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Hindawi Publishing CorporationInternational Journal of Mathematics and Mathematical SciencesVolume 2007, Article ID 72393, 5 pagesdoi:10.1155/2007/72393
Research ArticleOn Certain Multivalent Functions
Mamoru Nunokawa, Shigeyoshi Owa, Tadayuki Sekine, Rikuo Yamakawa,Hitoshi Saitoh, and Junichi Nishiwaki
Received 11 February 2007; Accepted 17 April 2007
Recommended by Linda R. Sons
Let �∗(p,α) be the class of functions f (z) which are analytic and p-valently starlike oforder α in the open unit disk E. The object of the present paper is to derive an interestingcondition for f (z) to be in the class �∗(p,α).
Copyright © 2007 Mamoru Nunokawa et al. This is an open access article distributed un-der the Creative Commons Attribution License, which permits unrestricted use, distribu-tion, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let �(p) denote the class of functions f (z) of the form
f (z)= zp +∞∑
n=p+1
anzn(p ∈N= {1,2,3, . . .}) (1.1)
which are analytic in the open unit disk E= {z : z ∈ C, |z| < 1}. A function f (z)∈�(p)is said to be p-valently starlike of order α(0 � α < p) in E if and only if
Rez f ′(z)f (z)
> α (z ∈ E). (1.2)
We denote by �∗(p,α) the subclass of �(p) consisting of functions which are p-valentlystarlike of order α in E. We only call a function f (z)∈�∗(p,0) to be p-valently starlike inE. Further, a function f (z)∈�(p) is said to be p-valently convex of order α(0 � α < p)in E if and only if
1 + Rez f ′′(z)f ′(z)
> α (z ∈ E). (1.3)
2 International Journal of Mathematics and Mathematical Sciences
We denote by �(p,α) the subclass of �(p) consisting of all p-valently convex functions oforder α in E. We also call a function f (z)∈�(p,0) p-valently convex function. From thedefinition, it is trivial that if f (z) is a p-valently convex function, then z f ′(z) is p-valentlystarlike in E.
2. Preliminaries
In this paper, we need the following lemmas.
Lemma 2.1. If M(z) = zp +∑∞
n=p+k anzn (1 � p and 1 � k) and N(z) = zp +
∑∞n=p+1 anz
n
are analytic in E and N(z) satisfies Re(N(z)/zN ′(z)) > δ (0 � δ < 1/p), then
Re(M′(z)N ′(z)
)> β implies Re
(M(z)N(z)
)>
2β+ kδ
2 + kδ. (2.1)
Remark 2.2. This lemma holds to be true for N(z) which is multivalently starlike in E.
We owe the above lemma to Ponnusamy and Karunakaran [1].
Lemma 2.3. Let f (z)∈�∗(p,0). Then
F(z)p+ 1
=∫ z
0f (t)dt ∈ S∗(p+ 1,0) (2.2)
or
RezF′(z)F(z)
> 0 (z ∈ E). (2.3)
The proof of this result can be found in [2].
3. Main result
Theorem 3.1. If f (z)∈�(p) satisfies the following condition:
Rez f (p)(z)f (p−1)(z)
> α (z ∈ E) (3.1)
for α(0 � α < 1), then f (z)∈�∗(p,α+ p− 1) or
Rez f ′(z)f (z)
> α+ p− 1 (z ∈ E). (3.2)
Proof. From assumption (3.1), f (p−1)(z) is univalently starlike of order α and it is trivialthat
(z f (p−1)(z)− f (p−2)(z)
)′ = z f (p)(z). (3.3)
Note that f (p−2)(z) is starlike in E by Lemma 2.3. Therefore, applying Lemma 2.1 and(3.3), we have
Rez f (p−1)(z)− f (p−2)(z)
f (p−2)(z)> α (z ∈ E). (3.4)
Mamoru Nunokawa et al. 3
From Lemma 2.3, f (p−2)(z) is 2-valently starlike in E. Now then, it is trivial that
(z f (p−2)(z)− 2 f (p−3)(z)
)′ = z f (p−1)(z)− f (p−2)(z). (3.5)
Then, from Lemma 2.1, 2-valently starlikeness of f (p−2)(z), and (3.5), we have
Rez f (p−2)(z)− 2 f (p−3)(z)
f (p−3)(z)> α (z ∈ E). (3.6)
Further, it is trivial that
(z f (p−3)(z)− 3 f (p−4)(z)
)′ = z f (p−2)(z)− 2 f (p−3)(z), (3.7)
and applying the same method and reason as above, we have
Rez f (p−3)(z)− 3 f (p−4)(z)
f (p−4)(z)> α (z ∈ E), (3.8)
where f (p−3)(z) is 3-valently starlike in E. Applying the mathematical induction, we have
Rez f ′(z)− (p− 1) f (z)
f (z)> α (z ∈ E), (3.9)
where f (z) is p-valently starlike in E. This shows that
Rez f ′(z)f (z)
> α+ p− 1 (z ∈ E), (3.10)
or f (z)∈�∗(p,α+ p− 1). �
Our main result shows the following.
