8. Gamma-Starlike Functions

Theorem 1. $\mathscr{L}_{\gamma} \subset S^*$ , for all real $\gamma$ . Proof. If $f(z) \in \mathcal{L}_{\gamma}$ and we set (3) $$\frac{1+w(z)}{1-w(z)} \equiv \frac{zf'(z)}{f(z)}.$$ for $z \in D$ , then w(0) = 0, $w(z) \neq \pm 1$ and w(z) is a meromorphic function. We will show that |w(z)| < 1, and this will imply that $$\operatorname{Re}\left[zf'(z)/f(z)\right]>0$$ . Let $w(z)=R(z)e^{i\varphi(z)},\,R\geqslant 0$ for $z\in D,$ and suppose that $z_0$ is a point of D such that (4) $$\max_{|z| \leq |z_0|} |w(z)| = |w(z_0)| = 1.$$ Then $\frac{\partial}{\partial^{\theta}}R(z_0)=0$ , and since $$\frac{zw'(z)}{w(z)} = \frac{\partial \Phi(z)}{\partial \theta} - i\frac{1}{R}\frac{\partial R(z)}{\partial \theta},$$ we must have $z_0w'(z_0)/w(z_0)=\partial\Phi(z_0)/\partial\theta$ , and hence $z_0w'(z_0)/w(z_0)$ must be a real number. A simple geometric argument can show even more. If we assume $\partial\Phi(z_0)/\partial\theta<0$ then w(z) would be locally univalent at $z_0$ and this would lead to a contradiction of (4). Thus we see that $\partial\Phi(z_0)/\partial\theta$ must be non-negative and so we can set (5) $$\frac{z_0 w'(z_0)}{w(z_0)} = B,$$ where $B \geqslant 0$ . Since $|w(z_0)| = 1$ and $w(z_0) \neq \pm 1$ , we must have (6) $$\frac{1+w(z_0)}{1-w(z_0)} = Ai,$$ where A is real and $A \neq 0$ . From (1) and (3) we have $$\operatorname{Re} I\big(\gamma, f(z)\big) = \ = \operatorname{Re} \left[ \left(\frac{1+w(z)}{1-w(z)}\right)^{1-\gamma} \left(\frac{1+w(z)}{1-w(z)} + \frac{zw'(z)}{w(z)} \left(\frac{w(z)}{1+w(z)} + \frac{w(z)}{1-w(z)}\right)\right)^{\gamma}\right]$$ where $$I\big(\gamma,f(z)\big) \equiv \bigg(\frac{zf'(z)}{f(z)}\bigg)^{1-\gamma}\bigg(1+z\frac{f''(z)}{f'(z)}\bigg)^{\gamma},$$ and thus at $z = z_0$ , by using (5) and (6) we obtain $$\operatorname{Re}I\big(\gamma,f(z_0)\big)=\operatorname{Re}\big[(Ai)^{1-\gamma}\big(Ai+\frac{B}{a}\big(A+\frac{1}{A}\big)i\big]^{\gamma}.$$ If we let C=A+B(A+1/A)/2, then since $B\geqslant 0$ and $A\neq 0$ we have AC>0 and obtain $$\operatorname{Re}I(\gamma,f(z_0))=\operatorname{Re}[(Ai)^{1-\gamma}(Ci)^{\gamma}]=\operatorname{Re}(|A|^{1-\gamma}|C|^{\gamma}i)=0.$$ This is a contradiction of (1) and so we must have |w(z)| < 1 for $z \in D$ and thus $f(z) \in S^*$ . Note that Theorem 1 shows that if $f \in \mathcal{L}_{\gamma}$ , then $f \in \mathcal{L}_{0} \equiv S^{*}$ . We can show more than this.

