A Novel Class of Starlike Functions

Theorem 2.1. Let $f \in \mathcal{S}_q^*(\alpha)$ for $0 \le \alpha < 1$ . Then $$\frac{zf'(z)}{f(z)} \prec q_{\alpha}(z) := \begin{cases} \frac{(1-2\alpha)z}{(1-z)(1-(1-z)^{1-2\alpha})}, & \alpha \neq 1/2\\ \frac{-z}{(1-z)\log(1-z)}, & \alpha = 1/2. \end{cases}$$ (2.2)

Theorem 2.2. Let $f \in \mathcal{S}$ of the form (2.5) belongs to $\mathcal{S}_q^*(\alpha)$ . Then $$\sum_{n=2}^{\infty} (n+\alpha-2)c_n < (1-\alpha).$$

Theorem 3.2. Let $f \in \mathcal{S}^*(q_\alpha)$ and $|z_0| = r < 1$ . Then (i) (Growth Theorem): $-f_{\alpha}(-r) \leq |f(z_0)| \leq f_{\alpha}(r)$ $$\frac{(1+r)^{2\alpha-1}-1}{2\alpha-1} \le |f(z_0)| \le \frac{1-(1-r)^{2\alpha-1}}{2\alpha-1} \ (\alpha \ne 1/2); \ \log(1+r) \le |f(z_0)| \le -\log(1-r) \ (\alpha = 1/2).$$ (ii) (Distortion Theorem): $f'_{\alpha}(-r) \leq |f'(z_0)| \leq f'_{\alpha}(r)$ $$(1+r)^{2(\alpha-1)} \le |f(z_0)| \le (1-r)^{2(\alpha-1)}.$$ (iii) (Rotation Theorem): $|\arg(f(z_0)/z_0)| \le \max_{|z|=r} \arg(f_\alpha(z)/z)$ . Equality holds at some $z_0 \neq 0$ if and only if f is a rotation of $f_{\alpha}$ . Remark 3.3. Interestingly, the above Growth, Distortion and Rotation results hold in the case of $\mathcal{K}(\alpha)$ [6, Theorem 1, p. 139] as well. It is due to the fact that the classes $\mathcal{S}^*(q_\alpha)$ and $\mathcal{K}(\alpha)$ have the same extremal function $f_\alpha(z)$ . 3.1. Coefficient Estimates. From [7], we get $|a_2| \leq 1 - \alpha$ for functions in $\mathcal{S}^*(q_\alpha)$ and the extremal function is $f_\alpha(z)$ , defined in (3.2). Using Fekete-Szegö bounds for Ma-Minda class [14] and [7, Remark 4.1, p. 12], we obtain the following:

Theorem 3.4. Let $f \in S^*(q_\alpha)$ . Then we have $$|a_3 - ta_2^2| \le \begin{cases} \frac{(3 - 2\alpha)(1 - \alpha)}{3} - t(1 - \alpha)^2, & t \le \frac{3 - 4\alpha}{6(1 - \alpha)} \\ \frac{(1 - \alpha)}{2}, & \frac{3 - 4\alpha}{6(1 - \alpha)} \le t \le \frac{9 - 4\alpha}{6(1 - \alpha)} \\ t(1 - \alpha)^2 - \frac{(3 - 2\alpha)(1 - \alpha)}{3}, & t \ge \frac{9 - 4\alpha}{6(1 - \alpha)}. \end{cases}$$ The result is sharp. Taking t = 0 and 1, respectively in the above result, we get the following

Corollary 3.5. Let $f \in \mathcal{S}^*(q_\alpha)$ . Then we have (i) $$|a_3| \le \begin{cases} \frac{(2\alpha - 3)(\alpha - 1)}{3}, & \alpha \le 3/4\\ \frac{1 - \alpha}{2}, & \alpha \ge 3/4. \end{cases}$$ The result is sharp and the extremal function is given by $f_{\alpha}(z)$ , given in (3.2) for $\alpha \leq 3/4$ and $\tilde{f}_{\alpha}(z) = \sqrt{\frac{(1-z^2)^{2a-1}-1}{2a-1}}$ for $\alpha \geq 3/4$ . (ii) $|a_3 - a_2^2| \le \frac{1-\alpha}{2}$ . The result is sharp for $\tilde{f}_{\alpha}(z)$ , defined in the above part (i) for $\alpha \ne 1/2$ and $f_{1/2}(z) := \sqrt{\log(1-z^2)}$ for $\alpha = 1/2$ . Recall that inverse of a function $f(z) = z + \sum_{n=2}^{\infty} a_n z^n \in \mathcal{S}$ is given by $f^{-1}(f(z)) = z$ , $z \in \mathbb{D}$ and $f(f^{-1}(w)) = w$ ( $|w| < r_0, r_0 > 1/4$ ), for which $$f^{-1}(w) = w - a_2 w^2 - (2a_2^2 - a_3)w^3 - (5a_3^2 - 5a_2 a_3 + a_4)w^4 + \cdots$$ (3.5)

