COEFFICIENT ESTIMATES FOR BOUNDED STARLIKE FUNCTIONS OF COMPLEX ORDER

Lemma 1. If $Q \neq 1$ and n and q are positive integers, then $$(1 - Q^{2}) \left\{ \left( \frac{1 + Q}{1 - Q} \right) |b|^{2} + \sum_{m=1}^{q} \left[ \left( \frac{1 + Q}{1 - Q} \right) |b|^{2} + \frac{2Q}{1 - Q} mn \operatorname{Re}\{b\} - m^{2} n^{2} \right] \cdot \left[ \frac{1}{m!} \prod_{j=0}^{m-1} u_{j} \right]^{2} \right\} = \left\{ \frac{n}{q!} \prod_{j=0}^{q} u_{j} \right\}^{2},$$ $$(2.1)$$ where $$u_j = \left| \left( \frac{1+Q}{n} \right) b + jQ \right|, \quad j = 0, 1, 2, \cdots$$ (2.2) The lemma can be proved by induction on q in the same way as the lemma in [7].

Theorem 1. If the function f(z) defined by (1.1) is in the class F(b, M, n), $M > \frac{1}{2}$ and $Q \neq 1$ , and if $$n^{2} - \frac{2Q}{1 - Q} nRe\{b\} - (\frac{1 + Q}{1 - Q})|b|^{2} \ge 0, \tag{2.3}$$ then $$\sum_{k=mn+1}^{(m+1)n} (k-1)^2 |a_k|^2 \le (1+Q)^2 |b|^2, \quad m=1,2,\cdots.$$ (2.4) If, on the other hand, $$n^{2} - \frac{2Q}{1 - Q}nRe\{b\} - (\frac{1 + Q}{1 - Q})|b|^{2} < 0,$$ (2.5) then $$\sum_{k=mn+1}^{(m+1)n} (k-1)^2 |a_k|^2 \le \left\{ \frac{n}{(m-1)!} \prod_{j=0}^{m-1} u_j \right\}^2$$ (2.6) for $m = 1, 2, \dots, q_0 + 1$ , and $$\sum_{k=mn+1}^{(m+1)n} (k-1)^2 |a_k|^2 \le \left\{ \frac{n}{q_0!} \prod_{j=0}^{q_0} u_j \right\}^2$$ (2.7) for $m = q_0 + 2$ , $q_0 + 3$ , $\cdots$ , where $u_j$ is given by (2.2) and $q_0$ is the natural number determined by $q_0 \in [c-1,c)$ , where $$c = \frac{\left(\frac{Q}{1-Q}\right)Re\{b\} + \sqrt{\left(\frac{Q}{1-Q}Re\{b\}\right)^2 + \left(\frac{1+Q}{1-Q}\right)|b|^2}}{n}.$$ (2.8) Also $$\sum_{k=n+1}^{\infty} \left[ (k-1)^2 - (\frac{1+Q}{1-Q})|b|^2 - \frac{2Q}{1-Q}(k-1)Re\{b\} \right] |a_k|^2 \le (\frac{1+Q}{1-Q})|b|^2. \tag{2.9}$$ The estimates in (2.4) and (2.6) are sharp. Proof. Since $f(z) \in F(b, M, n)$ , (1.8) gives $$[zf'(z) - f(z)] = \{[(1+Q)b - Q]f(z) + Qzf'(z)\} w(z).$$ Substituting the series expansions of f(z) and w(z), we obtain $$\sum_{k=n+1}^{\infty} (k-1)a_k z^k$$ $$= \left\{ (1+Q)bz + \sum_{k=n+1}^{\infty} \left[ (1+Q)b + Q(k-1) \right] a_k z^k \right\} \sum_{k=n}^{\infty} b_k z^k. \tag{2.10}$$ We now proceed by a method introduced by Clunie [5]. Equating the coefficients of $z^k$ on the two sides of (2.10) for $k = n + 1, \dots, 2n$ , we obtain $$(k-1)a_k = (1+Q)bb_{k-1}$$ for $k = n+1, \dots, 2n$ . Therefore, $$\sum_{k=n+1}^{2n} (k-1)^2 |a_k|^2 \le (1+Q)^2 |b|^2 \sum_{k=n+1}^{2n} |b_{k-1}|^2$$ $$\le (1+Q)^2 |b|^2, \tag{2.11}$$ since, we have, for 0 < r < 1, $$\sum_{k=n}^{\infty} |b_k|^2 r^{2k} = \frac{1}{2\pi} \int_0^{2\pi} |w(re^{i\varphi})|^2 