General Mathematics Vol. 17, No. 3 (2009), 125–133
On sufficient conditions for starlikeness of
p-valently Bazilevic˘ functions of the type β
and order γ 1
V.B.L Chaurasia, Vishal Saxena
Abstract
The aim of this paper is to establish certain sufficient conditions
for the class of starlikeness p-valently Bazilevic˘functions of the type
β and order γ. Our result is of general nature and capable of yielding
a number of unknown new and interesting results.
2000 Mathematics Subject Classification:30C45.
Key words and phrases: Multivalent functions, Bazilevic˘functions,
Starlikeness and convexity.
1
Received 15 September, 2008
Accepted for publication (in revised form) 10 October, 2008
125
126
1
V.B.L Chaurasia, V. Saxena
Introduction and definitions
Let An (p) be the class of normalized functions of the form
p
(1)
f (z) = z +
∞
X
ak z k
(n ∈ N := {1, 2, 3, ....}),
k=n+p
which are analytic in the unit disk ∆ := {z : |z| < 1}. A function f ∈ An (p)
is said to be in the class Sn∗ (p, α), if it satisfies
½ 0 ¾
zf (z)
(2)
Re
>α
(0 ≤ α < p; z ∈ ∆).
f (z)
A function in the class Sn∗ (α) is starlike of order α in ∆.
Let Kn (α) be subclass of An (p) consisting of functions f (z) which
satisfies
½
(3)
Re
zf 0 (z)
g(z)
¾
>α
(0 ≤ α < p; z ∈ ∆),
where g(z) ∈ Sn (p, 0) := Sn (p).
Also a function f ∈ An is said to be p-valently Bazilevic˘ function
of type β (β ≥ 0) and order γ (0 ≤ γ < p), if there exists a function
g(z) ∈ Sn∗ (p), such that
¾
½
zf 0 (z)
>γ
(4)
Re
[f (z)]1−β [g(z)]β
(0 ≤ γ < p; z ∈ ∆).
We denote the class of all such functions by Bn (β, γ). In particular,
when β = 1, a function f ∈ Kn (γ) := Bn (1, γ) is p-valently close-to-convex
of order γ in ∆. Moreover if β = 0, then Bn (0, γ) := Sn∗ (γ).
In order to prove our main results , we shall require the following lemma.
On sufficient conditions for starlikeness...
127
Lemma 1 ([3]) Let Ω be a set in the complex plane C and suppose that φ
is a mapping from C 2 × 4 to C which satisfies φ(ix; y; z) 6∈ Ω for z ∈ 4
and for all real x and y such that y ≥ −n(1 + x2 )/2. If the function P (z) =
1 + cn z n + ... , is analytic in 4 and φ(P (z); zp(z); z) ∈ Ω all z ∈ ∆ then
Re P (z) > 0.
2
Main Results
Applying Lemma 1, we now derive the following
Theorem 1 Let f (z) ∈ An (p), satisfies If
½
¾
0
00
0
00
0
αzf (z)
αzf (z)
zf (z)
zf (z)
zg (z)
Re
+ 0
−α (1−µ) 0
−αµ
+1
f (z)
f (z)
g (z)
[f (z)]1−µ [g (z)]µ
[f (z)]1−µ [g (z)]µ
(5)
³
´ ³
n
npα ´
> αβ βδ + − 1 + β −
(z ∈ 4, 0 ≤ α, β < p) ,
2
2
then f (z) ∈ Bn (p, µ, β).
Proof. Define P(z) by
0
zf (z)
= (p − β) P (z) + β,
[f (z)]1−µ [g (z)]µ
(6)
then P (z) = 1 + cn z n + ... is analytic in 4. Differentiate logarithmically (6)
with respect to z and making a little simplification, we get
00
0
0
0
zf (z)
zf (z)
zg (z)
δzf (z)
− (1 − µ)
−µ
0
1−µ
µ +
f (z)
f (z)
g (z)
[f (z)]
[g (z)]
(p − β) zP (z) + δ [(p − β) P (z) + β]2
=
(p − β) P (z) + β
0
(7)
128
V.B.L Chaurasia, V. Saxena
·
¸
0
00
0
0
0
αδzf (z)
zf (z)
zf (z)
zg (z)
zf (z)
+α
−α
(1−µ)
−αµ
+1
f 0 (z)
f (z)
g (z)
[f (z)]1−µ [g (z)]µ
[f (z)]1−µ [g (z)]µ
¡
¢
0
= α (p−β) zP (z)+(2βαδ−α+1) (p−β) P (z)+αδ (p−β)2 P 2 (z)+ δαβ 2 −αβ +β
³
´
0
= φ P (z), zP (z); z ,
(8)
where
(9)
¡
¢
φ(r, s; t) = α (p−β) δ+(2βαδ−α+1) (p−β) r+αδ (p−β)2 r2+ δαβ 2 −αβ +β
for all real x and y satisfying y = −n(1 + x2 )/2, we have
Reφ (ix, y; z) = α (p − β) y − αδ (p − β)2 x2 + β (αβδ − α + 1)
n
o
α
nα
≤ − (p − β) n − αδ (p − β)2 +
(p − β) x2 + β (αβδ − α + 1)
2
2
α
α (p − β)
= − (p − β) n −
(n + 2pδ − 2β) x2 + β (αβδ − α + 1)
2
2
(10)
Let
then
³
´ ³
n
npα ´
.