Theorem 2. If $$0 \leqslant \delta \leqslant \gamma$$ (or $\gamma \leqslant \delta \leqslant 0$ ) then $\mathscr{L}_{\gamma} \subset \mathscr{L}_{\sigma}$ . Proof. The case $\delta=0$ has been handled in Theorem 1, so we only need to consider the case $0<\delta/\gamma<1$ . If $f \in \mathscr{L}_{\gamma}$ , then there is a function $P_1(z) \in \mathscr{P} \equiv \{P(z) | P(0) = 1, P(z) \text{ is regular in } D \text{ and } \operatorname{Re} P(z) > 0\}$ satisfying $$\left(\frac{(zf'(z)}{f(z)}\right)^{1-\gamma}\left(1+\frac{zf''(z)}{f'(z)}\right)^{\gamma}\equiv P_{\mathbf{1}}(z)\,.$$ By Theorem 1 we have $f(z) \in S^*$ and hence there exists $P_2(z) \in \mathcal{P}$ such that $$\frac{zf'(z)}{f(z)} \equiv P_2(z).$$ If we raise sides of (7) to the $\delta/\gamma$ power we obtain $$\left(\frac{(zf'(z)}{f(z)}\right)^{\delta/\gamma-\delta}\left(1+\frac{zf''(z)}{f'(z)}\right)^{\delta}\equiv P_1(z)^{\delta/\gamma},$$ and if we raise both sides of (8) to the $(1-\delta/\gamma)$ power we obtain (10) $$\left(\frac{zf'(z)}{f(z)}\right)^{1-\delta/\gamma} = P_2(z)^{1-\delta/\gamma}.$$ Multiplying equation (9) by equation (10) we obtain $$(11) \qquad \left(\frac{zf'(z)}{f(z)}\right)^{1-\delta} \left(1+\frac{zf''(z)}{f'(z)}\right)^{\delta} \, \equiv P_{1}(z)^{\delta/\gamma} P_{2}(z)^{1-\delta/\gamma} \, \equiv P_{3}(z) \, .$$ Since $P_1(z) \in \mathscr{P}$ and $P_2(z) \in \mathscr{P},$ we have $P_3(0) = 1$ and $|{\rm arg} P_3(z)| \leqslant \frac{\delta}{\gamma}$ $|{\rm arg}\, P_1(z)| + \left(1-\frac{\delta}{\gamma}\right)|{\rm arg}\, P_2(z)| < \pi/2\,, \quad {\rm i.e.} \quad {\rm Re} \quad P_3(z) > 0 \quad {\rm and} \quad P_3(z) \in \mathscr{P},$ Consequently from (10) we have $f(z) \in \mathscr{L}_{\nu}$ . Note that the last theorem shows that if $f(z) \in \mathcal{L}$ and $\gamma \geqslant 1$ , then f(z) is a convex function. 3. It is possible to obtain bounds on the coefficients of gamma-starlike functions by using certain "standard" methods.

Theorem 3. If $f \in \mathscr{L}_{\gamma}$ , $f(z) = z + a_2 z^2 + n$ , and if $\mu$ is a complex constant, then $$|a_2(1+\gamma)|\leqslant 2\,,$$ $$|a_3(4+8\gamma)+a_2^2(\gamma^2-\neg\gamma-2)|\leqslant 4\,,$$ $$(14) \quad (1+\gamma)^2 |1+2\gamma| \, |a_3-\mu a_2| \leqslant \operatorname{Max} \left[ (1+\gamma)^2, \, |4\mu(1+2\gamma)-3(3\gamma+1)| \right]$$ all hold. Remarks. (i) $\gamma = 0$ (14) reduces to $$|a_3 - \mu a_2^2| \leq \text{Max}[1, |4\mu - 3|],$$ which is a result of Keogh and Merkes [2]. (ii) For $\gamma = 1$ (14) reduces to $$|a_3-\mu a_2^2|\leqslant \text{Max.}\left[\frac{1}{3},\,|\mu-1|\right].$$ (iii) If $f(z) \in \mathcal{L}_{\gamma}$ , $\gamma \geqslant 0$ , then $$|a_2| \leqslant 2/(1+\gamma)$$ and $$|a_3| \leqslant \begin{cases} \frac{3(3\gamma+1)}{(1+2\gamma)(1+\gamma)^2}, \ 0 \leqslant \gamma \leqslant \frac{7+\sqrt{57}}{2}, \\ \frac{1}{1+2\gamma}, \frac{7+\sqrt{57}}{2} < \gamma. \end{cases}$$ hold. Inequalities (12) - (16) may not be sharp. They would be sharp if it were possible to prove that the differential equation (17) $$\left(\frac{(zf'(z))}{f(z)}\right)^{1-\gamma} \left(1 + z\frac{f''(z)}{f'(z)}\right)^{\gamma} = \frac{1+z}{1-z},$$ with initial condition f(0) = 0, f'(0) = 1, has a solution that is a regular function in the unit disc. The authors have not been able to prove this for arbitrary $\gamma$ , but suspect that a solution exists for $\gamma \geqslant 0$ . The class $\mathcal{M}_a$ satisfies Theorems 1, 2 and inequality (13) and it is also true that $\mathcal{M}_0 \equiv \mathcal{Z}_0 \equiv S^*$ and $\mathcal{M}_1 \equiv \mathcal{L}_1 \equiv C$ . However, in general $\mathcal{M}_p \not\equiv \mathcal{L}_\beta$ ; This can be seen by considering $\beta = 1/2$ . By using infinite series, it is possible to show that (17) does have a regular solution for $\gamma = \beta = 1/2$ , and for this solution $a_3 = 5/3$ . For functions in $\mathcal{M}_{1/2}$ we must have $|a_3| \leq 29/18$ [3], and thus $\mathcal{M}_{1/2} \not\equiv \mathcal{L}_{1/2}$ . We conclude by indicating a refinement in the class of gamma-starlike functions.

