On the convex-exponent product of logharmonic mappings

Abstract

Sufficient conditions are obtained on two given logharmonic mappings $f_1$ and $f_2$ that ensure the product $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$, $0\le \lambda \le 1$, is a univalent starlike logharmonic mapping. Several illustrative examples are constructed from this product.

MSC: 30C35, 30C45, 35Q30.

1 Introduction

Let $U=\{z:|z|<1\}$ be the unit disk in the complex plane ℂ, and let B denote the set of bounded analytic functions a satisfying $|a(z)|<1$ in U. Also let $B_0$ denote its subclass consisting of $a\in B$ with $a(0)=0$. A logharmonic mapping f defined in U is a solution of the nonlinear elliptic partial differential equation

$$\frac{\overline{f_{\overline{z}}}}{\overline{f}}=a\frac{f_z}{f},$$

where the second dilatation function a lies in B. Thus the Jacobian

$$J_f=|f_z{|}^2(1-{|a|}^2)$$

is positive, and all non-constant logharmonic mappings are therefore sense-preserving and open in U. If f is a non-constant logharmonic mapping that vanishes only at $z=0$, then f admits the representation

$$f(z)=z^m{|z|}^{2\beta m}h(z)\overline{g(z)},$$

(1)

where m is a nonnegative integer, $Re\beta >-1/2$, while h and g are analytic functions in U satisfying $g(0)=1$ and $h(0)\ne 0$ (see [1]). The exponent β in (1) depends only on $a(0)$ and is given by

$$\beta =\overline{a(0)}\frac{1+a(0)}{1-{|a(0)|}^2}.$$

Note that $f(0)\ne 0$ if and only if $m=0$, and that a univalent logharmonic mapping in U vanishes at the origin if and only if $m=1$, that is, f has the form

$$f(z)=z{|z|}^{2\beta }h(z)\overline{g(z)},$$

where $Re\beta >-1/2$ and $0\notin (hg)(U)$. This class has been studied extensively in recent years, for instance, in [1–8], and [9].

In this case, $F(\zeta )=logf(e^{\zeta })$ is a univalent harmonic mapping of the half-plane $\{\zeta :Re(\zeta )<0\}$. Studies on univalent harmonic mappings can be found in [10–16], and [17]. Such mappings are closely related to the theory of minimal surfaces [18, 19].

In this work, emphasis is given on univalent and sense-preserving logharmonic mappings in U with respect to $a\in B_0$. These mappings are of the form

$$f(z)=zh(z)\overline{g(z)},$$

(2)

and the class consisting of such mappings is denoted by $S_{Lh}$. Also let $S_{Lh}^{\ast }$ denote its subclass of univalent starlike logharmonic mappings. The classical family $S^{\ast }$ of univalent analytic starlike functions is evidently a subclass of $S_{Lh}^{\ast }$. The representation in (2) is essential to the present work as it allows the treatment of logharmonic mappings f through their associated analytic representations h and g (see [3–5], and [6]). For example, Abdulhadi and Abu Muhanna [5] established a connection between starlike logharmonic mappings of order α and starlike analytic functions of order α.

It follows from (2) that the functions h, g and the dilatation a satisfy

$$\frac{z{g}^{\prime }(z)}{g(z)}=a(z)(1+\frac{z{h}^{\prime }(z)}{h(z)}).$$

(3)

Given an analytic function φ with a specified geometric property and $a\in B_0$, a common method to construct a logharmonic mapping $f(z)=zh(z)\overline{g(z)}$ is to solve for h and g via the equations

$$\frac{zh(z)}{g(z)}=\phi (z)\text{and}\frac{\frac{z{g}^{\prime }(z)}{g(z)}}{1+\frac{z{h}^{\prime }(z)}{h(z)}}=a(z).$$

Thus the solution is $f(z)=zh(z)\overline{g(z)}$ with

$$g(z)=exp{\int }_0^z\frac{a(s)}{1-a(s)}\frac{{\phi }^{\prime }(s)}{\phi (s)}ds\text{and}h(z)=\frac{\phi (z)g(z)}{z}.$$

In this paper, a new logharmonic mapping with a specified property is constructed by taking product combination of two functions possessing the given property. Specifically, if $f_1(z)=z{h}_1(z)\overline{g_1(z)}$ with respect to $a_1\in B_0$, and $f_2(z)=z{h}_2(z)\overline{g_2(z)}$ with respect to $a_2\in B_0$, we construct a new univalent logharmonic mapping $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$, $0\le \lambda \le 1$, with respect to $\mu \in B_0$. Sufficient conditions are obtained on $f_1$ and $f_2$ for the product combination $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$ to be starlike. We close the work by giving several examples of univalent starlike logharmonic mappings constructed from this product.

