Skip to main content

Further Results for Starlike Functions Related with Booth Lemniscate


In this paper we investigate an interesting subclass \(\mathcal {BS}(\alpha )\) (\(0\le \alpha <1\)) of starlike functions in the unit disc \(\Delta\). The class \(\mathcal {BS}(\alpha )\) was introduced by Kargar et al. (Anal Math Phys, 2017. which is strongly related to the Booth lemniscate. Some geometric properties of this class of analytic functions, including radius of starlikeness of order \(\gamma\) (\(0\le \gamma <1\)), the image of \(f(\{z:|z|<r\})\) when \(f\in \mathcal {BS}(\alpha )\), an special example and estimate of bounds for \(\mathrm{Re}\{f(z)/z\}\), are studied.


Let \({\mathcal {H}}\) denote the class of analytic functions in the open unit disc \(\Delta = \{z\in {\mathbb {C}} : |z| < 1\}\) on the complex plane \({\mathbb {C}}\). Also let \({\mathcal {A}}\) denote the subclass of \({\mathcal {H}}\) including of functions normalized by \(f(0)=f'(0)-1=0\). The subclass of \({\mathcal {A}}\) that consists of all univalent functions f(z) in \(\Delta\) is denoted by \({\mathcal {S}}\). We denote by \({\mathfrak {B}}\) the class of functions w(z) analytic in \(\Delta\) with \(w(0) = 0\) and \(|w(z)| < 1\), \((z \in \Delta )\). For two analytic and normalized functions f and g, we say that f is subordinate to g, written \(f \prec g\) in \(\Delta\), if there exists a function \(w\in {\mathfrak {B}}\) such that \(f (z) = g(w(z))\) for all \(z\in \Delta\). In special case, if the function g is univalent in \(\Delta\), then

$$\begin{aligned} f (z)\prec g(z) \Leftrightarrow \left( f (0) = g(0)\quad \mathrm{and}\quad f (\Delta )\subset g(\Delta ) \right) . \end{aligned}$$

It is easy to see that for any complex numbers \(\lambda \ne 0\) and \(\mu\), we have:

$$\begin{aligned} f (z)\prec g(z) \Rightarrow \lambda f(z)+\mu \prec \lambda g(z)+\mu . \end{aligned}$$

The set of all functions \(f\in {\mathcal {A}}\) that are starlike univalent in \(\Delta\) will be denoted by \({\mathcal {S}}^*\) and the set of all functions \(f\in {\mathcal {A}}\) that are convex univalent in \(\Delta\) will be denoted by \({\mathcal {K}}\). Robertson (see Robertson 1985) introduced and studied the class \({\mathcal {S}}^*(\gamma )\) of starlike functions of order \(\gamma \le 1\) as follows

$$\begin{aligned} {\mathcal {S}}^*(\gamma ):=\left\{ f\in {\mathcal {A}}:\ \ \mathrm{Re} \left\{ \frac{zf'(z)}{f(z)}\right\} > \gamma , \ z\in \Delta \right\} . \end{aligned}$$

We note that if \(\gamma \in [0,1)\), then a function in \({\mathcal {S}}^*(\gamma )\) is univalent. Also we say that \(f\in {\mathcal {K}}(\gamma )\) (the class of convex functions of order \(\gamma\)) if and only if \(zf'(z)\in {\mathcal {S}}^*(\gamma )\). In particular we put \({\mathcal {S}}^*(0)\equiv {\mathcal {S}}^*\) and \({\mathcal {K}}(0)\equiv {\mathcal {K}}\).

Recently, Kargar et al. (2017) introduced and studied a class functions related to the Booth lemniscate as follows.

Definition 1.1

(see Kargar et al. 2017) The function \(f\in {\mathcal {A}}\) belongs to the class \(\mathcal {BS}(\alpha )\), \(0\le \alpha <1\), if it satisfies the condition

$$\begin{aligned} \left( \frac{zf'(z)}{f(z)}-1\right) \prec \frac{z}{1-\alpha z^2}\quad (z\in \Delta ). \end{aligned}$$

