数理解析研究所講究録
第 1824 巻 2013 年 91-95

91

Strongly starlikeness criteria
for certain analytic functions
Yutaka Shimoda, and Shigeyoshi Owa
AbstractLetbe the subcless of analytic functions $f(z)$ in the open unit disk which
was introduced by S. Ponnusamy (Appl. Math. Lett. 24(2011), 381-385). For, some condition for the domain of such that $f(z)$ is strongly starlikeof order in$\mathcal{U}_{3}(\lambda)$$\mathbb{U}$$f(z)\in \mathcal{U}_{3}(\lambda)$$|z|$$\mathbb{U}.$$\gamma$1

Introduction
Let$\mathcal{A}$denote the class of functions $f(z)$ of the form

(1.1)

$f(z)=z+ \sum_{n=2}^{\infty}a_{n}z^{n}$that are analytic in the open unit disk $\mathbb{U}=\{z\in \mathbb{C} : |z|<1\}$ , and let be the subclass of$A$ consisting of $f(z)$which are univalent inObradovi\v{c} and Ponnusamy [2] define the classof $f(z)\in A$ satisfying the condition$\mathcal{S}$$\mathbb{U}.$$\mathcal{U}(\lambda)$(1.2)$|( \frac{z}{f(z)})^{2}f’(z)-1|<\lambda (z\in \mathbb{U})$for some real
The condition (1.2) is equivalent to$\lambda>0.$$|z^{2}( \frac{1}{f(z)}-\frac{1}{z})’|<\lambda (z\in \mathbb{U})$Ponnusamy [3] introduces the class
For some real$\gamma\in(0_{\}}1], a$(1.3)

We deote by$\mathcal{U}_{3}(\lambda)$

of function$f(z)\in \mathcal{U}(\lambda)$

for which $a_{3}-a_{2}^{2}=0.$function $f(z)\in A$ is called strongly starlike of order$| \arg\frac{zf’(z)}{f(z)}|<\frac{\pi\gamma}{2} (z\in \mathbb{U})$$SS(\gamma)$.

.

the set of all strongly starlike functions of order$\gamma$in$\mathbb{U}.$2010 Mathematics Subject Classification: Primary $30C45.$

Key $Wo7ds$ and Phrases : Analytic function, Strongly starlike of order

Schwarz lemma.

$\gamma,$$\gamma$if

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Ponnusamy [3] has shown the following theorem.

Theorem 1.1 Let$f(z)\in \mathcal{U}_{3}(\lambda),$$\gamma\in(0,1]$, and$\lambda.(\gamma, |a_{2}|)=\frac{-2(1+2\infty s_{2}^{g})|a_{2}|+2\sin_{2}^{\mathfrak{Q}}\sqrt{5+4\cos^{\frac{n}{2}}-4|a_{2}|^{4}}}{5+4coe_{2}^{\mathfrak{Q}}}.$

Then$f(z)\in S\mathcal{S}(\gamma)$

Theclass$\dot{u}m$for$0<\lambda\leq\lambda_{*}(\gamma, |a_{2}|)$of this paper is to derive a$\mathcal{S}\mathcal{S}(\gamma)$$\infty$.

ndition for the domain of$f(z)\in \mathcal{U}_{3}(\lambda)$

to be in the.

Main Result2Suppose that$f\in \mathcal{U}_{3}(\lambda)$(2.1). Then a simple calculation shows that$-z( \frac{z}{f(z)})’+(\frac{z}{f(z)})=(\frac{z}{f(z)})^{2}f’(z)$$=1+A_{3}z^{3}+\cdots=1+\lambda u,(z), w(z)\in \mathcal{B}_{3},$such that $w(O)=w(O)’=$
denotes the set of all analytic functions $w(z)$ in. From (2.1), we easily have the following$w(0)”=0,w”’(0)\neq 0$ and $|w(z)|<1$ for:
representation for

where

$U$

$\mathcal{B}_{3}$$z\in \mathbb{U}$$\frac{z}{f(z)}$(2.2)$\frac{z}{f(z)}-1=-a_{2}z-\lambda\int_{0}^{1}\frac{w(tz)}{t^{2}}dt.$

Since$w(z)\in \mathcal{B}_{3}$, from the Schwarz lemma(2.3)$|w(z)|\leq|z|^{3}$holds true. Thus, we have that
(2.4)$| \frac{z}{f(z)}-1|\leq|z|(|a_{2}|+\frac{\lambda}{2}|z|^{2}), z\in \mathbb{U}.$We have the following result.
Theorem 2. 1

If, thenof the equation$f(z)\in \mathcal{U}_{3}(\lambda)$root(2.5)$R_{0}$$f(z)\in \mathcal{S}\mathcal{S}(\gamma)$

for $|z|< \min\{1, r_{o}\}$ , where$r_{0}=\sqrt{R_{0}}$for the positive$\lambda^{2}(\frac{5}{4}+coe\frac{\pi\gamma}{2})X^{3}+\lambda|a_{2}|(1+2\cos\frac{\pi\gamma}{2})X^{2}+|a_{2}|^{2}X+\cos^{2}\frac{\pi\gamma}{2}-1=0.$93
ProofSuppose that $f(z)\in u_{3}(\lambda)$ . Then we can see from (2.1) and (2.3) that(2.6)$|( \frac{z}{f(z)})^{2}f’(z)-1|=\lambda|w(z)|\leq\lambda|z|^{3}.$Therefore, it follows from (2.4) and (2.6) that
(2.7)$| \arg\frac{zf’(z)}{f(z)}|\leq|\arg((\frac{z}{f(z)})^{2}f’(z))|+|\arg\frac{z}{f(z)}|$$\leq$

