ON SPIRAL-LIKE FUNCTIONS OF ORDER α AND TYPE β

Theorem 1. A function $f(z)=z+\sum_{n=2}^{\infty}a_nz^n$ , analytic in E, is in the class $S^{2}(\alpha,\beta)$ if and only if (2.3) $$f(z) = z \exp \left\{-2\beta(1-\alpha)e^{-i\lambda}\cos\lambda \int_0^z \frac{\phi(t)dt}{1+(2\beta-1)t\phi(t)}\right\},$$ for some $\phi \in A$ . Proof. First suppose $f(z) \in S^{2}(\alpha, \beta)$ . Noting that zf'(z)/f(z) satisfies the hypothesis of the first part of the Lemma, we see that $$\frac{zf'(z)}{f(z)} = \frac{1 + ((2\beta - 1) - 2\beta(1 - \alpha)e^{-i\lambda}\cos\lambda)z\phi(z)}{1 + (2\beta - 1)z\phi(z)}$$ for some \$\phi\in A\$ . Thus we have (2.4) $$\frac{f'(z)}{f(z)} - \frac{1}{z} = \frac{-2\beta(1-\alpha)e^{-i\lambda}\cos\lambda\cdot\phi(z)}{1+(2\beta-1)z\phi(z)}.$$ An integration from \$0\$ to \$z\$ in [\(2.4\)](#page-2-0) followed by an exponentiation leads to [\(2.3\).](#page-1-1) Conversely, if [\(2.3\)](#page-1-1) holds, then $$\frac{zf'(z)}{f(z)} = \frac{1 + ((2\beta - 1) - 2\beta(1 - \alpha)e^{-i\lambda}\cos\lambda)z\phi(z)}{1 + (2\beta - 1)z\phi(z)}.$$ Now the theorem follows by the converse part of the [Lemma.](#page-1-0) Remarks. (i) Putting \$\beta=1\$ , we obtain a representation formula determined by Libera [\[6\].](#page-10-0) - (ii) The representation formula determined by Juneja and Mogra [\[3\]](#page-10-1) for starlike functions of order \$\alpha\$ and type \$\beta\$ can be obtained by putting \$\lambda=0\$ in [Theorem](#page-1-2) 1. - (iii) \$\alpha=0\$ and \$\beta=(2-coe\lambda)/2\$ in [Theorem](#page-1-2) ¹ gives the result obtained by Goel [\[2\].](#page-10-4) - (iv) Replacing \$\alpha\$ by \$(1-\beta+2\alpha\beta)/(1+\beta)\$ and \$\beta\$ by \$(1+\beta)/2\$ in [Theorem](#page-1-2) 1, we get the corresponding result obtained by Mak\"owka [71. - (v) By taking appropriate values of \$\alpha,\$ \$\beta,\$ \$\lambda\$ in [Theorem](#page-1-2) 1, we obtain the corresponding representation formulae for the functions of the classes introduced by Spac\'ek [\[10\],](#page-11-0) Kulshrestha [\[5\],](#page-10-2) Padmanabhan [\[8\],](#page-11-1) and many others.

