Surveys in Mathematics and its Applications
ISSN 1842-6298 (electronic), 1843-7265 (print)
Volume 5 (2010), 35 – 47
SOME APPLICATIONS OF GENERALIZED
RUSCHEWEYH DERIVATIVES INVOLVING A
GENERAL FRACTIONAL DERIVATIVE
OPERATOR TO A CLASS OF ANALYTIC
FUNCTIONS WITH NEGATIVE COEFFICIENTS I
Waggas Galib Atshan and S. R. Kulkarni
Abstract. For certain univalent function f , we study a class of functions f as defined by making
use of the generalized Ruscheweyh derivatives involving a general fractional derivative operator,
satisfying
z(J1λ,µ f (z))0
Re
> β.
(1 − γ)J1λ,µ f (z) + γz 2 (J1λ,µ f (z))00
(
)
A necessary and sufficient condition for a function to be in the class Aλ,µ,ν
(n, β) is obtained.
γ
In addition, our paper includes distortion theorem, radii of starlikeness, convexity and close-toconvexity, extreme points. Also, we get some results in this paper.
1
Introduction
Let Ω denote the class of functions which are analytic in the unit disk U = {z ∈ C :
|z| < 1} and let A(n) denote the subclass of Ω consisting of functions of the form:
f (z) = z −
∞
X
ak z k ,
(ak ≥ 0, n ∈ IN ),
(1.1)
k=n+1
where f (z) is analytic and univalent in the unit disk U . Then the function f (z) ∈
A(n) is said to be in the class S(n, α), if and only if
0 zf (z)
Re
> α , (z ∈ U, 0 ≤ α < 1).
(1.2)
f (z)
A function f (z) ∈ S(n, α) is called starlike function of order α. A function f (z) ∈
A(n) is said to be in the class C(n, α) if and only if
zf 00 (z)
Re 1 + 0
> α, (z ∈ U, 0 ≤ α < 1).
(1.3)
f (z)
2000 Mathematics Subject Classification: 30C45.
Keywords: Distortion theorem; Radii of starlikeness; Extreme points.
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36
W. G. Atshan and S. R. Kulkarni
A function f (z) ∈ C(n, α) is called convex function of order α. It is observed that
f (z) ∈ C(n, α) if and only if zf 0 (z) ∈ S(n, α) ∀ n ∈ IN [2].
(1.4)
A function f (z) ∈ A(n) is said to be in the class K(n, α) if there is a convex function
g(z) such that
0 f (z)
> α, (∀ z ∈ U, 0 ≤ α < 1).
(1.5)
Re
g 0 (z)
A function f (z) ∈ K(n, α) is called close-to-convex of order α.
We shall need the fractional derivative operator ([10], [11]) in this paper.
Let a, b, c ∈ C with C 6= 0, −1, −2, · · · . The Gaussian hypergeometric function
F
2 1 is defined by
2 F1 (z) ≡
2 F1 (a, b; c; z) =
∞
X
(a)n (b)n z n
n=0
(c)n
n!
,
(1.6)
where (λ)n is the pochhammer symbol defined, in terms of the Gamma function, by
Γ(λ + n)
1
(n = 0)
=
(λ)n =
λ(λ
+
1)
·
·
·
(λ
+
n
−
1)
(n
∈ IN )
Γ(λ)
Definition 1. Let 0 ≤ λ < 1 and µ, ν ∈ IR. Then, in terms of familiar (Gauss’s)
λ,µ,ν
hypergeometric function 2 F1 , the generalized fractional derivative operator J0,z
of
a function f (z) is defined by:

λ−µ R z
1
d
−λ f (E) · F (µ − λ, 1 − ν; 1 − λ; 1 − E )dE

2 1
 Γ(1−λ) dz z
z
0 (z − E)
λ,µ,ν
f (z) =
J0,z
(0 ≤ λ < 1)

 dn J λ−n,µ,ν f (z),
(n ≤ λ < n + 1, n ∈ IN )
dz n 0,z
(1.7)
where the function f (z) is analytic in a simply-connected region of the z-plane containing the origin, with the order
f (z) = O(|z| ), (z → 0),
(1.8)
for > max{0, µ − ν} − 1, and the multiplicity of (z − E)−λ is removed by requiring
log(z − E) to be real, when z − E > 0.
The fractional derivative of order λ of a function f (z) is defined by
Z z
1
d
f (E)
Dzλ {f (z)} =
dE, 0 ≤ λ < 1,
(1.9)
Γ(1 − λ) dz 0 (z − E)λ
where f (z) it is chosen as in (1.7), and the multiplicity of (z − E)−λ is removed by
requiring log(z − E) to be real, when z − E > 0.
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37
Some applications of generalized Ruscheweyh derivatives
By comparing (e1.7) with (1.9), we find
λ,λ,ν
J0,z
f (z) = Dzλ {f (z)}, (0 ≤ λ < 1).
(1.10)
In terms of gamma function, we have
λ,µ,ν k
J0,z
z =
Γ(k + 1)Γ(1 − µ + ν + k)
z k−µ ,
Γ(1 − µ + k)Γ(1 − λ + ν + k)
(0 ≤ λ < 1, µ, ν ∈ IR and
(1.11)
k > max{0, µ − ν} − 1).
Definition 2. Let f (z) ∈ A(n) be given by (1.1). Then the class Aλ,µ,ν
(n, β) is
γ
defined by
(
(
)
z(J1λ,µ f (z))0
λ,µ,ν
Aγ (n, β) = f ∈ A(n) : Re
> β,
(1 − γ)J1λ,µ f (z) + γz 2 (J1λ,µ f (z))00
(z ∈ U, 0 ≤ γ < 1, n ∈ IN ; 0 ≤ β < 1; λ > −1)} ,
(1.12)
where J1λ,µ f is a generalized Ruscheweyh derivative defined by Goyal and Goyal [3,
p. 442] as
Γ(µ − λ + ν + 2) λ,µ,ν µ−1
(z
f (z))
zJ
Γ(ν + 2)Γ(µ + 1) 0,z
∞
X
= z−
ak C1λ,µ (k)z k ,
J1λ,µ f (z) =
(1.13)
(1.14)
k=n+1
where
C1λ,µ (k) =
Γ(k + µ)Γ(ν + 2 + µ − λ)Γ(k + ν + 1)
.
Γ(k)Γ(k + ν + 1 + µ − λ)Γ(ν + 2)Γ(1 + µ)
(1.15)
For µ = λ = α, ν = 1, the generalized Ruscheweyh derivatives reduces to ordinary
Ruscheweyh derivatives of f (z) of order α [7]:
Dα f (z) =
∞
X
z
Dα (z α−1 f (z)) = z −
ak Ck (α)z k ,
Γ(α + 1)
(1.16)
k=n+1
where
Ck (α) =
(α + 1)(α + 2) · · · (α + k − 1)
.
(k − 1)!
(1.17)
The class Aλ,µ,ν
(n, β) contains many well-known classes of analytic functions, for
γ
example:
(i) If µ = λ = α, ν = 1, n = 1, we get the class Aλ,λ,1
(1, β) was studied by Tehranchi
γ
and Kulkarni [12].
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38
W. G. Atshan and S. R. Kulkarni
(ii) If µ = λ = 0, ν = 1, α = β, γ = 0, we get the class of starlike function of order
α, (S(n, α)).
The same properties have been found for other classes in [4], [8] and [9].
Lemma 3. Let w = u + iν, then
Re w ≥ β if and only if |w − (1 + β)| ≤ |w + (1 − β)|.
Definition 4. Let f, h be analytic in U . Then h is said to be subordinate to f ,
written h ≺ f , if there exist function w that is analytic in U with |w(z)| < 1 in U
and h(z) = f (w(z)) in U for some analytic function w with w(0) = 0 and |w(z)| ≤ |z|
in U . If w is not merely a rotation of the disk (that is, if |w(z)| < |z|), then h is
said to be properly subordinate to f .
2
Main Results
The following theorem gives a necessary and sufficient condition for function to be
(n, β).
in Aλ,µ,ν
γ
(n, β) if and only if
Theorem 5. Let f (z) ∈ A(n), then f (z) ∈ Aλ,µ,ν
γ
∞
X
(γβ(1 + k − k 2 ) + k − β)C1λ,µ (k)ak < 1 − β(1 − γ),
(2.1)
k=n+1
where 0 ≤ γ < 1, 0 ≤ β < 1, λ > −1, n ∈ IN and C1λ,µ (k) is given by (1.15).
(n, β) so we have
Proof. Assume that f ∈ Aλ,µ,ν
γ
(
)
z(J1λ,µ f (z))0
>β
Re
(1 − γ)J1λ,µ f (z) + γz 2 (J1λ,µ f (z))00


∞


P




z−
kC1λ,µ (k)ak z k




k=n+1
!
! > β.
Re


∞
∞
P
P





C1λ,µ (k)ak z k + γ −
k(k − 1)C1λ,µ (k)ak z k 
 (1 − γ) z −

k=n+1
k=n+1
Hence
Re




1−
∞
P
k=n+1
kC1λ,µ (k)ak z k−1




∞
P




 (1 − γ) −
(1 − γ + γk(k − 1))C1λ,µ (k)ak z k−1 
> β,
k=n+1
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39
Some applications of generalized Ruscheweyh derivatives
or equivalently
8
P
>
>
<1 −
Re
>
>
:
∞
k=n+1
kC1λ,µ (k)ak z k−1 − β(1 − γ) + β
P
∞
(1 − γ) −
k=n+1
P
∞
k=n+1
(1 − γ + γk(k − 1))C1λ,µ (k)ak z k−1
(1 − γ + γk(k − 1))C1λ,µ (k)ak z k−1
9>
>=
>>; > 0.
This inequality is correct for all z ∈ U . Letting z → 1− yields


∞
P
λ,µ


1
−
β(1
−
γ)
−
(k
−
β(1
−
γ
+
γk(k
−
1)))C
(k)a

k
1


k=n+1
Re
>0
∞
P


λ,µ




(1 − γ + γk(k − 1))C1 (k)ak
(1 − γ) −
k=n+1
and so by the mean value theorem, we have
(
)
∞
X
Re 1 − β(1 − γ) −
(k − β(1 − γ + γk 2 − γk))C1λ,µ (k)ak > 0,
k=n+1
so we have
∞
X
(γβ(1 + k − k 2 ) + k − β)C1λ,µ (k)ak < 1 − β(1 − γ).
k=n+1
Conversely, let (2.1) hold. We will prove that (1.12) is correct and then f ∈
Aλ,µ,ν
(n, β).
γ
By Lemma 3 it is enough to prove that |w − (1 + β)| < |w + (1 − β)|, where
w=
z(J1λ,µ f (z))0
(1 − γ)J1λ,µ f (z) + γz 2 (J1λ,µ f (z))00
or show that
T
1
|z(J1λ,µ f (z))0 − (1 + β)(1 − γ)J1λ,µ f (z) − (1 + β)γz 2 (J1λ,µ f (z))00 |
|N (z)|
1
<
|z(J1λ,µ f (z))0 + (1 − β)(1 − γ)J1λ,µ f (z) + (1 − β)γz 2 (J1λ,µ f (z))00 |
|N (z)|
= Q,
=
where N (z) = (1−γ)J1λ,µ f (z)+γz 2 (J1λ,µ f (z))00 and it is easy to verify that Q−T > 0
and so the proof is complete.
Corollary 6. Let f ∈ Aλ,µ,ν
(n, β), then
γ
ak <
1 − β(1 − γ)
C1λ,µ (k)|γβ(1
+ k − k 2 ) + k − β|
, k = n + 1, n + 2, · · · .
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40
W. G. Atshan and S. R. Kulkarni
Theorem 7. The class Aλ,µ,ν
(n, β) is convex set.
γ
Proof. Let f (z), g(z) be the arbitrary elements of Aλ,µ,ν
(n, β), then for every t (0 <
γ
λ,µ,ν
t < 1), we show that (1 − t)f (z) + tg(z) ∈ Aγ (n, β) thus, we have
∞
X
(1 − t)f (z) + tg(z) = z −
[(1 − t)ak + tbk ]z k
k=n+1
and
∞ X
γβ(1 + k − k 2 ) + k − β
[(1 − t)ak + tbk ]C1λ,µ (k)
1 − β(1 − γ)
k=n+1
∞
X
γβ(1 + k − k 2 ) + k − β
ak C1λ,µ (k)
1 − β(1 − γ)
= (1 − t)
k=n+1
+t
∞
X
k=n+1
γβ(1 + k − k 2 ) + k − β
bk C1λ,µ (k) < 1.
1 − β(1 − γ)
This completes the proof.
(n, β), then the function
Corollary 8. Assume that f (z) and g(z) belong to Aλ,µ,ν
γ
λ,µ,ν
1
y(z) defined by y(z) = 2 (f (z) + g(z)) also belong to Aγ (n, β).
3
Distortion Theorem
In the next theorem, we will find distortion bound for J1λ,µ f (z).
(n, β), then
Theorem 9. Let f (z) ∈ Aλ,µ,ν
γ
1 − β(1 − γ)
|z|n+1 ≤ |J1λ,µ f (z)|
γβ(2 + n − (1 + n)2 ) + (n + 1) − β
1 − β(1 − γ)
≤ |z| +
|z|n+1 .
γβ(2 + n − (1 + n)2 ) + (n + 1) − β
|z| −
Proof. Let f (z) ∈ Aλ,µ,ν
(n, β). By Theorem 5, we have
γ
∞
X
C1λ,µ (k)ak <
k=n+1
1 − β(1 − γ)
, n ∈ IN = {1, 2, · · · }.
γβ(2 + n − (1 + n)2 ) + (n + 1) − β
Therefore
|J1λ,µ f (z)| ≤ |z|+|z|n+1
∞
X
k=n+1
C1λ,µ (k)ak < |z|+
1 − β(1 − γ)
|z|n+1
γβ(2 + n − (1 + n)2 ) + (n + 1) − β
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41
Some applications of generalized Ruscheweyh derivatives
and
∞
X
|J1λ,µ f (z)| ≥ |z|−|z|n+1
C1λ,µ (k)ak > |z|−
k=n+1
1 − β(1 − γ)
|z|n+1 .
γβ(2 + n − (n + 1)2 ) + (n + 1) − β
This completes the proof.
4
Radii of Starlikeness, Convexity and Close-to-convexity
In the next theorems, we will find the radii of starlikeness, convexity and close-toconvexity for the class Aλ,µ,ν
(n, β).
γ
Theorem 10. Let f (z) ∈ Aλ,µ,ν
(n, β). Then f (z) is a starlike of order α(0 ≤ α < 1)
γ
in |z| < r = r1 (λ, µ, ν, β, γ, α),where
(
r1 (λ, µ, ν, β, γ, α) = inf
k
(1 − α)[γβ(1 + k − k 2 ) + k − β]C1λ,µ (k)
(k − α)(1 − β(1 − γ))
)
1
k−1
(4.1)
and C1λ,µ (k) is given by (1.15).
(n, β). Then by Theorem 5
Proof. Let f (z) ∈ Aλ,µ,ν
γ
∞
X
(γβ(1 + k − k 2 ) + k − β)C1λ,µ (k)
ak < 1.
1 − β(1 − γ)
k=n+1
0
(z)
For 0 ≤ α < 1, we need to show that zff (z)
− 1 < 1 − α, we have to show that
∞
∞
P
P
k−1
(k − 1)ak |z|k−1
(k − 1)ak z
0
−
k=n+1
zf (z) − f (z) k=n+1
≤
=
< 1 − α.
∞
∞
P
f (z)
1 − P ak z k−1 k−1
a
|z|
1
−
k
k=n+1
Hence
∞
P
k=n+1
k−α
1−α
k=n+1
ak |z|k−1 ≤ 1. This is enough to consider
|z|k−1 ≤
(1 − α)[γβ(1 + k − k 2 ) + k − β]C1λ,µ (k)
,
(k − α)(1 − β(1 − γ))
therefore,
(
|z| ≤
(1 − α)[γβ(1 + k − k 2 ) + k − β]C1λ,µ (k)
(k − α)(1 − β(1 − γ))
)
1
k−1
.
(4.2)
Setting |z| = r1 (λ, µ, ν, β, γ, α) in (4.2), we get the radii of starlikeness, which completes the proof of the Theorem 10.
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42
W. G. Atshan and S. R. Kulkarni
By using (1.4) (Alexander’s theorem [2] : f is convex if and only if zf 0 is starlike),
we obtain:
Theorem 11. Let f (z) ∈ Aλ,µ,ν
(n, β). Then f (z) is convex of order α(0 ≤ α < 1)
γ
in |z| < r = r2 (λ, µ, ν, β, γ, α), where
r2 (λ, µ, ν, β, γ, α) = inf
k
1
k−1
(1 − α)[βγ(1 + k − k 2 ) + k − β] λ,µ
C1 (k)
k(k − α)(1 − β(1 − γ))
(4.3)
and C1λ,µ (k) is given by (1.15).
Proof. Let f (z) ∈ Aλ,µ,ν
(n, β). Then by Theorem 5
γ
∞
X
(γβ(1 + k − k 2 ) + k − β)C1λ,µ (k)
ak < 1.
1 − β(1 − γ)
k=n+1
00 (z) For 0 ≤ α < 1, we show that zff 0 (z)
< 1 − α,that is
∞
P
k(k − 1)ak z k−1 −
k=n+1
≤
∞
1 − P kak z k−1 k=n+1
or equivalently
∞
P
k
k=n+1
k−α
1−α
|z|k−1 ≤
∞
P
k(k − 1)ak |z|k−1
k=n+1
1−
∞
P
< 1 − α,
kak
|z|k−1
k=n+1
ak |z|k−1 ≤ 1. It is enough letting
(1 − α)[γβ(1 + k − k 2 ) + k − β] λ,µ
C1 (k).
k(k − α)(1 − β(1 − γ))
Therefore,
|z| ≤
1
k−1
(1 − α)[γβ(1 + k − k 2 ) + k − β] λ,µ
C1 (k)
.
k(k − α)(1 − β(1 − γ))
(4.4)
Setting |z| = r2 (λ, µ, ν, β, γ, α) in (4.4), we get the radii of convexity, which completes
the proof of Theorem 11.
∞
P
Theorem 12. If f (z) = z −
ak z k ∈ Aλ,µ,ν
(n, β), then f (z) is close-to-convex
γ
k=n+1
of order δ, 0 ≤ δ < 1 in |z| < r = r3 (λ, µ, ν, β, γ, δ),where
r3 (λ, µ, ν, β, γ, δ) = inf
k
1
k−1
(1 − δ)[γβ(1 + k − k 2 ) + k − β] λ,µ
C1 (k)
k(1 − β(1 − γ))
(4.5)
and C1λ,µ (k) is given by (1.15).
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43
Some applications of generalized Ruscheweyh derivatives
Proof. Let f (z) ∈ Aλ,µ,ν
(n, β). Then by Theorem 5
γ
∞
X
(γβ(1 + k − k 2 ) + k − β)C1λ,µ (k)
ak < 1,
1 − β(1 − γ)
k=n+1
for 0 ≤ δ < 1, we need to show that |f 0 (z)−1| ≤ 1−δ for |z| < r = r3 (λ, µ, ν, β, γ, δ),when
r3 (λ, µ, ν, β, γ, δ) is given by (4.5). Now
∞
∞
X
X
|f 0 (z) − 1| = kak z k−1 ≤
kak |z|k−1 .
k=n+1
∞
P
Thus |f 0 (z) − 1| < 1 − δ if
k=n+1
holds true if
|z|k−1 ≤
kak
k−1
(1−δ) |z|