Corollary 3.2. Let f (z)∈�(p), 2 � p, 0 � α < 1,
Rez f (p)(z)f (p−1)(z)
> α (z ∈ E), (3.11)
and put f (z)= zp−1 f1(z) where
f1(z)= z+∞∑
n=p+1
anzn−p+1. (3.12)
Then f1(z) is univalently starlike of order α in E.
Proof. From the definition of f1(z) and Theorem 3.1, it follows that
Rez f ′(z)f (z)
= p− 1 + Rez f ′1 (z)f1(z)
> α+ p− 1. (3.13)
This completes the proof. �
4 International Journal of Mathematics and Mathematical Sciences
Corollary 3.3. Let f (z)∈�(p), 2 � p, 0 � α < 1, and
1 + Rez f (p+1)(z)f (p)(z)
> α (z ∈ E). (3.14)
Then one has
Rez f ′(z)f (z)
> β(α) + p− 1 (z ∈ E), (3.15)
where
β(α)=
⎧⎪⎪⎪⎨⎪⎪⎪⎩
1− 2α22−2α
[1− 22α−1
] if α �= 12
,
12log2
if α= 12.
(3.16)
Proof. Putting
g(z)= f (p−1)(z)p!
= z+∞∑
n=2
bnzn, (3.17)
then from assumption (3.14), g(z) is univalently convex of order α, and therefore fromWilken-Feng result [3] and Theorem 3.1, we have
Rezg′(z)g(z)
= Rez f (p)(z)f (p−1)(z)
> β(α) (z ∈ E), (3.18)
and therefore it follows that
Rez f ′(z)f (z)
> β(α) + p− 1 (z ∈ E). (3.19)
�
We will give here an open problem.
Problem 3.4. Let f (z)∈�∗(p,α) and 0 � α < p.Then
F(z)p+ 1
=∫ z
0f (t)dt ∈ S∗
(p+ 1,β(p,α)
). (3.20)
What is the best β(p,α)?
Mamoru Nunokawa et al. 5
References
[1] S. Ponnusamy and V. Karunakaran, “Differential subordination and conformal mappings,”Complex Variables, vol. 11, no. 2, pp. 79–86, 1989.
[2] M. Nunokawa, “On the theory of multivalent functions,” Tsukuba Journal of Mathematics,vol. 11, no. 2, pp. 273–286, 1987.
[3] D. R. Wilken and J. Feng, “A remark on convex and starlike functions,” The Journal of the LondonMathematical Society, vol. 21, no. 2, pp. 287–290, 1980.
Mamoru Nunokawa: 798-8 Hoshikuki-Machi, Chuo-Ku, Chiba-Shi, Chiba 260-0808, JapanEmail address: mamoru [email protected]
Shigeyoshi Owa: Department of Mathematics, Kinki University, Higashi-Osaka,Osaka 577-8502, JapanEmail address: [email protected]
Tadayuki Sekine: Research unit of Mathematics, College of Pharmacy, Nihon University,7-1 Narashinodai 7-Chome, Funabashi-Shi, Chiba 274-8555, JapanEmail address: [email protected]
Rikuo Yamakawa: Department of Mathematics, Shibaura Institute of Technology, Minuma,Saitama-Shi, Saitama 337-8570, JapanEmail address: [email protected]
Hitoshi Saitoh: Department of Mathematics, Gunma National College of Technology,Maebashi 371-8530, Gunma, JapanEmail address: [email protected]
Junichi Nishiwaki: Department of Mathematics, Kinki University, Higashi-Osaka,Osaka 577-8502, JapanEmail address: [email protected]
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