Int. J. Nonlinear Anal. Appl. 1 (2010) No.1, 1–5
ISSN:2008-6822 (electronic)
http://www.ijnaa.com
UNIVALENT HOLOMORPHIC FUNCTIONS WITH FIXED
FINITELY MANY COEFFICIENTS INVOLVING SALAGEAN
OPERATOR
M. ACU1 AND SH. NAJAFZADEH2∗
Abstract. By using generalized Salagean differential operator a new class of
univalent holomorphic functions with fixed finitely many coefficients is defined.
Coefficient estimates, extreme points, arithmetic mean, and weighted mean properties are investigated.
1. Introduction and preliminaries
Let A denote the class of functions of the form
+∞
X
f (z) = z +
ak z k
(1.1)
k=t+1
which are holomorphic in the unit disk ∆ = {z : |z| < 1}.
We denote by N the subclass of A consisting of functions f (z) ∈ A which are
holomorphic univalent in ∆ and are of the form
f (z) = z −
+∞
X
ak z k ,
(ak ≥ 0).
(1.2)
k=t+1
The generalized Salagean operator is defined in [1] by
Dλ0 f (z) = f (z)
0
Dλ1 f (z) = (1 − λ)f (z) + λzf (z)
Dλn f (z) = Dλ1 (Dλn−1 f (z)),
λ≥0
see also [2].
If f (z) is given by (1.2), we see that
Dλn f (z)
=z−
+∞
X
[1 + (k − 1)λ]n ak z k .
(1.3)
k=t+1
Date: Received: March 2009; Revised: November 2009.
2000 Mathematics Subject Classification. Primary 30C45; Secondary 30C50.
Key words and phrases. Salagean operator; Univalent function; Coefficient estimate; Extreme
point; Arithmetic and weighted mean.
∗
Corresponding author.
1
2
M. ACU AND SH. NAJAFZADEH
When λ = 1, we get the classic Salagean differential operator [3].
A function f (z) ∈ N is said to be in Nn,λ (α, β, γ, θ) if and only if
[Dn+2 f (z)]0 − 1 Dn+1 f (z) λ
z λ
2α n+1
< γ,
z Dλ f (z) − β(1 + θ)α (1.4)
where α, β, γ, θ belong to [0,1].
cm
Now we introduce the class Nn,λ
(α, β, γ, θ), the subclass of Nn,λ (α, β, γ, θ) consisting
of functions with negative and fixed finitely many coefficient of the form
f (z) = z −
t
X
αγ(2 − β(1 + θ))cm
.
n+1 (m2 λ + m(1 − λ) − 1 + 2αλ)]
[(1
+
(m
−
1)λ)
m=2
(1.5)
Such type of work was recently carried out by Shams and Kulkani [4]. See also [5].
We need the following lemma for proving our main results.
Lemma 1.1. A function f (z) given by (1.2) is in the class Nn,λ (α, β, γ, θ) if and
only if
∞
X
[(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)ak ] ≤ αγ(2 − β(1 + θ)).
(1.6)
k=t+1
Proof. Let the inequality (1.6) holds true and suppose |z| = 1. Then we obtain
P
0
n+1
|(Dλn+2 f (z)) − z1 Dλn+1 f (z)| − γ|2α − 2α +∞
ak z k−1 − β(1 + θ)α|
k=t+1 (1 + (k − 1)λ)
=
P+∞
k=t+1 [(1
+ (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)ak − αγ(2 − β(1 + θ))] ≤ 0.
Hence, by maximum modulus theorem, we conclude that f (z) ∈ Nn,λ (α, β, γ, θ).
Conversely, let f (z) defined by (1.2) be in the class Nn,λ (α, β, γ, θ), so the condition
(1.4) yields
P
+∞ [(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1]a z k−1 k
k=t+1 P
< γ, z ∈ ∆.
n+1 − β(1 + θ)α
2α − ∞
2α(1
+
(k
−
1)λ)
k=t+1
Since for any z, |Re(z)| < |z|, then
( P+∞
)
n+1 2
k−1
[(1
+
(k
−
1)λ)
(k
λ
+
k(1
−
λ)
−
1)]a
z
k
k=t+1
P∞
Re
< γ.
n+1
α(2 − β(1 + θ)) − k=t+1 2α(1 + (k − 1)λ) ak z k−1
By letting z → 1 through real values, we get the required result.
