1 Introduction

A complex-valued function f(z) of class \(C^2\) is said to be a harmonic mapping, if it satisfies \(f_{z\bar{z}}=0\). Assume that f(z) is a harmonic mapping defined in a simply connected domain \(\Omega \subseteq {\mathbb {C}}\). Then f(z) has the canonical decomposition \(f(z)=h(z)+\overline{g(z)}\), where h(z) and g(z) are analytic in \(\Omega \). Let \({\mathbb {D}}=\{z: |z|<1\}\) be the unit disk; throughout this paper, we consider harmonic mappings f(z) in \({\mathbb {D}}\).

For \(z\in {\mathbb {D}}\), let

$$\begin{aligned} \Lambda _f(z)=\max \limits _{0\le \theta \le 2\pi }|f_z(z)+e^{-2i\theta }f_{\bar{z}}(z)|=|f_z(z)|+|f_{\bar{z}}(z)| \end{aligned}$$

and

$$\begin{aligned} \lambda _f(z)=\min \limits _{0\le \theta \le 2\pi }|f_z(z)+e^{-2i\theta }f_{\bar{z}}(z)|=||f_z(z)|-|f_{\bar{z}}(z)||. \end{aligned}$$

It is well known that f is locally univalent and sense-preserving in \({\mathbb {D}}\) if and only if its Jacobian satisfies

$$\begin{aligned} J_f(z)=\lambda _f(z)\Lambda _f(z)=|f_z(z)|^2-|f_{\bar{z}}(z)|^2>0\ \ ~\text{ for }~\ z \in {\mathbb {D}}. \end{aligned}$$

Let

$$\begin{aligned} \beta _h=\sup \limits _{z,w \in {\mathbb {D}},\ z\ne w}\frac{|f(z)-f(w)|}{\rho (z,w)} \end{aligned}$$

be the Bloch constant of f, where \(\rho \) denotes the hyperbolic distance in \({\mathbb {D}}\), and \(\rho (z,w)=\frac{1}{2}\ln \left( \frac{1+r}{1-r}\right) \) where r is the modulus of \(\frac{z-w}{1-\bar{z}w}.\) In [6], we see that the Bloch constant of \(f=h+\bar{g}\) can be expressed in terms of the modulus of the derivatives of h and g as follows:

$$\begin{aligned} \beta _h=\sup \limits _{z\in {\mathbb {D}}}(1-|z|^2)(|h{^\prime }(z)|+|g{^\prime }(z)|). \end{aligned}$$

For the extensive discussions on harmonic Bloch mappings, see [1–5, 10].

For \(v\in (0,\infty )\), a harmonic mapping f is called a harmonic v-Bloch mapping if and only if

$$\begin{aligned} \Vert f\Vert _{B_v}:=|f(0)|+\sup \limits _{z\in {\mathbb {D}}}\left( 1-|z|^2\right) ^{v}\Lambda _f(z)<\infty . \end{aligned}$$
(1)

Harmonic mappings are nature generalizations of analytic functions. Many classical results of analytic functions under some suitable restrictions can be extended to harmonic mappings. One of the well-known results is the Landau-type theorems for harmonic mappings. Many authors have considered such an active topic.

In [11], Liu proved the following theorems.

Theorem A

Suppose that f is a harmonic mapping of \({\mathbb {D}}\) with \(f(0)=\lambda _f(0)-1=0\). If \(\Lambda _f(z)\le \Lambda \) for all \(z\in {\mathbb {D}}\), then

$$\begin{aligned} |a_n|+|b_n|\le \frac{\Lambda ^2-1}{n\Lambda }\ \ n=2,3,\ldots . \end{aligned}$$
(2)

The above estimates are sharp for all \(n\ge 2\) with extremal functions \(f_n(z)=\Lambda ^2z-\int \limits _0^z\frac{(\Lambda ^3-\Lambda )dz}{\Lambda +z^{n-1}}\).