2 Product of logharmonic mappings

Let Ω be a simply connected domain in ℂ containing the origin. Then Ω is said to be α-spirallike, $-\pi /2<\alpha <\pi /2$, if $wexp(-t{e}^{i\alpha })\in \Omega$ for all $t\ge 0$ whenever $w\in \Omega$. Evidently, Ω is starlike (with respect to the origin) if $\alpha =0$.

The following result from [6] will be needed in the sequel.

Lemma 1 Let $f(z)=zh(z)\overline{g(z)}$ be logharmonic in U with $0\notin hg(U)$. Then $f\in S_{Lh}^{\ast }$ if and only if $\phi (z)=zh(z)/g(z)\in S^{\ast }$.

Theorem 1 Let $f(z)=zh(z)\overline{g(z)}\in S_{Lh}^{\ast }$ with respect to $a\in B_0$, and let γ be a constant with $Re\gamma >-1/2$. Then $F(z)=f(z){|f(z)|}^{2\gamma }$ is an α-spirallike logharmonic mapping with respect to

$$\hat{a}(z)=\frac{1+\overline{\gamma }}{1+\gamma }\frac{a(z)+\frac{\overline{\gamma }}{1+\overline{\gamma }}}{1+a(z)\frac{\gamma }{1+\gamma }}=\frac{(1+\overline{\gamma })a(z)+\overline{\gamma }}{1+\gamma +\gamma a(z)},$$

where $\alpha ={tan}^{-1}(2Im\gamma /(1+2Re\gamma ))$.

Proof The function $F=f{|f|}^{2\gamma }=f^{1+\gamma }{\overline{f}}^{\gamma }$ is logharmonic with respect to $\hat{a}=(\overline{F_{\overline{z}}}/\overline{F})/(F_z/F)$. Indeed,

$$\begin{array}{rl}\hat{a}(z)& =\frac{(1+\overline{\gamma })a(z)\frac{f_z}{f}+\overline{\gamma }\overline{(\frac{{\overline{f}}_{\overline{z}}}{\overline{f}})}}{(1+\gamma )\frac{f_z}{f}+\gamma \frac{\overline{f_{\overline{z}}}}{\overline{f}}}\\ =\frac{(1+\overline{\gamma })a(z)\frac{f_z}{f}+\overline{\gamma }\frac{f_z}{f}}{(1+\gamma )\frac{f_z}{f}+\gamma a(z)\frac{f_z}{f}}=\frac{1+\overline{\gamma }}{1+\gamma }\frac{a(z)+\frac{\overline{\gamma }}{1+\overline{\gamma }}}{1+a(z)\frac{\gamma }{1+\gamma }}.\end{array}$$

Thus

$$|\hat{a}(z)|=\left|\frac{a(z)+\frac{\overline{\gamma }}{1+\overline{\gamma }}}{1+a(z)\frac{\gamma }{1+\gamma }}\right|<1$$

provided ${|\gamma |}^2<{|1+\gamma |}^2$, which evidently holds since $Re\gamma >-1/2$.

Also $F=f{|f|}^{2\gamma }=f^{1+\gamma }{\overline{f}}^{\gamma }=z{|z|}^{2\gamma }{h}^{1+\gamma }{g}^{\gamma }\overline{h^{\overline{\gamma }}{g}^{1+\overline{\gamma }}}$. Let $H=h^{1+\gamma }{g}^{\gamma }$, $G=h^{\overline{\gamma }}{g}^{1+\overline{\gamma }}$, and $\psi (z)=zH(z)/G{(z)}^{e^{2i\alpha }}$. Then

$$e^{-i\alpha }\frac{z{\psi }^{\prime }(z)}{\psi (z)}=e^{-i\alpha }+((1+\gamma )e^{-i\alpha }-\overline{\gamma }{e}^{i\alpha })\frac{z{h}^{\prime }(z)}{h(z)}-((1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha })\frac{z{g}^{\prime }(z)}{g(z)}.$$

The condition on α ensures that

$$\frac{(1+\gamma )e^{-i\alpha }-\overline{\gamma }{e}^{i\alpha }}{cos\alpha }=\frac{(1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha }}{cos\alpha }=1.$$

Also Lemma 1 shows that $\phi (z)=zh(z)/g(z)\in S^{\ast }$. Thus

$$Re\left(e^{-i\alpha }\frac{z{\psi }^{\prime }(z)}{\psi (z)}\right)=(cos\alpha )Re\left(\frac{z{\phi }^{\prime }(z)}{\phi (z)}\right)>0,$$

and it follows from [[6], Theorem 2.1] that F is α-spirallike logharmonic whose dilatation is $\hat{a}(z)$. □

Remark 1 Observe that F in Theorem 1 is starlike if and only if $\gamma >-1/2$.

