数理解析研究所講究録
第 1824 巻 2013 年 74-81

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Extension theorems concerned with resultsby Ponnusamy and Karunakaran
Mamoru Nunokawa, Kazuo Kuroki, Janusz Sok\’ol and Shigeyoshi Owa

Introduction

1Let$\mathcal{A}(n, k)$

be the class of functions $f(z)$ of the form

(1.1)

$f(z)=z^{n}+ \sum_{m=n+k}^{x}a_{m}z^{m} (n\geqq 1, k\geqq 1)$: $|z|<1\}$ . For two functions $f(z)$ andwhich are analytic in the open unit disk$g(z)$ belonging to the class, Sakaguchi [5] has proved the following result.
$\mathbb{U}=\{z\in \mathbb{C}$$\mathcal{A}(1,1)$

Theorem Asatisfy

Let $f(z)\in \mathcal{A}(1,1)$ and $g(z)\in \mathcal{A}(1,1)$ be starlike in(1.2)${\rm Re}( \frac{f’(z)}{g(z)})>0 (z\in U)$$\mathbb{U}$. If $f(z)$ and $g(z)$

,then
(1.3)

$Re(\frac{f(z)}{g(z)})>0 (z\in \mathbb{U})$.

After Theorem , many mathematicians studying this field have applied this theorem toget some results. In 1989, Ponnusamy and Karunakaran [4] have improved Theorem A as
$A$following.

Theorem $B$ Let be a complex number Utth
and $g(z)\in A(n,j)$ $(j\geqq 1)$ satisfies$\alpha$${\rm Re}\alpha>0$

and $\beta<1$ . FUrther, let $f(z)\in$$\mathcal{A}(n, k)$(1.4)${\rm Re}( \frac{\alpha g(z)}{zg’(z)})>\delta (z\in \mathbb{U})$2000 Mathematics Subject Classification: Primary $30C45.$
Keywords and Phrases: Analytic, starlike, Jack’s lemma.

75
with

$0 \leqq\delta<\frac{{\rm Re}\alpha}{n}$(1.5). If $f(z)$ and $g(z)$ satisfy${\rm Re} \{(1-a)\frac{f(z)}{g(z)}+\alpha\frac{f^{l}(z)}{g(z)}\}>\beta$$(z\in \mathbb{U})$,then(1.6)${\rm Re}( \frac{f(z)}{g(z)})>\frac{2\beta+\delta k}{2+\delta k} (z\in \mathbb{U})$.

It is the purpose of the present paper is to discuss Theorem applying the lemma due toFukui and Sakaguchi [1]. To discuss our problems, we need the following lemmas.
$B$Lemma 1Let

$w(z)= \sum_{n=k}^{ac}a_{n}z^{n}$$(a_{k}\neq 0, k\geqq 1)$

be analytic in$\mathbb{U}$. If the maximum valueof $|w(z)|$ on the circle $|z|=r<1$ is attained at $z=z_{0}$ , then we have
(1.7)$\frac{z_{0}w’(z_{0})}{w(z_{0})}=\ell\geqq k,$which shows that$\frac{z_{0}w’(z_{0})}{w(z_{0})}$is a positive real number.

The proof of Lemma 1 can be found in [1] and we see that Lemma 1 is a generalization ofJack’s lemma given by Jack [2], Applying Lemma 1, we derive

Lemma 2$(z\in \mathbb{U})$Let$p(z)=1+ \sum_{n=k}^{\infty}c_{\eta}z^{n}$. If there exists a point$z_{0}\in \mathbb{U}$$(c_{k}\neq 0, k\geqq 1)$

be analytic insuch that${\rm Re} p(z)>0 (|z|<|z_{0}|)$and${\rm Re} p(z_{0})=0,$

then we have
(1.8)$-z_{0}p’(z_{0}) \geqq\frac{p}{2}(1+|p(z_{0})|^{2})$and so(1.9)$\frac{z_{0}p^{l}(z_{0})}{p(z_{0})}=i\ell,$

where(1.10)$k \leqq\frac{k}{2}(a+\frac{1}{a})\leqq\ell (\arg p(z_{0})=\frac{\pi}{2})$and(1.11)

with $p(z_{0})=\pm ia$$-k \geqq-\frac{k}{2}(a+\frac{1}{a})\geqq l (\arg p(z_{0})=-\frac{\pi}{2})$$(a>0)$ .

$\mathbb{U}$with $p(z)\neq 0$

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Proof.

Let us consider(1.12)$\phi(z)=\frac{1-p(z)}{1+p(z)}=\frac{c_{k}}{2}z^{k}+\cdots$

for $p(z)$ . Then, it follows that $\phi(0)=\phi^{l}(0)=\cdots=\phi^{(k-1)}(0)=0,$ $|\phi(z)|<1$
and $|\phi(z_{0})|=1$ . Therefore, applyin Lemma 1, we have that$(|z|<|z_{0}|)$$g$(1.13)$\frac{z_{0}\phi’(z_{0})}{\phi(z_{0})}=\frac{-2_{h}p’(z_{0})}{1-(p(z_{0}))^{2}}=\frac{-2_{h}p’(\eta\})}{1+|p(z_{0})|^{2}}=p\geqq k.$

This implies that $z_{0}p’(q))$ is a negative real number and(1.14)$-z_{0}p’(z_{0}) \geqq\frac{k}{2}(1+|p(z_{0})|^{2})$Let us use the same method by Nunokawa [3]. If argp$(a>0)$ . This gives us that.

$(z_{0})= \frac{\pi}{2}$, then we write $p(z_{0})=ia$${\rm Im}( \frac{z_{0}p’(z_{0})}{p(z_{0})})={\rm Im}(-\frac{iz_{0}p’(z_{0})}{a})\geqq\frac{k}{2}(a+\frac{1}{a})$If $\arg p(z_{O})=--$ ’ then we write $p(z_{0})=-ia$$\pi 2$.

$(a>0)$ . Thus we have that${\rm Im}( \frac{z_{0}p(z_{0})}{p(z_{0})})={\rm Im}(\frac{iz_{0}p’(z_{0})}{a})\leqq-\frac{k}{2}(a+\frac{1}{a})$.

This completes the proof of Lemma 2.

2

口Main results
With the help of Lemma 2, we derive the following theorem.

Theorem 1 Let be a$A(n, k)$ and $g(z)\in \mathcal{A}(n,j)$$\alpha$

(2.1)
$w$

伽$\omega$

mplex number with$(j\geqq 1)$${\rm Re}\alpha>0$

and $\beta<1.$, let $f(z)\in$$fb\hslash her$satisfies${\rm Re}( \frac{\alpha q(z)}{zg(z)})>\delta (z\in \mathbb{U})$$0 \leqq\delta<\frac{{\rm Re}\alpha}{n}$(2.2). If $f(z)$ and $g(z)$ sat 醜${\rm Re} \{(1-\alpha)\frac{f(z)}{g(z)}+\alpha\frac{f’(z)}{g(z)}\}+\frac{\delta k}{2(1-\beta_{i})}|\frac{f(z)}{g(z)}-\beta_{1}|^{2}>\beta (z\in \mathbb{U})$then(2.3)

where${\rm Re}( \frac{f(z)}{g(z)})>\beta_{1} (z\in \mathbb{U})$$\beta_{1}=\frac{2\beta+\delta k}{2+\delta k}.$,77
Proof.

Dcfining thc function $p(z)$ by$p(z)= \frac{\frac{f(z)}{g(z)}-\beta_{1}}{1-/\beta_{1}},$(2.4)

we see that $p(O)=1$ and(2.5)$(1- \alpha)\frac{f(z)}{g(z)}+\alpha\frac{f’(z)}{g(z)}-\beta$$=( \beta_{1}-\beta)+(1-\beta_{1})(p(z)+\frac{\alpha g(z)}{zg(z)}zp’(z))$$>- \frac{\delta k}{2(1-\prime\theta_{1})}|\frac{f(z)}{g(z)}-\beta_{1}|^{2}$for all$z\in \mathbb{U}$. Let us suppose that there exists a point$z_{0}\in \mathbb{U}$such that$| \arg p(z_{0})|<\frac{\pi}{2} (|z|<|z_{0}|)$and$| \arg p(z_{0})|=\frac{\pi}{2}.$Then, by means of Lemma 2, we have that
-勧$p’(z_{0}) \geqq\frac{k}{2}(1+|p(z_{0})|^{2})$ .

(2.6)

If follows from the above that${\rm Re} \{(1-\alpha)\frac{f(z_{0})}{g(z_{0})}+\alpha\frac{f^{l}(z_{0})}{g’(z_{0})}-\beta\}$$=( \beta_{1}-\beta)+(1-\beta_{1}){\rm Re}\{p(z_{0})+\frac{\alpha g(z_{0})}{z_{0}g^{l}(z_{0})}z_{0}p’(z_{0})\}$$=( \beta_{1}-\beta)-(1-\beta_{1}){\rm Re}\{\frac{\alpha g(z_{0})}{z09(z_{0})}(-z_{0}p’(z_{0}))\}$$\delta k$$\leqq(\beta_{1}-\beta)-(1-\beta_{1})_{\overline{2}}(1+|p(z_{0})|^{2})$$=- \frac{\delta k}{2(1-\beta_{1})}|\frac{f(z_{0})}{g(z_{0})}-\beta_{1}|^{2}$

which contradicts (2.5). This completes the proof of the theorem.

$\square$78

Remark 1
Theorem$B$

If $f(z)$ and $g(z)$ satisfy $f(z_{0})=\beta_{1}’g(z_{0})$ in Theorem 1, then Theorem 1 becomesgiven by Ponnusamy and Kamnakaran [4]. We also have

and $\beta<1$ . Further, let $f(z)\in$
be a complex number with. If $f(z)$and $g(z)\in \mathcal{A}(n,j)$ $(j\geqq 1)$ satisfies the condition (2.1) withand $g(z)$ satisfy

Theorem 2Let${\rm Re}\alpha>0$$\alpha$$0 \leqq\delta<\frac{{\rm Re}\alpha}{n}$$\mathcal{A}(n, k)$(2.7)$| \arg\{(1-\alpha)\frac{f(z)}{g(z)}+\alpha\frac{f’(z)}{g’(z)}-\beta\}|<\frac{\pi}{2}+Tan^{-1}(\frac{\delta k|p(z)|}{2(\frac{2r}{1-r^{2}}+\frac{|{\rm Im}\alpha|}{n}+1)})$

for $|z|=r<1$ , then
(2.8)$| \arg(\frac{f(z)}{g(z)}-\beta_{1})|<\frac{\pi}{2} (z\in U)$

$or$

(2.9)

where

${\rm Re}( \frac{f(z)}{g(z)})>\beta_{1}’ (z\in \mathbb{U})$$\beta_{1}=\frac{2,9+\delta k}{2+\delta k}$,and$p(z)= \frac{\frac{f(z)}{g(z)}-\beta_{1}}{1-\beta_{1}}.$

Proof.

Note that the function $p(z)$ is analytic in$\mathbb{U}$and $p(0).=1$ . It follows that$\arg\{(\beta_{i}-\beta)+(1-\beta_{1})(p(z)+\frac{\alpha g(z)}{zg(z)}zp’(z))\}|$$| \arg\{(1-\alpha)\frac{f(z)}{g(z)}+\alpha\frac{f’(z)}{\phi(z)}-\beta$$< \frac{\pi}{2}+$for $|z|=r<1$ . If there exists a pointTan$z_{0}\in \mathbb{U}$$-1( \frac{\delta k|p(z)|}{2(\frac{2r}{1-r^{2}}+\frac{|{\rm Im}\alpha|}{n}+1)}1$

such that$| \arg p(z_{0})|<\frac{\pi}{2} (|z|<|z_{0}|)$and$| \arg p(z_{0})|=\frac{\pi}{2},$79
then, by Lemma 2, we have that$\frac{z_{0}p’(z_{0})}{p(z_{0})}=i\ell,$where$\frac{k}{2}(a+\frac{1}{a})\leqq\ell (\arg p(z_{0})=\frac{\pi}{2})$and$- \frac{k}{2}(a+\frac{1}{a})\geqq\ell (\arg p(z_{0})=-\frac{\pi}{2})$with $p(z_{0})=\pm ia$$(a>0)$ . If $\arg p(z_{0})=\frac{\pi}{2}$ , then it follows that$\arg\{(1-\alpha)\frac{f(z_{0})}{g(z_{0})}+\alpha\frac{f’(z_{0})}{g(z_{0})}-\beta\}$$= \arg p(z_{0})(\frac{\beta_{i}-\beta}{p(z_{0})}+(1-j?_{1})(1+\frac{\alpha 9(z_{0})}{z_{0}g’(z_{0})}\frac{z_{0}p’(z_{0})}{p(z_{O})})\}$$= \frac{\pi}{2}+\arg\{-(\frac{j\prime;_{1}-\beta}{a})i+(1-\beta_{1})(1+i\ell\frac{\alpha q(z_{0})}{z_{0}g(z_{0})})\}$$= \frac{\pi}{2}+\arg I(z_{0})$

,

where
(2.9)

$I(z_{0})=-( \frac{\beta_{1}-\beta}{a})i+(1-\beta_{i})(1+i\ell\frac{\alpha g(z_{0})}{z_{0}g(z_{0})})$.

Note that(2.10)${\rm Im} I(z_{0})= \frac{\beta-\beta_{1}}{a}+(1-\beta_{1})\ell{\rm Re}\frac{\alpha q(z_{0})}{z_{0}g(z_{0})}$$>(1- \beta_{1})\delta l+\frac{\beta-\beta_{1}}{a}$$\geqq\frac{\delta k}{2}(1-\beta_{1})(a+\frac{1}{a})+\frac{\prime\prime}{a}$$= \frac{\delta k}{2}(1-\beta_{1})a>0$

and(2.11)${\rm Re} I(z_{0})=(1- \prime/d_{1})(1-\ell{\rm Im}(\frac{\alpha g(z_{0})}{z_{0}g’(z_{0})}))\leqq(1-\beta_{1})(1+\ell|{\rm Im}(\frac{\alpha g(z_{0})}{z_{0}g(z_{0})})|)$.

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Letting(2.12)$q(z)= \frac{\alpha g(z)}{zg(z)}+1-\frac{\alpha}{n},$we know that $q(z)$ is analytic in(2.13)$\mathbb{U}$with $q(O)=1$ . This gives us that$|{\rm Im} q(z)|=|{\rm Im}( \frac{\alpha g(z)}{zg’(z)}+1-\frac{\alpha}{n})|\leqq\frac{2r}{1-r^{2}}$for $|z|=r<1$ . Thus we have that
(2.14)$|{\rm Im}( \frac{\alpha g(z_{\theta})}{z_{0}g(z_{0})})|\leqq\frac{2r}{1-r^{2}}+\frac{|{\rm Im}\alpha|}{n} (|z|=r<1)$.

Using (2.11) and (2.14), we obtain that$\arg I(z_{0})=$Tan$-1( \frac{{\rm Im} I(z_{0})}{{\rm Re} I(z_{0})})\geqq$

Tan$- i(\frac{\delta ka}{2(\frac{2r}{1-r^{2}}+\frac{|{\rm Im}\alpha|}{n}+1)}1,$

which contradicts our condition (2.7).
, using the same way, we also have that
If$\arg p(z_{0})=-\frac{\pi}{2}$$\arg\{(1-\alpha)\frac{f(z_{0})}{g(z_{0})}+\alpha\frac{f’(z_{0})}{g’(z_{0})}-\beta\}\leqq-\{\frac{\pi}{2}+Tan^{-1}(\frac{\delta ka}{2(\frac{2r}{1-r^{2}}+\frac{|{\rm Im}\alpha|}{n}+i)}1\},$

which contradicts (2.7).

口

Remark 2 If $f(z)$ satisfies the conditions in Theorem , then $f(z)$ satisfies the conditionsof Theorem 2. In this case, we see that Theorem 2 becomes Theorem B.
$B$References[1] S. Fukui and K. Sakaguchi, An extension of a theorem of S. Ruscheweyh, Bull. Fac. Edu.
Wakayama Univ. Nat. Sci. 30 (1980), 1–3.
[2] I. S. Jack, Rmctions starlike and convex of order469-474.

$\alpha$, J. London Math. Soc. 2 (1971),[3] M. Nunokawa, On properties of non-Camtheodory functions, Proc. Japan Acad. 68(1992), 152-153.

81[4] S. Ponnusamy and V. Karunakaran, Differential $sub_{07}dinationar\iota d$ conformul $mappir\iota gs,$
Complex Variables. 11 (1989), 79-86.

[5] K. Sakaguchi, On a certain univalent mapping, J. Math. Soc. Japan. 11 (1959), 72-75.

Mamoru Nunokawa
Emeritus Professor
University of Gunma
798-8 Hoshikuki,Chuou-Ward, Chiba 260-0808

Japan$E$-mail: mamoru-nuno@doctor.nifty.jp

Kazuo Kuroki
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502
Japan
$E$

-mail: freedom@sakai.zaq.ne.jp

Janusz Sok\’olDepartment of Mathematics
Rzeszow University of TechnologyAl.
, Warszawy 12, 35-959 Rzesz\’ow
Poland-mail: jsokol@prz.edu.pl$Powst’a\iota\iota c\’{o} w$$E$Shigeyoshi Owa
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502
Japan-mail: shige21@ican.zaq.ne.jp$E$