Theorem B

Let f be a harmonic mapping of \({\mathbb {D}}\) with \(f(0)=\lambda _f(0)-1=0\), and \(\Lambda _f(z)\le \Lambda \) for all \(z\in {\mathbb {D}}\). Then f is univalent in the disk \(D_{r_1}\) with \(r_1=\frac{1}{1+\Lambda -\frac{1}{\Lambda }}\) and \(f(D_{r_1})\) contains a schlicht disk \(D_{\sigma _1}\) with

$$\begin{aligned} \sigma _1=\left\{ \begin{array} {ll} 1+\left( \Lambda -\frac{1}{\Lambda }\right) \ln \frac{\Lambda -\frac{1}{\Lambda }}{1+\Lambda -\frac{1}{\Lambda }} &{} \quad \Lambda >1\\ \\ 1 &{}\quad \Lambda =1. \end{array}\right. \end{aligned}$$

The result is sharp when \(\Lambda =1\).

Subsequently, in 2011, Chen et al. [4] proved the following theorems.

Theorem C

Let \(f=h+\bar{g}\) be a harmonic v-Bloch mapping, where h and g are analytic in \({\mathbb {D}}\) with the expansions

$$\begin{aligned} h(z)=\sum \limits _{n=1}^{\infty }a_nz^n\ \ ~\text{ and }~\ \ g(z)=\sum \limits _{n=1}^{\infty }b_nz^n. \end{aligned}$$
(3)

If \(\lambda _f(0)=\alpha \) for some \(\alpha \in (0,1)\) and \(\Vert f\Vert _{B_v}\le M\) for \(M>0.\) Then for \(n\ge 2\),

$$\begin{aligned} |a_n|+|b_n|\le A_n(\alpha , v,M)=\inf \limits _{0<r<1}\mu (r) \end{aligned}$$

where

$$\begin{aligned} \mu (r)=\frac{M^2-\alpha ^2\left( 1-r^2\right) ^{2v}}{nr^{n-1}\left( 1-r^2\right) ^vM}. \end{aligned}$$

Particularly, if \(v=M=\alpha =1\), then \(A_2(1,1,1)=0\), \(A_3(1,1,1)=\frac{1}{3}\) and for \(n\ge 4\), \(A_n(1,1,1)<\frac{(n+1)eM}{2n}.\) The above results are sharp for \(n=2\) and \(n=3\).

Theorem D

Let f be a harmonic mapping with \(f(0)=\lambda _{f}(0)-\alpha =0\) and \(\Vert f\Vert _{B_v}\le M\), where M and \(\alpha \in (0,1]\) are constants. Then f is univalent in \({\mathbb {D}}_{\rho _0}\), where

$$\begin{aligned} \rho _0=\psi (r_0)=\max \limits _{0<r<1}\psi (r),\quad \psi (r)=\frac{\alpha r\left( 1-r^2\right) M}{\alpha M\left( 1-r^2\right) ^v-\alpha ^2\left( 1-r^2\right) ^{2v}+M^2}. \end{aligned}$$

Moreover, \(f({\mathbb {D}}_{\rho _0})\) contains a univalent disk \({\mathbb {D}}_{R_0}\) with

$$\begin{aligned} R_0=r_0\left[ \alpha +\frac{M^2-\alpha ^2\left( 1-r^2\right) ^{2v}}{M\left( 1-r_0^2\right) ^v}\log \frac{M^2-\alpha ^2\left( 1-r^2\right) ^{2v}}{\alpha M\left( 1-r^2_0\right) ^v-\alpha ^2\left( 1-r^2_0\right) ^{2v}+M^2}\right] . \end{aligned}$$

The coefficient estimates are crucial in obtaining Landau-type theorems. In the second part of this paper, by using Parseval equation, we first obtain the coefficient estimates for harmonic v-Bloch mappings, and then for \(0<v<\frac{1}{2}\), we obtain its Landau-type theorems.

Assume that

$$\begin{aligned} f(z)=P[F](z)=\int \limits _0^{2\pi }P(r,x-\varphi )F(x)\mathrm{d}x \end{aligned}$$

is a sense-preserving univalent harmonic mapping of \({\mathbb {D}}\) with the boundary function \(F(x)=e^{i\gamma (x)}\) where

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

is the Poisson kernel and \(z=re^{i\varphi }\in {\mathbb {D}}\). Then f(z) is called a harmonic K-quasiconformal mapping if there exists a constant k such that

$$\begin{aligned} \sup \limits _{z\in {\mathbb {D}}}\left| \frac{f_{\bar{z}}(z)}{f_z(z)}\right| \le k=\frac{K-1}{K+1}. \end{aligned}$$

For harmonic K-quasiconformal mappings defined in \({\mathbb {D}}\), there are many interesting results (See [7, 9, 14] and [16–19]). In [13], Partyka and Sakan proved the following theorem:

Theorem E

Given \(K\ge 1\) and let \(f(z)=P[F](z)\) be a harmonic K-quasiconformal mapping of \({\mathbb {D}}\) onto itself, with the boundary function F(t). If \(f(0)=0\), then for a.e. \(z=e^{it}\in \partial {\mathbb {D}}\)

$$\begin{aligned} \frac{2^{5(1-K^2)/2}}{(K^2+K-1)^K}\le |F\,'(t)|\le K^{3K}2^{5(K-\frac{1}{K})/2}. \end{aligned}$$
(4)

Using this theorem, we obtain the coefficient estimates for \(f=P[F]\) as follows:

$$\begin{aligned} |a_n|+|b_n|\le B_n(K)=\frac{4}{n\pi }K^{3K}2^{5(K-\frac{1}{K})/2},\ \ \ n=1,2,\ldots . \end{aligned}$$

2 Coefficient Estimates for Harmonic v-Bloch Mappings

Theorem 2

Assume that \(f(z)=h(z)+\overline{g(z)}\) is a harmonic v-Bloch mapping such that \(f(0)=0\) and \(\Vert f\Vert _{B_v}\le M\) for some constants \(M>0\), where h(z) and g(z) are given by (3). Then the following inequality

$$\begin{aligned} |a_n|^2+|b_n|^2\le A_n(v,M) \end{aligned}$$
(5)

holds for all \(n=1, 2, 3,\ldots \), where

$$\begin{aligned} A_n(v,M)=\left\{ \begin{array} {ll} \frac{M^2}{n}\inf \limits _{0<t<1}\frac{1-(1-t^2)^{1-2v}}{t^{2n}(1-2v)} \ \ &{} v\ne \frac{1}{2}\\ \\ \frac{M^2}{n}\inf \limits _{0<t<1}\frac{-\ln (1-t^2)}{t^{2n}} \ \ &{} v=\frac{1}{2}. \end{array}\right. \end{aligned}$$

Furthermore, if \(0<v<1\), then \(\lim \limits _{n\rightarrow \infty }A_n(v,M)=0\). If \(v\ge 1\), then \(A_n(v,M)\le \frac{M^2}{2v-1}\frac{(n+1)^{2v-1}-1}{n}(1+\frac{1}{n})^n\).

Proof

Using the assumption that \(f(0)=0\) and \(\Vert f\Vert _{B_v}\le M\), according to (1), we have

$$\begin{aligned} \Lambda _f(z)=|h{^\prime }(z)|+|g{^\prime }(z)|\le \frac{M}{\left( 1-|z|^2\right) ^v}:=\Lambda _r \end{aligned}$$

holds for any \(z=re^{i\theta }\in {\mathbb {D}}\). Using \(f_{\theta }(z)=i\left[ zh{^\prime }(z)-\overline{zg{^\prime }(z)}\right] \) and applying Parseval equation, then

$$\begin{aligned} \frac{1}{2\pi }\int \limits _{0}^{2\pi }|f_{\theta }(re^{i\theta })| ^2\mathrm{d}\theta= & {} \frac{1}{2\pi }\int \limits _{0}^{2\pi } \left| \sum \limits _{n=1}^{\infty }na_nr^ne^{in\theta }-\sum \limits _{n=1}^{\infty }n\overline{b_n}r^ne^{-in\theta }\right| ^2\mathrm{d}\theta \\= & {} \sum \limits _{n=1}^{\infty }n^2\left( |a_n|^2+|b_n|^2\right) r^{2n}. \end{aligned}$$

Applying \(|f_{\theta }(z)|\le |z|\Lambda _f(z)\le r\Lambda _r\), we have

$$\begin{aligned} \sum \limits _{n=1}^{\infty }n^2\left( |a_n|^2+|b_n|^2\right) r^{2n}\le r^2\Lambda ^2_r\le \frac{r^2M^2}{\left( 1-r^2\right) ^{2v}}. \end{aligned}$$

This implies that

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

For any \(0<t<1\), integrals from both sides give

$$\begin{aligned} \sum \limits _{n=1}^{\infty }n\left( |a_n|^2+|b_n|^2\right) \frac{t^{2n}}{2}\le M^2\int \limits _0^t\frac{r}{\left( 1-r^2\right) ^{2v}}\mathrm{d}r :=M^2\varphi (t). \end{aligned}$$
(6)
  1. (i)