Corollary 3.6. Let $f \in \mathcal{S}^*(q_\alpha)$ and the inverse be given by $f^{-1}(w) = w + \sum_{n=2}^{\infty} b_n w^n$ . Then (i) $|b_2| \le 1 - \alpha$ , (ii) $$|b_3| \le \begin{cases} 2(1-\alpha)^2 - \frac{(3-2\alpha)(1-\alpha)}{3}, & 0 \le \alpha \le 3/8\\ \frac{1-\alpha}{2}, & 3/8 \le \alpha \le 1. \end{cases}$$ The inequalities are sharp.

Theorem 3.7. Let $f \in S^*(q_\alpha)$ $(\alpha \in E(\alpha))$ . Then, we have $$|\beta_n| \le \frac{1-\alpha}{2n}.$$ The inequality is sharp.

Theorem 3.8. A function $f \in \mathcal{S}$ belongs to $\mathcal{S}^*(q_\alpha)$ if and only if it satisfies $$\frac{1}{z}\left(f(z)*\frac{z-\lambda z^2}{(1-z)^2}\right) \neq 0 \quad (z \in \mathbb{D}),\tag{3.6}$$ where $$\lambda := \lambda(\theta) = \begin{cases} \frac{1 - 2\alpha}{(1 - 2\alpha) - (e^{-i\theta} - 1)(1 - (1 - e^{i\theta})^{1 - 2\alpha})}, & \alpha \neq 1/2\\ (1 - (1 - e^{-i\theta})\log(1 - e^{i\theta}))^{-1}, & \alpha = 1/2, \end{cases}$$ $(\theta \in [-\pi, \pi]).$

Corollary 3.9. A function f ∈ S belongs to S ∗ (qα) if and only if it satisfies $$1 \neq \sum_{n=2}^{\infty} \rho(\theta) a_n z^{n-1} := \begin{cases} \sum_{n=2}^{\infty} \frac{(e^{-i\theta} - 1)(1 - (1 - e^{i\theta})^{1-2\alpha})n - (1 - 2\alpha)}{(1 - 2\alpha) - (e^{-i\theta} - 1)(1 - (1 - e^{i\theta})^{1-2\alpha})} a_n z^{n-1}, & \alpha \neq 1/2 \\ \sum_{n=2}^{\infty} \frac{(1 - e^{-i\theta})\log(1 - e^{i\theta})n - 1}{1 - (1 - e^{-i\theta})\log(1 - e^{i\theta})} a_n z^{n-1}, & \alpha = 1/2. \end{cases}$$ Consequently, the next result follows.

Corollary 3.10. Let f ∈ S satisfies $$\sum_{n=2}^{\infty} |\rho(\theta)| |a_n| < 1,$$ then f ∈ S^∗ (qα).

Theorem 4.1. Let 0 ≤ α < 1, then S ∗ q (α) ⊂ K(α). Proof. Let f ∈ S^∗ q (α), then inequality [\(1.2\)](#page-1-0) yields $$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) > -\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} - \alpha\right),$$ which upon a straightforward calculation yields $$\operatorname{Re}\left(1 + \frac{zf''(z)}{f'(z)}\right) > \alpha.$$ This completes the proof. Silverman [21] introduced the class $$\mathcal{G}_b := \left\{ f \in \mathcal{A} : \left| \frac{1 + z f''(z) / f'(z)}{z f'(z) / f(z)} - 1 \right| < b, \ 0 < b \le 1 \right\}.$$ In the following result we show the relation between the classes $\mathcal{G}_b$ and $\mathcal{S}_q^*$ .

