Dziok Boundary Value Problems 2013, 2013:98

Lemma 1. [3] If $$\omega \in \Omega$$ , $\omega(z) = \sum_{n=1}^{\infty} c_n z^n$ $(z \in \mathcal{U})$ , then $$|c_n| \le 1$$ $(n = 1, 2),$ $|c_2| \le 1 - |c_1|^2,$ $|c_2 - \mu c_1^2| \le \max\{1, |\mu|\}$ $(\mu \in \mathbb{C}).$ <span id="page-2-4"></span>The result is sharp. The functions $$\omega(z) = z,$$ $\omega_a(z) = z \frac{z+a}{1+\overline{a}z}$ $(z \in \mathcal{U}, |a| < 1)$ are the extremal functions. <span id="page-2-3"></span><span id="page-2-1"></span>Theorem 1 Let $$(1-\alpha)(\beta_k-\alpha_k)+\alpha(\delta_k-\gamma_k)\neq 0 \quad (k=2,3).$$ <span id="page-2-2"></span>If $f \in \mathcal{W}_{\alpha}(\Phi, \Psi; p)$ , then $$|a_2| \le \frac{|p_1|}{|(1-\alpha)(\beta_2 - \alpha_2) + \alpha(\delta_2 - \gamma_2)|},$$ (2) $$|a_3| \le \frac{|p_1|}{|(1-\alpha)(\beta_3 - \alpha_3) + \alpha(\delta_3 - \gamma_3)|} \max\{1, |\beta|\},$$ (3) $$|a_3 - \mu a_2^2| \le \frac{|p_1|}{|(1 - \alpha)(\beta_3 - \alpha_3) + \alpha(\delta_3 - \gamma_3)|} \max\{1, |\gamma|\} \quad (\mu \in \mathbb{C}), \tag{4}$$ <span id="page-3-3"></span>where $$\beta = \frac{p_2}{p_1} + \frac{(1 - \alpha)\alpha_2(\beta_2 - \alpha_2) + \alpha\gamma_2(\delta_2 - \gamma_2)}{[(1 - \alpha)(\beta_2 - \alpha_2) + \alpha(\delta_2 - \gamma_2)]^2} p_1,\tag{5}$$ <span id="page-3-0"></span> $$\gamma = \frac{(1-\alpha)(\beta_3 - \alpha_3) + \alpha(\delta_3 - \gamma_3)}{[(1-\alpha)(\beta_2 - \alpha_2) + \alpha(\delta_2 - \gamma_2)]^2} p_1 \mu - \beta. \tag{6}$$ The results are sharp. <span id="page-3-4"></span>Proof Let $f \in \mathcal{W}_{\alpha}(\Phi, \Psi; p)$ . Then there exists a function $\omega \in \Omega$ , $\omega(z) = \sum_{n=1}^{\infty} c_n z^n$ $(z \in \mathcal{U})$ , such that $$(1-\alpha)\frac{\phif}{\varphif} + \alpha \frac{\psif}{\chif} = p \circ \omega. \tag{7}$$ It is easy to verify that $$(p \circ \omega)(z) = 1 + (p_1c_1)z + (p_1c_2 + p_2c_1^2)z^2 + \cdots \quad (z \in \mathcal{U}),$$ (8) $$(1 - \alpha) \frac{(\phi f)(z)}{(\varphi f)(z)} + \alpha \frac{(\psi f)(z)}{(\chi f)(z)} = 1 + A_2 z + A_3 z^2 + \dots \quad (z \in \mathcal{U}), \tag{9}$$ <span id="page-3-1"></span>where $$A_2 = \left[ (1 - \alpha)(\beta_2 - \alpha_2) + \alpha(\delta_2 - \gamma_2) \right] a_2,$$ $$A_3 = \left[ (1 - \alpha)(\beta_3 - \alpha_3) + \alpha(\delta_3 - \gamma_3) \right] a_3 - \left[ (1 - \alpha)\alpha_2(\beta_2 - \alpha_2) + \alpha\gamma_2(\delta_2 - \gamma_2) \right] a_2^2.$$ <span id="page-3-2"></span>Thus, by (7), we have $$a_2 = \frac{p_1 c_1}{(1 - \alpha)(\beta_2 - \alpha_2) + \alpha(\delta_2 - \gamma_2)},\tag{10}$$ $$a_3 = \frac{p_1 c_2 + p_2 c_1^2 + [(1 - \alpha)\alpha_2(\beta_2 - \alpha_2) + \alpha \gamma_2(\delta_2 - \gamma_2)]a_2^2}{(1 - \alpha)(\beta_3 - \alpha_3) + \alpha(\delta_3 - \gamma_3)},$$ (11) which by Lemma 1 gives sharp estimation (2). Let $\mu$ be a complex number. Then, by (10) and (11) we obtain $$a_3 - \mu a_2^2 = \frac{p_1}{(1 - \alpha)(\beta_3 - \alpha_3) + \alpha(\delta_3 - \gamma_3)} \{c_2 - \gamma c_1^2\},$$ where $\gamma$ is defined by (6). Thus, by Lemma 1, we have (4). Let functions $f_1, f_2 \in A$ satisfy the conditions $$(1 - \alpha) \frac{(\phi f_1)(z)}{(\varphi f_1)(z)} + \alpha \frac{(\psi f_1)(z)}{(\chi f_1)(z)} = p(z) \quad (z \in \mathcal{U}),$$ $$(1 - \alpha) \frac{(\phi f_2)(z)}{(\varphi f_2)(z)} + \alpha \frac{(\psi f_2)(z)}{(\chi f_2)(z)} = p(z^2) \quad (z \in \mathcal{U}).$$ Then the functions belong to the class $W_{\alpha}(\Phi, \Psi; p)$ and they realize the equality in the estimation (4). Thus, the results are sharp. Putting $\mu = 0$ in (4) we get the sharp estimation (3).