d\varphi \le 1$$ and letting r tends to 1 we obtain $\sum_{k=n}^{\infty} |b_k|^2 \leq 1$ . Again, for $p \ge n + 1$ , (2.10) can be put in the form $$G(z) = H(z)w(z), \quad z \in U$$ with $$G(z) = \sum_{k=n+1}^{n+p} (k-1)a_k z^k + \sum_{k=n+p+1}^{\infty} d_k z^k$$ and $$H(z) = (1+Q)bz + \sum_{k=n+1}^{p} [(1+Q)b + Q(k-1)] a_k z^k,$$ where $\sum_{k=n+p+1}^{\infty} d_k z^k$ converges in U. Since |w(z)| < 1 for $z \in U$ , we obtain, for 0 < r < 1, $$\frac{1}{2\pi} \int_0^{2\pi} |G(re^{i\varphi})|^2 d\varphi \le \frac{1}{2\pi} \int_0^{2\pi} |H(re^{i\varphi})|^2 d\varphi,$$ so that $$\sum_{k=n+1}^{n+p} (k-1)^2 |a_k|^2 r^{2k} + \sum_{k=n+p+1}^{\infty} |d_k|^2 r^{2k}$$ $$\leq (1+Q)^2 |b|^2 r^2 + \sum_{k=n+1}^{p} |(1+Q)b + Q(k-1)|^2 |a_k|^2 r^{2k}.$$ Letting r tends to 1 and rearranging, we obtain for $p \ge n + 1$ , $$\sum_{k=p+1}^{p+n} (k-1)^2 |a_k|^2 \le (1-Q^2) \left\{ \left(\frac{1+Q}{1-Q}\right) |b|^2 + \sum_{k=n+1}^p \left[ \left(\frac{1+Q}{1-Q}\right) |b|^2 + \frac{2Q}{1-Q} (k-1)Re\{b\} - (k-1)^2 \right] |a_k|^2 \right\}.$$ (2.12) Let the inequality (2.3) hold true. Then, for $k \geq n+1$ , if $Q \cdot Re\{b\} > 0$ , we have $$\begin{split} &(\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}(k-1)Re\{b\} - (k-1)^2 \\ = &(k-1)^2 \left[ (\frac{1+Q}{1-Q}) \frac{|b|^2}{(k-1)^2} + \frac{2Q}{1-Q} \frac{Re\{b\}}{(k-1)} - 1 \right] \\ \leq &(k-1)^2 \left[ (\frac{1+Q}{1-Q}) \frac{|b|^2}{n^2} + \frac{2Q}{1-Q} \frac{Re\{b\}}{n} - 1 \right] \\ = &\frac{(k-1)^2}{n^2} \left[ (\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}nRe\{b\} - n^2 \right] \\ \leq &0. \end{split}$$ If $Q \cdot Re\{b\} < 0$ , we have $$(\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}(k-1)Re\{b\} - (k-1)^2$$ $$\leq (\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q} \cdot n \cdot Re\{b\} - n^2$$ $$\leq 0.$$ Hence (2.12) yields $$\sum_{k=p+1}^{p+n} (k-1)^2 |a_k|^2 \le (1+Q)^2 |b|^2 \quad \text{for} \quad p \ge n+1.$$ Putting $p = mn, m \ge 2$ and combining with (2.11) we obtain (2.4). Suppose now the inequality (2.5) is true. Let $q_0$ be as defined in the statement of the theorem. Then $q_0$ is the largest of the natural numbers k for which $$k^2n^2 - \frac{2Q}{1-Q}knRe\{b\} - (\frac{1+Q}{1-Q})|b|^2 < 0.$$ We now establish by an inductive argument inequalities (2.6) for $m = 1, 2, \dots, q_0 + 1$ and the inequalities $$\sum_{k=mn+1}^{(m+1)n} \left[ \left( \frac{1+Q}{1-Q} \right) |b|^2 + \frac{2Q}{1-Q} (k-1) Re\{b\} - (k-1)^2 \right] |a_k|^2 \\ \leq \left[ \left( \frac{1+Q}{1-Q} \right) |b|^2 + \frac{2Q}{1-Q} mn Re\{b\} - m^2 n^2 \right] \left[ \frac{1}{m!