R E S E A R C H
Open Access
On some inequalities of a certain class of
analytic functions
Mohsan Raza
1*, Sarfraz Nawaz Malik
1and Khalida Inayat Noor
2*Correspondence: [email protected] 1Department of Mathematics, GC University Faisalabad, Faisalabad, Pakistan
Full list of author information is available at the end of the article
Abstract
The aim of this paper is to study the properties of a subclass of analytic functions related top-valent Bazilevic functions by using the concept of differential subordination. We investigate some results concerned with coefficient bounds, inclusion results, radius problem, covering theorem, angular estimation of a certain integral operator, and some other interesting properties.
MSC: 30C45; 30C50
Keywords: Bazilevic functions; differential subordination;p-valent
1 Introduction and preliminaries
LetApbe the class of analytic functions
f(z) =zp+
∞
n=p+ anzn
p∈N={, , , . . .}, (.)
defined in the open unit discE={z:|z|< }. A functionf ∈Apis ap-valent starlike
func-tion of orderρif and only if
Rezf (z)
f(z) >ρ, ≤ρ<p,z∈E.
This class of functions is denoted byS∗p(ρ). It is noted thatS∗p() =S∗p. Iff(z)∈Apsatisfies
argzf
(z)
pf(z)
<π η
(.)
for someη∈(, ] and for allz∈E, then the functionf is called strongly starlikep-valent of orderηinE. We denote this class byS˜p∗(η). Letf(z) andf(z) be analytic inE. We say f(z) is subordinate tof(z), writtenf≺forf(z)≺f(z), if there exists a Schwarz function w(z),w() = , and|w(z)|< inE, thenf(z) =f(w(z)). A functionf inApis said to belong
to the classS∗p[A,B], –≤B<A≤, if and only if
zf(z) f(z) ≺p
+Az
+Bz, z∈E.
Forp= , we obtain the classS∗[A,B] of Janowski starlike functions. Janowski functions have extensively been studied by several researchers; see for example [–]. It is clear that
f ∈S∗p[A,B] if and only if
zff(z)(z)–p –AB –B
<pA–B
–B (– <B<A≤,z∈E), (.)
and
Rezf (z)
f(z) >p –A
(B= –,z∈E).
A functionf ∈Apisp-valent Bazilevic function of type (α,β) and orderρif and only if
Rezf (z)
f(z)
f(z) g(z)
α
f(z) zp
iβ
>ρ, ≤ρ<p,z∈E,
whereα≥,β∈Randg∈Sp∗. Forp= andρ= , this class was introduced by Bazilevic and these functions are univalent forα≥,β∈R. This class of functions is studied by many authors; for some details, see [–].
Using this concept, we generalize and define a subclass ofp-valent Bazilevic functions of type (α,β) as follows.
Definition . A functionf ∈Mp(α,β,μ,A,B) if it satisfies the condition
zf(z) f(z)
f(z) g(z)
α
f(z) zp
iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –ipβ ≺p +Az
+Bz, z∈E,
whereα≥,μ> ,g∈Sp∗, –≤B<A≤ andβis any real. We have the following special cases.
(i) Forβ= , we have the subclass of Bazilevic functions defined by Patel []. (ii) Forβ= ,p= ,g(z) =z,A= – ρ,B= –, we obtain the subclass of Bazilevic
functions defined in [].
ForA= – ρ,B= –, we have the following subclass of analytic functions.
Definition . A functionf ∈Bp(α,β,ρ) if it satisfies the condition
zf(z) f(z)
f(z) g(z)
α
f(z) zp
iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –ipβ ≺p
+ ( – ρ)z –z ,
whereα≥,μ> ,g∈Sp∗, ≤ρ< ,β is any real andz∈E. In other words, a function f ∈Bp(α,β,ρ) if it satisfies the condition
Re
p
zf(z) f(z)
f(z) g(z)
α
f(z) zp
iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
We need the following definition and lemmas which will be used in our main results.
Definition . Let:C×E→Cbe analytic in a domainDandhbe univalent inE. Ifp is analytic inEwith (p(z),zp(z);z)∈Dwhenz∈E, then we say thatpsatisfies a first-order differential subordination if
p(z),zp(z);z≺h(z), z∈E. (.)