Theorem 2. Let \$f(z)=z+\sum\_{=2}^{\infty}a\_{n}z^{n}\$ be analytic in E. Then \$f(z)\in S^{\lambda}(\alpha, \beta)\$ , iffor some \$a\in[0,1\$) and \$\lambda\in(-\pi/2, \pi/2)\$ , (3.1) $$\sum_{n=2}^{\infty} \{2n(1-\beta)-1+|(1-2\beta)+2\beta(1-\alpha)e^{-i\lambda}\cos\lambda|\}|a_n| \leq 2\beta(1-\alpha)\cos\lambda,$$ whenever $\beta \in (0, 1/2]$ , (3.2) $$\sum_{n=2}^{\infty} \{(n-1) + |(2\beta-1)(n-1) + 2\beta(1-\alpha)e^{-i\lambda}\cos\lambda|\} |a_n| \leq 2\beta(1-\alpha)\cos\lambda,$$ whenever $\beta \in [1/2, 1]$ , holds. Proof. Let \$|z|=r<1\$ . Noting that (3.3) $$|zf'(z)-f(z)| < \sum_{n=0}^{\infty} (n-1)|a_n|r$$ , and $$(3.4) |2\beta((zf'(z)-f(z))+(1-\alpha)e^{-i\lambda}\cos\lambda\cdot f(z))-(zf'(z)-f(z))| \\ \ge \{2\beta(1-\alpha)\cos\lambda-\sum_{n=0}^{\infty}(1-2\beta)n|a_n|-\sum_{n=0}^{\infty}|(1-2\beta)+2\beta(1-\alpha)e^{-i\lambda}\cos\lambda| |a_n|\},$$ we see that $$|zf'(z)-f(z)|-|2\beta((zf'(z)-f(z))+(1-\alpha)e^{-i\lambda}\cos\lambda\cdot f(z))-(zf'(z)-f(z))|$$ $$\leq \sum_{n=2}^{\infty} \{2n(1-\beta)-1+|(1-2\beta)+2\beta(1-\alpha)e^{-i\lambda}\cos\lambda|\}|a_n|-2\beta(1-\alpha)\cos\lambda,$$ provided $0 < \beta \le 1/2$ . The last quantity is $\le 0$ by (3.1), so that $f(z) \in S^1(\alpha, \beta)$ . For the second part, we assume that (3.2) holds for $\beta \in [1/2, 1]$ . In this case, $$(3.5) |2\beta((zf'(z)-f(z))+(1-\alpha)e^{-i\lambda}\cos\lambda\cdot f(z))-(zf'(z)-f(z))| \\ \ge \{2\beta(1-\alpha)\cos\lambda-\sum_{n=2}^{\infty}|(2\beta-1)(n-1)+2\beta(1-\alpha)e^{-i\lambda}\cos\lambda| |a_n|\}r.$$ Now the theorem follows, as before, from (3.3), (3.5) and (3.2).

Corollary 1. A function $f(z)=z+\sum_{n=2}^{\infty}a_nz^n$ , analytic in E, is a $\lambda$ -spiral-like function of order $\alpha$ if $$\sum_{n=2}^{\infty} \{ (n-1) + |(n-1) + 2(1-\alpha)e^{-i\lambda}\cos\lambda| \} |a_n| \leq 2(1-\alpha)\cos\lambda ,$$ for some $\alpha \in [0, 1)$ , $\lambda \in (-\pi/2, \pi/2)$ .

Corollary 2. A function $f(z)=z+\sum_{n=2}^{\infty}a_nz^n$ , analytic in E, is in the class $H(\lambda)$ if $$\sum_{n=2}^{\infty} \{(n-1)+|(1-\cos\lambda)(n-1)+(2-\cos\lambda)e^{-i\lambda}\cos\lambda|\}|a_n| \leq (2-\cos\lambda)\cos\lambda \ ,$$ for some $\lambda \in (-\pi/2, \pi/2)$ .