Theorem 2. Let $\alpha \neq -\frac{1}{2}$ , -1. If $f \in \mathcal{M}_{\alpha}(\varphi, p)$ , then $$|a_2| \le \frac{|p_1|}{(1+\alpha)|\alpha_2|}, \qquad |a_3| \le \frac{|p_1|}{2(1+2\alpha)|\alpha_3|} \max\{1, |\beta|\},$$ $$|a_3 - \mu a_2^2| \le \frac{|p_1|}{2(1+2\alpha)|\alpha_3|} \max\{1, |\gamma|\} \quad (\mu \in \mathbb{C}),$$ where $$\beta = \frac{p_2}{p_1} + \frac{(1+3\alpha)}{(1+\alpha)^2} p_1, \qquad \gamma = \frac{2(1+2\alpha)\alpha_3 p_1}{(1+\alpha)^2 \alpha_2^2} \mu - \beta.$$ The results are sharp. Proof Let $f \in \mathcal{M}_{\alpha}(\varphi, p) = \mathcal{W}_{\alpha}(\Phi, \Psi; p)$ , where $$\chi(z) = \phi(z) = z\varphi'(z), \qquad \psi(z) = z(z\varphi'(z))' \quad (z \in \mathcal{U}).$$ <span id="page-4-5"></span>Since $$\beta_n = \gamma_n = n\alpha_n, \qquad \delta_n = n^2\alpha_n \quad (n = 2, 3),$$ the results follow from Theorem 1. If we put $\alpha = 0$ in Theorem 1, then we obtain the following theorem.

Theorem 3. Let $\beta_k \neq \alpha_k$ (k = 2, 3). If $f \in \mathcal{W}(\Phi; p)$ , then $$|a_2| \le \frac{|p_1|}{|\beta_2 - \alpha_2|}, \qquad |a_3| \le \frac{|p_1|}{|\beta_3 - \alpha_3|} \max\{1, |\beta|\},$$ $$|a_3 - \mu a_2^2| \le \frac{|p_1|}{|\beta_3 - \alpha_3|} \max\{1, |\gamma|\} \quad (\mu \in \mathbb{C}),$$ <span id="page-4-3"></span>where $$\beta = \frac{p_2}{p_1} + \frac{\alpha_2 p_1}{\beta_2 - \alpha_2}, \qquad \gamma = \frac{\beta_3 - \alpha_3}{(\beta_2 - \alpha_2)^2} p_1 \mu - \beta.$$ <span id="page-4-1"></span><span id="page-4-0"></span>The results are sharp.