Recall that (Piejko and Sokół 2013), a one-parameter family of functions given by

$$\begin{aligned} F_{\alpha }(z):=\frac{z}{1-\alpha z^2}=\sum _{n=1}^{\infty }\alpha ^{n-1}z^{2n-1}\quad (z\in \Delta ,~0\le \alpha \le 1). \end{aligned}$$

are starlike univalent when \(0\le \alpha \le 1\) and are convex for \(0 \le \alpha \le 3-2\sqrt{2}\approx 0.1715\). We have also \(F_{\alpha }(\Delta )=D(\alpha )\), where

$$\begin{aligned} D(\alpha )=\left\{ x+iy\in {\mathbb {C}}: ~ \left( x^2+y^2\right) ^2-\frac{x^2}{(1-\alpha )^2}-\frac{y^2}{(1+\alpha )^2}<0, (0\le \alpha <1)\right\} \end{aligned}$$


$$\begin{aligned} D(1)=\left\{ x+iy\in {\mathbb {C}}: ~ \left( \forall t\in (-\infty ,-i/2]\cup [i/2,\infty )\right) [x+iy\ne it]\right\} . \end{aligned}$$

It is clear that the curve

$$\begin{aligned} \left( x^2+y^2\right) ^2-\frac{x^2}{(1-\alpha )^2}-\frac{y^2}{(1+\alpha )^2}=0\quad (x, y)\ne (0, 0), \end{aligned}$$

is the Booth lemniscate of elliptic type (see Fig. 1, for \(\alpha =1/3\)). For more details, see Kargar et al. (2017).

Fig. 1

\(\left( x^2+y^2\right) ^2-9x^2/4-9 y^2/16=0\)

Lemma 1.1

(see Kargar et al. 2017) Let\(F_{\alpha }(z)\)be given by (1.3). Then for\(0\le \alpha <1\), we have

$$\begin{aligned} \frac{1}{\alpha -1}< \mathrm{Re}\left\{ F_{\alpha }(z)\right\} <\frac{1}{1-\alpha }\quad (z\in \Delta ). \end{aligned}$$

Therefore by definition of subordination and by Lemma 1.1, \(f\in {\mathcal {A}}\) belongs to the class \(\mathcal {BS}(\alpha )\), if it satisfies the condition

$$\begin{aligned} \frac{\alpha }{\alpha -1}< \mathrm{Re}\left\{ \frac{zf'(z)}{f(z)}\right\} <\frac{2-\alpha }{1-\alpha }\quad (z\in \Delta ). \end{aligned}$$

The following lemma will be useful.

Lemma 1.2

(see Ruscheweyh and Stankiewicz 1985) Let\(F,G\in {\mathcal {H}}\)be any convex univalent functions in\(\Delta\). If\(f\prec F\)and\(g\prec G\), then

$$\begin{aligned} f*g\prec F*G\quad {in\ \ \Delta }. \end{aligned}$$

In this work, some geometric properties of the class \(\mathcal {BS}(\alpha )\) are investigated.

Main Results

We start with the following lemma that gives the structural formula for the function of the considered class.

Lemma 2.1

The function\(f\in {\mathcal {A}}\)belongs to the class\(\mathcal {BS}(\alpha )\), \(0\le \alpha <1\), if and only if there exists an analytic functionq\(q(0)=0\)and\(q\prec {F_{\alpha }}\)such that

$$\begin{aligned} f(z) = z \exp \left( \int _0^z \frac{q(t)}{t}\mathrm{d}t \right) . \end{aligned}$$

The proof is easy. Putting \(q={F_{\alpha }}\) in Lemma 2.1 we obtain the function

$$\begin{aligned} {\tilde{f}}(z) = z\left( \frac{1+z\sqrt{\alpha }}{1-z\sqrt{\alpha }} \right) ^{\frac{1}{2\sqrt{\alpha }}}, \end{aligned}$$

which is extremal function for several problems in the class \(\mathcal {BS}(\alpha )\). Moreover, we consider

$$\begin{aligned} F(z):=\frac{{\tilde{f}}(z)}{z} =\left( \frac{1+z\sqrt{\alpha }}{1-z\sqrt{\alpha }} \right) ^{\frac{1}{2\sqrt{\alpha }}} =1+z+\frac{1}{2}z^2+\frac{1}{3}\left( \alpha +\frac{1}{2}\right) z^3\cdots . \end{aligned}$$

