Sim and Kwon Journal of Inequalities and Applications 2013, 2013:370

Lemma 1. (Kuroki and Owa [1]) Let $f \in A$ and $0 \le \alpha < 1 < \beta$ . Then $f \in S(\alpha, \beta)$ if and only if <span id="page-1-5"></span> $$\frac{zf'(z)}{f(z)} \prec 1 + \frac{\beta - \alpha}{\pi} i \log \left( \frac{1 - e^{2\pi i \frac{1 - \alpha}{\beta - \alpha}} z}{1 - z} \right) \quad (z \in \mathbb{U}).$$ Lemma 1 means that the function p defined by $$p(z) = 1 + \frac{\beta - \alpha}{\pi} i \log \left( \frac{1 - e^{2\pi i \frac{1 - \alpha}{\beta - \alpha}} z}{1 - z} \right)$$ (2) <span id="page-1-4"></span>maps the unit disk $\mathbb{U}$ onto the strip domain w with $\alpha < \text{Re}(w) < \beta$ . We note that the function $f \in \mathcal{A}$ , given by <span id="page-1-3"></span> $$f(z) = z \exp\left\{\frac{\beta - \alpha}{\pi} i \int_0^z \frac{1}{t} \log\left(\frac{1 - e^{2\pi i \frac{1 - \alpha}{\beta - \alpha}} t}{1 - t}\right) dt\right\},\,$$ is in the class $S(\alpha, \beta)$ .

Theorem 1. For given $0 \le \alpha < 1 < \beta$ , let A and B be real numbers such that $$\frac{2 - \alpha - \beta}{\beta - \alpha} \le B < A \le \frac{\beta - 2\alpha\beta + \alpha}{\beta - \alpha}.$$ (3) Then $S^*(A, B) \subset S(\alpha, \beta)$ . Proof At first, we note that $$-1 < \frac{2-\alpha-\beta}{\beta-\alpha}$$ and $\frac{\beta-2\alpha\beta+\alpha}{\beta-\alpha} < 1$ . For $f \in S^*(A, B)$ , we know that the following inequality holds: <span id="page-1-2"></span><span id="page-1-1"></span> $$\frac{1-A}{1-B} < \operatorname{Re} \left\{ \frac{zf'(z)}{f(z)} \right\} < \frac{1+A}{1+B} \quad (z \in \mathbb{U}).$$ Therefore, it suffices to show that $\alpha$ and $\beta$ satisfy the following inequalities: $$\alpha \le \frac{1-A}{1-B}$$ and $\frac{1+A}{1+B} \le \beta$ . (4) Using inequality (4), we can derive that $$1 + B \ge \frac{2(1 - \alpha)}{\beta - \alpha}$$ and $1 + A \le \frac{2\beta(\beta - \alpha)}{\beta - \alpha}$ . (5) Also <span id="page-2-1"></span><span id="page-2-0"></span> $$1 - B \le \frac{2(\beta - 1)}{\beta - \alpha}$$ and $1 - A \ge \frac{2\alpha(\beta - a)}{\beta - \alpha}$ . (6) By the above inequalities (5) and (6), we can easily obtain the inequalities (3), so the proof of Theorem 1 is completed. $\Box$

Lemma 2. (Miller and Mocanu [2]) Let $\Xi$ be a set in the complex plane $\mathbb C$ and let b be a complex number such that $\operatorname{Re}(b) > 0$ . Suppose that a function $\psi : \mathbb C^2 \times \mathbb U \to \mathbb C$ satisfies the condition $$\psi(i\rho,\sigma;z)\notin\Xi$$ <span id="page-2-3"></span>for all real $\rho$ , $\sigma \le -|b-i\rho|^2/(2\operatorname{Re}(b))$ and all $z \in \mathbb{U}$ . If the function p(z) defined by $p(z) = b + b_1 z + b_2 z^2 + \cdots$ is analytic in $\mathbb{U}$ and if <span id="page-2-2"></span> $$\psi(p(z), zp'(z); z) \in \Xi$$ , then Re(p(z)) > 0 in $\mathbb{U}$ .

Theorem 2. Let $f \in A$ , $1/2 \le \alpha < 1$ and $Re\{zf'(z)/f(z)\} > \alpha$ in $\mathbb{U}$ . Then $$\operatorname{Re}\left\{\frac{f(z)}{z}\right\} > \gamma(\alpha) := \frac{1}{3 - 2\alpha} \quad (z \in \mathbb{U}). \tag{7}$$ Proof Write $\gamma(\alpha) := \gamma$ and note that $\frac{1}{2} \le \gamma < 1$ for $\frac{1}{2} \le \alpha < 1$ . Let p be defined by $$p(z) = \frac{1}{1 - \nu} \left( \frac{zf'(z)}{f(z)} - \gamma \right).$$ Then p is analytic in $\mathbb{U}$ , p(0) = 1 and <span id="page-2-4"></span> $$\frac{zf'(z)}{f(z)} = 1 + \frac{(1-\gamma)zp'(z)}{(1-\gamma)p(z) + \gamma} = \psi(p(z), zp'(z)),$$ where $$\psi(r,s) = 1 + \frac{(1-\gamma)s}{(1-\gamma)r + \gamma}.$$ (8) Also, $$\{\psi(p(z), zp'(z)): z \in \mathbb{U}\} \subset \{w \in \mathbb{C}: \operatorname{Re}(w) > \alpha\} := \Omega_{\alpha}.$$ Now for all real $\rho$ , $\sigma \leq -\frac{1}{2}(1+\rho^2)$ , $$\begin{split} \operatorname{Re} \left( \psi(i\rho, \sigma) \right) &= \operatorname{Re} \left( 1 + \frac{(1 - \gamma)\sigma}{(1 - \gamma)i\rho + \gamma} \right) = 1 + \frac{\gamma(1 - \gamma)\sigma}{\gamma^2 + (1 - \gamma)^2\rho^2} \\ &\leq 1 - \frac{1}{2}\gamma(1 - \gamma)\frac{1 + \rho^2}{\gamma^2 + (1 - \gamma)^2\rho^2}. \end{split}$$ Now, we let <span id="page-3-0"></span> $$g(\rho) = \frac{1 + \rho^2}{\nu^2 + (1 - \nu)^2 \rho^2}.$$ (9) Then $$g'(\rho) = \frac{2(2\gamma - 1)\rho}{(\gamma^2 + (1 - \gamma)^2 \rho^2)^2},$$ hence $g'(\rho) = 0$ occurs at only $\rho = 0$ and g satisfies $$g(0) = \frac{1}{\gamma^2}$$ and $$\lim_{\rho\to\infty}g(\rho)=\frac{1}{(1-\gamma)^2}.$$ Since $1/2 \le \gamma < 1$ , we have $$\frac{1}{\gamma^2} \leq g(\rho) < \frac{1}{(1-\gamma)^2},$$ <span id="page-3-1"></span>hence we get <span id="page-3-2"></span> $$\operatorname{Re}(\psi(i\rho,\sigma)) \leq 1 - \frac{1}{2}\gamma(1-\gamma)\frac{1}{\gamma^2} = \frac{3\gamma-1}{2\gamma} = \alpha.$$ This shows that $\text{Re}\{\psi(i\rho,\sigma)\}\notin\Omega_{\alpha}$ . By Lemma 2, we get Re(p(z))>0 in $\mathbb{U}$ , and this shows that inequality (7) holds and the proof of Theorem 2 is completed.

Theorem 3. Let $f \in A$ , $1 < \beta < 3/2$ and $Re\{zf'(z)/f(z)\} < \beta$ in $\mathbb{U}$ . Then $$\operatorname{Re}\left\{\frac{f(z)}{z}\right\} < \delta(\beta) := \frac{1}{3 - 2\beta} \quad (z \in \mathbb{U}). \tag{10}$$ Proof Note that $\delta := \delta(\beta) = \frac{1}{3-2\beta} > 1$ for $\beta > 1$ . Let p be defined by $$p(z) = \frac{1}{1 - \delta} \left( \frac{zf'(z)}{f(z)} - \delta \right).$$ Then p is analytic in $\mathbb{U}$ , p(0) = 1 and $$\frac{zf'(z)}{f(z)} = \psi(p(z), zp'(z)),$$ where $\psi$ is given in (8). Also $$\{\psi(p(z), zp'(z)) : z \in \mathbb{U}\} \subset \{w \in \mathbb{C} : \operatorname{Re}(w) < \beta\} := \Omega_{\beta}.$$ Now, for all real ^ρ, ^σ ^≤ – ( + ρ), $$\operatorname{Re}(\psi(i\rho,\sigma)) \geq 1 - \frac{1}{2}\delta(1-\delta)g(\rho),$$ where g(ρ) is given [\(](#page-3-0)). Since $$\frac{1}{(1-\delta)^2} < g(\rho) \le \frac{1}{\delta^2}$$ for all δ > , we have $$\operatorname{Re}(\psi(i\rho,\sigma)) \geq \frac{3\delta-1}{2\delta} = \beta.$$ This shows that Re{ψ(iρ, ^σ)} ^∈/ Ωβ . By Lemma , we get Re(p(z)) > in ^U, and this is equivalent to $$\operatorname{Re}\left\{\frac{f(z)}{z}\right\} < \delta \quad (z \in \mathbb{U}),$$ and the proof of Theorem is completed. - Combining the above Theorems and [,](#page-3-1) we can obtain the following result: Theorem Let f ∈ A, / ≤ α << β < / and α < Re{zf (z)/^f (z)} ^&lt; ^β in ^U. Then $$\gamma(\alpha) < \operatorname{Re}\left\{\frac{f(z)}{z}\right\} < \delta(\beta) \quad (z \in \mathbb{U}),$$ where γ (α) and δ(β) is given ([\)](#page-2-2) and [\(\)](#page-3-2).

Lemma 3. (Keogh and Merkers [6]) Let $p(z) = 1 + c_1 z + c_2 z^2 + \cdots$ be a function with positive real part in $\mathbb{U}$ . Then, for any complex number v, $$|c_2 - \nu c_1^2| \le 2 \max\{1, |1 - 2\nu|\}.$$ The following result holds for the coefficient of $f \in S(\alpha, \beta)$ .

Theorem 5. Let $0 \le \alpha < 1 < \beta$ and let the function f given by $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ be in the class $S(\alpha, \beta)$ . Then, for a complex number $\mu$ , $$\left| a_3 - \mu a_2^2 \right| \le \frac{\beta - \alpha}{2\pi} \sqrt{2 - 2\cos\left(\frac{1 - \alpha}{\beta - \alpha} \cdot 2\pi\right)} \cdot \max\left\{ 1; \left| \frac{1}{2} + (1 - 2\mu)\frac{\beta - \alpha}{\pi}i + \left(\frac{1}{2} - (1 - 2\mu)\frac{\beta - \alpha}{\pi}i\right)e^{2\pi i\frac{1 - \alpha}{\beta - \alpha}} \right| \right\}. \tag{11}$$ Proof Let us consider a function q given by q(z) = zf'(z)/f(z). Then, since $f \in \mathcal{S}(\alpha, \beta)$ , we have $q(z) \prec p(z)$ , where $$p(z) = 1 + \frac{\beta - \alpha}{\pi} i \log \left( \frac{1 - e^{2\pi i \frac{1 - \alpha}{\beta - \alpha}} z}{1 - z} \right).$$ Let <span id="page-5-0"></span> $$h(z) = \frac{1 + p^{-1}(q(z))}{1 - p^{-1}(q(z))} = 1 + h_1 z + h_2 z^2 + \cdots \quad (z \in \mathbb{U}).$$ Then h is analytic and has positive real part in the open unit disk $\mathbb{U}$ . We also have $$q(z) = p\left(\frac{h(z) - 1}{h(z) + 1}\right) \quad (z \in \mathbb{U}). \tag{12}$$ We find from equation (12) that $$a_2 = \frac{1}{2}B_1h_1$$ and $$a_3 = \frac{1}{4}B_1h_2 - \frac{1}{8}B_1h_1^2 + \frac{1}{8}B_2h_1^2 + \frac{1}{8}B_1^2h_1^2,$$ which imply that $$a_3 - \mu a_2^2 = \frac{1}{4} B_1 (h_2 - \nu h_1^2),$$ where $$\nu = \frac{1}{2} \left( 1 - \frac{B_2}{B_1} - B_1 + 2\mu B_1 \right).$$ <span id="page-6-5"></span><span id="page-6-4"></span><span id="page-6-2"></span><span id="page-6-1"></span> <span id="page-6-0"></span>Applying Lemma 3, we can obtain $$|a_3 - \mu a_2^2| = \frac{1}{4} |B_1| |h_2 - \nu h_1^2|$$ $$\leq \frac{1}{2} \cdot \max\{1; |1 - 2\nu|\}. \tag{13}$$ And substituting $$B_1 = \frac{\beta - \alpha}{\pi} i \left( 1 - e^{2\pi i \frac{1 - \alpha}{\beta - \alpha}} \right) \tag{14}$$ <span id="page-6-3"></span>and $$B_2 = \frac{\beta - \alpha}{2\pi} i \left( 1 - e^{4\pi i \frac{1 - \alpha}{\beta - \alpha}} \right) \tag{15}$$ in (13), we can obtain the result as asserted. Using the above Theorem 5, we can get the following result.

Corollary 1. Let the function f, given by $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ , be in the class $S(\alpha, \beta)$ . Also let the function $f^{-1}$ , defined by $$f^{-1}(f(z)) = z = f(f^{-1}(z)),$$ (16) be the inverse of f. If $$f^{-1}(w) = w + \sum_{n=2}^{\infty} b_n w^n \quad \left( |w| < r_0; r_0 > \frac{1}{4} \right), \tag{17}$$ then $$|b_2| \le \frac{2(\beta - \alpha)}{\pi} \sin\left(\frac{1 - \alpha}{\beta - \alpha}\pi\right)$$ and $$|b_3| \le \frac{\beta - \alpha}{2\pi} \sqrt{2 - 2\cos\left(\frac{1 - \alpha}{\beta - \alpha} \cdot 2\pi\right)} \cdot \max\left\{1; \left|\frac{1}{2} - 3\frac{\beta - \alpha}{\pi}i + \left(\frac{1}{2} + 3\frac{\beta - \alpha}{\pi}i\right)e^{2\pi i\frac{1-\alpha}{\beta - \alpha}}\right|\right\}.$$ Proof Relations (16) and (17) give $$b_2 = -a_2$$ and $b_3 = 2a_2^2 - a_3$ . Thus, we can get the estimate for $|b_2|$ by $$|b_2| = |a_2| \le \frac{2(\beta - \alpha)}{\pi} \sin\left(\frac{1 - \alpha}{\beta - \alpha}\pi\right),$$ <span id="page-7-6"></span>immediately. An application of Theorem 5 (with $\mu$ = 2) gives the estimates for $|b_3|$ , hence the proof of Corollary 1 is completed. Next, we shall estimate on some initial coefficient for the bi-univalent functions $f \in S_{\sigma}(\alpha, \beta)$ .

Theorem 6. Let f be given by $f(z) = z + \sum_{n=2}^{\infty} a_n z^n$ be in the class $S_{\sigma}(\alpha, \beta)$ . Then <span id="page-7-5"></span><span id="page-7-4"></span> $$|a_2| \le \frac{|B_1|\sqrt{|B_1|}}{|B_1^2 + B_1 - B_2|} \tag{18}$$ and $$|a_3| < |B_1| + |B_2 - B_1|, \tag{19}$$ where $B_1$ and $B_2$ are given by (14) and (15). Proof If $f \in S_{\sigma}(\alpha, \beta)$ , then $f \in S(\alpha, \beta)$ and $g \in S(\alpha, \beta)$ , where $g = f^{-1}$ . Hence $$Q(z) := \frac{zf'(z)}{f(z)} \prec p(z)$$ and $L(z) := \frac{zg'(z)}{g(z)} \prec p(z)$ , where p(z) is given by (2). Let $$h(z) = \frac{1 + p^{-1}(Q(z))}{1 - p^{-1}(Q(z))} = 1 + h_1 z + h_2 z^2 + \cdots$$ and $$k(z) = \frac{1 + p^{-1}(L(z))}{1 - p^{-1}(L(z))} = 1 + k_1 z + k_2 z^2 + \cdots$$ Then h and k are analytic and have positive real part in $\mathbb{U}$ . Also, we have <span id="page-7-2"></span><span id="page-7-1"></span><span id="page-7-0"></span> $$Q(z) = p\left(\frac{h(z) - 1}{h(z) + 1}\right) \quad \text{and} \quad L(z) = p\left(\frac{k(z) - 1}{k(z) + 1}\right).$$ By suitably comparing coefficients, we get $$a_2 = \frac{1}{2}B_1h_1,\tag{20}$$ <span id="page-7-3"></span> $$2a_3 - a_2^2 = \frac{1}{2}B_1h_2 - \frac{1}{4}B_1h_1^2 + \frac{1}{4}B_2h_1^2, \tag{21}$$ $$-a_2 = \frac{1}{2}B_1k_1 \tag{22}$$ and $$3a_2^2 - 2a_3 = \frac{1}{2}B_1k_2 - \frac{1}{4}B_1k_1^2 + \frac{1}{4}B_2k_1^2,\tag{23}$$ where $B_1$ and $B_2$ are given by (14) and (15), respectively. Now, considering (20) and (22), we get <span id="page-8-7"></span><span id="page-8-6"></span> $$h_1 = -k_1. (24)$$ Also, from (21), (22), (23) and (24), we find that $$a_2^2 = \frac{B_1^3(h_2 + k_2)}{4(B_1^2 + B_1 - B_2)}. (25)$$ Therefore, we have $$\left|a_2^2\right| \leq \frac{|B_1|^3}{4|B_1^2 + B_1 - B_2|} \left(|h_2| + |k_2|\right) \leq \frac{|B_1|^3}{|B_1^2 + B_1 - B_2|}.$$ This gives the bound on $|a_2|$ as asserted in (18). Now, further computations from (21), (23), (24) and (25) lead to $$a_3 = \frac{1}{8} (B_1(h_2 + 3k_2) + 2h_1^2(B_2 - B_1)).$$ This equation, together with the well-known estimates: $$|h_1| \le 2$$ , $|h_2| \le 2$ and $|k_2| \le 2$ lead us to inequality (19). Therefore, the proof of Theorem 6 is completed. $\Box$

Definition 1. Let $\alpha$ and $\beta$ be real numbers such that $0 \le \alpha < 1 < \beta$ . The function $f \in \mathcal{A}$ belongs to the class $S(\alpha, \beta)$ if f satisfies the following inequality: $$\alpha < \operatorname{Re} \left\{ \frac{zf'(z)}{f(z)} \right\} < \beta \quad (z \in \mathbb{U}).$$ We remark that for given $\alpha$ , $\beta$ ( $0 \le \alpha < 1 < \beta$ ), $f \in \mathcal{S}(\alpha, \beta)$ if and only if f satisfies the following two subordination equations: $$\frac{zf'(z)}{f(z)} < \frac{1 + (1 - 2\alpha)z}{1 - z} \quad \text{and} \quad \frac{zf'(z)}{f(z)} < \frac{1 + (1 - 2\beta)z}{1 - z},$$ (1)  <span id="page-1-0"></span>since the functions $(1+(1-2\alpha)z)/(1-z)$ and $(1+(1-2\beta)z)/(1-z)$ map $\mathbb U$ onto the right half plane, having real part greater than $\alpha$ , and the left half plane, having real part smaller than $\beta$ , respectively. The above class $\mathcal{S}(\alpha,\beta)$ is introduced by Kuroki and Owa [1]. They investigated coefficient estimates for $f \in \mathcal{S}(\alpha,\beta)$ and found the necessary and sufficient condition for $f \in \mathcal{S}(\alpha,\beta)$ using the following subordination.