Theorem 4. Let $\beta_2\beta_3 \neq 0$ . If $f \in \mathcal{CW}(\Phi; P)$ , then $$\left|a_3 - \mu a_2^2\right| \le \frac{1}{2|\beta_3|} \left(D + \max\{0, A_\mu\} + \max\{0, B_\mu\}\right),$$ (12) $$|a_3| \le \frac{1}{2|\beta_2|} \left( D + \max\{0, A_0\} + \max\{0, B_0\} \right), \tag{13}$$ $$|a_2| = 2\frac{|p_1| + |\alpha_2||q_1|}{|\beta_2|},\tag{14}$$ <span id="page-5-3"></span><span id="page-5-2"></span>where $$A_{\mu} = \left| \alpha_3 \left( q_2 + q_1^2 \right) - 2\mu \frac{\alpha_2^2 \beta_3 q_1^2}{\beta_2^2} \right| + C_{\mu} - |\alpha_3| |q_1|, \qquad D = 2|p_1| + |\alpha_3| |q_1|, \tag{15}$$ $$B_{\mu} = 2 \left| p_2 - \mu \frac{\beta_3 p_1^2}{\beta_2^2} \right| + C_{\mu} - 2|p_1|, \qquad C_{\mu} = |\alpha_2||p_1||q_1| \left| 1 - 2\mu \frac{\beta_3}{\beta_2^2} \right|. \tag{16}$$ The results (12) and (13) are sharp for $A_{\mu}B_{\mu} \geq 0$ and $A_0B_0 \geq 0$ , respectively. Proof Let $f \in \mathcal{CW}(\Phi; P)$ . Then there exists a function $g \in S^*(q)$ and functions $\omega, \eta \in \Omega$ , $$\omega(z) = \sum_{n=1}^{\infty} c_n z^n, \qquad \eta(z) = \sum_{n=1}^{\infty} d_n z^n \quad (z \in \mathcal{U}),$$ <span id="page-5-1"></span><span id="page-5-0"></span>such that $$\frac{\phi f}{\varphi g} = p \circ \omega, \qquad \frac{zg'(z)}{g(z)} = (q \circ \eta)(z) \quad (z \in \mathcal{U}). \tag{17}$$ Thus, by (8), we have <span id="page-5-5"></span> $$b_2 = q_1 d_1,$$ $2b_3 = q_1 d_2 + (q_2 + q_1^2)d_1^2,$ $a_2 = \frac{c_1 p_1 + \alpha_2 d_1 q_1}{\beta_2},$ (18) $$\beta_3 a_3 = \frac{\alpha_3}{2} \left\{ q_1 d_2 + \left( q_2 + q_1^2 \right) d_1^2 \right\} + \alpha_2 p_1 q_1 c_1 d_1 + p_1 c_2 + p_2 c_1^2, \tag{19}$$ and by Lemma 1, we obtain the sharp estimation (14). Let $\mu$ be a complex number. Then, by (18), (19) and Lemma 1 we have <span id="page-5-4"></span> $$2|\beta_3||a_3 - \mu a_2^2| \le (A - C)|d_1|^2 + (B - C)|c_1|^2 + 2C|d_1||c_1| + D,$$ (20) or equivalently $$2|\beta_3||a_3 - \mu a_2^2| \le A|d_1|^2 + B|c_1|^2 - C(|d_1| - |c_1|)^2 + D,$$ (21) where $A = A_{\mu}$ , $B = B_{\mu}$ , $C = C_{\mu}$ , D are defined by (15) and (16). Thus, we obtain $$2|\beta_3||a_3 - \mu a_7^2| \le A|d_1|^2 + B|c_1|^2 + D, (22)$$ and, in consequence, by Lemma 1 we have (12). It is easy to verify that the equality in (22) is attained by choosing $c_1 = d_1 = 1$ , $c_2 = d_2 = 0$ if $A \ge 0$ , $B \ge 0$ or $c_1 = d_1 = 0$ , $c_2 = d_2 = 1$ if $A \le 0$ , $B \le 0$ . Therefore, we consider functions $f_1, f_2 \in A$ such that $$\frac{(\phi f_1)(z)}{(\varphi g)(z)} = p(z), \qquad \frac{zg'(z)}{g(z)} = q(z) \quad (z \in \mathcal{U})$$ and $$\frac{(\phif_2)(z)}{(\phig)(z)}=p\big(z^2\big),\qquad \frac{zg'(z)}{g(z)}=p\big(z^2\big) \quad (z\in\mathcal{U}),$$ <span id="page-6-2"></span>respectively. Then the functions belong to the class $\mathcal{CW}(\Phi; p)$ and they realize the equality in the estimation (12) for $AB \ge 0$ . Putting $\mu = 0$ in (12), we get the sharp estimation (13). $\square$ The following theorem gives the complete sharp estimation of the Fekete-Szegö functional in the class $\mathcal{CW}(\Phi; P)$ .

