On certain subclasses of analytic and harmonic mappings

Theorem 2.1. Let M > 0, $\alpha \in (0,1]$ and $f = h + \overline{g} \in \mathcal{D}^0_{\mathcal{H}}(\alpha, M)$ be of the form (1.1). For $n \geq 2$ , we have $|a_n| \leq M/\left(n + (n^2 - 2n)\alpha\right)$ and $|b_n| \leq M/\left(n + (n^2 - 2n)\alpha\right)$ . The result is sharp for the functions $f_1$ and $f_2$ , where the functions are given by $f_1(z) = z + Mz^n/\left(n + (n^2 - 2n)\alpha\right)$ and $f_2(z) = z + M\overline{z^n}/\left(n + (n^2 - 2n)\alpha\right)$ for $n \geq 2$ .

Theorem 2.2. The harmonic map $f = h + \overline{g}$ belongs to $\mathcal{D}^0_{\mathcal{H}}(\alpha, M)$ if, and only if, the function $F_{\varepsilon} = h + \varepsilon g$ belongs to $\mathcal{D}(\alpha, M)$ for each $\varepsilon$ with $|\varepsilon| = 1$ .

Theorem 2.3. Let M > 0, $\alpha \in (0,1]$ and $f = h + \overline{g} \in \mathcal{D}^0_{\mathcal{H}}(\alpha, M)$ be of the form (1.1). Then, $$|z| - \frac{M|z|^2}{2} \le |f(z)| \le |z| + \frac{M|z|^2}{2}.$$ (2.2) For each $z \in \mathbb{D}$ , $z \neq 0$ , equality occurs for the function f given by $f(z) = z + Mz^2/2$ or its suitable rotations.

Theorem 2.4. If $f \in \mathcal{D}^0_{\mathcal{H}}(\alpha, M)$ for M > 0 and $\alpha \in (0, 1]$ , then $J_f(z) \leq (1 + M|z|)^2$ , with equality for the function $f(z) = z + Mz^2/2$ .

Theorem 2.5. Let $f = h + \overline{g} \in \mathcal{H}$ with g'(0) = 0 be given by (1.1). If $$\sum_{n=2}^{\infty} (n + (n^2 - 2n)\alpha) (|a_n| + |b_n|) \le M, \tag{2.10}$$ then $f \in \mathcal{D}^0_{\mathcal{H}}(\alpha, M)$ .

Lemma 2.1. [6] Let $f = h + \overline{g}$ be given by (5.2). If $\sum_{n=2}^{\infty} n(|a_n| + |b_n|) \le 1$ , then f is starlike in $\mathbb{D}$ .

Theorem 2.6. Let M > 0, $\alpha \in (0,1]$ and $f = h + \overline{g} \in \mathcal{D}^0_{\mathcal{H}}(\alpha, M)$ be given by (1.1). Then f is starlike in $|z| \leq r_1$ , where $r_1 \in (0,1)$ is the smallest root of the equation $$2Mr_{2}F_{1}\left(1,\frac{1}{\alpha};1+\frac{1}{\alpha};r\right)-1=0.$$

Lemma 4.1. If p ∈ P is of the form [\(3.8\)](#page-8-2) with c¹ ≥ 0, then $$c_1 = 2p_1, \ c_2 = 2p_1^2 + 2(1 - p_1^2)p_2,$$ (4.1) $$c_3 = 2p_1^3 + 4(1 - p_1^2)p_1p_2 - 2(1 - p_1^2)p_1p_2^2 + 2(1 - p_1^2)(1 - |p_2|^2)p_3$$ (4.2) for some p¹ ∈ [0, 1] and p2, p³ ∈ D := {z ∈ C : |z| ≤ 1}. For p¹ ∈ T := {z ∈ C : |z| = 1}, there is a unique function p ∈ P with c¹ as in [\(4.1\)](#page-9-0), namely $$p(z) = \frac{1 + p_1 z}{1 - p_1 z}, \ z \in \mathbb{D}.$$ For p¹ ∈ D and p² ∈ T, there is a unique function p ∈ P with c1, c² as in [\(4.1\)](#page-9-0), namely $$p(z) = \frac{1 + (p_1 + \overline{p_1}p_2)z + p_2z^2}{1 - (p_1 - \overline{p_1}p_2)z - p_2z^2}, \ z \in \mathbb{D}.$$ (4.3) For p1, p² ∈ D and p³ ∈ T, there is a unique function p ∈ P with c1, c2, c³ as in [\(4.1\)](#page-9-0) and [\(4.2\)](#page-9-0), namely $$p(z) = \frac{1 + (\overline{p_2}p_3 + \overline{p_1}p_2 + p_1)z + (\overline{p_1}p_3 + p_1\overline{p_2}p_3 + p_2)z^2 + p_3z^3}{1 + (\overline{p_2}p_3 + \overline{p_1}p_2 - p_1)z + (\overline{p_1}p_3 - p_1\overline{p_2}p_3 - p_2)z^2 - p_3z^3}, \ z \in \mathbb{D}.$$

Lemma 4.2. [\[12\]](#page-20-29) Let A, B, C ∈ R and $$Y(A, B, C) := \max_{z \in \overline{\mathbb{D}}} (|A + Bz + Cz^2| + 1 - |z|^2).$$ (i) If AC ≥ 0, then $$Y(A, B, C) = \begin{cases} |A| + |B| + |C|, & |B| \ge 2(1 - |C|), \\ 1 + |A| + \frac{B^2}{4(1 - |C|)}, & |B| < 2(1 - |C|). \end{cases}$$ (ii) If AC < 0, then $$Y(A,B,C) = \begin{cases} 1 - |A| + \frac{B^2}{4(1-|C|)}, & -4AC\left(C^{-2} - 1\right) \le B^2 \land |B| < 2\left(1 - |C|\right), \\ 1 + |A| + \frac{B^2}{4(1+|C|)}, & B^2 < \min\left\{4\left(1 + |C|\right)^2, -4AC\left(C^{-2} - 1\right)\right\}, \\ R(A,B,C), & Otherwise, \end{cases}$$ where $$R(A,B,C) = \begin{cases} |A| + |B| + |C|, & |C| (|B| + 4|A|) \le |AB|, \\ -|A| + |B| + |C|, & |AB| \le |C| (|B| - 4|A|), \\ (|A| + |C|) \sqrt{1 - \frac{B^2}{4AC}}, & Otherwise. \end{cases}$$

Theorem 5.1. Let $f \in \mathcal{P}(M)$ with $0 < M \le 1/\log 4 (\approx 0.7213475)$ . Then $$\left| H_{2,1}\left(F_{f^{-1}}/2\right) \right| \le \begin{cases} M^2/36, & 0 < M \le \frac{1}{39}(6 + \sqrt{114}) \\ \frac{M^2}{144}\left(39M^2 - 12M + 2\right), & \frac{1}{39}(6 + \sqrt{114}) < M \le 1/\log 4. \end{cases}$$ (5.1) The inequality (5.1) is sharp.