General Mathematics Vol. 15, No. 4 (2007), 83-100
Argument Estimates of certain Multivalent
Functions involving Dziok–Srivastava
Operator1
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
Abstract
The purpose of this paper is to derive some argument properties
using multivalent functions in the open unit disc involving DziokSrivastava operator. We also investigate their integral preserving
property.
2000 Mathematical Subject Classification: 30C45
Key words:Strongly starlike functions, subordination, univalent functions
1
Introduction
Let Ap be the class of functions of the form
(1.1)
p
f (z) = z +
∞
X
an+p z n+p
(p ∈ N = {1, 2, 3, ...})
n=1
which are analytic in the open unit disc U := {z ∈ C : |z| < 1}. A function
f ∈ Ap is said to be p−valently starlike of order α in U, if it satisfies
½ ′ ¾
zf (z)
(1.2)
Re
> α (0 ≤ α < p; z ∈ U).
f (z)
1
Received 13 August, 2007
Accepted for publication (in revised form) 4 December, 2007
83
84
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
The class of all p-valently starlike functions of order α is denoted by Sp∗ (α).
A function f ∈ Ap is said to be p−valently convex of order α in U, if it
satisfies
½
¾
zf ′′ (z)
(1.3)
Re 1 + ′
> α (0 ≤ α < p; z ∈ U).
f (z)
The class of all p-valently convex functions of order α is denoted by Kp (α).
It follows from (1.2) and (1.3) that
(1.4)
f ∈ Kp (α) is equivalent with zf ′ ∈ Sp∗ (α).
Further, a function f ∈ Ap is said to be a p-valently close to convex function
of order β and type α, if there exists a function g ∈ Sp∗ (α) such that
½ ′ ¾
zf (z)
(1.5)
Re
> β (0 ≤ α, β < p; z ∈ U).
g(z)
It is well known ( see [13]) that every p- valently close-to-convex function is
p -valent in U. For arbitrary fixed real numbers A and B (−1 ≤ B < A ≤ 1),
let P(A, B) denote the class of functions of the form
φ(z) = 1 + c1 z + c2 z 2 + ...
(1.6)
which are analytic in U and satisfies the condition
(1.7)
φ(z) ≺
1 + Az
1 + Bz
(z ∈ U),
where the symbol ≺ stands for subordination. The class P(A, B) was introduced and studied by Janowski [11].
We note that a function φ ∈ P(A, B), if and only if
¯
¯
¯
¯ A−B
1
−
AB
¯φ(z) −
¯<
(1.8)
(B 6= −1, z ∈ U).
¯
1 − B2 ¯ 1 − B2
For any φ ∈ P(A, B),
(1.9)
Re {φ(z)} >
1−A
2
(B 6= −1, z ∈ U).
Argument Estimates of certain ...
85
For a function f ∈ Ap , given by (1.1), the generalized Bernardi-LiberaLivingston integral operator F [1] is defined by
(1.10)
γ+p
F (z) =
zγ
Zz
tγ−1 f (t)dt (γ > −p;
z ∈ U).
0
A simple computation shows that
(1.11)
p
F (z) = z +
∞
X
n=1
γ+p
an+p z n+p
γ+p+n
