New Criteria for Convex-Exponent Product of Log-Harmonic Functions

Abstract

In this study, we consider different types of convex-exponent products of elements of a certain class of log-harmonic mapping and then find sufficient conditions for them to be starlike log-harmonic functions. For instance, we show that, if f is a spirallike function, then choosing a suitable value of $\gamma$, the log-harmonic mapping $F\left(z\right)={f\left(z\right)\left|f\left(z\right)\right|}^{2\gamma }$ is $\alpha$-$spiralike$ of order $\rho$. Our results generalize earlier work in the literature.

1. Introduction

Let E be the open unit disk $E=\{z\in \mathbb{C}:|z|<1\}$ and $\mathcal{H}\left(E\right)$ denote the linear space of all analytic functions defined on E. Additionally, let $\mathcal{A}$ be a subclass consisting of $f\in \mathcal{H}\left(E\right)$ such that $f\left(0\right)=f^{\prime }\left(0\right)-1=0$.

A $C^2$-function defined in E is said to be harmonic if $\Delta f=0$, and a log-harmonic function f is a solution of the nonlinear elliptic partial differential equation

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

(1)

where the second dilation function $a\in \mathcal{H}\left(E\right)$ is such that $\left|a\right(z\left)\right|<1$ for all $z\in E$. In the above formula, ${\overline{f}}_{\overline{z}}$ means $\overline{\left(f_{\overline{z}}\right)}$. Observe that f is log-harmonic if $logf$ is harmonic. The authors in [1] have proven that, if f is a non-constant log-harmonic mapping that vanishes only at $z=0$, then f should be in the form

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

(2)

where m is a nonnegative integer, $\mathrm{Re}\beta >-\frac{1}{2}$, while h and g are analytic functions in $\mathcal{H}\left(E\right)$ satisfying $g\left(0\right)=1$ and $h\left(0\right)\ne 0$. The exponent $\beta$ in (2) depends only on $a\left(0\right)$ and is given by

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

(3)

We remark that $f\left(0\right)\ne 0$ if and only if $m=0$ and that a univalent log-harmonic mapping in E vanishes at the origin if and only if $m=1$, that is, f has the form

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

where $\mathrm{Re}\beta >-\frac{1}{2}$ and $0\notin hg\left(E\right)$.

Recently, the class of log-harmonic functions has been extensively studied by many authors; for instance, see [1,2,3,4,5,6,7,8,9,10].

The Jacobian of log-harmonic function f is given by

$$J_f\left(z\right)=|f_z{|}^2{(1-|a\left(z\right)|}^2)$$

(4)

and is positive. Therefore, all non-constant log-harmonic mappings are sense-preserving in the unit disk E. Let B denote the class of functions $a\in \mathcal{H}\left(E\right)$ with $\left|a\right(z\left)\right|<1$ and $B_0$ denote $a\in B$ such that $a\left(0\right)=0$.

It is easy to see that, if $f\left(z\right)=zh\left(z\right)\overline{g\left(z\right)}$, then the functions h and g, and the dilation a satisfy

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

(5)

Definition 1.

(See [2].) Let $f={z\left|z\right|}^{2\beta }h\left(z\right)\overline{g\left(z\right)}$ be a univalent log-harmonic mapping. We say that f is a starlike log-harmonic mapping of order α if

$$\frac{\partial argf\left(r{e}^{i\theta }\right)}{\partial \theta }=\mathrm{Re}\frac{z{f}_z-\overline{z}{f}_{\overline{z}}}{f}>\alpha ,0\le \alpha <1$$

for all $z\in E$. Denote by $S{T}_{LH}\left(\alpha \right)$ the class of all starlike log-harmonic mappings.

By taking $\beta =0$ and $g\left(z\right)=1$ in Definition 1, we obtain the class of starlike analytic functions in $\mathcal{A}$, which we denote by $S^{*}\left(\alpha \right)$.

The following lemma shows the relationship of the classes $S{T}_{LH}\left(\alpha \right)$ and $S^{*}\left(\alpha \right)$.

Lemma 1.

(See [2].) Let $f\left(z\right)={z\left|z\right|}^{2\beta }h\left(z\right)\overline{g\left(z\right)}$ be a log-harmonic mapping on E, $0\notin hg\left(E\right)$. Then, $f\in S{T}_{LH}\left(\alpha \right)$ if and only if $\phi \left(z\right)=\frac{zh\left(z\right)}{g\left(z\right)}\in S^{*}\left(\alpha \right)$.

In [2], the authors studied the class of $\alpha -spirallike$ functions and proved that, if $f\left(z\right)={z\left|z\right|}^{2\beta }h\left(z\right)\overline{g\left(z\right)}$ is a log-harmonic mapping on E, $0\notin hg\left(E\right)$, then f is $\alpha -spirallike$ if

$$\mathrm{Re}\left(e^{-i\alpha }\frac{z{f}_z-\overline{z}{f}_{\overline{z}}}{f}\right)>0,0\le \alpha <1$$

for all $z\in E$. We remark that a simply connected domain $\Omega$ in $\mathbb{C}$ containing the origin is said to be $\alpha -spirallike$, $-\frac{\pi }{2}<\alpha <\frac{\pi }{2}$ if $wexp(-t{e}^{i\alpha })\in \Omega$ for all $t\ge 0$ whenever $w\in \Omega$ and that f is an $\alpha -spirallike$ function, if $f\left(E\right)$ is an $\alpha$-$spiralike$ domain. Motivated by this, we define the class of $\alpha -spirallike$ log-harmonic mappings of order $\rho$ as follows:

Definition 2.

Let $f\left(z\right)={z\left|z\right|}^{2\beta }h\left(z\right)\overline{g\left(z\right)}$ be a univalent log-harmonic mapping on E, with $0\notin hg\left(E\right)$. Then, we say that f is an $\alpha -spirallike$ log-harmonic mapping of order $\rho (0\le \rho <1)$ if

$$\mathrm{Re}\left(e^{-i\alpha }\frac{z{f}_z-\overline{z}{f}_{\overline{z}}}{f\left(z\right)}\right)>\rho cos\alpha (z\in E)$$

for some real $\alpha \left(\right|\alpha |<\frac{\pi }{2})$. The class of these functions is denoted by $S_{LH}^{\alpha }\left(\rho \right)$. Furthermore, we define $S_{LH}^{\alpha }\left(1\right)={\bigcap }_{0\le \rho <1}{S}_{LH}^{\alpha }\left(\rho \right)$.