Corollary 3. A function $f(z)=z+\sum_{n=2}^{\infty}a_nz^n$ , analytic in E, is in the class $F_{\lambda,M}$ if for some $\lambda\in(-\pi/2,\pi/2)$ , $$\sum_{n=2}^{\infty} \{ (n-M) + \sqrt{(1-M)^2 + (2M-1)\cos^2 \lambda} \} |a_n| \leq (2M-1)\cos \lambda ,$$ whenever $1/2 < M \le 1$ . $$\sum_{n=2}^{\infty} \{ (n-1)M + |(M-1)(n-1) + (2M-1)e^{-i\lambda}\cos\lambda| \} |a_n| \leq (2M-1)\cos\lambda \; ,$$ whenever $M \ge 1$ , holds. Remarks. (i) For \$\lambda=0\$ in [Theorem](#page-2-2) 2, we obtain the sufficient condition determined by Juneja and Mogra [\[3\].](#page-10-1) - (ii) Replacing \$\alpha\$ by \$(1-\beta+\&\beta)/(1+\beta),\$ \$\beta\$ by \$(1+\beta)/2\$ in \$Th\ovalbox{\tt\small REJECT} rem2\$ we obtain the corresponding sufficient condition for a function \$f(z)\$ to be in the class introduced by Mak\"owka [\[7\].](#page-10-3) - (iii) By fixing the parameters \$\alpha,\$ \$\beta\$ , and \$\lambda\$ in [Theorem](#page-2-2) 2, we can obtain sufficient conditions for a function to be in the classes introduced by Padmanabhan [\[8\],](#page-11-1) Wright [\[11\],](#page-11-2) and others.

Theorem 3. Let \$f(z)=z+\sum\_{=2}^{\infty}a\_{n}z^{\prime}\*\$ be analytic in E. If \$f(z)\in S^{\lambda}(\alpha, \beta)\$ , then for \$|z|=r\_{1}0<r<1\$ , and for all \$a\in[0,1\$ ), \$\beta\in(0,1/2)\cup(1/2,1\$], \$\lambda\in(-\pi/2, \pi/2)\$ , $$|f(z)| \leq r \left[ \frac{(1+(2\beta-1)r)^{(1-\cos\lambda)}}{(1-(2\beta-1)r)^{(1+\cos\lambda)}} \right]^{\beta(1-\alpha)\cos\lambda/(2\beta-1)},$$ and $$(4.2) |f(z)| \ge r \left[ \frac{(1-(2\beta-1)r)^{(1-\cos\lambda)}}{(1+(2\beta-1)r)^{(1+\cos\lambda)}} \right]^{\beta(1-\alpha)\cos\lambda/(2\beta-1)};$$ whereas for \$\alpha\in[0,1\$ ), \$\beta=1/2,\$ \$\lambda\in(-\pi/2, \pi/2)\$ , $$(4.3) |f(z)| \leq r \exp\left((1-\alpha)\cos\lambda \cdot r\right),$$ and $$(4.4) |f(z)| \ge r \exp\left(-(1-\alpha)\cos\lambda \cdot r\right).$$ All these estimates are sharp for all admissiue values of \$\alpha,\$ \$\beta,\$ \$\lambda\$ . Proof. Since \$f(z)\in S^{\lambda}(\alpha, \beta)\$ , the condition [\(1.1\)](#page-0-0) coupled with an application of Schwarz's [Lemma,](#page-1-0) implies \$|zf^{\nu}(z)/f(z)-\xi\{<R\$, where $$\xi = \frac{1 - (2\beta - 1)((2\beta - 1) - 2\beta(1 - \alpha)\cos^2\lambda)r^2 - i\beta(2\beta - 1)(1 - \alpha)\sin2\lambda \cdot r^2}{1 - (2\beta - 1)^2r^2},$$ and $$R = \frac{2\beta(1-\alpha)\cos\lambda \cdot r}{1-(2\beta-1)^2r^2}, \quad (|z|=r).