Volume 10 (2009), Issue 2, Article 53, 11 pp.
ON APPLICATION OF DIFFERENTIAL SUBORDINATION FOR CERTAIN
SUBCLASS OF MEROMORPHICALLY p-VALENT FUNCTIONS WITH POSITIVE
COEFFICIENTS DEFINED BY LINEAR OPERATOR
WAGGAS GALIB ATSHAN AND S. R. KULKARNI
D EPARTMENT OF M ATHEMATICS
C OLLEGE OF C OMPUTER S CIENCE A ND M ATHEMATICS
U NIVERSITY OF A L -Q ADISIYA
D IWANIYA - I RAQ
waggashnd@yahoo.com
D EPARTMENT OF M ATHEMATICS
F ERGUSSON C OLLEGE , P UNE - 411004, I NDIA
kulkarni_ferg@yahoo.com
Received 06 January, 2008; accepted 02 May, 2009
Communicated by S.S. Dragomir
A BSTRACT. This paper is mainly concerned with the application of differential subordinations
for the class of meromorphic multivalent functions with positive coefficients defined by a linear
operator satisfying the following:
1 + Az
z p+1 (Ln f (z))0
≺
(n ∈ N0 ; z ∈ U ).
p
1 + Bz
In the present paper, we study the coefficient bounds, δ-neighborhoods and integral representations. We also obtain linear combinations, weighted and arithmetic means and convolution
properties.
−
Key words and phrases: Meromorphic functions, Differential subordination, convolution (or Hadamard product), p-valent
functions, Linear operator, δ-Neighborhood, Integral representation, Linear combination, Weighted
mean and Arithmetic mean.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let L(p, m) be a class of all meromorphic functions f (z) of the form:
(1.1)
f (z) = z
−p
+
∞
X
ak z k for any m ≥ p,
p ∈ N = {1, 2, . . . },
ak ≥ 0,
k=m
which are p-valent in the punctured unit disk
U ∗ = {z : z ∈ C, 0 < |z| < 1} = U/{0}.
The first author, Waggas Galib, is thankful of his wife (Hnd Hekmat Abdulah) for her support of him in his work.
005-08
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WAGGAS G ALIB ATSHAN
AND
S. R. K ULKARNI
Definition 1.1. Let f, g be analytic in U . Then g is said to be subordinate to f, written g ≺ f ,
if there exists a Schwarz function w(z), which is analytic in U with w(0) = 0 and |w(z)| <
1 (z ∈ U ) such that g(z) = f (w(z)) (z ∈ U ). Hence g(z) ≺ f (z) (z ∈ U ), then g(0) = f (0)
and g(U ) ⊂ f (U ). In particular, if the function f (z) is univalent in U , we have the following
(e.g. [6]; [7]):
g(z) ≺ f (z)(z ∈ U ) if and only if g(0) = f (0)
and g(U ) ⊂ f (U ).
Definition 1.2. For functions f (z) ∈ L(p, m) given by (1.1) and g(z) ∈ L(p, m) defined by
∞
X
bk z k , (bk ≥ 0, p ∈ N, m ≥ p),
(1.2)
g(z) = z −p +
k=m
we define the convolution (or Hadamard product) of f (z) and g(z) by
∞
X
(1.3)
(f ∗ g)(z) = z −p +
ak bk z k , (p ∈ N, m ≥ p, z ∈ U ).
k=m
