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Differential subordination theorems for new classes of meromorphic multivalent Quasi-Convex functions and some applications
Research Article
Abbas Kareem Wanas∗
Department of Mathematics,College of Computer Science and Mathematics, University of Al-Qadisiya,Diwaniya, Iraq
Received 15 December 2014; accepted (in revised version) 09 March 2015
Abstract: In the present paper, we study new classes of meromorphic multivalent quasi-convex functions, we obtain some subordination theorems for such classes in punctured unit disk. Also we give some applications of firstâ ˘A¸Sorder differential subordination.
MSC: 34G10• 26A33 • 30C45
Keywords: Meromorphic multivalent quasi-convex functions• Differential subordination • Derivative operator.
2015 IJAAMM all rights reserved.c
1. Introduction
Let Lp(λ) denotes the class of all functions f of the form:
f(z ) = z−p+ X∞ n=0
anzn−λ 0< λ < 1,p ∈ N = {1,2,···} , (1)
which are analytic in the punctured unit disk U = {z ∈ C : 0 < |z | < 1}.
Also, let Tp(λ) denotes the class of all functions f of the form:
f(z ) = z−p− X∞ n=0
anzn−λ an> 0,0 < λ < 1,p ∈ N = {1,2,···} , (2)
which are analytic in the punctured unit disk U .
For two functions f and g analytic in∆ = {z ∈ C : |z | < 1}, we say f is subordinate to g in ∆, denote by f ≺ g or f(z ) ≺ g (z )(z ∈ ∆), if there exists a Schwarz function w analytic in U with w (0) = 0 and |w (z )| < 1(z ∈ ∆) such that f(z ) = g (w (z )),(z ∈ ∆). In particular, if the function g is univalent in ∆ , then f ≺ g if and only if f (0) = g (0) and f(∆) ⊂ g (∆).
Letψ : C3× U → C . and let h be univalent in ∆. Assume that k , ψ are analytic and univalent in ∆ if k satisfies the differential subordination
ψ(k(z ), z k0(z ), z2k00(z ); z ) ≺ h(z ), (3)
then k is called a solution of the differential subordination. The univalent function q is called a dominant of the solutions of the differential subordination, or more simply dominant if k≺ q for all k satisfying (3) . A dominant ˇq
∗ Corresponding author.
E-mail address:[email protected]
Abbas Kareem Wanas/ Int. J. Adv. Appl. Math. and Mech. 2(3) (2015) 126 - 133 that satisfies ˇq≺ q for all dominants q of (3) is said to be the best dominant of (3).
Let Lpbe the class of all functionsΦ of the form:
Φ(z ) = z−p+ X∞ n=0
anzn p∈ N = {1, 2, · · · } , which are analytic in the punctured unit disk U . Also, let Tpbe the class of all functionsΦ of the form:
Φ(z ) = z−p− X∞ n=0
anzn an> 0,p ∈ N = {1,2,···} ,
which are analytic in the punctured unit disk U .
A function f ∈ Lp(λ) Tp(λ) is meromorphic multivalent starlike if f (z ) 6= 0 and
−R e
§z f0(z ) f(z )
ª
> 0, z ∈ U .
Similar, f ∈ Lp(λ) Tp(λ) is meromorphic multivalent convex if f0(z ) 6= 0 and
−R e
§
1+z f00(z ) f0(z )
ª
> 0, z ∈ U .
A function f ∈ Lp(λ) Tp(λ) is called meromorphic multivalent Quasi-convex function if there exists a meromorphic multivalent convex function g such that g(z ) 6= 0 and
−R e
¨ z f0(z )0 g0(z )
«
> 0, z ∈ U .
A functionΦ ∈ Lp Tp is meromorphic multivalent starlike if Φ(z ) 6= 0 and
−R e
§zΦ0(z ) Φ(z )
ª
> 0, z ∈ U .
Similar,a functionΦ is meromorphic multivalent convex if Φ0(z ) 6= 0 and
−R e
§
1+zΦ00(z ) Φ0(z )
ª
> 0, z ∈ U .
Moreover, a functionΦ is called meromorphic multivalent Quasi-convex function if there exists a meromorphic multivalent convex functionΨ such that Ψ0(z ) 6= 0 and
−R e
(z Φ0(z ))0 Ψ0(z )
> 0, z ∈ U .
2. Preliminaries
Definition 2.1 (Srivastava and Owa[11]).
The fractional derivative of orderλ,(0 < λ < 1) of a function f is defined by Dzλf(z ) = 1
Γ (1 − λ) d d z
Z z 0
f(")
(z − ")λ d", (4)
where f is an analytic function in a simply-connected region of the z-plane containing the origin, and the multi- plicity of(z − ")−λis removed by requiring log(z − ") to be real, when (z − ") > 0.
