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Abstract

In this paper, we introduce a family of analytic functions given by $$ψ_{A,B}(z):= \dfrac{1}{A-B}\log{\dfrac{1+Az}{1+Bz}},$$ which maps univalently the unit disk onto either elliptical or strip domains, where either $A=-B=α$ or $A=αe^{iγ}$ and $B=αe^{-iγ}$ ($α\in(0,1]$ and $γ\in(0,π/2]$). We study a class of non-univalent analytic functions defined by \begin{equation*} \mathcal{F}[A,B]:=\left\{f\in\mathcal{A}:\left( \dfrac{zf'(z)}{f(z)}-1\right)\precψ_{A,B}(z)\right \}. \end{equation*} Furth

Results & Lemmas (14)

Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.

Lemma 2.2 Lemma 2.2. The function is convex and univalent on. Proof. Let. By [12, Corollary 3], it is enough to show that Re H(z) > 0. Since, using…
Lemma 2.2. The function $\psi_{A,B}(z)$ is convex and univalent on $\mathbb{D}$ . Proof. Let $H(z) := 1 + z \psi''_{A,B}(z) / \psi'_{A,B}(z)$ . By [12, Corollary 3], it is enough to show that Re H(z) > 0. Since $|A| = |B| = \alpha \le 1$ , using basic calculation with [4, Theorem 1], we obtain Re $$H(z)$$ = Re $\left(-1 + \frac{1}{1 + Az} + \frac{1}{1 + Bz}\right)$ > $-1 + \frac{2}{1 - \alpha} > 0$ . Thus the result holds.
Theorem 2.3 Theorem 2.3. Let, then in the disk, (i) For, we have (2.2) (ii) For and, we have where and
Theorem 2.3. Let $p \in \mathcal{L}[A, B]$ , then in the disk $\mathbb{D}_r = \{z \in \mathbb{C} : |z| \le r < 1\}$ , (i) For $A = -B = \alpha$ , we have $$|\operatorname{Re} p(z)| \le \frac{1}{2\alpha} \log \left( \frac{1 + \alpha r}{1 - \alpha r} \right)$$ (2.2) $$|\operatorname{Im} p(z)| \le \frac{1}{2\alpha} \sin^{-1} \left( \frac{2\alpha r}{1 + \alpha^2 r^2} \right). \tag{2.3}$$ (ii) For $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , we have $$\frac{-1}{2\alpha \sin \gamma} (\eta - \tau) \le \operatorname{Re} p(z) \le \frac{1}{2\alpha \sin \gamma} (\eta + \tau). \tag{2.4}$$ $$\frac{1}{2\alpha \sin \gamma} \log T_2 \le \operatorname{Im} p(z) \le \frac{1}{2\alpha \sin \gamma} \log T_1. \tag{2.5}$$ where $$\eta := \sin^{-1}\left(\frac{2\alpha r \sin \gamma}{\sqrt{1 + \alpha^4 r^4 - 2\alpha^2 r^2 \cos 2\gamma}}\right),$$ $$\tau := \tan^{-1}\left(\frac{-\alpha^2 r^2 \sin 2\gamma}{1 - \alpha^2 r^2 \cos 2\gamma}\right)$$ and $$T_j := \left(\frac{\sqrt{1 + \alpha^4 r^4 - 2\alpha^2 r^2 \cos 2\gamma} + (-1)^j 2\alpha r \sin \gamma}{1 - \alpha^2 r^2}\right)^{-1} \quad (j = 1, 2).$$
Lemma 2.5 Lemma 2.5. Let and, then we have (i) for, (ii) for and, Further for, we have (iii) for, (iv) for and,