Definition 1.3 ([9]). Let f (z) be a function in the class L(p, m) given by (1.1). We define a
linear operator Ln by
L0 f (z) = f (z),
1
L f (z) = z
−p
+
∞
X
(p + k + 1)ak z k =
k=m
(z p+1 f (z))0
zp
and in general
(1.4)
Ln f (z) = L(Ln−1 f (z))
∞
X
= z −p +
(p + k + 1)n ak z k
=
(z
p+1
k=m
n−1
L
zp
f (z))0
,
(n ∈ N).
It is easily verified from (1.4) that
(1.5)
z(Ln f (z))0 = Ln+1 f (z) − (p + 1)Ln f (z),
(f ∈ L(p, m),
n ∈ N0 = N ∪ {0}).
(1) Liu and Srivastava [4] introduced recently the linear operator when m = 0, investigating several inclusion relationships involving various subclasses of meromorphically
p-valent functions, which they defined by means of the linear operator Ln (see [4]).
(2) Uralegaddi and Somanatha [10] introduced the linear operator Ln when p = 1 and
m = 0.
(3) Aouf and Hossen [2] obtained several results involving the linear operator Ln when
m = 0 and p ∈ N.
We introduce a subclass of the function class L(p, m) by making use of the principle of
differential subordination as well as the linear operator Ln .
Definition 1.4. Let A and B (−1 ≤ B < A ≤ 1) be fixed parameters. We say that a function f (z) ∈ L(p, m) is in the class L(p, m, n, A, B), if it satisfies the following subordination
condition:
z p+1 (Ln f (z))0
1 + Az
(1.6)
≺
(n ∈ N0 ; z ∈ U ).
p
1 + Bz
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
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A PPLICATION O F D IFFERENTIAL S UBORDINATION
3
By the definition of differential subordination, (1.6) is equivalent to the following condition:
p+1 n
z (L f (z))0 + p Bz p+1 (Ln f (z))0 + pA < 1, (z ∈ U ).
We can write
2β
, −1 = L(p, m, n, β),
L p, m, n, 1 −
p
where L(p, m, n, β) denotes the class of functions in L(p, m) satisfying the following:
Re{−z p+1 (Ln f (z))0 } > β
(0 ≤ β < p; z ∈ U ).
2. C OEFFICIENT B OUNDS
Theorem 2.1. Let the function f (z) of the form (1.1), be in L(p, m). Then the function f (z)
belongs to the class L(p, m, n, A, B) if and only if
∞
X
(2.1)
k(1 − B)(p + k + 1)n ak < (A − B)p,
k=m
where −1 ≤ B < A ≤ 1, p ∈ N, n ∈ N0 , m ≥ p.
The result is sharp for the function f (z) given by
f (z) = z −p +
(A − B)p
zm,
k(1 − B)(p + k + 1)n
m ≥ p.
Proof. Assume that the condition (2.1) is true. We must show that f ∈ L(p, m, n, A, B), or
equivalently prove that
p+1 n
z (L f (z))0 + p (2.2)
Bz p+1 (Ln f (z))0 + Ap < 1.
We have
∞
P
p+1
−(p+1)
n
k−1
p+1 n
z (−pz
+
k(p + k + 1) ak z ) + p z (L f (z))0 + p k=m
=
∞
Bz p+1 (Ln f (z))0 + Ap P
n a z k−1 ) + Ap Bz p+1 (−pz −(p+1) +
k(p
+
k
+
1)
k
k=m
∞
P
k(p + k + 1)n ak z k+p
k=m
= ∞
P
(A − B)p + B
k(p + k + 1)n ak z k+p k=m