Additionally, we denote by $S^{\alpha }\left(\rho \right)$ the subclass of all $f\in \mathcal{A}$ such that f is $\alpha$-$spiralike$ of order $\rho$ and $S^{\alpha }\left(1\right)={\bigcap }_{0\le \rho <1}{S}^{\alpha }\left(\rho \right)$.

Lemma 2.

([2]) If $f\left(z\right)={z\left|z\right|}^{2\beta }h\left(z\right)\overline{g\left(z\right)}$ is log-harmonic on E and $0\notin hg\left(E\right)$, with $\mathrm{Re}\beta >-\frac{1}{2}$, then $f\in S_{LH}^{\alpha }\left(\rho \right)$ if and only if $\psi \left(z\right)=\frac{zh\left(z\right)}{g{\left(z\right)}^{e^{2i\alpha }}}\in S^{\alpha }\left(\rho \right)$.

In the celebrated paper [11], the authors introduce a new way of studying harmonic functions in Geometric Function Theory. Additionally, many authors investigated the linear combinations of harmonic functions in a plane; see, for example, [12,13,14]. In Section 2 of this paper, taking the convex-exponent product combination of two elements, a specified class of new log-harmonic functions is constructed. Indeed, we show that, if $f\left(z\right)=zh\left(z\right)\overline{g}\left(z\right)$ is spirallike log-harmonic of order $\rho$, then by choosing suitable parameters of $\alpha$ and $\gamma$, the function $F\left(z\right)=f\left(z\right)\left|f\right(z{|}^{2\gamma }$ is log-harmonic spirallike of order $\alpha$. Additionally, in Section 3, we provide some examples that are constructed from Section 2.

2. Main Results

Theorem 1.

Let $f\left(z\right)=zh\left(z\right)\overline{g\left(z\right)}\in S{T}_{LH}\left(\rho \right),(0\le \rho <1)$ with respect to $a\in B_0$, $\varphi \in S^{*}\left(\gamma \right),(0\le \gamma <1)$ and $\alpha ,\beta$ be real numbers with $\alpha +\beta =1$. Then, $F\left(z\right)=f{\left(z\right)}^{\alpha }K{\left(z\right)}^{\beta }$ is starlike log-harmonic mapping of order $\alpha \rho +\beta \gamma$ with respect to a, where

$$K\left(z\right)=\varphi \left(z\right)exp\left\{2\mathrm{Re}{\int }_0^z\frac{a\left(s\right)}{1-a\left(s\right)}\frac{{\varphi }^{\prime }\left(s\right)}{\varphi \left(s\right)}ds\right\}.$$

Proof. 

By definition of F, we have

$$\frac{F_z}{F}=\alpha \frac{f_z}{f}+\beta \frac{K_z}{K}and\frac{F_{\overline{z}}}{F}=\alpha \frac{f_{\overline{z}}}{f}+\beta \frac{K_{\overline{z}}}{K}.$$

(6)

Additionally direct computations show that

$$\frac{K_z}{K}=\frac{1}{1-a\left(z\right)}\frac{{\varphi }^{\prime }\left(z\right)}{\varphi \left(z\right)},and\frac{\overline{K_{\overline{z}}}}{\overline{K}}=\frac{a\left(z\right)}{1-a\left(z\right)}\frac{{\varphi }^{\prime }\left(z\right)}{\varphi \left(z\right)}.$$

(7)

Now, in view of Equations (6) and (7),

$$\widehat{a}\left(z\right)=\frac{\frac{\overline{F_{\overline{z}}}}{\overline{F}}}{\frac{F_z}{F}}=\frac{\alpha \frac{\overline{f_{\overline{z}}}}{\overline{f}}+\beta \overline{\frac{K_{\overline{z}}}{\overline{K}}}}{\alpha \frac{f_z}{f}+\beta \frac{K_z}{K}}=a\left(z\right)\frac{\alpha \frac{f_z}{f}+\beta \frac{K_z}{K}}{\alpha \frac{f_z}{f}+\beta \frac{K_z}{K}}=a\left(z\right).$$

On the other hand,

$$\begin{array}{cc}\hfill \mathrm{Re}\frac{z{F}_z-\overline{z}{F}_{\overline{z}}}{F}& =\mathrm{Re}\left(\alpha \frac{z{f}_z}{f}+\beta \frac{z{K}_z}{K}\right)-\mathrm{Re}\left(\alpha \frac{z\overline{f_{\overline{z}}}}{\overline{f}}+\beta \frac{z\overline{K_{\overline{z}}}}{\overline{K}}\right)\hfill \\ \hfill & =\alpha \mathrm{Re}\left(\frac{z{f}_z}{f}-\frac{z\overline{f_{\overline{z}}}}{\overline{f}}\right)+\beta \mathrm{Re}\left(\frac{z{K}_z}{K}-\frac{z\overline{K_{\overline{z}}}}{\overline{K}}\right)\hfill \\ \hfill & >\alpha \rho +\beta \gamma .\hfill \end{array}$$

The above relation shows that F is a log-harmonic starlike function of order $\alpha \rho +\beta \gamma$, and the proof is complete. □

Theorem 2.

Let $f\left(z\right)=zh\left(z\right)\overline{g\left(z\right)}\in S_{LH}^{\beta }\left(\rho \right)$ with respect to $a\in B_0$ and γ be a constant with $\mathrm{Re}\gamma >-\frac{1}{2}$. Then, $F\left(z\right)={f\left(z\right)\left|f\left(z\right)\right|}^{2\gamma }$ is an $\alpha -spirallike$ log-harmonic mapping of order ρ with respect to

$$\widehat{a}\left(z\right)=\frac{(1+\overline{\gamma })a\left(z\right)+\overline{\gamma }}{1+\gamma +\gamma a\left(z\right)},$$

where $\left|\beta \right|<\frac{\pi }{2}$ and $\alpha ={tan}^{-1}\left(\frac{tan\beta +2\mathrm{Im}\gamma }{1+2\mathrm{Re}\gamma }\right).$

Proof. 

