数理解析研究所講究録 1341 巻 2003 年 77-84

77

ARGUMENT ESTIMATES FOR CERTAIN ANALYTIC FUNCTIONSMAMORU NUNOKAWA, SHIGEYOSHI OWA , HITOSHI SAITOHAND NICOLAE N. PASCU

ABSTRACT. Let $p(z)$ be analytic in the open unit disk with $p(0)=1$ and $p’(0)=$. S.S.Miller and P.T.Mocanu (J. Math. Anal. Appl. 276(2002)) have shown someinteresting subordination theorems for such functions $p(z)$ . The object of the present
paper is to discuss some sufficient conditions for arguments of $p(z)$ to be.
for$\mathrm{U}$$0$$| \arg p(z)|<\frac{\pi}{2}\rho$$z\in \mathrm{U}$1. INTRODUCTION
Let $p(z)$ be analytic in the open unit disk $\mathrm{u}=\{z\in \mathbb{C}:|z|<1\}$ with $p(0)=1$ and$p’(0)=0$ . For such functions $p(z)$ , Miller and Mocanu [3] have showm some interesting

subordination theorems.

ForTheorem A. ([3])$\frac{1}{2}<\rho\leqq 1$

define the function $q(z)$ by$q(z)=q_{\rho}(z)=( \frac{1+z}{1-z})^{\rho}$フ

and let $t_{0}\in(0,1)$ be the unique solution of$t^{\rho} \{(1-\rho)t^{2}\cos(\frac{\pi}{2}\rho)+t\sin(\frac{\pi}{2}\rho)-(1-\rho)\cos(\frac{\pi}{2}\rho)\}+t^{2}-1=0$.

If $p(z)$ is analytic in , with $p(0)=1,p’(0)=0$ and$\mathrm{u}$$| \arg(zp’(z)+p(z)^{2}+p(z))|<\frac{\pi}{2}(\rho+1)-\mathrm{T}\mathrm{a}\mathrm{n}$ $-1( \frac{t_{0}}{1+\rho-(1-\rho)t_{0}^{2}})$,rneans the subordinations.
then $p(z)\prec q_{\rho}(z),$ where the simbol
To discuss our problems for functions $p(z)$ , we need the following lemma due to Hallenbeck and Ruscheweyh [2] which is the same as one by Fukui and Sakaguchi [1].
$nn\prec$.
2000 Mathematics Subject Classification. Primary
Key Words and Phrases. analytic function, argument estimate, subordination.
$30\mathrm{C}45$

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M.Nunokawa,S.Owa,H.Saitoh,N.N.Pascu

Let $p(z)$ be analytic in $|z|<R$ and $p^{(k)}(0)=0(0\leqq k\leqq n)$ . Then if$|p(z)|$ attains its maximum value on the circle $|z|=r<R$ at a point, we have

Lemma 1.1.

$z_{0}$(1.1)$\frac{z_{0}p’(z_{0})}{p(z_{0})}\geqq n+1$.

Applying the above lemma, we derive
Lemma 1.2.
Let $p(z)$ be analytic in $\mathrm{U},p(0)=1,p’(0)=0$ , and let $p(z)\neq 0(z\in \mathrm{U})$ .
such that
If there enists a point$z_{0}\in \mathrm{u}$$| \arg p(z)|<\frac{\pi}{2}\alpha$$(|z|<|z_{0}|)$and$| \arg p(z_{0})|=\frac{\pi}{2}\alpha$

for some$\alpha>0$, then we have

(1.2)$\frac{z_{0}p’(z_{0})}{p(z_{0})}=i\alpha k$

,

where
$k \geqq(a+\frac{1}{a})\geqq 2$

when$\arg p(z_{0})=\frac{\pi}{2}\alpha$and$k \leqq-(a+\frac{1}{a})\leqq-2$

when$\arg p(z_{0})=-\frac{\pi}{2}\alpha$,whem $p(z_{0})^{1/\alpha}=\pm ia$ and $a>0$ .
We use the same manner which was used by Nunokawa [4] for the proof of thelemma. Let us putPrvof.

(1.3)$q(z)=p(z)^{1/\alpha}$

Then we see that ${\rm Re} q(z)>0(|z|<|z_{0}|),{\rm Re} q(z_{0})=0,$ $q(0)=1$ and $q’(0)=0$ . Definingbythe function$\phi(z)$(1.4)$\phi(z)=\frac{1-q(z)}{1+q(z)}$,we have that $\phi(0)=0,$ $|\phi(z)|<1(|z|<|z_{0}|)$ , and $|\phi(z_{0})|=1$ .
In view of Lemma 1.1, we know that
(1.5)$\frac{z_{0}\phi’(z_{0})}{\phi(z_{0})}=\frac{-2z_{0}q’(z_{0})}{1-q(z_{0})^{2}}$79
Argument estimates for certain analytic functions.

$= \frac{-2z_{0}q’(z_{0})}{1+|q(z_{0})|^{2}}\geqq 2$

It follows from (1.5) that(1.6)$-z_{0}q’(z_{0})\geqq(1+|q(z_{0})|^{2})$is anon-vanishing pure imaginary
is anegative real number. Since
andnumber, we can put $q(z_{0})=ia$ , where is anon-vanishing real number.
We have, for $a>0$ ,$q(z_{0})$$z_{0}q’(z_{0})$

$a$

(1.7)

${\rm Im}( \frac{z_{0}q’(z_{0})}{q(z_{0})})={\rm Im}(-\frac{iz_{0}q’(z_{0})}{|q(z_{0})|})\geqq(\frac{1+a^{2}}{a})\geqq 2$

and, for $a<0$ ,(1.8)${\rm Im}( \frac{z_{0}q’(z_{0})}{q(z_{0})})={\rm Im}(\frac{iz_{0}q’(z_{0})}{|q(z_{0})|})\leqq-(\frac{1+a^{2}}{a})\leqq-2$

On the other hand, it follows that
(1.9)$\frac{z_{0}q’(z_{0})}{q(z_{0})}=\frac{1}{\alpha}(\frac{z_{0}p’(z_{0})}{p(z_{0})})$

This completes the proof of Lemma 1.2.

$\square$

2. ARGUMENT ESTIMATES
Our first property for argument estimates of analytic function $p(z)$ is contained inetyith $p(0)=1$ ated $p’(0)=0$ . If $p(z)$ satisfies

Theorem 2.1.

Let $p(z)$ be analytic in(2.1)$|\arg(zp’(z)+p(z)^{2}+\alpha p(z))|<\pi\rho$

for some$\alpha(\alpha>0),$ $\rho(0<\rho\leqq\rho_{0})$$\mathrm{u}$$(z\in \mathrm{U})$, where $\rho_{0}(0<\rho_{0}<1)$ is given by$\tan(\frac{\pi}{2}\rho 0)=\frac{2}{\alpha}\rho_{0}$,then(2.2)

Proof.

point$| \arg p(z)|<\frac{\pi}{2}\rho$$(z\in \mathrm{u})$.

Let afunction $p(z)$ satisB the conditions of the theorem. If there exists asuch that$z_{0}\in \mathrm{u}$$| \arg p(z)|<\frac{\pi}{2}\rho$$(|z|<|z_{0}|)$and$| \arg p(z_{0})|=\frac{\pi}{2}\rho$then applying Lemma 1.2, we have that,

80M. Nunokawa,S.Owa,H.Saitoh, N.N.Pascu(2.3)$\frac{z_{0}p’(z_{0})}{p(z_{0})}=i\rho k$

,

where
$k \geqq a+\frac{1}{a}\geqq 2$

when$\arg p(z_{0})=\frac{\pi}{2}\rho$

and$k \leqq-(a+\frac{1}{a})\leqq-2$

whenwith $p(z_{0})^{1/\rho}=Atia$ $(a>0)$ . It follows that, for(2.4)$\arg p(z_{0})=-\frac{\pi}{2}\rho$$\arg p(z_{0})=\frac{\pi}{2}\rho$

and$k \geqq a+\frac{1}{a}\geqq 2$,$\arg(z_{0}p’(z_{0})+p(z_{0})^{2}+\alpha p(z_{0}))=\arg p(z_{0})(\frac{z_{0}p’(z_{0})}{p(z_{0})}+p(z_{0})+\alpha)$$= \frac{\pi}{2}\rho+\arg(i\rho k+a^{\rho}e^{1\frac{}{l}p}..+\alpha)=\frac{\pi}{2}\rho+\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{\rho k+a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho)})$Since, by $0<\rho\leqq\rho_{0}<1$ and $k\geqq 2$ ,(2.5)$\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{\rho k+a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho)})\geqq \mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{2\rho+a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{n}{2}\rho)})>0$

we define $g(a)$ by(2.6)$g(a)= \frac{2\rho+a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho)}$$(a>0)$.

Noting that(2.7)

we define(2.8)$g’(a)= \frac{\alpha\rho a^{\rho-1}\cos(\frac{\pi}{2}\rho)(\tan(\frac{\pi}{2}\rho)-\frac{2\rho}{\alpha})}{(\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho))^{2}}$$h(\rho)$,by$h( \rho)=\tan(\frac{\pi}{2}\rho)-\frac{2\rho}{\alpha}$$(0<\rho\leqq\rho_{0}<1)$.

Then $h(0)=0,h(\rho_{0})=0,$ and(2.9)$h”( \rho)=\frac{\pi^{2}}{2}\sec^{2}(\frac{\pi}{2}\rho)\tan(\frac{\pi}{2}\rho)>0$.

This shows that $g’(a)\leqq 0$ for $a>0$ , that is, that
(2.10)$\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{\rho k+a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho)})\geqq \mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\tan(\frac{\pi}{2}\rho))=\frac{\pi}{2}\rho$.

,

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Therefore, we conclude that(2.11)

when$\arg(z_{0}p’(z_{0})+p(z_{0})^{2}+\alpha p(z_{0}))\geqq\pi\rho$$\arg p(z_{0})=\frac{\pi}{2}\rho$

Similarly, for(2.12).

$\arg p(z_{0})=-\frac{\pi}{2}\rho$

and $k \leqq-(a+\frac{1}{a})\leqq-2$ , we have that$\arg(z_{0}p’(z_{0})+p(z_{0})^{2}+\alpha p(z_{0}))=-\frac{\pi}{2}\rho+\arg(i\rho k+a^{\rho}e^{-:\frac{*}{2}\beta}+\alpha)$$=- \frac{\pi}{2}\rho+\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{\rho k-a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho)})$$\leqq-\frac{\pi}{2}\rho+\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{-2\rho-a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho)})$$=- \frac{\pi}{2}\rho-\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{2\rho+a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho)})$$\leqq-\frac{\pi}{2}\rho-\frac{\pi}{2}\rho=-\pi\rho$Thus, for such apoint(2.13)$z_{0}\in \mathrm{u}$.

. we see that$|\arg(z_{0}p’(z_{0})+p(z_{0})^{2}+\alpha p(z_{0}))|\geqq\pi\rho$,which contradicts our condition for $p(z)$ .
Consequently, we conclude that$| \arg p(z)|<\frac{\pi}{2}\rho$$(z\in \mathrm{U})$.
$\square$Let us consider the function $p(z)$ defined by

Example 2.1.

$p(z)=1+ \frac{1}{5}z^{2}$.

Then we see that$zp’(z)+p(z)^{2}+ \frac{1}{2}p(z)=\frac{3}{2}+\frac{9}{10}z^{2}+\frac{1}{25}z^{4}$

Letting$\alpha=\frac{1}{2}$.

and$\rho=\frac{1}{\pi}\mathrm{S}\mathrm{i}\mathrm{n}^{-1}(\frac{19}{30})$

in Theorem 2.1, we have that$| \arg(zp’(z)+p(z)^{2}+\frac{1}{2}p(z))|<\pi\rho=\mathrm{S}\mathrm{i}\mathrm{n}^{-1}(\frac{19}{30})$

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and$| \arg p(z)|<\mathrm{S}\mathrm{i}\mathrm{n}^{-1}(\frac{1}{5})<\frac{\pi}{2}\rho$

If we take$\alpha=1$

in Theorem 2.1, thenurith $p(0)=1$ and $p’(0)=0$ . I $p(z)$Corollary 2.1.

Let $p(z)$ be analytic in(2.14)$|\arg(zp’(z)+p(z)^{2}+p(z))|<\pi\rho$satisfies

for some.

$\rho(0<\rho\leqq\frac{1}{2})$$| \arg p(z)|<\frac{\pi}{2}\rho$

(1)

$f$

$(z\in \mathrm{u})$, then(2.15)

Remark 2.1.

$\mathrm{u}$I$\alpha=\frac{4}{5}$, then.

$(z\in \mathrm{U})$$0<\rho\leqq\beta 0$

and $0.647873<\rho_{0}<0.647874$ .

(2)I, then$0<\rho\leqq\rho_{0}$and $0.809251<\rho_{0}<0.809252$ .

(3)I, then$0<\rho\leqq\rho_{0}$and $0.880966<\rho_{0}<0.880967$ .

(4)I, then$0<\rho\leqq\rho_{0}$and $0.913417<\rho_{0}<0.913418$ .

(5)

I $\alpha=1.1$ , then $0<\rho\leqq n$ and $0.401247<\rho_{0}<0.491248$ .

(6)

I $\alpha=1.2$ , then

$f$ $\alpha=\frac{1}{2}$$f$ $\alpha=\frac{1}{3}$$f$ $\alpha=\frac{1}{4}$

$f$

$f$

$0<\rho\leqq\rho_{0}$and $0.262943<\rho_{0}<0.262944$ .

I $\alpha=1.3$ , then there is no $\rho_{0}>0$ such that$0<\alpha<1.3$ in Theorem 2.1.
(7)

$f$

$\tan(\frac{\pi}{2}\rho 0)=\frac{2}{\alpha}\rho$.