The univalent functionqis called dominant of the differential subordination (.) ifp≺q for allpsatisfies (.). Ifq˜≺qfor all dominants of (.), then we say thatq˜ is the best dominant of (.).
Lemma .([]) If–≤B<A≤,λ> and the complex numberγ satisfiesRe{γ} ≥
–λ( –A)/( –B),then the differential equation
q(z) + zq
(z)
λq(z) +γ =
+Az
+Bz, z∈E,
has a univalent solution in E given by
q(z) =
⎧ ⎨ ⎩
zλ+γ(+Bz)λ(A–B)/B
λztλ+γ–(+Bt)λ(A–B)/Bdt–
γ
λ, B= ,
zλ+γeλAz
λztλ+γ–eλAtdt–
γ
λ, B= .
If h(z) = +cz+cz+· · · is analytic in E and satisfies
h(z) + zh
(z)
λh(z) +γ ≺
+Az
+Bz, z∈E,
then
h(z)≺q(z)≺ +Az +Bz,
and q(z)is the best dominant.
Lemma .([]) Letεbe a positive measure on[, ].Let g be a complex-valued func-tion defined on E×[, ]such that g(·,t)is analytic in E for each t∈[, ]and g(z,·)is
ε-integrable on[, ]for all z∈E.In addition,suppose thatReg(z,t) > ,g(–r,t)is real and Re{/g(z,t)} ≥/g(–r,t) for|z| ≤r< and t ∈[, ].If g(z) =g(z,t)dε(t), then
Re{/g(z)} ≥/g(–r).
Lemma .([, Chapter ]) Let a,band c= , –, –, . . .be complex numbers.Then, forRec>Reb> ,
(i) F(a,b,c;z) =
(c)
(c–b)(b)
tb–( –t)c–b–( –tz)–adt,
(ii) F(a,b,c;z) =F(b,a,c;z),
(iii) F(a,b,c;z) = ( –z)–aF
a,c–b,c; z z–
Lemma .([]) Let–≤B≤B<A≤A≤.Then
+Az +Bz ≺
+Az +Bz .
Lemma .([]) Let F be analytic and convex in E.If f,g∈Apand f,g≺F,then
μf + ( –μ)g≺F, ≤μ≤.
Lemma .([]) Let f(z) =∞k=akzkbe analytic in E and F(z) =
∞
k=bkzkbe analytic
and convex in E.If f ≺F,then
|ak| ≤ |b| (k∈N).
Lemma .([]) Let h(z) = +dz+dz+· · ·be analytic in E and h(z)= in E.If there exists a point z∈E such that|argh(z)|<πη(|z|<|z|)and|argh(z)|=πη( <η≤), then we have zh(z)
h(z) =ikη,where
⎧ ⎨ ⎩
k≥(x+x), whenargh(z) =πη, k≤–
(x+
x), whenargh(z) = –
π
η,
and(h(z))/η=±ix(x> ).
Lemma . Let g∈S∗[A,B].Then the function
G(z) =
c+α+iβ
zc+iβ
z
tc+iβ–gα(t)dt
α
(.)
belongs to S∗[A,B]for c≥–α––AB.
Proof is straightforward by using Lemma ..
Throughout this paper,α≥,β∈R,μ> , and –≤B<A≤ unless otherwise stated.
2 Main results
Theorem . If f ∈Mp(α,β,μ,A,B),then
zf(z) pf(z)
f(z) g(z)
α
f(z) zp
iβ
≺q(z), (.)
where q(z) =pQμ(z) and
Q(z) =
⎧ ⎨ ⎩
t
p μ–(+Bzt
+Bz) p
μ(A–B)/Bdt, B= ,
t
p μ–e
p
μ(t–)Azdt, B= .
(.)
In hypergeometric function form,
q(z) =
⎧ ⎨ ⎩
[F(,pμ( –
A B);
p
μ+ ;
Bz Bz+) ]
–, B= ,
[F(,μp+ ; –
p
μAz) ]
and if A< –μpB, –≤B< ,then Mp(α,β,μ,A,B)⊂Bp(α,β,ρ),where
ρ=p
F
,p
μ
–A B
;p
μ+ ;
B B–
–
. (.)
This result is best possible.