} \prod_{j=0}^{m-1} u_j \right]^2 \tag{2.13}$$ for $m=1,\cdots,q_0$ . For m = 1, (2.6) reduces to (2.11) whereas the left member of (2.13) $$\leq \left[ \frac{\left(\frac{1+Q}{1-Q}\right)|b|^2 + \frac{2Q}{1-Q}nRe\{b\} - n^2}{n^2} \right] \sum_{k=n+1}^{2n} (k-1)^2 |a_k|^2$$ $$\leq \left[ \left(\frac{1+Q}{1-Q}\right)|b|^2 + \frac{2Q}{1-Q}nRe\{b\} - n^2 \right] \left(\frac{1+Q}{n}\right)^2 |b|^2$$ by (2.11), so that (2.13) holds for m = 1. Suppose that (2.6) and (2.13) hold for $m=1,\dots,q-1$ , where $2 \leq q \leq q_0$ . For p=qn, (2.12) yields $$\begin{split} &\sum_{k=qn+1}^{(q+1)n} (k-1)^2 |a_k|^2 \leq (1-Q)^2 \left\{ (\frac{1+Q}{1-Q})|b|^2 \\ &+ \sum_{m=1}^{q-1} \sum_{k=mn+1}^{(m+1)n} \left[ (\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}(k-1)Re\{b\} - (k-1)^2 \right] |a_k|^2 \right\} \\ &\leq & (1-Q^2) \left\{ (\frac{1+Q}{1-Q})|b|^2 + \sum_{m=1}^{q-1} \left[ (\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}mnRe\{b\} - m^2n^2 \right] \cdot \left[ \frac{1}{m!} \prod_{j=0}^{m-1} u_j \right]^2 \right\}, \end{split}$$ by (2.13) for $m = 1, \dots, q - 1$ . Hence by Lemma 1, $$\sum_{k=qn+1}^{(q+1)n} (k-1)^2 |a_k|^2 \le \left[ \frac{n}{(q-1)!} \prod_{j=0}^{q-1} u_j \right]^2$$ so that (2.6) holds for m = q. Now, $$\sum_{k=qn+1}^{(q+1)n} \left[ \left( \frac{1+Q}{1-Q} \right) |b|^2 + \frac{2Q}{1-Q} (k-1)Re\{b\} - (k-1)^2 \right] |a_k|^2$$ $$\leq \left[ \frac{\left( \frac{1+Q}{1-Q} \right) |b|^2 + \frac{2Q}{1-Q} nRe\{b\} - q^2 n^2}{q^2 n^2} \right] \sum_{k=qn+1}^{(q+1)n} (k-1)^2 |a_k|^2$$ $$\leq \left[ \frac{\left( \frac{1+Q}{1-Q} \right) |b|^2 + \frac{2Q}{1-Q} qnRe\{b\} - q^2 n^2}{q^2 n^2} \right] \left[ \frac{n}{(q-1)!} \prod_{j=0}^{q-1} u_j \right]^2,$$ using (2.6) with m = q (since $(\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}qnRe\{b\} - q^2n^2 > 0$ because $q \le q_0$ ). Thus (2.13) holds for m = q. Hence (2.6) and (2.13) hold for $m = 1, \dots, q_0$ . It follows now, by the argument used above to show that (2.6) holds for m = q, that (2.6) holds for $m = q_0 + 1$ . By the definition of $q_0$ , we have $$\left(\frac{1+Q}{1-Q}\right)|b|^2 + \frac{2Q}{1-Q}n(q_0+1)Re\{b\} - (q_0+1)^2n^2 \le 0.$$ Hence $(\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}(k-1)Re\{b\} - (k-1)^2 \le 0$ for $k > (q_0+1)n$ . Thus, for $p \ge (q_0+1)n$ , (2.12) yields $$\begin{split} &\sum_{k=p+1}^{p+n} (k-1)^2 |a_k|^2 \\ &\leq (1-Q^2) \left\{ (\frac{1+Q}{1-Q})|b|^2 + \sum_{m=1}^{q_0} \sum_{k=mn+1}^{(m+1)n} \left[ (\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}(k-1)Re\{b\} - (k-1)^2 \right] |a_k|^2 \right\} \\ &\leq (1-Q^2) \left\{ (\frac{1+Q}{1-Q})|b|^2 + \sum_{m=1}^{q_0} \left[ (\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}mnRe\{b\} - m^2n^2 \right] \left[ \frac{1}{m!