Starlikeness of certain non-univalent functions

Abstract

We consider three classes of functions defined using the class ${\mathcal {P}}$ of all analytic functions $p(z)=1+cz+\cdots$ on the open unit disk having positive real part and study several radius problems for these classes. The first class consists of all normalized analytic functions f with $f/g\in {\mathcal {P}}$ and $g/(zp)\in {\mathcal {P}}$ for some normalized analytic function g and $p\in {\mathcal {P}}$. The second class is defined by replacing the condition $f/g\in {\mathcal {P}}$ by $|(f/g)-1|<1$ while the other class consists of normalized analytic functions f with $f/(zp)\in {\mathcal {P}}$ for some $p\in {\mathcal {P}}$. We have determined radii so that the functions in these classes to belong to various subclasses of starlike functions. These subclasses includes the classes of starlike functions of order $\alpha$, parabolic starlike functions, as well as the classes of starlike functions associated with lemniscate of Bernoulli, reverse lemniscate, sine function, a rational function, cardioid, lune, nephroid and modified sigmoid function.

1 Introduction

Let ${\mathbb {D}}$ denote the open unit disc in ${\mathbb {C}}$. Let ${\mathcal {A}}:=\{f: f \text { is analytic on } {\mathbb {D}}, f(0)= 0, \text { and } f'(0)=1\}$. Let ${\mathcal {S}}:=\{f\in {\mathcal {A}}: f \text { is univalent on } {\mathbb {D}}\}$. An analytic function f is subordinate to another analytic function g, written $f\prec g$, if $f(z)=g(w(z))$ for some analytic function $w:{\mathbb {D}}\rightarrow {\mathbb {D}}$ that fixes the origin. If $g \in {\mathcal {S}}$, then $f\prec g$ if and only if the functions f and g takes the origin to the same point as well as the range of f is a subset of the range of g: $f({\mathbb {D}})\subseteq g({\mathbb {D}})$. Several well known subclasses of starlike and convex functions were characterized by subordination of $zf'(z)/f(z)$ or $1+(zf''(z)/f'(z)$ to some function in ${\mathcal {P}}$. Using a function $\varphi$, univalent in unit disc ${\mathbb {D}}$ with ${\text {Re}} \varphi (z)>0$, starlike with respect to $\varphi (0)=1$, symmetric about the real axis and $\varphi '(0)>0$, Ma and Minda [22] gave a unified treatment for several subclasses of starlike and convex functions by studying the classes ${\mathcal {S}}^*(\varphi )=\{f \in {\mathcal {A}}: zf'(z)/f(z)\prec \varphi (z)\}$ and ${\mathcal {K}}(\varphi )=\{f \in {\mathcal {A}}: 1+zf''(z)/f'(z)\prec \varphi (z)\}$. Convolution theorems for these two classes in a more general setting was previously studied by Shanmugham [35] with the stronger assumption that $\varphi$ is convex. Several authors considered these classes for various choices of the function $\varphi$. For $-1< B<A \le 1$, let $\varphi$ be the bilinear transform that maps the unit disc ${\mathbb {D}}$ onto the disc whose diametric end points are $(1+A)/(1+B)$ and $(1-A)/(1-B)$; if we impose $\varphi (0)=1$, then this map** is given by $\varphi (z)=(1+Az)/(1+Bz)$. For this function $\varphi$, the classes ${\mathcal {S}}^*(\varphi )$ and ${\mathcal {K}}(\varphi )$ reduce respectively to the classes ${\mathcal {S}}^*[A, B]$ and ${\mathcal {K}}[A, B]$ of Janowski starlike and convex functions. Other well-known choices for $\varphi (z)$ include $\sqrt{1+z}, \ e^z, \ 1+\sin z, \ z+\sqrt{1+z^2}$. Readers may refer to [11, 39] for brief survey of these classes.

The class ${\mathcal {P}}$ of analytic functions $p(z)=1+cz+\cdots$ on ${\mathbb {D}}$ having positive real part is known as the class of Carathéodory functions or the class of functions with positive real part. For $0\le \alpha <1$, let ${\mathcal {P}}(\alpha ):=\{p \in {\mathcal {P}} : {\text {Re}} p(z)>\alpha \}$. For any two subclasses ${\mathcal {F}}$ and ${\mathcal {G}}$ of ${\mathcal {A}}$, the ${\mathcal {G}}-$radius for the class ${\mathcal {F}}$, denoted by $R_{{\mathcal {G}}}({\mathcal {F}})$, is the largest number $R_{{\mathcal {G}}}\in (0, 1)$ such that $r^{-1}f(rz)\in {\mathcal {G}}$, for all $f\in {\mathcal {F}}$ and for $0<r<R_{{\mathcal {G}}}$. Among the several studies available on radius problems, a major focus has been on ratio between two classes of functions, where one of them belong to some particular subclasses of ${\mathcal {A}}$ and can be seen in [23,24,25, 30, 31]. Let $\varPhi$ be a function defined on ${\mathbb {D}}$. For $\varPhi (z)=1, 1/(1+z), 1/(1-z)^2, 1/(1-z^2), 1+(z/2)$, various authors have considered radius problems for classes of functions f satisfying the following conditions: (i) ${\text {Re}}f(z)/g(z)> 0$ and ${\text {Re}} g(z)/(z\varPhi (z))> 0$, (ii) $\left| (f(z)/g(z))-1\right| < 1$ and ${\text {Re}} g(z)/(z\varPhi (z)) > 0$, (iii) ${\text {Re}} f(z)/(z\varPhi (z)) > 0$. See [2, 7, 14, 15, 33]. Also, in 2019, Lecko and Sim [20] considered two functions $\varPhi (z)= 1/(1-z)^2, 1/(1-z^2)$ where $z\varPhi (z)$ are starlike functions with integer coefficients and studied on ${\text {Re}} f(z)/(z\varPhi (z)) > 0$. Similar studies can be seen in [16,17,18].

Motivated by the aforestated works, we define three classes of functions by making use of the class ${\mathcal {P}}$ as follows:

$$\begin{aligned} {\mathcal {G}}_1&:=\left\{ f\in {\mathcal {A}}: \frac{f}{g}\in {\mathcal {P}}, \frac{g}{z p}\in {\mathcal {P}} \text { for some } g \in {\mathcal {A}}, p\in {\mathcal {P}}\right\} , \end{aligned}$$

(1)

$$\begin{aligned} {\mathcal {G}}_2&:=\left\{ f\in {\mathcal {A}}: \frac{f}{g}\in {\mathcal {P}}, \frac{g}{z p}\in {\mathcal {P}} (1/2) \text { for some } g \in {\mathcal {A}}, p\in {\mathcal {P}}\right\} , \end{aligned}$$

(2)

and

$$\begin{aligned} {\mathcal {G}}_3&:=\left\{ f\in {\mathcal {A}}:\frac{f}{z p}\in {\mathcal {P}} \text { for some } p\in {\mathcal {P}}\right\} . \end{aligned}$$

(3)

We determine radii for functions in the classes ${\mathcal {G}}_1, {\mathcal {G}}_2, {\mathcal {G}}_3$ to belong to several subclasses of ${\mathcal {A}}$ like starlike functions of order $\alpha$, starlike functions associated with lemniscate of Bernoulli, reverse lemniscate, sine function, exponential function, cardioid, lune, nephroid, a particular rational function, modified sigmoid function and parabolic starlike functions. The disc that contains the image of unit disc ${\mathbb {D}}$ under the map** $zf'(z)/f(z)$ aids in determining the radius of various classes and we discuss this map** in the following section.