arcsin$(\lambda|z|^{3})+$

Now, we have to find the range of(2.8)

arcsin$|z|$

forarcsin$(|z|(|a_{2}|+ \frac{\lambda}{2}|z|^{2}))$$f(z)\in S\mathcal{S}(\gamma)$

such that$( \lambda|z|^{8\sqrt{1-|z|^{2}(|a_{2}|+\frac{\lambda}{2}|z|^{2})^{2}}+}\sqrt{1-\lambda^{2}|z|^{6}}|z|(|a_{2}|+\frac{\lambda}{2}|z|^{2}))<\frac{\pi\gamma}{2},$which is equivalent to(2.9)$\lambda|z|^{3}\sqrt{1-|z|^{2}(|a_{2}|+\frac{\lambda}{2}|z|^{2})^{2}}+\sqrt{1-\lambda^{2}|z|^{6}}|z|(|a_{2}|+\frac{\lambda}{2}|z|^{2})<\sin\frac{\pi\gamma}{2}.$Putting(2. 10)$F(X)= \lambda^{2}(\frac{5}{4}+\cos\frac{\pi\gamma}{2})X^{3}+\lambda|a_{2}|(1+2\cos\frac{\pi\gamma}{2})X^{2}+|a_{2}|^{2}X-\sin^{2}\frac{\pi\gamma}{2}$and(2. 11)$G(X)= \lambda^{2}(\frac{5}{4}-\cos\frac{\pi\gamma}{2})X^{3}+\lambda|a_{2}|(1-2\cos\frac{\pi\gamma}{2})X^{2}+|a_{2}|^{2}X-\sin^{2}\frac{\pi\gamma}{2},$(2.9) can be written as $F(X)G(X)>0$ with $X=|z|^{2}$ . Since $F(O)<0$ and $G(O)<0,$$F(X)G(X)>0$ is equivalent to $F(X)<0$ and $G(X)<0$ . Comparing the coefficients of$F(X)$ and $G(X)$ , we easily find the inequality $G(X)<F(X)$ .
In order to find the condition ofcondition for $F(X)<0.$
Since

$|z|$

such that$f(z)\in \mathcal{U}_{3}(\lambda)$

to be in$\mathcal{S}\mathcal{S}(\gamma)$, we consider the$F’(X)=3 \lambda^{2}(\frac{5}{4}+\cos\frac{\pi\gamma}{2})X^{2}+2\lambda|a_{2}|(1+2\cos\frac{\pi\gamma}{2})X+|a_{2}|^{2}>0$

and $\lim_{Xarrow\infty}F’(X)=\infty,$

94$F(X)$ is

an increaeing function for . Thus $F(X)$ has a positive root $R_{0}>0$ . Therefore, for$X$$|z|< \min\{1, R_{0}\}$, inequality (2.9) holds.

RemarkSubstituting $X=1$ and solving the equation (2.5) as the equation of , we haveof Theoreml. 1.
$\lambda$

3$\lambda_{2}(\gamma, |a_{2}|)$Example

We give an example which shows the existance of satisfying Theorem 2.1. Since $F(X)$for $F(1)>0.$
has a unique solution for $0<X<1$ if $F(1)>0$ , we consider a condition ofFrom$r_{0}$$|a_{2}|$$F(1)= \lambda^{2}(\frac{5}{4}+\cos\frac{\pi\gamma}{2})+\lambda|a_{2}|(1+2c\infty\frac{\pi\gamma}{2})+|a_{2}|^{2}-sin^{2_{\frac{\pi\gamma}{2}}}>0,$

we have$(|a_{2}|+ \frac{\lambda(1+2coe_{2}^{\mathfrak{Q}})}{2})^{2}>\sin^{2}\frac{\pi\gamma}{2}(1-\lambda^{2})$,This gives us that.

(3.1)

If$|a_{2}|$$|a_{2}|> \sin\frac{\pi\gamma}{2}\sqrt{1-\lambda^{2}}-\frac{\lambda(1+2\cos^{\underline{\pi}_{2}}1)}{2} (0<\lambda<1)$satisfies the condition (3.1), $F(X)$ has a poeitive root for $0<X<1.$

Let us take$\lambda=\frac{1}{2}$and$\gamma=\frac{2}{3}$. Then$|a_{2}|> \frac{1}{4}$from (3.1). Thus, we may take $|a_{2}|=1$ and$F(X)=7X^{3}+16X^{2}+16X-12.$(3.2)

Since $F(O)=-12<0$ , and $F(1)=27>0,$ $F(X)$ has a real positive root $0<X<1.$of (3.2) satisfies $0.47605<r_{0}<0.47615.$

Actually, the root

$r_{0}$References[1] P. L. Duren, Univalent Functions, Springer-Verlag, New York, Berlin, Heiderberg, Tokyo,

1983[2] M. Obradovi\v{c} and S. Ponnusamy, Radius properties for subclasses of univalent functions,
Analysis 25(2005), 183–188[3] S.Ponnusamy, Starlikeness criteria for a certain class of analytic functions, Appl. Math.
Lett. 24(2011) 381–385.

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Yutaka Shimoda and Shigeyoshi Owa
Department of MathematicsKinki University
Higashi-Osaka, Osaka 577-8502
Japan-mail: yutakashimoda119@gmail.com
shige21@ican.zaq.ne.jp$E$