1 Introduction

Assume that \(f=u+iv\) is a complex-valued harmonic function defined on the unit disk \(\mathbb {D}=\{z\in \mathbb {C}:\, |z|<1\}\), i.e., u and v are real harmonic in \({\mathbb D}\). Then f admits the decomposition \(f=h+\overline{g}\), where h and g are analytic in \({\mathbb D}\). Often h and g are referred to as the analytic and co-analytic parts of f, respectively. If in addition f is univalent in \({\mathbb D}\), then f has a non-vanishing Jacobian in \({\mathbb D}\), where the Jacobian of f is given by

$$\begin{aligned} J_f(z)= |f_z(z)|^2-|f_{\bar{z}}(z)|^2=|h'(z)|^2-|g'(z)|^2. \end{aligned}$$

We say that f is sense-preserving in \({\mathbb D}\) if \(J_f(z)>0\) in \({\mathbb D}\). Moreover, the converse is also true, see [15]. If f is sense-preserving, then the complex dilatation \(\omega :=g'/h'\) is analytic in \({\mathbb D}\) and maps \(\mathbb {D}\) into \(\mathbb {D}.\)

A four times continuously differentiable function F in \(\mathbb {D}\) is called biharmonic if the Laplacian of F is harmonic in D, i.e., F satisfies the biharmonic equation \(\Delta (\Delta F)=0\). Moreover, every biharmonic mapping F in \(\mathbb {D}\) has the representation

$$\begin{aligned} F=|z|^{2}G+H, \end{aligned}$$

where G and H are complex-valued harmonic functions in \(\mathbb {D}\), see [1, 2]. We say that a biharmonic mapping F is sense-preserving in \(\mathbb {D}\) if

$$\begin{aligned} J_F(z)=|F_z(z)|^2-|F_{\overline{z}}(z)|^2>0 \end{aligned}$$

for \(z\in \mathbb {D}\setminus \{0\}\), F is sense-reversing in \(\mathbb {D}\) if

$$\begin{aligned} J_F(z)=|F_z(z)|^2-|F_{\overline{z}}(z)|^2<0 \end{aligned}$$

for \(z\in \mathbb {D}\setminus \{0\}\). Clearly, every harmonic mapping is biharmonic but the converse is not necessarily true. Biharmonic mappings arise in many physical situations, particularly in fluid dynamics and elasticity problems, and have many important applications in engineering and biology, see [11, 13, 14].

Denote by \(\mathcal {S}_H\) the class of all univalent sense-preserving harmonic mappings \(f=h+\overline{g}\) where h and g have power series expansions about the origin given by

$$\begin{aligned} h(z)=z+\sum _{n=2}^{\infty }a_n z^n~ \text{ and } ~ g(z)=\sum _{n=1}^{\infty }b_n z^n, \quad z\in {\mathbb D}, \end{aligned}$$
(1)

where we write for convenience \(a_0=0\) and \(a_1=1\). Also, let

$$\begin{aligned} \mathcal {S}_H^0=\{f\in \mathcal {S}_H:\,f_{\bar{z}}(0)=0\}, \end{aligned}$$

so that \(\mathcal {S}_H^0\subset \mathcal {S}_H\), and let \(\mathcal {S}=\{f=h+\overline{g}\in \mathcal {S}_H^0:\, g(z)\equiv 0\}. \) Just like the class \(\mathcal {S}\) has been a central object in the study of univalent function theory, \(\mathcal {S}_H^0\) plays a vital role in the study of harmonic univalent mappings (see [5, 7]). The Bieberbach conjecture had been a driving force to develop the theory of univalent functions for a long time [6, 9, 17] and was finally solved in the affirmative by Louis de Branges in 1985. On the other hand, the corresponding coefficient conjecture for the class \(\mathcal {S}_H^0\) has not been solved even for the second coefficient of the analytic part h of \(f\in \mathcal {S}_H^0\) [5, 7]. We say that a harmonic function \(f=h+\overline{g}\) in \({\mathbb D}\) has integer coefficients if all the Taylor coefficients \(a_n\) of h and \(b_n\) of g are (rational) integers, and a biharmonic mapping \(F=|z|^2G+H\) has integer coefficients if both G and H have integer coefficients. A similar convention applies when we say \(f=h+\overline{g}\) or \(F=|z|^2G+H\) has (rational) half-integer coefficients or integer coefficients in complex quadratic fields.