By definition of F, we have

$$F\left(z\right)={f\left(z\right)\left|f\left(z\right)\right|}^{2\gamma }=z^{1+\gamma }{\overline{z}}^{\gamma }H\left(z\right)\overline{G\left(z\right)},$$

where

$$H\left(z\right)=h^{1+\gamma }\left(z\right)g^{\gamma }\left(z\right)\mathrm{and}G\left(z\right)=h^{\overline{\gamma }}\left(z\right)g^{1+\overline{\gamma }}\left(z\right).$$

With a straightforward calculation and using Equation (5),

$$\frac{z{F}_z}{F}=(1+\gamma )\left(1+\frac{z{h}^{\prime }\left(z\right)}{h\left(z\right)}\right)+\gamma \frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}=\left(1+\frac{z{h}^{\prime }\left(z\right)}{h\left(z\right)}\right)((1+\gamma )+\gamma a\left(z\right)),$$

and

$$\frac{\overline{z}{F}_{\overline{z}}}{F}=\gamma \left(1+\frac{\overline{z{h}^{\prime }\left(z\right)}}{\overline{h\left(z\right)}}\right)+(1+\gamma )\frac{\overline{z{g}^{\prime }\left(z\right)}}{\overline{g\left(z\right)}}=\left(1+\frac{\overline{z{h}^{\prime }\left(z\right)}}{\overline{h\left(z\right)}}\right)(\gamma +(1+\gamma )\overline{a\left(z\right)}).$$

If we consider

$$\widehat{a}\left(z\right)=\frac{\overline{\left(\frac{\overline{z}{F}_{\overline{z}}\left(z\right)}{F\left(z\right)}\right)}}{\frac{z{F}_z\left(z\right)}{F\left(z\right)}},$$

then

$$\widehat{a}\left(z\right)=\frac{\overline{\gamma }+(1+\overline{\gamma })a\left(z\right)}{(1+\gamma )+\gamma a\left(z\right)}.$$

Now, in view of $\left|a\right(z\left)\right|<1$, it easy to see that $|\widehat{a}\left(z\right)|<1$ provided that $\left|\frac{\overline{\gamma }}{1+\overline{\gamma }}\right|<1$, which evidently holds ${\left|\gamma \right|}^2<{|1+\overline{\gamma }|}^2$ since $\mathrm{Re}\gamma >-\frac{1}{2},$ and this means that F is a log-harmonic function.

Additionally, by putting

$$\psi \left(z\right)=\frac{zH\left(z\right)}{G{\left(z\right)}^{e^{2i\alpha }}},$$

we have

$$\psi \left(z\right)=\frac{zH\left(z\right)}{G{\left(z\right)}^{e^{2i\alpha }}}=\frac{zh{\left(z\right)}^{1+\gamma }g{\left(z\right)}^{\gamma }}{{\left(h^{\overline{\gamma }}\left(z\right)g^{1+\overline{\gamma }}\left(z\right)\right)}^{e^{2i\alpha }}}.$$

Then, we obtain

$$\begin{array}{cc}\hfill & e^{-i\alpha }\frac{z{\psi }^{\prime }\left(z\right)}{\psi \left(z\right)}=e^{-i\alpha }+[(1+\gamma )e^{-i\alpha }-\overline{\gamma }{e}^{i\alpha }]\frac{z{h}^{\prime }\left(z\right)}{h\left(z\right)}-[(1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha }]\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\hfill \\ \hfill & =(-\gamma e^{-i\alpha }+\overline{\gamma }{e}^{i\alpha })+[(1+\gamma )e^{-i\alpha }-\overline{\gamma }{e}^{i\alpha }]\left(1+\frac{z{h}^{\prime }\left(z\right)}{h\left(z\right)}\right)\hfill \\ \hfill & -[(1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha }]\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}.\hfill \end{array}$$

The condition on $\alpha$ ensures that

$$(1+\gamma )e^{-i\alpha }-\overline{\gamma }{e}^{i\alpha }=\frac{cos\alpha }{cos\beta }{e}^{-i\beta }\mathrm{and}(1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha }=\frac{cos\alpha }{cos\beta }{e}^{i\beta },$$

because by letting $\gamma ={\gamma }_1+i{\gamma }_2$, the first equality holds true if and only if

$$cos\beta cos\alpha -i(1+2{\gamma }_1)sin\alpha cos\beta +i2{\gamma }_{2}cos\beta cos\alpha =cos\alpha cos\beta -icos\alpha sin\beta$$

or, equivalently, after simplification

$$2{\gamma }_{2}cot\beta -(1+2{\gamma }_1)tan\alpha cot\beta =-1$$

or

$$\alpha ={tan}^{-1}\left(\frac{tan\beta +2\mathrm{Im}\gamma }{1+2\mathrm{Re}\gamma }\right).$$

Thus, by hypothesis,

$$\mathrm{Re}\left\{e^{-i\alpha }\frac{z{\psi }^{\prime }\left(z\right)}{\psi \left(z\right)}\right\}=\frac{cos\alpha }{cos\beta }\mathrm{Re}\left(e^{-i\beta }(1+\frac{z{h}^{\prime }\left(z\right)}{h\left(z\right)})-e^{i\beta }\frac{z{g}^{\prime }\left(z\right)}{g\left(z\right)}\right)>\rho cos\alpha$$

and it follows that F is an $\alpha$-spirallike log-harmonic mapping of order $\rho$ in which the dilation is $\widehat{a}\left(z\right)$. □

Theorem 3.

Let $f_k\left(z\right)=z{h}_k\left(z\right)\overline{g_k}\left(z\right)\in S_{LH}^{\beta }\left(\rho \right)$ with $k=1,2$ and with respect to the same $a\in B_0$ and γ be a constant with $\mathrm{Re}\gamma >-\frac{1}{2}$. Moreover, let

$$F_1\left(z\right)=f_1\left(z\right)|f_1{\left(z\right)|}^{2\gamma }and{F}_2\left(z\right)=f_2\left(z\right){\left|f_2\left(z\right)\right|}^{2\gamma }.$$

Then, $F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)$ is an α-spirallike log-harmonic mapping of order ρ with respect to

$$\widehat{a}\left(z\right)=\frac{(1+\overline{\gamma })a\left(z\right)+\overline{\gamma }}{1+\gamma +\gamma a\left(z\right)},$$

where $\left|\beta \right|<\frac{\pi }{2}$ and $\alpha ={tan}^{-1}\left(\frac{tan\beta +2\mathrm{Im}\gamma }{1+2\mathrm{Re}\gamma }\right).$

Proof. 

