NOTES ON CERTAIN SUBCLASS OF p-VALENTLY BAZILEVI ˇC FUNCTIONS

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p-Valently Bazilevi ˇc Functions Zhi-Gang Wang and Yue-Ping Jiang

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NOTES ON CERTAIN SUBCLASS OF p-VALENTLY BAZILEVI ˇ C FUNCTIONS

ZHI-GANG WANG YUE-PING JIANG

School of Mathematics and Computing Science School of Mathematics and Econometrics Changsha University of Science and Technology Hunan University

Changsha 410076, Hunan, Changsha 410082, Hunan,

People’s Republic of China People’s Republic of China EMail:zhigangwang@foxmail.com EMail:ypjiang731@163.com

Received: 04 February, 2007

Accepted: 07 July, 2008

Communicated by: A. Sofo 2000 AMS Sub. Class.: Primary 30C45.

Key words: Analytic functions, Multivalent functions, Bazileviˇc functions, subordination be- tween analytic functions, Briot-Bouquet differential subordination.

Abstract: In the present paper, we discuss a subclassMp(λ, µ, A, B)ofp-valently Bazile- viˇc functions, which was introduced and investigated recently by Patel [5]. Such results as inclusion relationship, coefficient inequality and radius of convexity for this class are proved. The results presented here generalize and improve some earlier results. Several other new results are also obtained.

Acknowledgements: The present investigation was supported by the National Natural Science Foun- dation under Grant 10671059 of People’s Republic of China. The first-named author would like to thank Professors Chun-Yi Gao and Ming-Sheng Liu for their continuous support and encouragement. The authors would also like to thank the referee for his careful reading and making some valuable comments which have essentially improved the presentation of this paper.

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1. Introduction

LetA_pdenote the class of functions of the form:

f(z) =z^p+

∞

X

n=p+1

a_nzⁿ (p∈N:={1,2,3, . . .}), which are analytic in the open unit disk

U:={z :z ∈C and |z|<1}.

For simplicity, we write

A₁ =:A.

For two functionsfandg, analytic inU, we say that the functionfis subordinate tog inU, and write

f(z)≺g(z) (z ∈U),

if there exists a Schwarz functionω, which is analytic inUwith ω(0) = 0 and |ω(z)|<1 (z ∈U) such that

f(z) = g ω(z)

(z ∈U).

Indeed it is known that

f(z)≺g(z) (z ∈U) =⇒f(0) =g(0) and f(U)⊂g(U).

Furthermore, if the functiongis univalent in U, then we have the following equiva- lence:

f(z)≺g(z) (z ∈U)⇐⇒f(0) =g(0) and f(U)⊂g(U).

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LetM_p(λ, µ, A, B)denote the class of functions in A_p satisfying the following subordination condition:

zf⁰(z)

f^1−µ(z)g^µ(z) +λ

1 + zf⁰⁰(z)

f⁰(z) −(1−µ)zf⁰(z)

f(z) −µzg⁰(z) g(z)

≺p1 +Az 1 +Bz (1.1)

(−15B < A51; z ∈U)

for some realµ(µ=0), λ(λ=0)andg ∈ S_p^∗, whereS_p^∗denotes the usual class of p-valently starlike functions inU.

For simplicity, we write M_p

λ, µ,1− 2α p ,−1

=M_p(λ, µ, α) :=

f(z)∈ A_p : <

zf⁰(z)

f^1−µ(z)g^µ(z) +λ

1 + zf⁰⁰(z)

f⁰(z) −(1−µ)zf⁰(z)

> α

, for someα(05α < p)andz ∈U.

The classM_p(λ, µ, A, B)was introduced and investigated recently by Patel [5].

The author obtained some interesting properties for this class in the caseλ > 0, he also proved the following result:

Theorem 1.1. Let

µ=0, λ >0 and −15B < A51.

Iff ∈ Mp(λ, µ, A, B), then

(1.2) zf⁰(z)

pf^1−µ(z)g^µ(z) ≺ λ

pQ(z) =q(z)≺ 1 +Az

1 +Bz (z ∈U),

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where

Q(z) =





 R1

0 s^p^λ⁻¹ ^1+Bsz_1+Bz^p(A−B)_λB

ds (B 6= 0), R1

0 s^p^λ⁻¹exp ^p_λ(s−1)Az

ds (B = 0), andq(z)is the best dominant of (1.2).

In the present paper, we shall derive such results as inclusion relationship, coefficient inequality and radius of convexity for the classM_p(λ, µ, A, B)by making use of the techniques of Briot-Bouquet differential subordination. The results presented here generalize and improve some known results. Several other new results are also obtained.

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2. Preliminary Results

In order to prove our main results, we shall require the following lemmas.

Lemma 2.1. Let

µ=0, λ=0 and −15B < A51.

Then

M_p(λ, µ, A, B)⊂ M_p(0, µ, A, B).

Proof. Suppose thatf ∈ M_p(λ, µ, A, B). By virtue of (1.2), we know that zf⁰(z)

f^1−µ(z)g^µ(z) ≺p1 +Az

1 +Bz (z∈U),

which implies that f ∈ M_p(0, µ, A, B). Therefore, the assertion of Lemma 2.1 holds true.

