On Booth lemniscate of starlike functions

Corollary 1.1. We have that f ∈ BS(α) if and only if (1.9) $$f(z) = z \exp \int_0^z \frac{F_{\alpha}(w(t)) - 1}{t} dt \qquad (z \in \Delta),$$ for some function w(z), analytic in ∆, with |w(z)| ≤ |z| in ∆. Proof. From [\(1.8\)](#page-2-2) it follows that there exists a function w(z), analytic in ∆, with |w(z)| ≤ |z| in ∆, such that <span id="page-3-0"></span> $$z\left(\frac{f'(z)}{f(z)} - \frac{1}{z}\right) = F_{\alpha}(w(z)) \qquad (z \in \Delta),$$ or $$z\left(\log\frac{f(z)}{z}\right)' = F_{\alpha}(w(z)) \qquad (z \in \Delta).$$ This gives [\(1.9\)](#page-3-0). On the other hand, it is a easy calculation that a function having the form [\(1.9\)](#page-3-0) satisfies condition [\(1.8\)](#page-2-2). Applying formula [\(1.9\)](#page-3-0) for w(z) = z gives that (1.10) $$f_0(z) = z \left(\frac{1 + \sqrt{\alpha}z}{1 - \sqrt{\alpha}z}\right)^{1/\sqrt{\alpha}} \qquad (z \in \Delta),$$ is in the class BS(α). <span id="page-3-1"></span>Lemma 1.1. Let Fα(z) be given by [\(1.2\)](#page-0-0). Then $$(1.11) \frac{1}{\alpha - 1} < \mathfrak{Re}\left\{F_{\alpha}(z)\right\} < \frac{1}{1 - \alpha} (0 \le \alpha < 1).$$ Proof. If α = 0, then we have −1 < Re{Fα} = Re(z) < 1. For 0 < α < 1, the function {Fα} does not have any poles in ∆ and is analytic in ∆, thus looking for the min{Re{Fα(z)} : |z| < 1} it is sufficient to consider it on the boundary ∂Fα(∆) = {Fα(e iϕ) : ϕ ∈ [0, 2π]}. A simple calculation give us $$\Re\left\{F_{\alpha}(e^{i\varphi})\right\} = \frac{(1-\alpha)\cos\varphi}{1+\alpha^2 - 2\alpha\cos2\varphi} \qquad (\varphi \in [0,2\varphi]).$$ So we can see that Re {Fα(z)} is well defined also for ϕ = 0 and ϕ = 2π. Define $$g(x) = \frac{(1-\alpha)x}{1+\alpha^2 - 2\alpha(2x^2 - 1)} \qquad (-1 \le x \le 1),$$ then for 0 < α < 1, we have g 0 (x) > 0. Thus for −1 ≤ x ≤ 1, we have $$\frac{1}{\alpha - 1} = g(-1) \le g(x) \le g(1) = \frac{1}{1 - \alpha}.$$ This completes the proof. We note that from Lemma [1.1](#page-3-1) and by definition of subordination, the function f ∈ A belongs to the class BS(α), 0 ≤ α < 1, if it satisfies the condition <span id="page-3-2"></span> $$\frac{1}{\alpha - 1} < \Re \left( \frac{zf'(z)}{f(z)} - 1 \right) < \frac{1}{1 - \alpha} \qquad (z \in \Delta),$$ or equivalently (1.12) $$\frac{\alpha}{\alpha - 1} < \Re \left(\frac{zf'(z)}{f(z)}\right) < \frac{2 - \alpha}{1 - \alpha} \qquad (z \in \Delta).$$ It is clear that BS(0) ≡ S(0, 2) ⊂ S^∗ , where the class S(α, β), α < 1 and β > 1, was recently considered by K. Kuroki and S. Owa in [\[6\]](#page-8-5). <span id="page-4-2"></span>Corollary 1.2. If $f \in \mathcal{BS}(\alpha)$ , then <span id="page-4-0"></span>(1.13) $$\frac{zf'(z)}{f(z)} \prec P_{\alpha}(z) \qquad (z \in \Delta),$$ where <span id="page-4-1"></span>(1.14) $$P_{\alpha}(z) = 1 + \frac{2}{\pi(1-\alpha)} i \log\left(\frac{1 - e^{\pi i(1-\alpha)^2}z}{1-z}\right) \qquad (z \in \Delta),$$ is convex univalent in $\Delta$ .

