An investigation on a new class of harmonic mappings

Theorem 1.1. [5] Let $s(z) \in \mathcal{S}_{\alpha}^*$ , then $$rF(\cos\alpha, -r) \le |s(z)| \le rF(\cos\alpha, r)$$ (1.4) and $$[(1-r)\cos\alpha - (1+r)\sin\alpha]F(\cos\alpha, -r)$$ $$\leq |s'(z)| \leq [(1+r)\cos\alpha + (1-r)\sin\alpha]F(\cos\alpha, r), \tag{1.5}$$ where $$F(\cos\alpha, r) = \frac{1}{(1+r)^{\cos\alpha(\cos\alpha-1)}(1-r)^{\cos\alpha(\cos\alpha+1)}}$$ (1.6) for all |z| = r < 1 and $|\alpha| < \pi/2$ .

Lemma 1.2. [6] Let $\phi(z)$ be regular in the unit disc $\mathbb{D}$ with $\phi(0) = 0$ , then if $|\phi(z)|$ attains its maximum value on the circle |z| = r at the point $z_1$ , one has $z_1\phi'(z_1) = k\phi(z_1)$ for some $k \ge 1$ .

Lemma 1.3. [7] If $s_1(z)$ and $s_2(z)$ are regular in $\mathbb{D}$ , $s_1(0) = s_2(0)$ , $s_2(z)$ maps $\mathbb{D}$ onto a many-sheeted region which is starlike with respect to the origin, and $s_1'(z)/s_2'(z) \in \mathcal{P}$ , then $s_1(z)/s_2(z) \in \mathcal{P}$ .

Lemma 1.4. [8] Let $f = h(z) + \overline{g(z)}$ be an element of $S_{\mathcal{HPST}(\alpha)^*}$ , then $$\frac{|b_1| - r}{1 - |b_1|r} \le \left| \frac{g'(z)}{h'(z)} \right| \le \frac{|b_1| + r}{1 + |b_1|r} \tag{1.7}$$ for all |z| = r < 1. This inequality is sharp because the extremal function is $$e^{i\alpha}\frac{g'(z)}{h'(z)}=\frac{z+b}{1+\bar{b}z},$$ where $b = e^{i\alpha}b_1$ .

Theorem 2.1. $s(z) \in \mathcal{S}_{\alpha}^*$ if and only if $$z\frac{s'(z)}{s(z)} - 1 < \frac{2\cos\alpha e^{-i\alpha}z}{1 - z} \tag{2.1}$$ for all $z \in \mathbb{D}$ . Proof Let s(z) in $\mathcal{S}_{\alpha}^*$ , then we have $$e^{i\alpha}z\frac{s'(z)}{s(z)} = \cos\alpha\frac{1+\phi(z)}{1-\phi(z)} + i\sin\alpha$$ or $$e^{i\alpha}z\frac{s'(z)}{s(z)} = e^{i\alpha}\frac{1+\phi(z)}{1-\phi(z)}$$ for some $\phi(z) \in \Omega$ and all $z \in \mathbb{D}$ . Thus $$z\frac{s'(z)}{s(z)} - 1 = \frac{1 + e^{-2i\alpha}\phi(z)}{1 - \phi(z)} - 1$$ $$= \frac{1 + (\cos 2\alpha - i\sin 2\alpha)\phi(z) - 1 + \phi(z)}{1 - \phi(z)}$$ $$= \frac{2\cos \alpha e^{-i\alpha}\phi(z)}{1 - \phi(z)}$$ for some $\phi(z) \in \Omega$ and all $z \in \mathbb{D}$ . Since $\phi(z) \in \Omega$ , we have that (2.1) is true. The sufficient part of the proof can be seen by following the above steps in the opposite direction by considering the subordination principle.