2 Map** of $zf'(z)/f(z)$ for ${\mathcal {G}}_1$, ${\mathcal {G}}_2$, ${\mathcal {G}}_3$

In this section, we show that the classes ${\mathcal {G}}_1, {\mathcal {G}}_2, {\mathcal {G}}_3$ are non-empty and contain non-univalent functions. We also determine the disk containing the image of the disc ${\mathbb {D}}$ under the function $zf'/f$ when f belongs to each of the classes. We shall use the function $p_0:{\mathbb {D}}\rightarrow {\mathbb {C}}$ defined by

$$\begin{aligned} p_0(z)=\frac{1+z}{1-z}. \end{aligned}$$

(4)

This function $p_0$ belongs to ${\mathcal {P}}$ and maps the unit disc ${\mathbb {D}}$ onto the right half-plane. For functions $p \in {\mathcal {P}}(\alpha )$, we shall use the following inequality [34, Lemma 2]:

$$\begin{aligned} \left| \frac{zp'(z)}{p(z)}\right| \le \frac{2(1-\alpha )r}{(1-r)(1+(1-2\alpha )r)}, \quad |z|\le r. \end{aligned}$$

(5)

2.1 The class ${\mathcal {G}}_1$

Define the functions $f_1, g_1:{\mathbb {D}}\longrightarrow {\mathbb {C}}$ by

$$\begin{aligned} f_1(z)=z\left( \frac{1+z}{1-z}\right) ^3, \quad \text {and} \quad g_1(z)=z\left( \frac{1+z}{1-z}\right) ^2. \end{aligned}$$

(6)

The functions $f_1$ and $g_1$ satisfy $f_1/g_1 \in {\mathcal {P}}$ and $g_1/zp_0 \in {\mathcal {P}}$ where $p_0\in {\mathcal {P}}$ is given in (4). Thus the function $f_1\in {\mathcal {G}}_1$ and the class ${\mathcal {G}}_1$ is non-empty. It is an extremal function for the class ${\mathcal {G}}_1$. From the coefficients of Taylor series expansion of functions $f_1$ given by

$$\begin{aligned} f_1(z)=z+6z^2+18z^3+38z^4+\dots , \end{aligned}$$

it is evident that the functions $f_1$ is not univalent. Hence, the class ${\mathcal {G}}_1$ contains non-univalent functions. As

$$\begin{aligned} f_1'(z)=\frac{(1+z)^2(1+6z-z^2)}{(1-z)^4} \end{aligned}$$

the function $f_1'$ vanishes at $z=-\rho =-(\sqrt{10}-3)$. From Theorem 1, it is apparent that the radii of univalence of the class ${\mathcal {G}}_1$ is $\sqrt{10}-3$ and it coincides with its radius of starlikeness.

If the function $f\in {\mathcal {G}}_1$, then there exists an element $g \in {\mathcal {A}}$ and $p\in {\mathcal {P}}$ such that

$$\begin{aligned} \frac{f}{g}\in {\mathcal {P}} \quad \text {and}\quad \frac{g}{z p}\in {\mathcal {P}}. \end{aligned}$$

(7)

Let $p_1, \ p_2$ be two functions defined on unit disc ${\mathbb {D}}$ by

$$\begin{aligned} p_1(z)=\frac{f(z)}{g(z)} \quad \text {and}\quad p_2(z)=\frac{g(z)}{zp(z)}. \end{aligned}$$

(8)

From (7) and (8), we have $f(z)=zp_1(z)p_2(z)p(z)$. Then a calculation shows that

$$\begin{aligned} \left| \frac{zf'(z)}{f(z)}-1\right| \le \left| \frac{zp_1'(z)}{p_1(z)}\right| +\left| \frac{zp_2'(z)}{p_2(z)}\right| +\left| \frac{zp'(z)}{p(z)}\right| . \end{aligned}$$

(9)

Since $p_1, \ p_2, \ p \in {\mathcal {P}}$, it follows from (5) and (9) that

$$\begin{aligned} \left| \frac{zf'(z)}{f(z)}-1\right| \le \frac{6r}{1-r^2}, \quad |z|\le r. \end{aligned}$$

(10)

2.2 The class ${\mathcal {G}}_2$

The functions $f_2$ and $g_2$ are defined on the unit disc ${\mathbb {D}}$ as follows:

$$\begin{aligned} f_2(z)=z\frac{(1+z)^2}{(1-z)^3} \quad \text {and}\quad g_2(z)=z\frac{(1+z)}{(1-z)^2}. \end{aligned}$$

(11)

The functions $f_2$ and $g_2$ satisfy $f_2/g_2 \in {\mathcal {P}}$ and $g_2/(zp_0) \in {\mathcal {P}}(1/2)$ where $p_0\in {\mathcal {P}}$ is given in (4). Thus the function $f_2\in {\mathcal {G}}_2$ and the class ${\mathcal {G}}_2$ is non empty. It is an extremal function for the class ${\mathcal {G}}_2$. Note that the Taylor series expansion of

$$\begin{aligned} f_2(z)=z+5z^2+13z^3+25z^4+\dots \end{aligned}$$

and it is clear that the function $f_2$ is not univalent function, as it does not satisfy de Brange’s theorem. Thus the class ${\mathcal {G}}_2$ contains non-univalent functions. Since

$$\begin{aligned} f_2'(z)=\frac{1+6z+5z^2}{(1-z)^4}, \end{aligned}$$

the function $f_2'$ vanishes at $z=-\zeta =- 1/ 5$ and it follows that the radius of univalence of the class of functions ${\mathcal {G}}_2$ is 1/5 by the Theorem 1. It also coincides with radius of starlikeness.

If the function $f \in {\mathcal {G}}_2$, then there exists a normalized analytic function g and $p\in {\mathcal {P}}$ such that

$$\begin{aligned} \frac{f}{g} \in {\mathcal {P}} \quad \text {and}\quad \frac{g}{zp}\in {\mathcal {P}}\left( 1/2\right) . \end{aligned}$$

(12)

Let $h_1$ and $h_2$ be two functions, $h_1, h_2:{\mathbb {D}}\longrightarrow {\mathbb {C}}$ defined as

$$\begin{aligned} h_1(z)=\frac{f(z)}{g(z)} \quad \text {and}\quad h_2(z)=\frac{g(z)}{zp(z)}. \end{aligned}$$

(13)

Using (12) and (13), we have $f(z)=zh_1(z)h_2(z)p(z)$, it can be shown that

$$\begin{aligned} \frac{zf'(z)}{f(z)}-1= \frac{zh_1'(z)}{h_1(z)}+\frac{zh_2'(z)}{h_2(z)}+\frac{zp'(z)}{p(z)}. \end{aligned}$$

(14)

As $h_1, p \in {\mathcal {P}}$ and $h_2 \in {\mathcal {P}}(1/2)$, it follows from (5) and (14) that

$$\begin{aligned} \left| \frac{zf'(z)}{f(z)}-1\right| \le \frac{r(r+5)}{1-r^2}, \quad |z|\le r. \end{aligned}$$

(15)

It would be interesting to find the boundary of the set $\cup _{f\in {\mathcal {G}}_2}\{zf'(z)/f(z): |z|\le r\}$. This would help in finding sharp radii to some of the problems where we have got only a lower bound.

