Surveys in Mathematics and its Applications
ISSN 1842-6298 (electronic), 1843-7265 (print)
Volume 4 (2009), 77 – 88
APPLICATIONS OF GENERALIZED
RUSCHEWEYH DERIVATIVE TO UNIVALENT
FUNCTIONS WITH FINITELY MANY
COEFFICIENTS
Abdul Rahman S. Juma and S. R. Kulkarni
Abstract. By making use of the generalized Ruscheweyh derivative, the authors investigate
several interesting properties of certain subclasses of univalent functions having the form
X
m
f (z) = z −
n=2
1
X
∞
(1 − β)en
zn −
ak z k .
(1 − nβ + α(1 − n))B1λ,µ (n)
k=m+1
Introduction
Let A(n, p) denote the class of functions f normalized by f (z) = z p −
∞
P
ak z k , (p, n ∈
k=n+p
IN = {1, 2, 3, 4, ....}), which are analytic and multivalent in the unit disk U = {z :
|z| < 1}. The
function f (z) is said to be starlike of order δ(0 ≤ δ < p) if and only
zf 0 (z)
if Re f (z) > δ, (z ∈ U). On the other hand f (z) is said to be convex of order
00 (z)
δ(0 ≤ δ < p) if and only if Re 1 + zff 0 (z)
> δ, (z ∈ U).
Now for α ≥ 0, 0 ≤ β < 1 and λ > −1, µ, ν ∈ IR, the following classes
(
(
)
)
J λ,µ f (z)
Jpλ,µ f (z)
p
λ,µ
Mp (α, β) = f (z) ∈ A(n, p) : Re
> α
− p + β
z(Jpλ,µ f (z))0
z(Jpλ,µ f (z))0
(1.1)
(
Mλ,µ
1,1 (α, β)
=
f (z) ∈ A(1, 1) : Re
(
J1λ,µ f (z)
z(J1λ,µ f (z))0
)
)
J λ,µ f (z)
1
> α
− 1 + β
z(J λ,µ f (z))0
1
(1.2)
2000 Mathematics Subject Classification: 30C45
Keywords: Ruscheweyh derivative; Integral operator; Univalent function.
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78
A. R. S. Juma and S. R. Kulkarni
where J1λ,µ f (z) is the generalized Ruscheweyh derivative of f defined by
J1λ,µ f (z) =
Γ(µ − λ + ν + 2) λ,µ,ν µ−1
zJ
(z
f (z))
Γ(ν + 2)Γ(µ + 1) 0,z
(1.3)
λ,µ,ν
where J0,z
f (z) generalized fractional derivative operator of order λ (see eg.[1], [5],
[7], [8]).
We can write (1.3) as
J1λ,µ f (z)
=z−
∞
X
ak B1λ,µ (k)z k
(1.4)
k=n+1
where
B1λ,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)
(1.5)
λ,µ
Consider the class Mλ,µ
1,1 (α, β, em ) is subclass of M1,1 (α, β) consisting of functions of the form
f (z) = z −
m
X
(1 − β)en
λ,µ
n=2 (1 − nβ + α(1 − n))B1 (n)
zn −
∞
X
ak z k ,
(1.6)
k=m+1
where
en =
[1 − nβ + α(1 − n)]B1λ,µ (n)
an .
1−β
Many authors have studied the different cases (e.g. [2], [4], [6]), and for λ =
µ, ν = 1, we get the case was studied by [3].
2
Coefficient Bounds
In order to prove our results, we need the following Lemma due to A.R.S.Juma and
S. R. Kulkarni [2].
Lemma 1. Let f (z) = z p −
∞
P
ak z k . Then f (z) ∈ Mλ,µ
p (α, β) if and only if
k=m+p
∞
X
[(1 + α) − k(αp + β)] λ,µ
B (k)ak < 1.
[(1 + α) − p(αp + β)] p
(2.1)
k=m+p
First we prove
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79
Applications of generalized Ruscheweyh derivative
Theorem 2. Let the function f (z) defined by (1.6). Then f ∈ Mλ,µ
1,1 (α, β, em ) if
and only if
∞
m
X
X
(1 − kβ) + α(1 − k) λ,µ
B1 (k)ak < 1 −
en .