From (1.7) we conclude that \(f \in \mathcal {BS}(\alpha )\) is starlike of order \(\frac{\alpha }{\alpha -1} < 0,\) hence f may not be univalent in \(\Delta\). It may therefore be interesting to consider a problem to find the radius of starlikeness of order \(\gamma\), \(\gamma \in [0,1)\) (hence univalence) of the class \(\mathcal {BS}(\alpha )\), i.e. the largest radius \(r_s(\alpha ,\gamma )\) such that each function \(f \in \mathcal {BS}(\alpha )\) is starlike of order \(\gamma\) in the disc \(|z|< r_s(\alpha ,\gamma )\). For this purpose we recall the following property of the class \({\mathfrak {B}}.\)

Lemma 2.2

(Schwarz lemma) (see Duren 1983) Letwbe analytic in the unit disc\(\Delta\), with\(w(0)=0\)and\(|w(z)|<1\)in\(\Delta .\)Then\(|w'(0)|\le 1\)and\(|w(z)|\le |z|\)in\(\Delta .\)Strict inequality holds in both estimates unlesswis a rotation of the disc:\(w(z)=e^{i\theta }z.\)

Theorem 2.1

Let\(\alpha \in (0,1)\)and\(\gamma \in [0,1)\)be given numbers. If\(f\in \mathcal {BS}(\alpha )\), thenfis starlike of order\(\gamma\)in the disc\(|z|< r_s(\alpha ,\gamma )=\frac{\sqrt{1+4\alpha (1-\gamma )}-1}{2\alpha (1-\gamma )}.\)The result is sharp.


Let \(f\in \mathcal {BS}(\alpha ),\)\(\alpha \in (0,1).\) Then through (1.2) we have \(\left( \frac{zf'(z)}{f(z)}-1\right) \prec \frac{z}{1-\alpha z^2}\) so there exists \(w\in {\mathfrak {B}}\) such that

$$\begin{aligned} \mathrm{Re} \left\{ \frac{zf'(z)}{f(z)}\right\} =\mathrm{Re}\left\{ \frac{1+w(z)-\alpha w^2(z)}{1-\alpha w^2(z)}\right\} \end{aligned}$$

for all \(z\in \Delta\). Applying the Schwarz lemma we obtain

$$\begin{aligned} \mathrm{Re} \left\{ \frac{zf'(z)}{f(z)}\right\}&=\mathrm{Re}\left\{ 1+\frac{w(z)}{1-\alpha w^2(z)}\right\} =1+\mathrm{Re}\left\{ \frac{w(z)}{1-\alpha w^2(z)}\right\} \\&\ge 1-\frac{|w(z)|}{1-\alpha |w(z)|^2}\ge 1-\frac{|z|}{1-\alpha |z|^2}=1-\frac{r}{1-\alpha r^2}, \end{aligned}$$

where \(r=|z|<1\). Let us consider a function \(h(r)=1-\frac{r}{1-\alpha r^2},\)\(r \in (0,1).\) Note that \(h'(r)=-\frac{1+\alpha r^2}{(1-\alpha r^2)^2}<0\) for all \(r \in [0,1)\) hence h is a strictly decreasing function and it decreases from 1 to \(\frac{\alpha }{\alpha -1}<0.\) Therefore the equation \(h(r)=\gamma\) has for given \(\alpha\) and \(\gamma\) the smallest positive root \(r_s(\alpha ,\gamma )\) in (0, 1). Therefore f is starlike of order \(\gamma\) in \(|z|<r\le r_s(\alpha ,\gamma )\). Note that for the function \({\tilde{f}}\) given in (2.2) we obtain

$$\begin{aligned} \mathrm{Re} \frac{z{\tilde{f}}'(z)}{{\tilde{f}}(z)}=\mathrm{Re}\left\{ 1+\frac{z}{1-\alpha z^2}\right\} =:A(z) \end{aligned}$$

and \(A(-r_s(\alpha ,\gamma ))=\gamma\). \(\square\)

Putting \(\gamma =0\) in Theorem 2.1, we obtain.

Corollary 2.1

Let\(\alpha \in (0,1)\). If\(f\in \mathcal {BS}(\alpha )\)thenfis starlike univalent in the disc\(|z|< r_s(\alpha )=\frac{\sqrt{1+4\alpha }-1}{2\alpha }\). The result is sharp.