} \prod_{j=0}^{m-1} u_j \right]^2 \right\}, \end{split}$$ by (2.13) for $m = 1, 2, \dots, q_0$ . Hence, by Lemma 1, $$\sum_{k=p+1}^{p+n} (k-1)^2 |a_k|^2 \le \left[ \frac{n}{q_0!} \prod_{j=0}^{q_0} u_j \right] \text{ for } p \ge (q_0+1)n.$$ Putting $p = mn, m = q_0 + 2, q_0 + 3, \dots$ , we obtain (2.7). Finally, we obtain from (2.12), $$\sum_{k=n+1}^{p} \left[ (k-1)^2 - (\frac{1+Q}{1-Q})|b|^2 - \frac{2Q}{1-Q}(k-1)Re\{b\} \right] |a_k|^2 \le (\frac{1+Q}{1-Q})|b|^2$$ for $p \geq n+1$ and letting p tends to $\infty$ , we obtain (2.9). This completes the proof of Theorem 1. The estimates (2.4) and (2.6) are sharp with equality holding in (2.4) for a given mfor the funnction $f_{\varepsilon}(z)(|\varepsilon|=1)$ defined by $$f_{\varepsilon}(z) = \begin{cases} z[1 - \varepsilon Q z^{mn}]^{-(\frac{1+Q}{Q})\frac{b}{mn}}, & Q \neq 0, \\ z \exp\left[\frac{\varepsilon b z^{mn}}{mn}\right], & Q = 0, \end{cases}$$ if $n^2 - \frac{2Q}{1-Q}nRe\{b\} - (\frac{1+Q}{1-Q})|b|^2 \ge 0$ . Also the equality in (2.6) holds for the function $f_{\varepsilon}(z)(|\varepsilon| = 1)$ defined by $$f_{\varepsilon}(z) = \begin{cases} \left[1 - \varepsilon Q z^{n}\right]^{-\left(\frac{1+Q}{Q}\right)\frac{b}{n}}, & Q \neq 0, \\ z \exp\left[\frac{\varepsilon b z^{n}}{n}\right], & Q = 0, \end{cases}$$ if $$n^2 - \frac{2Q}{1-Q}nRe\{b\} - (\frac{1+Q}{1-Q})|b|^2 < 0$$ . Remark 1. Under the hypothesis of Theorem 1, since Q tends to 1 as M tends to $\infty$ , it follows that $n^2 - \frac{2Q}{1-Q}nRe\{b\} - (\frac{1+Q}{1-Q})|b|^2 < 0$ for all sufficiently large M and hence (2.6) and (2.7) hold for all sufficiently large M. Also since $q_o$ tends to $\infty, 1+Q$ tends to 2 as M tends to $\infty$ , we obtain the following corollaries:

Corollary 1. If $f(z) = z + \sum_{k=n+1}^{\infty} a_k z^k \in F((1-\alpha)e^{-i\lambda}\cos\lambda, M, n) = F_M(\lambda, \alpha, n)$ , $|\lambda| < \frac{\pi}{2}$ and $0 \le \alpha < 1$ , then $$\sum_{k=mn+1}^{(m+1)n} (k-1)^2 |a_k|^2 \le \left\{ \frac{n}{(m-1)!} \prod_{j=0}^{m-1} \left| (\frac{1+Q}{n})(1-\alpha)e^{-i\lambda} \cos \lambda + jQ \right| \right\}^2$$ for $m=1,2,\cdots$ . The estimates are sharp for the function $f_{\varepsilon}(z)(|\varepsilon|=1)$ given by $$f_{\varepsilon}(z) = \begin{cases} z \left[1 - \varepsilon Q z^{n}\right]^{-\left(\frac{1+Q}{Q}\right) \frac{(1-\alpha)e^{-i\lambda}\cos\lambda}{n}}, & Q \neq 0, \\ z \exp\left[\frac{\varepsilon(1-\alpha)e^{-i\lambda}z^{n}\cos\lambda}{n}\right], & Q = 0. \end{cases}$$