≤ αβ βδ + − 1 + β −
2
2
´ ³
n
³
n
npα ´o
,
Ω = w; Rew > 0, αβ βδ + − 1 + β −
2
2
³
´
0
φ P (z), zP (z); z ∈ Ω and φ (ix, y; z) ∈
/ Ω.
For all real x and y = −n(1 + x2 )/2, z ∈ 4, By application of lemma (1),
the result (5) follows at once.
On taking δ = 1, µ = 0 in Theorem 1, we get
On sufficient conditions for starlikeness...
129
Corollary 1 If f (z) ∈ An (p) satisfies
½ 0
µ
¶¾
00
³
´ ³
zf (z)
n
f (z)
npα ´
(11) Re
αz 0
+1
> αβ β + − 1 + β +
,
f (z)
f (z)
2
2
(z ∈ 4, 0 ≤ α, 0 ≤ β < p) then f (z) ∈ Sn (p, β).
On taking β = 0, n = 1,p = 1 and β = α/2, n = 1, p = 1 respectively we
have a known result obtained by Ravichandran et al. [1]. On taking µ = 1
and δ = 1, we find an interesting result contained in the following corollary.
Corollary 2 If f (z) ∈ An (p), satisfies
½
¾ 0
0
00
0
αzf (z) αzf (z)
zf (z)
zg (z)
Re
+
−α
+1
0
g (z)
f (z)
g (z)
g (z)
(12)
³
´ ³
n
npα ´
> αβ β + − 1 + β −
2
2
(z ∈ ∆, 0 ≤ α, β < p) ,
then f (z) ∈ Kn (p, β).
Theorem 2 Let 0 ≤ β < p ,
³
nn
¡
¢o2
n ´2
λ= p−β+
(p − β)2 , η =
(p − β) − δβ 2 − β
2
2
¡
¢
v = (p − β)2 + δβ 2 − β and σ = (p − β)2 (2β − 1)2
Also, suppose that t0 to be the smallest positive root of the equation
(13)
and
©
ª
2λ (p−β)2 t3 + (p−β)2 (2λ+η−v+σ)+3λβ 2 t2
+2β 2 (2λ+η−v+σ) t+(λ+2η−v+σ)β 2 −(p−β)2 η = 0
¤
(p − β)2 (1 + t0 ) £ 2
+
(λ
+
η
−
v
+
σ)
t
+
η
= ρ2 .
λt
0
0
2
2
(p − β) t0 + β
130
V.B.L Chaurasia, V. Saxena
Now, if
¯·
¸·
¸¯
0
0
0
0
00
¯
δzf (z)
zf (z)
zg (z) ¯¯
δzf (z)
zf (z)
¯
¯ [f (z)]1−µ [g (z)]µ −p [f (z)]1−µ [g (z)]µ + f 0 (z) −(1−µ) f (z) −µ g (z) ¯ ≤ ρ, z ∈ ∆
then f (z) ∈ Bn (p, µ, β).