$\mathfrak{M}us$

we see that

Next, we derive

Theorem 2.2.
(2.16)

for some

Let $p(z)$ be analytic in $\mathrm{U}w\cdot.thp(0)=1$ and $p’(0)=0$ . I$fp(z)$ satisfies$| \arg(zp’(z)+p(z)^{2}+\alpha p(z))|<\frac{\pi}{2}\rho+\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{2\rho}{\alpha})$$\alpha(\alpha>0),\rho(\rho_{0}\leqq\rho<1),$$(z\in \mathrm{U})$where $\rho_{0}(0<\rho_{0}<1)$ is given by$\tan(\frac{\pi}{2}\rho 0)=\frac{2}{\alpha}\rho_{0}$then(1.7)

Pmof.

$| \arg p(z)|<\frac{\pi}{2}\rho$$(z\in \mathrm{U})$.

Using the same technique as in the proof of Theorem 2.1, we know that$\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{2\rho+a^{\rho}\sin(\frac{\pi}{2}\rho)}{\alpha+a^{\rho}\cos(\frac{\pi}{2}\rho)})$,83
Argument estimates for certain analytic functions

is increasing for $a>0$ . Thus, we obtain(2.18)

for$| \arg(z_{0}p’(z_{0})+p(z_{0})^{2}+\alpha p(z_{0}))|\geqq\frac{\pi}{2}\rho+\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(\frac{2\rho}{\alpha})$$z_{0}\in \mathrm{u}$such that$(|z|<|z_{0}|)$$| \arg p(z)|<\frac{\pi}{2}\rho$

and$| \arg p(z_{0})|=\frac{\pi}{2}\rho$.

This contradicts our condition of the theorem. Therefore,$(z\in \mathrm{U})$$| \arg p(z)|<\frac{\pi}{2}\rho$.
$\square$

Letting $\alpha=1$ in Theorern 2.2, we obtain

Let $p(z)$ be analytic in

Corollary 2.2.

satisfies(2.19)$\mathrm{u}$with $p(0)=1$ and $p’(0)=0$ .

$| \arg(zp’(z)+p(z)^{2}+p(z))|<\frac{\pi}{2}\rho+\mathrm{T}\mathrm{a}\mathrm{n}^{-1}(2\rho)$

for some$\rho(\frac{1}{2}\leqq\rho<1)$

If $p(z)$$(z\in \mathrm{u})$, then(2.20)$| \arg p(z)|<\frac{\pi}{2}\rho$$(z\in \mathrm{U})$.

Finally, we note that

Remark 2.2.

(1)

If

$\alpha=\frac{4}{5}$$\alpha=\frac{1}{2}$, then$0<\rho\leqq\rho_{0}$$\alpha=\frac{1}{3}$, then$0<\rho\leqq\rho_{0}$$\alpha=\frac{1}{4}$, then$0<\rho_{rightarrow}\leq\rho_{0}$and $0.809251<\rho_{0}<0.809252$ .

(2)

If

(3)

If(4)
(6)

If
If
If(7)

If $\alpha=1.3$ , then there is no(5), then$\alpha=1.2$ , then$\alpha=1.1$

and $0.880966<\rho_{0}<0.880967$ .
and $0.913417<\rho_{0}<0.913418$ .

$0<\rho\leqq\rho_{0}$$0<\rho\leqq\rho 0$$0<\alpha<1.3$ in Theorem

2.2.

, then $0<\rho\leqq\rho 0$ and $0.647873<\rho_{0}<0.647874$ .

and $0.401247<\rho 0<0.491248$ .

and $0.262943<\rho_{0}<0.262944$ .
$\rho_{0}>0$such that$\tan(\frac{\pi}{2}\rho 0)=\frac{2}{\alpha}\rho$. Thus we see that

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REFERENCES[1] S. Fukui and K. Sakaguchi, An extension of a theorem of St. Ruscheeueyh, Bull. Fac. Edu. Wakayama
Univ. Nat. Sci., 29 (1980), 1–3.

[2] D. J. Hallenbeck and St. Ru8cheweyh, Subordinations by convez functions, Proc. Amer. Math. Soc.,
52 (1975), 191-195.
[3] S. S. Miller and P. T. Mocanu, Libera transform of finctions with bounded turning, J. Math. Anal.
APPI., 276 (2002), 90-97.
[4] M. Nunokawa, On the order of strongly starlikeneas of strvngly convex functions, Proc. Japan Acad.,
69 (1993), 2U-237.

Mamoru Nunokawa
Emeritus Professor
Department of Mathernatics
University of Gunma
Aramaki, Maebashi, Gunma 371-8510
JapanShigeyoshi $Owa$
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-850flJapan

Hitoshi Saitoh
Department of Mathernatics
National Gunma College of Technology
Toriba, Maebashi, Gunrna 371-8530
Japan
Nicolae N. Pascu

Department of Mathematics
University of Brcnsov$\pi u\mathrm{n}silvania$$R$-fl200Brasov