According to the definitions of $F_1$ and $F_2$, we have

$$\begin{array}{cc}\hfill F_1^{\lambda }\left(z\right)& =(f_1\left(z\right)|f_1\left(z\right){{|}^{2\gamma })}^{\lambda }\hfill \\ \hfill & ={\left(z\right|z{|}^{2\gamma }{h}_1^{1+\gamma }\left(z\right)g_1^{\gamma }\left(z\right)\overline{h_1^{\overline{\gamma }}\left(z\right)g_1^{1+\overline{\gamma }}\left(z\right)})}^{\lambda }\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill F_2^{1-\lambda }\left(z\right)& =(f_2\left(z\right)|f_2\left(z\right){{|}^{2\gamma })}^{1-\lambda }\hfill \\ \hfill & ={\left(z\right|z{|}^{2\gamma }{h}_2^{1+\gamma }\left(z\right)g_2^{\gamma }\left(z\right)\overline{h_2^{\overline{\gamma }}\left(z\right)g_2^{1+\overline{\gamma }}\left(z\right)})}^{1-\lambda }.\hfill \end{array}$$

Putting the values of $F_1^{\lambda }$ and $F_2^{1-\lambda }$ on F, we obtain

$$\begin{array}{cc}\hfill F\left(z\right)& ={\left(z\right|z{|}^{2\gamma }{h}_1^{1+\gamma }\left(z\right)g_1^{\gamma }\left(z\right)\overline{h_1^{\overline{\gamma }}\left(z\right)g_1^{1+\overline{\gamma }}\left(z\right)})}^{\lambda }{\left(z\right|z{|}^{2\gamma }{h}_2^{1+\gamma }\left(z\right)g_2^{\gamma }\left(z\right)\overline{h_2^{\overline{\gamma }}\left(z\right)g_2^{1+\overline{\gamma }}\left(z\right)})}^{1-\lambda }\hfill \\ \hfill & ={z\left|z\right|}^{2\gamma }H\left(z\right)\overline{G\left(z\right)},\hfill \end{array}$$

where

$$H\left(z\right)=h_1{\left(z\right)}^{\lambda (1+\gamma )}{g}_1{\left(z\right)}^{\lambda \gamma }{h}_2{\left(z\right)}^{(1-\lambda )(1+\gamma )}{g}_2{\left(z\right)}^{(1-\lambda )\gamma }$$

(8)

and

$$G\left(z\right)={h_1\left(z\right)}^{\lambda \overline{\gamma }}{g_1\left(z\right)}^{\lambda (1+\overline{\gamma })}{h_2\left(z\right)}^{(1-\lambda )\overline{\gamma }}{g_2\left(z\right)}^{(1-\lambda )(1+\overline{\gamma })}.$$

(9)

Now, we show that the second dilation of F i.e., $\mu \left(z\right)$ satisfies the condition $\left|\mu \right(z\left)\right|<1.$ For this, since

$$\mu \left(z\right)=\frac{\frac{{\overline{F}}_{\overline{z}}\left(z\right)}{\overline{F}\left(z\right)}}{\frac{F_z\left(z\right)}{F\left(z\right)}},$$

we have

$$\begin{array}{cc}\hfill & \mu \left(z\right)=\frac{\lambda \frac{\overline{F_{1\overline{z}}\left(z\right)}}{{\overline{F}}_1\left(z\right)}+(1-\lambda )\frac{\overline{F_{2\overline{z}}\left(z\right)}}{{\overline{F}}_2\left(z\right)}}{\lambda \frac{F_{1z}\left(z\right)}{F_1\left(z\right)}+(1-\lambda )\frac{F_{2z}\left(z\right)}{F_2\left(z\right)}}\hfill \\ \hfill & =\frac{\lambda [\overline{\gamma }(1+\frac{z{h}_1^{\prime }}{h_1})+(1+\overline{\gamma })\frac{z{g}_1^{\prime }}{g_1}]+(1-\lambda )[\overline{\gamma }(1+\frac{z{h}_2^{\prime }}{h_2})+(1+\overline{\gamma })\frac{z{g}_2^{\prime }}{g_2}]}{\lambda [(1+\gamma )(1+\frac{z{h}_1^{\prime }}{h_1})+\gamma \frac{z{g}_1^{\prime }}{g_1}]+(1-\lambda )[(1+\gamma )(1+\frac{z{h}_2^{\prime }}{h_2})+\gamma \frac{z{g}_2^{\prime }}{g_2}]}\hfill \\ \hfill & =\frac{\lambda (1+\frac{z{h}_1^{\prime }}{h_1})[\overline{\gamma }+(1+\overline{\gamma })a\left(z\right)]+(1-\lambda )(1+\frac{z{h}_2^{\prime }}{h_2})[\overline{\gamma }+(1+\overline{\gamma })a\left(z\right)]}{\lambda (1+\frac{z{h}_1^{\prime }}{h_1})[(1+\gamma )+\gamma a\left(z\right)]+(1-\lambda )(1+\frac{z{h}_2^{\prime }}{h_2})[(1+\gamma )+\gamma a\left(z\right)]}\hfill \\ \hfill & =\frac{[\lambda (1+\frac{z{h}_1^{\prime }}{h_1})+(1-\lambda )(1+\frac{z{h}_2^{\prime }}{h_2})][\overline{\gamma }+(1+\overline{\gamma })a\left(z\right)]}{[\lambda (1+\frac{z{h}_1^{\prime }}{h_1})+(1-\lambda )(1+\frac{z{h}_2^{\prime }}{h_2})][(1+\gamma )+\gamma a\left(z\right)]}\hfill \\ \hfill & =\frac{[\overline{\gamma }+(1+\overline{\gamma })a\left(z\right)]}{\left[\right(1+\gamma )+\gamma a(z\left)\right]}\hfill \\ \hfill & =\frac{(1+\overline{\gamma })}{(1+\gamma )}\frac{a\left(z\right)+\frac{\overline{\gamma }}{1+\overline{\gamma }}}{1+\frac{a\left(z\right)\gamma }{1+\gamma }},\hfill \end{array}$$

(10)

and the condition $\mathrm{Re}\gamma >-\frac{1}{2}$ ensures that $\left|\mu \right(z\left)\right|<1$ in E, which implies that F is a locally univalent log-harmonic mapping. Now, to prove

