International Journal of Scientific & Engineering Research, Volume 3, Issue 10, October-2012 1
ISSN 2229-5518
Department of Mathematics, University of Rajasthan
Jaipur-302004, India
AND
Department of Mathematics, Govt. Post Graduate College
Neemuch-458441, India
The aim of this paper is to analysis the subclass SC(,,) pertaining to the
Hadamard product of pq-function ([12]) with negative coefficients in unit disc
= {z : | z | < 1}.
Further, coefficient estimates, distortion theorem and radius of convexity for this class are also established. In addition we discuss closure properties and integral operator for function belonging to the class SC(,,).
Key words and Phrases: pq-function, Hadamard product, coefficient estimates,
Distortion theorem, closure properties.
*2000 Mathematics Subject Classification. 30 C 45, 33 C 20.
+E-mail: drvblc@yahoo.com
++E-mail: meghwal66@gmail.com
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Let A denote the class of the function of the form
f(z) z k2
a z k
k
…(1.1)
which are analytic in the unit disc = {z : | z | < 1 }.
A function f A is said to belong to the class A of starlike functions of order
(0 < 1), if it satisfies, for z , the conditions
z f ' z)
Re
f(z)
…(1.2)
We denote this class by S*(). Further, f A is said to be convex function of order
in , if it satisfies
z f '' z)
Re 1
f ' z)
z
…(1.3)
for some (0 < 1). We denote this class k(). Let T denote subclass of A,
consisting functions of the form
f(z)
z
k2
a z k
k
a 0
k
…(1.4)
The function
S z) z (1 z) 21
0 1
…(1.5)
is the familiar extremal function for the class S*(), setting
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k
i 2
C( k) i2
k 1
k 2,
…(1.6)
Using (1.5) and (1.6), we can write
S z) z
k2
C ( k) z k
…(1.7)
Clearly, C(,k) is a decreasing function in , and that
1
, 2
lim
k
C( k) 1
1
2
1
…(1.8)
1
2
By the definition of differential operator Dn, introduced by Slagean [8], we know
that
Dn f(z) z k2
k n a
k
z k
…(1.9)
Therefore Hadamard product of two analytic functions given by (1.7) and (1.9) can
be written as
Dn
f S z) z
k2
k n C( k) a z k
k
…(1.10)
Here we use the condition which is satisfied by the subclass SC(,,)
1 z[z(D n f S
' z) (1 Dn f S
z)'
Re 1
…(1.11)
z(D n f S
' z) (1 Dn f S
z)
0 1 0 1 C, z
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The Fox-Wright function [12, p.50, equation 1.5] appearing in this paper is
defined by
a
p
j1
a
j
k
z k j
j j 1,p
z)
p q p q b
z q
…(1.12)
j j 1,q
k 0
j1
b k k !
j j
where
q
j 1,..., p) and j 1,..., q) are real and pos itiveand1
j j
j1
p
j j j1
Now we can write
z {
p
z)}]
q
p
j1
q
a
j
z
p
j1
q
a k 1 z k j j
…(1.13)
j-1
b
j
k 2
j1
b
j
k 1 k 1
j
and
A [z{
p
z)}] A (z
q
p q z A
p
j1
q
a k 1 z k j j
…(1.14)
k 2
j1
b
j
k 1 k 1
j
where
q
A j1 p
j1
b
j
a
j
…(1.15)
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A[z{
p
z)}]
q
is in the class SC(,,) iff
A k n k 1 k 1 1 C( k)
p
j1
q
a k 1
j j
1
k 2
j1
b
j
k 1 k 1
j
…(2.1)
1 z [z{D n A(z
S
p q
' z) (1 Dn A(z
S
p q
z)] '
1
1
z {D n A(z
S
p q
' z) (1 Dn A(z
S
p q
z)
p
a
j
k 1 C( k)
j
A
k n j1 q
k 1 1 k)z k 1
k 2
j1
b
j
p
k 1 k 1
j
1
1 A
k n j1 q
a k 1 C( k)
j j
k 1 z k 1
k 2
j1
b
j
k 1 k 1
j
Hence, by using the maximum modulus principle,
A [z{
p
z)}]
q
is in the class
SC(,,). Conversely, assume that the function
A [z{
p
z)}]
q
defined by (1.14) is
in the class SC(,,). Then we will have
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z[z{D n A(z
p q
S
' z) (1 Dn A(z
S
p q
z)] '
Re 1
z {D
n A(z S ' z) (1 Dn
p q
1
A(z S z)
p q
p
j1
a
j
k 1 C( k)
j
A
k n k 1 1 k)z k 1
1
Re 1
k 2
j1
b
j
p
k 1 k 1
j
1 A
k n j1
q
a
j
k 1 C( k)
j
k 1 z
k 1
k 2
j1
b
j
k 1 k 1
j
and now when
z 1
we obtain
A k n k 1 1 k) C( k)
p
j1
q
a k 1
j j
k 2
j1
p
b
j
k 1 k 1
j
1
1 A
k n k 1
j1
q
a k 1 C( k)
j j
k 2
j1
b
j
k 1 k 1
j
and finally
A k n k 1 k 1 1 C( k)
p
j1
q
a k 1
j j
1
k 2
j1
b
j
k 1 k 1
j
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A [z{
p
z)}]
q
defined by (1.14) be in the class
SC(,,). Then
p
j1
q
a
j
k 1
j
1
A k n k 1 k 1 1 C( k)
k
2).