$$ Hence we have $$(4.5) \frac{1-2\beta(1-\alpha)\cos\lambda \cdot r + (2\beta-1)(2\beta(1-\alpha)\cos^2\lambda - (2\beta-1))r^2}{1-(2\beta-1)^2r^2} \le \operatorname{Re}\left(\frac{zf'(z)}{f(z)}\right) \le \frac{1+2\beta(1-\alpha)\cos\lambda \cdot r + (2\beta-1)(2\beta(1-\alpha)\cos^2\lambda - (2\beta-1))r}{1-(2\beta-1)^2r^2}$$ Noting $$\log\left(\left|\frac{f(z)}{z}\right|\right) = \operatorname{Re}\int_{0}^{z} \left(\frac{f'(s)}{f(s)} - \frac{1}{s}\right) ds = \int_{0}^{r} \frac{1}{t} \operatorname{Re}\left(te^{i\theta} \frac{f'(te^{i\theta})}{f(te^{i\theta})} - 1\right) dt$$ and using (4.5), we see that (4.6) $$\log\left(\left|\frac{f(z)}{z}\right|\right) \leq 2\beta(1-\alpha)\cos\lambda \cdot \int_0^r \frac{1+(2\beta-1)\cos\lambda \cdot t}{1-(2\beta-1)^2t^2}dt.$$ Now suppose \$\alpha\in[0,1\$ ), \$\beta\in(0,1/2)\cup(1/2,1\$ ] and \$\lambda\in(-\pi/2, \pi/2)\$ . Then from [\(4.6\),](#page-5-0) we get $$\log\left(\left|\frac{f(z)}{z}\right|\right) \leq \frac{\beta(1-\alpha)\cos\lambda}{2\beta-1}\log\left\{\frac{(1+(2\beta-1)r)^{(1-\cos\lambda)}}{(1-(2\beta-1)r)^{(1+\cos\lambda)}}\right\},\,$$ which gives [\(4.1\).](#page-4-0) For the case when \$\alpha\in[0,1\$ ), \$\beta=1/2\$ , and \$\lambda\in(-\pi/2, \pi/2),\$ \$(4.6)\$ immediately proves [\(4.3\).](#page-4-1) In view of $$\log\left(\left|\frac{f(z)}{z}\right|\right) = \operatorname{Re}\left(\log\frac{f(z)}{z}\right) = \int_{0}^{r} \operatorname{Re}\left(\frac{\partial}{\partial t}\left(\log\frac{f(t)}{t}\right)\right) dt$$ $$= \int_{0}^{r} \frac{1}{t} \operatorname{Re}\left(t\frac{f'(t)}{f(t)} - 1\right) dt,$$ and with the aid of (4.5) we may write $$(4.7) \qquad \log\left(\left|\frac{f(z)}{z}\right|\right) \ge -2\beta(1-\alpha)\cos\lambda \cdot \int_0^\tau \frac{1-(2\beta-1)\cos\lambda \cdot t}{1-(2\beta-1)^2t^2}dt.$$ If \$\beta\neq 1/2\$ , then carrying out the integration in [\(4.7\),](#page-5-1) we obtain [\(4.2\).](#page-4-2) Further, when \$\beta=1/2\$ , then we immediately get [\(4.4\)](#page-4-3) from [\(4.7\).](#page-5-1) The extremal function for all the inequalities is given by $$f(z) = \begin{cases} z/\{(1-(2\beta-1)e^{i\theta}z\}^{2\beta(1-\alpha)\cos\lambda\cdot e^{-i\lambda/(2\beta-1)}}, & \beta \neq 1/2 \\ z\exp\{(1-\alpha)\cos\lambda\cdot e^{i(\theta-\lambda)}z\}, & \beta = 1/2 \end{cases}$$ where \$0\leqq\alpha<1,\$ \$-\pi/2<\lambda<\pi/2\$ ; and \$\theta(0\leqq\theta\leqq 2\pi)\$ is determined by $$\tan (\theta/2) = ((1-(2\beta-1)r)/(1+(2\beta-1)r)) \cot (\pi/2-\lambda/2)$$ for the equality in [\(4.1\)](#page-4-0) and [\(4.3\),](#page-4-1) and by the equation $$\tan (\theta/2) = ((1-(2\beta-1)r)/(1+(2\beta-1)r)\cot (-\lambda/2)$$ for the equality in [\(4.2\)](#page-4-2) and [\(4.4\).](#page-4-3)

Corollary 4. If \$f(z)=z+\sum\_{=2}^{\infty}a\_{n}z^{n}\$ , analytic in \$E\$, is ^a \$\lambda\$ -spiral-like function of order \$\alpha\$ , then for \$|z|=r<1\$ , $$r\left[\frac{(1-r)^{(1-\cos\lambda)}}{(1+r)^{(1+\cos\lambda)}}\right]^{(1-\alpha)\cos\lambda} \leq |f(z)| \leq r\left[\frac{(1+r)^{(1-\cos\lambda)}}{(1-r)^{(1+\cos\lambda)}}\right]^{(1-\alpha)\cos\lambda}.$$ By choosing $\alpha=0$ and $\beta=(2M-1)/2M$ in Theorem 3, we get the following result.