2. Main Results
We begin by proving a necessary and sufficient conditions for a function belonging
cm
to the class Nn,λ
(α, β, γ, θ).
UNIVALENT HOLOMORPHIC FUNCTIONS WITH ...
3
cm
Theorem 2.1. Let f (z) defined by (1.2), then f (z) ∈ Nn,λ
(α, β, γ, θ) if and only if
t
+∞
X
X
[(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αλ)]ak
<1−
cm .
αγ(2
−
β(1
+
θ))
m=2
k=t+1
(2.1)
Proof. By letting
am =
αγ(2 − β(1 + θ))cm
,
(1 + (m − 1)λ)n+1 (m2 λ + m(1 − λ) − 1 + 2αγ)
(2.2)
cm
cm
since Nn,λ
(α, β, γ, θ) ⊂ Nn,λ (α, β, γ, θ), so f (z) ∈ Nn,λ
(α, β, γ, θ) if and only if
t
X
(1 + (m − 1)λ)n+1 (m2 λ + m(1 − λ) − 1 + 2αγ)
am
αγ(2
−
β(1
+
θ))
m=2
+∞
X
(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)
+
ak < 1
αγ(2
−
β(1
+
θ))
k=t+1
or
+∞
t
X
X
(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)
<1−
cm ,
αγ(2
−
β(1
+
θ))
m=2
k=t+1
and this gives the result.
cm
Corollary 2.2. If f (z) defined by (1.2) be in Nn,λ
(α, β, γ, θ) then for k ≥ t + 1 we
have
P
αγ(2 − β(1 + θ))(1 − tm=2 cm )
ak ≤
,
(2.3)
(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)
and result is best possible for the function
t
X
αγ(2 − β(1 + θ))cm
zm
n+1 (m2 λ + m(1 − λ) − 1 + 2αγ)
(1
+
(m
−
1)λ)
m=2
P
αγ(2 − β(1 + θ))(1 − tm=2 cm )
−
zk
(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)
g(z) = z −
(2.4)
3. Extreme points and arithmetic mean structure
cm
Now we find Extreme points and convolution structure for functions in Nn,λ
(α, β, γ, θ).
Theorem 3.1. Let
ft (z) = z −
t
X
αγ(2 − β(1 + θ))cm
zm
n+1
2
(1 + (m − 1)λ) (m λ + m(1 − λ) − 1 + 2αγ)
m=2
and for k ≥ t + 1
t
X
αγ(2 − β(1 + θ))cm
zm
n+1 (m2 λ + m(1 − λ) − 1 + 2αγ)
(1
+
(m
−
1)λ)
m=2
P
αγ(2 − β(1 + θ))(1 − tm=2 cm )
−
zk .
n+1
2
(1 + (k − 1)λ) (k λ + k(1 − λ) − 1 + 2αγ)
fk (z) = z −
4
M. ACU AND SH. NAJAFZADEH
cm
Then F (z) ∈ Nn,λ
(α, β, γ, θ) if and only if it can be expressed in the form
F (z) =
+∞
X
σk fk (z)
k=t
where σk ≥ 0
(k ≥ t) and
Proof. Let F (z) =
P+∞
k=t
P+∞
k=t
σk = 1.
σk fk (z), then
t
X
αγ(2 − β(1 + θ))cm
zm
n+1 (m2 λ + m(1 − λ) − 1 + 2αγ)
(1
+
(m
−
1)λ)
m=2
P
+∞
X
(1 − +∞
m=2 cm )αγ(2 − β(1 + θ))σk
−
zk .
n+1
2
(1 + (k − 1)λ) (k λ + k(1 − λ) − 1 + 2αγ)
k=t+1
F (z) = z −
Finally we have
P
∞
X
(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)(1 − +∞
m=2 cm )αγ(2 − β(1 + θ))σk
n+1
2
αγ(2 − β(1 + θ))(1 + (k − 1)λ) (k λ + k(1 − λ) − 1 + 2αγ)
k=t+1
= (1 −
t
X
cm )
m=2
+∞
X
σk = (1 −
k=t+1
t
X
cm )(1 − σt ) < 1 −
m=2
t
X
cm .