Remark 2.1

Note that \(\lim _{\alpha \longrightarrow 0^+} r_s(\alpha )= \lim _{\alpha \longrightarrow 0^+} \frac{2}{\sqrt{1+4\alpha }+1}=1.\) Moreover, it is worth mentioning that \(\lim _{\alpha \longrightarrow 1^-} r_s(\alpha )=\frac{\sqrt{5}-1}{2}= 0,618\dots\), i.e. this limit is a reciprocal of the golden ratio \(\frac{\sqrt{5}+1}{2}.\)

Now we consider the following question:

For a given number \(r \in (0,1]\) find \(\alpha (r)\) such that for each function \(f\in \mathcal {BS}(\alpha (r))\) the image \(f(\{z:|z|<r\})\) is a starlike domain.

Theorem 2.2

Let\(r \in (0,1]\)be the given number. If \(0\le \alpha < \frac{1-r}{r^2}\), then each function\(f\in \mathcal {BS}(\alpha )\)maps a disc\(|z|<r\)onto a starlike domain.


After using the same argument as in the proof of Theorem 2.1 we conclude that \(f \in \mathcal {BS}(\alpha )\) satisfies the equality

$$\begin{aligned} \mathrm{Re} \left\{ \frac{zf'(z)}{f(z)}\right\} =\mathrm{Re}\left\{ \frac{1+w(z)-\alpha w^2(z)}{1-\alpha w^2(z)}\right\} \end{aligned}$$

for all \(z \in \Delta\) with some \(w\in {\mathfrak {B}}\). Then we have by Schwarz’s lemma that

$$\begin{aligned} \mathrm{Re} \left\{ \frac{zf'(z)}{f(z)}\right\} \ge 1-\frac{|w(z)|}{1-\alpha |w(z)|^2}\ge 1-\frac{|z|}{1-\alpha |z|^2}. \end{aligned}$$

Consequently, for \(|z|<r\), we obtain \(\mathrm{Re} \left\{ \frac{zf'(z)}{f(z)}\right\} >1-\frac{r}{1-\alpha r^2}.\) Let us note that a function \(g(\alpha )=1-\frac{r}{1-\alpha r^2},\)\(\alpha \in [0,1),\) has positive values for \(0\le \alpha < \frac{1-r}{r^2}\). Therefore the image of the disc \(|z|<r\) is a starlike domain. \(\square\)

Theorem 2.3

Let\(n\ge 2\)be integer. If one of the following conditions holds


\(\frac{1}{\alpha +n(1-\alpha )}< |c|<1,\)


\(n > \frac{3-\alpha }{1-\alpha }\)   and    \(\frac{1}{\alpha -2+n(1-\alpha )}< |c|<1,\)


\(n\ge \frac{2-\alpha }{1-\alpha }\)   and    \(|c|>1,\)


\(n<\frac{2-\alpha }{1-\alpha }\)   and    \(1<|c|<\frac{1}{2-\alpha +n(\alpha -1)},\)

then the function\(g_n(z)=z+cz^n\)does not belong to the class\(\mathcal {BS}(\alpha )\).


Let us put \(G(z)=\frac{zg'_n(z)}{g_n(z)}-1=\frac{1+cnz^{n-1}}{1+cz^{n-1}}-1.\) To prove our assertion it suffices to show that the function G is not subordinate to \(F_{\alpha }\) or equivalently, because of the univalence of the dominant function \(F_{\alpha }\), that the set \(G(\Delta )\) is not included in \(F_{\alpha }(\Delta )=D(\alpha ).\) Upon performing simple calculation we find that the set \(G(\Delta )\) is the disc with the diameter from the point \(x_1=\frac{|c|(n-1)}{|c|-1}\) to the point \(x_2=\frac{|c|(n-1)}{|c|+1}.\) The set \(D(\alpha )\) is bounded by the curve

$$\begin{aligned} \left( x^2+y^2 \right) ^2-\frac{x^2}{(1-\alpha )^2}-\frac{y^2}{(1+\alpha )^2}=0,\quad (x,y)\ne (0,0). \end{aligned}$$