Theorem 5. Let $\beta_2\beta_3 \neq 0$ . If $f \in \mathcal{CW}(\Phi; P)$ , then <span id="page-6-0"></span> $$|a_{3} - \mu a_{2}^{2}| \leq \begin{cases} \frac{1}{2|\beta_{3}|} (A + B + D) & if 0 \leq A \leq C \vee 0 \leq B \leq C \\ & \vee (A \geq C \wedge B \geq C), \end{cases}$$ $$\frac{1}{2|\beta_{3}|} (D + B - C + \frac{C^{2}}{C - A}) & if A < 0 \wedge B \geq C,$$ $$\frac{1}{2|\beta_{3}|} (D + A - C + \frac{C^{2}}{C - B}) & if B < 0 \wedge A \geq C,$$ $$(23)$$ where $A = A_{\mu}$ , $B = B_{\mu}$ , $C = C_{\mu}$ , D are defined by (15) and (16). The result is sharp. Proof From Theorem 4, we have sharp estimation (23) for $AB \ge 0$ . Let now $A \ge C$ and B < 0. Then, by (20) and Lemma 1 we have $$2|\beta_3||a_3-\mu a_2^2|\leq \nu(|c_1|),$$ where $$\nu(x) := -(C - B)x^2 + 2Cx + (A - C) + D.$$ Simply calculations give that the function $\nu$ attains a maximum in the interval [0,1] at the point $x = \frac{C}{C-B} \le 1$ . Thus, we have (23) for $A \ge C$ and B < 0. Moreover, the equality in (20) is attained by choosing the functions $\eta(z) = z$ , $\omega(z) = z \frac{z+a}{1+\overline{a}z}$ , for $a = \frac{C}{C-B}$ , i.e. $c_1 = \frac{C}{C-B}$ , $d_1 = 1$ and $c_2 = 1 - |a|^2$ , $d_2 = 0$ . Therefore, the result is sharp for $A \ge C$ and B < 0. Next, let A < 0 and $C \le B$ . Then, by (20) and Lemma 1 we have $$2|\beta_3||a_3-\mu a_2^2|<\widetilde{\nu}(|d_1|),$$ where <span id="page-6-1"></span> $$\widetilde{\nu}(x) := -(C - A)x^2 + 2Cx + (B - C) + D.$$ Since the function $\widetilde{v}$ attains a maximum in the interval [0,1] at the point $x=\frac{C}{C-A}\leq 1$ , we have the estimation (23) for A<0 and $C\leq B$ . The equality in (20) is attained by choosing the functions $\omega(z)=z$ , $\eta(z)=z\frac{z+a}{1+\overline{a}z}$ , for $a=\frac{C}{C-A}$ , i.e. $c_1=1$ , $d_1=\frac{C}{C-A}$ and $c_2=0$ , $d_2=1-|a|^2$ . Finally, let us assume $(0\leq A\leq C\land B\leq 0)\lor (0\leq B\leq C\land A\leq 0)$ . Then, by (20) we have $$2|\beta_3||a_3 - \mu a_2^2| \le F(|c_1|, |d_1|),\tag{24}$$ where $$F(x, y) = -(C - A)x^2 - (C - B)y^2 + 2Cxy + D.$$ Since F is the continuous function on $T := [0,1] \times [0,1]$ , by (24) we have <span id="page-7-0"></span> $$2|\beta_3||a_3 - \mu a_2^2| \le \max F(T) = \max F(\partial T \cup K),\tag{25}$$ where K is the set of critical points of the function F in T. It is easy to verify that $$K \setminus \partial T = \begin{cases} \emptyset & \text{if } C^2 \neq (C - A)(C - B) \lor A = C, \\ \{(x, y) \in \text{int } T : x = \frac{C}{C - A}y\} & \text{if } C^2 = (C - A)(C - B) \land A \neq C. \end{cases}$$ If $C^2 = (A - C)(B - C) \neq 0$ , then $$F\left(\frac{C}{C-A}y, y\right) = \frac{C^2 - (C-A)(C-B)}{A-C}y^2 + D = D \quad (y \in [0,1]).$$ Moreover, we have $$F(x,0) = -(C-A)x^{2} + D \le D, \qquad F(0,y) = -(C-B)y^{2} + D \le D \quad (x,y \in [0,1]),$$ $$F(x,1) = -(C-A)x^{2} + 2Cx + B - C + D \le F(1,1) = A + B + D \quad (x \in [0,1]),$$ $$F(1,y) = -(C-B)y^{2} + 2Cy + A - C + D \le F(1,1) = A + B + D \quad (y \in [0,1]).$$ Thus, we obtain $$\max F(\partial T \cup K) = A + B + D$$ <span id="page-7-1"></span>which by (25) gives (23) for $(0 \le A \le C \land B \le 0) \lor (0 \le B \le C \land A \le 0)$ . The equality in (24) is attained by choosing $c_1 = d_1 = 1$ and $c_2 = d_2 = 0$ . Therefore, the result is sharp and the proof is completed. Putting $\mu$ = 0 in Theorem 5 we obtain the following theorem.