Corollary 5. If $f(z)=z+\sum_{n=0}^{\infty}a_nz^n$ , analytic in E, is in $F_{\lambda,M}$ , then for |z|= $r<1, M>1/2, (M\neq 1)$ $$r \left[ \frac{(1 - ((M-1)/M)r)^{(1-\cos\lambda)}}{(1 + ((M-1)/M)r)^{(1+\cos\lambda)}} \right]^{(2M-1)\cos\lambda/2(M-1)} \\ \leq |f(z)| \leq r \left[ \frac{(1 + ((M-1)/M)r)^{(1-\cos\lambda)}}{(1 - ((M-1)/M)r)^{(1+\cos\lambda)}} \right]^{(2M-1)\cos\lambda/2(M-1)};$$ whereas for M=1, $$r \exp(-\cos \lambda \cdot r) \le |f(z)| \le r \exp(\cos \lambda \cdot r)$$ . Remarks. (i) Taking $\lambda=0$ , Theorem 3 coincides with the result of Juneja and Mogra [3]. - (ii) Putting $\alpha=0$ , $\beta=(2-\cos\lambda)/2$ ; and replacing $\alpha$ by $(1-\beta+2\alpha\beta)/(1+\beta)$ , $\beta$ by $(1+\beta)/2$ in Theorem 3 we get, respectively, the distortion theorems obtained by Goel [2], and Maköwka [7]. - (iii) Different values of $\alpha$ , $\beta$ , and $\lambda$ in Theorem 3 lead to the corresponding distortion theorems for the respective classes defined by Padmanabhan [8]. Wright [11] and others.

Theorem 4. Let $f \in S^{2}(\alpha, \beta)$ , and $f(z) = z + \sum_{n=2}^{\infty} a_{n}z^{n}$ , $z \in E$ . (a) If $\beta(1-\alpha)(2-\alpha)\cos^{2}\lambda > (1-\beta)(1+(1-\alpha)\cos^{2}\lambda)$ , let (a) If $$\beta(1-\alpha)(2-\alpha)\cos^2\lambda > (1-\beta)(1+(1-\alpha)\cos^2\lambda)$$ , let $$N = \left[ \frac{\beta(1-\alpha)(2-\alpha)\cos^2\lambda}{(1-\beta)(1+(1-\alpha)\cos^2\lambda)} \right].$$ Then (5.1) $$|a_n| \leq \frac{1}{(n-1)!} \prod_{k=2}^{n} |(2\beta-1)(k-2)+2\beta(1-\alpha)e^{-i\lambda} \cos \lambda|,$$ for $n=2, 3, \dots, N+2$ ; and $$|a_n| \leq \frac{1}{(N+1)! (n-1)} \prod_{k=2}^{N+3} |(2\beta-1)(k-2)+2\beta(1-\alpha)e^{-i\lambda} \cos \lambda| , \quad n > N+2.$$ (b) If $$\beta(1-\alpha)(2-\alpha)\cos^2\lambda \leq (1-\beta)(1+(1-\alpha)\cos^2\lambda)$$ , then $$(5.3) |a_n| \leq \frac{2\beta(1-\alpha)\cos\lambda}{n-1}, for n \geq 2.