Let \(Q(\sqrt{d}i)\), where d is a positive square-free integer, be the complex quadratic field. The integers of \(Q(\sqrt{d}i)\) are known [4, P\(_{187}\)] to have the form \(m+n \alpha \) for integral m and n, where

$$\begin{aligned} \alpha= & {} \left\{ \begin{array}{l@{\quad }l} \displaystyle \sqrt{d}i &{} d\ne 3( \text{ mod } 4), \\ \displaystyle \frac{1+\sqrt{d}i}{2}&{} d=3 (\text{ mod } 4). \end{array}\right. \end{aligned}$$

We note that the integers of Q(i) are the Gaussian integers.

In 1946, Friedman [8] proved the following interesting result and for a simple proof of it, we refer to [16] (see also [22]).

Theorem A

If \(f\in \mathcal {S}\) has integer coefficients, then f is one of the nine functions from \(\mathcal {S}_{{\mathbb Z}}\), where

$$\begin{aligned} \mathcal {S}_{{\mathbb Z}} =\left\{ z,\; \frac{z}{1\pm z},\; \frac{z}{1\pm z^2},\; \frac{z}{(1\pm z)^2},\; \frac{z}{1\pm z+z^2}\right\} . \end{aligned}$$
(2)

In [12], Jenkins presented a different proof of Theorem A extending the results also to functions with coefficients in an imaginary quadratic extension of the rationals, see also [23, 24]. In 1985, authors [3] gave the following results by allowing the coefficients of \(f\in \mathcal {S}\) to be integers in a complex quadratic field.

Theorem B

If \(f\in \mathcal {S}\) has integer coefficients in Q(i) (Gaussian integers), then f must have one of the twenty-one forms from \(\mathcal {S}_{\mathbb {Z}_1}\), where

$$\begin{aligned} \mathcal {S}_{\mathbb {Z}_1}= & {} \Bigg \{ z,\; \frac{z}{1\pm z},\; \frac{z}{1\pm z^2},\; \frac{z}{(1\pm z)^2},\; \frac{z}{1\pm z+z^2},\nonumber \\&\frac{z}{1\pm iz}, \frac{z}{1\pm iz^2}, \frac{z}{(1\pm iz)^2}, \frac{z}{1\pm iz -z^2},\nonumber \\&\frac{z}{1\pm (1+i)z+iz^2}, \frac{z}{1\pm (1-i)z-iz^2}\Bigg \}. \end{aligned}$$
(3)

Theorem C

If \(f\in \mathcal {S}\) has integer coefficients in \(Q(\sqrt{3}i)\), then f is one of the thirty-one functions from \(\mathcal {S}_{\mathbb {Z}_2}\), where

$$\begin{aligned} \mathcal {S}_{\mathbb {Z}_2}= & {} \Bigg \{ z,\; \frac{z}{1\pm z},\; \frac{z}{1\pm z^2},\; \frac{z}{(1\pm z)^2},\; \frac{z}{1\pm z+z^2},\nonumber \\&\frac{z}{1\pm \omega z}, \frac{z}{1\pm \overline{\omega }z}, \frac{z}{1\pm \sqrt{3}i z-z^2}, \frac{z}{1\pm (1+\omega )z +\omega z^2},\nonumber \\&\frac{z}{1\pm (1+\overline{\omega })z +\overline{\omega } z^2}, \frac{z}{1\pm \omega z^2}, \frac{z}{1\pm \overline{\omega } z^2}, \frac{z}{1\pm \overline{\omega } z -\omega z^2},\nonumber \\&\frac{z}{1\pm {\omega } z -\overline{\omega } z^2}, \frac{z}{1\pm 2\overline{\omega } z-\omega z^2}, \frac{z}{1\pm 2{\omega } z-\overline{\omega } z^2}\Bigg \} \end{aligned}$$
(4)