(γ > −p;
z ∈ U).
It readily follows from (1.11) that
(1.12)
f ∈ Ap implies F ∈ Ap .
For any two analytic functions f (z) =
∞
P
an z n and g(z) =
∞
P
bn z n , the
n=0
n=0
Hadamard product or convolution of f (z) and g(z), written as (f ∗ g)(z) is
defined by
∞
X
(f ∗ g)(z) =
an b n z n .
n=0
For complex parameters α1 , α2 , ..., αq and β1 , β2 , ..., βs with
(βj 6= 0, −1, −2, ...; j = 1, 2, ..., s), we define the generalized hypergeometric
function q Fs (z) by
(1.13)
q Fs (α1 , α2 , ..., αq ; β1 , β2 , ..., βs ; z)
=
∞
X
n=0
(α1 )n (α2 )n ...(αq )n
(β1 )n (β2 )n ...(βs )n (1)n
zn
( q ≤ s + 1; q, s ∈ N0 = N ∪ {0} ; z ∈ U)
where (λ)n is the Pochhammer symbol defined by
(
1
for n = 0
.
(1.14)
(λ)n =
λ (λ + 1) ... (λ + n − 1) for n = 1, 2, 3...
Corresponding to a function hp (α1 , α2 , ...αq ; β1 , β2 , ...βs ; z) defined by
hp (α1 , α2 , ...αq ; β1 , β2 , ...βs ; z) = z p q Fs (α1 , α2 , ..., αq ; β1 , β2 , ..., βs ; z),
86
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
we consider the Dziok–Srivastava operator [7]
Hp (α1 , α2 , ..., αq ; β1 , β2 , ..., βs )f (z) : Ap −→ Ap ,
defined by the convolution
Hp (α1 , α2 , ...αq ; β1 , β2 , ...βs )f (z) = hp (α1 , α2 , ...αq ; β1 , β2 , ...βs ; z) ∗ f (z).
We observe that, for a function f of the form (1.1), we have
(1.15)
p
Hp (α1 , α2 , ..., αq ; β1 , β2 , ..., βs )f (z) = z +
∞
X
Γn a n z n
n=k
where
(1.16)
Γn =
(α1 )n−p (α2 )n−p , ..., (αq )n−p
.
(β1 )n−p (β2 )n−p , ..., (βs )n−p (1)n−p
For convenience, we write
(1.17)
Hp (α1 , α2 , ..., αq ; β1 , β2 , ..., βs ) := Hp,q,s (α1 )
Thus, through a simple calculations, we obtain
(1.18) z (Hp,q,s (α1 )f (z))′ = α1 Hp,q,s (α1 + 1)f (z) − (α1 − p)Hp,q,s (α1 )f (z).
The Dziok–Srivastava operator Hp (α1 , α2 , ..., αq ; β1 , β2 , ..., βs ) includes various other linear operators which were considered in earlier works in the
literature. In particular, for p = 2, q = 1, the Dziok–Srivastava operator
reduces to the operator Lp (α1 , α2 ; β1 )f (z), studied by Saitoh and Nunokawa
[19]. For p = s = 1 and q = 2, we obtain the linear operator:
F(α1 , α2 ; β1 )f (z) = H1 (α1 , α2 ; β1 )f (z),
which was introduced by Hohlov [9]. Moreover, putting α2 = 1, we obtain
the Carlson-Shaffer operator [3]:
L(α1 , β1 )f (z) = H1 (α1 , 1; β1 )f (z).
Argument Estimates of certain ...
87
Ruscheweyh [18] introduced an operator
(1.19)
Dλ f (z) =
z
∗ f (z) (λ ≥ −1; f ∈ Ap ).
(1 − z)λ
From the equation (1.18), we have
Dλ f (z) = H1 (λ + 1, 1; 1)f (z).
(1.20)
A detailed investigation on argument estimates using Lp (a, c) was discussed