Let a , b , c∈ C with c 6= 0, −1, −2, · · · . The Gaussian hypergeometric function2F1( see[12] ) is defined by
2F1(z ) = 2F1(a, b, c ; z ) = X∞ n=0
(a)n(b )n
(c )n
zn
n ! (z ∈ ∆),
where(x )nis the Pochhammer symbol defined, in terms of the Gamma function, by (x )n = Γ (x + n)
Γ (x ) = § 1 (n = 0),
x(x + 1)···(x + n − 1) (n ∈ N ).
127
Definition 2.2 (Goyal and Goyal[4]).
Let 0≤ λ < 1 and µ, ν ∈ R . Then, in terms of familiar (Gauss) hypergeometric function2F1, the generalized fractional derivative operator J0,zλ,µ,νof a function f is defined by:
J0,zλ,µ,νf(z ) =
Γ (1−λ)1 d
d zzλ−µRz
0(z − ")−λf(").2F1(µ − λ,−ν;1 − λ;1 −z")d " ,(0 ≤ λ < 1)
dn
d znJ0,zλ−n,µ,νf(z ) (n ≤ λ < n + 1,n ∈ N ).
(5) where the function f is analytic in a simply-connected region of the z-plane containing the origin, with the order
f(z ) = O(|z |ε), (z → 0)
forε > ma x {0,µ − ν} − 1 , and the multiplicity of (z − ε)−λis removed by requiring l o g(z − ε) > 0 to be real, when (z − ε) > 0 .
By comparing (4) with (5), we find J0,zλ,λ,νf(z ) = Dzλf(z ),(0 ≤ λ < 1).
In terms of gamma function, we have J0,zλ,µ,νzn = Γ (n + 1)Γ (n − µ + ν + 1)
Γ (n − µ + 1)Γ (n − λ + ν + 1)zn−µ,(0 ≤ λ < 1,µ,ν ∈ R,n > ma x0,µ − ν − 1). (6)
Lemma 2.1 (Miller and Mocanu[8]).
Let q be univalent in the unit disk4 and θ and φ be analytic in a domain D containing q (4) with φ (w ) 6= 0 when w∈ q (4). Set Q (z ) = z q0(z )φ q (z ) and h (z ) = θ q (z ) +Q (z ). Suppose that
1- Q(z ) is starlike univalent in 4, and 2- R e
nz h0(z ) Q(z )
o> 0 for z ∈ 4.
Ifθ (k (z )) + z k0(z )φ (k (z )) ≺ θ q (z ) + z q0(z )φ q (z ) , then k ≺ q and q is the best dominant.
Lemma 2.2 (Shanmugam and et al.[9]).
Let q be convex univalent in the unit disk4 and ψ and γ ∈ C with R e n
1+z qq000(z )(z )+ψγo
> 0. If k is analytic in 4 and ψk (z ) + γz k0(z ) ≺ ψq (z ) + γz q0(z ), then k (z ) ≺ q (z ) and q is the best dominant.
Such type of study was carried out by various authors for another classes, like, Ibrahim and Darus[5–7] , Darus and Ibrahim[3], Singh et al. [10], Billing [2] and Atshan and Wanas [1].
3. Subordination results
In this section, we obtain some sufficient conditions for subordination of analytic functions in the classes Lp(λ) and Tp(λ).
Theorem 3.1.
Let the function q be univalent in U , q(z ) 6= 0 and assume that R e
1+ p r +s(1 − t )
t q(z )s−r
+ (r − 1)z q0(z )
q(z ) +z q00(z ) q0(z )
> 0, (7)
where r, s ∈ C , t ∈ C \ {0}. Suppose that z q (z )r−1
q0(z ) is starlike univalent in U . If f ∈ Lp(λ) satisfies the subordi- nation
(1 − t ) − zpf0(z )0 g0(z )
!αs
+ t − zpf0(z )0 g0(z )
!αr
p+ α z zpf0(z )00
zpf0(z )0 −z g00(z ) g0(z )
!
≺ (1 − t ) q (z )s
+ t q (z )r
p+z q0(z ) q(z )
, (8)
then
− zpf0(z )0 g0(z )
!α
≺ q (z ) , (z ∈ U , α ∈ C \ {0}) and q is the best dominant.
Abbas Kareem Wanas/ Int. J. Adv. Appl. Math. and Mech. 2(3) (2015) 126 - 133
Proof. Define the function k by
k(z ) = − zpf0(z )0 g0(z )
!α
, z∈ U . (9)
Note that
(1 − t )(k (z ))s+ t (k (z ))r
p+z k0(z ) k(z )
= (1 − t ) − zpf0(z )0 g0(z )
!αs
+ t − zpf0(z )0 g0(z )
!αr
×
×
p+ α z zpf0(z )00
zpf0(z )0 −z g00(z ) g0(z )
!