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(\rho \right),$$

we have to show that $\psi \left(z\right)=\frac{zH\left(z\right)}{G{\left(z\right)}^{e^{2i\alpha }}}\in S^{\alpha }\left(\rho \right)$. However, a direct calculation shows that

$$\begin{array}{cc}\hfill & \psi \left(z\right)=\frac{zH\left(z\right)}{G{\left(z\right)}^{e^{2i\alpha }}}=\frac{\left[z{h}_1^{\lambda (1+\gamma )}\left(z\right)g_1^{\lambda \gamma }\left(z\right)h_2^{(1-\lambda )(1+\gamma )}\left(z\right)g_2^{(1-\lambda )\gamma }\left(z\right)\right]}{{\left[h_1^{\lambda \overline{\gamma }}\left(z\right)g_1^{\lambda (1+\overline{\gamma })}\left(z\right)h_2^{(1-\lambda )\overline{\gamma }}\left(z\right)g_2^{(1-\lambda )(1+\overline{\gamma })}\left(z\right)\right]}^{e^{2i\alpha }}}.\hfill \end{array}$$

Now,

$$\begin{array}{cc}\hfill & e^{-i\alpha }\frac{z{\psi }^{\prime }\left(z\right)}{\psi \left(z\right)}\hfill \\ \hfill & =e^{-i\alpha }\left[1+\lambda (((1+\gamma )-e^{2i\alpha }\overline{\gamma })\frac{z{h}_1^{\prime }\left(z\right)}{h_1\left(z\right)}-((1+\overline{\gamma })e^{2i\alpha }-\gamma )\frac{z{g}_1^{\prime }\left(z\right)}{g_1\left(z\right)})\right]\hfill \\ \hfill & +e^{-i\alpha }\left[(1-\lambda )(((1+\gamma )-e^{2i\alpha }\overline{\gamma })\frac{z{h}_2^{\prime }\left(z\right)}{h_2\left(z\right)}-((1+\overline{\gamma })e^{2i\alpha }-\gamma )\frac{z{g}_2^{\prime }\left(z\right)}{g_2\left(z\right)})\right]\hfill \\ \hfill & =-\gamma e^{-i\alpha }+e^{i\alpha }\overline{\gamma }\hfill \\ \hfill & +\lambda \left[((1+\gamma )e^{-i\alpha }-e^{i\alpha }\overline{\gamma })(1+\frac{z{h}_1^{\prime }\left(z\right)}{h_1\left(z\right)})-((1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha })\frac{z{g}_1^{\prime }\left(z\right)}{g_1\left(z\right)}\right]\hfill \\ \hfill & +(1-\lambda )\left[((1+\gamma )e^{-i\alpha }-e^{i\alpha }\overline{\gamma })(1+\frac{z{h}_2^{\prime }\left(z\right)}{h_2\left(z\right)})-((1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha })\frac{z{g}_2^{\prime }\left(z\right)}{g_2\left(z\right)}\right].\hfill \end{array}$$

By hypothesis, we know that

$$\begin{array}{c}\hfill (1+\gamma )e^{-i\alpha }-\overline{\gamma }{e}^{i\alpha }=\frac{cos\alpha }{cos\beta }{e}^{-i\beta }\mathrm{and}(1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha }=\frac{cos\alpha }{cos\beta }{e}^{i\beta },\end{array}$$

so

$$\begin{array}{c}\hfill \mathrm{Re}\left\{e^{-i\alpha }\frac{z{\psi }^{\prime }\left(z\right)}{\psi \left(z\right)}\right\}\\ \hfill & =\lambda \frac{cos\alpha }{cos\beta }\mathrm{Re}\left(e^{-i\beta }(1+\frac{z{h}_1^{\prime }\left(z\right)}{h_1\left(z\right)})-e^{i\beta }\frac{z{g}_1^{\prime }\left(z\right)}{g_1\left(z\right)}\right)\hfill \\ \hfill & +(1-\lambda )\frac{cos\alpha }{cos\beta }\mathrm{Re}\left(e^{-i\beta }(1+\frac{z{h}_2^{\prime }\left(z\right)}{h_2\left(z\right)})-e^{i\beta }\frac{z{g}_1^{\prime }\left(z\right)}{g_1\left(z\right)}\right)\hfill \\ \hfill & >\rho cos\alpha \hfill \end{array}$$

and the proof is completed. □

Theorem 4.

Let $f_k\left(z\right)=z{h}_k\left(z\right){\overline{g}}_k\left(z\right)\in S_{LH}^{\beta }\left(\rho \right)$ with respect to $a{}_k\in B_0(k=1,2)$. Moreover, suppose that $\mathrm{Re}\gamma >-\frac{1}{2}$,

$$F_1\left(z\right)=f_1\left(z\right)|f_1{\left(z\right)|}^{2\gamma }and{F}_2\left(z\right)=f_2\left(z\right){\left|f_2\left(z\right)\right|}^{2\gamma }.$$

If

$$\mathrm{Re}\left[(1-a_1\left(z\right){\overline{a}}_2\left(z\right))\left(1+\frac{z{h}_1^{\prime }\left(z\right)}{h_1\left(z\right)}\right)\overline{\left(1+\frac{z{h}_2^{\prime }\left(z\right)}{h_2\left(z\right)}\right)}\right]\ge 0(foranyz\in E),$$

then

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(\rho \right),$$

where $\left|\beta \right|<\frac{\pi }{2},0\le \lambda \le 1$ and $\alpha ={tan}^{-1}\left(\frac{tan\beta +2Im\gamma }{1+2Re\gamma }\right).$

Proof. 

Using the same argument as in Theorem 3, we have

$$F\left(z\right)={z\left|z\right|}^{2\gamma }H\left(z\right)\overline{G\left(z\right)},$$

where $H\left(z\right)$ and $G\left(z\right)$ are defined by Equations (8) and (9). Now, we show that the second dilation of F, i.e., $\mu \left(z\right)$, satisfies the condition $\left|\mu \right(z\left)\right|<1.$ For this, since

$$\mu \left(z\right)=\frac{\frac{{\overline{F}}_{\overline{z}}\left(z\right)}{\overline{F}\left(z\right)}}{\frac{F_z\left(z\right)}{F\left(z\right)}},$$

using a similar argument to the relation Equation (10) of Theorem 3, we have

$$\begin{array}{cc}\hfill & \left|\mu \left(z\right)\right|=\left|\frac{\lambda (1+\frac{z{h}_1^{\prime }}{h_1})[\overline{\gamma }+(1+\overline{\gamma })a_1\left(z\right)]+(1-\lambda )(1+\frac{z{h}_2^{\prime }}{h_2})[\overline{\gamma }+(1+\overline{\gamma })a_2\left(z\right)]}{\lambda (1+\frac{z{h}_1^{\prime }}{h_1})[(1+\gamma )+\gamma a_1\left(z\right)]+(1-\lambda )(1+\frac{z{h}_2^{\prime }}{h_2})[(1+\gamma )+\gamma a_2\left(z\right)]}\right|.\hfill \end{array}$$