k=n+1
≤ 1 but, by Theorem 5 above inequality
(1 − δ)[βγ(1 + k − k 2 ) + k − β] λ,µ
C1 (k).
k(1 − β(1 − γ))
This completes the proof.
5
Extreme Points
In the next theorem, we will find extreme points for the class Aλ,µ,ν
(n, β).
γ
Theorem 13. Let fn (z) = z and
fk (z) = z−
1 − β(1 − γ)
[γβ(1 + k −
k2 )
+k−
β]C1λ,µ (k)
z k , (k = n+1, n+2, · · · , n ∈ IN = {1, 2, · · · }),
(n, β) if and only if it can be exwhere C1λ,µ (k) is given by (1.15). Then f ∈ Aλ,µ,ν
γ
∞
∞
P
P
pressed in the form f (z) =
σk fk (z),where σk ≥ 0 and
σk = 1. In particular,
k=n
k=n
the extreme points of Aλ,µ,ν
(n, β) are the functions fn (z) = z and
γ
fk (z) = z −
(1 − β(1 − γ))
[βγ(1 + k − k 2 ) + k − β]C1λ,µ (k)
z k , k = n + 1, n + 2, · · · .
Proof. Firstly, let us express f as in the above theorem, therefore we can write
"
#
∞
∞
X
X
1 − β(1 − γ)
k
f (z) =
σk fk (z) = σn z +
σk z −
z
2 ) + k − β]C λ,µ (k)
[γβ(1
+
k
−
k
1
k=n
k=n+1
= z(σn +
= z−
∞
X
k=n+1
∞
X
σk ) −
∞
X
1 − β(1 − γ)
λ,µ
2
k=n+1 [βγ(1 + k − k ) + k − β]C1 (k)
σk z k
gk z k ,
k=n+1
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44
W. G. Atshan and S. R. Kulkarni
where
gk =
1 − β(1 − γ)
[γβ(1 + k − k 2 ) + k − β]C1λ,µ (k)
σk .
Therefore, f ∈ Aλ,µ,ν
(n, β), since
γ
∞
∞
X
X
gk [γβ(1 + k − k 2 ) + k − β]C1λ,µ (k)
=
σk = 1 − σn < 1.
1 − β(1 − γ)
k=n+1
k=n+1
Conversely, assume that f ∈ Aλ,µ,ν
(n, β), then by (2.1) we may set
γ
σk =
∞
X
[γβ(1 + k − k 2 ) + k − β]C1λ,µ (k)
ak , k ≥ n + 1 and 1 −
σk = σn .
1 − β(1 − γ)
k=n+1
Then
∞
X
∞
X
(1 − β(1 − γ))σk
zk
2 ) + k − β]C λ,µ (k)
[γβ(1
+
k
−
k
1
k=n+1
k=n+1
!
∞
∞
∞
X
X
X
σk +
σk fk (z)
= z−
σk (z − fk (z)) = z 1 −
f (z) = z −
k
ak z = z −
k=n+1
∞
X
= σn z +
k=n+1
σk fk (z) =
∞
X
k=n+1
σk fk (z).
k=n
k=n+1
This completes the proof.
6
Subordination
Theorem 14. For n = 1, let f (z) ∈ Aλ,µ,ν
(1, β) and h(z) be an arbitrary element
γ
of A(1) such that h ≺ f , defined in Definition 4, and if
1 dk (f (w(z)))
hk =
k!
dz k
z=0
(6.1)
also if
∞
P
[β(1 − γ) + βγk(k − 1) − k]|hk |
k=2
|h1 |
< (1 − β(1 − γ)).
(6.2)
Then h ∈ Aλ,µ,ν
(1, β).
γ
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Surveys in Mathematics and its Applications 5 (2010), 35 – 47
http://www.utgjiu.ro/math/sma
45
Some applications of generalized Ruscheweyh derivatives
Proof. Since h ≺ f by definition of subordination there is analytic function w(z)
such that |w(z)| ≤ |z| and h(z) = f (w(z)). But h(z) is the composition of two
analytic functions in the unit disk, therefore we can expand this function in terms
of Taylor series at origin as below
h(z) =
∞
X
hk z k ,
k=0
where hk is defined in (6.1). Hence
h0 =
f (w(0))
w0 (0)f 0 (0)
= 0, h1 =
= w0 (0).
0!
1!
Therefore, we can write
h(z) = h1 z +
∞
X
hk z k
k=2
and
J1λ,µ h(z) = h1 z +
∞
X
C1λ,µ (k)hk z k ,
k=2
Aλ,µ,ν
(1, β)
γ
we must prove h(z) ∈
in other words we show that
(
)
z(J1λ,µ h(z))0 − β(1 − γ)J1λ,µ h(z) − βγz 2 (J1λ,µ h(z))00
Re
>0
(1 − γ)J1λ,µ h(z) + γz 2 (J1λ,µ h(z))00
or
∞
∞
X
X
hk C1λ,µ (k)z k
Re{[h1 z +
khk C1λ,µ (k)z k − β(1 − γ)h1 z − β(1 − γ)
k=2
k=2
−βγ
∞
X
k(k − 1)hk C1λ,µ (k)z k ]/[(1 − γ)h1 z + (1 − γ)
+γ
C1λ,µ (k)hk z k
k=2
k=2
∞
X
∞
X
k(k − 1)hk C1λ,µ (k)z k ]} > 0,
k=2
or we prove


∞
P
λ,µ
k−1



hk C1 (k)z
(β(1 − γ) + βγk(k − 1) − k) 
 h1 (1 − β(1 − γ)) −

k=2
Re
> 0.
∞
P






(1 − γ)h1 +
hk C1λ,µ (k)((1 − γ) + γk(k − 1))z k−1
k=2
Letting z →
(
1− ,
and using the mean value theorem, we have to prove
Re h1 (1 − β(1 − γ)) −
∞
X
)
(β(1 − γ) + βγk(k − 1) −
k)hk C1λ,µ (k)
>0
k=2
by (6.2) the last inequality is true and the result can be obtained.
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Surveys in Mathematics and its Applications 5 (2010), 35 – 47
http://www.utgjiu.ro/math/sma
46
W. G. Atshan and S. R. Kulkarni
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47
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Waggas Galib Atshan
S. R. Kulkarni
Department of Mathematics,
Department of Mathematics,
College of Computer Science and Mathematics,
Fergusson College, Pune - 411004,
University of AL-Qadisiya, Diwaniya, Iraq.
India.
e-mail: waggashnd@yahoo.com
e-mail: kulkarni− ferg@yahoo.com
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Surveys in Mathematics and its Applications 5 (2010), 35 – 47
http://www.utgjiu.ro/math/sma