    For \(v=\frac{1}{2}\). In this case, \(\varphi (t)=\frac{-\ln (1-t^2)}{2}\). It follows from (6) that

    $$\begin{aligned} |a_n|^2+|b_n|^2\le \frac{M^2}{n}\frac{-\ln \left( 1-t^2\right) }{t^{2n}}. \end{aligned}$$

If \(n=1\), then \(\min \limits _{0<t<1}\frac{M^2}{n}\frac{-\ln (1-t^2)}{t^{2}}=M^2\). For \(n>1\), since \(\lim \limits _{t\rightarrow 0}\frac{-\ln (1-t^2)}{t^{2n}}=\infty =\lim \limits _{t\rightarrow 1}\frac{-\ln (1-t^2)}{t^{2n}}\), we see that \(\inf \limits _{0<t<1}\frac{-\ln (1-t^2)}{t^{2n}}\) exists. Hence,

$$\begin{aligned} |a_n|^2+|b_n|^2\le A_n\left( {\frac{1}{2}, M}\right) =\frac{M^2}{n}\inf \limits _{0<t<1}\frac{-\ln \left( 1-t^2\right) }{t^{2n}}. \end{aligned}$$

Let \(t_0=\sqrt{\frac{n}{n+1}}\). Then

$$\begin{aligned} A_n\left( {\frac{1}{2}, M}\right) \le \frac{M^2}{n}\frac{-\ln \left( 1-t_0^2\right) }{t_0^{2n}} =\frac{M^2\ln (n+1)}{n}\left( 1+\frac{1}{n}\right) ^n. \end{aligned}$$
(7)

This implies that \(\lim \limits _{n\rightarrow \infty }A_n(\frac{1}{2}, M)=0\).

  1. (ii)

    For \(v\ne \frac{1}{2}\). In this case, \(\varphi (t)=\frac{1-(1-t^2)^{1-2v}}{2(1-2v)}\). It follows from (6) that

    $$\begin{aligned} |a_n|^2+|b_n|^2\le \frac{M^2}{n}\frac{1-(1-t^2)^{1-2v}}{(1-2v)t^{2n}}:=\frac{M^2}{n}m(t). \end{aligned}$$

If \(v<\frac{1}{2}\), then \(\inf \limits _{0<t<1} m(t)=\frac{1}{1-2v}\). Hence,

$$\begin{aligned} A_n(v,M)\le \frac{M^2}{n(1-2v)},\ \ \left( v<\frac{1}{2}\right) . \end{aligned}$$
(8)

For \(v>\frac{1}{2}\), \(m(t)=\frac{1-(1-t^2)^{2v-1}}{(1-t^2)^{2v-1}(2v-1)t^{2n}}>0\). If \(n=1\), then \(\inf \limits _{0<t<1}m(t)=2v-1\). Else if \(n>1\), then since \(\lim \limits _{t\rightarrow 0}m(t)=\infty =\lim \limits _{t\rightarrow 1}m(t)\) we see that \(\inf \limits _{0<t<1}m(t)\) exists. Therefore \(A_n(v,M)=\frac{M^2}{n}\inf \limits _{0<t<1}m(t)\) and

$$\begin{aligned} A_n(v,M)\le \frac{M^2}{n}m(t_0)=\frac{M^2}{2v-1}\frac{(n+1)^{2v-1}-1}{n}\left( 1+\frac{1}{n}\right) ^n,\quad \left( v>\frac{1}{2}\right) . \end{aligned}$$
(9)

It follows from (7), (8) and (9) that if \(v<1\), then \(\lim \limits _{n\rightarrow \infty }A_n(v,M)=0\). If \(v=1\), then \(A_n(1,M)\le M^2(1+\frac{1}{n})^{n}\). If \(v>1\), then \(A_n(v,M)\le \frac{M^2}{2v-1}\frac{(n+1)^{2v-1}-1}{n}(1+\frac{1}{n})^n=O(n^{2v-2}).\)

This completes the proof. \(\square \)

Remark 1

We point out that \(|a_n|+|b_n|\le \sqrt{2(|a_n|^2+|b_n|^2)}\le \sqrt{2A_n(v,M)}\). This implies that for \(0< v<1\), the coefficients of harmonic v-Bloch mappings would close to 0 as \(n\rightarrow \infty \). Furthermore, our results show that for \(v\ge 1\), \(|a_n|+|b_n|\le O(n^{v-1})\). The following example shows that Theorem 1 is sharp for \(v=1\).

