F]\) be a harmonic mapping of \({\mathbb {D}}\) satisfies \(w(0)=w_{\bar{z}}(0)=w_z(0)-1=0\) and \(|w(z)|<M\) for \(z\in {\mathbb {D}}\), where \(M>1\) and \(F\) is the boundary function. Then

$$\begin{aligned} \Lambda _w(re^{i\theta })\le 1+\frac{2Mr\sqrt{2-r^2}}{1-r^2} \quad \text{ for }\ \ z=re^{i\theta }\in {\mathbb {D}}\end{aligned}$$
(3.4)

and

$$\begin{aligned} \lambda _w(re^{i\theta })\ge 1-\frac{2Mr\sqrt{2-r^2}}{1-r^2} \quad \text{ for }\ \ z=re^{i\theta }\in {\mathbb {D}}. \end{aligned}$$
(3.5)

Proof

Let \(z=re^{i\theta }\in {\mathbb {D}}.\) According to (2.2) and Lemma 2.1, we have

$$\begin{aligned} \left| e^{i\alpha }w_z(z) +e^{-i\alpha }w_{\bar{z}}(z)\right|= & {} \left| e^{i\alpha } +\frac{1}{\pi }\int \limits _0^{2\pi }\mathrm{Re\,}\left\{ e^{i\alpha }\frac{2z-e^{-it}z^2}{(e^{it}-z)^2}\right\} F(t)dt\right| \\\le & {} 1 +\frac{M}{\pi }\int \limits _0^{2\pi } \frac{|z||2-e^{-it}z|}{|e^{it}-z|^2}dt\\= & {} 1+\frac{2Mr\sqrt{2-r^2}}{1-r^2}. \end{aligned}$$

This shows that \(\Lambda _w\le 1+\frac{2Mr\sqrt{2-r^2}}{1-r^2}\). Similarly,

$$\begin{aligned} \left| e^{i\alpha }w_z(z)+e^{-i\alpha }w_{\bar{z}}(z)\right|\ge & {} 1-\frac{M}{\pi }\int \limits _0^{2\pi } \frac{|z||2-e^{-it}z|}{|e^{it}-z|^2}dt\\= & {} 1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}. \end{aligned}$$

This implies that

$$\begin{aligned} \lambda _w(z)\ge 1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}. \end{aligned}$$

It follows from (3.3) that \(1-\frac{2Mr\sqrt{2-r^2}}{1-r^2}>0\) holds for \(0\le r<r_0\).

The proof is completed. \(\square \)