∞
P


n


k(p + k + 1) ak


k=m
≤
< 1.
∞
P



k(k + p + 1)n ak 

 (A − B)p + B
k=m
The last inequality by (2.1) is true.
Conversely, suppose that f (z) ∈ L(p, m, n, A, B). We must show that the condition (2.1)
holds true. We have
p+1 n
z (L f (z))0 + p Bz p+1 (Ln f (z))0 + Ap < 1,
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
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4
WAGGAS G ALIB ATSHAN
AND
S. R. K ULKARNI
hence we get
∞
P
k(p + k + 1)n ak z k+p
k=m
< 1.
∞
(A − B)p + B P k(p + k + 1)n ak z k+p k=m
Since Re(z) < |z|, so we have


∞
P


n
k+p


k(p + k + 1) ak z


k=m
Re
< 1.
∞
P


n
k+p


k(p + k + 1) ak z 
 (A − B)p + B
k=m
We choose the values of z on the real axis and letting z → 1− , then we obtain


∞
P


n


k(p + k + 1) ak


k=m
< 1,
∞
P



k(p + k + 1)n ak 

 (A − B)p + B
k=m
then
∞
X
k(1 − B)(p + k + 1)n ak < (A − B)p
k=m
and the proof is complete.
Corollary 2.2. Let f (z) ∈ L(p, m, n, A, B), then we have
ak ≤
(A − B)p
, k ≥ m.
k(1 − B)(p + k + 1)n
Corollary 2.3. Let 0 ≤ n2 < n1 , then L(p, m, n2 , A, B) ⊆ L(p, m, n1 , A, B).
3. N EIGHBOURHOODS AND PARTIAL S UMS
Definition 3.1. Let −1 ≤ B < A ≤ 1, m ≥ p, n ∈ N0 , p ∈ N and δ ≥ 0. We define the δ neighbourhood of a function f ∈ L(p, m) and denote Nδ (f ) such that
(
∞
X
−p
(3.1) Nδ (f ) = g ∈ L(p, m) : g(z) = z +
bk z k , and
k=m
∞
X
k(1 − B)(p + k + 1)n
|ak − bk | ≤ δ
(A − B)p
k=m
)
.
Goodman [3], Ruscheweyh [8] and Altintas and Owa [1] have investigated neighbourhoods for
analytic univalent functions, we consider this concept for the class L(p, m, n, A, B).
Theorem 3.1. Let the function f (z) defined by (1.1) be in L(p, m, n, A, B). For every complex
−p
∈ L(p, m, n, A, B), then Nδ (f ) ⊂ L(p, m, n, A, B),
number µ with |µ| < δ, δ ≥ 0, let f (z)+µz
1+µ
δ ≥ 0.
Proof. Since f ∈ L(p, m, n, A, B), f satisfies (2.1) and we can write for γ ∈ C, |γ| = 1, that
p+1 n
z (L f (z))0 + p
(3.2)
6= γ.
Bz p+1 (Ln f (z))0 + pA
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
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A PPLICATION O F D IFFERENTIAL S UBORDINATION
5
Equivalently, we must have
(f ∗ Q)(z)
6= 0,
z −p
(3.3)
z ∈ U ∗,
where
Q(z) = z
−p
+
∞
X
ek z k ,
k=m
such that ek =
Since
γk(1−B)(p+k+1)n
(A−B)p
f (z)+µz −p
1+µ
, satisfying |ek | ≤
k(1−B)(p+k+1)n
(A−B)p
and k ≥ m, p ∈ N, n ∈ N0 .
∈ L(p, m, n, A, B), by (3.3),
1
f (z) + µz −p
∗ Q(z) 6= 0,
z −p
1+µ
and then
1
(3.4)
z −p
(f ∗ Q)(z) + µz −p
1+µ
6= 0.
Now assume that (f ∗Q)(z)
< δ. Then, by (3.4), we have
z −p 1 f ∗Q
(f ∗ Q)(z) |µ| − δ
µ
|µ|
1
>
1 + µ z −p + 1 + µ ≥ |1 + µ| − |1 + µ| |1 + µ| ≥ 0.
z −p
This is a contradiction as |µ| < δ. Therefore (f ∗Q)(z)
≥ δ.
z −p Letting
∞
X
−p
g(z) = z +
bk z k ∈ Nδ (f ),
k=m
then
(g ∗ Q)(z) ((f − g) ∗ Q)(z) δ−
≤
z −p
z −p
∞
X
≤
(ak − bk )ek z k k=m
≤
∞
X
|ak − bk ||ek ||z|k
k=m
m
< |z|
∞ X
k(1 − B)(p + k + 1)n
k=m
(A − B)p
|ak − bk |
≤ δ,
therefore
(g∗Q)(z)
z −p
6= 0, and we get g(z) ∈ L(p, m, n, A, B), so Nδ (f ) ⊂ L(p, m, n, A, B).