with

$$\begin{aligned} \omega =\frac{1+\sqrt{3}i}{2}. \end{aligned}$$

Theorem D

If \(f\in \mathcal {S}\) has integer coefficients in \(Q(\sqrt{d}i)\), \(d\ne 1\) and \(d\ne 3\), then f is one of the eleven functions from \(\mathcal {S}_{\mathbb {Z}_3}\), where

$$\begin{aligned} \mathcal {S}_{\mathbb {Z}_3}=\Bigg \{ z,\; \frac{z}{1\pm z},\; \frac{z}{1\pm z^2},\; \frac{z}{(1\pm z)^2},\; \frac{z}{1\pm z+z^2}, \frac{z}{1\pm \sqrt{2}i z-z^2}\Bigg \}. \end{aligned}$$
(5)

A univalent biharmonic (resp. harmonic or analytic) function F in \({\mathbb D}\) is said to be convex (resp. starlike) if F is univalent and maps \({\mathbb D}\) onto a convex (resp. starlike with respect to the origin) domain (see [6, 7, 9, 17]). Moreover, if F is univalent, \(F(0)=0\) and \(\frac{\partial }{\partial \theta }(arg\frac{\partial }{\partial \theta }F(re^{i\theta }))>0\) for \(z=re^{i\theta }\ne 0\), then F is convex (cf. [17]).

In [3], the authors pointed that all the univalent functions listed Theorems B, C, and D are starlike with respect to the origin. In fact there are nine convex functions: z, \(\frac{z}{1\pm z}\), \(\frac{z}{1\pm iz}\), \(\frac{z}{1\pm \omega z}\) \(\frac{z}{1\pm \overline{\omega } z}\), where \(\omega =\frac{1+\sqrt{3}i}{2}\), all but \(f(z) = z\) map \(\mathbb {D}\) onto half-planes. Each of the remaining thirty-six functions maps \(\mathbb {D}\) onto a domain whose complement is either one or two slits.

Recently, we determined all functions \(f=h+\overline{g}\) in \(\mathcal {S}_H\) such that h and g have integer coefficients or half-integer coefficients, see [19, 20]. It is a natural question to determine all functions \(f=h+\overline{g}\) in \(\mathcal {S}_H\) such that h and g have integer coefficients in complex quadratic fields. We obtain the following surprising results:

Theorem 1

If \(f\in \mathcal {S}_H\) has integer coefficients in Q(i) (Gaussian integers), then f is one of the twenty-one functions from \(\mathcal {S}_{\mathbb {Z}_1}\), where \(\mathcal {S}_{\mathbb {Z}_1}\) is given by (3).

Theorem 2

If \(f\in \mathcal {S}_H\) has integer coefficients in \(Q(\sqrt{3}i)\), then f is one of the thirty-one functions from \(\mathcal {S}_{\mathbb {Z}_2}\), where \(\mathcal {S}_{\mathbb {Z}_2}\) is given by (4).

Theorem 3

If \(f\in \mathcal {S}_H\) has integer coefficients in \(Q(\sqrt{d}i)\), \(d\ne 1\) and \(d\ne 3\), then f is one of the eleven functions from \(\mathcal {S}_{\mathbb {Z}_3}\), where \(\mathcal {S}_{\mathbb {Z}_3}\) is given by (5).

The key ingredient in the proof of Theorem A is the Area Theorem due to Gronwall [10]. We refer to the work of Jenkins [12] for some information that led to Theorems B, C, D, and some related ideas. Unfortunately, there is no corresponding area theorem for the harmonic case as in the lines of proofs of Theorems B, C, and D. So, it becomes necessary to obtain a suitable method to obtain a proof of Theorems 1, 2, and 3. In Section 2, we present proofs of Theorems 1, 2, and 3. The proofs uses Theorems B, C, D, respectively, and a result of Rogosinski on subordination.

As a generalization, we consider univalent sense-preserving biharmonic mappings \(F=|z|^2G=|z|^2(h+\overline{g})\) with integer coefficients in complex quadratic fields. Under the condition that F is sense-preserving in \(\mathbb {D}\) and \(h'(z)\ne 0\), or F is sense-reversing in \(\mathbb {D}\) and \(g(z)\equiv 0\), we find all univalent biharmonic mappings \(F=|z|^2G=|z|^2(h+\overline{g})\) that have integer coefficients in a fixed complex quadratic field \(Q(\sqrt{d}i)\), where d is a positive square-free integer.

Theorem 4

Let \(F(z)=|z|^2G(z)=|z|^2(h(z)+\overline{g(z)})\) be a univalent biharmonic mapping with \(h(0)=g(0)=g'(0)=h'(0)-1=0\), where G has integer coefficients in Q(i) (Gaussian integers). If F is sense-preserving in \(\mathbb {D}\) and \(h'(z)\ne 0\), or F is sense-reversing in \(\mathbb {D}\) and \(g(z)\equiv 0\), then F is one of the following thirty-one functions:

$$\begin{aligned}&|z|^2z,\; |z|^2\frac{z}{1\pm z},\; |z|^2\frac{z}{1\pm z^2},\; |z|^2\frac{z}{(1\pm z)^2},\; |z|^2\frac{z}{1\pm z+z^2},\nonumber \\&\quad |z|^2\frac{z}{1\pm iz}, |z|^2\frac{z}{1\pm iz^2}, |z|^2\frac{z}{(1\pm iz)^2}, |z|^2\frac{z}{1\pm iz -z^2},\nonumber \\&\quad |z|^2\frac{z}{1\pm (1+i)z+iz^2}, |z|^2\frac{z}{1\pm (1-i)z-iz^2}. \end{aligned}$$

Theorem 5

Let \(F(z)=|z|^2G(z)=|z|^2(h(z)+\overline{g(z)})\) be a univalent biharmonic mapping with \(h(0)=g(0)=g'(0)=h'(0)-1=0\), where G has integer coefficients in \(Q(\sqrt{3}i)\). If F is sense-preserving in \(\mathbb {D}\) and \(h'(z)\ne 0\), or F is sense-reversing in \(\mathbb {D}\) and \(g(z)\equiv 0\), then F is one of the following twenty-one functions:

$$\begin{aligned}&|z|^2 z,\; |z|^2\frac{z}{1\pm z},\; |z|^2\frac{z}{1\pm z^2},\; |z|^2\frac{z}{(1\pm z)^2},\; |z|^2\frac{z}{1\pm z+z^2},\nonumber \\&\quad |z|^2\frac{z}{1\pm \omega z}, |z|^2\frac{z}{1\pm \overline{\omega }}, |z|^2\frac{z}{1\pm \sqrt{3}iz-z^2}, |z|^2\frac{z}{1\pm (1+\omega )z +\omega z^2},\nonumber \\&\quad |z|^2\frac{z}{1\pm (1+\overline{\omega })z +\overline{\omega } z^2}, |z|^2\frac{z}{1\pm \omega z^2}, |z|^2\frac{z}{1\pm \overline{\omega } z^2}, |z|^2\frac{z}{1\pm \overline{\omega } z -\omega z^2},\nonumber \\&\quad |z|^2\frac{z}{1\pm {\omega } z -\overline{\omega } z^2}, |z|^2\frac{z}{1\pm 2\overline{\omega } z-\omega z^2}, |z|^2\frac{z}{1\pm 2{\omega } z-\overline{\omega } z^2}, \end{aligned}$$

where

$$\begin{aligned} \omega =\frac{1+\sqrt{3}i}{2}. \end{aligned}$$

Theorem 6

Let \(F(z)=|z|^2G(z)=|z|^2(h(z)+\overline{g(z)})\) be a univalent biharmonic mapping with \(h(0)=g(0)=g'(0)=h'(0)-1=0\), where G has integer coefficients in \(Q(\sqrt{d}i)\), \(d\ne 1\), \(d\ne 3\). If F is sense-preserving in \(\mathbb {D}\) and \(h'(z)\ne 0\), or F is sense-reversing in \(\mathbb {D}\) and \(g(z)\equiv 0\), then F is one of the following eleven functions:

$$\begin{aligned} |z|^2 z,\; |z|^2\frac{z}{1\pm z},\; |z|^2\frac{z}{1\pm z^2},\; |z|^2\frac{z}{(1\pm z)^2},\; |z|^2\frac{z}{1\pm z+z^2}, |z|^2\frac{z}{1\pm \sqrt{2}iz-z^2}. \end{aligned}$$

2 The Proofs of Theorems 1, 2 and 3

We begin this section with the two lemmas:

Lemma E

([7, p. 87, Theorem]) For all functions \(f\in \mathcal {S}_H^0\), the sharp inequality \(|b_2|\le \frac{1}{2}\) holds, with equality if and only if \(\omega (z)=e^{i\alpha }z\) for some real \(\alpha \).

The notion of subordination is an important property in analytic function theory, see [6, 21]. For analytic functions f and g in \({\mathbb D}\), we say that f is subordinate to g, written \(f(z)\prec g(z)\) or simply \(f\prec g\), if there exists a Schwarz’ function \(\varphi \) (i.e., \(\varphi \) is analytic in \({\mathbb D}\) with \(\varphi (0)=0\) and \(|\varphi (z)|<1\) for \(z\in {\mathbb D}\)) such that \(f(z)=g(\varphi (z)).\) The condition implies that \(f(0)=g(0)\) and \(|f'(0)|\le |g'(0)|\). If, in addition, g is univalent, then \(f\prec g\) if and only if \(f(\mathbb {D})\subset g(\mathbb {D})\) and \(f(0)=g(0)\).

The following result due to Rogosinski [21] (see also Duren [6, p. 195, Theorem 6.4]) is crucial in the proof of our main results.

Lemma F

If \(g(z)=\sum _{n=1}^{\infty }b_nz^n\) is analytic in \(\mathbb {D}\) and \(g\prec f\) for some convex function from \(f\in \mathcal {S}\), then \(|b_n|\le 1\) for \(n\ge 1\).

The method of proof of Theorem 1 is exactly same as from [19, 20], and we include here relevant details for the sake of completeness.

2.1 The Proof of Theorem 1

Let \(f=h+\overline{g}\in \mathcal {S}_H\), where h and g have the standard form given by (1):

$$\begin{aligned} h(z)=z+\sum _{n=2}^{\infty }a_n z^n~ \text{ and } ~ g(z)=\sum _{n=1}^{\infty }b_n z^n, \end{aligned}$$

\(a_n,b_n\) are integers in Q(i). Since f is sense-preserving, we have \(J_f = |h'|^2-|g'|^2> 0 \) so that \(\varphi '(z)\ne 0\) in \(\mathbb {D}\) for \(\varphi (z)=h(z)-g(z)\), and \(|b_1|=|g'(0)|<|h'(0)|=1\) which implies that \(b_1=0\). Thus, \(f\in \mathcal {S}_H^0\). But then, by Lemma E, \(|b_2|\le 1/2\). Since \(b_2\) is an integer in Q(i), we must have \(b_2=0\).

Now, we claim that \(g(z) \equiv 0\) in \({\mathbb D}\). Suppose on the contrary that g is not identically zero. Because f is sense-preserving, we have \(|h'|=|g'+\varphi '|>|g'|\), that is

$$\begin{aligned} \left| \frac{g'}{\varphi '}+1\right| >\left| \frac{g'}{\varphi '}\right| , ~ \text{ i.e., } ~ \mathrm{Re}\left\{ \frac{g'(z)}{\varphi '(z)}\right\} >-\frac{1}{2} ~ \text{ for } z\in {\mathbb D}\text{. } \end{aligned}$$