Corollary 4.2. $\mathcal{S}_q^* \subset \mathcal{G}_1$ .

Corollary 4.3. Let $f \in \mathcal{S}_q^*$ , then f is -1-convex function.

Theorem 4.4. We have $S_q^(\alpha) \subset S^(q_\alpha) \subset S^*(\gamma)$ whenever $0 \le \gamma \le \frac{2\alpha-1}{2(1-2^{1-2\alpha})}$ for $\alpha \ne 1/2$ , or whenever $0 \le \gamma \le 1/\log 4$ for $\alpha = 1/2$ .

Theorem 4.5. Let $f \in \mathcal{S}^*(q_\alpha)$ . Then we have $$\operatorname{Re} \frac{f(z)}{z} > \gamma(\alpha) := \begin{cases} \frac{3 - 2\alpha - 2^{2(1-\alpha)}}{4 - 2\alpha - 3 \cdot 2^{1-2\alpha}}, & \alpha \neq 1/2\\ \frac{2(1 - \log 4)}{2 - 3\log 4}, & \alpha = 1/2, \end{cases}$$ (4.1) which also implies $S_q^(\alpha) \subset S^(q_\alpha) \subset \mathcal{Q}(\gamma(\alpha))$ .

Theorem 5.1. Let $f \in \mathcal{SL}$ . Then $f \in \mathcal{S}_q^(\alpha)$ whenever $|z| < \tilde{r}(\alpha) < 1$ , where $\tilde{r}(\alpha)$ is the smallest positive root of $$2(1-r)(\sqrt{1-r}-\alpha) - r = 0 \quad (0 \le \alpha < 1). \tag{5.1}$$ The result is sharp.

Theorem 5.2. Let $f \in \mathcal{S}_l$ . Then $f \in \mathcal{S}_q^(\alpha)$ whenever $|z| < \tilde{r}(\alpha) < 1$ , where $\tilde{r}(\alpha)$ is the smallest positive root of $$(1 - \log(1+r))(1-r)(1 - \log(1+r) - \alpha) - r = 0 \quad (0 \le \alpha < 1).$$

Theorem 5.3. Let $f \in \mathcal{S}_e$ . We have $f \in \mathcal{S}_q^(\alpha)$ in $|z| < \tilde{r}(\alpha)$ , where $\tilde{r}(\alpha)$ is the smallest positive root of $$\begin{cases} 4(1-r^2)(e^{-r}-\alpha) - (1+r^2)^2 = 0, & 0 \le \alpha < 0.246646 \\ e^{-r}-\alpha - r = 0, & 0.246646 \le \alpha < 1. \end{cases}$$ (5.5)

Theorem 5.4. Let $f \in \mathcal{S}_{SG}$ . We have $f \in \mathcal{S}_q^(\alpha)$ in $|z| < \tilde{r}(\alpha)$ , where $\tilde{r}(\alpha)$ is the smallest positive root of $$\begin{cases} 4(1-r^2)(1+e^{-r})(2-\alpha(1+e^r)) - (1+r^2)^2(1+e^r) = 0, & 0 \le \alpha < 0.546407 \\ (1+e^{-r})(2-\alpha(1+e^r)) - r(1+e^r) = 0, & 0.546407 \le \alpha < 1. \end{cases}$$