Proof Let
h(z) = zf
(z)
pf(z)
f(z) g(z)
α
f(z) zp
iβ
,
whereh(z) is analytic inEwithh() = . Differentiating logarithmically, we obtain
zf(z) f(z)
f(z) g(z)
α
f(z) zp
iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –ipβ
=ph(z) +μzh
(z)
h(z) ≺p +Az
+Bz. (.)
Using Lemma . forλ=pμandγ = , we have
h(z)≺q(z)≺ +Az +Bz,
where q(z) is given in (.) and is the best dominant of (.). Next, in order to prove Mp(α,β,μ,A,B)⊂ Bp(α,β,ρ), we show that inf|z|<{Req(z)}=q(–). Now, we set a=
p
μ(B–A)/B,b=
p
μandc=
p
μ+ , then it is clear thatc>b> ; therefore, forB= it follows
from (.) by using Lemma . that
Q(z) = ( +Bz)a
tb–( +Btz)–adt
=(b)
(c)F
,a,c; Bz Bz+
. (.)
To prove thatinf|z|<{Req(z)}=q(–), we need to show that
Re/Q(z)≥/Q(–).
SinceA< –μpB with –≤B< implies thatc>a> , therefore, by using Lemma ., (.) yields
Q(z) =
g(z,t)dε(t),
where
g(z,t) = +Bz
+ ( –t)Bz (≤t≤),
dε(t) = (b)
(a)(c–a)t
which is a positive measure on [, ]. For –≤B< it is clear thatReg(z,t) > andg(–r,t) is real for ≤ |z| ≤r< andt∈[, ]. Also,
Re
g(z,t) =Re
+ ( –t)Bz +Bz ≥
– ( –t)Br –Br =
g(–r,t)
for|z| ≤r< . Therefore, using Lemma ., we have
Re/Q(z)≥/Q(–r).
Now, lettingr→–, it follows
Re/Q(z)≥/Q(–).
Therefore,Mp(α,β,μ,A,B)⊂Bp(α,β,ρ).
Forβ= , we have the following result proved in [].
Corollary . If f ∈Mp(α,μ,A,B),then
zf(z) pf(z)
f(z) g(z)
α
≺
⎧ ⎨ ⎩
[F(,μp( –
A B);
p
μ+ ;
Bz Bz+) ]
–, B= ,
[F(,μp + ; –
p
μAz) ]–, B= ,
and if A< –μpB, –≤B< ,then Mp(α,μ,A,B)⊂Bp(α,ρ),where
ρ=p
F
,p
μ
–A B
;p
μ+ ;
B B–
– .
This result is best possible.
Forp= , we have the classM(α,β,μ,A,B) =M(α,β,μ,A,B). We denote the class of functionsf∈A, having Taylor series representation of the form
f(z) =z+
∞
k=n+ akzk
and satisfying the condition
zf(z) f(z)
f(z) g(z)
α
f(z) z
iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –iβ ≺ +Az
+Bz, z∈E, (.)
byM∗(α,β,μ,A,B), whereg(z) =z+∞k=n+bkzksuch thatRezg (z)
Theorem . Let f ∈M∗(α,β,μ,A,B).Then
|an+| ≤
(A–B) +α(n+ )( +μn)
|α+iβ+n|( +μn) . (.)
Proof Sincef ∈M∗(α,β,μ,A,B), therefore,
zf(z) f(z)
f(z) g(z)
α
f(z) z
iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –iβ ≺ +Az +Bz.
Now, using the fact thatf(z) =z+∞k=n+akzkandg(z) =z+
∞
k=n+bkzk, we obtain
+(α+iβ+n)( +μn)an+–α( +μn)bn+
zn+· · · ≺ +Az +Bz.
By a well-known result due to Janowski and Lemma ., we have
(α+iβ+n)( +μn)an+–α( +μn)bn+≤A–B.
By the triangle inequality, we obtain
(α+iβ+n)( +μn)an+–α( +μn)bn+≤A–B.
Using the coefficient bound for the classS∗, we have the required result.
Forβ= andg(z) =z, we have the following result proved in [].
Corollary . Let f ∈M(α,μ,A,B).Then
|an+| ≤
(A–B) (α+n)( +μn).
Forβ= ,p= ,g(z) =z,A= – ρandB= –, we have the following result proved in [].
