On sufficient condition for starlikeness
1D. O. Makinde, T. O. Opoola
Abstract
In this paper,we give a condition for starlikeness of the integral oper- ator of the formF(z) =
Z z
0 k
Y
ı=1
fı(s) s
α1
ds.
2000 Mathematics Subject Classification:30C45
Key words and phrases: Starlike, Convex, Briot-Bouquet subordination.
1 Introduction
LetA be the class of all analytic functions f(z) defined in the open unit disk U ={z∈C :|z|<1}andSthe subclass ofAconsisting of univalent functions
(1) f(z) =z+
X∞ k=2
akzk
S∗={f ∈S:Re(zf0(z)
f(z) )>0, z∈U}, Mα={f ∈S : f(z)f0(z)
z 6= 0, ReJ(α, f;z)>0, z∈U}
1Received 14 February, 2009
Accepted for publication (in revised form) 17 March, 2009
35
where J(α, f;z) = (1−α)zff(z)0(z) +α(1 + zff000(z)(z)) be the class of starlike and α−convexfunctions respectively.
Letp(z) be the class of functions that are regular in U and of the form :
(2) p(z) = 1 +
X∞ k=1
bkzk Furthermore, let h(z) = 1+z1−z.
Let T be the univalent [5] subclass of Aconsisting of functionsf(z) satisfying
|zf(z)2f0(z)2 −1|<1,(z∈U)
LetTn be the subclass ofT for which fk(0) = 0 (k= 2,3, ..., n).
Let Tn,µ be the subclass of Tn consisting of functions of the form Rz
0
Qk
ı=1(fıs(s))α1dssatisfying: |zf(z)2f0(z)2 −1|< µ,(z∈U) for someµ(0< µ≤1).
2 Preliminaries
Theorem 1 [1] LetM andN be analytic inU withM(0) =N(0) = 0.IfN(z) maps onto a many sheeted region which is starlike with respect to the origin and Re{MN00(z)(z)}>0 in U, then Re{MN(z)(z)}>0 in U.
Theorem 2 [6] Letfı ∈Tn,µı(ı= 1,2, ..., k;k∈ N∗) be defined by
(3) fi(z) =z+
X∞ n=2
ainzn
for all i = 1,2, ..., k;α, β ∈ C;R{β} ≥ γ and γ = Xk i=1
1 + (1 +µi)M
|α| (M ≥ 1,0< µi <1, k∈N∗). If |fi(z)| ≤ M(z∈U), i= 1,2, ..., k then, the integral operator
(4) Fα,β(z) ={β
Z z 0
tβ−1 Yk i=1
(fi(t) t )α1dt}β1 is univalent.
Theorem 3 [2] Let h be convex in U and Re{βh(z) +γ} >0, z ∈U.If p ∈ H(U) whereH(U) is the class of functions which are analytic in the unit disk, with p(0) =h(0)andpsatisfies the Briot-Bouquet differential subordinations:
p(z) +βp(z)+γzp0(z) ≺h(z), z∈U. Then, p(z)≺h(z), z∈U.
3 Main Results
We now give the proof of the following results:
Theorem 4 Let Fα(z) be the function in U defined by
(5) Fα(z) =
Z z 0
Yk i=1
(fi(s)
s )α1ds, α∈C.
If fi ∈S∗ then, F(z)∈S∗ where fi is as in equation (3) above.
Proof. By differentiating (5), we obtain: F0(z) = Qk
i=1(fi(z)z )1α. Thus, zF0(z)
F(z) = Qk
i=1(fiz(z))α1 Rz
0
Qk
i=1(fi(s)s )α1ds. Let
(6) M =zF0(z), N(z) =F(z)
From (5) and (6) we have:
M0(z)
N0(z) = 1 + zF00(z)
F0(z) , M0(z) N0(z) = 1 +
Pk i=1 1
α(zff(z)i0(z)−1) Qk
i=1(fiz(z))α1
|M0(z)
N0(z) −1|= |Pk i=1 1
α(zff(z)i0(z)−1)|
|Qk
i=1(fiz(z))α1| ≤ Pk
i=1|α1||zff(z)i0(z)−1|
|Qk
i=1(fiz(z))α1| .
By hypothesis fi ∈ S∗. This means that |zff(z)i0(z)−1| <1, which implies that
|MN00(z)(z) −1|<1. Thus Re{MN00(z)(z)}>0 and by Theorem 1,Re{MN(z)(z)}>0. This implies thatRe{zFF(z)0(z)}>0. Hence F ∈S∗.
Remark 1 The integral in (5) is equivalent to that in (4) of section 2 with β = 1.
Let S={f :U →C} ∩S. LetF(z)∈U be defined by
(7) F(z) =
Z z 0
Yk i=1
(fi(s) s )α1ds.
Theorem 5 Letz∈U, α∈C, Reα >0andmα=Mα∩s.IfF ∈mα, then F ∈ S∗that is mα⊂S∗.
Proof. From (6) above, we have F(z)Fz0(z) 6= 0 and forF ∈mα,we have (8) ReJ(α, f;z) =Re{(1−α)zF0(z)
F(z) +α(1 + zF0(z) F(z) )}
forp(z) = zFF(z)0(z), zpp(z)0(z) = 1 +zFF0”(z)(z) −p(z).This implies that
(9) 1 +zF00(z)
F0(z) = zp0(z)
p(z) +p(z) using (7) and (9) in (8), we obtain
(10) ReJ(α, f;z) =Re{(1−α)p(z) +α(zp0(z)
p(z) +p(z))}.
Simplifying (10), we obtainReJ(α, f;z) =Re{p(z) +α(zpp(z)0(z))}
p(0) + αzpp(0)0(0) = 1 and p(0) = h(0) = 1. Thus, using Theorem 3 with β = 1 andγ = 0,we havep(z)+αzpp(z)0(z) < h(z) = 1+z1−z. This implies thatp(z)≺h(z).
That isRe{p(z)}>0. Thus, Re{zFF(z)0(z)}>0. Hence, F ∈S∗.
References
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D. O. Makinde
Department of Mathematics Obafemi Awolowo University Ile Ife 220005, Nigeria
e-mail: [email protected], [email protected] Timothy O. Opoola
University of Ilorin Mathematics
Department of Mathematics, University of Ilorin, Ilorin,Nigeria e-mail: [email protected]