Proof. Define P(z) by
·
0
zf (z)
(p − β) P (z) + β =
[f (z)]1−µ [g (z)]µ
¸
then P (z) = 1 + cn z n + ..., is analytic in 4. A computation shows that
0
δzf (z)
[f (z)]1−µ [g(z)]µ
00
0
0
(z)
zf (z)
(z)
− µ zgg(z)
− (1 − µ) zff (z)
0
f (z)
0
(z)+β}2 −{(p−β)P (z)+β}
= (p−β)zP (z)+δ{(p−β)P
(p−β)P (z)+β
+
³
´
0
= ϕ P (z), zP (z); z ,
where
ϕ (r, s; t) =
¤
(p − β) (r − 1) £
(p − β) s + δ {(p − β) r + β}2 − {(p − β) r + β}
(p − β) r + β
For all real x and y = −n(1 + x2 )/2, we have
ª2
(p−β)2 (1+x2) h©
2 2
2
|ϕ (ix, y; z)| =
(p−β)
y−(p−β)
x
δ+δβ
−β
(p−β)2 x2 +β 2
¤
¡
¢
+(p−β)2 x2 (2δβ −1)2 = g x2 , y
2
Now
¤
∂g
2 (p − β)2 (1 + x2 ) £
2 2
2
(p
−
β)
y
−
(p
−
β)
x
δ
+
δβ
−
β
<0
=
∂y
(p − β)2 x2 + β 2
then we have
h(t) = g(t, −n(1 + t)/2) ≤ g(t, y)
On sufficient conditions for starlikeness...
131
where
h
n
¢
¡
¢2 ¡
¢2
2¡
(p−β)2 (1+t)
n 2 2
h(t) = (p−β)
(p−β)
p−β
+
t
+
(p−β)2 p−β + n2 + n2 (p−β)−(δβ 2 −β)
2
2
t+β 2
o
¡
¢2
¡
¢2 i
− (p − β)2 + (δβ 2 − β) + (p − β)2 (2β − 1)2 t + n2 (p − β) − (δβ 2 − β)
=
(p−β)2 (1+t)
(p−β)2 t+β 2
[λt2 + (λ + η − v + σ) t + η] ,
where
¡
¢2
λ = p − β + n2 (p − β)2 ,
η=
©n
2
ª2
(p − β) − (δβ 2 − β)
and
©
ª2
υ = (p − β)2 + (δβ 2 − β) and σ = (p − β)2 (2β − 1)2
Now
0
h (t) = ©
(p − β)2
(p − β)2 t + β 2
£
©
ª 2
2
2
3
2
ª2 2 (p − β) λt + (p − β) (2λ + η − v + σ) + 3β λ t
+2β 2 (2λ + η − v + σ) t + (λ + 2η − v + σ) β 2 − (p − β)2 η
¤
Taking h(t) = 0, we obtain (13) which is cubic in t. Let t0 be the smallest
positive root of the equation, then we have h(t) = h(t0 ), and hence
|ϕ (ix, y; z)|2 ≥ h (t0 ) = ρ2
Define Ω = w : |w| < ρ then φ(P (z), zP (z); z) ∈ Ω for all real x and
y = −n(1 + x2 )/2, z ∈ ∆. Therefore by application of Lemma 1 the result
follows On taking β = 0 and µ = 1,δ = 1, n = 1, p = 1 in Theorem 1, we
obtain the following interesting result.
132
V.B.L Chaurasia, V. Saxena
Corollary 3 If f (z) ∈ A1 (p, 0) = A1 (p) satisfies
¯µ 0
¶µ 0
¶¯
00
0
¯ zf (z)
zf (z) zf (z) zg (z) ¯¯
¯
+ 0
−
≤ ρ, (z ∈ ∆)
¯ g (z) − p
g (z)
f (z)
g (z) ¯
where
·
¸
(1 + t0 ) 9 2 5
1
ρ =
t + t0 +
t0
4 0 2
4
2
t0 is the smallest positive root by the equation
3t3 +
19 2 1
t − =0
4
4
then f (z) ∈ K1 (p, 0).
Remark 1 On taking β = 0 and µ = 0, δ = 1, n = 1, p = 1 in Theorem ,
we obtain a known result due to Ravichandran et. Al. [1].
References
[1] V. Ravichandran, C. Selvaraj, Rajagopal Rajalakshmi, Sufficient Conditions for Starlike Functions of Order α, Journal Inq. Pure Appl.
Math., Volume 3, Article 81 [2003], 471-476.
[2] H. Irmak, K. Piejko and J. Stankiewicz, A note involving p-valently
Bazilevic˘ functions, Int. J. Math. Math. Sci., Vol. 2005 (2005), 11491153.
On sufficient conditions for starlikeness...
133
[3] S.S. Miller and P.T. Mocanu, Differential subordinations and inequalities in the complex plane, J. Differ. Equations, 67 (1987), 199211.
V.B.L Chaurasia, Vishal Saxena
Department of Mathematics
University of Rajasthan
Jaipur - 302055, India
e-mail: saxenavishal13@rediffmail.com