2.3 The class ${\mathcal {G}}_3$

The function $g_1$, defined by (6), belongs to the class ${\mathcal {G}}_3$ and therefore the class ${\mathcal {G}}_3$ is non-empty. It is an extremal function for the class ${\mathcal {G}}_3$. From the coefficients of Taylor series expansion of functions $g_1$ given by

$$\begin{aligned} g_1(z)=z+4z^2+8z^3+12z^4+\dots \end{aligned}$$

it is evident that the function $g_1$ is not univalent. Hence, the class ${\mathcal {G}}_3$ contains non-univalent functions. As

$$\begin{aligned} g_1'(z)=\frac{1+5z+3z^2-z^3}{(1-z)^3}, \end{aligned}$$

the function $g_1'$ vanishes at $z=-\eta =-(\sqrt{5}-2)$. From Theorem 1, it is apparent that the radii of univalence of the class ${\mathcal {G}}_3$ coincide with their radii of starlikeness. We now discuss the map** $zf'(z)/f(z)$ when the function $f \in {\mathcal {G}}_3$. Let $p\in {\mathcal {P}}$ and $p_1$ be a function defined on unit disc ${\mathbb {D}}$ such that

$$\begin{aligned} p_1(z)=\frac{f(z)}{zp(z)}. \end{aligned}$$

(16)

It is clear from (16) that $f(z)=zp_1(z)p(z)$. Then it follows that

$$\begin{aligned} \left| \frac{zf'(z)}{f(z)}-1\right| \le \left| \frac{zp_1'(z)}{p_1(z)}\right| +\left| \frac{zp'(z)}{p(z)}\right| . \end{aligned}$$

(17)

Applying (5) in (17), we obtain

$$\begin{aligned} \left| \frac{zf'(z)}{f(z)}-1\right| \le \frac{4r}{1-r^2}, \quad |z|\le r. \end{aligned}$$

(18)

Using the equations (10), (15), (18), we investigate several radius problems associated with functions in the classes ${\mathcal {G}}_1$, ${\mathcal {G}}_2$ and ${\mathcal {G}}_3$ in the next section.

3 Radius of starlikeness

In this section, we determine the radii of the classes ${\mathcal {G}}_1, \ {\mathcal {G}}_2, \ {\mathcal {G}}_3$ to belong to various Ma-Minda starlike classes of functions. For $0\le \alpha \le 1$, the class ${\mathcal {S}}^*(\alpha )={\mathcal {S}}^*[1-2\alpha , -1]=\{f\in {\mathcal {A}}: {\text {Re}} zf'(z)/f(z)>\alpha \}$ is the class of starlike functions of order $\alpha$. These classes were studied extensively in [6, 11,12,13].

Theorem 1

The following sharp results hold for the class ${\mathcal {S}}^{*}(\alpha )$:

• (i) $R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_1)= (1-\alpha )/ \left( 3+\sqrt{10-2\alpha +\alpha ^2}\right)$.

• (ii) $R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_2)= 2(1-\alpha )/ \left( 5+\sqrt{25-4\alpha +\alpha ^2}\right)$.

• (iii) $R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_3)= (1-\alpha )/ \left( 2+\sqrt{5-2\alpha +\alpha ^2}\right)$.

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_1)$ be the root of the equation $m(r)=\alpha$. From (10), it follows that

  $$\begin{aligned} {\text {Re}}\frac{zf'(z)}{f(z)}\ge \frac{1-6r-r^2}{1-r^2}=m(r)\ge m(\rho )=\alpha . \end{aligned}$$

  This shows that $R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_1)$ is at least $\rho$. At $z=-R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_1)=-\rho$, the function $f_1$ defined in (6) satisfies

  $$\begin{aligned} {\text {Re}}\frac{zf_1'(z)}{f_1(z)}=\frac{1-6\rho -\rho ^2}{1-\rho ^2}=\alpha . \end{aligned}$$

  Thus the radius is sharp.

• (ii) The function $n(r):[0,1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r)(1-r^2)^{-1}$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_2)$ be the root of the equation $n(r)=\alpha$. Let $f\in {\mathcal {G}}_2$ and let $h_1, p \in {\mathcal {P}}$ and $h_2 \in {\mathcal {P}}(1/2)$ be the functions defined in Sect. 2.2. For $h\in {\mathcal {P}}$, by [4, Lemma 2.3], we have

  $$\begin{aligned} {\text {Re}}\frac{zh'(z)}{h(z)}\ge - \frac{2r}{1-r^2},\quad |z|\le r. \end{aligned}$$

  (19)

  For $k\in {\mathcal {P}}(1/2)$, by [4, Lemma 2.4], we have

  $$\begin{aligned} {\text {Re}}\frac{zk'(z)}{k(z)}\ge - \frac{r}{1+r},\quad |z|\le r. \end{aligned}$$

  (20)

  Using (19) and (20) in (14), it follows that

  $$\begin{aligned} {\text {Re}}\frac{zf'(z)}{f(z)}\ge 1-\frac{4r}{1-r^2}-\frac{r}{1+r}=\frac{1-5r}{1-r^2}=n(r)\ge n(\rho )=\alpha . \end{aligned}$$

  This shows that $R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_2)$ is at least $\rho$. At $z=-R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_1)=-\rho$, the function $f_2$ defined in (11) satisfies

  $$\begin{aligned} {\text {Re}}\frac{zf_2'(z)}{f_2(z)}=\frac{1-5\rho }{1-\rho ^2}=\alpha . \end{aligned}$$

  Thus the radius is sharp.

• (iii) The function defined by $s(r)=(1-4r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho = R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_3)$ be the root of the equation $s(r)=\alpha$. From (18), it follows that

  $$\begin{aligned} {\text {Re}}\frac{zf'(z)}{f(z)}\ge \frac{1-4r-r^2}{1-r^2}=s(r)\ge s(\rho )=\alpha . \end{aligned}$$

  This shows that $R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_3)$ is at least $\rho$. For the function $g_1$ defined in (6), at $z=-R_{{\mathcal {S}}^{*}(\alpha )}({\mathcal {G}}_2)=-\rho$,

  $$\begin{aligned} {\text {Re}}\frac{zg_1'(z)}{g_1(z)}=\frac{1-4\rho -\rho ^2}{1-\rho ^2}=\alpha . \end{aligned}$$

  Thus the radius is sharp.

$\square$

The class ${\mathcal {S}}_{L}^{*}={\mathcal {S}}^{*}(\sqrt{1+z})$ and it represents the collection of functions in the class ${\mathcal {A}}$ whose $zf'(z)/f(z)$ lies in the region bounded by the lemniscate of Bernoulli $|w^2-1|=1$. Various studies on ${\mathcal {S}}_{L}^{*}$ can be seen in [3, 28, 38]. Ali et al. [1, Lemma 2.2] had proved, for $2\sqrt{2}/3<a<\sqrt{2}$, that

$$\begin{aligned} \{w:|w-a|<\sqrt{2}-a\} \subset \{w:|w^2-1|<1\}.\end{aligned}$$

(21)

Using this lemma, we obtain radii results for the classes ${\mathcal {G}}_1, \ {\mathcal {G}}_2, \ {\mathcal {G}}_3$ to be in the class ${\mathcal {S}}_{L}^{*}$ in the following theorem.