j1
b
j
k 1 k 1
j
…(2.2)
and the equality is attained for the function
A [z{
p
z)}]
q
given by
A [z{
p
z)}] z
q
1
k n k 1 k 1 1 C( k)
z k
…(2.3)
A [z{
p
z)}]
q
be in class SC(,,) then for
0 | z | = r
r
1
r k A [z{
z)}] |
k n k 1 k 1 1 C( k) p q
r
1
k n k 1 k 1 1 C ( k)
r k
…(3.1)
z |
1
k n k 1 k 1 1 C( k)
| z |k A [z{
p
z)}] |
q
z |
1
k n k 1 k 1 1 C ( k)
| z |k
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Now as we have assumed | z | = r < 1, we get the required result easily.
A [z{
p
z)}]
q
is in the class SC(,,) then
A [z{
p
z)}]
q
is included in a disc with centre at the origin and radius r, where
r 1
1
k n k 1 k 1 1 C ( k)
…(3.2)
A [z{
p
z)}]
q
be in the class SC(,,) then
1 1
k n1 k 1 k 1 1 C( k)
r k1 A [z{
p
z)}] |
q
1
1
k n1k 1 k 1 1 C ( k)
r k1
where equality holds for the function
A [z{
p
z)}]
q
given by (1.14).
1 k1
k n k 1 k 1 1 C( k)
| z |k1 A [z{
p
z)}] |
q
1
k1
k n k 1 k 1 1 C ( k)
| z |k1
Again by assuming | z | = r, we get the desired result easily.
A [z{
p
z)}]
q
is in the class SC(,,) then
A [z{
p
z)}] is
q
convex in | z | < R, where
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R Inf. k n 2 k 1 k 1 1 C( k)
p
j1
q
a
j
k 1
j
1
k 1
…(4.1)
j1
b
j
k 1 k 1
j
The result is sharp.
z Az{
p
z)}] '' q
1
z | R
[Az{
p
z)}] '
q
In view of (1.4), we have
A
k 2
p k(k 1 j1
q
a
j
k 1
j
z |k 1
z Az{
p
z)}] ' '
q
j1
b k 1 k 1
j j
[Az{
p
z)}] ' q
1 A
p
k j1 q
a
j
k 1 z |k 1 j
k 2
j1
b
j
k 1 k 1
j
Hence, we get
A k 2
p
j1
q
a
j
k 1 z |k 1 j
1
…(4.2)
k 2
j1
b
j
k 1 k 1
j
But from Theorem 1, we have
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p
k n k 1 k 1 1 C( k)
A j1 q
a
j
k 1
j
1
…(4.3)
k 2
1
j1
b
j
k 1 k 1
j
and thus from (4.2) and (4.3), we obtain
1
k n2 k 1 k 1 1 C ( k) k 1
z |
1
Hence
A [z{
p
z)}]
q
is convex in | z | < R. The result is sharp and given by (4.1).