Theorem 6. Let $\beta_2\beta_3 \neq 0$ . If $f \in \mathcal{CW}(\Phi; P)$ , then $$|a_{3}| \leq \begin{cases} \frac{1}{2|\beta_{3}|}(A+B+D) & if \ 0 \leq A \leq C \lor 0 \leq B \leq C \\ & \lor (A \geq C \land B \geq C), \end{cases}$$ $$\begin{cases} \frac{D}{2|\beta_{3}|} & if \ A \leq 0 \land B \leq 0, \\ \frac{1}{2|\beta_{3}|}(D+B-C+\frac{C^{2}}{C-A}) & if \ A < 0 \land B \geq C, \\ \frac{1}{2|\beta_{3}|}(D+A-C+\frac{C^{2}}{C-B}) & if \ B < 0 \land A \geq C, \end{cases}$$ where $A = A_0$ , $B = B_0$ , $C = C_0$ , D are defined by (15) and (16). The result is sharp.

Corollary 1. Let $\alpha_2\alpha_3 \neq 0$ . If $f \in S^c(\varphi, p)$ , then $$|a_2| \leq \frac{1}{2} \left| \frac{p_1}{\alpha_2} \right|, \qquad |a_3| \leq \frac{1}{6} \left| \frac{p_1}{\alpha_3} \right| \max \left\{ 1, \left| \frac{p_2}{p_1} + p_1 \right| \right\},$$ $$\left|a_3 - \mu a_2^2\right| \le \frac{1}{6} \left|\frac{p_1}{\alpha_3}\right| \max\left\{1, |\gamma|\right\} \quad (\mu \in \mathbb{C}),$$ where $$\gamma = \frac{3\alpha_3 p_1}{2\alpha_2^2} \mu - p_1 - \frac{p_2}{p_1}.$$ The results are sharp.

Corollary 2. Let $\alpha_2\alpha_3 \neq 0$ . If $f \in S^*(\varphi, p)$ , then $$|a_2| \le \left| \frac{p_1}{\alpha_2} \right|, \qquad |a_3| \le \frac{1}{2} \left| \frac{p_1}{\alpha_3} \right| \max \left\{ 1, \left| \frac{p_2}{p_1} + p_1 \right| \right\},$$ $$|a_3 - \mu a_2^2| \le \frac{1}{2} \left| \frac{p_1}{\alpha_3} \right| \max \left\{ 1, |\gamma| \right\} \quad (\mu \in \mathbb{C}),$$ where $$\gamma = 2 \frac{\alpha_3 p_1}{\alpha_2^2} \mu - p_1 - \frac{p_2}{p_1}.$$ The results are sharp. Choosing the function p in Theorems 1-6, we can obtain several new results. Let a, b be complex number, |b| < 1, $a \ne b$ , and let $$p(z) = \frac{1+az}{1+hz} \quad (z \in \mathcal{U}).$$ It is clear, that $$p(z) = 1 + (a - b)z - b(a - b)z^{2} + \cdots \quad (z \in \mathcal{U}).$$ Thus, by Theorems 1-3 and 5, we obtain the following four corollaries.