Theorem 3.2. Let f (z) be defined by (1.1) and the partial sums S1 (z) and Sq (z) be defined by
S1 (z) = z −p and
m+q−2
Sq (z) = z
−p
+
X
ak z k ,
q > m, m ≥ p, p ∈ N.
k=m
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
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WAGGAS G ALIB ATSHAN
Also suppose that
P∞
k=m
AND
S. R. K ULKARNI
Ck ak ≤ 1, where
Ck =
k(1 − B)(p + k + 1)n
.
(A − B)p
Then
f ∈ L(p, m, n, A, B)
f (z)
1
Re
>1−
,
Sq (z)
Cq
(i)
(3.5)
(ii)
Re
(3.6)
Sq (z)
f (z)
>
Cq
,
1 + Cq
z ∈ U, q > m.
Proof.
−p +µz −p
(i) Since z 1+µ
= z −p ∈ L(p, m, n, A, B), |µ| < 1, then by Theorem 3.1, we have
N1 (z −p ) ⊂ L(p, m, n, A, B), p ∈ N(N1 (z −p ) denoting the 1-neighbourhood). Now since
∞
X
Ck ak ≤ 1,
k=m
then f ∈ N1 (z −p ) and f ∈ L(p, m, n, A, B).
(ii) Since {Ck } is an increasing sequence, we obtain
m+q−2
X
(3.7)
∞
X
ak + Cq
k=m
ak ≤
k=q+m−1
∞
X
Ck ak ≤ 1.
k=m
Setting
G1 (z) = Cq
f (z)
1
− 1−
Sq (z)
Cq
∞
P
Cq
ak z k+p
k=q+m−1
=
1+
m+q−2
P
+ 1,
ak z k+p
k=m
from (3.7) we get
∞
P
Cq
ak z k+p
G1 (z) − 1 k=q+m−1
=
G1 (z) + 1 m+q−2
∞
P
P
k+p + C
k+p 2 + 2
a
z
a
z
k
q
k
k=m
k=q+m−1
Cq
∞
P
ak
k=q+m−1
≤
2−2
m+q−2
P
k=m
ak − Cq
∞
P
≤ 1.
ak
k=q+m−1
o
n
> 1−
This proves (3.5). Therefore, Re(G1 (z)) > 0 and we obtain Re Sfq(z)
(z)
the same manner, we can prove the assertion (3.6), by setting
Sq (z)
Cq
G2 (z) = (1 + Cq )
−
.
f (z)
1 + Cq
This completes the proof.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
1
.
Cq
Now, in
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A PPLICATION O F D IFFERENTIAL S UBORDINATION
7
4. I NTEGRAL R EPRESENTATION
In the next theorem we obtain an integral representation for Ln f (z).
Theorem 4.1. Let f ∈ L(p, m, n, A, B), then
Z z
p(Aψ(t) − 1)
n
L f (z) =
dt,
p+1 (1 − Bψ(t))
0 t
where |ψ(z)| < 1, z ∈ U ∗ .
Proof. Let f (z) ∈ L(p, m, n, A, B). Letting − z
p+1 (Ln f (z))0
p
= y(z), we have
1 + Az
1 + Bz
y(z) ≺
y(z)−1 or we can write By(z)−A < 1, so that consequently we have
y(z) − 1
= ψ(z), |ψ(z)| < 1, z ∈ U.
By(z) − A
We can write
−z p+1 (Ln f (z))0
1 − Aψ(z)
=
,
p
1 − Bψ(z)
which gives
(Ln f (z))0 =
Hence
Z
n
L f (z) =
0
z
p(Aψ(z) − 1)
.
− Bψ(z))
z p+1 (1
p(Aψ(t) − 1)
dt,
tp+1 (1 − Bψ(t))
and this gives the required result.
5. L INEAR C OMBINATION
In the theorem below, we prove a linear combination for the class L(p, m, n, A, B).
Theorem 5.1. Let
fi (z) = z
−p
+
∞
X
ak,i z k ,
(ak,i ≥ 0, i = 1, 2, . . . , `, k ≥ m, m ≥ p)
k=m
belong to L(p, m, n, A, B), then
F (z) =
`
X
ci fi (z) ∈ L(p, m, n, A, B),
i=1
where
P`
i=1 ci
= 1.
Proof. By Theorem 2.1, we can write for every i ∈ {1, 2, . . . , `}
∞
X
k(1 − B)(p + k + 1)n
ak,i < 1,
(A
−
B)p
k=m
therefore
F (z) =
`
X
i=1
ci z −p +
∞
X
!
ak,i z k
k=m
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
= z −p +
∞
X
`
X
k=m
i=1
!
ci ak,i z k .
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WAGGAS G ALIB ATSHAN
AND
S. R. K ULKARNI
However,
∞
X
k(1 − B)(p + k + 1)n
(A − B)p
k=m
`
X
!
ci ak,i
i=1
" ∞
#
`
X
X k(1 − B)(p + k + 1)n
=
ak,i ci ≤ 1,
(A − B)p
i=1 k=m