Lemma 2.2 (see [3]). Let

−15B₁ 5B₂ < A₂ 5A₁ 51.

Then 1 +A₂z

1 +B₂z ≺ 1 +A₁z 1 +B₁z. Lemma 2.3 (see [4]). LetF be analytic and convex inU. If

f, g∈ A and f, g ≺F, then

λf + (1−λ)g ≺F (05λ51).

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Lemma 2.4 (see [6]). Let

f(z) =

∞

X

k=0

a_kz^k

be analytic inUand

g(z) =

∞

X

k=0

bkz^k be analytic and convex inU. Iff ≺g, then

|a_k|5|b₁| (k∈N).

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3. Main Results

We begin by stating our first inclusion relationship given by Theorem3.1below.

µ=0, λ₂ =λ₁ =0 and −15B₁ 5B₂ < A₂ 5A₁ 51.

Then

M_p(λ₂, µ, A₂, B₂)⊂ M_p(λ₁, µ, A₁, B₁).

Proof. Suppose thatf ∈ M_p(λ₂, µ, A₂, B₂). We know that zf⁰(z)

f^1−µ(z)g^µ(z)+λ₂

1 + zf⁰⁰(z)

f⁰(z) −(1−µ)zf⁰(z)

≺p1 +A₂z 1 +B₂z. Since

−15B₁ 5B₂ < A₂ 5A₁ 51, it follows from Lemma2.2that

(3.1) zf⁰(z)

f^1−µ(z)g^µ(z)+λ₂

1 + zf⁰⁰(z)

f⁰(z) −(1−µ)zf⁰(z)

≺p1 +A1z 1 +B₁z. that is, thatf ∈ Mp(λ2, µ, A1, B1). Thus, the assertion of Theorem3.1 holds true for

λ₂ =λ₁ =0.

If

λ₂ > λ₁ =0,

by virtue of Lemma2.1and (3.1), we know thatf ∈ M_p(0, µ, A₁, B₁), that is

(3.2) zf⁰(z)

f^1−µ(z)g^µ(z) ≺p1 +A₁z 1 +B1z.

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At the same time, we have (3.3) zf⁰(z)

f^1−µ(z)g^µ(z) +λ₁

1 + zf⁰⁰(z)

f⁰(z) −(1−µ)zf⁰(z)

=

1−λ₁ λ₂

zf⁰(z)

f^1−µ(z)g^µ(z)+ λ₁ λ₂

zf⁰(z) f^1−µ(z)g^µ(z) +λ₂

1 + zf⁰⁰(z)

f⁰(z) −(1−µ)zf⁰(z)

. It is obvious that

h₁(z) = 1 +A₁z 1 +B₁z

is analytic and convex inU. Thus, we find from Lemma2.3, (3.1), (3.2) and (3.3) that

zf⁰(z)

f^1−µ(z)g^µ(z)+λ₁

1 + zf⁰⁰(z)

f⁰(z) −(1−µ)zf⁰(z)

≺p1 +A₁z 1 +B1z, that is, thatf ∈ M_p(λ₁, µ, A₁, B₁). This implies that

M_p(λ₂, µ, A₂, B₂)⊂ M_p(λ₁, µ, A₁, B₁).

Remark 1. SettingA₁ = A₂ = Aand B₁ = B₂ = B in Theorem 3.1, we get the corresponding result obtained by Guo and Liu [2].

Corollary 3.2. Let

µ=0, λ2 =λ1 =0 and p > α2 =α1 =0.

Then

Mp(λ2, µ, α2)⊂ Mp(λ1, µ, α1).

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Theorem 3.3. Iff ∈ A_p satisfies the following conditions:

<

f(z) z^p

>0 and

zf⁰(z)

f^1−µ(z)g^µ(z)−p

< νp

05µ < 1

2; 0< ν 51; z ∈U

forg ∈ S_p^∗, thenf isp-valently convex (univalent) in|z|< R(p, µ, ν), where R(p, µ, ν) = 2pµ+ 2µ−ν−2 +p

(2pµ+ 2µ−ν−2)²+ 4(ν+p)(p−2pµ)

2(ν+p) .

Proof. Suppose that

h(z) := zf⁰(z)

pf^1−µ(z)g^µ(z) −1 (z ∈U).

Thenhis analytic inUwith

h(0) = 0 and |h(z)|<1 (z ∈U).

Thus, by applying Schwarz’s Lemma, we get

h(z) =νzψ(z) (0< ν 51), whereψ is analytic inUwith

|ψ(z)|51 (z ∈U).

Therefore, we have

(3.4) zf⁰(z) = pf^1−µ(z)g^µ(z)(1 +νzψ(z)).

Differentiating both sides of (3.4) with respect toz logarithmically, we obtain (3.5) 1 + zf⁰⁰(z)

f⁰(z) = (1−µ)zf⁰(z)

f(z) +µzg⁰(z)

g(z) + νz(ψ(z) +zψ⁰(z)) 1 +νzψ(z) .