by Cho et al. [6]. In this paper, motivated by the work of Cho et al.
[6], we give some argument properties of function in certain subclasses of
Ap involving the Dziok–Srivastava operator Hp,q,s (α1 ). An application of
certain integral operator is also considered . The results obtained here,
besides extending the works of Bulboacă [2], Nunokawa [16], Chichra [4],
Libera [12] and Sakaguchi [20], also yields a number of new results.
2
Main Results
To establish the main results we need the following lemmas.
Lemma 2.1. [14]. Let h(z) be convex univalent in U and let ψ(z) be analytic
in U with Re {ψ(z)} ≥ 0. If φ(z) is analytic in U and ψ(0) = φ(0), then
φ(z) + ψ(z)zφ′ (z) ≺ h(z) (z ∈ U)
(2.1)
implies
(2.2)
φ(z) ≺ h(z) (z ∈ U).
Lemma 2.2. [17]. Let p(z) be analytic in U with p(0) = 1 and p(z) 6= 0.
If there exists two points z1 , z2 ∈ U such that
(2.3)
−
η2 π
η1 π
= arg p(z1 ) < arg p(z) < arg p(z2 ) =
2
2
88
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
for η1 > 0, η2 > 0, and for |z| < |z1 | = |z2 |, then we have
η1 + η2
z1 p′ (z1 )
= −i
m
p(z1 )
2
(2.4)
and
z2 p′ (z2 )
η1 + η2
=i
m
p(z2 )
2
(2.5)
where
1 − |a|
m≥
1 + |a|
(2.6)
and a := i tan
µ
π η1 − η2
4 η1 + η2
¶
.
Theorem 2.3. Let α1 > 0, −1 ≤ B < A ≤ 1, f ∈ Ap , and suppose that
g ∈ Ap satisfies
1 + Az
Hp,q,s (α1 + 1)g(z)
≺
Hp,q,s (α1 )g(z)
1 + Bz
(2.7)
(z ∈ U).
If
(2.8)
½
¾
Hp,q,s (α1 )f (z)
Hp,q,s (α1 + 1)f (z)
π
π
− δ1 < arg (1 − λ)
+λ
− β < δ2
2
Hp,q,s (α1 )g(z)
Hp,q,s (α1 + 1)g(z)
2
(λ ≥ 0;
0 ≤ β < 1;
0 < δ1 , δ2 ≤ 1;
z ∈ U),
then
(2.9)
π
− η1 < arg
2
½
Hp,q,s (α1 )f (z)
−β
Hp,q,s (α1 )g(z)
¾
<
π
η2
2
(z ∈ U)
where η1 (0 < η1 ≤ 1) and η2 (0 < η2 ≤ 1) are the solutions of the equations
(2.10)(
i
h
λ(η1 +η2 )(1−|a|) sin π2 (1−t1 )
, f or B 6= −1
η1 + π2 tan−1 2α1 (1+A)/(1+B)(1+|a|)+λ(η
π
1 +η2 )(1−|a|) cos 2 (1−t1 )
δ1 =
η1 , for B = −1
(2.11)(
δ2 =
η2 +
2
π
tan−1
h
λ(η1 +η2 )(1−|a|) sin π2 (1−t1 )
2α1 (1+A)/(1+B)(1+|a|)+λ(η1 +η2 )(1−|a|) cos
η2 , for B = −1
π
(1−t1 )
2
i
, f or B 6= −1
Argument Estimates of certain ...
89
and t1 is given by
(2.12)
2
t1 = sin−1
π
½
A−B
1 − AB
¾
.
Proof. Let
(2.13)
Hp,q,s (α1 )f (z)
= β + (1 − β)φ(z).
Hp,q,s (α1 )g(z)
Then the function is analytic φ(z) analytic in U with φ(0) = 1. On differentiating with respect to z both sides of (2.13) and using the identity (1.18),
we get
¾
½
Hp,q,s (α1 + 1)f (z)
Hp,q,s (α1 )f (z)
+λ
−β =
(2.14)
(1 − λ)
Hp,q,s (α1 )g(z)
Hp,q,s (α1 + 1)g(z)
n
o
′ (z)
= (1 − β) φ(z) + λzφ
α1 r(z)
where,
(2.15)
r(z) =
Hp,q,s (α1 + 1)g(z)