then F (z) ∈ L(p, m, n, A, B), so the proof is complete.
6. W EIGHTED M EAN
AND
A RITHMETIC M EAN
Definition 6.1. Let f (z) and g(z) belong to L(p, m), then the weighted mean hj (z) of f (z) and
g(z) is given by
1
hj (z) = [(1 − j)f (z) + (1 + j)g(z)].
2
In the theorem below we will show the weighted mean for this class.
Theorem 6.1. If f (z) and g(z) are in the class L(p, m, n, A, B), then the weighted mean of
f (z) and g(z) is also in L(p, m, n, A, B).
Proof. We have for hj (z) by Definition 6.1,
"
!
!#
∞
∞
X
X
1
hj (z) =
(1 − j) z −p +
ak z k + (1 + j) z −p +
bk z k
2
k=m
k=m
=z
−p
∞
X
1
+
((1 − j)ak + (1 + j)bk )z k .
2
k=m
Since f (z) and g(z) are in the class L(p, m, n, A, B) so by Theorem 2.1 we must prove that
∞
X
1
n 1
k(1 − B)(p + k + 1)
(1 − j)ak + (1 + j)bk
2
2
k=m
∞
∞
X
X
1
1
n
= (1 − j)
k(1 − B)(p + k + 1) ak + (1 + j)
k(1 − B)(p + k + 1)n bk
2
2
k=m
k=m
1
1
≤ (1 − j)(A − B)p + (1 + j)(A − B)p.
2
2
The proof is complete.
Theorem 6.2. Let f1 (z), f2 (z), . . . , f` (z) defined by
(6.1)
fi (z) = z
−p
+
∞
X
ak,i z k ,
(ak,i ≥ 0, i = 1, 2, . . . , `, k ≥ m, m ≥ p)
k=m
be in the class L(p, m, n, A, B), then the arithmetic mean of fi (z) (i = 1, 2, . . . , `) defined by
`
1X
h(z) =
fi (z)
` i=1
(6.2)
is also in the class L(p, m, n, A, B).
Proof. By (6.1), (6.2) we can write
`
∞
1 X −p X
h(z) =
z +
ak,i z k
` i=1
k=m
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
!
= z −p +
∞
X
k=m
!
`
1X
ak,i z k .
` i=1
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A PPLICATION O F D IFFERENTIAL S UBORDINATION
9
Since fi (z) ∈ L(p, m, n, A, B) for every i = 1, 2, . . . , `, so by using Theorem 2.1, we prove
that
!
∞
`
X
X
1
k(1 − B)(p + k + 1)n
ak,i
`
i=1
k=m
!
`
∞
`
1X
1X X
n
=
k(1 − B)(p + k + 1) ak,i ≤
(A − B)p.
` i=1 k=m
` i=1
The proof is complete.
7. C ONVOLUTION P ROPERTIES
Theorem 7.1. If f (z) and g(z) belong to L(p, m, n, A, B) such that
∞
∞
X
X
k
−p
−p
(7.1)
f (z) = z +
ak z ,
g(z) = z +
bk z k ,
k=m
k=m
then
T (z) = z
−p
+
∞
X
(a2k + b2k )z k
k=m
is in the class L(p, m, n, A1 , B1 ) such that A1 ≥ (1 − B1 )µ2 + B1 , where
√
2(A − B)
µ= p
.
m(m + 2)n (1 − B)
Proof. Since f, g ∈ L(p, m, n, A, B), Theorem 2.1 yields
∞ X
k(1 − B)(p + k + 1)n
k=m
and
(A − B)p
∞ X
k(1 − B)(p + k + 1)n
k=m
(A − B)p
2
ak
≤1
2
bk
≤ 1.
We obtain from the last two inequalities
2
∞
X
1 k(1 − B)(p + k + 1)n
(7.2)
(a2k + b2k ) ≤ 1.
2
(A
−
B)p
k=m
However, T (z) ∈ L(p, m, n, A1 , B1 ) if and only if
∞ X
k(1 − B1 )(p + k + 1)n
(7.3)
(a2k + b2k ) ≤ 1,
(A
1 − B1 )p
k=m
where −1 ≤ B1 < A1 ≤ 1, but (7.2) implies (7.3) if
2
k(1 − B1 )(p + k + 1)n
1 k(1 − B)(p + k + 1)n
<
.
(A1 − B1 )p
2
(A − B)p
Hence, if
1 − B1
k(p + k + 1)n 2
1−B
<
α , where α =
.
A1 − B1
2p
A−B
In other words,
1 − B1
k(k + 2)n 2
<
α .
A1 − B1
2
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
http://jipam.vu.edu.au/
10
WAGGAS G ALIB ATSHAN AND S. R. K ULKARNI
This is equivalent to
A1 − B1
2
>
.
1 − B1
k(k + 2)n α2
So we can write
2(A − B)2
A1 − B1
>
= µ2 .
1 − B1
m(m + 2)n (1 − B)2