Lemma 1.2. (See [9]) Let $q(z) = \sum_{n=1}^{\infty} C_n z^n$ be analytic and univalent in $\Delta$ , and suppose that q(z) maps $\Delta$ onto a convex domain. If $p(z) = \sum_{n=1}^{\infty} A_n z^n$ is analytic in $\Delta$ and satisfies the following subordination $$p(z) \prec q(z) \qquad (z \in \Delta),$$ then $$|A_n| \le |C_1| \qquad n \ge 1.$$

Theorem 2.1. Let $f \in A$ and $0 < \alpha < 1$ . If $f \in BS(\alpha)$ then (2.1) $$\log \frac{f(z)}{z} \prec \int_0^z \frac{P_{\alpha}(t) - 1}{t} dt \qquad (z \in \Delta),$$ where <span id="page-4-5"></span> $$P_{\alpha}(z) - 1 = \frac{2}{\pi(1 - \alpha)} i \log \left( \frac{1 - e^{\pi i(1 - \alpha)^2} z}{1 - z} \right) \qquad (z \in \Delta)$$ and <span id="page-4-4"></span> $$\widetilde{P}_{\alpha}(z) = \int_{0}^{z} \frac{P_{\alpha}(t) - 1}{t} dt \qquad (z \in \Delta),$$ are convex univalent in $\Delta$ .

Corollary 2.1. If $f \in \mathcal{BS}(\alpha)$ and |z| = r < 1, then <span id="page-5-1"></span>(2.4) $$\min_{|z|=r} \left| \exp \widetilde{P}_{\alpha}(z) \right| \le \left| \frac{f(z)}{z} \right| \le \max_{|z|=r} \left| \exp \widetilde{P}_{\alpha}(z) \right|.$$

Theorem 2.2. Assume that the function f of the form (1.1) belongs to the class $\mathcal{BS}(\alpha)$ where $0 \le \alpha \le 3 - 2\sqrt{2}$ . then $|a_2| \le 1$ and <span id="page-5-7"></span>(2.6) $$|a_n| \le \frac{1}{n-1} \prod_{k=2}^{n-1} \left( \frac{k}{k-1} \right) \qquad (n=3,4,\ldots).$$

Lemma 2.1. Let the function g(z) given by $$g(z) = 1 + c_1 z + c_2 z^2 + \cdots,$$ be in the class $\mathcal{P}$ . Then, for any complex number $\mu$ <span id="page-6-6"></span> $$|c_2 - \mu c_1^2| \le 2 \max\{1, |2\mu - 1|\}.$$ The result is sharp.

Theorem 2.3. Let $0 \le \alpha < 1$ , $f \in \mathcal{BS}(\alpha)$ and $\mathfrak{F}$ is the kth root transform of f defined by (2.13). Then, for any complex number $\mu$ , $$(2.14) |b_{2k+1} - \mu b_{k+1}^2| \le \frac{1}{2k} \max \left\{ 1, \left| \frac{2(\mu - 1)}{k} + 1 \right| \right\}.$$ The result is sharp.

Corollary 2.2. (Fekete-Szegö inequality) Suppose that $f \in \mathcal{BS}(\alpha)$ and $0 \le \alpha < 1$ . Then, for any complex number $\mu$ , $$\left| a_3 - \mu a_2^2 \right| \le \frac{1}{2} \max \left\{ 1, |2\mu - 1| \right\}.$$ The result is sharp. It is well known that every function $f \in \mathcal{S}$ has an inverse $f^{-1}$ , defined by $f^{-1}(f(z)) = z, z \in \Delta$ and $$f(f^{-1}(w)) = w$$ $(|w| < r_0; r_0 < 1/4),$ where <span id="page-7-2"></span> $$(2.26) f^{-1}(w) = w - a_2 w^2 + (2a_2^2 - a_3)w^3 - (5a_2^3 - 5a_2a_3 + a_4)w^4 + \cdots$$ <span id="page-7-4"></span>Corollary 2.3. Let the function f, given by (1.1), be in the class $\mathcal{BS}(\alpha)$ where $0 \le \alpha < 1$ . Also let the function $f^{-1}(w) = w + \sum_{n=2}^{\infty} b_n w^n$ be inverse of f. Then $$(2.27) |b_2| \le 1,$$ and $$|b_3| \le \frac{3}{2}.$$ Proof. Relation [\(2.26\)](#page-7-2) give us $$b_2 = -a_2$$ and $b_3 = 2a_2^2 - a_3$ . Thus, we can get the estimate for |b2| by $$|b_2| = |a_2| \le 1.$$ For estimate of |b3|, it suffices in Corollary [2.2,](#page-7-3) we put µ = 2. Hence the proof of Corollary [2.3](#page-7-4) is completed.