However, by hypothesis, we obtain

$$\begin{array}{cc}\hfill & {\left|\lambda (1+\frac{z{h}_1^{\prime }}{h_1})[(1+\gamma )+\gamma a_1\left(z\right)]+(1-\lambda )(1+\frac{z{h}_2^{\prime }}{h_2})[(1+\gamma )+\gamma a_2\left(z\right)]\right|}^2\hfill \\ \hfill & -{\left|\lambda (1+\frac{z{h}_1^{\prime }}{h_1})[\overline{\gamma }+(1+\overline{\gamma })a_1\left(z\right)]+(1-\lambda )(1+\frac{z{h}_2^{\prime }}{h_2})[\overline{\gamma }+(1+\overline{\gamma })a_2\left(z\right)]\right|}^2\hfill \\ \hfill & =(2\mathrm{Re}\gamma +1)\left({\lambda }^2{\left|1+\frac{z{h}_1^{\prime }}{h_1}\right|}^2(1-|a_1{|}^2{)+(1-\lambda )}^2{\left|1+\frac{z{h}_2^{\prime }}{h_2}\right|}^2(1-|a_2{|}^2)\right)\hfill \\ \hfill & +(2\mathrm{Re}\gamma +1)\left(2\lambda (1-\lambda )\mathrm{Re}\left[(1-a_1{\overline{a}}_2)(1+\frac{z{h}_1^{\prime }}{h_1})\overline{(1+\frac{z{h}_2^{\prime }}{h_2})}\right]\right)>0.\hfill \end{array}$$

Therefore, $\left|\mu \right(z\left)\right|<1$ in E, which implies that F is a locally univalent mapping. Moreover, by following a similar proof to that in Theorem 3, we observe that

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(\rho \right),$$

and the proof is completed. □

Theorem 5.

Let $f_k\left(z\right)=z{h}_k\left(z\right){\overline{g}}_k\left(z\right)$ be univalent log-harmonic functions with respect to $a{}_k\in B_0(k=1,2)$ and $\mathrm{Re}\gamma >-\frac{1}{2}$. Moreover, suppose that $z{h}_k{g}_k={\varphi }_k\left(z\right)$, where

$${\varphi }_k\left(z\right)=zexp\left\{2{\int }_0^z\frac{a_k\left(t\right)}{t(1-a_k\left(t\right))}dt\right\}$$

and

$$F_1\left(z\right)=f_1\left(z\right)|f_1{\left(z\right)|}^{2\gamma }and{F}_2\left(z\right)=f_2\left(z\right){\left|f_2\left(z\right)\right|}^{2\gamma }.$$

Then,

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(1\right)$$

where $0\le \lambda \le 1$ and $\alpha ={tan}^{-1}\left(\frac{2\mathrm{Im}\gamma }{1+2\mathrm{Re}\gamma }\right)$.

Proof. 

Since $z{h}_k{g}_k={\varphi }_k\left(z\right)$, by definition of $a_k\left(z\right)$ and ${\varphi }_k\left(z\right)$, we obtain

$$1+\frac{z{h}_k^{\prime }\left(z\right)}{h_k\left(z\right)}=\frac{1}{1-a_k\left(z\right)}(k=1,2).$$

Let

$$\mu \left(z\right)=\frac{\frac{{\overline{F}}_{\overline{z}}\left(z\right)}{\overline{F}\left(z\right)}}{\frac{F_z\left(z\right)}{F\left(z\right)}}.$$

Using a similar argument to the relation in Equation (10) of Theorem 3, we obtain

$$\begin{array}{cc}\hfill & \left|\mu \left(z\right)\right|=\left|\frac{\lambda (1-a_2\left(z\right))[\overline{\gamma }+(1+\overline{\gamma })a_1\left(z\right)]+(1-\lambda )((1-a_1\left(z\right))[\overline{\gamma }+(1+\overline{\gamma })a_2\left(z\right)]}{\lambda (1-a_2\left(z\right))[(1+\gamma )+\gamma a_1\left(z\right)]+(1-\lambda )(1-a_1\left(z\right))[(1+\gamma )+\gamma a_2\left(z\right)]}\right|.\hfill \end{array}$$

Now, $\left|\mu \right(z\left)\right|<1$ is equivalent to

$$\begin{array}{cc}\hfill & \psi \left(\lambda \right):={\left|\lambda (1-a_2\left(z\right))[(1+\gamma )+\gamma a_1\left(z\right)]+(1-\lambda )(1-a_1\left(z\right))[(1+\gamma )+\gamma a_2\left(z\right)]\right|}^2\hfill \\ \hfill & -{\left|\lambda (1-a_2\left(z\right))[\overline{\gamma }+(1+\overline{\gamma })a_1\left(z\right)]+(1-\lambda )((1-a_1\left(z\right))[\overline{\gamma }+(1+\overline{\gamma })a_2\left(z\right)]\right|}^2\hfill \\ \hfill & =(2\mathrm{Re}\gamma +1)[{\lambda }^2|1-a_2{\left(z\right)|}^2(1-|a_1\left(z\right){|}^2)\hfill \\ \hfill & +2\lambda (1-\lambda )\mathrm{Re}\left[(1-a_2\left(z\right))(1-\overline{a_1\left(z\right)})(1-a_1\left(z\right)\overline{a_2\left(z\right)})\right]\hfill \\ \hfill & +{(1-\lambda )}^2|1-a_1{\left(z\right)|}^2(1-|a_2{\left(z\right)|}^2\left)\right]>0.\hfill \end{array}$$