Corollary 3. Let $$(1-\alpha)(\beta_k-\alpha_k)+\alpha(\delta_k-\gamma_k)\neq 0$$ $(k=2,3)$ . If $f\in\mathcal{W}_{\alpha}(\Phi,\Psi;\frac{1+az}{1+bz})$ , then $$|a_{2}| \leq \frac{|a-b|}{|(1-\alpha)(\beta_{2}-\alpha_{2})+\alpha(\delta_{2}-\gamma_{2})|},$$ $$|a_{3}| \leq \frac{|a-b|}{|(1-\alpha)(\beta_{3}-\alpha_{3})+\alpha(\delta_{3}-\gamma_{3})|} \max\{1,|\beta|\},$$ $$|a_{3}-\mu a_{2}^{2}| \leq \frac{|a-b|}{|(1-\alpha)(\beta_{3}-\alpha_{3})+\alpha(\delta_{3}-\gamma_{3})|} \max\{1,|\gamma|\} \quad (\mu \in \mathbb{C}),$$ where $$\beta = -b + \frac{(1-\alpha)\alpha_2(\beta_2 - \alpha_2) + \alpha\gamma_2(\delta_2 - \gamma_2)}{[(1-\alpha)(\beta_2 - \alpha_2) + \alpha(\delta_2 - \gamma_2)]^2}(a-b),$$ $$\gamma = \frac{(1-\alpha)(\beta_3 - \alpha_3) + \alpha(\delta_3 - \gamma_3)}{[(1-\alpha)(\beta_2 - \alpha_2) + \alpha(\delta_2 - \gamma_2)]^2}(a-b)\mu - \beta.$$ The results are sharp.

Corollary 4. Let $\alpha \neq -\frac{1}{2}$ , -1. If $f \in \mathcal{M}_{\alpha}(\varphi, \frac{1+az}{1+hz})$ , then $$|a_{2}| \leq \frac{|a-b|}{|1+\alpha||\alpha_{2}|}, \qquad |a_{3}| \leq \frac{|a-b|}{2|1+2\alpha||\alpha_{3}|} \max\{1, |\beta|\},$$ $$|a_{3} - \mu a_{2}^{2}| \leq \frac{|a-b|}{2|1+2\alpha||\alpha_{3}|} \max\{1, |\gamma|\} \quad (\mu \in \mathbb{C}),$$ where $$\beta = -b + \frac{(1+3\alpha)}{(1+\alpha)^2}(a-b), \qquad \gamma = \frac{2(1+2\alpha)(a-b)\alpha_3}{(1+\alpha)^2\alpha_2^2}\mu - \beta.$$ The results are sharp.

Corollary 5. Let $\beta_k \neq \alpha_k$ (k = 2, 3). If $f \in \mathcal{W}(\Phi; \frac{1+az}{1+hz})$ , then $$|a_2| \le \frac{|a-b|}{|\beta_2 - \alpha_2|}, \qquad |a_3| \le \frac{|a-b|}{|\beta_3 - \alpha_3|} \max\{1, |\beta|\},$$ $$\left|a_3 - \mu a_2^2\right| \le \frac{|a-b|}{|\beta_3 - \alpha_3|} \max\{1, |\gamma|\} \quad (\mu \in \mathbb{C}),$$ where $$\beta = -b + \frac{(a-b)\alpha_2}{\beta_2 - \alpha_2}, \qquad \gamma = \frac{(\beta_3 - \alpha_3)(a-b)}{(\beta_2 - \alpha_2)^2}\mu - \beta.$$ The results are sharp.