Corollary . Let f satisfy the condition
Re
zf(z) f(z)
f(z) z
α
+μ
+zf
(z)
f(z) + ( –α)
–zf
(z)
f(z) >ρ.
Then
|an+| ≤
( –ρ) (n+α)( +μn).
Theorem . Forμ≥μ≥and–≤B≤B<A≤A≤,
Proof Letf ∈Mp(α,β,μ,A,B). Then
zf(z) f(z) f(z) g(z) α f(z) zp iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –ipβ ≺p +Az +Bz .
Since –≤B≤B<A≤A≤, therefore by Lemma ., we have
zf(z) f(z) f(z) g(z) α f(z) zp iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –ipβ ≺p +Az +Bz .
Hence, we have f ∈Mp(α,β,μ,A,B). For μ=μ ≥, we have the required result. Whenμ>μ≥, Theorem . implies that
zf(z) f(z) f(z) g(z) α f(z) zp iβ
≺p +Az +Bz .
Now
zf(z) f(z) f(z) g(z) α f(z) zp iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –ipβ
=
–μ
μ
zf(z) f(z) f(z) g(z) α f(z) zp iβ
+μ
μ
zf(z) f(z) f(z) g(z) α f(z) zp iβ
+μ
+zf
(z)
f(z) + (α+iβ– ) zf(z)
f(z) –α zg(z)
g(z) –ipβ
.
Using Lemma ., we have the required result.
Forβ= , we have the following result.
Corollary . Forμ≥μ≥and–≤B≤B<A≤A≤,
Mp(α,μ,A,B)⊂Mp(α,μ,A,B).
This result is proved in[].
Forβ= ,p= ,g(z) =z,A= – ρand B= –,we have the class M(α,μ,ρ)defined as
Re
zf(z) f(z) f(z) z α +μ
+zf
(z)
f(z) – zf(z)
f(z) +α
zf(z)
f(z) – >ρ (.)
Corollary . Forα≥,μ≥μ≥and >ρ≥ρ≥,
M(α,μ,ρ)⊂M(α,μ,ρ).
Theorem . Let f ∈Apsatisfy
Re
f(z)
zp > and
zff(z)(z)g(z)f(z)αfz(z)p
iβ
–p<σp, <σ≤,
for g∈S∗p.Then f is p-valent convex in|z|<Rα,β,σ,where
Rα,β,σ =
| –α–iβ|+ αp+σ
–| –α–iβ|+ αp+σ– p(αp–p–σ)/(αp–p–σ). (.) Proof Let
h(z) = zf
(z)
pf(z)
f(z) g(z)
α
f(z) zp
iβ
– ,
whereh(z) is analytic inEwithh() = and|h(z)|< . By using the Schwarz lemma, we get
h(z) =σzψ(z),
whereψ(z) is analytic inEwith|ψ(z)|< . Differentiating logarithmically, we have
+zf
(z)
f(z) = ( –α–iβ) zf(z)
f(z) +α zg(z)
g(z) +
σz(zψ(z) +ψ(z)) +σzψ(z) +ipβ.
SinceRe{fz(zp)}> , therefore,
zf(z) f(z) =p+
zϕ(z)
ϕ(z) , Reϕ(z) > .
This implies that
Re
+zf
(z)
f(z) ≥( –α)p+αRe zg(z)
g(z) –| –α–iβ|
zϕϕ(z)(z)–σz(zψ
(z) +ψ(z))
+σzψ(z)
.
Now, using the well-known results for classesSp∗,Pand the Schwarz function [], we have
Re
+zf
(z)
f(z) ≥( –α)p+αp –r
+r–| –α–iβ| r ( –r)–
σr
( –r) ( <σ≤)
= (αp–p–σ)r
– (| –α–iβ|+ αp+σ)r+p
( –r) .
Forσ= andβ= , we have the following result which is proved in [].
Corollary . Let f ∈Apsatisfy
Re
f(z)
zp > and
zff(z)(z)g(z)f(z)α–p<p,
g∈S∗p.Then f is p-valent convex in|z|<Rα,where
Rα=
+ α(p– ) –( + α(p– ))– p(αp–p– )
(αp–p– ) .