Theorem 2 Let $f_k(z)=z{h}_k(z)\overline{g_k(z)}\in S_{Lh}^{\ast }$ ($k=1,2$) with respect to the same $a\in B_0$. Then $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$, $0\le \lambda \le 1$, is a univalent starlike logharmonic mapping with respect to the same a.

Proof Let $\mu =(\overline{F_{\overline{z}}}/\overline{F})/(F_z/F)$. It follows from (3) that

$$\begin{array}{rl}\mu & =\frac{\lambda \frac{\overline{f_{1\overline{z}}}}{\overline{f_1}}+(1-\lambda )\frac{\overline{f_{2\overline{z}}}}{\overline{f_2}}}{\lambda \frac{f_{1z}}{f_1}+(1-\lambda )\frac{f_{2z}}{f_2}}=\frac{\lambda \frac{g_1^{\prime }}{g_1}+(1-\lambda )\frac{g_2^{\prime }}{g_2}}{\lambda \frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}+(1-\lambda )\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}}\\ =\frac{\lambda a\frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}+(1-\lambda )a\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}}{\lambda \frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}+(1-\lambda )\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}}=a.\end{array}$$

(4)

Hence $|\mu (z)|<1$ in U, which implies that F is a locally univalent logharmonic mapping.

Next F is shown to have the form (2). Since $f_1=z{h}_1\overline{g_1}$, and $f_2=z{h}_2\overline{g_2}$, then

$$\begin{array}{rl}F(z)& =f_1^{\lambda }(z)f_2^{1-\lambda }(z)={(z{h}_1(z)\overline{g_1(z)})}^{\lambda }{(z{h}_2(z)\overline{g_2(z)})}^{1-\lambda }\\ =z{h}_1^{\lambda }{h}_2^{1-\lambda }\overline{g_1^{\lambda }(z)g_2^{1-\lambda }(z)}=zh(z)\overline{g(z)}\end{array}$$

(5)

with $h=h_1^{\lambda }{h}_2^{1-\lambda }$ and $g=g_1^{\lambda }{g}_2^{1-\lambda }$.

Since $f_k$ is starlike, that is, each ${\phi }_k=z{h}_k/g_k$ satisfies the condition $Rez{\phi }_k^{\prime }(z)/{\phi }_k(z)>0$ in U, direct computations show that

$$\begin{array}{rl}\frac{\partial arg(F(r{e}^{i\theta }))}{\partial \theta }=& Re(\frac{z{F}_z}{F}-\frac{\overline{z}{F}_{\overline{z}}}{F})=\lambda Re(\frac{z{f}_{1z}}{f_1}-\frac{\overline{z}{f}_{1\overline{z}}}{f_1})\\ +(1-\lambda )Re(\frac{z{f}_{2z}}{f_2}-\frac{\overline{z}{f}_{2\overline{z}}}{f_2})\\ =& \lambda Re\left(\frac{z{\phi }_1^{\prime }(z)}{{\phi }_1(z)}\right)+(1-\lambda )Re\left(\frac{z{\phi }_2^{\prime }(z)}{{\phi }_2(z)}\right)>0.\end{array}$$

Thus F is starlike. □

The following corollary is an immediate consequence of Theorem 2.

Corollary 1 Let $f_k(z)=z{h}_k(z)\overline{g_k(z)}\in S_{Lh}^{\ast }$ ($k=1,2,\dots ,n$) with respect to the same $a\in B_0$. Then $F=f_1^{{\lambda }_1}{f}_2^{{\lambda }_2}\cdots f_n^{{\lambda }_n}$ is a univalent starlike logharmonic mapping with respect to the same a, where $0\le {\lambda }_k\le 1$ and ${\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n=1$.