In terms of subordination, we may rewrite the last inequality as

$$\begin{aligned} \frac{g'(z)}{\varphi '(z)}\prec \frac{z}{1-z} ~ \text{ for } z\in {\mathbb D}\text{. } \end{aligned}$$

Let \(n_0=\min \{n:\, b_n\ne 0\}\) and observe that \(\varphi '(z)=1+\sum _{n=2}^{\infty }n(a_n-b_n) z^{n-1}\ne 0\) in \({\mathbb D}\) so that \(1/\varphi '\) has the form

$$\begin{aligned} \frac{1}{\varphi '(z)}=1+\sum _{n=1}^{\infty }c_n z^n. \end{aligned}$$

Then \(b_{n_0}\ne 0\) for some \(n_0> 2\), and therefore, we have the representation

$$\begin{aligned} \frac{g'(z)}{\varphi '(z)}=n_0 b_{n_0}z^{n_0-1}+\sum _{n=n_0}^{\infty }d_n z^n ~ \text{ for } z\in {\mathbb D}\text{. } \end{aligned}$$

By Lemma F, we deduce that \(|n_0b_{n_0}|\le 1\). Since \(b_{n_0}\) is an integer in Q(i) and \(n_0>2\), it follows that \(b_{n_0}=0\) which is a contradiction. Thus, we conclude that \(g(z)\equiv 0\). Finally, the proof of the theorem follows from Theorem B.

2.2 The Proof of Theorem 2

By using the same argument as that of Theorems 1 and C, the proof of Theorem 2 follows. Here we omit it.

2.3 The Proof of Theorem 3

The proof of Theorem 2 follows from the same argument as that of Theorems 1 and D.

3 The Proofs of Theorems 4, 5 and 6

Lemma 1

([18, Theorem 2]) Let \(F(z)=|z|^2G(z)=|z|^2(h(z)+\overline{g(z)})\) be a univalent and sense-preserving biharmonic mapping with \(h'(z)\ne 0\) \((z\in \mathbb {D})\). Then G is univalent in \({\mathbb D}\).

Similar to Lemma 1, we have

Lemma 2

Let \(F(z)=|z|^2h(z)\) be a univalent and sense-reserving biharmonic mapping, where

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

Then h is univalent in \({\mathbb D}\).

Proof

It is easy to verify that

$$\begin{aligned} F_z(z)=\overline{z}h(z)+|z|^2h'(z) \;\,\text{ and }\;\,F_{\overline{z}}(z)=zh(z). \end{aligned}$$

Suppose that F is sense-reversing, that is \(|F_z(z)|<|F_{\overline{z}}(z)|\) in \(\mathbb {D}\setminus \{0\}\), it follows that

$$\begin{aligned} |\overline{z}h(z)+|z|^2h'(z)|<|zh(z)| \end{aligned}$$

for \(z\ne 0\). Hence,

$$\begin{aligned} \Big |1+\frac{zh'(z)}{h(z)}\Big |<1\;\, \text{ in }\;\, \mathbb {D}/ \{0\}, \end{aligned}$$

which implies

$$\begin{aligned} \mathrm{Re}\frac{z h'(z)}{h(z)}<0. \end{aligned}$$
(6)

Clearly, h has a simple zero at the origin and no zeros elsewhere in the disk. For each \(z=\rho e^{i\theta }\),

$$\begin{aligned} \frac{\partial }{\partial \theta }\{arg h(\rho e^{i\theta })\}=\mathrm{Im} \Big \{\frac{\partial }{\partial \theta }\{\log h(\rho e^{i\theta })\}\Big \}=\mathrm{Im} \Big \{\frac{izh'(z)}{h(z)}\Big \}=\mathrm{Re}\Big \{\frac{zh'(z)}{h(z)} \Big \}. \end{aligned}$$

Therefore,

$$\begin{aligned} \frac{\partial }{\partial \theta }\{arg h(\rho e^{i\theta })\}<0, \;\,0\le \theta \le 2\pi . \end{aligned}$$