Theorem 2

The following results for the class ${\mathcal {S}}_{L}^{*}$ are sharp.

• (i) $R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_1)= (3-2\sqrt{2})/ (\sqrt{2}-1)\left( 3+\sqrt{12-2\sqrt{2}}\right) \approx 0.0687$.

• (ii) $R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_2)= 2(\sqrt{2}-1)/ \left( 5+\sqrt{33-4\sqrt{2}}\right) \approx 0.0809$.

• (iii) $R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_3)= (3-2\sqrt{2})/ (\sqrt{2}-1)\left( 2+\sqrt{7-2\sqrt{2}}\right) \approx 0.1025$.

Proof

• (i) The function defined by $m(r)=6r(1-r^2)^{-1}+1, \ 0\le r<1$ is an increasing function. Let $\rho =R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_1)$ be the root of the equation $m(r)=\sqrt{2}$. For $0<r\le R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_1)$, we have $m(r)\le \sqrt{2}$. That is,

  $$\begin{aligned} \frac{6r}{1-r^2}+1\le \sqrt{2}=m(\rho ). \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (21), the disc obtained in (10) is contained in the region bounded by lemniscate. For the function $f_1$ defined in (6), at $z=R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_1)=\rho$,

  $$\begin{aligned} \left| \left( \frac{zf'(z)}{f(z)}\right) ^2-1\right| =\left| \left( \frac{1+6\rho -\rho ^2}{1-\rho ^2}\right) ^2-1\right| =1. \end{aligned}$$

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(5r+r^2)(1-r^2)^{-1}+1$ is an increasing function. Let $\rho =R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=\sqrt{2}$. For $0<r\le R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_2)$, we have $n(r)\le \sqrt{2}$. That is,

  $$\begin{aligned} \frac{5r+r^2}{1-r^2}+1\le \sqrt{2}=n(\rho ). \end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (21), the disc obtained in (15) is contained in the region bounded by lemniscate. For the function $f_2$ defined in (11), at $z=R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_2)=\rho$,

  $$\begin{aligned} \left| \left( \frac{zf'(z)}{f(z)}\right) ^2-1\right| =\left| \left( \frac{1+5\rho }{1-\rho ^2}\right) ^2-1\right| =1. \end{aligned}$$

• (iii) The function defined by $s(r)=4r(1-r^2)^{-1}+1, \ 0\le r<1$ is an increasing function. Let $\rho = R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=\sqrt{2}$. For $0<r\le R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_3)$, we have $s(r)\le \sqrt{2}$. That is,

  $$\begin{aligned} \frac{4r}{1-r^2}+1\le \sqrt{2}=s(\rho ). \end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (21), the disc obtained in (18) is contained in the region bounded by lemniscate. For the function $g_1$ defined in (6), at $z=R_{{\mathcal {S}}_{L}^{*}}({\mathcal {G}}_3)=\rho$,

  $$\begin{aligned} \left| \left( \frac{zf'(z)}{f(z)}\right) ^2-1\right| =\left| \left( \frac{1+4\rho -\rho ^2}{1-\rho ^2}\right) ^2-1\right| =\left| (\sqrt{2})^2-1\right| =1. \end{aligned}$$

$\square$

Let $\varphi _{PAR}(z):=1+\left( 2/\pi ^2\left( \log (1+\sqrt{z})/(1-\sqrt{z}) \right) ^2 \right)$. Since $\varphi _{PAR}({\mathbb {D}})=\{w:{\text {Re}}w>|w-1|\}$ is a parabolic region, the functions in the class ${\mathcal {S}}_{p}:={\mathcal {S}}^{*}(\varphi _{PAR})$ are known as parabolic starlike functions. These functions are studied by authors in [9, 21, 32]. Shanmugam and Ravichandran [36, pp.321] had proved, for $1/2< a<3/2$, that

$$\begin{aligned} \{w:|w-a|<a-1/2\}\subset \{w:{\text {Re}}w>|w-1|\}. \end{aligned}$$

(22)

The following theorem gives the radius of parabolic starlikeness of the three classes ${\mathcal {G}}_1$, ${\mathcal {G}}_2$ and ${\mathcal {G}}_3$.

Theorem 3

The following results hold for the class ${\mathcal {S}}_{p}$:

• (i) $R_{{\mathcal {S}}_{p}}({\mathcal {G}}_1)= \sqrt{37}-6 \approx 0.0827$.

• (ii) $R_{{\mathcal {S}}_{p}}({\mathcal {G}}_2)\ge (2\sqrt{7}-5)/ 3 \approx 0.0972$.

• (iii) $R_{{\mathcal {S}}_{p}}({\mathcal {G}}_3)= \sqrt{17}-4 \approx 0.1231$.

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{p}}({\mathcal {G}}_1)$ be the root of the equation $m(r)=1/2$. For $0<r\le R_{{\mathcal {S}}_{p}}({\mathcal {G}}_1)$, we have $m(r)\ge 1/2$. That is,

  $$\begin{aligned} \frac{6r}{1-r^2}\le \frac{1}{2}=m(\rho ). \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (22), the disc obtained in (10) is contained in the region bounded by parabola. For the function $f_1$ defined in (6), at $z=R_{{\mathcal {S}}_{p}}({\mathcal {G}}_1)=\rho$,

  $$\begin{aligned} {\text {Re}}\frac{zf_1'(z)}{f_1(z)}=\frac{1+6\rho -\rho ^2}{1-\rho ^2}=\frac{1}{2}=\left| \frac{zf_1'(z)}{f_1(z)}-1\right| . \end{aligned}$$

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{p}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=1/2$. For $0<r\le R_{{\mathcal {S}}_{p}}({\mathcal {G}}_2)$, we have $n(r)\ge 1/2$. That is,

  $$\begin{aligned} \frac{r(r+5)}{1-r^2}\le \frac{1}{2}=n(\rho ). \end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (22), the disc obtained in (15) is contained in the region bounded by parabola. This shows that $R_{{\mathcal {S}}_{p}}({\mathcal {G}}_2)$ is at least $\rho$.