$$ The bounds in (5.1) and (5.3) are sharp for all admissible \$\alpha,\$ \$\beta,\$ \$\lambda\$ , and for each \$n\$ . Proof. Sinoe \$f\in S^{\lambda}(\alpha, \beta),\$ \$(2.3)\$ gives $$\frac{zf'(z)}{f(z)} = \frac{1 + ((2\beta - 1) - 2\beta(1 - \alpha)e^{-i\lambda}\cos\lambda)w(z)}{1 + (2\beta - 1)w(z)},$$ where \$w(z)=z\phi(z)\$ is analytic in \$E\$ and satisfies the \$\ovalbox{\tt\small REJECT} nditiooew(O)=0\$, and \$|w(z)|<1\$ for \$z\in E\$ . Now [\(5.4\)](#page-7-0) may be written as $$\{2\beta(1-\alpha)e^{-i\lambda}\cos\lambda\cdot z+\sum_{k=2}^{\infty}((2\beta-1)(k-1)+2\beta(1-\alpha)e^{-i\lambda}\cos\lambda)a_kz^k\}w(z)=\sum_{k=2}^{\infty}(1-k)a_kz^k$$ which is equivalent to $$\begin{aligned} \{2\beta(1-\alpha)e^{-i\lambda}\cos\lambda \cdot z + \sum_{k=2}^{n-1}((2\beta-1)(k-1) + 2\beta(1-\alpha)e^{-i\lambda}\cos\lambda)a_k z^k\}w(z) \\ &= \sum_{k=2}^{n}(1-k)a_k z^k + \sum_{k=n+1}^{\infty}b_k z^k \;, \end{aligned}$$ where \$\sum\_{k='+1}^{\infty}b\_{k}z^{k}\$ converges in \$E\$ . Then, sinoe \$|w(z)|<1\$ , (5.5) $$|2\beta(1-\alpha)e^{-i\lambda}\cos\lambda \cdot z + \sum_{k=2}^{n-1} ((2\beta-1)(k-1) + 2\beta(1-\alpha)e^{-i\lambda}\cos\lambda)a_k z^k|$$ $$\geq |\sum_{k=2}^{n} (1-k)a_k z^k + \sum_{k=n+1}^{\infty} b_k z^k|.$$ Writing \$z=re^{\theta},\$ \$r<1\$ , squaring both sides of [\(5.5\),](#page-7-1) and then integrating, we get $$\begin{aligned} 4\beta^2(1-\alpha)^2\cos^2\lambda\cdot r^2 + \sum\limits_{k=2}^{n-1}|(2\beta-1)(k-1)+2\beta(1-\alpha)e^{-i\lambda}\cos\lambda|^2|a_k|^2r^{2k} \ &\geq \sum\limits_{k=2}^n(k-1)^2|a_k|^2r^{2k} + \sum\limits_{k=n+1}^\infty|b_k|^2r^{2k} \ . \end{aligned}$$ Let \$r\rightarrow 1\$ , then on some simplification we obtain (5.6) $$(n-1)^{2}|a_{n}|^{2} \leq 4\beta^{2}(1-\alpha)^{2}\cos^{2}\lambda$$ $$+ \sum_{k=0}^{n-1} \{|(2\beta-1)(k-1)+2\beta(1-\alpha)e^{-i\lambda}\cos\lambda|-(k-1)^{2}\}|a_{k}|^{2}, \quad n \geq 2.$$ Now there may be following two cases: Let \$\beta(1-\alpha)(2-\alpha)\$ cos2 \$\lambda\geqq(1-\beta)\$ ( \$1+(1-a)\$ cos2 \$\lambda\$ ). Suppose that \$n\leqq N+2\_{j}\$ then for \$n=2,\$ \$(5.6)\$ gives $$|a_2| \leq 2\beta(1-\alpha)\cos\lambda$$ , which proves (5.1) for \$n=2\$. We establish (5.1) for \$n\leqq N+2\$, from (5.6), by mathematical induction. Suppose (5.1) is valid for \$k=2,3,\$ \$\cdots,\$ \$n-1\$ . Then it follows from (5.6), $$\begin{split} (n-1)^{2}|a_{n}|^{2} & \leq 4\beta^{2}(1-\alpha)^{2}\cos^{2}\lambda \\ & + \sum_{k=2}^{n-1} \left\{ \{|(2\beta-1)(k-1)+M|^{2}-(k-1)^{2}\} \frac{1}{((k-1)!)