Theorem 3 Let $f_k(z)=z{h}_k(z)\overline{g_k(z)}\in S_{Lh}^{\ast }$ ($k=1,2$) with respect to $a_k\in B_0$. Suppose also that

$$Re(1-a_1\overline{a_2})\frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}\overline{\left(\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}\right)}\ge 0.$$

Then $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$, $0\le \lambda \le 1$, is a univalent starlike logharmonic mapping.

Proof The argument is similar to the proof of Theorem 2. From (5), evidently F has the form (2).

Let $(F_z/F)\mu (z)=\overline{F_{\overline{z}}}/\overline{F}$. Since $|a_k|<1$, it follows from (3) and (4) that

$$\begin{array}{rl}|\mu (z)|& =\left|\frac{\lambda \frac{g_1^{\prime }}{g_1}+(1-\lambda )\frac{g_2^{\prime }}{g_2}}{\lambda \frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}+(1-\lambda )\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}}\right|\\ =\left|\frac{\lambda a_1\frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}+(1-\lambda )a_2\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}}{\lambda \frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}+(1-\lambda )\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}}\right|.\end{array}$$

(6)

By assumption,

$$\begin{array}{r}|\lambda \frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}+(1-\lambda )\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}{|}^2-|\lambda a_1\frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}+(1-\lambda )a_2\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}{|}^2\\ ={\lambda }^2(1-|a_1{|}^2)|\frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}{|}^2+{(1-\lambda )}^2(1-|a_2{|}^2)|\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}{|}^2\\ +2\lambda (1-\lambda )Re((1-a_1\overline{a_2})\frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}\overline{\left(\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}\right)})>0,\end{array}$$

whence $|\mu (z)|<1$, which implies that F is locally univalent.

Now the associated analytic function for F is given by $\phi =(z{h}_1^{\lambda }{h}_2^{1-\lambda })/(g_1^{\lambda }{g}_2^{1-\lambda })$. Let $\phi (U)=\Omega$. From Lemma 1, ${\phi }_k=z{h}_k/g_k\in S^{\ast }$, and thus

$$Re\left(\frac{z{\phi }^{\prime }(z)}{\phi (z)}\right)=\lambda Re\left(\frac{z{\phi }_1^{\prime }(z)}{{\phi }_1(z)}\right)+(1-\lambda )Re\left(\frac{z{\phi }_2^{\prime }(z)}{{\phi }_2(z)}\right)>0.$$

Hence Ω is a starlike domain, and we deduce that F is a univalent starlike logharmonic mapping. □

Theorem 4 Let $f_k=z{h}_k\overline{g_k}\in S_{Lh}$ with respect to $a_k\in B$, $k=1,2$, satisfying $z{h}_k{g}_k=z$. Then $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$, $0\le \lambda \le 1$, is a univalent starlike logharmonic mapping.

Proof Since

$$\frac{{(z{h}_k)}^{\prime }}{(z{h}_k)}+\frac{g_k^{\prime }}{g_k}=\frac{1}{z},$$

it follows from (3) that

$$\frac{{(z{h}_k)}^{\prime }}{(z{h}_k)}=\frac{1}{z(1+a_k)}.$$

(7)

With $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$, (6) and (7) readily yield

$$|\mu (z)|=\left|\frac{\lambda a_1+(1-\lambda )a_2+a_1{a}_2}{1+(1-\lambda )a_1+\lambda a_2}\right|.$$

Evidently $|\mu (z)|<1$ is equivalent to $\psi (\lambda )=|1+(1-\lambda )a_1+\lambda a_2{|}^2-|\lambda a_1+(1-\lambda )a_2+a_1{a}_2{|}^2>0$.

Now

$$\begin{array}{rl}\psi (\lambda )=& 2\lambda ((1-{|a_1|}^2)Re{a}_2-(1-{|a_2|}^2)Re{a}_1-({|a_1|}^2-{|a_2|}^2))\\ +(1-{|a_2|}^2){|1+a_1|}^2\end{array}$$

is a continuous monotonic function of λ in the interval $[0,1]$. Since

$$\psi (0)=(1-{|a_2|}^2)|1+a_1{|}^2>0$$

and

$$\psi (1)=(1-{|a_1|}^2)|1+a_2{|}^2>0,$$

we deduce that $\psi (\lambda )>0$ for all $\lambda \in [0,1]$, and thus F is locally univalent.