(2.2)
1−β
n=2
k=m+1
Proof. Let
an =
(1 − β)en
(1 − nβ + α(1 − n))B1λ,µ (n)
.
(2.3)
λ,µ
We can say that f ∈ Mλ,µ
1,1 (α, β, em ) ⊂ M1,1 (α, β) if and only if
m
X
[(1 − nβ) + α(1 − n)]B λ,µ (n)
1
1−β
n=2
or
an +
∞
X
[1 − kβ + α(1 − k)]B1λ,µ (k)
ak < 1
1−β
k=m+1
∞
m
X
X
[1 − kβ + α(1 − k)]B1λ,µ (k)
ak < 1 −
en .
1−β
n=2
k=m+1
Corollary 3. Let f (z) defined by (1.6) be in Mλ,µ
1,1 (α, β, em ). Then for k ≥ m + 1
we have
∞
P
en )
(1 − β)(1 −
n=2
ak ≤
,
(2.4)
(1 − kβ + α(1 − k))B1λ,µ (k)
this result is sharp for
h(z) = z −
m
X
(1 − β)(1 −
(1 − β)en
n
λ,µ
n=2 [(1 − nβ) + α(1 − n)]B1 (n)
z −
∞
P
en )
n=2
zk .
[(1 − kβ) + α(1 − k)]B1λ,µ (k)
(2.5)
Corollary 4. Let f defined by (1.6). Then f ∈ M0,0
1,1 (0, β, em ), that is,
f (z) = z −
m
X
en (1 − β)
n=2
1 − nβ
n
z −
∞
X
ak z k ,
k=m+1
if and only if
∞
m
X
X
1 − kβ
ak < 1 −
en .
1−β
k=m+1
n=2
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80
A. R. S. Juma and S. R. Kulkarni
Corollary 5. Let f defined by (1.6). Then f ∈ M1,0
1,1 (0, β, em ), that is,
m
∞
X
X
(1 − β)(ν + 1)en
n
f (z) = z −
=z −
ak z k
(1 − nβ)Γ(n + ν)
n=2
if and only if
k=m+1
m
∞
X
X
(1 − kβ)Γ(k + ν)
ak < 1 −
en .
(1 − β)(ν + 1)
n=2
k=m+1
Corollary 6. Let f defined by (1.6). Then f ∈ M0,0
1,1 (α, 0, em ), that is,
f (z) = z −
m
X
n=2
if and only if
∞
X
∞
X
en
zn −
ak z k
1 + α(1 − n)
k=m+1
(1 + α(1 − k))ak < 1 −
m
X
en .
n=2
k=m+1
Corollary 7. Let f defined by (1.6). Then f ∈ M1,0
1,1 (α, 0, em ) , that is,
f (z) = z −
m
X
n=2
if and only if
∞
X
en (ν + 1)
n
ak z k
z −
(1 + α(1 − n))Γ(n + ν)
k=m+1
∞
m
X
X
[1 + α(1 − k)]Γ(k + ν)
ak < 1 −
en .
(ν + 1)
n=2
k=m+1
We claim that all these results are entirely new.
3
Extreme points and other results
Theorem 8. Let f1 (z), f2 (z), ..., f` (z) defined by
fi (z) = z −
m
X
(1 − β)en
n
λ,µ
n=2 [1 − nβ + α(1 − n)]B1 (n)
z −
∞
X
ak,i z k ,
(3.1)
k=m+1
(i = 1, · · · , `) be in the class Mλ,µ
1,1 (α, β, em ). Then G(z) defined by
G(z) =
`
X
i=1
λi fi and
`
X
i=1
λi = 1, 0 ≤
m
X
en ≤ 1, 0 ≤ en ≤ 1
n=2
is also in this class.
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81
Applications of generalized Ruscheweyh derivative
Proof. In view of Theorem 2, we have
∞
m
X
X
[1 − kβ − α(1 − k)]B1λ,µ (k)
ak,i < 1 −
en
1−β
n=2
k=m+1
for every i = 1, 2, · · · , `. Here
G(z) =
`
X
λi fi = z −
i=1
m
X
(1 − β)en
n=2
(1 − nβ + α(1 − n))B1λ,µ (n)
∞
X
−
`
X
(
λi ak,i )z k .
k=m+1 i=1
Thus,
`
∞
X
(1 − kβ + α(1 − k))B1λ,µ (k) X
(
λi ak,i )
1−β
i=1
k=m+1
!
`
∞
X
X
(1 − kβ + α(1 − k))B1λ,µ (k)
=
ak,i λi
1−β
i=1 k=m+1
<
`
X
i=1
(1 −
m
X
en )λi = 1 −
n=2
Remark 9. The function G(z) =
m
X
en .
n=2
1
2 [f1 (z)
+ f2 (z)] belongs to Mλ,µ
1,1 (α, β, em ) if
f1 (z), f2 (z) are in the class Mλ,µ
1,1 (α, β, em ).
Remark 10. The class Mλ,µ
1,1 (α, β, em ) is convex set.
Theorem 11. Let fi (z), (i = 1, · · · , `) defined by (3.1) be in Mλ,µ
1,1 (α, β, en ). Then
the function
F (z) = z −
m
X
en (1 − β)
λ,µ
n=2 (1 − nβ + α(1 − n))B1 (n)
is also in the class Mλ,µ
1,1 (α, β, em ), where bk =
∞
X
n
1
`
z −
P̀
bk z k
(bk ≥ 0)
k=m+1
ak,i .
i=1
Proof. It is clear that
∞
∞
`
X
X
(1 − kβ + α(1 − k))B1λ,µ (k)
(1 − kβ + α(1 − k))B1λ,µ (k) X
bk =
(
ak,i )
1−β
`(1 − β)
i=1
k=m+1
k=m+1
!
`
∞
λ,µ
X (1 − kβ + α(1 − k))B (k)
1X
1
=
ak,i ,
`
1−β
i=1
k=m+1
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82
A. R. S. Juma and S. R. Kulkarni
by Theorem 2, we have the last expression is less than
m
P
1
`
P̀
m
P
(1 −
i=1
en ) = 1 −
n=2
en .
n=2
The next Theorem is very useful to obtain the extreme points of the class
Mλ,µ
1,1 (α, β, em ).
Theorem 12. Let
fm (z) = z −
m
X
en (1 − β)
λ,µ
n=2 (1 − nβ + α(1 − n))B1 (n)
zn
(3.2)
and for k ≥ m + 1
fk (z) = z −
m
X
n=2
(1 − β)(1 −
en (1 − β)
(1 − nβ + α(1 −
n))B1λ,µ (n)
zn −
m
P
en )
n=2
(1 − kβ + α(1 −
k))B1λ,µ (k)
zk .
(3.3)
Then the function f (z) ∈
if and only if it can be expressed in the
∞
∞
P
P
form f (z) =
δk fk (z), where δk ≥ 0, (k ≥ m) and
δk = 1.
Mλ,µ
1,1 (α, β, em )
k=m
k=m
Proof. Let
f (z) =
∞
X
δk fk (z) + δm fm (z)
k=m+1
= δm z − δm
+
∞
X
m
X
en (1 − β)
λ,µ
n=2 (1 − nβ + α(1 − n))B1 (n)
δk z −
∞
X
δk
m
X
zn
(1 − nβ + α(1 −

(1 −
en )(1 − β)
∞
X


n=2
k
−
δk 
z
 (1 − kβ + α(1 − k))B λ,µ (k) 
1
k=m+1
k=m+1
∞
X
δk )z − (δm +
k=m+1
−
∞
X
k=m+1
∞
X
k=m+1
(1 −
n))B1λ,µ (n)
zn
m
P

= (δm +
n=2
k=m+1
!
en (1 − β)
m
P
δk )
m
X
en (1 − β)
λ,µ
n=2 (1 − nβ + α(1 − n))B1 (n)
zn
en )(1 − β)
n=2
(1 − kβ + α(1 −
δk z
k))B1λ,µ (k)
k
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83
Applications of generalized Ruscheweyh derivative
= z−
m
X
en (1 − β)
n=2 (1 − nβ + α(1 −
n))B1λ,µ (n)
(1 −
∞
X
zn −
m
P
en )(1 − β)
n=2
k=m+1 (1 − kβ + α(1 −
P e )(1−β)
δk z
k))B1λ,µ (k)
k
,
m
therefore at the end we can write
(1−kβ+α(1−k))(1−
∞
P
(1−β)(1−kβ+α(1−k))B1λ,µ (k)
k=m+1
= (1 −
m
X
∞
X
en )
n=2
n
n=2
δk = (1 −
m
X
en )(1 − δm ) < 1 −
n=2
k=m+1
δk B1λ,µ (k)
m
X
en .
n=2
Thus f ∈ Mλ,µ
1,1 (α, β, em ).
Conversely, let f ∈ Mλ,µ
1,1 (α, β, em ), that is,
f (z) = z −
m
X
en (1 − β)
λ,µ
n=2 (1 − nβ + α(1 − n))B1 (n)
zn −
∞
X
ak z k ,
k=m+1
put
δk =
(1 − kβ + α(1 − k))B1λ,µ (k)
ak (k ≥ m + 1)
m
P
(1 − β)(1 −
en )
n=2
∞
P
we have δk ≥ 0 and if we set δm = 1 −
δk , then we have
k=m+1
f (z) = z −
m
X
en (1 − β)
n
λ,µ
n=2 (1 − nβ + α(1 − n))B1 (n)
= fm (z) −
= fm (z) −
= (1 −
∞
X
k=m+1
∞
X
k=m+1
∞
X
(1 −
m
P
en )(1 − β)
n=2
λ,µ
k=m+1 (1 − kβ + α(1 − k))B1 (k)
!
m
X
en (1 − β)
z−
z n − fk (z) δk
λ,µ
(1
−
β
+
α(1
−
n))B
(n)
1
n=2
(fm (z) − fk (z))δk
δk )fm (z) +
k=m+1
z −
∞
X
∞
X
k=m+1
δk fk (z) =
∞
X
δk fk (z).
k=m+1
Corollary 13. The extreme points of the class Mλ,µ
1,1 (α, β, em ) are the functions
fk (z), (k ≥ m), defined by (3.2), (3.3).
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δk z k
84
A. R. S. Juma and S. R. Kulkarni
Theorem 14. Let f (z) defined by (1.6) be in the class Mλ,µ
1,1 (α, β, em ). Then f (z)
is starlike of order δ(0 ≤ δ < 1) in |z| < r, where r is the largest value such that
m
X
n=2
<
en
(1 − nβ + α(1 −
n))B1λ,µ (n)
r
n−1
(1 −
∞
X
+
k=m+1
m
P
en )
n=2
(1 − kβ + α(1 −
k))B1λ,µ (k)
1
, (k ≥ m + 1).
1−β
rk−1
(3.4)
Proof. It is sufficient to show that
0
zf (z)
f (z) − 1 < 1 − δ.
(3.5)
Therefore
m
∞
P
P
(1−β)en
n−
k
z
n
ka
z
z −
0
k
λ,µ
(1−nβ+α(1−n))B
(n)
zf (z)
1
n=2
k=m+1
− 1
m
∞
f (z) − 1 = P
(1−β)en
z− P
zn −
ak z k
(1−nβ+α(1−n))B1λ,µ (n)
n=2
k=m+1
∞
P
(n−1)(1−β)en
|z|n−1 +
(k − 1)ak |z|k−1
λ,µ
(1−nβ+α(1−n))B
(n)
1
n=2
k=m+1
m
∞
P
P
(1−β)en
1−
|z|n−1 −
ak |z|k−1
λ,µ
n=2 (1−nβ+α(1−n))B1 (n)
k=m+1
∞
m
P
P
(k−1)(1−β)(1−m
(n−1)(1−β)en
n=2 en )
rn−1 +
rk−1
λ,µ
λ,µ
(1−nβ+α(1−n))B
(1−kβ+α(1−k))B
(n)
(k)
1
1
n=2
k=m+1
m
P
≤
<
Pe )
m
1−
m
P
(1−β)en
rn−1
λ,µ
n=2 (1−nβ+α(1−n))B1 (n)
−
∞
P
k=m+1
(1−β)(1−
.
n
n=2
(1−kβ+α(1−k))B1λ,µ (k)
rk−1
Thus (3.5) holds true if the last expression is less than 1 − δ or,
m
X
n=2
+
∞
X
k=m+1
(n − δ)(1 − β)en
n))B1λ,µ (n)
(1 − δ)(1 − nβ + α(1 −
m
P
(k − δ)(1 − β)(1 −
en )
n=2
(1 − δ)(1 − kβ + α(1 −
k))B1λ,µ (k)
rn−1
rk−1 < 1
at the end we find (3.4).
Making use the following Theorem we obtain the next corollary.
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85
Applications of generalized Ruscheweyh derivative
Theorem 15. [Alexander’s Theorem] Let f be analytic in U with f (0) = f 0 (0) −
1 = 0. Then f is convex function if and only if zf 0 is starlike function.
Corollary 16. Let f ∈ Mλ,µ
1,1 (α, β, em ). Then f (z) is convex of order δ(0 ≤ δ < 1)
in |z| < r where r is the largest value such that
m
X
n=2
k(1 −
nen
(1 − nβ + α(1 −
n))B1λ,µ (n)
rn−1 +
m
P
en )
n=2
(1 − kβ + α(1 −
k))B1λ,µ (k)
rk−1 <
1
.
1−β
Theorem 17. Let f ∈ Mλ,µ
1,1 (α, β, em ) and λ > 0, if
dn =
(1 − β)e2n
(2 ≤ n ≤ m),
(1 − nβ + α(1 − n))B1λ,µ (n)
(3.6)
then the function
H(z) = z −
m
X
(1 − β)dn
λ,µ
n=2 ((1 − nβ) + α(1 − n))B1 (n)
zn −
∞
X
ak z k
k=m+1
is also in the class Mλ,µ
1,1 (α, β, em ).
Proof. By assumption we have (1 − nβ + α(1 − n))B1λ,µ (n) > 1 therefore,
dn =
So, 0 ≤
m
P
n=2
dn <
m
P
(1 − β)e2n
(1 − nβ + α(1 − n))B1λ,µ (n)
< en ≤ 1.
en ≤ 1, thus
n=2
∞
∞
X
X
(1 − kβ + α(1 − k))B1λ,µ (k)
(1 − kβ + α(1 − k))B1λ,µ (k)
a
<
< 1.
k
m
m
P
P
k=m+1
k=m+1
(1 − β)(1 −
dn )
(1 − β)(1 −
en )
n=2
n=2
Theorem 18. Let f, g ∈ Mλ,µ
1,1 (α, β, em ) and λ > 0. Then
(f ∗ g)(z) = z −
m
X
(1 − β)2 e2n
zn −
λ,µ
2
2
n=2 (1 − nβ + α(1 − n)) (B1 (n))



is also in the class Mλ,µ
1,1 (α, β, dm ) if λ1 ≤ inf
k
∞
X
ak bk z k
k=m+1
(B1λ,µ (k))2
P d ) − 1, where dn(2 ≤ n ≤ m)
m
(1−
n
n=2
are defined by (3.6).
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86
A. R. S. Juma and S. R. Kulkarni
Proof. By making use of (3.6), we have
(f ∗ g)(z) = z −
m
X
(1 − β)dn
λ,µ
n=2 (1 − nβ + α(1 − n))B1 (n)
∞
X
zn −
ak bk z k .
k=m+1
By Theorem 17, we have
∞
X
(1 − kβ + α(1 − k))B1λ,µ (k)
ak < 1
m
P
k=m+1
(1 − β)(1 −
dn )
n=2
and
∞
X
(1 − kβ + α(1 − k))B1λ,µ (k)
bk < 1.
∞
P
k=m+1
(1 − β)(1 −
dn )
n=2
Therefore, by Cauchy-Schwarz inequality we obtain
∞
X
(1 − kβ + α(1 − k))B1λ,µ (k) p
ak bk < 1.
m
P
k=m+1
dn )
(1 − β)(1 −
(3.7)
n=2
Now, we want to prove that
∞
X
(1 − kβ + α(1 − k))B1λ,µ (k)
ak bk < 1.
m
P
k=m+1
(1 − β)(1 −
dn )
(3.8)
n=2
In view of (3.7) the inequality (3.8) holds true if
p
ak bk
B1λ1 ,µ (k)
B1λ,µ (k)
< 1.
(3.9)
But we have
B1λ,µ (k) p
(1 − kβ + α(1 − k))B1λ,µ (k) p
a
b
<
ak bk < 1.
k k
m
m
P
P
1−
dn
(1 − β)(1 −
dn )
n=2
n=2
Therefore (3.9) holds true if
1−
m
P
dn
n=2
B1λ,µ (k)
<
B1λ,µ (k)
B1λ1 ,µ (k)
, or B1λ1 ,µ (k) <
(B1λ,µ (k))2
.
m
P
1−
dn
n=2
Then λ1 <
(B1λ,µ (k))2
m
1−
dn
n=2
P
− 1.
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Surveys in Mathematics and its Applications 4 (2009), 77 – 88
http://www.utgjiu.ro/math/sma
Applications of generalized Ruscheweyh derivative
87
Theorem 19. Let f (z) ∈ Mλ,µ
1,1 (α, β, em ). Then
Z z
E(t) − α
λ,µ
J1 f (z) = exp
dt , |E(z)| < 1, z ∈ U.
0 (βE(t) − α)t
Proof. The case α = 0 is obvious. Therefore, suppose that α 6= 0. Then for f ∈
J λ,µ f (z)
1
Mλ,µ
1,1 (α, β, em ) and let w = z(J λ,µ f (z))0 we have Re(w) > α|w − 1| + β, therefore,
1
w−1 1
w−1
= E(z)
,
where
|E(z)| < 1, z ∈ U. This yields
w−β < α or w−β
α
Thus
J1λ,µ f (z)
z(J1λ,µ f (z))0
−1
J1λ,µ f (z)
z(J1λ,µ f (z))0
−β
(J1λ,µ f (z))0
(J1λ,µ f (z))
=
=
E(z)
E(z)
J1λ,µ f (z) − z(J1λ,µ f (z))0
=
or
.
λ,µ
λ,µ
0
α
α
J1 f (z) − βz(J1 f (z))
E(z) − α
.
(βE(z) − α)z
By the integration we get the result.
References
[1] S. P. Goyal and Ritu Goya, On a class of multivalent functions defined
by generalized Ruscheweyh derivatives involving a general fractional derivative operator, Journal of Indian Acad. Math., 27 (2) (2005), 439-456.
MR2259538(2007d:30005). Zbl 1128.30008.
[2] A. R. S. Juma and S. R. Kulkarni, Application of generalized Ruscheweyh derivative to multivalent functions, J.Rajasthan Acad. Phy. Sci. 6 (2007), no.3, 261-272.
MR2355701.
[3] Sh. Najafzadeh and S. R. Kulkarni, Application of Ruscheweyh derivative on
univalent functions with negative and fixed many coefficients, J. Rajasthan Acad.
Phy. Sci., 5 (1) (2006), 39-51. MR2213990. Zbl 1137.30005.
[4] S. Owa, M. Nunokawa and H. M. Srivastava, A certain class of multivalent
functions, Appl. Math. Lett., 10 (2) (1997), no.2, 7-10. MR1436490.
[5] R. K. Raina and T. S. Nahar, Characterization properties for starlikeness and
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MR1796782(2001h:30014).
[6] S. Shams and S. R. Kulkarni, class of univalent functions with negative and
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Surveys in Mathematics and its Applications 4 (2009), 77 – 88
http://www.utgjiu.ro/math/sma
88
A. R. S. Juma and S. R. Kulkarni
[7] H. M. Srivastava, Distortion inequalities for analytic and univalent functions associated with certain fractional calculus and other linear operators
(In analytic and Geometric Inequalities and Applications, eds. T. M. Rassias and H.M. Srivastava), Kluwar Academic Publishers, 478 (1999), 349-374.
MR1785879(2001h:30016).
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their applications, Applied Math. and Computation, 118 (2001), no. 1, 1-52.
MRMR1805158(2001m:26016). Zbl 1022.26012.
Abdul Rahman S. Juma
S. R. Kulkarni
Department of Mathematics
Department of Mathematics
Al-Anbar University,
Fergusson College,
Ramadi, Iraq.
Pune 411004, India
e-mail: absa662004@yahoo.com
e-mail: kulkarni− ferg@yahoo.com
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Surveys in Mathematics and its Applications 4 (2009), 77 – 88
http://www.utgjiu.ro/math/sma