Definition 1. Let $f(z)=z+\sum\limits_{a}^{\infty}a_{n}z^{n}$ be regular in the unit disc D, with $f(z),\ f'(z)$ and $[1+zf''(z)/f'(z)]\neq 0$ in 0<|z|<1. Suppose $\gamma$ is real and (1) $$\operatorname{Re}\left[\left(\frac{zf'(z)}{f(z)}\right)^{1-\gamma}\left(1+\frac{zf''(z)}{f'(z)}\right)^{\gamma}\right]>0,$$ for $z \in D$ , where the powers appearing in (1) are meant as principal values. ^&lt;sup>1</sup> This work was carried out while the second author was an IREX Scholar in Poland. Then we say that f(z) is a gamma-starlike function and we denote the class of such functions by $\mathscr{L}^r$ . Remarks. (i) Condition (1) is equivalent to the following condition: (2) $$\left| (1-\gamma)\arg\frac{zf'(z)}{f(z)} + \gamma \arg\left(1 + \frac{f''(z)}{f'(z)}\right) \right| < \frac{\pi}{2}$$ - (ii) If $\gamma = 0$ , $\mathscr{L}_0 \equiv S^*$ , the class of starlike functions, while if $\gamma = 1$ , $\mathscr{L}_1 \equiv C$ , the class of convex functions. - 2. We now show that if f(z) is a gamma-starlike function then f(z) is starlike and univalent.

Definition 2. Let $f(z)=z+\sum_{n=0}^{\infty}a_nz^n$ be regular in the unit disc D with f(z), f'(z), $1+zf''(z)/f'(z)\neq 0$ in 0<|z|<1, and suppose $\gamma$ is a real constant, $0\leqslant \gamma<1$ . If (18) $$\operatorname{Re}\left[\left(\frac{zf'(z)}{f(z)}\right)^{1-\gamma}\left(\left(1+z\frac{f''(z)}{f'(z)}\right)^{\gamma}\right]>\alpha.$$ for $z \in D$ , then we say that f(z) is a gamma-starlike function of order a, and we denote the class of such functions by $\mathcal{L}_{\nu}(a)$ . If (18) is replaced by $$\left| (1-\gamma) \arg \left( \frac{z f(z)}{f(z)} \right) + \gamma \arg \left( 1 + \frac{z f^{\prime\prime}(z)}{f^\prime(z)} \right) \right| < \frac{\pi}{2} a$$ for $z \in D$ , then we say that f(z) is a strongly gamma-starlike function of order a, and we denote the class of such functions by $\mathcal{L}_{\nu}^{*}(a)$ . Note that $\mathcal{L}^(a)$ and $\mathcal{L}_1^(a)$ are respectively the classes of strongly-starlike and strongly—convex functions of order a introduced by Brannan and Kirwan [1].