^{2}} \prod_{p=2}^{k} |(2\beta-1)(p-2)+M|^{2} \right\} \\ & = \frac{1}{((n-2)!)^{2}} \prod_{k=2}^{n} |(2\beta-1)(k-2)+M|^{2} , \end{split}$$ where \$M=2\beta(1-a)e^{-\ell\lambda}\$ cos \$\lambda\$ . Thus we get $$|a_n| \leq \frac{1}{(n-1)!} \prod_{k=2}^n |(2\beta-1)(k-2)+M|$$ , which completes the proof of (5.1). Next, we suppose \$n>N+2\$. Then (5.6) gives $$\begin{split} (n-1)^2|a_n|^2 & \leq 4\beta^2(1-\alpha)^2 \cos^2 \lambda \\ &+ \sum_{k=2}^{N+2} \{|(2\beta-1)(k-1)+M|^2-(k-1)^2\}|a_k|^2 \\ &+ \sum_{k=N+3}^{n-1} \{|(2\beta-1)(k-1)+M|^2-(k-1)^2\}|a_k|^2 \\ & \leq 4\beta^2(1-\alpha)^2 \cos^2 \lambda + \sum_{k=2}^{N+2} \{|(2\beta-1)(k-1)+M|^2-(k-1)^2\}|a_k|^2 \ . \end{split}$$ On substituting upper estimates for \$a\_{2},\$ \$a\_{s},\$ \$\cdots,\$ \$a\_{N+2}\$ obtained above, and simplifying, we obtain (5.2). (b) Let $$\beta(1-\alpha)(2-\alpha)\cos^2\lambda \le (1-\beta)(1+(1-\alpha)\cos^2\lambda)$$ , then it follows from (5.6) $$(n-1)^2|a_n|^2 \le 4\beta^2(1-\alpha)^2\cos^2\lambda \ , \quad (n\ge 2)$$ which proves (5.3). The bound in (5.1) is sharp for the function given by (5.7) $$f(z) = z/(1-(2\beta-1)z)^{2\beta(1-\alpha)e^{-i\lambda\cos\lambda\cdot(2\beta-1)^{-1}}}$$ The bounds in (5.3) are sharp for the functions given by $$f_n(z) = z/(1-(2\beta-1)z^{n-1})^{2\beta(1-\alpha)e^{-i\lambda\cos\lambda\cdot(2\beta-1)^{-1}(n-1)^{-1}}$$ for \$\beta\neq 1/2\$ ; whereas for \$\beta=1/2\$ , $$f_n(z) = z \exp \{((1-\alpha)e^{-i\lambda}\cos\lambda \cdot (n-1)^{-1})z^{n-1}\}, \quad (n \ge 2).$$ Remarks. (i) Taking \$\beta=1\$ , we obtain a theorem of Libera [\[6\].](#page-10-0) - (ii) By choosing $\alpha=0$ , $\beta=(2-\cos\lambda)/2$ in Theorem 4, we get the result of Goel [2]. - (iii) $\lambda=0$ leads to the result obtained by Juneja and Mogra [3]. - (iv) Replacing $\alpha$ by $(1-\beta+2\alpha\beta)/(1+\beta)$ and $\beta$ by $(1+\beta)/2$ in Theorem 4, we obtain the corresponding results of Maköwka [7]. - (v) The coefficient estimates determined by Kulshrestha [5], Zamorski [12], Wright [11], and many others can be obtained from Theorem 4 by taking different values of $\alpha$ , $\beta$ , and $\lambda$ .