Example 1

For \(v=1\), we consider harmonic function:

$$\begin{aligned} f(z)=\sum \limits _{n=1}^{\infty }z^{2^n}. \end{aligned}$$

Then

$$\begin{aligned} \frac{|zf{^\prime }(z)|}{1-|z|}\le \sum \limits _{n=1}^{\infty }\left( \sum \limits _{2^k\le n}2^k\right) |z|^n\le \sum \limits _{n=1}^{\infty }2n|z|^n=\frac{2|z|}{(1-|z|)^2}, \end{aligned}$$

Hence,

$$\begin{aligned} (1-|z|^2)|f{^\prime }(z)|\le 4 \ \ (|z|<1). \end{aligned}$$

It follows from (1) that f(z) is a 1-Bloch harmonic function. Moreover, its coefficients do not tend to 0.

Theorem 3

Let \(f(z)=h(z)+\overline{g(z)}\) be a harmonic v-Bloch mapping of \({\mathbb {D}}\) satisfying \(f(0)=\lambda _f(0)-1=0\) and \(0<v<\frac{1}{2}\). Then f is univalent in the disk \({\mathbb {D}}_{r_*}:=\{z: |z|<r_*\}\), where \(r_*\) is the root of the following equation:

$$\begin{aligned} 1-M\sqrt{\frac{2}{1-2v}}\Phi (r)=0 \end{aligned}$$
(10)

and \(\Phi (r):=\sum \limits _{n=1}^{\infty }\sqrt{n+1}r^n\).

Proof

Let \(z_1=r_1e^{i\theta _1}\in {\mathbb {D}}_r\) and \(z_2=r_2e^{i\theta _2}\in {\mathbb {D}}_r\), where \(0<r<r_*\) and \(z_1\ne z_2\). For \(0<v<\frac{1}{2}\), applying Theorem 1, we have

$$\begin{aligned} |a_n|+|b_n|\le \sqrt{2(|a_n|^2+|b_n|^2)}\le \sqrt{\frac{2}{1-2v}}\frac{M}{\sqrt{n}}. \end{aligned}$$

Then

$$\begin{aligned} |f(z_1)-f(z_2)|\ge & {} \lambda _f(0)|z_1-z_2|-|z_1-z_2|\sum \limits _{n=2}^{\infty }(|a_n|+|b_n|)nr^{n-1}\\\ge & {} |z_1-z_2|\left( 1-M\sqrt{\frac{2}{1-2v}}\sum \limits _{n=2}^{\infty }\sqrt{n}r^{n-1}\right) \\= & {} |z_1-z_2|\left( 1-M\sqrt{\frac{2}{1-2v}}\Phi (r)\right) \\:= & {} |z_1-z_2|\varphi (r). \end{aligned}$$

Since \(\varphi (r)\) is a continuous decreasing function satisfying \(\varphi (0)=1\), \(\lim \limits _{r\rightarrow 1^{-}}\varphi (r)=-\infty \), we see that equation \(\varphi (r)=0\) has the root \(0<r_*<1\). Then for any \(0<r<r_{*}\), we have \(|f(z_1)-f(z_2)|>0\). This shows that f(z) is univalent in the disk \(D_{r_{*}}\).

The proof is completed. \(\square \)

For \(M=1\) and some constants \(v\in \left( 0, \frac{1}{2}\right) \), when calculated by computer, we obtain some \(r_{*}\) which were shown by the following table:

M

v

\(r_*\)

1

1/5

0.264534

1

1/4

0.248227

1

1/3

0.214222

1

49/100

0.0650995

3 Coefficient Estimates for Harmonic K-Quasiconformal Mappings

Theorem 3 Given \(K\ge 1\), let \(f(z)=P[F](z)=h(z)+\overline{g(z)}\) be a harmonic K-quasiconformal self-mapping of \({\mathbb {D}}\) satisfying \(f(0)=0\) with the boundary function F, where

$$\begin{aligned} h(z)=\sum \limits _{n=1}^{\infty }a_nz^n\ \ \ and \ \ \ g(z)=\sum \limits _{n=1}^{\infty }b_nz^n \end{aligned}$$

are analytic in \({\mathbb {D}}\). Then

$$\begin{aligned} |a_n|+|b_n|\le B_n(K):= \frac{4}{n\pi }K^{3K}2^{5(K-1/K)/2}\ \ \ n=1,2,\ldots . \end{aligned}$$
(11)