Theorem 2.2. Let $f = h(z) + \overline{g(z)}$ be an element of $S_{\mathcal{H}(\alpha)}$ . If $w(z) = \frac{g'(z)}{h'(z)} \in \mathcal{P}$ , then $\frac{g(z)}{h(z)} \in \mathcal{P}$ for all $z \in \mathbb{D}$ . Proof A version of this theorem was proved by Sakaguchi for a univalent starlike function [7, 9]. Since $f = h(z) + \overline{g(z)} \in \mathcal{S}_{\mathcal{H}(\alpha)}$ , then h(z) and g(z) are regular in $\mathbb{D}$ and h(0) = g(0) = 0. On the other hand, we have $$w(z) = \frac{g'(z)}{h'(z)} \in \mathcal{P} \quad \text{if and only if} \quad \frac{g'(z)}{h'(z)} < \frac{1+z}{1+z}$$ (2.2) for all $z \in \mathbb{D}$ . Geometrically, this means that $\frac{g'(z)}{h'(z)}$ maps $\mathbb{D}$ inside the open disc centered on the real axis with diameter end points $\frac{1-r}{1+r}$ and $\frac{1+r}{1-r}$ . Now we define a function $\phi(z)$ by $$\frac{g(z)}{h(z)} = \frac{1 + \phi(z)}{1 - \phi(z)} \quad (z \in \mathbb{D}). \tag{2.3}$$ Then $\phi(z)$ is analytic in $\mathbb{D}$ , and $\phi(0) = 0$ . On the other hand, $$w(z) = \frac{g'(z)}{h'(z)} = \frac{2z\phi'(z)}{1 - \phi(z)} \frac{1}{e^{i\alpha}(1 + e^{-2i\alpha})\phi(z)} + \frac{1 + \phi(z)}{1 - \phi(z)} \quad (z \in \mathbb{D}).$$ (2.4) Now, it is easy to realize that the subordination (2.2) is equivalent to $|\phi(z)| < 1$ in (2.3) for all $z \in \mathbb{D}$ . Indeed, assume to the contrary that there exists $z_1 \in \mathbb{D}$ such that $|\phi(z_1)| = 1$ . Then by Jack's lemma (Lemma 1.2), $z_1\phi'(z_1) = k\phi(z_1)$ , $k \ge 1$ , for such $z_1$ we have $$w(z_1) = \frac{g'(z_1)}{h'(z_1)} = \frac{2k\phi(z_1)}{1-\phi(z_1)} \frac{1}{e^{i\alpha}(1+e^{-2i\alpha})\phi(z_1)} + \frac{1+\phi(z_1)}{1-\phi(z_1)} = w\big(\phi(z_1)\big) \notin w(\mathbb{D}),$$ since $|\phi(z_1)| = 1$ and $k \ge 1$ . But this is a contradiction to the condition $w(z) = \frac{g'(z)}{h'(z)} < \frac{1+z}{1-z}$ , and so the assumption is wrong, i.e., $|\phi(z)| < 1$ for all $z \in \mathbb{D}$ . Remark 2.3 Theorem 2.2 is an extension of Lemma 1.3 to the harmonic mappings.

Corollary 2.4. Let $f = h(z) + \overline{g(z)}$ be an element of $S_{\mathcal{H}(\alpha)}$ , then $$\frac{r(1-r)}{1+r}F(\cos\alpha, -r) \le \left| g(z) \right| \le \frac{r(1+r)}{1-r}F(\cos\alpha, r) \tag{2.5}$$ and $$\left[ (1-r)\cos\alpha - (1+r)\sin\alpha \right] \frac{1-r}{1+r} F(\cos\alpha, -r) \leq \left| g'(z) \right| \leq \left[ (1+r)\cos\alpha + (1-r)\sin\alpha \right] \frac{1+r}{1-r} F(\cos\alpha, r), \tag{2.6}$$ where $F(\cos \alpha, r)$ is given by (1.6) for all |z| = r < 1 and $|\alpha| < \pi/2$ . Proof The proof of this theorem is a simple consequence of Theorem 1.1 and Theorem 2.2 since $$\operatorname{Re} w(z) = \operatorname{Re} \frac{g'(z)}{h'(z)} \quad \Rightarrow \quad \frac{g'(z)}{h'(z)} \prec \frac{1+z}{1-z} \quad \Rightarrow \quad \frac{1-r}{1+r} \leq \left| \frac{g'(z)}{h'(z)} \right| \leq \frac{1+r}{1-r},$$ then $$\left|h'(z)\right| \frac{1-r}{1+r} \le \left|g'(z)\right| \le \left|h'(z)\right| \frac{1+r}{1-r}$$ and $$\operatorname{Re} \frac{g(z)}{h(z)} > 0 \quad \Rightarrow \quad \frac{g(z)}{h(z)} \prec \frac{1+z}{1-z} \quad \Rightarrow \quad \frac{1-r}{1+r} \le \left| \frac{g(z)}{h(z)} \right| \le \frac{1+r}{1-r},$$ then $$\left|h(z)\right| \frac{1-r}{1+r} \le \left|g(z)\right| \le \left|h(z)\right| \frac{1+r}{1-r}.$$

Corollary 2.5. Let $f = h(z) + \overline{g(z)}$ be an element of $S_{\mathcal{H}(\alpha)}$ , then $$\begin{split} & \frac{[(1-r)\cos\alpha - (1+r)\sin\alpha]^2 (F(\cos\alpha, -r))^2 (1-r^2)(1-|b_1|^2)}{(1+|b_1|r)^2} \\ & \leq J_{f(z)} \leq \frac{[(1+r)\cos\alpha + (1-r)\sin\alpha]^2 (F(\cos\alpha, r))^2 (1-r^2)(1-|b_1|^2)}{(1-|b_1|r)^2}, \end{split}$$ where $F(\cos \alpha, r)$ is given by (1.6) for all |z| = r < 1, $|\alpha| < \pi/2$ and $J_{f(z)}$ is the Jacobian of f defined by $J_{f(z)} = |h'(z)|^2 - |g'(z)|^2$ for all $z \in \mathbb{D}$ . Proof The proof of this corollary is a simple consequence of Lemma 1.1 and Lemma 1.4.