However, by taking the derivative of $\psi \left(\lambda \right)$, we have

$$\begin{array}{cc}\hfill {\psi }^{\prime }\left(\lambda \right)& =2(2\mathrm{Re}\gamma +1)\hfill \\ \hfill & \left[\mathrm{Re}\left[(1-a_2\left(z\right))(1-\overline{a_1\left(z\right)})(1-a_1\left(z\right)\overline{a_2\left(z\right)})\right]-|1-a_1\left(z\right){|}^2(1-|a_2\left(z\right){|}^2)\right],\hfill \end{array}$$

which shows that $\psi$ is a continuous monotonic function of $\lambda$ in the interval $[0,1]$. Since

$$\psi \left(0\right)=(2\mathrm{Re}\gamma +1){\left|1-a_2\left(z\right)\right|}^2(1-|a_1\left(z\right){|}^2)>0$$

and

$$\psi \left(1\right)=(2\mathrm{Re}\gamma +1){\left|1-a_1\left(z\right)\right|}^2(1-|a_2\left(z\right){|}^2)>0,$$

we deduce that $\psi \left(\lambda \right)>0$ for all $\lambda \in [0,1]$, which implies that F is a locally univalent mapping. Now, to prove

$$F=F_1^{\lambda }{F}_2^{1-\lambda }\in S_{LH}^{\alpha }$$

(11)

we have to show that $\psi \left(z\right)=\frac{zH\left(z\right)}{G{\left(z\right)}^{e^{2i\alpha }}}\in S^{\alpha }\left(1\right)$, where $H\left(z\right)$ and $G\left(z\right)$ are defined by Equations (8) and (9). A direct computation such as that in Theorem 3 shows that

$$\begin{array}{c}\hfill \frac{(1+\gamma )e^{-i\alpha }-\overline{\gamma }{e}^{i\alpha }}{cos\alpha }=\frac{(1+\overline{\gamma })e^{i\alpha }-\gamma e^{-i\alpha }}{cos\alpha }=1.\end{array}$$

Additionally, we note that

$$1+\frac{z{h}_1^{\prime }}{h_1}-\frac{z{g}_1^{\prime }}{g_1}=1+\frac{z{h}_2^{\prime }}{h_2}-\frac{z{g}_2^{\prime }}{g_2}=1.$$

Using these relation and the same argument as that made in Theorem 3, we obtain $\psi \left(z\right)=\frac{zH\left(z\right)}{G{\left(z\right)}^{e^{2i\alpha }}}\in S^{\alpha }\left(1\right)$, and the proof is complete. □

Theorem 6.

Let $f_k\left(z\right)=z{h}_k\left(z\right){\overline{g}}_k\left(z\right)(k=1,2)$ be log-harmonic functions with respect to $a{}_k\in B_0$. Moreover, suppose that $z{h}_k{g}_k=z$ and

$$F_1\left(z\right)=f_1\left(z\right)|f_1{\left(z\right)|}^{2\gamma }and{F}_2\left(z\right)=f_2\left(z\right){\left|f_2\left(z\right)\right|}^{2\gamma }.$$

Then,

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(1\right),$$

where $0\le \lambda \le 1$ and $\alpha ={tan}^{-1}\left(\frac{2\mathrm{Im}\gamma }{(1+2\mathrm{Re}\gamma )}\right)$.

Proof. 

Since $z{h}_k{g}_k=z$, by definition of $a_k\left(z\right)$, we obtain

$$1+\frac{z{h}_k^{\prime }\left(z\right)}{h_k\left(z\right)}=\frac{1}{1+a_k\left(z\right)}(k=1,2).$$

Using the same argument as that in Theorem 5, we obtain our result, but we omit the details. □

3. Examples

We provide several examples in this section.

Example 1.

Let $\mathrm{Re}\gamma >-\frac{1}{2}$ and

$$f\left(z\right)=z\frac{{(1+z)}^{[cos\beta (1-\rho )e^{i\beta }-1]}}{{(1-z)}^{(1-\rho )cos\beta e^{i\beta }}}{(1+\overline{z})}^{[(1-\rho )cos\beta e^{i\beta }-e^{2i\beta }]}{(1-\overline{z})}^{(1-\rho )cos\beta e^{i\beta }}.$$

Then, it is easy to see that f is a β-spirallike log-harmonic mapping of order ρ with respect to $a\left(z\right)=-z{e}^{-2i\beta }.$ Now, Theorem 2 implies that the function $F\left(z\right)={f\left(z\right)\left|f\left(z\right)\right|}^{2\gamma }$ is a α-spirallike log-harmonic mapping of order ρ with respect to

$$\widehat{a}\left(z\right)=\frac{-(1+\overline{\gamma })z{e}^{-2i\beta }+\overline{\gamma }}{(1+\gamma )-\gamma e^{-2i\beta }z},$$

where

$$\alpha ={tan}^{-1}\left(\frac{tan\beta +2\mathrm{Im}\gamma }{1+2\mathrm{Re}\gamma }\right).$$

The image in Example 1 is shown in Figure 1.

Figure 1. Image of $F\left(z\right)$ for $\beta =0.5$, $\rho =1$, and $\gamma =0.25$ in Example 1.

Example 2.

Let $\mathrm{Re}\gamma >-\frac{1}{2},0<a<1$, $f_1$ be the function defined in Example 1 and

$$f_2\left(z\right)=z\frac{{(1+z)}^{[cos\beta \frac{(1+a-2\rho )}{1+a}{e}^{i\beta }-1]}}{{(1-az)}^{\frac{(1+a-2\rho )}{1+a}cos\beta e^{i\beta }}}{(1+\overline{z})}^{[\frac{(1+a-2\rho )}{1+a}cos\beta e^{i\beta }-e^{2i\beta }]}{(1-a\overline{z})}^{\frac{(1+a-2\rho )}{a(1+a)}cos\beta e^{i\beta }}.$$

Then, it is easy to see that $f_1$ and $f_2$ are β-spirallike log-harmonic mappings of order ρ with respect to $a_2\left(z\right)=a_1\left(z\right)=-z{e}^{-2i\beta }.$ Additionally, suppose that

$$F_1\left(z\right)=f_1\left(z\right)|f_1{\left(z\right)|}^{2\gamma }and{F}_2\left(z\right)=f_2\left(z\right){\left|f_2\left(z\right)\right|}^{2\gamma }.$$

Then, Theorem 3 shows that

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(\rho \right),$$

where $0\le \lambda \le 1$ and $\alpha ={tan}^{-1}\left(\frac{tan\beta +2\mathrm{Im}\gamma }{(1+2\mathrm{Re}\gamma )}\right)$.

Example 3.