Hp,q,s (α1 )g(z)
Let
(2.16)
r(z) = ρe(πθ/2)i
then from (2.7) followed by (1.8) and (1.9), it follows that
1−A
1+A
<ρ<
,
1−B
1+B
−t1 < θ < t1 for B 6= −1,
where t1 is given by (2.12), and
1−A
< ρ < ∞,
2
(2.17)
−1 < θ < 1 for B 6= −1.
90
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
Let h(z) be the function which maps the open unit disc U onto the angular
©
ª
domain ω : − π2 δ1 < arg(ω) < − π2 δ2 with h(0) = 1. Applying Lemma 2.1
λ
for this h(z) with ψ(z) = αr(z)
, we see that Re (φ(z)) > 0 in U and hence
φ(z) 6= 0 in U. Suppose there exists points z1 and z2 such that (2.3) is satisfied. Then by Lemma 2.2, we obtain (2.4) and (2.5) with the restrictions
(2.6). For the case B 6= −1, we have
¾
½
Hp,q,s (α1 + 1)f (z1 )
Hp,q,s (α1 )f (z1 )
+λ
−β
arg (1 − λ)
Hp,q,s (α1 )g(z1 )
Hp,q,s (α1 + 1)g(z1 )
½
¾
λ z1 φ′ (z1
= arg φ(z1 ) + arg 1 +
α1 r(z1 ) φ(z1
"
#
−πθ
λei( 2 ) η1 + η2
−π
η1 + arg 1 − i
m
=
2
ρα1
2
¸
·
λ(η1 + η2 )m sin π2 (1 − θ))
−π
−π
−1
≤
≤
η1 − tan
η1 −
π
2
2α1 ρ + λ(η1 + η2 )m cos 2 (1 − θ)
2
¸
·
λ(η1 + η2 )(1 − |a|) sin π2 (1 − t1 )
−1
− tan
2α1 (1 + A)/(1 + B)(1 + |a|) + λ(η1 + η2 )(1 − |a|) cos π2 (1 − t1 )
π
= − δ1
2
and½
¾
Hp,q,s (α1 + 1)f (z2 )
Hp,q,s (α1 )f (z2 )
+λ
−β
arg (1 − λ)
Hp,q,s (α1 )g(z2 )
Hp,q,s (α1 + 1)g(z2 )
½
¾
λ z2 φ′ (z2
π
= arg φ(z2 ) + arg 1 +
≥ η2 +
α1 r(z2 ) φ(z2
2
·
¸
λ(η1 + η2 )(1 − |a|) sin π2 (1 − t1 )
−1
+ tan
2α1 (1 + A)/(1 + B)(1 + |a|) + λ(η1 + η2 )(1 − |a|) cos π2 (1 − t1 )
π
δ2 .
=
2
Similarly for the case B = –1, we have
¾
½
Hp,q,s (α1 + 1)f (z)
π
Hp,q,s (α1 )f (z)
+λ
− β ≤ − δ1
arg (1 − λ)
Hp,q,s (α1 )g(z)
Hp,q,s (α1 + 1)g(z)
2
Argument Estimates of certain ...
91
and
¾
½
Hp,q,s (α1 + 1)f (z2 )
π
Hp,q,s (α1 )f (z2 )
+λ
− β ≥ δ2
arg (1 − λ)
Hp,q,s (α1 )g(z2 )
Hp,q,s (α1 + 1)g(z2 )
2
where we have used the inequality (2.6) with δ1 , δ2 and t1 are as given
in (2.10), (2.11) and (2.12) respectively. These obviously contradict the
assumption of Theorem 2.3. The proof of Theorem 2.3 is thus completed.
If we let δ1 = δ2 = δ in Theorem 2.3, we readily obtain the following.
Corollary 2.4. Let α1 > 0,
that g ∈ Ap satisfies
(2.18)
−1 ≤ B < A ≤ 1,
1 + Az
Hp,q,s (α1 + 1)g(z)
≺
Hp,q,s (α1 )g(z)
1 + Bz
f ∈ Ap , and suppose
(z ∈ U).
If
(2.19)
¯
¾¯
½
¯
¯ π
H
(α
+
1)f
(z)
H
(α
)f
(z)
p,q,s
1
p,q,s
1
¯arg (1 − λ)
¯< δ
+
λ
−
β
¯
¯ 2
Hp,q,s (α1 )g(z)
Hp,q,s (α1 + 1)g(z)
(λ ≥ 0;
0 ≤ β < 1;
0 < δ ≤ 1;
z ∈ U),
then
(2.20)
¯
½
¾¯
¯
¯
¯arg Hp,q,s (α1 )f (z) − β ¯ < π η
¯
¯ 2
Hp,q,s (α1 )g(z)
(z ∈ U)
where η (0 < η ≤ 1) is the solution of the equation