Theorem 5.5. Let $f \in \mathcal{S}_{S}^{}$ . Then $f \in \mathcal{S}_{q}^{}(\alpha)$ in $|z| < \tilde{r}(\alpha)$ , where $\tilde{r}(\alpha)$ is the smallest positive root of $$(1 - \sin r)(1 - \sin r \cosh r - \alpha) - r \cosh r = 0 \quad (0 \le \alpha < 1).$$ Proof. Let $f \in \mathcal{S}_S^*$ . Then for some Schwarz function $v(z) = Re^{it}$ $(0 \le R \le r = |z| < 1; 0 \le t \le 2\pi)$ , we have $zf'(z)/f(z) = 1 + \sin(v(z))$ . To prove the result, we show (1.2) holds for f considered here. For this we consider $$\operatorname{Re}(1 + \sin(v(z))) - \left| \frac{zv'(z)\cos(v(z))}{1 + \sin(v(z))} - \alpha \right| \ge \operatorname{Re}(1 + \sin(v(z))) - \alpha - \frac{|z||v'(z)||\cos(v(z))|}{|1 + \sin(v(z))|}$$ (5.9) A calculation shows that $\operatorname{Re}(1+\sin(\upsilon(z)))=1+\sin(R\cos t)\cosh(R\sin t)$ and $\sin(R\cos t)\geq -\sin R, \quad \cosh(R\sin t)\leq \cosh(R).$ Thus we have $$\operatorname{Re}(1+\sin(v(z))) \ge 1-\sin R \cosh R \ge 1-\sin r \cosh r.$$ Also $$\max_{|\upsilon(z)|=R}|\cos(\upsilon(z))|=\cosh R\leq \cosh r \text{ and } \min_{|\upsilon(z)|=R}|1+\sin(\upsilon(z))|=1-\sin R\geq 1-\sin r.$$ Using these inequalities in right side of the inequality (5.9) and further applying (5.6) with |z| = r, (5.9) finally reduces to $$\operatorname{Re}\left(1+\sin(v(z))\right) - \left|\frac{zv'(z)\cos(v(z))}{1+\sin(v(z))} - \alpha\right| \ge 1-\sin r \cosh r - \alpha - \left(\frac{r\cosh r}{1-\sin r}\right) > 0,$$ provided $r < \tilde{r}(\alpha) \le \sqrt{2} - 1$ for every $0 \le \alpha < 1$ .

Theorem 5.6. Let $f \in \mathcal{S}_C$ and $0 \le \alpha < 3/4$ . Then $f \in \mathcal{S}_q^(\alpha)$ in $|z| < \tilde{r}(\alpha)$ , where $\tilde{r}(\alpha)$ is the smallest positive root of $$(3 - 2r^2)(1 - r^2 - 2\alpha) - 6\sqrt{3}r(1+r) = 0 \quad (0 \le \alpha < 3/4).$$

Theorem 5.8. Let $f \in S^*(q_\alpha)$ $(0 \le \alpha < 1)$ . Then the following holds: - (i) f is starlike of order $\gamma$ in $|z| < \tilde{r}$ whenever - (a) $\frac{2\alpha-1}{2(1-2^{1-2\alpha})} < \gamma < 1$ for $\alpha \neq 1/2$ , where $\tilde{r}$ is the smallest such r < 1 satisfying the equation $$(2\alpha - 1)r - \gamma(1+r)(1 - (1+r)^{1-2\alpha}) = 0,$$ or whenever (b) $\frac{1}{\log 4} < \gamma < 1$ for $\alpha = 1/2$ , where $\tilde{r}$ is the smallest such r < 1 satisfying the equation $$r - \gamma(1+r)\log(1+r) = 0.$$ (ii) Let $\beta > 1$ then $f \in \mathcal{M}(\beta)$ in $|z| < r_0$ if there exists $r_0$ , the smallest such r < 1 satisfying the equation $$(1-2\alpha)r - \beta(1-r)(1-(1-r)^{1-2\alpha}) = 0 \ (\alpha \neq 1/2) \quad or \quad r+\beta(1-r)(\log(1-r)) = 0 \ (\alpha = 1/2),$$ $else \ r_0 = 1.$

Theorem 5.10. Let $f \in S^*(q_\alpha)$ and $\beta \geq 1$ . Then for $|z| < \tilde{r}$ , we have $$\left| \frac{zf'(z)}{f(z)} - 1 \right| < \beta,$$ where $\gamma := \gamma(\alpha)$ given by (4.1) and $\tilde{r}$ is the smallest such r < 1 satisfying the equation $$2(1-\gamma)r - \beta(1-r)(1-|1-2\gamma|r) = 0.$$