Corollary 6. Let $\beta_2\beta_3 \neq 0$ . If $f \in \mathcal{CW}(\Phi; \frac{1+az}{1+bz})$ , then $$|a_{3} - \mu a_{2}^{2}| \leq \begin{cases} \frac{|b-a|}{2|\beta_{3}|} (D+A+B) & \text{if } 0 \leq A \leq C \vee 0 \leq B \leq C \\ & \vee (A \geq C \wedge B \geq C), \end{cases}$$ $$\frac{|b-a|}{2|\beta_{3}|} D & \text{if } A \leq 0 \wedge B \leq 0,$$ $$\frac{|b-a|}{2|\beta_{3}|} (D+B-C+\frac{C^{2}}{C-A}) & \text{if } A < 0 \wedge B \geq C,$$ $$\frac{|b-a|}{2|\beta_{3}|} (D+A-C+\frac{C^{2}}{C-B}) & \text{if } B < 0 \wedge A \geq C,$$ where $$A = \left| \alpha_3(2b - a) - 2\mu \frac{\alpha_2^2 \beta_3(b - a)}{\beta_2^2} \right| + C - |\alpha_3|, \qquad D = 2 + |\alpha_3|,$$ $$B = 2 \left| b - \mu \frac{\beta_3(b - a)}{\beta_2^2} \right| + C - 2, \qquad C = |\alpha_2| |b - a| \left| 1 - 2\mu \frac{\beta_3}{\beta_2^2} \right|.$$ The result is sharp. Let $0 < \theta \le 1$ and let $$p(z) = \left(\frac{1+z}{1-z}\right)^{\theta} \quad (z \in \mathcal{U}).$$ It is easy to verify, that $$p(z) = 1 + 2\theta z + \theta(\theta + 1)z^2 + \cdots \quad (z \in \mathcal{U}).$$ Thus, by Theorems 1-3 and 5, we obtain the following four corollaries. Corollary 7 Let $(1-\alpha)(\beta_k-\alpha_k)+\alpha(\delta_k-\gamma_k)\neq 0$ (k=2,3). If $f\in\mathcal{W}_{\alpha}(\Phi,\Psi;(\frac{1+z}{1-z})^{\theta})$ , then $$|a_{2}| \leq \frac{2\theta}{|(1-\alpha)(\beta_{2}-\alpha_{2})+\alpha(\delta_{2}-\gamma_{2})|},$$ $$|a_{3}| \leq \frac{2\theta}{|(1-\alpha)(\beta_{3}-\alpha_{3})+\alpha(\delta_{3}-\gamma_{3})|} \max\{1,|\beta|\},$$ $$|a_{3}-\mu a_{2}^{2}| \leq \frac{2\theta}{|(1-\alpha)(\beta_{3}-\alpha_{3})+\alpha(\delta_{3}-\gamma_{3})|} \max\{1,|\gamma|\} \quad (\mu \in \mathbb{C}),$$ where $$\begin{split} \beta &= \frac{1+\theta}{2} + 2\frac{(1-\alpha)\alpha_2(\beta_2-\alpha_2) + \alpha\gamma_2(\delta_2-\gamma_2)}{[(1-\alpha)(\beta_2-\alpha_2) + \alpha(\delta_2-\gamma_2)]^2}\theta, \\ \gamma &= 2\frac{(1-\alpha)(\beta_3-\alpha_3) + \alpha(\delta_3-\gamma_3)}{[(1-\alpha)(\beta_2-\alpha_2) + \alpha(\delta_2-\gamma_2)]^2}\theta\mu - \beta. \end{split}$$ The results are sharp.

Corollary 8. Let $\alpha \neq -\frac{1}{2}$ , -1. If $f \in \mathcal{M}_{\alpha}(\varphi, (\frac{1+z}{1-z})^{\theta})$ , then $$|a_2| \le \frac{2\theta}{|(1+\alpha)\alpha_2|}, \qquad a_3 \le \frac{\theta}{|(1+2\alpha)\alpha_3|} \max\{1, |\beta|\},$$ $$|a_3 - \mu a_2^2| \le \frac{\theta}{|(1+2\alpha)\alpha_3|} \max\{1, |\gamma|\} \quad (\mu \in \mathbb{C}),$$ where $$\beta = \frac{1+\theta}{2} + \frac{2(1+3\alpha)}{(1+\alpha)^2}\theta, \qquad \gamma = \frac{4(1+2\alpha)\theta\alpha_3}{(1+\alpha)^2\alpha_2^2}\mu - \beta.$$ The results are sharp.

Corollary 9. Let $\beta_k \neq \alpha_k$ (k = 2, 3). If $f \in \mathcal{W}(\Phi; (\frac{1+z}{1-z})^{\theta})$ , then $$|a_2| \le \frac{2\theta}{|\beta_2 - \alpha_2|}, \qquad |a_3| \le \frac{2\theta}{|\beta_3 - \alpha_3|} \max\{1, |\beta|\},$$ $$|a_3 - \mu a_2^2| \le \frac{2\theta}{|\beta_3 - \alpha_3|} \max\{1, |\gamma|\} \quad (\mu \in \mathbb{C}),$$ where $$\beta = \frac{1+\theta}{2} + \frac{2\alpha_2\theta}{\beta_2 - \alpha_2}, \qquad \gamma = \frac{2(\beta_3 - \alpha_3)\theta}{(\beta_2 - \alpha_2)^2}\mu - \beta.$$ The results are sharp.