. (10)
From (8) and (10), we have (1 − t )(k (z ))s+ t (k (z ))r
p+z k0(z ) k(z )
≺ (1 − t ) q (z )s
+ t q (z )r
p+z q0(z ) q(z )
. (11)
By setting
θ (w ) = (1 − t ) ws+ t p wr a n d φ (w ) = t wr−1, w6= 0,
we see thatθ (w ) is analytic in C , φ (w ) is analytic in C \{0} and that φ (w ) 6= 0, w ∈ C \{0}.
Also, we get
Q(z ) = z q0(z )φ q (z ) = t z q (z )r−1q0(z ) and
h(z ) = θ q (z ) +Q (z ) = (1 − t ) q (z )s+ t q (z )r
p+z q0(z ) q(z )
. It is clear that Q(z ) is starlike univalent in U ,
R e z h0(z ) Q(z )
= Re
1+ p r +s(1 − t )
t q(z )s−r
+ (r − 1)z q0(z )
q(z ) +z q00(z ) q0(z )
> 0. (12)
From (7) and (12), we have R e z h0(z )
Q(z )
> 0.
Therefore, by Lemma2.1, we get k(z ) ≺ q (z ). By using (9), we obtain the result.
By fixingα = p = 1 in Theorem3.1, we obtain the following corollary:
Corollary 3.1.
Let the function q be univalent in U , q(z ) 6= 0 and assume that R e
1+ p r +s(1 − t )
t q(z )s−r
+ (r − 1)z q0(z )
q(z ) +z q00(z ) q0(z )
> 0,
where r, s ∈ C , t ∈ C \ {0}. Suppose that z q (z )r−1
q0(z ) is starlike univalent in U . If f ∈ Lp(λ) satisfies the subordi- nation
(1 − t ) − z f0(z )0 g0(z )
!s
+ t − z f0(z )0 g0(z )
!r
1+ α z z f0(z )00
z f0(z )0 −z g00(z ) g0(z )
!
≺ (1 − t ) q (z )s
+ t q (z )r
1+z q0(z ) q(z )
, then
− z f0(z )0 g0(z ) ≺ q (z ) and q is the best dominant.
129
By taking q(z ) = − 1+B z1+Az (−1 ≤ B < A ≤ 1) in Corollary3.1, we obtain the following corollary:
Corollary 3.2.
Let the function q be convex univalent in U , and assume that
R e
1+ p r +s(1 − t ) t
−
1+ Az 1+ B z
s−r
+1+ r (A − B) z − AB z2 (1 + Az )(1 + B z )
> 0,
where r, s∈ C , t ∈ C \ {0}. If f ∈ Lp(λ) satisfies the subordination
(1 − t ) − z f0(z )0 g0(z )
!s
+ t − z f0(z )0 g0(z )
!r
1+ α z z f0(z )00
z f0(z )0 −z g00(z ) g0(z )
!
≺ (1 − t )
−
1+ Az 1+ B z
s
+ t
−
1+ Az 1+ B z
r 1+ 2Az + AB z2 (1 + Az )(1 + B z )
, then
− z f0(z )0 g0(z ) ≺ −
1+ Az 1+ B z
,(−1 ≤ B < A ≤ 1) and q(z ) = − 1+B z1+Az is the best dominant.
Theorem 3.2.
Let the function q be convex univalent in U , q0(z ) 6= 0 and assume that R e
1+z q00(z ) q0(z ) +1
γ
> 0, (13)
whereγ ∈ C \{0}.
Suppose that
−
zpf0(z )0 g0(z )
α
is analytic in U . If f ∈ Tp(λ) satisfies the subordination
− zpf0(z )0 g0(z )
!α
+ αγ − zpf0(z )0 g0(z )
!α
z zpf0(z )00
zpf0(z )0 −z g00(z ) g0(z )
!
≺ q (z ) + γz q0(z ), (14)
then
− zpf0(z )0 g0(z )
!α
≺ q (z ) , (z ∈ U , α ∈ C \ {0})
and q is the best dominant.
Proof. Define the function k by
k(z ) = − zpf0(z )0 g0(z )
!α
, z∈ U . (15)
Note that
k(z ) + γz k0(z ) = − zpf0(z )0 g0(z )
!α
+ αγ − zpf0(z )0 g0(z )
!α
z zpf0(z )00
zpf0(z )0 −z g00(z ) g0(z )
!
. (16)
From (14) and (16), we have
k(z ) + γz k0(z ) ≺ q (z ) + γz q0(z ). (17)
By settingψ = 1 in Lemma2.2, we get k(z ) ≺ q (z ). By using (15), we obtain the result.
By fixingα = p = 1 in Theorem3.2, we obtain the following corollary:
Abbas Kareem Wanas/ Int. J. Adv. Appl. Math. and Mech. 2(3) (2015) 126 - 133
Corollary 3.3.
Let the function q be convex univalent in U , q0(z ) 6= 0 and assume that (3.2). Suppose that−
zpf0(z )0
g0(z ) is analytic in U . If f ∈ Tp(λ) satisfies the subordination
− z f0(z )0
g0(z ) + γ − z f0(z )0 g0(z )
! z z f0(z )00
z f0(z )0 −z g00(z ) g0(z )
!
≺ q (z ) + γz q0(z ),
then
− z f0(z )0
g0(z ) ≺ q (z ) , (z ∈ U ).
and q is the best dominant.
By taking q(z ) = − 1+z1−z in Corollary3.3, we obtain the following corollary:
Corollary 3.4.
Let the function q be convex univalent in U and assume that
R e
z2+ 1 (1 − z )(1 + z )+1
γ
> 0.
If f ∈ Tp(λ) satisfies the subordination
− z f0(z )0
g0(z ) + γ − z f0(z )0 g0(z )
! z z f0(z )00
z f0(z )0 −z g00(z ) g0(z )
!
≺ −
1+ z 1− z
− 2γz
(1 − z )(1 + z ), then
− z f0(z )0 g0(z ) ≺ −
1+ z 1− z
, (z ∈ U ) and q is the best dominant.
4. Applications of fractional derivative operator
In this section, we introduce some applications of section3containing fractional derivative operators. Assume that Φ(z ) =
X∞ n=0
σnzn.
By Definition2.1, we have DzλΦ(z ) =
X∞ n=0
Γ (n + 1)
Γ (n + 1 − λ)σnzn−λ= X∞ n=0
anzn−λ,
where
an= Γ (n + 1)
Γ (n + 1 − λ)σn, n= 0,1,2,... .
Thus z−p+ DzλΦ(z ) ∈ Lp(λ) and z−p− DzλΦ(z ) ∈ Tp(λ)(σn≥ 0), then we have the following results:
Theorem 4.1.
Let the assumptions of Theorem3.1hold. Then
−
zp z−p+ DzλΦ(z )00 z−p+ DzλΨ (z )0
α
≺ q (z ) , z ∈ U
and q is the best dominant.
131
Proof. Define the function f by f (z ) = z−p+ DzλΦ(z ) (z ∈ U ),
it can easily observed that f ∈ Lp(λ). Thus by using Theorem3.1, we obtain the result.
Theorem 4.2.
Let the assumptions of Theorem3.2hold. Then
−
zp z−p− DzλΦ(z )00 z−p− DzλΨ (z )0
α
≺ q (z ) , z ∈ U
and q is the best dominant.
Proof. Define the function f by f (z ) = z−p− DzλΦ(z ) (z ∈ U ),
it can easily observed that f ∈ Tp(λ). Thus by using Theorem3.2, we obtain the result.
By using (6), we have J0,zλ,µ,νΦ(z ) =
X∞ n=0
Γ (n + 1)Γ n − µ + ν + 1
Γ n − µ + 1 Γ (n − λ + ν + 1)σnzn−µ= X∞ n=0
anzn−µ, where
an= Γ (n + 1)Γ n − µ + ν + 1
Γ n − µ + 1 Γ (n − λ + ν + 1)σn, n= 0,1,2,... .
Letµ = λ. Then z−p+ J0,zλ,µ,νΦ(z ) ∈ Lp(λ) and z−p− J0,zλ,µ,νΦ(z ) ∈ Tp(λ) (σn≥ 0), then we have the following results:
Theorem 4.3.
Let the assumptions of Theorem3.1hold. Then
−
zp
z−p+ J0,zλ,µ,νΦ(z )00
z−p+ J0,zλ,µ,νΨ (z )0
α
≺ q (z ) , z ∈ U
and q is the best dominant.
Proof. Define the function f by f (z ) = z−p+ J0,zλ,µ,νΦ(z ) (z ∈ U ),
it can easily observed that f ∈ Lp(λ). Thus by using Theorem3.1, we obtain the result.
Theorem 4.4.
Let the assumptions of Theorem3.2hold. Then
−
zp
z−p− J0,zλ,µ,νΦ(z )00
z−p− J0,zλ,µ,νΨ (z )0
α
≺ q (z ) , z ∈ U
and q is the best dominant.
Proof. Define the function f by f (z ) = z−p− J0,zλ,µ,νΦ(z ) (z ∈ U ),
it can easily observed that f ∈ Tp(λ). Thus by using Theorem3.2, we obtain the result.
Abbas Kareem Wanas/ Int. J. Adv. Appl. Math. and Mech. 2(3) (2015) 126 - 133
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