We have \(\min _{|z|=1}\mathrm{Re}\{F_{\alpha }(z)\}=F_{\alpha }(-1)=\frac{1}{\alpha -1}\) and \(\max _{|z|=1}\mathrm{Re}\{F_{\alpha }(z)\}=F_{\alpha }(1)=\frac{1}{1-\alpha }.\) If one of the conditions \((i)-(iv)\) is satisfied then \(\min \{x_1,x_2\}<\frac{1}{\alpha -1}\) or \(\max \{x_1,x_2\}>\frac{1}{1-\alpha },\) and then \(G(\Delta )\) is not included in \(D(\alpha ).\) The proof of theorem is completed. \(\square\)

Recently, one of the interesting problems for mathematician is to find bounds for \(\mathrm{Re}\{f(z)/z\}\) (see Kargar et al. 2016; Sim and Kwon 2013). In the sequel, we obtain lower and upper bounds for \(\mathrm{Re}\{f(z)/z\}\). We first get the following result for the function F(z) given by (2.3).

Theorem 2.4

The functionF(z) of the form (2.3) is convex univalent in\(\Delta\).


A simple calculation gives us

$$\begin{aligned} 1+\frac{zF''(z)}{F'(z)}=1+\left( \frac{1}{2\sqrt{\alpha }}-1\right) \left( \frac{2\sqrt{\alpha }z}{1-\alpha z^2}\right) +\frac{2\sqrt{\alpha }z}{1-\sqrt{\alpha }z}. \end{aligned}$$

It is sufficient to show that (2.4) has positive real part in the unit disc. From Lemma 1.1 we obtain

$$\begin{aligned} \mathrm{Re}\left\{ 1+\frac{zF''(z)}{F'(z)}\right\}= & {} \mathrm{Re}\left\{ 1+\left( \frac{1}{2\sqrt{\alpha }}-1\right) \left( \frac{2\sqrt{\alpha }z}{1-\alpha z^2}\right) +\frac{2\sqrt{\alpha }z}{1-\sqrt{\alpha }z}\right\} \\= & {} 1+2\sqrt{\alpha }\left( \frac{1}{2\sqrt{\alpha }}-1\right) \mathrm{Re}\left\{ \frac{z}{1-\alpha z^2}\right\} +2\sqrt{\alpha }\mathrm{Re}\left\{ \frac{z}{1-\sqrt{\alpha }z}\right\} \\> & {} 1+\left( 1-2\sqrt{\alpha }\right) \left( \frac{1}{\alpha -1}\right) -\frac{2\sqrt{\alpha }}{1+\sqrt{\alpha }}=:K(\alpha )\quad (0\le \alpha <1). \end{aligned}$$

It is easily seen that \(K'(\alpha )=\frac{1}{(\alpha -1)^2}>0\). Thus \(K(\alpha )\ge K(0)=0\), and hence F(z) is convex univalent function. \(\square\)

In the proof of the next theorem we will use the following result concerning the convexity of the boundary of \(D(\alpha )\).

Lemma 2.3

(see Piejko and Sokół 2013) Suppose that\(F_\alpha\)is given by (1.3). If \(0 \le \alpha \le 3-2\sqrt{2}\approx 0.1715\), then the curve\(F_\alpha (e^{i\varphi })\), \(\varphi \in [0,2\pi )\), is convex. If \(\alpha \in (3-2\sqrt{2},1)\), then the curve\(F_\alpha (e^{i\varphi })\), \(\varphi \in [0,2\pi )\), is concave.Moreover, in both cases this curve is symmetric with respect to both axes.

Theorem 2.5

If a functionfbelongs to the class\(\mathcal {BS}(\alpha )\), \(0 \le \alpha \le 3-2\sqrt{2}\), then

$$\begin{aligned} \frac{f(z)}{z}\prec F(z)\quad (z\in \Delta ), \end{aligned}$$

whereF(z) is given by (2.3).


Let \(0 \le \alpha \le 3-2\sqrt{2}\) and let f be in the class \(\mathcal {BS}(\alpha )\). Then we have

$$\begin{aligned} \phi (z):=\frac{zf'(z)}{f(z)}-1\prec F_\alpha (z)\quad (z\in \Delta ), \end{aligned}$$

where \(F_\alpha\) is given by (1.3). It is well known that the normalized function

$$\begin{aligned} l(z)=\log \frac{1}{1-z}=\sum _{n=1}^{\infty }\frac{z^n}{n}\quad (z\in \Delta ), \end{aligned}$$

belongs to the class \({\mathcal {K}}\) and for \(f\in {\mathcal {A}}\) we get

$$\begin{aligned} \phi (z)*l(z)=\int _{0}^{z}\frac{\phi (t)}{t}\mathrm{d}t\quad \mathrm{and}\quad F_\alpha (z)*l(z)=\int _{0}^{z}\frac{F_\alpha (t)}{t}\mathrm{d}t. \end{aligned}$$