i
h
λη sin π2 (1−t1 )
2
−1


 η + π tan
α1 (1+A)/(1+B)+λη cos π2 (1−t1 )
(2.21)
δ=


 η
for B 6= −1
for B = −1
and t1 is given by (2.12).
For s = q = 1 in Theorem 2.3 , we have the following corollary.
92
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
Corollary 2.5. Let α1 > 0, −1 ≤ B < A ≤ 1, f ∈ Ap , and suppose that
g ∈ Ap satisfies
Lp (α1 + 1, β1 )f (z)
1 + Az
≺
Lp (α1 , β1 )g(z)
1 + Bz
(2.22)
(z ∈ U).
If
(2.23)
½
¾
π
Lp (α1 , β1 )f (z)
Lp (α1 + 1, β1 )f (z)
π
− δ1 < arg (1 − λ)
+λ
− β < δ2
2
Lp (α1 , β1 )g(z)
Lp (α1 + 1, β1 )g(z)
2
(λ ≥ 0;
0 ≤ β < 1;
0 < δ1 , δ2 ≤ 1;
z ∈ U),
then
π
− η1 < arg
2
(2.24)
½
Lp (α1 , β1 )f (z)
−β
Lp (α1 , β1 )g(z)
¾
<
π
η2
2
(z ∈ U)
where η1 (0 ≤ η1 ≤ 1) and η2 (0 ≤ η2 ≤ 1) are the solutions of the equations
(2.25)(
h
i
λ(η1 +η2 )(1−|a|) sin π2 (1−t1 )
η1 + π2 tan−1 2α1 (1+A)/(1+B)(1+|a|)+λ(η
, f orB 6= −1
π
1 +η2 )(1−|a|) cos 2 (1−t1 )
δ1 =
η1 , for B = −1
(2.26)(
δ2 =
η2 +
2
π
−1
tan
h
λ(η1 +η2 )(1−|a|) sin π2 (1−t1 )
2α1 (1+A)/(1+B)(1+|a|)+λ(η1 +η2 )(1−|a|) cos
π
(1−t1 )
2
η2 , for B = −1
i
, f or B 6= −1
and t1 is given by (2.12).
For δ1 = δ2 = δ,
s = q = 1 in Theorem 2.3, we get the following result.
Corollary 2.6. Let α1 > 0, −1 ≤ B < A ≤ 1, f ∈ Ap , and suppose that
g ∈ Ap satisfies
(2.27)
Lp (α1 + 1, β1 )f (z)
1 + Az
≺
Lp (α1 , β1 )g(z)
1 + Bz
(z ∈ U).
If
(2.28)
¯
½
¾¯
¯ π
¯
L
(α
,
β
)f
(z)
L
(α
+
1,
β
)f
(z)
p
1
1
p
1
1
¯arg (1 − λ)
+λ
− β ¯¯ < δ
¯
Lp (α1 , β1 )g(z)
Lp (α1 + 1, β1 )g(z)
2
Argument Estimates of certain ...
(λ ≥ 0;
0 ≤ β < 1;
93
0 < δ ≤ 1;
z ∈ U),
then
(2.29)
¯
¾¯
½
¯ π
¯
L
(α
,
β
)f
(z)
p
1
1
¯arg
− β ¯¯ < η
¯
Lp (α1 , β1 )g(z)
2
(z ∈ U)
where η (0 < η ≤ 1) is the solution of the equation

h
i
λη sin π2 (1−t1 )
2
−1


 η + π tan
α1 (1+A)/(1+B)(1+|a|)+λη cos π2 (1−t1 )
(2.30) δ =


 η
for B 6= −1
for B = −1
and t1 is given by (2.12).
Remark 2.7. For δ1 = δ2 , s = q = 1, α1 = β1 = p, A = 1, B = −1
and λ = 1 in Theorem 2.3, we get the result obtained by Nunokawa [16].
Taking s = q = 1, α1 = µ + p, β1 = 1,
Theorem 2.3, we have the following corollary.
A = 1, and B = 0 in
Corollary 2.8. Let −1 ≤ B < A ≤ 1, f ∈ Ap , and suppose that g ∈ Ap
satisfies
Dµ+p g(z)
≺1+z
Dµ+p−1 g(z)
(2.31)
(z ∈ U).
If
(2.32)
½
¾
Dµ+p−1 f (z)
Dµ+p f (z)
π
π
+ λ µ+p
− β < δ2
− δ1 < arg (1 − λ) µ+p−1
2
D
g(z)
D g(z)
2
(λ ≥ 0;
0 ≤ β < 1;
0 < δ1 , δ2 ≤ 1;
z ∈ U),
then
(2.33)
π
− η1 < arg
2
½
Dµ+p−1 f (z)
−β
Dµ+p−1 g(z)
¾
<
π
η2
2
(z ∈ U)
94
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
where η1 (0 < η1 ≤ 1) and η2 (0 < η2 ≤ 1) are the solutions of the equations
(2.34) 
i
h
λ(η1 +η2 )(1−|a|) sin π2 (1−t1 )
2
−1


η
+
tan
π
 1 π
2(µ+p)(1+|a|)+λ(η1 +η2 )(1−|a|) cos 2 (1−t1 )
δ1 =
for B 6= −1


 η
for B = −1
1
(2.35) 