With ${\phi }_k=z{h}_k/g_k$, then

$$Re\left(\frac{z{\phi }_k^{\prime }(z)}{{\phi }_k(z)}\right)=Re((1-a_k)\frac{z{(z{h}_k)}^{\prime }}{(z{h}_k)})=Re\left(\frac{1-a_k}{1+a_k}\right)>0.$$

Hence ${\phi }_k$ is starlike univalent, and from Lemma 1, $f_k(z)=z{h}_k(z)\overline{g_k(z)}$ is starlike univalent logharmonic.

The associated analytic function for F is given by $\phi (z)=(z{h}_1^{\lambda }{h}_2^{1-\lambda })/(g_1^{\lambda }{g}_2^{1-\lambda })$. Further

$$Re\left(\frac{z{\phi }^{\prime }(z)}{\phi (z)}\right)=\lambda Re\left(\frac{z{\phi }_1^{\prime }(z)}{{\phi }_1(z)}\right)+(1-\lambda )Re\left(\frac{z{\phi }_2^{\prime }(z)}{{\phi }_2(z)}\right)>0,$$

and thus F is starlike. □

The proof of Theorem 4 evidently gives the following result of [[6], Lemma 3.1 and Theorem 3.2].

Corollary 2 Let $f_k=z{h}_k\overline{g_k}\in S_{Lh}$ ($k=1,2$) with respect to $a_k\in B_0$, and suppose that $z{h}_k{g}_k=z$. Then $\phi (z)=z{(h_k(z))}^2\in S^{\ast }$.

3 Examples

We give several illustrative examples in this section.

Example 1 Let

$$f(z)=z\left(\frac{1-\overline{z}}{1-z}\right).$$

Then f is a univalent logharmonic mapping with respect to $a(z)=-z$, and it maps U onto U [6].

Now the function $F(z)=f(z){|f(z)|}^{2\gamma }$ is an α-spirallike logharmonic mapping with respect to

$$\hat{a}(z)=\frac{1+\overline{\gamma }}{1+\gamma }\frac{-z+\frac{\overline{\gamma }}{1+\overline{\gamma }}}{1-z\frac{\gamma }{1+\gamma }},$$

where $\alpha ={tan}^{-1}(2Im\gamma /(1+2Re\gamma ))$. In particular, if $\gamma =i$, then $\alpha ={tan}^{-1}(2)=0.352\pi$, and

$$\hat{a}(z)=\frac{-i-(1-i)z}{1+i-iz}.$$

The image of circles in the unit disk under f is shown in Figure 1, and Figure 2 shows the image of the radial slits in U by F.

Figure 1

Graph of circles in U by $\mathit{f}\mathbf{(}\mathit{z}\mathbf{)}\mathbf{=}\frac{\mathit{z}\mathbf{(}\mathbf{1}\mathbf{-}\overline{\mathit{z}}\mathbf{)}}{\mathbf{1}\mathbf{-}\mathit{z}}$ .

Figure 2

Graph of radial slits by $\mathit{F}\mathbf{(}\mathit{z}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{(}\mathit{z}\mathbf{)}{\mathbf{|}\mathit{f}\mathbf{(}\mathit{z}\mathbf{)}\mathbf{|}}^{\mathbf{2}\mathit{i}}$ , $\mathit{f}\mathbf{(}\mathit{z}\mathbf{)}\mathbf{=}\frac{\mathit{z}\mathbf{(}\mathbf{1}\mathbf{-}\overline{\mathit{z}}\mathbf{)}}{\mathbf{1}\mathbf{-}\mathit{z}}$ .

Example 2 Consider the functions

$$f_1(z)=z\left(\frac{1-\overline{z}}{1-z}\right)exp\{Re\frac{4z}{1-z}\}\text{and}{f}_2(z)=z\left(\frac{1+\overline{z}}{1+z}\right).$$

Since ${\phi }_1(z)=z/{(1-z)}^2$ and ${\phi }_2(z)=z/{(1+z)}^2$ are starlike analytic functions, it follows from [[5], Theorem 1] that $f_1$ and $f_2$ are starlike logharmonic mappings with respect to $a(z)=z$. Theorem 2 shows that $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$, $0\le \lambda \le 1$, is a starlike univalent logharmonic mapping.

The image of F is shown in Figure 3 for $\lambda =1/3$.

Figure 3

Graph of circles in U by $\mathit{F}\mathbf{(}\mathit{z}\mathbf{)}\mathbf{=}\mathit{z}{\mathbf{(}\frac{\mathbf{1}\mathbf{-}\overline{\mathit{z}}}{\mathbf{1}\mathbf{-}\mathit{z}}\mathbf{exp}\mathbf{\{}\mathbf{Re}\frac{\mathbf{4}\mathit{z}}{\mathbf{1}\mathbf{-}\mathit{z}}\mathbf{\}}\mathbf{)}}^{\mathbf{1}\mathbf{/}\mathbf{3}}{\mathbf{(}\frac{\mathbf{1}\mathbf{+}\overline{\mathit{z}}}{\mathbf{1}\mathbf{+}\mathit{z}}\mathbf{)}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$ .