Corollary 2. If $f(z) = z + \sum_{k=n+1}^{\infty} a_k z^k \in F(b, \infty, n) = S(1-b, n)$ , then $$\sum_{k=mn+1}^{(m+1)n} (k-1)^2 |a_k|^2 \le \left\{ \frac{n}{(m-1)!} \prod_{j=0}^{m-1} \left| \frac{2b}{n} + j \right| \right\}^2$$ for $m = 1, 2, \cdots$ . The estimates are sharp for the function $f_{\varepsilon}(z)(|\varepsilon| = 1)$ given by $$f_{\varepsilon}(z) = (1 - \varepsilon z^n)^{\frac{-2b}{n}}.$$ Corollary 3 [14]. If $f(z) = z + \sum_{k=n+1}^{\infty} a_k z^k \in F(e^{-i\lambda}\cos\lambda, \frac{1}{\cos\lambda}, n) = (H^*)_n^{\lambda}$ , then $$|a_k| \le \frac{(2-\cos\lambda)\cos\lambda}{k-1}, \qquad k \ge n+1.$$ The estimates are sharp for each k > n + 1. Corollary 4 [7,8,14]. If $f(z) = z + \sum_{k=n+1}^{\infty} a_k z^k \in F((1-\alpha)e^{-i\lambda}\cos\lambda, \infty, n) = S^{\lambda}(\alpha, n), 0 \le \alpha < 1, |\lambda| < \frac{\pi}{2}$ , then $$\sum_{k=mn+1}^{(m+1)n} (k-1)^2 |a_k|^2 \le \left\{ \frac{n}{(m-1)!} \prod_{j=0}^{m-1} \left| \frac{2(1-\alpha)e^{-i\lambda} \cos \lambda}{n} + j \right| \right\}^2$$ for $m = 1, 2, \cdots$ . The estimates are sharp. Remark 2. - (1) Choosing (i) $\alpha = \lambda = 0$ , (ii) $\lambda = 0$ in Corollary 4 we get, respectively, results of MacGregor [13] and Boyd [4]. - (2) Choosing (i) n = 1, (ii) n = 1, α = 0 in Corollary 1 we get, respectively, results of Aouf [1] and Kulshrestha [11]. - (3) Choosing n = 1 in Corollary 2 we get results of Nasr and Aouf [15]. - (4) Choosing n = 1 in Corollary 3 we get a result of Goel [6]. - (5) Choosing $n = 1, b = (1 \alpha)e^{-i\lambda}\cos\lambda, 0 \le \alpha < 1, |\lambda| < \frac{\pi}{2}, M \ge 1$ in Theorem 1 we obtain a result of Plaskota [17].

Theorem 2. If the function f(z) defined by (1.9) is in the class $F^*(b, M, n)$ and $M \ge 1$ , then $$\sum_{k=mn}^{(m+1)n-1} (k+1)^2 |a_k|^2 \le (1+Q)^2 |b|^2, \quad Q = 1 - \frac{1}{M}, \tag{2.14}$$ for $m = 1, 2, \dots, and$ $$\sum_{k=n}^{\infty} \left[ (k+1)^2 - (\frac{1+Q}{1-Q})|b|^2 + \frac{2Q}{1-Q}(k+1)Re\{b\} \right] |a_k|^2 \le (\frac{1+Q}{1-Q})|b|^2. \tag{2.15}$$ Estimate (2.14) is sharp with equality holding for given m for the function $$f_{\varepsilon}(z) = \begin{cases} z^{-1} [1 + \varepsilon Q z^{mn+1}]^{-(\frac{1+Q}{Q}) \frac{b}{mn+1}}, & Q \neq 0, \\ z^{-1} \exp[\frac{-\varepsilon b z^{mn+1}}{mn+1}], & Q = 0, \end{cases}$$ where $|\varepsilon| = 1$ . The proof is analogous to that of Theorem 1 and is omitted. Remark 3. Choosing $n = 1, b = (1-a)e^{-i\beta}\cos\beta, 0 \le a < 1, |\beta| < \frac{\pi}{2}, M = m \ge 1$ , the above theorems yields results of Jakubouski [9].