• (iii) The function defined by $s(r)=1-(4r(1-r^2)^{-1}), \ 0\le r<1$ is a decreasing function. Let $\rho = R_{{\mathcal {S}}_{p}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=1/2$. For $0<r\le R_{{\mathcal {S}}_{p}}({\mathcal {G}}_3)$, we have $s(r)\ge 1/2$. That is,

  $$\begin{aligned} \frac{4r}{1-r^2}\le \frac{1}{2}=s(\rho ). \end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (22), the disc obtained in (18) is contained in the region bounded by parabola. For the function $g_1$ defined in (6), at $z=-R_{{\mathcal {S}}_{p}}({\mathcal {G}}_3)=-\rho$,

  $$\begin{aligned} {\text {Re}}\frac{zg_1'(z)}{g_1(z)}=\frac{1-4\rho -\rho ^2}{1-\rho ^2}=\frac{1}{2}=\left| \frac{zg_1'(z)}{g_1(z)}-1\right| . \end{aligned}$$

$\square$

In 2015, Mendiratta et al. [27] introduced the class of starlike functions associated with the exponential function as ${\mathcal {S}}_{e}^{*}={\mathcal {S}}^{*}(e^z)$ and it satisfies the condition $\left| \log zf'(z)/f(z)\right| <1$. They had also proved, for $e^{-1}\le a\le (e+e^{-1})/2$, that

$$\begin{aligned} \{w\in {\mathbb {C}}:|w-a|<a-e^{-1}\}\subseteq \{w\in {\mathbb {C}}:|\log w|<1 \}. \end{aligned}$$

(23)

Theorem 4

The following results hold for the class ${\mathcal {S}}_{e}^{*}$:

• (i) $R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_1)= (e-1)/\left( 3e+\sqrt{10e^{2}-2e+1 }\right) \approx 0.1042$.

• (ii) $R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_2)\ge 2(e^2+e-2)/(2+e)\left( 5e+\sqrt{8+4e+29e^{2} }\right) \approx 0.1213$.

• (iii) $R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_3)= 2(e-1)^2/(e+1)\left( 4e+\sqrt{4-8e+20e^{2}}\right) \approx 0.1543$.

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_1)$ be the root of the equation $m(r)=1/e$. For $0<r\le R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_1)$, we have $m(r)\ge 1/e$. That is,

  $$\begin{aligned} \frac{6r}{1-r^2}\le 1- \frac{1}{e}. \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (23), the disc obtained in (10) is contained in the region bounded by exponential function. For the function $f_1$ defined in (6), at $z=R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_1)=\rho$,

  $$\begin{aligned} \left| \log \frac{zf_1'(z)}{f_1(z)}\right| =\left| \log \frac{1+6\rho -\rho ^2}{1-\rho ^2}\right| =1. \end{aligned}$$

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=1/e$. For $0<r\le R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_2)$, we have $n(r)\ge 1/e$. That is,

  $$\begin{aligned} \frac{r(r+5)}{1-r^2}\le 1-\frac{1}{e}. \end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (23), the disc obtained in (15) is contained in the region bounded by the exponential function. This shows that $R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_2)$ is at least $\rho$.

• (iii) The function defined by $s(r)=1-(4r(1-r^2)^{-1}), \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=1/e$. For $0<r\le R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_3)$, we have $s(r)\ge 1/e$. That is,

  $$\begin{aligned} \frac{4r}{1-r^2}\le \frac{e-1}{e}. \end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (23), the disc obtained in (18) is contained in the region bounded by the exponential function. For the function $g_1$ defined in (6), at $z=R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_3)=\rho$,

  $$\begin{aligned} \left| \log \frac{zg_1'(z)}{g_1(z)}\right| =\left| \log \frac{1+4\rho -\rho ^2}{1-\rho ^2}\right| =1. \end{aligned}$$

$\square$

Theorem 5 provides radii results for starlike functions associated with a cardioid. Sharma et al. [37] studied various properties of the class ${\mathcal {S}}_{c}^{*}={\mathcal {S}}^{*}(1+(4/3)z+(2/3)z^2)$. Geometrically, if a function $f\in {\mathcal {S}}_{c}^{*}$ then $zf'(z)/f(z)$ lies in the region bounded by the cardioid

$$\begin{aligned}\varOmega _{c}=\{u+iv: (9u^{2}+9v^{2}-18u+5)^{2}-16(9u^{2}+9v^{2}-6u+1)=0\}. \end{aligned}$$

They had also proved, for $1/3 < a \le 5/3$, that

$$\begin{aligned} \{w \in {\mathbb {C}}: \left| w-a \right| <(3a-1)/3 \}\subseteq \varOmega _{c}. \end{aligned}$$

(24)

Theorem 5

The following results hold for the class ${\mathcal {S}}_{c}^{*}$:

• (i) $R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_1)= (\sqrt{85}-9)/2 \approx 0.1097$.

• (ii) $R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_2)\ge (\sqrt{265}-15)/10 \approx 0.1279$.

• (iii) $R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_3)= \sqrt{10}-3 \approx 0.1623$.

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_1)$ be the root of the equation $m(r)=1/3$. For $0<r\le R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_1)$, we have $m(r)\ge 1/3$. That is,

  $$\begin{aligned} \frac{6r}{1-r^2}\le 1- \frac{1}{3}. \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (24), the disc obtained in (10) is contained in the region bounded by the cardioid. For the function $f_1$ defined in (6), at $z=R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_1)=\rho$,

  $$\begin{aligned} \left| \frac{zf_1'(z)}{f_1(z)}\right| =\left| \frac{1+6\rho -\rho ^2}{1-\rho ^2}\right| =\frac{1}{3}=\varOmega _{c}(-1). \end{aligned}$$

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=1/3$. For $0<r\le R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_2)$, we have $n(r)\ge 1/3$. That is,

  $$\begin{aligned} \frac{r(r+5)}{1-r^2}\le 1-\frac{1}{3}. \end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (24), the disc obtained in (15) is contained in the region bounded by the cardioid. This shows that $R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_2)$ is at least $\rho$.

• (iii) The function defined by $s(r)=1-(4r(1-r^2)^{-1}), \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=1/3$. For $0<r\le R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_3)$, we have $s(r)\ge 1/3$. That is,

  $$\begin{aligned} \frac{4r}{1-r^2}\le \frac{2}{3}. \end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (24), the disc obtained in (18) is contained in the region bounded by the cardioid. For the function $g_1$ defined in (6), at $z=R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_3)=\rho$,

  $$\begin{aligned} \left| \frac{zf_1'(z)}{f_1(z)}\right| =\left| \frac{1+4\rho -\rho ^2}{1-\rho ^2}\right| =\frac{1}{3}=\varOmega _{c}(-1). \end{aligned}$$

$\square$

In 2019, Cho et al. [5] considered the class of starlike functions associated with sine function where the class ${\mathcal {S}}_{sin}^{*}$ is defined as ${\mathcal {S}}_{sin}^{*}=\{f\in {\mathcal {A}}: zf'(z)/f(z)\prec 1+ \sin z:=q_0(z)\}$ for $z \in {\mathbb {D}}$. For $\left| a-1\right| \le \sin 1$, they had established the following inclusion:

$$\begin{aligned} \{w \in {\mathbb {C}}: \left| w-a \right| < \sin 1-\left| a-1\right| \}\subseteq \varOmega _{s}. \end{aligned}$$

(25)

Here $\varOmega _{s}:=q_{0}({\mathbb {D}})$ is the image of the unit disk ${\mathbb {D}}$ under the map**s $q_{0}(z)=1+\sin z$.

Theorem 6

The following results are sharp for the class ${\mathcal {S}}_{sin}^{*}$.

• (i) $R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_1)= \sin 1 /\left( 3+\sqrt{9+ \sin ^2 1 }\right) \approx 0.1375$.

• (ii) $R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_2)= 2\sin 1 /\left( 5+\sqrt{25+ 4\sin 1+4\sin ^2 1 }\right) \approx 0.1589$.