Definition 1.1. For $f \in \mathcal{S}$ , we define below a class $\mathcal{S}_{q}^{*}(\alpha)$ if f satisfies: $$\operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) > \left|1 + \frac{zf''(z)}{f'(z)} - \frac{zf'(z)}{f(z)} - \alpha\right| \quad (0 \le \alpha < 1). \tag{1.2}$$ We observe that the class $\mathcal{S}_q^(\alpha) \subseteq \mathcal{S}$ . The identity function f(z) = z satisfies the inequality (1.2) for all $0 \le \alpha < 1$ , hence $\mathcal{S}_q^(\alpha) \ne \emptyset$ . For $\alpha = 0$ , we set $\mathcal{S}_q^(0) =: \mathcal{S}_q$ . We also obtain $\mathcal{S}_q^(\alpha)$ as a subclass of another fascinating class of starlike functions, which is introduced in the following section. The paper is structured as follows: In the next section, we show $f \in \mathcal{S}_q^*(\alpha)$ implies $$\frac{zf'(z)}{f(z)} \prec q_{\alpha}(z),$$ where $q_{\alpha}$ is given in (2.2). This function $q_{\alpha}(z)$ which is brought to day light by studying the class (1.2), is first introduced by MacGregor [9]. Using $q_{\alpha}(z)$ , MacGregor [9] also proved that convex functions of order $\alpha$ is starlike of order $$\frac{2\alpha - 1}{2 - 2^{2(1-\alpha)}} \ (\alpha \neq 1/2) \ \text{ and } \ \frac{1}{\log 4} \ (\alpha = 1/2).$$ Apart from this, he [9] also proved $q_{\alpha}(z)$ to be univalent and $$\min_{|z| < 1} \operatorname{Re} q_{\alpha}(z) = q_{\alpha}(-1),$$ can refer [12] also. We study this function in detail and obtain certain sharp coefficient bounds for functions in $\mathcal{S}^(q_\alpha)$ . In the sequel we discuss many inclusion and radius results pertaining to the classes $\mathcal{S}_q^(\alpha)$ and $\mathcal{S}^*(q_\alpha)$ . 2. Results related to $$\mathcal{S}_q^*(\alpha)$$ We begin this section by showing the existence of some analytic function belonging to $\mathcal{S}_q^*(\alpha)$ apart from f(z) = z. Example 1. Let $f_{\gamma}(z) = z + \gamma z^2$ . If $|\gamma| < r_0$ , where $r_0$ is the smallest such r < 1 satisfying the equation: $$\alpha^{2}(2r-1)^{2}(r-1)^{3} - 33r^{2} + 57r^{3} - 48r^{4} + 16r^{5} + 2\alpha(r-1)^{2}r(2r-1) + (2r-1)(r-1)(\alpha+r-3\alpha r + 2\alpha r^{2})(2-4r) + 9r = 1 \quad (0 \le \alpha < 1),$$ then $f_{\gamma} \in \mathcal{S}_{q}^{*}(\alpha)$ . Moreover, $r_{0} \in (0, 1/4]$ .