(7.4)
Hence we get A1 ≥ (1 − B1 )µ2 + B1 .
Theorem 7.2. Let f (z) and g(z) of the form (7.1) belong to L(p, m, n, A, B). Then the convolution (or Hadamard product) of two functions f and g belong to the class, that is, (f ∗ g)(z) ∈
L(p, m, n, A1 , B1 ), where A1 ≥ (1 − B1 )v + B1 and
v=
(A − B)2
.
m(1 − B)2 (m + 2)n
Proof. Since f, g ∈ L(p, m, n, A, B), by using the Cauchy-Schwarz inequality and Theorem
2.1, we obtain
(7.5)
∞
X
k(1 − B)(p + k + 1)n p
ak b k
(A
−
B)p
k=m
≤
∞
X
k(1 − B)(p + k + 1)n
ak
(A − B)p
k=m
! 12
∞
X
k(1 − B)(p + k + 1)n
bk
(A − B)p
k=m
! 21
≤1.
We must find the values of A1 , B1 so that
(7.6)
∞
X
k(1 − B1 )(p + k + 1)n
ak bk < 1.
(A
1 − B1 )p
k=m
Therefore, by (7.5), (7.6) holds true if
p
(1 − B)(A1 − B1 )
(7.7)
ak b k ≤
, k ≥ m, m ≥ p, ak 6= 0, bk 6= 0.
(1 − B1 )(A − B)
√
(A−B)p
By (7.5), we have ak bk < k(1−B)(p+k+1)
n , therefore (7.7) holds true if
2
k(1 − B1 )(p + k + 1)n
k(1 − B)(p + k + 1)n
≤
,
(A1 − B1 )p
(A − B)p
which is equivalent to
(1 − B1 )
k(1 − B)2 (p + k + 1)n
<
.
(A1 − B1 )
(A − B)2 p
Alternatively, we can write
(1 − B1 )
k(1 − B)2 (k + 2)n
<
,
(A1 − B1 )
(A − B)2
to obtain
A1 − B1
(A − B)2
>
= v.
1 − B1
m(1 − B)2 (m + 2)n
Hence we get A1 > v(1 − B1 ) + B1 .
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
http://jipam.vu.edu.au/
A PPLICATION O F D IFFERENTIAL S UBORDINATION
11
R EFERENCES
[1] O. ALTINTAS AND S. OWA, Neighborhoods of certain analytic functions with negative coefficients, IJMMS, 19 (1996), 797–800.
[2] M.K. AOUF AND H.M. HOSSEN, New criteria for meromorphic p-valent starlike functions,
Tsukuba J. Math., 17 (1993), 481–486.
[3] A.W. GOODMAN, Univalent functions and non-analytic curves, Proc. Amer. Math. Soc., 8 (1957),
598–601.
[4] J.-L. LIU AND H.M. SRIVASTAVA, Classes of meromorphically multivalent functions associated
with the generalized hypergeometric functions, Math. Comput. Modelling, 39 (2004), 21–34.
[5] J.-L. LIU AND H.M. SRIVASTAVA, Subclasses of meromorphically multivalent functions associated with a certain linear operator, Math. Comput. Modelling, 39 (2004), 35–44.
[6] S.S. MILLER AND P.T. MOCANU, Differential subordinations and univalent functions, Michigan
Math. J., 28 (1981), 157–171.
[7] S.S. MILLER AND P.T. MOCANU, Differential Subordinations : Theory and Applications, Series
on Monographs and Textbooks in Pure and Applied Mathematics, Vol. 225, Marcel Dekker, New
York and Basel, 2000.
[8] St. RUSCHEWEYH, Neighborhoods of univalent functions, Proc. Amer. Math. Soc., 81 (1981),
521–527.
[9] H.M. SRIVASTAVA AND J. PATEL, Applications of differential subordination to certain subclasses
of meromorphically multivalent functions, J. Ineq. Pure and Appl. Math., 6(3) (2005), Art. 88.
[ONLINE: http://jipam.vu.edu.au/article.php?sid=561]
[10] B.A. URALEGADDI AND C. SOMANATHA, New criteria for meromorphic starlike univalent
functions, Bull. Austral. Math. Soc., 43 (1991), 137–140.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 53, 11 pp.
http://jipam.vu.edu.au/