By Lemma 2.3 we deduce that the function \(F_\alpha\) is convex. Thus applying Lemma 1.2 in (2.6) we obtain

$$\begin{aligned} \phi (z)*l(z)\prec F_\alpha (z)*l(z)\quad (z\in \Delta ). \end{aligned}$$

Now from (2.7) and (2.8), we can obtain

$$\begin{aligned} \int _{0}^{z}\frac{\phi (t)}{t}\mathrm{d}t\prec \int _{0}^{z}\frac{F_\alpha (t)}{t}\mathrm{d}t\quad (z\in \Delta ). \end{aligned}$$


$$\begin{aligned} \frac{f(z)}{z}= \exp \int _{0}^{z}\frac{\phi (t)}{t}\mathrm{d}t\prec \int _{0}^{z}\frac{F_\alpha (t)}{t}\mathrm{d}t=\frac{{\tilde{f}}(z)}{z}. \end{aligned}$$

This completes the proof of theorem. \(\square\)

Here by combining Theorem 2.4, Theorem 2.5 and (1.1), we get:

Theorem 2.6

Let \(f\in \mathcal {BS}(\alpha )\), \(0 \le \alpha \le 3-2\sqrt{2}\)and\(|z|=r<1\). Then

$$\begin{aligned} \left( \frac{1-r\sqrt{\alpha }}{1+\sqrt{\alpha }} \right) ^{\frac{1}{2\sqrt{\alpha }}}\le \mathrm{Re}\left( \frac{f(z)}{z}\right) \le \left( \frac{1+r\sqrt{\alpha }}{1-r\sqrt{\alpha }} \right) ^{\frac{1}{2\sqrt{\alpha }}}\quad (z\in \Delta ). \end{aligned}$$

The result is sharp.


By the subordination principle, we have:

$$\begin{aligned} f(z)\prec g(z)\Rightarrow f(|z|<r)\subset g(|z|<r)\quad (0\le r<1). \end{aligned}$$

From Theorem 2.4, since F(z) is convex univalent in \(\Delta\), and it is real for real z, thus it maps the disc \(|z|=r<1\) onto a convex set symmetric which respect to the real axis laying between \(F(-r)-1\) and \(F(r)-1\). Now the assertion is obtained from Theorem 2.5. \(\square\)


  1. Duren PL (1983) Univalent functions. Springer, Berlin

    MATH  Google Scholar 

  2. Kargar R, Ebadian A, Sokół J (2016) On subordination of some analytic functions. Sib Math J 57:599–604

    MathSciNet  Article  MATH  Google Scholar 

  3. Kargar R, Ebadian A, Sokół J (2017) On Booth lemniscate and starlike functions. Anal Math Phys.

    MATH  Google Scholar 

  4. Piejko K, Sokół J (2013) Hadamard product of analytic functions and some special regions and curves. J Inequal Appl 2013:420

    MathSciNet  Article  MATH  Google Scholar 

  5. Robertson MS (1985) Certain classes of starlike functions. Mich Math J 32:135–140

    MathSciNet  Article  MATH  Google Scholar 

  6. Ruscheweyh St, Stankiewicz J (1985) Subordination under convex univalent function. Bull Pol Acad Sci Math 33:499–502

    MathSciNet  MATH  Google Scholar 

  7. Sim YJ, Kwon OS (2013) Notes on analytic functions with a bounded positive real part. J Inequal Appl 2013:370.

    MathSciNet  Article  MATH  Google Scholar 

Download references

Author information



Corresponding author

Correspondence to Lucyna Trojnar-Spelina.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (, which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Kargar, R., Ebadian, A. & Trojnar-Spelina, L. Further Results for Starlike Functions Related with Booth Lemniscate. Iran J Sci Technol Trans Sci 43, 1235–1238 (2019).

Download citation


  • Booth lemniscate
  • Radius of starlikeness
  • Starlike function
  • Convex function
  • Subordination

Mathematics Subject Classification

  • 30C45