A [z{
pr
z)}],
qr
(r = 1,2,…,m) be defined by
A [z{
pr
z)}] z A
r
p
j1
q
a k 1z k jr jr
…(5.1)
k 2
j1
b
jr
k 1 k 1
jr
for z , be in the class SC(,,) then the function h(z) defined by
h(z) z k2
b z k k
also belongs to the class SC(,,), where
1 m
b a
k m r1 kr
…(5.2)
where
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a A
kr
p
j1
q
a k 1
jr jr
…(5.3)
j1
b k 1 k 1
jr jr
A [z{
pr
z)}]
qr
belongs to SC(,,), it follows from Theorem 1, that
A k n k 1 k 1 1 C ( k)
p
j1
q
a k 1
jr jr
k 2
. 1
k 1
1
r 1,2,..., m)
j1
b
jr
k 1
jr
Therefore
k2
k n k 1 k 1 1 C ( k) b
k
1 m
k n k 1 k 1 1 C( k)
m
a
kr
k2
r1
1 m
m r1
k2
k n k 1 k 1 1 C ( k) a
1
kr
where akr is given by (5.3).
Hence by Theorem 1,
h(z) SC(
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A [z{
p
z)}]
q
defined by (1.14) be in the class
SC(,,) then pq (z), defined by
a z
z)
j j 1,p z
x)
dx.
…(6.1)
p q p
q b j j 1,q
0 p q
also belongs to the class SC(,,).
Proof. From the representation of pq(z), it is obtained that
z)
p q
p
j1
q
a
j
z
p
j1
q
a k z k j j j
j1
b
j
k 2
j1
b
j
k
j
k !
j
and
A [
p q
z)] z A
p
j1
q
a k 1 z k j j
…(6.2)
k 2
j1
b
j
k 1 k !
j
where A is given by (1.15).
Therefore
A k n k 1 k 1 1 C ( k)
p
j1
q
a k
j j j
1
k 2
j1
b
j
k
j
k !
j
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Since
A [z{
p
z)}]SC(
q
so by virtue of Theroem 1,
{A
p q
z)}
is in the
class SC(,,).
Then pq(z) is univalent in | z | < R*, where
A
p q
z)} is in the classSC(
and defined by
equation (6.2).Then
is univalent in | z | < R*, where
p q
k n
R * Inf.
1
k 1 k 1 1 C( k) k 1 k 2
…(6.3)
1
The result is sharp.
A {
p
z)}' 1 1
q
for | z | R *
Now since
A{
p
z)}' 1 A
q
p
j1
q
a
j
k
j
z |k 1 j
1
…(6.4)
k 2
j1
b
j
k
j
k 1
j
But from Theorem 1, we know that
p
k n k 1 k 1 1 C( k)
A j1 q
a k 1
j j
1
…(6.5)
k 2
1 )
j1
b
j
k 1 k 1
j
From equation (6.4) and (6.5), we have
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z |
k n
1
k 1 k 1 1 C( k) k 1 k
2).
1
The result is sharp and given by (6.3).
On putting
j 1,..., p) 1 and j 1,..., q) 1
j j
in the result (2.1), (3.1) and
(4.1), the coefficient estimates, Distortion Theorem and radius of convexity will also applicable for Generalized Hypergeometric function pFq(z). [1,p.73, equation 2].
We obtain the following results:
(I) Let the function [z{pFq(z)}] is in the class SC(,,) iff
A k n k 1 k 1 1 C ( k)
p
j1
q
a k 1
j
1
k 2
j1
b
j
k 1 k 1!
…(7.1)
(II) Let the function z{
p
F z)}]
q
be in the class SC(,,) then for 0 | z | < r
r 1 r k
k n k 1 k 1 1 C( k)
z{
p
F z)}] |
q
r
(1
k n k 1 k 1 1 C( k)
r k
…(7.2)
(III) If
z{
p
F z)}]
q
is in the class SC(,,) then
z{
p
F z)}]
q
is convex in
| z | < R, where
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R Inf. k n 2 k 1 k 1 1 C ( k)
p
j1
q
a
j
k 1
1
k 1
j1
b
j
k 1 k 1!
…(7.3)
The result is sharp.
(IV) Closure property and integral operator for the function pFq(z) can also be examine to the class SC(,,).
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The authors are highly thankful to Professor H.M. Srivastava of the University of Victoria, Victoria, Canada, for his kind help and many valuable suggestions in the preparation of this paper.
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