• (iii) $R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_3)= \sin 1 /\left( 2+\sqrt{4+ \sin ^2 1 }\right) \approx 0.2018$.

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_1)$ be the root of the equation $m(r)=1-\sin 1$.For $0<r\le R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_1)$, we have $m(r)\ge 1-\sin 1$. That is,

  $$\begin{aligned}\frac{6r}{1-r^2}\le \sin 1.\end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (25), the disc obtained in (10) is contained in the region $\varOmega _{s}$ bounded by the sine function. For the function $f_1$ defined in (6), at $z=-R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_1)=-\rho$,

  $$\begin{aligned} \left| \frac{zf_1'(z)}{f_1(z)}\right| =\left| \frac{1-6\rho -\rho ^2}{1-\rho ^2}\right| =1+\sin 1=q_0(1). \end{aligned}$$

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=1-\sin 1$. For $0<r\le R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_2)$, we have $n(r)\ge 1-\sin 1$. That is,

  $$\begin{aligned} \frac{r(r+5)}{1-r^2}\le \sin 1.\end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (25), the disc obtained in (15) is contained in the region bounded by the sine function. For the function $f_2$ defined in (11), at $z=R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_1)=\rho$,

  $$\begin{aligned} \left| \frac{zf_2'(z)}{f_2(z)}\right| =\left| \frac{1+5\rho }{1-\rho ^2}\right| =1+\sin 1=q_0(1). \end{aligned}$$

• (iii) The function defined by $s(r)=1-(4r(1-r^2)^{-1}), \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=1-\sin 1$. For $0<r\le R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_3)$, we have $s(r)\ge 1-\sin 1$. That is,

  $$\begin{aligned} \frac{4r}{1-r^2}\le \sin 1.\end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (25), the disc obtained in (18) is contained in the region bounded by the sine function. For the function $g_1$ defined in (6), at $z=-R_{{\mathcal {S}}_{sin}^{*}}({\mathcal {G}}_3)=-\rho$,

  $$\begin{aligned} \left| \frac{zf_1'(z)}{f_1(z)}\right| =\left| \frac{1-4\rho -\rho ^2}{1-\rho ^2}\right| =1+\sin 1=q_0(1). \end{aligned}$$

$\square$

In 2015, Raina and Sokół[29] introduced the class . They showed that a function if and only if $zf'(z)/f(z)$ belongs to a lune shaped region ${\mathcal {L}}:=\{w \in {\mathbb {C}}: \left| w^{2}-1\right| < 2 |w|\}$. Gandhi and Ravichandran [8, Lemma 2.1] proved that

$$\begin{aligned} \{w \in {\mathbb {C}}: \left| w-a \right|< 1-|\sqrt{2}-a |\}\subseteq \{w \in {\mathbb {C}}: \left| w^{2}-1 \right| < 2\left| w\right| \}. \end{aligned}$$

(26)

Theorem 7

The following results hold for the class :

• (i) .

• (ii) .

• (iii) .

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let be the root of the equation $m(r)=\sqrt{2}-1$. For , we have $m(r)\ge \sqrt{2}-1$. That is,

  $$\begin{aligned} \frac{6r}{1-r^2}\le 2-\sqrt{2}. \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (26), the disc obtained in (10) is contained in the region bounded by the intersection of disks $\left\{ w:|w - 1| < \sqrt{2}\right\}$ and $\left\{ w:|w + 1| < \sqrt{2}\right\}$. For the function $f_1$ defined in (6), at ,

  $$\begin{aligned} \left| \left( \frac{zf_1'(z)}{f_1(z)}\right) ^2-1\right| =\left| \left( \frac{1-6\rho -\rho ^2}{1-\rho ^2}\right) ^2-1\right| =2\left| \frac{1-6\rho -\rho ^2}{1-\rho ^2}\right| . \end{aligned}$$

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let be the root of the equation $n(r)=\sqrt{2}-1$. For , we have $n(r)\ge \sqrt{2}-1$. That is,

  $$\begin{aligned} \frac{r(r+5)}{1-r^2}\le 2-\sqrt{2}. \end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (26), the disc obtained in (15) is contained in the region bounded by the lune. This shows that is at least $\rho$.

• (iii) The function $s(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $s(r)=1-(4r(1-r^2)^{-1})$ is a decreasing function. Let be the root of the equation $s(r)=\sqrt{2}-1$. For , we have $s(r)\ge \sqrt{2}-1$. That is,

  $$\begin{aligned} \frac{4r}{1-r^2}\le 2-\sqrt{2}. \end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (26), the disc obtained in (18) is contained in the region bounded by the lune. For the function $g_1$ defined in (6), at ,

  $$\begin{aligned} \left| \left( \frac{zg_1'(z)}{g_1(z)}\right) ^2-1\right| =\left| \left( \frac{1-4\rho -\rho ^2}{1-\rho ^2}\right) ^2-1\right| =2\left| \frac{1-4\rho -\rho ^2}{1-\rho ^2}\right| . \end{aligned}$$

$\square$

In the next theorem, we provide radii for starlike functions associated with a rational function. Kumar and Ravichandran [19] introduced the class of starlike functions associated with a rational function

$$\begin{aligned} \psi (z)=1+\frac{z(k+z)}{k(k-z)} \end{aligned}$$

where $k=\sqrt{2}+1$, defined by ${\mathcal {S}}_{R}^{*}={\mathcal {S}}^{*}(\psi (z))$. For $2(\sqrt{2}-1) < a \le \sqrt{2}$, they had proved that

$$\begin{aligned} \{w \in {\mathbb {C}}: \left| w-a \right| < a-2(\sqrt{2}-1)\}\subseteq \psi ({\mathbb {D}}). \end{aligned}$$

(27)

Theorem 8

The following results hold for the class ${\mathcal {S}}_{R}^{*}$:

• (i) $R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_1)= (3-2\sqrt{2})/\left( 3+\sqrt{26-12\sqrt{2}}\right) \approx 0.0285$.

• (ii) $R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_2)\ge (20-14\sqrt{2})/(2-\sqrt{2})\left( 5+\sqrt{105-56\sqrt{2}}\right) \approx 0.0340$.

• (iii) $R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_3)= (3-2\sqrt{2})/\left( 2+\sqrt{21-12\sqrt{2}}\right) \approx 0.0428$.

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_1)$ be the root of the equation $m(r)=2(\sqrt{2}-1)$. For $0<r\le R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_1)$, we have $m(r)\ge 2(\sqrt{2}-1)$. That is,

  $$\begin{aligned} \frac{6r}{1-r^2}\le 1-2(\sqrt{2}-1). \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (27), the disc obtained in (10) is contained in the region bounded by the rational function. For the function $f_1$ defined in (6), at $z=-R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_1)=-\rho$,

  $$\begin{aligned} \left| \frac{zf_1'(z)}{f_1(z)}\right| =\left| \frac{1-6\rho -\rho ^2}{1-\rho ^2}\right| =2(\sqrt{2}-1)=\psi (1). \end{aligned}$$

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=2(\sqrt{2}-1)$. The function defined by

  $$\begin{aligned} n(r)=\frac{1-5r-2r^2}{1-r^2}+1 \end{aligned}$$

  is a decreasing function. For $0<r\le R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_2)$, we have $n(r)\ge 2(\sqrt{2}-1)$. That is,

  $$\begin{aligned} \frac{r(r+5)}{1-r^2}\le 3-2\sqrt{2}. \end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (27), the disc obtained in (15) is contained in the region bounded by the rational function. This shows that $R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_2)$ is at least $\rho$.