Theorem 5. $\gamma$ -s.r. $S^{\lambda}(\alpha, \beta)$ is the smallest positive root r of the equation (6.3) $$(2\beta-1)(2\beta(1-\alpha)\cos(\gamma-\lambda)\cdot\cos\lambda-(2\beta-1)\cos\gamma)r^2-2\beta(1-\alpha)\cos\lambda\cdot r+\cos\gamma=0$$ The result is sharp for the extremal function given in (5.7). Proof. Let $f \in S^{\lambda}(\alpha, \beta)$ . Then by Lemma, (6.4) $$\frac{zf'(z)}{f(z)} = \frac{1 + ((2\beta - 1) - 2\beta(1 - \alpha)e^{-i\lambda}\cos\lambda)w(z)}{1 + (2\beta - 1)w(z)} ,$$ where w(z) satisfies the conditions w(0)=0 and |w(z)|<1. If $B(z)=e^{i\tau}zf'(z)/f(z)$ , then (6.4) may be written $$w(z) = \frac{e^{i\tau} - B(z)}{(2\beta - 1)B(z) - e^{i\tau}((2\beta - 1) - 2\beta(1 - \alpha)e^{-i\lambda}\cos\lambda)}, \quad (z \in E).$$ Now by applying Schwarz's Lemma, it follows that B(z) maps the disk $|z| \le r$ onto a disk $|B(z) - \eta| < R$ , where $$\eta = \frac{e^{i7}(1 - (2\beta - 1)(2\beta - 1 - 2\beta(1 - \alpha)e^{-i\lambda}\cos\lambda)r^2)}{1 - (2\beta - 1)^2r^2},$$ and $$R = \frac{2\beta(1-\alpha)\cos\lambda \cdot r}{1-(2\beta-1)^2r^2}.$$ Hence Re $(e^{i\tau}zf'(z)/f(z)) \ge 0$ if and only if $$\operatorname{Re}\left\{\frac{e^{i\tau}(1-(2\beta-1)(2\beta-1-2\beta(1-\alpha)e^{-i\lambda}\cos\lambda)r^{2})}{1-(2\beta-1)^{2}r^{2}}\right\} \geq \frac{2\beta(1-\alpha)\cos\lambda \cdot r}{1-(2\beta-1)^{2}r^{2}},$$ which, on simplification, and with the aid of (6.2) concludes the proof of the theorem. By choosing appropriate values of $\alpha$ , $\beta$ , $\lambda$ and $\gamma$ in Theorem 5 we obtain the corresponding results for several subclasses of S.

Corollary 6. $\gamma$ —s.r. $F_{\lambda,M}$ is the smallest positive root r of the equation $(M-1)((2M-1)\cos(\gamma-\lambda)\cdot\cos\lambda-(M-1)\cos\gamma)r^2-(2M-1)M\cos\lambda\cdot r+M^2\cos\gamma=0 \ ,$ for M>1/2. Taking $\alpha=0$ and $\beta=(2-\cos\lambda)/2$ in Theorem 5, we get the following result.

Corollary 7. $\gamma$ -s.r. $H(\lambda)$ is the smallest positive root of the equation $(1-\cos\lambda)((2-\cos\lambda)\cos(\gamma-\lambda)\cdot\cos\lambda-(1-\cos\lambda)\cos\gamma)r^2-(2-\cos\lambda)\cos\lambda\cdot r+\cos\gamma=0$ . When $\gamma=0$ in Corollary 7, we get the radius of starlikeness of $H(\lambda)$ as obtained by Goel [2]. Remarks. Different values of the parameters $\alpha$ , $\beta$ , $\lambda$ , and $\gamma$ lead to $\gamma$ -spiral radius and radius of starlikeness of the classes studied by Robertson [9], Libera [6], Maköwka [7], and others. The authors are thankful to Dr. O. P. Juneja for some helpful suggestions.