m=2
cm
Thus F (z) ∈ Nn,λ
(α, β, γ, θ).
cm
Conversely suppose F (z) ∈ Nn,λ
(α, β, γ, θ), so
F (z) = z −
t
X
+∞
X
αγ(2 − β(1 + θ))cm
−
ak z k .
n+1
2
(1 + (k − 1)λ) (k λ + k(1 − λ) − 1 + 2αγ) k=t+1
m=2
By putting
(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)
ak , (k ≥ t + 1)
P
αγ(2 − β(1 + θ))(1 − tm=2 cm )
P
we have σk ≥ 0 and if we put σt = 1− +∞
k=t+1 σk , we conclude the required result. σk =
Theorem 3.2. Let fj (z)
(j = 1, 2, ..., l) defined by
l
X
+∞
X
αγ(2 − β(1 + θ))
m
fj (z) = z −
z −
ak,j z k
n+1 (m2 λ + m(1 − λ) − 1 + 2αγ)
(1
+
(m
−
1)λ)
m=2
k=t+1
j=1,...,l
be in
cm
Nn,λ
(α, β, γ, θ),
+∞
X
then the function
+∞
X
αγ(2 − β(1 + θ))Cm
m
H(z) = z −
z −
dk z k ,
n+1 (m2 λ + m(1 − λ) − 1 + 2αγ)
(1
+
(m
−
1)λ)
m=2
k=t+1
(ck ≥ 0)
P
cm
is also in Nn,λ
(α, β, γ, θ), where dk = 1l lj=1 ak,j .
UNIVALENT HOLOMORPHIC FUNCTIONS WITH ...
5
Proof. We have
+∞
X
(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ)
dk
αγ(2
−
β(1
+
θ))
k=t+1
l
+∞
X
(1 + (k − 1)λ)n+1 (k 2 λ + k(1 − λ) − 1 + 2αγ) X
=
(
ak,j )
lαγ(2
−
β(1
+
θ))
j=1
k=t+1
#
" +∞
l
n+1
2
X
X
1
(1 + (k − 1)λ) (k λ + k(1 − λ) − 1 + 2αγ)
ak,j
=
l j=1 k=t+1
αγ(2 − β(1 + θ))
!
l
t
t
X
X
1X
<
1−
cm = 1 −
cm
l j=1
m=2
m=2
cm
and the proof by Theorem 2.1 is complete. So Nn,λ
(α, β, γ, θ) is closed under arithmetic mean.
cm
Remark 3.3. with the same calculation with theorem 3.2 we can prove that Nn,λ
(α, β, γ, θ)
is closed under weighted mean.
Acknowledgements: The authors would like to thank the referees for their
valuable comments.
References
1. F.M. Al-Oboudi, On univalent functions by a generalized Salagean operator, Int. J. Math.
Math. Sci. 27 (2004), 1429-1436.
2. D. Raducanu, On the Fekete Szego inequality for a class of analytic functions defined by using
the generalized Salagean operator, General Math. Vol. 10 (16) (2008), 19-27.
3. G. S. Salagean, Subclasses of univalent functions, Lect. Notes Math. 1013, (1983), 362-372
4. S, Shams and S. R. Kulkarni, A class of univalent functions with negative and fixed finitely
many coefficients, Acta Ciencia Indica, XXIXM (3), (2003), 587-594.
5. Sh. Najafzadeh and S.R. Kulkarni, Application of Ruscheweyh derivative on univalent functions
with negative and fixed finitely many coefficients, Rajasthan Acad. Phy. Sci. 5(1), (2006), 39-51.
1
Department of Mathematics,Lucian Blaga University, 2400 Sibiu, Romania
E-mail address: Muger [email protected]
2
Department of Mathematics, Faculty of Science,University of Maragheh, MaraghehIran
E-mail address: [email protected]