• (iii) The function defined by $s(r)=1-(4r(1-r^2)^{-1}), \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=2(\sqrt{2}-1)$. For $0<r\le R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_3)$, we have $s(r)\ge 2(\sqrt{2}-1)$. That is,

  $$\begin{aligned} \frac{4r}{1-r^2}\le 2-\sqrt{2}. \end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (27), the disc obtained in (18) is contained in the region bounded by the rational function. For the function $g_1$ defined in (6), at $z=-R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_3)=-\rho$,

  $$\begin{aligned} \left| \frac{zg_1'(z)}{g_1(z)}\right| =\left| \frac{1-4\rho -\rho ^2}{1-\rho ^2}\right| =2(\sqrt{2}-1)=\psi (1). \end{aligned}$$

$\square$

Mendiratta et al. [26] studied the subclass of starlike function associated with left half of shifted lemniscate of Bernoulli, given by $\left| (w-\sqrt{2})^{2}-1\right| < 1$. The class ${\mathcal {S}}_{RL}^{*}$ is defined as

$$\begin{aligned} {\mathcal {S}}_{RL}^{*}={\mathcal {S}}^{*}\left( \sqrt{2}-(\sqrt{2}-1)\sqrt{\frac{1-z}{1+2(\sqrt{2}-1)z}} \right) . \end{aligned}$$

For $\sqrt{2}/3 \le a < \sqrt{2}$, they had proved the following inclusion:

$$\begin{aligned} \{w \in {\mathbb {C}}: \left| w-a \right|< r_{RL}\}\subseteq \{w \in {\mathbb {C}}: | (w-\sqrt{2})^{2}-1 |< 1 \}, \end{aligned}$$

(28)

where

$$\begin{aligned} r_{RL}= \left( \left( 1-\left( \sqrt{2}-a\right) ^{2}\right) ^{1/2}- \left( 1-\left( \sqrt{2}-a\right) ^{2}\right) \right) ^{1/2}. \end{aligned}$$

Using this result, we obtain ${\mathcal {S}}_{RL}^{*}$-radii of the classes ${\mathcal {G}}_1, \ {\mathcal {G}}_2, \ {\mathcal {G}}_3$ in the following theorem.

Theorem 9

Let $\eta =\sqrt{2(\sqrt{2}-1)}-2(\sqrt{2}-1)$. Then the following sharp results hold for the class ${\mathcal {S}}_{RL}^{*}$.

• (i) $R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_1)$ is the smallest positive root $(\approx 0.0475)$ in (0, 1) of the equation $(36+2\eta )r^2-\eta r^2-\eta =0$.

• (ii) $R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_2)$ is the smallest positive root $(\approx 0.0567)$ in (0, 1) of the equation $(1-\eta )r^4+10r^3+(25-2\eta ) r^2-\eta =0$.

• (iii) $R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_3)$ is the smallest positive root $(\approx 0.0711)$ in (0, 1) of the equation $\eta r^4-(16+2\eta )r^2+\eta =0$.

Proof

• (i) The function defined by $m(r)=(6r(1-r^2)^{-1})+1, \ 0\le r<1$ is an increasing function. Let $\rho =R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_3)$ be the root of the equation $m(r)=1+\sqrt{\eta }$. For $0<r\le R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_1)$, we have $m(r)\le \sqrt{2}$. That is,

  $$\begin{aligned} \left( \frac{6r}{1-r^2}\right) ^2 \le \eta =(m(\rho )-1)^2. \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (28), the disc obtained in (10) is contained in the region bounded by the reverse lemniscate. This shows that $R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_1)$ is at least $\rho$. For the function $f_1$ defined in (6), the radius is sharp.

• (ii) The function defined by $n(r)=(5r+r^2)(1-r^2)^{-1}+1, \ 0\le r<1$ is an increasing function. Let $R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=1+\sqrt{\eta }$. For $0<r\le R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_2)$, we have $n(r)\le \sqrt{2}$. That is,

  $$\begin{aligned} \left( \frac{5r+r^2}{1-r^2}\right) ^2\le \eta =(n(\rho )-1)^2. \end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (28), the disc obtained in (15) is contained in the region bounded by reverse lemniscate. This shows that $R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_2)$ is at least $\rho$. The obtained radius is sharp for the function $f_2$ defined in (11).

• (iii) The function defined by $s(r)=4r(1-r^2)^{-1}+1, \ 0\le r<1$ is an increasing function. Let $R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=1+\sqrt{\eta }$. For $0<r\le R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_3)$, we have $s(r)\le \sqrt{2}$. That is,

  $$\begin{aligned} \left( \frac{4r}{1-r^2}\right) ^2 \le \eta =(s(\rho )-1)^2. \end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (28), the disc obtained in (18) is contained in the region bounded by reverse lemniscate. This shows that $R_{{\mathcal {S}}_{RL}^{*}}({\mathcal {G}}_3)$ is at least $\rho$. The obtained radius is sharp for the function $g_1$ defined in (6).

The sharpness of the results can be seen by looking at the graphs. $\square$

In 2020, Wani and Swaminathan [40, Lemma 2.2] introduced the class ${\mathcal {S}}_{Ne}^{*}={\mathcal {S}}^{*}(1+z-(z^3/3))$ that maps the open unit disc ${\mathbb {D}}$ onto the interior of a two cusped kidney shaped curve $\varOmega _{Ne}:=\{u+iv: ((u-1)^2+v^2-4/9)^3-4v^2/3<0\}$. For $1/3 < a \le 1$, they had proved that

$$\begin{aligned} \{w \in {\mathbb {C}}: \left| w-a \right| < a-1/3\}\subseteq \varOmega _{Ne}. \end{aligned}$$

(29)

Our next theorem determines the ${\mathcal {S}}_{Ne}^{*}$-radii results for the classes ${\mathcal {G}}_1$, ${\mathcal {G}}_2$ and ${\mathcal {G}}_3$.

Theorem 10

The following sharp results hold for the class ${\mathcal {S}}_{Ne}^{*}$.

• (i) $R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_1)= (\sqrt{85}-9)/2 \approx 0.1097$.

• (ii) $R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_2)=(\sqrt{265}-15)/10 \approx 0.1278$.

• (iii) $R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_3)=\sqrt{10}-3\approx 0.1622$.

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho = R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_1)$ be the root of the equation $m(r)=1/3$. For $0<r\le R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_1)$, we have $m(r)\ge 1/3$. That is,

  $$\begin{aligned} \frac{6r}{1-r^2}\le 1-\frac{1}{3}. \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (29), the disc obtained in (10) is contained in the region bounded by the nephroid. For the function $f_1$ defined in (6), at $z=R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_1)=\rho$,

  $$\begin{aligned} \left| \frac{zf_1'(z)}{f_1(z)}\right| =\left| \frac{1+6\rho -\rho ^2}{1-\rho ^2}\right| =\frac{1}{3}\in \partial \varOmega _{Ne} \end{aligned}$$

  where $\partial \varOmega _{Ne}$ denotes the boundary of nephroid domain.