In particular, if \(K=1\) then \(|a_n|+|b_n|\le B_n(1)=\frac{4}{n\pi }.\)

Proof

For every \(z=re^{i\theta }\in {\mathbb {D}}\),

$$\begin{aligned} f(re^{i\theta })=\sum \limits _{n=1}^{\infty }a_nr^ne^{in\theta }+\sum \limits _{n=1}^{\infty }\overline{b_n}r^ne^{-in\theta }. \end{aligned}$$

We find that

$$\begin{aligned} a_nr^n= & {} \frac{1}{2\pi }\int \limits _{0}^{{2\pi }}f(re^{i\theta })e^{-in\theta }\mathrm{d}\theta ,\ n=1,2,\ldots ,\\ \overline{b_n}r^n= & {} \frac{1}{2\pi }\int \limits _{0}^{{2\pi }}f(re^{i\theta })e^{in\theta }\mathrm{d}\theta ,\ n=1,2,\ldots . \end{aligned}$$

For every n (see [12] and [15]), we set \(a_n=|a_n|e^{i\alpha _n}\) , \(b_n=|b_n|e^{i\beta _n}\) and \(\theta _n=\frac{\alpha _n+\beta _n}{2n}\). Then

$$\begin{aligned} (|a_n|+|b_n|)r^n= & {} \left| \frac{1}{2\pi }\int \limits _{0}^{{2\pi }}f(re^{i\theta }) [e^{-i\alpha _n}e^{-in\theta }+ e^{i\beta _n}e^{in\theta }]\mathrm{d}\theta \right| \\= & {} \left| \frac{1}{2\pi }\int \limits _{0}^{{2\pi }} f(re^{i\theta })[e^{-in(\theta +\theta _n)}+ e^{in(\theta +\theta _n))}]\mathrm{d}\theta \right| \\= & {} \left| \frac{1}{\pi }\int \limits _{0}^{{2\pi }}f(re^{i\theta }) \cos n(\theta +\theta _n)\mathrm{d}\theta \right| . \end{aligned}$$

Integrating by parts, we have

$$\begin{aligned} (|a_n|+|b_n|)r^n=\left| \frac{1}{n\pi }\int \limits _{0}^{{2\pi }}f_{\theta }(re^{i\theta }) \sin n(\theta +\theta _n)\mathrm{d}\theta \right| . \end{aligned}$$
(12)

In [8, Theorem 2.8], Kalaj proved that the radial limits of \(f_{\theta }\) and \(f_r\) exist almost everywhere and

$$\begin{aligned} \lim \limits _{r\rightarrow 1^{-}}f_{\theta }(re^{i\theta })=F{^\prime }(\theta ), \end{aligned}$$

for almost every \(z=re^{i\theta }\in {\mathbb {D}}\). Here F is the boundary function of f. Hence, tending \(r\rightarrow 1^-\) in (12) and also using (4), we obtain:

$$\begin{aligned} |a_n|+|b_n|\le \frac{1}{n\pi }\int _0^{2\pi }|F{^\prime }(\theta )||\sin n(\theta +\theta _n)|\mathrm{d}\theta \le \frac{4K^{3K}2^{5(K-1/K)/2}}{n\pi }. \end{aligned}$$

This completes the proof. \(\square \)

Remark 2

Given the boundary function \(F(t)=\rho (t)e^{i\gamma (t)}\) of \({\mathbb {R}}\) onto a convex Jordan curve \(\gamma \in C^{1,\mu } (0<\mu \le 1)\), suppose that \(f(z)=P[F](z)\) is a harmonic K-quasiconformal mapping of \({\mathbb {D}}\) onto the convex domain bounded by \(\gamma \). According to [8, Theorem 3.1], we know that \(\Vert F{^\prime }(t)\Vert _{\infty }<\infty \). Using (12), we can see that \(|a_n|+|b_n|\le \frac{4\Vert F{^\prime }\Vert _{\infty }}{n\pi }\rightarrow 0\), as \(n\rightarrow \infty \).