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We now suppose that

(3.6) φ(z) := f(z)

z^p = 1 +c₁z+c₂z²+· · · , by hypothesis, we know that

(3.7) <(φ(z))>0 (z ∈U).

It follows from (3.6) that

(3.8) zf⁰(z)

f(z) =p+ zφ⁰(z) φ(z) . Upon substituting (3.8) into (3.5), we get

(3.9) 1 + zf⁰⁰(z)

f⁰(z) = (1−µ)p+ (1−µ)zφ⁰(z)

φ(z) +µzg⁰(z)

g(z) + νz(ψ(z) +zψ⁰(z)) 1 +νzψ(z) . Now, by using the following well known estimates (see [1]):

<

zφ⁰(z) φ(z)

=− 2r

1−r², <

zg⁰(z) g(z)

=−p1−r 1 +r, and

<

z(ψ(z) +zψ⁰(z)) 1 +zψ(z)

=− r 1−r for|z|=r <1in (3.9), we obtain

<

1 + zf⁰⁰(z) f⁰(z)

=(1−µ)p− 2(1−µ)r

1−r² −pµ(1−r)

1 +r − νr 1−νr

=(1−µ)p− 2(1−µ)r

1−r² −pµ(1−r)

1 +r − νr 1−r

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=−(ν+p)r²−(2pµ+ 2µ−ν−2)r−(p−2pµ)

1−r² ,

which is certainly positive ifr < R(p, µ, ν).

Puttingν= 1in Theorem3.3, we get the following result.

Corollary 3.4. Iff ∈ A_p satisfies the following conditions:

<

f(z) z^p

>0 and

zf⁰(z)

f^1−µ(z)g^µ(z) −p

< p

05µ < 1

2; z ∈U

forg ∈ S_p^∗, thenf isp-valently convex (univalent) in|z|< R(p, µ), where R(p, µ) = 2pµ+ 2µ−3 +p

(2pµ+ 2µ−3)²+ 4(1 +p)(p−2pµ)

2(1 +p) .

Remark 2. Corollary3.4corrects the mistakes of Theorem 3.8 which was obtained by Patel [5].

µ=0, λ=0 and −15B < A51.

If

f(z) =z+

∞

X

k=n+1

a_kz^k (z ∈U) satisfies the following subordination condition:

(3.10) f⁰(z)

f(z) z

^µ−1 +λ

zf⁰⁰(z)

f⁰(z) + (1−µ)

1− zf⁰(z) f(z)

≺ 1 +Az 1 +Bz,

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then

(3.11) |a_n+1|5 A−B

(1 +nλ)(n+µ) (n ∈N).

Proof. Suppose that

f(z) =z+

∞

X

k=n+1

akz^k (z ∈U) satisfies (3.10). It follows that

(3.12) f⁰(z)

f(z) z

^µ−1 +λ

zf⁰⁰(z)

f⁰(z) + (1−µ)

1− zf⁰(z) f(z)

= 1 + (1 +nλ)(n+µ)an+1zⁿ+· · · ≺ 1 +Az

1 +Bz (z ∈U).

Therefore, we find from Lemma2.4, (3.12) and−15B < A51that (3.13) |(1 +nλ)(n+µ)a_n+1|5A−B.

The assertion (3.11) of Theorem3.5can now easily be derived from (3.13).

Taking A = 1−2α (0 5 α < 1) and B = −1 in Theorem 3.5, we get the following result.

Corollary 3.6. Let

µ=0, λ=0 and 05α <1.

If

f(z) = z+

∞

X

k=n+1

a_kz^k

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satisfies the following inequality:

< f⁰(z)

f(z) z

µ−1

+λ

zf⁰⁰(z)

f⁰(z) + (1−µ)

1−zf⁰(z) f(z)

!

> α,

then

|a_n+1|5 2(1−α)

(1 +nλ)(n+µ) (n ∈N).

Remark 3. Corollary3.6provides an extension of the corresponding result obtained by Guo and Liu [2].

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References

[1] W.M. CAUSEYANDE.P. MERKES, Radii of starlikeness for certain classes of analytic functions, J. Math. Anal. Appl., 31 (1970), 579–586.

[2] D. GUO AND M.-S. LIU, On certain subclass of Bazileviˇc functions, J. In- equal. Pure Appl. Math., 8(1) (2007), Art. 12. [ONLINE: http://jipam.

vu.edu.au/article.php?sid=825].

[3] M.-S. LIU, On a subclass ofp-valent close-to-convex functions of orderβ and typeα, J. Math. Study, 30 (1997), 102–104 (in Chinese).

[4] M.-S. LIU, On certain subclass of analytic functions, J. South China Normal Univ., 4 (2002), 15–20 (in Chinese).

[5] J. PATEL, On certain subclass ofp-valently Bazileviˇc functions, J. Inequal. Pure Appl. Math., 6(1) (2005), Art. 16. [ONLINE:http://jipam.vu.edu.au/

article.php?sid=485].

[6] W. ROGOSINSKI, On the coefficients of subordinate functions, Proc. London Math. Soc. (Ser. 2), 48 (1943), 48–82.