 η2 +
δ2 =


 η
2
2
π
tan−1
h
λ(η1 +η2 )(1−|a|) sin π2 (1−t1 )
2(µ+p)(1+|a|)+λ(η1 +η2 )(1−|a|) cos π2 (1−t1 )
i
for B 6= −1
for B = −1
and t1 is given by
(2.36)
2
t1 = sin−1
π
½
A−B
1 − AB
¾
.
Letting B → A and g(z) = z p in Theorem 2.3, we get the following
corollary
Corollary 2.9. Let f ∈ Ap . If
(2.37)
½
¾
π
Hp,q,s (α1 )f (z)
Hp,q,s (α1 + 1)f (z)
π
− δ1 < arg (1 − λ)
+λ
− β < δ2
p
p
2
z
z
2
(2.38)
(α1 > 0,
λ ≥ 0;
0 ≤ β < 1;
0 < δ1 , δ2 ≤ 1;
z ∈ U),
then,
(2.39)
π
− η1 < arg
2
½
Hp,q,s (α1 )f (z)
−β
zp
¾
<
π
η2 (z ∈ U)
2
where η1 (0 < η1 ≤ 1) and η2 (0 < η2 ≤ 1) are the solutions of the equations
·
¸
2
−1 λ(η1 + η2 )(1 − |a|)
(2.40)
δ1 = η1 + tan
π
2α1 (1 + |a|)
and
(2.41)
·
¸
2
−1 λ(η1 + η2 )(1 − |a|)
δ2 = η2 + tan
.
π
2α1 (1 + |a|)
Argument Estimates of certain ...
95
Corollary 2.10. Under the hypothesis of Corollary 2.9, we have
(2.42)
π
π
− η1 < arg {G′ (z) − β} < η2
2
2
(z ∈ U)
where the function G(z) is defined in U by
Z z
Hp,q,s (α1 )f (t)
dt
(2.43)
G(z) =
tp
0
and η1 (0 < η1 ≤ 1) and η2 (0 < η2 ≤ 1) are the solutions of the equations
(2.40) and (2.41).
For q = s = 1,
following corollary.
B → A and g(z) = z p , in Corollary 2.9, we have the
Corollary 2.11. If f ∈ Ap , satisfies
(2.44)
½
¾
π
Lp (α1 + 1, β1 )f (z)
Lp (α1 + 1, β1 )f (z)
π
− δ1 < arg (1 − λ)
δ2
+
λ
−
β
<
2
zp
zp
2
(2.45)
(α1 > 0,
λ ≥ 0;
0 ≤ β < 1;
0 < δ1 , δ2 ≤ 1;
z ∈ U),
then
(2.46)
π
− η1 < arg
2
½
Lp (α1 , β1 )f (z)
−β
zp
¾
<
π
η2 (z ∈ U)
2
where η1 (0 < η1 ≤ 1) and η2 (0 < η2 ≤ 1) are the solutions of the equations
·
¸
2
−1 λ(η1 + η2 )(1 − |a|)
(2.47)
δ1 = η1 + tan
π
2α1 (1 + |a|)
(2.48)
·
¸
2
−1 λ(η1 + η2 )(1 − |a|)
δ2 = η2 + tan
.
π
2α1 (1 + |a|)
Remark 2.12. Taking q = s, p = α1 = β1 , λ = 1, and β = 0 in Corollary
2.9 and q = s = p = α1 = α2 , and β = 0 in Corollary 2.10, we get the
results obtained by Cho et al. [5]
96
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
(0 ≤ α < p), B = −1 and δ1 = δ2 = 1 in Theorem
Setting A = 1 − 2α
p
2.3, we get the following corollary.
Corollary 2.13. Let α1 > 0, f ∈ Ap , and suppose that g ∈ Kp (α). If
½
¾
Hp,q,s (α1 )f (z)
Hp,q,s (α1 + 1)f (z)
(2.49)
Re (1 − λ)
+λ
>β
Hp,q,s (α1 )g(z)
Hp,q,s (α1 + 1)g(z)
(λ ≥ 0;
0 ≤ β < 1;
z ∈ U),
then
(2.50)
Re
½
Hp,q,s (α1 )f (z)
Hp,q,s (α1 )g(z)
¾
>β
(z ∈ U).
Corollary 2.14. Let α1 > 0, −1 ≤ B < A ≤ 1, f ∈ Ap , and suppose that
g ∈ Kp (α). If
½
¾
Lp (α1 , β1 )f (z)
Lp (α1 + 1, β1 )f (z)
(2.51)
Re (1 − λ)
+λ
−β >β
Lp (α1 , β1 )g(z)
Lp (α1 , β1 )g(z)
(λ ≥ 0;
0 ≤ β < 1;
z ∈ U),
then,
(2.52)
Re
½
Lp (α1 , β1 )f (z)
Lp (α1 , β1 )g(z)
¾
>β
(z ∈ U).
Remark 2.15. For q = s = 1, α1 = β1 = p = 1 and α = 0, Corollary
2.13, is the result obtained by Bulboacă [2]. If we put q = s = 1, α1 =
β1 = p = 1 and β = 0,and g(z) = z in Corollary 2.13, then we have the
result due to Chichra [4]. Further,taking q = s = 1, α1 = β1 = p, λ = 1
and α = β = 0 in 2.13, we get the corresponding results obtained by Libera
[12] and Sakaguchi [20].
Theorem 2.16. If f ∈ Ap , satisfies
½
¾
π
Hp,q,s (α1 )f (z)
π
−
β
<
δ2
(2.53)
− δ1 < arg
2
zp
2
Argument Estimates of certain ...
(α1 > 0,
then
(2.54)
λ ≥ 0;
0 ≤ β < 1;
97
0 < δ1 , δ2 ≤ 1;