Lemma 2.5. Let $\alpha \in (0,1)$ and $\gamma \in (0,\pi/2]$ , then we have (i) for $$A = -B = \alpha$$ , $$\{w \in \mathbb{C} : |w-1| < h_2\} \subset 1 + \psi_{A,B}(\mathbb{D}) \subset \{w \in \mathbb{C} : |w-1| < h_1\}.$$ (ii) for $$A = \alpha e^{i\gamma}$$ and $B = \alpha e^{-i\gamma}$ , $$\{w \in \mathbb{C} : |w-1| < k_1 + k\} \subset 1 + \psi_{AB}(\mathbb{D}) \subset \{w \in \mathbb{C} : |w-1| < k_2\}.$$ Further for $\alpha = 1$ , we have (iii) for $$A = -B = 1$$ , $$\left\{w \in \mathbb{C} : |w-1| < \frac{\pi}{4}\right\} \subset 1 + \psi_{A,B}(\mathbb{D}).$$ (iv) for $$A = e^{i\gamma}$$ and $B = e^{-i\gamma}$ , $$\left\{ w \in \mathbb{C} : |w - 1| < \frac{\gamma}{2 \sin \gamma} \right\} \subset 1 + \psi_{A,B}(\mathbb{D}).$$
Lemma 2.7 Lemma 2.7. Let and p(z):= (1 + Cz)/(1 + Dz). Then if and only if (i) for (2.9) (ii) for and, (2.10)
Lemma 2.7. Let $-1 < D < C \le 1$ and p(z) := (1 + Cz)/(1 + Dz). Then $p(z) \prec \mathcal{L}[A, B]$ if and only if (i) for $$A = -B = \alpha$$ $$C \le \begin{cases} h_2 + (1 - h_2)D, & when (1 - CD)/(1 - D^2) \le 1, \\ h_2 + (1 + h_2)D, & when (1 - CD)/(1 - D^2) \ge 1. \end{cases}$$ (2.9) (ii) for $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , $$C \le \begin{cases} k_1 - k + (1 - k_1 + k)D, & when (1 - CD)/(1 - D^2) \le 1 + k, \\ k_1 + k + (1 + k_1 + k)D, & when (1 - CD)/(1 - D^2) \ge 1 + k. \end{cases}$$ (2.10)
Corollary 2.8 Corollary 2.8. Let and be such that zf'(z)/f(z) = (1 + Cz)/(1 + Dz), then if and only if, the condition (2.9) or (2.10) holds. Now from…
Corollary 2.8. Let $-1 < D < C \le 1$ and $f \in A$ be such that zf'(z)/f(z) = (1 + Cz)/(1 + Dz), then $f \in \mathcal{F}[A, B]$ if and only if, the condition (2.9) or (2.10) holds. Now from (1.4), we have $f \in \mathcal{F}[A, B]$ if and only if there exists an analytic function $p(z) \prec \psi_{A,B}(z)$ such that $$f(z) = z \exp\left(\int_0^z \frac{p(t)}{t} dt\right). \tag{2.11}$$ If we take $p(z) = \psi_{A,B}(z)$ , then we obtain from (2.11) $$f_{A,B}(z) := z \exp \frac{Li_2(-Bz) - Li_2(-Az)}{A - B},$$ (2.12) where $Li_2(x) = \sum_{n=1}^{\infty} x^n/n^2$ , denotes the Spence's (or dilogarithm) function. The function $f_{A,B}(z)$ is nonunivalent but have the extremal properties for many problems for the class $\mathcal{F}[A,B]$ (see Fig:1). Kumar and Gangania [5] studied the class $\mathcal{F}(\psi)$ as defined in (1.2) and obtained growth and covering theorem for the case when $1 + \psi(z) \not\prec (1+z)/(1-z)$ . As a consequence of the same and Remark 2.4, we obtain the following result:
Lemma 2.9 Lemma 2.9. Let and be defined as in (2.12), then for |z| = r - (i) Growth Theorem:. - (ii) Covering Theorem: Either f is a rotation of or.…
Lemma 2.9. Let $f \in \mathcal{F}[A, B]$ and $f_{A,B}$ be defined as in (2.12), then for |z| = r - (i) Growth Theorem: $-f_{A,B}(-r) \le |f(z)| \le f_{A,B}(r)$ . - (ii) Covering Theorem: Either f is a rotation of $f_{A,B}$ or $\{w \in \mathbb{C} : |w| \le -f_{A,B}(-1)\} \subset f(\mathbb{D})$ . ![](_page_5_Figure_15.jpeg) FIGURE 1. (A) The images of $\partial \mathbb{D}_{f_{A,B}(1)}$ , $f_{A,B}(\partial \mathbb{D})$ and $\partial \mathbb{D}_{f_{A,B}(-1)}$ , for $A = 0.5e^{i\pi/3}$ and $B = 0.5e^{-i\pi/3}$ and (B) zoomed image of cusp. For $z = re^{i\theta}$ , where $\theta$ is fixed but arbitrary, as a consequence of growth theorem and $\psi_{A,B}(-r) \le \text{Re } \psi_{A,B}(re^{i\theta}) \le \psi_{A,B}(r)$ , we obtain $$\log \frac{f(z)}{z} = \int_0^r \frac{p(te^{i\theta})}{t} dt,$$ where $p(z) := \psi_{A,B}(w(z))$ and w is a Schwarz function. Further $$\frac{f(z)}{z} = \exp \int_0^r \frac{p(te^{i\theta})}{t} dt = \exp \left( \int_0^r \operatorname{Re} \frac{p(te^{i\theta})}{t} dt + i \int_0^r \operatorname{Im} \frac{p(te^{i\theta})}{t} dt \right)$$ $$\left| \frac{f(z)}{z} \right| \le \exp \int_0^r \frac{\operatorname{Re} \psi_{A,B}(te^{i\theta})}{t} dt$$ $$\exp \int_0^r \frac{\psi_{A,B}(-t)}{t} dt \le \left| \frac{f(z)}{z} \right| \le \exp \int_0^r \frac{\psi_{A,B}(t)}{t} dt.$$ Here $L(f,r) := \int_0^{2\pi} |zf'(z)| d\theta$ is the length of the boundary curve f(|z|=r). Now we obtain the following result:
Corollary 2.10 Corollary 2.10. Let and, then for |z| = r, we have (i). - (ii). - (iii) - (iv).
Corollary 2.10. Let $f \in \mathcal{F}[A, B]$ and $M(r) = \exp \int_0^r \frac{\psi_{A,B}(t)}{t} dt$ , then for |z| = r, we have (i) $$M(-r) \le \left| \frac{f(z)}{z} \right| \le M(r)$$ . - (ii) $(1 + \max_{|z| \le r} |\psi_{A,B}(z)|) M(-r) \le |f'(z)| \le (1 + \max_{|z| \le r} |\psi_{A,B}(z)|) M(r)$ . - (iii) $2\pi r(1+\max_{|z|\leq r}|\psi_{A,B}(z)|)M(-r)\leq L(f,r)\leq 2\pi r(1+\max_{|z|< r}|\psi_{A,B}(z)|)M(r).$ - (iv) $f(z)/z \prec f_{A,B}(z)/z$ .
Theorem 3.1 Theorem 3.1. Let,, where and be given numbers. If, then f is starlike of order in the disc, where (i) For, (3.1) (ii) For and, is the…
Theorem 3.1. Let $\alpha \in (0,1]$ , $\gamma \in (0,\gamma_0)$ , where $\gamma_0 \simeq 1.2461...$ and $\delta \in [0,1)$ be given numbers. If $f \in \mathcal{F}[A,B]$ , then f is starlike of order $\delta$ in the disc $|z| < r(\delta)$ , where (i) For $$A = -B = \alpha$$ , $$r(\delta) = \frac{\exp(2\alpha(1-\delta)) - 1}{\alpha(\exp(2\alpha(1-\delta)) + 1)}.$$ (3.1) (ii) For $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , $r(\delta)$ is the smallest positive root of $$\tan(2\alpha\sin\gamma(\delta-1)) = \frac{\alpha^4 r^4 \sin 2\gamma + 2\alpha^3 r^3 \sin\gamma\cos 2\gamma - \alpha^2 r^2 \sin 2\gamma - 2\alpha r \sin\gamma}{\alpha^4 r^4 \cos 2\gamma - 2\alpha^3 r^3 \sin\gamma\sin 2\gamma - \alpha^2 r^2 \cos 2\gamma - \alpha^2 r^2 + 1}.$$ (3.2) The result is sharp.
Theorem 3.2 Theorem 3.2. Let and. If, for the case when and, then there is an unique such that - (i) when, f is starlike of order in, where the root of…
Theorem 3.2. Let $\alpha \in (0,1]$ and $\delta \in [0,1)$ . If $f \in \mathcal{F}[A,B]$ , for the case when $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , then there is an unique $\gamma' \in (\gamma_0, \pi/2)$ such that - (i) when $\gamma \in (\gamma_0, \gamma']$ , f is starlike of order $\delta$ in $\mathbb{D}_{r(\delta)}$ , where $r(\delta)$ the root of the equation (3.2). - (ii) when $\gamma \in (\gamma', \pi/2)$ , f is starlike of order $\delta \in [0, g(\alpha, \pi/2))$ in $\mathbb{D}$ and starlike of order $\delta \in [g(\alpha, \pi/2), 1)$ in $\mathbb{D}_{r(\delta)}$ .
Corollary 3.3 Corollary 3.3. Let and. If then (i) For, f is starlike univalent in the disc, where (ii) For and,, whenever, and f is starlike univalent in…
Corollary 3.3. Let $\alpha \in (0,1]$ and $\gamma \in (0,\pi/2]$ . If $f \in \mathcal{F}[A,B]$ then (i) For $A = -B = \alpha$ , f is starlike univalent in the disc $|z| < r_0$ , where $$r_0 = \frac{\exp(2\alpha) - 1}{\alpha(\exp(2\alpha) + 1)}.$$ (ii) For $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , $f \in \mathcal{S}^*$ , whenever $\gamma \in (\gamma', \pi/2)$ , and f is starlike univalent in the disc $|z| < r_0$ , whenever $\gamma \in (0, \gamma_0) \cup (\gamma_0, \gamma']$ , where $\gamma_0$ , $\gamma'$ are given in Theorems 3.1 and 3.2, respectively and $r_0$ is the smallest positive root of $$\tan(2\alpha\sin\gamma) + \frac{\alpha^4r^4\sin2\gamma + 2\alpha^3r^3\sin\gamma\cos2\gamma - \alpha^2r^2\sin2\gamma - 2\alpha r\sin\gamma}{\alpha^4r^4\cos2\gamma - 2\alpha^3r^3\sin\gamma\sin2\gamma - \alpha^2r^2\cos2\gamma - \alpha^2r^2 + 1} = 0.$$ The result is sharp.
Corollary 3.4 Corollary 3.4. Let. Then for the disc, where is given in Corollary 3.3, we have (i) For. (ii) For and. where, and are given in Theorem 2.3.
Corollary 3.4. Let $f \in \mathcal{F}[A, B]$ . Then for the disc $|z| < r \le r_0$ , where $r_0$ is given in Corollary 3.3, we have (i) For $A = -B = \alpha$ . $$\left| \arg \left( \frac{zf'(z)}{f(z)} \right) \right| < \tan^{-1} \left( \frac{\sin^{-1} \left( \frac{2\alpha r}{1+\alpha^2 r^2} \right)}{2\alpha - \log \left( \frac{1+\alpha r}{1-\alpha r} \right)} \right)$$ (ii) For $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ . $$\left| \arg \left( \frac{zf'(z)}{f(z)} \right) \right| < \tan^{-1} \left( \frac{\log T_2}{2\alpha \sin \gamma - \eta + \tau} \right),$$ where $T_2$ , $\eta$ and $\tau$ are given in Theorem 2.3.
Corollary 3.5 · radius Corollary 3.5. Let. Then for the disc, where is given in Corollary 3.3,, where is the smallest positive root of the following equation: (i)…
Corollary 3.5. Let $f \in \mathcal{F}[A, B]$ . Then for the disc $|z| < r_s \le r_0$ , where $r_0$ is given in Corollary 3.3, $f \in \mathcal{SS}^*(\beta)$ , where $r_s \in (0, r_0]$ is the smallest positive root of the following equation: (i) For $A = -B = \alpha$ , $$\sin^{-1}\left(\frac{2\alpha r}{1+\alpha^2 r^2}\right) = \tan\left(\frac{\beta\pi}{2}\right)\left(2\alpha - \log\left(\frac{1+\alpha r}{1-\alpha r}\right)\right)$$ (ii) For $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , $$\frac{\log T_2}{2\alpha\sin\gamma - \eta + \tau} = \tan\left(\frac{\beta\pi}{2}\right),\,$$ where $T_2$ , $\eta$ and $\tau$ are given in Theorem 2.3. We now discuss the radius estimates for the classes $\mathcal{BS}^(\alpha)$ and $\mathcal{S}_{cs}$ , which are not contained in $\mathcal{S}$ . Further we derive radii for functions in $\mathcal{BS}^(\alpha)$ and $\mathcal{S}_{cs}^*$ to be in $\mathcal{F}[A, B]$ .
Theorem 3.6 · radius Theorem 3.6. Let, and is the smallest solution of where. If then for the disc,, where Moreover the radii and are sharp. Proof. Let. Since…
Theorem 3.6. Let $\alpha \in (0,1)$ , $r_0 = (-1 + \sqrt{1 + 4\alpha h_1^2})/(2\alpha h_1)$ and $\alpha_0 \in (0,1)$ is the smallest solution of $$1 - \frac{r_0^2 \cos^2 \theta (1 - \alpha r_0^2)^2}{(1 + \alpha^2 r_0^4 - 2\alpha r_0^2 \cos 2\theta)^2 h_1^2} - \frac{r_0^2 \sin^2 \theta (1 + \alpha r_0^2)^2}{(1 + \alpha^2 r_0^4 - 2\alpha r_0^2 \cos 2\theta)^2 h_2^2} = 0,$$ where $\theta \in (0, \pi/2)$ . If $f \in \mathcal{BS}^*(\alpha)$ then for the disc $|z| \leq r_b < 1$ , $f \in \mathcal{F}[A, B]$ , where $$r_b = \begin{cases} r_0, & \text{for } A = -B = \alpha \leq \alpha_0 \\ r_1 := \frac{-1 + \sqrt{1 + 4\alpha h_2^2}}{2\alpha h_2}, & \text{for } A = -B = \alpha > \alpha_0 \\ r_2 := \frac{-1 + \sqrt{1 + 4\alpha (k_1 + k)^2}}{2\alpha (k_1 + k)}, & \text{for } A = \alpha e^{i\gamma} \text{ and } B = \alpha e^{-i\gamma}. \end{cases}$$ Moreover the radii $r_0$ and $r_2$ are sharp. Proof. Let $f \in \mathcal{BS}^*(\alpha)$ . Since for $A = -B = \alpha$ , $\max_{|z|=1} \operatorname{Re} \psi_{A,B}(z) = h_1$ . Thus to find such r < 1 for which the image of zf'(z)/f(z) - 1 under the disc |z| < r lies inside $\psi_{A,B}(\mathbb{D})$ , it is necessary that $$\max_{|z|=r<1} \operatorname{Re}\left(\frac{z}{1-\alpha z^2}\right) = \frac{r}{1-\alpha r^2} \le h_1 \tag{3.3}$$ must hold. Clearly for $|z| \le r_0$ , the inequality (3.3) holds. Now to see that for $\alpha \le \alpha_0$ , radius $r_0$ is also sufficient for $zf'(z)/f(z) - 1 \prec z/(1 - \alpha z^2) \in \psi_{A,B}(\mathbb{D})$ in the disc $|z| \le r_0$ . For $\zeta = re^{i\theta}$ $(\theta \in [0, 2\pi))$ , we have $$B_r(\theta) := \frac{\zeta}{1 - \alpha \zeta^2} = \frac{r \cos \theta (1 - \alpha r^2)}{1 + \alpha^2 r^4 - 2\alpha r^2 \cos 2\theta} + i \frac{r \sin \theta (1 + \alpha r^2)}{1 + \alpha^2 r^4 - 2\alpha r^2 \cos 2\theta}.$$ Since $\operatorname{Re} B_r(\theta) = \operatorname{Re} B_r(-\theta)$ , $\operatorname{Re} B_r(\theta) = -\operatorname{Re} B_r(\pi - \theta)$ and $\operatorname{Im} B_r(\theta) = \operatorname{Im} B_r(\pi - \theta)$ , therefore the curve $B_r(\theta)$ is symmetric about real and imaginary axis thus it is sufficient to consider for $\theta \in [0, \pi/2]$ . Now for $r = r_0$ , the square of the distance from the origin to the points of $B_{r_0}(\theta)$ is given by $$Dist(0; B_{r_0}(\theta)) := \frac{r_0^2}{1 + \alpha^2 r_0^4 - 2\alpha r_0^2 \cos 2\theta}.$$ Since $Dist(0; B_{r_0}(\theta))' < 0$ , thus $Dist(0; B_{r_0}(\theta))$ is a decreasing function of $\theta$ . Hence the farthest point of $B_{r_0}(\theta)$ from origin is $(r_0/(1 - \alpha r_0^2), 0)$ , which lies on the boundary of $\Omega_1$ . Now for $A = -B = \alpha > \alpha_0$ and $A = \alpha e^{i\gamma}$ , $B = \alpha e^{-i\gamma}$ with |z| = r, we have $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{r}{1 - \alpha r^2}.$$ Therefore, by Lemma 2.5, we see that $\mathcal{F}[A,B]$ -radius for the class $\mathcal{BS}^*(\alpha)$ are the smallest positive roots $r_1$ and $r_2$ of the equations $r/(1-\alpha r^2)=h_2$ and $r/(1-\alpha r^2)=k_1+k$ , respectively.
Theorem 3.7 · radius Theorem 3.7. Let and be the smallest solution of where and and If then for the disc,, where Moreover the radii and are sharp. Proof. Let.…
Theorem 3.7. Let $\alpha \in (0,1)$ and $\alpha_0 \in (0,1)$ be the smallest solution of $$\frac{r_0^2((\alpha - 1)r_0 + \cos\theta(1 - \alpha r_0^2))^2}{h_1^2} + \frac{r_0^2(1 + \alpha r_0^2)^2 \sin^2\theta}{h_2^2} = N,$$ where $\theta \in (0, \pi/2)$ and $$N = (1 + r_0^2 - 2r_0\cos\theta)^2(1 + \alpha^2r_0^2 + 2\alpha r_0\cos\theta)^2$$ and $$r_0 = (-(1 + (1 - \alpha)h_1) + \sqrt{(1 + (1 - \alpha)h_1)^2 + 4\alpha h_1^2})/(2\alpha h_1).$$ If $f \in \mathcal{S}_{cs}^*(\alpha)$ then for the disc $|z| \leq r_{cs} < 1$ , $f \in \mathcal{F}[A, B]$ , where $$r_{cs} = \begin{cases} r_0, & for \ A = -B = \alpha \leq \alpha_0 \\ r_1 := \frac{-(1 + (1 - \alpha)h_2) + \sqrt{(1 + (1 - \alpha)h_2)^2 + 4\alpha h_2^2}}{2\alpha h_2}, & for \ A = -B = \alpha > \alpha_0 \\ r_2 := \frac{-(1 + (1 - \alpha)(k_1 + k)) + \sqrt{(1 + (1 - \alpha)(k_1 + k))^2 + 4\alpha (k_1 + k)^2}}{2\alpha (k_1 + k)}, & for \ A = \alpha e^{i\gamma}, B = \alpha e^{-i\gamma}. \end{cases}$$ Moreover the radii $r_0$ and $r_2$ are sharp. Proof. Let $f \in \mathcal{S}_{cs}^*(\alpha)$ . Since for $A = -B = \alpha$ , $\max_{|z|=1} \operatorname{Re} \psi_{A,B}(z) = h_1$ . Thus to find such r < 1 for which the image of zf'(z)/f(z) - 1 under the disc |z| < r lies inside $\psi_{A,B}(\mathbb{D})$ , it is necessary that $$\max_{|z|=r<1} \text{Re}\left(\frac{z}{(1-z)(1+\alpha z)}\right) = \frac{r}{(1-r)(1+\alpha r)} \le h_1$$ (3.4) must hold. Clearly for $|z| \le r_0$ , the equation (3.4) holds. Now to see that for $\alpha \le \alpha_0$ radius $r_0$ is also sufficient for $zf'(z)/f(z)-1 \prec z/((1-z)(1+\alpha z)) \in \psi_{A,B}(\mathbb{D})$ in the disc $|z| \le r_0$ . For $\zeta = re^{i\theta}$ $(\theta \in [0, 2\pi))$ , we have $$CS_r(\theta) := \frac{\zeta}{(1-\zeta)(1+\alpha\zeta)} = \frac{r((\alpha-1)r + \cos\theta(1-\alpha r_0^2))}{(1+r^2 - 2r\cos\theta)(1+\alpha^2 r^2 + 2\alpha r\cos\theta)} + \frac{r(1+\alpha r_0^2)\sin\theta}{(1+r^2 - 2r\cos\theta)(1+\alpha^2 r^2 + 2\alpha r\cos\theta)}.$$ Since Re $CS_r(\theta)$ = Re $CS_r(-\theta)$ , therefore the curve $CS_r(\theta)$ is symmetric about real axis thus it is sufficient to consider for $\theta \in [0, \pi]$ . Now for $r = r_0$ , the square of the distance from the origin to the points of $CS_{r_0}(\theta)$ is given by $$Dist(0; CS_{r_0}(\theta)) := \frac{r_0^2}{(1 + r^2 - 2r\cos\theta)^2 (1 + \alpha^2 r^2 + 2\alpha r\cos\theta)^2}.$$ Since $Dist(0; CS_{r_0}(\theta))' = 0$ for $\theta = 0, \theta_0$ and $\pi$ with $Dist(0; CS_{r_0}(\theta))' < 0$ whenever $\theta \in (0, \theta_0)$ and $Dist(0; CS_{r_0}(\theta))' > 0$ whenever $\theta \in (\theta_0, \pi)$ . And $Dist(0; CS_{r_0}(0)) - Dist(0; CS_{r_0}(\pi)) > 0$ , hence the farthest point of $CS_{r_0}(\theta)$ from origin is equal to $h_1$ obtained at $\theta = 0$ . Now for $A = -B = \alpha > \alpha_0$ and $A = \alpha e^{i\gamma}$ , $B = \alpha e^{-i\gamma}$ with |z| = r, we have $$\left| \frac{zf'(z)}{f(z)} - 1 \right| \le \frac{r}{(1-r)(1+\alpha r)}.$$ Therefore, by Lemma 2.5, we see that $\mathcal{F}[A,B]$ -radius for the class $\mathcal{S}_{cs}^*(\alpha)$ are the smallest positive roots $r_1$ and $r_2$ of the equations $r/((1-r)(1+\alpha r)) = h_2$ and $r/((1-r)(1+\alpha r)) = k_1 + k$ , respectively. ![](_page_10_Figure_12.jpeg) FIGURE 3. $f_{\alpha}(z) = z/(1-\alpha z^2)$ and $g_{\alpha}(z) = z/((1-z)(1+\alpha z))$ , (A) $f_{0.5}(\mathbb{D}_{r_0=0.77}) \subset \psi_{0.5,-0.5}(\partial \mathbb{D})$ and (B) $g_{0.5}(\mathbb{D}_{r_0=0.59}) \subset \psi_{0.5,-0.5}(\partial \mathbb{D})$ .

Definitions (2)

Def 1.1 Definition 1.1. Let. Then if and only if.
Definition 1.1. Let $p \in \mathcal{S}$ . Then $p \in \mathcal{L}[A, B]$ if and only if $$p(z) \prec \psi_{A,B}(z)$$ .
Def 1.2 Definition 1.2. Let. Then if and only if In section 2 we investigate various characteristic properties of functions in the classes and. We…
Definition 1.2. Let $f \in \mathcal{A}$ . Then $f \in \mathcal{F}[A, B]$ if and only if $$\frac{zf'(z)}{f(z)} - 1 \prec \psi_{A,B}(z). \tag{1.4}$$ In section 2 we investigate various characteristic properties of functions in the classes $\mathcal{L}[A, B]$ and $\mathcal{F}[A, B]$ . We also obtain the extremal function of the class $\mathcal{F}[A, B]$ which is non-univalent and study its various geometric properties. Further, in section 3 we derive the sharp radius of starlikeness of order $\delta$ , univalence for the functions in $\mathcal{F}[A, B]$ and also find the sharp radii for functions in $\mathcal{BS}(\alpha)$ , $\mathcal{S}_{cs}(\alpha)$ and others to be in the class $\mathcal{F}[A, B]$ .
Function classes studied:

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The second and third Hankel determinants for starlike MA--Minda subclass associa
2026
Sharp Estimates of Logarithmic Coefficients for a Certain Class of Starlike Func
2026
Moduli difference of initial inverse logarithmic coefficients for starlike and c
2026
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