On a Class of certain Non-Univalent Functions

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. 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 <span id="page-2-1"></span> $$|\operatorname{Re} p(z)| \le \frac{1}{2\alpha} \log \left( \frac{1 + \alpha r}{1 - \alpha r} \right)$$ (2.2) <span id="page-2-2"></span> $$|\operatorname{Im} p(z)| \le \frac{1}{2\alpha} \sin^{-1} \left( \frac{2\alpha r}{1 + \alpha^2 r^2} \right). \tag{2.3}$$ (ii) For $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , we have <span id="page-2-3"></span> $$\frac{-1}{2\alpha \sin \gamma} (\eta - \tau) \le \operatorname{Re} p(z) \le \frac{1}{2\alpha \sin \gamma} (\eta + \tau). \tag{2.4}$$ <span id="page-2-4"></span> $$\frac{1}{2\alpha \sin \gamma} \log T_2 \le \operatorname{Im} p(z) \le \frac{1}{2\alpha \sin \gamma} \log T_1. \tag{2.5}$$ 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. 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. 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 $$A = -B = \alpha$$ <span id="page-4-2"></span> $$C \le \begin{cases} h_2 + (1 - h_2)D, & when (1 - CD)/(1 - D^2) \le 1, \\ h_2 + (1 + h_2)D, & when (1 - CD)/(1 - D^2) \ge 1. \end{cases}$$ (2.9) (ii) for $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , <span id="page-4-3"></span> $$C \le \begin{cases} k_1 - k + (1 - k_1 + k)D, & when (1 - CD)/(1 - D^2) \le 1 + k, \\ k_1 + k + (1 + k_1 + k)D, & when (1 - CD)/(1 - D^2) \ge 1 + k. \end{cases}$$ (2.10)

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 $\{w \in \mathbb{C} : |w| \le -f_{A,B}(-1)\} \subset f(\mathbb{D})$ .  <span id="page-5-1"></span>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. 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. 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 $$A = -B = \alpha$$ , <span id="page-6-0"></span> $$r(\delta) = \frac{\exp(2\alpha(1-\delta)) - 1}{\alpha(\exp(2\alpha(1-\delta)) + 1)}.$$ (3.1) (ii) For $A = \alpha e^{i\gamma}$ and $B = \alpha e^{-i\gamma}$ , $r(\delta)$ is the smallest positive root of <span id="page-6-1"></span> $$\tan(2\alpha\sin\gamma(\delta-1)) = \frac{\alpha^4 r^4 \sin 2\gamma + 2\alpha^3 r^3 \sin\gamma\cos 2\gamma - \alpha^2 r^2 \sin 2\gamma - 2\alpha r \sin\gamma}{\alpha^4 r^4 \cos 2\gamma - 2\alpha^3 r^3 \sin\gamma\sin 2\gamma - \alpha^2 r^2 \cos 2\gamma - \alpha^2 r^2 + 1}.$$ (3.2) The result is sharp.

Theorem 3.2. 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 3.4. 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. 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. 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 <span id="page-9-0"></span> $$\max_{|z|=r<1} \operatorname{Re}\left(\frac{z}{1-\alpha z^2}\right) = \frac{r}{1-\alpha r^2} \le h_1 \tag{3.3}$$ 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. 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 <span id="page-10-0"></span> $$\max_{|z|=r<1} \text{Re}\left(\frac{z}{(1-z)(1+\alpha z)}\right) = \frac{r}{(1-r)(1+\alpha r)} \le h_1$$ (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.  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) $g_{0.5}(\mathbb{D}_{r_0=0.59}) \subset \psi_{0.5,-0.5}(\partial \mathbb{D})$ .

Definition 1.1. Let $p \in \mathcal{S}$ . Then $p \in \mathcal{L}[A, B]$ if and only if $$p(z) \prec \psi_{A,B}(z)$$ .

Definition 1.2. Let $f \in \mathcal{A}$ . Then $f \in \mathcal{F}[A, B]$ if and only if <span id="page-1-1"></span> $$\frac{zf'(z)}{f(z)} - 1 \prec \psi_{A,B}(z). \tag{1.4}$$ 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 $\mathcal{F}[A, B]$ .