Let $\mathrm{Re}\gamma >-\frac{1}{2}$,

$$f_1\left(z\right)=\frac{z}{|1+z|}\sqrt{\frac{1-\overline{z}}{1-z}}$$

and

$$f_2\left(z\right)=\frac{z}{1-z}{e}^{\mathrm{Re}\frac{1}{1-z}}.$$

Firstly, we show that $f_1$ and $f_2$ are log-harmonic starlike functions of order $1/2$ with respect to $a_1\left(z\right)=-z$ and $a_2\left(z\right)=\frac{z}{2-z}$, respectively. A direct computation shows that

$$\frac{z{\left(f_1\right)}_z}{f_1}=\frac{1}{1-z^2},\overline{\left(\frac{\overline{z}{\left(f_1\right)}_{\overline{z}}}{f_1}\right)}=\frac{-z}{1-z^2}$$

$$\frac{z{\left(f_2\right)}_z}{f_2}=\frac{2-z}{2(1-z^2)},\overline{\left(\frac{\overline{z}{\left(f_2\right)}_{\overline{z}}}{f_2}\right)}=\frac{z}{2(1-z^2)}.$$

Therefore, we obtain

$$\overline{\left(\frac{\overline{z}{\left(f_1\right)}_{\overline{z}}}{f_1}\right)}=a_1\left(z\right)\frac{z{\left(f_1\right)}_z}{f_1}and\overline{\left(\frac{\overline{z}{\left(f_2\right)}_{\overline{z}}}{f_2}\right)}=a_2\left(z\right)\frac{z{\left(f_2\right)}_z}{f_2},$$

and this means that $f_1$ and $f_2$ are locally univalent log-harmonic functions. Additionally,

$$Re\frac{z{\left(f_1\right)}_z-\overline{z}{\left(f_1\right)}_{\overline{z}}}{f_1}=Re\left(\frac{1}{1-z^2}+\frac{z}{1-z^2}\right)=Re\frac{1}{1-z}>\frac{1}{2},$$

and

$$Re\frac{z{\left(f_2\right)}_z-\overline{z}{\left(f_2\right)}_{\overline{z}}}{f_2}=Re\left(\frac{2-z}{2(1-z^2)}-\frac{z}{2(1-z^2)}\right)=Re\frac{1}{1+z}>\frac{1}{2}.$$

Hence, $f_1$ and $f_2$ are starlike log-harmonic functions of order $1/2$. Additionally, let

$$F_1\left(z\right)=f_1\left(z\right)|f_1{\left(z\right)|}^{2\gamma }and{F}_2\left(z\right)=f_2\left(z\right){\left|f_2\left(z\right)\right|}^{2\gamma }.$$

Since for $z=r{e}^{i\theta }$,

$$\begin{array}{cc}\hfill & \mathrm{Re}(1-a_1{\overline{a}}_2)(1+\frac{z{h}_1^{\prime }}{h_1})\overline{(1+\frac{z{h}_2^{\prime }}{h_2})}\hfill \\ \hfill & ={(1-|z|}^2)\mathrm{Re}\frac{1}{{(1-\overline{z})}^2}\frac{1}{1-z^2}=\frac{1-{\left|z\right|}^2}{{|1-z|}^2}\mathrm{Re}\frac{1}{(1-\overline{z})(1+z)}\hfill \\ \hfill & =\frac{1-r^2}{|1-r{e}^i{\theta |}^2}(1-r^2)>0.\hfill \end{array}$$

Theorem 4 implies that

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(\frac{1}{2}\right),$$

where $0\le \lambda \le 1$ and $\alpha ={tan}^{-1}\left(\frac{2\mathrm{Im}\gamma }{1+2\mathrm{Re}\gamma }\right)$.

The images in Examples 2–4 are shown in Figure 2, Figure 3 and Figure 4.

Figure 2. Images of $f_1\left(z\right)$ and $f_2\left(z\right)$ in Example 3.

Figure 3. Images of $F_1\left(z\right)$ and $F_2\left(z\right)$ for $\gamma =1+i$ in Example 3.

Figure 4. Image of $F\left(z\right)$ for $\gamma =1+i$ and $\lambda =0.5$ in Example 3.

Example 4.

Let $\mathrm{Re}\gamma >\frac{1}{2}$, $a_1\left(z\right)=z$, and $h_1\left(z\right)=g_1\left(z\right)=\frac{1}{1-z}$. Moreover, let $a_2\left(z\right)=-z$ and $h_2\left(z\right)=g_2\left(z\right)=\frac{1}{1+z}$. Then, it is easy to verify that all conditions of Theorem 5 are satisfied. Hence, according to Theorem 5, by taking

$$F_1\left(z\right)=\frac{{z\left|z\right|}^{2\gamma }}{{(1-z)}^{1+2\gamma }{(1-\overline{z})}^{1+2\gamma }}$$

and

$$F_2\left(z\right)=\frac{{z\left|z\right|}^{2\gamma }}{{(1+z)}^{1+2\gamma }{(1+\overline{z})}^{1+2\gamma }},$$

we have

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(1\right),$$

where $0\le \lambda \le 1$ and $\alpha ={tan}^{-1}\left(\frac{2\mathrm{Im}\gamma }{1-\rho +2\mathrm{Re}\gamma }\right)$.

Example 5.

Let $\mathrm{Re}\gamma >-\frac{1}{2}$, $a_1\left(z\right)=-z$ and $h_1\left(z\right)=\frac{1}{1-z},g\left(z\right)=1-z$. Moreover, let $a_2\left(z\right)=z$ and $h_2\left(z\right)=\frac{1}{1+z},g_2\left(z\right)=1+z$. Then, it is easy to verify that all conditions of Theorem 6 are satisfied. Hence, according to Theorem 6, by taking

$$F_1\left(z\right)=\frac{{z\left|z\right|}^{2\gamma }(1-\overline{z})}{(1-z)}and{F}_2\left(z\right)=\frac{{z\left|z\right|}^{2\gamma }(1+\overline{z})}{(1+z)},$$

we have

$$F\left(z\right)=F_1^{\lambda }\left(z\right)F_2^{1-\lambda }\left(z\right)\in S_{LH}^{\alpha }\left(1\right),$$

where $0\le \lambda \le 1$ and $\alpha ={tan}^{-1}\left(\frac{2\mathrm{Im}\gamma }{1-\rho +2\mathrm{Re}\gamma }\right)$.

4. Conclusions

In this paper, we have shown that, if $f\left(z\right)=zh\left(z\right)\overline{g}\left(z\right)$ is spirallike log-harmonic of order $\rho$, then by choosing suitable parameters of $\alpha$ and $\gamma$, the function $F\left(z\right)=f\left(z\right)\left|f\right(z{|}^{2\gamma }$ is log-harmonic spirallike of order $\alpha$. Moreover, we provide some examples for the obtained results.