Definition 2.3. Let $\mathcal{S}^*(q_\alpha)$ denote the class of analytic functions $f \in \mathcal{S}$ , satisfying $$\frac{zf'(z)}{f(z)} \prec q_{\alpha}(z) \quad (z \in \mathbb{D}, \ 0 \le \alpha < 1). \tag{2.6}$$ We infer from Theorem 2.1, $\mathcal{S}_q^(\alpha) \subset \mathcal{S}^(q_\alpha)$ , so is non-empty. This class generalizes subclass of $\mathcal{S}$ , such as for $\alpha = 0$ , it reduces to $\mathcal{S}^(1/2)$ . 3. About $$S^*(q_\alpha)$$ A function $f \in \mathcal{S}^*(q_\alpha)$ if and only if there exists an analytic function $s, s(z) \prec q_\alpha(z)$ such that $$f(z) = z \exp \int_0^z \frac{s(t) - 1}{t} dt.$$ (3.1) Specifically, for $s(z)=q_{\alpha}(z),$ the structural formula (3.1) yields $$f_{\alpha}(z) = \begin{cases} \frac{1 - (1 - z)^{2\alpha - 1}}{2\alpha - 1}, & \alpha \neq 1/2; \\ -\log(1 - z), & \alpha = 1/2. \end{cases}$$ (3.2) Taylor series of $f_{\alpha}(z)$ $(0 \le \alpha < 1)$ is given as follows $$f_{\alpha}(z) = z + (1 - \alpha)z^{2} + (3 - 5\alpha + 2\alpha^{2})\frac{z^{3}}{3} + (6 - 13\alpha + 9\alpha^{2} - 2\alpha^{3})\frac{z^{4}}{6} + \cdots$$ $$= z + \sum_{n=2}^{\infty} \left(\frac{\prod_{j=2}^{n} (j - 2\alpha)}{n!} z^{n}\right) \quad (0 \le \alpha < 1),$$ plays an extremal function for many cases in $\mathcal{S}^*(q_\alpha)$ . Interestingly, this function $f_\alpha(z)$ is also the extremal function for the class $\mathcal{K}(\alpha)$ see [4,12].  FIGURE 1. Image of unit disk $\mathbb{D}$ under the function $q_{\alpha}(z)$ for various $\alpha$ . Remark 3.1. Using the results in the proof of Theorem 2.1, we deduce $$p(z) + \frac{zp'(z)}{p(z)} \prec \frac{1 + (1 - 2\alpha)z}{1 - z} \Rightarrow p(z) \prec q_{\alpha}(z),$$ which further yields $$1 + \frac{zf''(z)}{f'(z)} \prec \frac{1 + (1 - 2\alpha)z}{1 - z} \Rightarrow \frac{zf'(z)}{f(z)} \prec q_{\alpha}(z).$$ (3.3) The function $f_{\alpha}(z)$ , given by (3.2) is an extremal function for the above differential subordination implication. This result (3.3) is initially proved by MacGregor [9] and later in [12, p. 113–115], using the following: $$\min_{|z|<1} (\operatorname{Re} q_{\alpha}(z)) = q_{\alpha}(-1) = \begin{cases} \frac{2\alpha - 1}{2 - 2^{2(1-\alpha)}}, & \alpha \neq 1/2\\ \frac{1}{\log 4}, & \alpha = 1/2, \end{cases}$$ (3.4) and $\min_{|z|=r} \operatorname{Re} q_{\alpha}(z) = q_{\alpha}(-r), \, \max_{|z|=r} \operatorname{Re} q_{\alpha}(z) = q_{\alpha}(r)$ We first show that an analytic univalent function $q_{\alpha}(z)$ is a Ma-Minda function. We have $q_{\alpha}(0) = 1$ and $\operatorname{Re} q_{\alpha}(z) > 0$ ( $0 \le \alpha < 1$ ). A further calculation reveals $q'_{\alpha}(0) > 0$ . Now the Taylor series of $q_{\alpha}(z)$ is given as follows: $$q_{\alpha}(z) = 1 + (1 - \alpha)z + (3 - 4\alpha + \alpha^2)\frac{z^2}{3} + (2 - 3\alpha + \alpha^2)\frac{z^3}{2} + (45 - 72\alpha + 26\alpha^2 + 2\alpha^3 - \alpha^4)\frac{z^4}{45} + \cdots,$$ shows the function is symmetric with respect to the real axis as it has real coefficients. For detail analysis for the geometry of functions defined on $\mathbb{D}$ , see [7]. The function $q_{\alpha}(z)$ is starlike with respect to $q_{\alpha}(0) = 1$ as we get $$\operatorname{Re}\left(\frac{e^{i\theta}q_{\alpha}'(e^{i\theta})}{q_{\alpha}(e^{i\theta})-1}\right) > 0, \qquad (-\pi \le \theta \le \pi) \quad (0 \le \alpha < 1),$$ by performing a highly complex computation in Mathematica 11.0, which otherwise is not possible manually. Therefore $q_{\alpha}(z)$ is a Ma-Minda function as it satisfies all the conditions to be one. Figure 1 depicts $q_{\alpha}(\mathbb{D})$ for various $\alpha \in [0,1)$ . Observation: It is clear that $q_0(z)$ is a convex function but it is not the case for every $q_{\alpha}$ . In fact, using Mathematica 11.0, graph of Re $\left(1 + \frac{zq_{\alpha}''(z)}{q_{\alpha}'(z)}\right)\Big|_{z=e^{i\theta}}$ $(-\pi \le \theta \le \pi)$ , is not positive for some $\alpha \in (0,1)$ . Thus we coin below a problem which is open at present. Open Problem: Find the range of $\alpha$ in (0,1) for which $q_{\alpha}(\mathbb{D})$ is convex. Based on certain subordination results proved in [8], we have $f(z)/z \prec f_{\alpha}(z)/z$ and the following: