General Mathematics Vol. 18, No. 3 (2010), 35–39
On sufficient condition for starlikeness
1
D. O. Makinde, T. O. Opoola
Abstract
In this paper,we give a condition for starlikeness of the integral oper1
Z zY
k fı (s) α
ator of the form F (z) =
ds.
s
0 ı=1
2000 Mathematics Subject Classification:30C45
Key words and phrases: Starlike, Convex, Briot-Bouquet subordination.
1
Introduction
Let A be the class of all analytic functions f (z) defined in the open unit disk
U = {z ∈ C :| z |< 1} and S the subclass of A consisting of univalent functions
(1)
f (z) = z +
∞
X
ak z k
k=2
S ∗ = {f ∈ S : Re(
Mα = {f ∈ S :
zf 0 (z)
) > 0, z ∈ U },
f (z)
f (z)f 0 (z)
6= 0, ReJ(α, f ; z) > 0, z ∈ U }
z
1
Received 14 February, 2009
Accepted for publication (in revised form) 17 March, 2009
35
36
D. O. Makinde, T. O. Opoola
00
0
(z)
(z)
+ α(1 + zff 0 (z)
) be the class of starlike and
where J(α, f ; z) = (1 − α) zff (z)
α − convex functions respectively.
Let p(z) be the class of functions that are regular in U and of the form :
(2)
p(z) = 1 +
∞
X
bk z k
k=1
1+z
1−z .
Furthermore, let h(z) =
Let T be the univalent [5] subclass of A consisting of functions f (z) satisfying
2 0
| zf f(z)(z)
2 − 1| < 1, (z ∈ U )
Let Tn be the subclass of T for which f k (0) = 0 (k = 2, 3, ..., n).
Let Tn,µ be the subclass of Tn consisting of functions of the form
R z Qk fı (s) 1
z 2 f 0 (z)
α
ı=1 ( s ) ds satisfying: | f (z)2 − 1| < µ, (z ∈ U ) for some µ(0 < µ ≤ 1).
0
2
Preliminaries
Theorem 1 [1] Let M and N be analytic in U with M (0) = N (0) = 0. If N (z)
maps onto a many sheeted region which is starlike with respect to the origin
0 (z)
M (z)
and Re{ M
0
N (z) } > 0 in U , then Re{ N (z) } > 0 in U .
Theorem 2 [6] Let fı ∈ Tn,µı (ı = 1, 2, ..., k; k ∈ N ∗ ) be defined by
(3)
fi (z) = z +
∞
X
ain z n
n=2
for all i = 1, 2, ..., k; α, β ∈ C; R{β} ≥ γ and γ =
k
X
1 + (1 + µi )M
(M ≥
|α|
i=1
1, 0 < µi < 1, k ∈ N ∗ ). If |fi (z)| ≤ M (z ∈ U ), i = 1, 2, ..., k then, the integral
operator
Z z
k
Y
1
fi (t) 1
β−1
t
(4)
Fα,β (z) = {β
(
) α dt} β
t
0
i=1
is univalent.
Theorem 3 [2] Let h be convex in U and Re{βh(z) + γ} > 0, z ∈ U.If p ∈
H(U ) where H(U ) is the class of functions which are analytic in the unit disk,
with p(0) = h(0) and p satisfies the Briot-Bouquet differential subordinations:
zp0 (z)
p(z) + βp(z)+γ
≺ h(z), z ∈ U. Then, p(z) ≺ h(z), z ∈ U.
On sufficient condition for starlikeness
3
37
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
k
zY
0 i=1
(
fi (s) 1
) α ds, α ∈ C.
s
If fi ∈ S ∗ then, F (z) ∈ S ∗ where fi is as in equation (3) above.
Proof. By differentiating (5), we obtain: F 0 (z) =
Qk fi (z) 1
( z )α
zF 0 (z)
.
= R z Qi=1
fi (s) 1
k
F (z)
α ds
(
)
i=1
s
0
Let
Qk
i=1 (
fi (z) 1
α
z ) .
Thus,
M = zF 0 (z), N (z) = F (z)
(6)
From (5) and (6) we have:
M 0 (z)
zF 00 (z) M 0 (z)
=
1
+
,
=1+
N 0 (z)
F 0 (z) N 0 (z)
Pk
0
1 zfi (z)
i=1 α ( f (z) −
Qk fi (z) 1
α
i=1 ( z )
1)
0
P
Pk
zf 0 (z)
1 zfi (z)
| ki=1 α1 ( f i(z) − 1)|
M 0 (z)
i=1 | α || f (z) − 1|
| 0
− 1| =
≤
.
Q
Q
1
1
N (z)
| ki=1 ( fiz(z) ) α |
| ki=1 ( fiz(z) ) α |
By hypothesis fi ∈ S ∗ . This means that |
0 (z)
|M
N 0 (z)
− 1| < 1. Thus
0 (z)
Re{ M
N 0 (z) }
zfi0 (z)
f (z)
− 1| < 1, which implies that
(z)
> 0 and by Theorem 1, Re{ M
N (z) } > 0. This
0
(z)
∗
implies that Re{ zF
F (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. Let F (z) ∈ U be defined by
(7)
F (z) =
Z
k
zY
0 i=1
(
fi (s) 1
) α ds.
s
38
D. O. Makinde, T. O. Opoola
Theorem 5 Let z ∈ U, α ∈ C, Reα > 0 and mα = Mα ∩s. If F ∈ mα , then F ∈
S ∗ that is mα ⊂ S ∗ .
Proof. From (6) above, we have
(8)
F (z)F 0 (z)
z
ReJ(α, f ; z) = Re{(1 − α)
for p(z) =
zF 0 (z) zp0 (z)
F (z) , p(z)
(9)
zF ”(z)
F 0 (z)
=1+
1+
6= 0 and for F ∈ mα , we have
zF 0 (z)
zF 0 (z)
+ α(1 +
)}
F (z)
F (z)
− p(z). This implies that
zF 00 (z)
zp0 (z)
=
+ p(z)
F 0 (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)
0
(z)
Simplifying (10), we obtain ReJ(α, f ; z) = Re{p(z) + α( zpp(z)
)}
p(0) +
αzp0 (0)
p(0)
= 1 and p(0) = h(0) = 1. Thus, using Theorem 3 with β =
0
(z)
1 and γ = 0,we have p(z)+ αzp
p(z) < h(z) =
That is Re{p(z)} > 0. Thus,
0 (z)
Re{ zF
F (z) }
1+z
1−z .
This implies that p(z) ≺ h(z).
> 0. Hence, F ∈ S ∗ .
References
[1] S.A. Halim, On a class of Analytic Functions involing, Tamkang, Journal
of Mathematics vol.23 Number 1, Spring 1992.
[2] M. Acu and S. Owa, On some subclasses of univalent function, Journal
of inequalities in pure and applied mathematics, vol. 6,(3) art 70, 2005.
[3] S. Kanas and F. Ronning, Uniformly Starlike and convex functions and
other related classes of univalent functions, Ann University, Marie CurieSklodowska Section A, 53, 1999, 95-105.
[4] S.S. Miller and P.T. Mocanu, Univalent solutions of Briot-Bouquet differential equations, Journal of Differential equations, 56, 1985, 297-309.
[5] S. Ozaki and M.Nunokawa, The Schwarzian derivative and Univalent
functions,, Pro. Ameri.Math. Soc., 33, 1972, 302-394.
On sufficient condition for starlikeness
39
[6] N. Seenivasagan, Sufficient Conditions for Univalence, Applied Mathematics E-Notes, 8, 2008, 30-35.
D. O. Makinde
Department of Mathematics
Obafemi Awolowo University
Ile Ife 220005, Nigeria
e-mail: dmakinde@oauife.edu.ng, makindemyiuv@yahoo.com
Timothy O. Opoola
University of Ilorin
Mathematics
Department of Mathematics, University of Ilorin, Ilorin,Nigeria
e-mail: opoolato@unilorin.edu.ng