Rz γ−1


 (γ + p) t Hp,q,s (α1 )f (t)dt
π
− η1 < arg

2


0
z γ+p
−β




<



π
η2
2
z ∈ U)
(γ > −p; z ∈ U)
where η1 (0 ≤ η1 ≤ 1) and η2 (0 ≤ η2 ≤ 1) are the solutions of the equations
·
¸
2
−1 (η1 + η2 )(1 − |a|)
(2.55)
δ1 = η1 + tan
π
2(γ + p)(1 + |a|)
(2.56)
·
¸
2
−1 (η1 + η2 )(1 − |a|)
δ2 = η2 + tan
.
π
2(γ + p)(1 + |a|)
Proof Consider the function φ(z) defined in U by
(γ + p)
(2.57)
Rz
tγ−1 Hp,q,s (α1 )f (t)dt
0
z γ+p
= β + (1 − β)φ(z).
Then φ(z) is analytic in U with φ(0) = 1. Differentiating both sides of (2.57)
and simplifying, we get
½
¾
zφ′ (z)
Hp,q,s (α1 )f (z)
− β = (1 − β) φ(z) +
.
(2.58)
zp
γ+p
Now, by using Lemma 2.1 and a similar method in the proof of Theorem
2.3, we get (2.54).
Theorem 2.17. Let f ∈ Ap . If
½
¾
π
Hp,q,s (α1 + 1)f (z) α1 − p − γ
π
−
< δ2
(2.59)
− δ1 < arg
2
Hp,q,s (α1 )f (z)
α1
2
(α1 > 0;
p + γ > 0; 0 < δ1 , δ2 ≤ 1;
z ∈ U)
98
T.N.Shanmugam, S.Sivasubramanian, Shigeyoshi Owa
then








z γ Hp,q,s (α1 )f (z)
π
π
< η2
(2.60) − η1 < arg Rz

 2
2

 tγ−1 Hp,q,s (α1 )f (t)dt 

0
(γ > −p;
z ∈ U)
where η1 (0 < η1 ≤ 1) and η2 (0 < η2 ≤ 1) are the solutions of the equations
·
¸
2
−1 (η1 + η2 )(1 − |a|)
(2.61)
δ1 = η1 + tan
π
2(γ + p)(1 + |a|)
(2.62)
·
¸
2
−1 (η1 + η2 )(1 − |a|)
δ2 = η2 + tan
.
π
2(γ + p)(1 + |a|)
Proof. The proof of the theorem is much akin to the proof of Theorem
2.16 and hence we omit the details involved.
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T.N.Shanmugam
Department of Mathematics
College of Engineering, Anna university
Chennai 600 025, India
E-mail: shan@annauniv.edu
S.Sivasubramanian
Department of Mathematics
Easwari Engineering College
Ramapuram, Chennai-600 089, India
Shigeyoshi Owa
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577, Japan
E-mail: owas@kindai.ac.jp