Corollary 10. Let $\beta_2\beta_3 \neq 0$ . If $f \in \mathcal{CW}(\Phi; (\frac{1+z}{1-z})^{\theta})$ , then $$|a_{3} - \mu a_{2}^{2}| \leq \begin{cases} \frac{\theta}{|\beta_{3}|} (D + A + B) & \text{if } 0 \leq A \leq C \lor 0 \leq B \leq C \\ & \lor (A \geq C \land B \geq C), \end{cases}$$ $$\frac{\theta D}{|\beta_{3}|} & \text{if } A \leq 0 \land B \leq 0,$$ $$\frac{\theta}{|\beta_{3}|} (D + B - C + \frac{C^{2}}{C - A}) & \text{if } A < 0 \land B \geq C,$$ $$\frac{\theta}{|\beta_{3}|} (D + A - C + \frac{C^{2}}{C - B}) & \text{if } B < 0 \land A \geq C,$$ where $$A = \left| \alpha_3 \frac{1+5\theta}{2} - 4\mu \frac{\alpha_2^2 \beta_3 \theta}{\beta_2^2} \right| + C - |\alpha_3|, \qquad D = 2 + |\alpha_3|,$$ $$B = \left| 1 + \theta - 4\mu \frac{\beta_3 \theta}{\beta_2^2} \right| + C - 2, \qquad C = 2\theta |\alpha_2| \left| 1 - 2\mu \frac{\beta_3}{\beta_2^2} \right|.$$ The result is sharp. Let $\Omega_k = \{u + iv : u > k\sqrt{(u-1)^2 + v^2}\}$ , k > 0. Note that $\Omega_k$ is the convex domain contained in the right half plane, with $1 \in \Omega_k$ . More precisely, it is the elliptic domain for k > 1, the hyperbolic domain for 0 < k < 1 and the parabolic domain for k = 1. Let us denote by $h_k$ the univalent function, which maps the unit disc $\mathcal{U}$ onto the conic domain $\Omega_k$ with $h_k(0) = 1$ . Obviously, the function $h_k$ is convex in $\mathcal{U}$ . It is easy to check that $f \in \mathcal{W}(\Phi; h_k)$ if and only if $$\operatorname{Re}\left(\frac{(\phif)(z)}{(\varphig)(z)}\right) > k \left| \frac{(\phif)(z)}{(\varphig)(z)} - 1 \right| \quad (z \in \mathcal{U}).$$ The following lemma gives coefficients estimates for the function.

Lemma 2. [13] Let $$h_k = 1 + \sum_{n=1}^{\infty} p_n z^n$$ ( $z \in U$ ). Then $$p_{1} = \begin{cases} \frac{2D^{2}(k)}{1-k^{2}} & for \ 0 \leq k < 1, \\ \frac{8}{\pi^{2}} & for \ k = 1, \\ \frac{\pi^{2}}{4\sqrt{t(1+t)(k^{2}-1)}\mathcal{K}^{2}(t)} & for \ k > 1, \end{cases}$$ $$p_{2} = \begin{cases} \frac{D^{2}(k)+2}{3}p_{1} & for \ 0 \leq k < 1, \\ \frac{2}{3}p_{1} & for \ k = 1, \\ \frac{4(t^{2}+6t+1)\mathcal{K}^{2}(t)-\pi^{2}}{24\sqrt{t(1+t)\mathcal{K}^{2}(t)}}p_{1} & for \ k > 1, \end{cases}$$ where $D(k) = \frac{2}{\pi} \arcsin k$ and $K^2(t)$ is the complete elliptic integral of first kind. Using Lemma 1 in Theorems 1-5 we obtain the solutions of the Fekete-Szegö problem for the classes $W_{\alpha}(\Phi, \Psi; h_k)$ , $\mathcal{M}_{\alpha}(\varphi, h_k)$ , $\mathcal{W}_{\alpha}(\Phi; h_k)$ , $\mathcal{CW}_{\alpha}(\Phi; h_k)$ . Remark 1 The classes $W_{\alpha}(\Phi, \Psi; h)$ , $\mathcal{M}_{\alpha}(\varphi, h)$ , $\mathcal{CW}_{\alpha}(\Phi; P)$ reduced to well-known subclasses by judicious choices of the parameters; see, for example [1–28]. In particular, they generalize several well-known classes defined by linear operators, which were investigated in earlier works. Also, the obtained results generalize several results obtained in these classes of functions.