Paper 18 20 2009 PDF
Paper 18 20 2009 PDF
Paper 18 20 2009 PDF
Pramila Vijaywargiya
1. Introduction
For a fixed p ∈ N := {1, 2, 3, ...} , let An,p denote the class of all analytic functions
of the form ∞
X
f (z) = z p + ak z k , (n ∈ N ), (1.1)
k=n+p
which are p - valent in the open unit disc U = {z : z ∈ C, |z| < 1}. Upon differentiat-
ing both sides of (1.1) q - times with respect to z, the following differential operator
is obtained : ∞ X
f (q) (z) = α(p; q) z p−q + α(k; q) ak z k−q , (1.2)
k=n+p
where
p!
α(p; q) = (p ≥ q; p ∈ N ; q ∈ N ∪ {0}). (1.3)
(p − q)!
Several researchers have investigated higher-order derivatives of multivalent func-
tions, (see, e.g.[8,12]).
175
P. Vijayvargiya - On some results for λ−spirallike functions of complex order...
We shall require the following results due to Miller and Mocanu in order to prove
our main results of the next section:
Lemma 1 [6]. Let q be univalent in the unit disk U, and let θ and φ be analytic
in a domain D containing q(U), with φ(w) 6= 0 when w ∈ q(U).
Set Q(z) = zq 0 (z)φ(q(z)), h(z) = θ(q(z)) + Q(z) and suppose that
176
P. Vijayvargiya - On some results for λ−spirallike functions of complex order...
0
(i) Q is starlike (univalent) in U with Q(0) = 0 and Q (0) 6= 0.
(ii) Re {zh0 (z)/Q(z)} > 0 for z ∈ U.
If p is analytic in U, with p(0) = q(0), p(U) ⊂ D and
then
Re {p(z)} > 0 (z ∈ U).
Theorem 1. Let f ∈ Snλ,q (ρ; b), (|λ| < π2 , b 6= 0, complex, 0 ≤ ρ < 1, p >
q),then !γ
f (q) (z) 1
p−q
≺ 2γ(p−q)(1−ρ)b cos λe−iλ
(3.1)
α(p; q) z (1 − z)
where γ 6= 0 is complex and either |2γ(p − q)(1 − ρ)b cos λe−iλ + 1| ≤ 1 or
|2γ(p − q)(1 − ρ)b cos λe−iλ − 1| ≤ 1 and this is the best dominant.
−iλ
−1
Proof. Let q(z) = (1 − z)−2γ(p−q)(1−ρ)b cos λe , φ(w) = γb(p − q) cos λe−iλ w−1
and θ(w) = 1 in Lemma 1. Then it is easy to verify the conditions (i) and (ii) of the
Lemma 1. Namely q is univalent in U by Lemma 3, while
1 + (1 − 2ρ)z
h(z) = θ(q(z)) + zq 0 (z)φ(q(z)) = .
1−z
177
P. Vijayvargiya - On some results for λ−spirallike functions of complex order...
Consequently, for p(z) = 1 + pn z n + pn+1 z n+1 + ..., analytic in U with p(z) 6= 0 for
0 < |z| < 1, from (2.1) we get
⇒ p(z) ≺ q(z)
γ
f (q) (z)
Now, if in (3.2) we choose p(z) = α(p ; q) z p−q , then we have
!γ
f (q) (z) 1
≺ 2γ(p−q)(1−ρ)b cos λe−iλ
α(p; q) z p−q (1 − z)
Corollary 2. Let f (z) ∈ Snλ,q (ρ; b), (|λ| < π2 , b 6= 0, complex, 0 ≤ ρ < 1, p >
q), then
! eiλ
α(p; q) z p−q 2(p−q)(1−ρ)b cos λ
≺ (1 − z) (3.3)
f (q) (z)
and this is the best dominant.
From (3.3), we have the following inequality for f (z) ∈ Snλ,q (ρ; b)
eiλ
!
α(p; q) z p−q 2(p−q)(1−ρ)b cos λ
− 1 ≤ |z| (z ∈ U) (3.4)
(q) (z)
f
178
P. Vijayvargiya - On some results for λ−spirallike functions of complex order...
Then we have
!βeiλ
f (q) (z) n
Re > (z ∈ U),
α(p; q) z p−q 2β(p − q)(1 − ρ)b cos λ + n
(1 − B)zp0 (z)
Re 1 − ρ + > 0, (z ∈ U) (3.10)
β(p − q)b cos λ {(1 − B)p(z) + B}
(1 − B)v
φ(u, v) = 1 − ρ + ,
β(p − q)b cos λ [(1 − B)u + B]
n o
−B
then φ(u, v) is continuous in D = C − 1−B × C.
Also, (1, 0) ∈ D and Re {φ(1, 0)} = 1 − ρ > 0.
−n(1+u22 )
Furthermore, for all (iu2 , v1 ) ∈ D such that v1 ≤ 2 , we have
(1 − B)v1
Re {φ(iu2 , v1 )} = 1 − ρ + Re
β(p − q)b cos λ {(1 − B)iu2 + B}
179
P. Vijayvargiya - On some results for λ−spirallike functions of complex order...
B(1 − B)v1
=1−ρ+ n o
β(p − q)b cos λ (1 − B)2 u22 + B 2
Corollary 4. Let the function f (z) defined by (1.1) be in the Snλ,q (0; b). Then
neiλ
!
f (q) (z) 2(p−q)b cos λ
1
Re > (z ∈ U).
α(p; q)
z p−q
2
References
180
P. Vijayvargiya - On some results for λ−spirallike functions of complex order...
Pramila Vijaywargiya
Department of Mathematics
University of Rajasthan
Jaipur (INDIA)- 3020055
email: pramilavijay1979@gmail.com
181