Example 3 In this example, let

$$f_1(z)=z\frac{(1-\overline{z})}{(1-z)}\text{and}{f}_2(z)=z\frac{(1-\overline{z})}{(1-z)}exp\{Re\frac{4z}{1-z}\},$$

and

$$F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z),0\le \lambda \le 1.$$

Simple calculations show that $f_1$ and $f_2$ are respectively starlike logharmonic with dilatations $a_1(z)=-z$ and $a_2(z)=z$. Also F is logharmonic with respect to $\mu (z)=z((1-2\lambda )+z)/(1+(1-2\lambda )z)$.

Since

$$\begin{array}{rl}Re((1-a_1\overline{a_2})\frac{{(z{h}_1)}^{\prime }}{(z{h}_1)}\overline{\left(\frac{{(z{h}_2)}^{\prime }}{(z{h}_2)}\right)})& =Re\left((1+{|z|}^2)\frac{1}{z(1-z)}\frac{1+\overline{z}}{\overline{z}{(1-\overline{z})}^2}\right)\\ =\frac{(1+{|z|}^2)}{{|z|}^2{|1-z|}^2}Re\frac{1+z}{1-z}>0,\end{array}$$

the conditions of Theorem 3 are satisfied, and thus F is starlike univalent.

The image of circles in U under F for $\lambda =1/3$ is shown in Figure 4.

Figure 4

Graph of $\mathit{F}\mathbf{(}\mathit{z}\mathbf{)}\mathbf{=}\mathit{z}{\mathbf{(}\frac{\mathbf{1}\mathbf{-}\overline{\mathit{z}}}{\mathbf{1}\mathbf{-}\mathit{z}}\mathbf{)}}^{\mathbf{1}\mathbf{/}\mathbf{3}}{\mathbf{(}\frac{\mathbf{1}\mathbf{-}\overline{\mathit{z}}}{\mathbf{1}\mathbf{-}\mathit{z}}\mathbf{exp}\mathbf{\{}\mathbf{Re}\frac{\mathbf{4}\mathit{z}}{\mathbf{1}\mathbf{-}\mathit{z}}\mathbf{\}}\mathbf{)}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$ .

Example 4 Let $f_1(z)=z{h}_1(z)\overline{g_1(z)}$, where $z{h}_1(z)g_1(z)=z$, $a_1(z)=z$, and

$$h_1(z)=\frac{1}{1+z},g_1(z)=1+z.$$

Thus

$$f_1(z)=\frac{z(1+\overline{z})}{(1+z)}.$$

Further, let $f_2(z)=z{h}_2(z)\overline{g_2(z)}$, where $z{h}_2(z)g_2(z)=z$, $a_2(z)=z^2$, and

$$h_2(z)=\frac{1}{\sqrt{1+z^2}},g_2(z)=\sqrt{1+z^2}.$$

In this case,

$$f_2(z)=\frac{z\sqrt{1+{\overline{z}}^2}}{\sqrt{1+z^2}}.$$

Since $f_1$ and $f_2$ satisfy the conditions of Theorem 4, we deduce that $F(z)=f_1^{\lambda }(z)f_2^{1-\lambda }(z)$, $0\le \lambda \le 1$, is a univalent starlike logharmonic mapping. The image of U under F for $\lambda =1/3$ is shown in Figure 5.

Figure 5

Graph of $\mathit{F}\mathbf{(}\mathit{z}\mathbf{)}\mathbf{=}\mathit{z}{\mathbf{(}\frac{\mathbf{1}\mathbf{+}\overline{\mathit{z}}}{\mathbf{1}\mathbf{+}\mathit{z}}\mathbf{)}}^{\mathbf{1}\mathbf{/}\mathbf{3}}{\mathbf{(}\frac{\sqrt{\mathbf{1}\mathbf{+}{\overline{\mathit{z}}}^{\mathbf{2}}}}{\sqrt{\mathbf{1}\mathbf{+}{\mathit{z}}^{\mathbf{2}}}}\mathbf{)}}^{\mathbf{2}\mathbf{/}\mathbf{3}}$ .