Thus, as z runs around the circle \(|z|=\rho \) in the clockwise direction, the point h(z) traverses a closed curve \(C_\rho \) with decreasing argument. Because h has exactly one zero inside the circle \(|z|=\rho \), the argument principle tells us that \(C_\rho \) surrounds the origin exactly once. But if \(C_\rho \) winds about the origin only once with decreasing argument, it can have no self-intersection. Thus, \(C_\rho \) is a simple closed curve, and h assumes each values \(w\in h(\mathbb {D}_\rho )\) (\(\mathbb {D}_\rho =\{z, |z|< \rho \}\)) exactly once in the disk \(|z|<\rho \). Since this is true for every \(\rho <1\), it follows that h is univalent in \(\mathbb {D}\). \(\square \)

3.1 The Proof of Theorem 4

Suppose that \(F(z)=|z|^2G(z)=|z|^2(h(z)+\overline{g(z)})\) is a univalent biharmonic mapping with \(h(0)=g(0)=g'(0)=h'(0)-1=0\), and G has integer coefficients in Q(i). If F is sense-preserving and \(h'(z)\ne 0\), then by using Lemma 1, G is univalent in \(\mathbb {D}\), and thus, the theorem follows from Theorem 1. Supposing that F is sense-reversing and \(g(z)\equiv 0\) \((z\in \mathbb {D})\), we have that h is univalent by Lemma 2. By using Theorem B, the theorem follows.

3.2 The Proof of Theorem 5

By using similar argument as that in the proof of Theorem 4, Lemmas 1 and 2, Theorems C and 2, we obtain this theorem.

3.3 The Proof of Theorem 6

By using similar argument as that in the proof of Theorem 4, Lemmas 1 and 2, Theorems D and 3, this theorem follows.

Remark 1

In [1], authors proved that \(F=|z|^2G\) is univalent and starlike in \({\mathbb D}\) whenever G is harmonic and starlike in \({\mathbb D}\). By using this result, we know that all biharmonic mappings listed Theorems 4, 5, and 6 are starlike with respect to the origin.

It is known that for the analytic function h, h is convex in \(\mathbb {D}\) if and only if

$$\begin{aligned} \frac{\partial }{\partial \theta }\Big (arg\big (\frac{\partial }{\partial \theta }h(re^{i\theta })\big )\Big )\ge 0. \end{aligned}$$

Suppose that h is analytic and convex, and \(F=|z|^2h\). Let \(F_r=r^2h\). Then \(F_r\) is analytic and convex, that is, for each \(r_1\in (0,1)\) and \(\theta \in [0, 2\pi )\),

$$\begin{aligned} \frac{\partial }{\partial \theta } \Big (arg\big (\frac{\partial }{\partial \theta }(r^2h(r_1e^{i\theta }))\big )\Big )\ge 0. \end{aligned}$$

By letting \(r_1=r\), it follows that

$$\begin{aligned} \frac{\partial }{\partial \theta } \Big (arg\big (\frac{\partial }{\partial \theta }(r^2h(re^{i\theta }))\big )\Big )\ge 0, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{\partial }{\partial \theta } \Big (arg\big (\frac{\partial }{\partial \theta }(F(re^{i\theta }))\big )\Big )\ge 0. \end{aligned}$$

Since r is arbitrary, it follows that F is convex. From this, we obtain the following remarks:

Remark 2

The following nine biharmonic mappings, \(|z|^2z\), \(|z|^2\frac{z}{1\pm z}\), \(|z|^2\frac{z}{1\pm iz}\), \(|z|^2\frac{z}{1\pm \omega z}\) \(|z|^2\frac{z}{1\pm \overline{\omega } z}\), where \(\omega =\frac{1+\sqrt{3}i}{2}\), are convex, and all but \(f(z)=|z|^2z\) map \(\mathbb {D}\) onto half-planes.

Remark 3

Except the nine convex biharmonic mappings, each of the remaining thirty-six biharmonic mappings in Theorems 4, 5, and 6 maps \(\mathbb {D}\) onto a domain whose complement is either one or two slits.