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=1/3$. For $0<r\le R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_2)$, we have $n(r)\ge 1/3$. That is,

  $$\begin{aligned} \frac{r(r+5)}{1-r^2}\le \frac{2}{3}. \end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (29), the disc obtained in (15) is contained in the region bounded by the nephroid. For the function $f_2$ defined in (11), at $z=R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_2)=\rho$,

  $$\begin{aligned} \left| \frac{zf_2'(z)}{f_2(z)}\right| =\left| \frac{1+5\rho }{1-\rho ^2}\right| =\frac{5}{3}\in \partial \varOmega _{Ne}. \end{aligned}$$

• (iii) The function defined by $s(r)=1-(4r(1-r^2)^{-1}), \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=1/3$. For $0<r\le R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_3)$, we have $s(r)\ge 1/3$. That is,

  $$\begin{aligned} \frac{4r}{1-r^2}\le \frac{2}{3}. \end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (29), the disc obtained in (18) is contained in the region bounded by the nephroid. For the function $g_1$ defined in (6), at $z=R_{{\mathcal {S}}_{Ne}^{*}}({\mathcal {G}}_3)=\rho$,

  $$\begin{aligned} \left| \frac{zg_1'(z)}{g_1(z)}\right| =\left| \frac{1+4\rho -\rho ^2}{1-\rho ^2}\right| =\frac{1}{3}\in \partial \varOmega _{Ne}. \end{aligned}$$

$\square$

In 2020, Goel and Kumar [10] introduced the class ${\mathcal {S}}_{SG}^{*}$ that maps the open unit disc ${\mathbb {D}}$ onto a domain $\varDelta _{SG}:=\{w\in {\mathbb {C}}:|\log ( w/(2-w))|<1\}$ and ${\mathcal {S}}_{SG}^{*}={\mathcal {S}}^{*}(2/(1+e^{-z}))$. For $2/(1+e)<a<2e/(1+e)$, they had proved the following inclusion:

$$\begin{aligned} \{w\in {\mathbb {C}}:|w-a|<r_{SG}\}\subset \varDelta _{SG}, \end{aligned}$$

(30)

provided $r_{SG}=((e-1)/(e+1))-|a-1|$. Theorem 11 provides ${\mathcal {S}}_{SG}^{*}$-radii of the classes ${\mathcal {G}}_1, \ {\mathcal {G}}_2, \ {\mathcal {G}}_3$.

Theorem 11

The following sharp results hold for the class ${\mathcal {S}}_{SG}^{*}$.

• (i) $R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_1)= 2(e-1)/\left( (6+6e)+\sqrt{40+64e+40e^2}\right) \approx 0.0766$.

• (ii) $R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_2)=2(e-1)/\left( (5+5e)+\sqrt{25+42e+33e^2}\right) \approx 0.0901$.

• (iii) $R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_3)=2(e-1)/\left( (4+4e)+\sqrt{20+24e+20e^2}\right) \approx 0.1140$.

Proof

• (i) The function defined by $m(r)=(1-6r-r^2)(1-r^2)^{-1}, \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_1)$ be the root of the equation $m(r)=2/(1+e)$. For $0<r\le R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_1)$, we have $m(r)\ge 2/(1+e)$. That is,

  $$\begin{aligned} \frac{6r}{1-r^2}\le \frac{e-1}{e+1}. \end{aligned}$$

  For the class ${\mathcal {G}}_1$, the centre of the disc is 1, therefore, by (30), the disc obtained in (10) is contained in the region bounded by the modified sigmoid. For the function $f_1$ defined in (6), at $z=R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_1)=\rho$,

  $$\begin{aligned} \left| \log \frac{zf_1'(z)/f_1(z)}{2-(zf_1'(z)/f_1(z))}\right| =\left| \log \frac{(1+6\rho -\rho ^2)/(1-\rho ^2)}{2-((1+6\rho -\rho ^2)/(1-\rho ^2))}\right| =1. \end{aligned}$$

• (ii) The function $n(r):[0, 1)\longrightarrow {\mathbb {R}}$ defined by $n(r)=(1-5r-2r^2)(1-r^2)^{-1}+1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_2)$ be the root of the equation $n(r)=2/(1+e)$. For $0<r\le R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_2)$, we have $n(r)\ge 2/1+e$. That is,

  $$\begin{aligned}\frac{r(r+5)}{1-r^2}\le \frac{e-1}{e+1}.\end{aligned}$$

  For the class ${\mathcal {G}}_2$, the centre of the disc is 1, therefore, by (30), the disc obtained in (15) is contained in the region bounded by the modified sigmoid. For the function $f_2$ defined in (11), at $z=R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_2)=\rho$,

  $$\begin{aligned} \left| \log \frac{zf_2'(z)/f_2(z)}{2-(zf_2'(z)/f_2(z))}\right| =\left| \log \frac{(1+5\rho )/(1-\rho ^2)}{2-((1+5\rho )/(1-\rho ^2))}\right| =1. \end{aligned}$$

• (iii) The function defined by $s(r)=1-(4r(1-r^2)^{-1}), \ 0\le r<1$ is a decreasing function. Let $\rho =R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_3)$ be the root of the equation $s(r)=2/(1+e)$. For $0<r\le R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_3)$, we have $s(r)\ge 2/(1+e)$. That is,

  $$\begin{aligned}\frac{4r}{1-r^2}\le \frac{e-1}{e+1}.\end{aligned}$$

  For the class ${\mathcal {G}}_3$, the centre of the disc is 1, therefore, by (30), the disc obtained in (18) is contained in the region bounded by the modified sigmoid. For the function $g_1$ defined in (6), at $z=R_{{\mathcal {S}}_{SG}^{*}}({\mathcal {G}}_3)=\rho$,

  $$\begin{aligned} \left| \log \frac{zg_1'(z)/g_1(z)}{2-(zg_1'(z)/g_1(z))}\right| =\left| \log \frac{(1+4\rho -\rho ^2)/(1-\rho ^2)}{2-((1+4\rho -\rho ^2)/(1-\rho ^2))}\right| =1. \end{aligned}$$

$\square$

Though we have no proof, we believe that the sharp radii for the class ${\mathcal {G}}_2$ are the following:

• 1. $R_{{\mathcal {S}}_{p}}({\mathcal {G}}_2)= 5-2\sqrt{6} \approx 0.1010$.

• (i) $R_{{\mathcal {S}}_{e}^{*}}({\mathcal {G}}_2)= 2(e^2+e+2)/(2+e)\left( 5e+\sqrt{-8+4e+29e^{2} }\right) \approx 0.1276$.

• (ii) $R_{{\mathcal {S}}_{c}^{*}}({\mathcal {G}}_2)= (15-\sqrt{217})/2 \approx 0.1345$.

• (iii) .

• (iv) $R_{{\mathcal {S}}_{R}^{*}}({\mathcal {G}}_2)= (10\sqrt{2}-14)/(\sqrt{2}-1)\left( 5+\sqrt{81-40\sqrt{2}}\right) \approx 0.0345$.

Our estimate for these radii are respectively 0.0972, 0.1213, 0.1279, 0.1131, 0.0340 and are very much close to the above mentioned values.