Theorem . Let f ∈Apsatisfy
zff(z)(z)g(z)f(z)α–p<σp,
for g∈S∗p.Then,forα> ,f is p-valent
α-convex in|z|<Rα,σ,where
Rα,σ=
αp+σ–(αp+σ)– αp(αp–σ)
(αp–σ) . (.)
Proof Let
h(z) = zf
(z)
pf(z)
f(z) g(z)
α
– ,
whereh(z) is analytic inEwithh() = and|h(z)|< . By using the Schwarz lemma, we get
h(z) =σzψ(z),
whereψ(z) is analytic inEwith|ψ(z)|< . Differentiating logarithmically, we have
α
+zf
(z)
f(z)
+
–
α
zf(z) f(z) =
zg(z) g(z) +
σ α
z(zψ(z) +ψ(z)) +σzψ(z) .
This implies that
Re
α
+zf
(z)
f(z)
+
–
α
zf(z) f(z) ≥Re
zg(z) g(z) –
σ α
(zψ +(z) +σzψψ(z)(z)).
Now, using the well-known results for classesS∗p, and the Schwarz function, we have
Re
p
α
+zf
(z)
f(z)
+
–
α
zf(z) f(z) ≥
–r +r–
σr
αp( –r) ( <σ≤)
= (αp–σ)r
– (αp+σ)r+αp
LetQ(r) = (αp–σ)r– (αp+σ)r+αp. Then forp∈N,α> and <σ≤,Q() =αp> andQ() = –σ< . It shows that the root lies in (, ). This implies thatRe{α( +zff((zz))) +
( –α)zff(z(z))}> ifr<Rα,σ, whereRα,σ is given by (.).
Theorem . Let f ∈M(α,β,μ,A,B).Then E is mapped by f on a domain that contains the disc|w|<Rα,β,μ∈(, ),where
Rα,β,μ= |
α+iβ+ |( +μ)
|α+iβ+ |( +μ) + (A–B) + α( +μ). (.)
Proof Letwbe any complex number such thatf(z)=w. Then
wf(z) w–f(z)
=z+
a+ w
z+· · ·,
is univalent inE, so that
a+
w
≤.
Therefore,
w–|a| ≤.
Hence,
w≥
|α+iβ+ |( +μ)
|α+iβ+ |( +μ) + (A–B) + α( +μ)=Rα,β,μ.
Forβ= ,g(z) =z,A= – ρ,B= –, we have the following result proved in [].
Corollary . Let f ∈M(α,μ,ρ).Then E is mapped by f on a domain that contains the disc|w|<Rα,μ,where
Rα,β,μ=
( +α)( +μ) ( +α)( +μ) + ( –ρ)
.
Theorem . Letα> ,c≥–α––AB and let f∈A.If
argzf
(z)
f(z)
f(z) g(z)
α
f(z) z
iβ
–ρ<π
υ ( <υ≤, ≤ρ< ), (.)
for some g∈S∗,then
argzF
(z)
F(z)
F(z) G(z)
α
F(z) z
iβ
–ρ<π
η ( <η≤), (.)
where
F(z) =
c+α+iβ
zc
z
tc–fα+iβ(t)dt
α+iβ
with
υ=
⎧ ⎨ ⎩
η+
πtan
–( (+B)ηsin(π(–t(α,β,c,A,B))/)
|c+iβ|(+B)+α(+A)+η(+B)cos(π(–t(α,β,c,A,B))/)), B= –,
η, B= –, (.)
and
t(α,β,c,A,B) =
πsin
– α(A–B) +β( –B)
|c+iβ|( –B) +α( –AB). (.)
Proof Since
Fα+iβ(z) =c+α+iβ zc
z
tc–fα+iβ(t)dt,
therefore,
z–iβFα+iβ–(z)F(z) =
α+iβ
(c+α+iβ)z–iβfα+iβ(z) –cz–iβFα+iβ(z).
Now using (.), we have
zF(z) F(z)
F(z) G(z)
α
F(z) z
iβ
=
α+iβ{(c+α+iβ)z
cfα+iβ(z) –czcFα+iβ(z)}
(c+α+iβ)ztc+iβ–gα(t)dt .
Let
h(z) = –ρ
zF(z) F(z)
F(z) G(z)
α
F(z) z
iβ
–ρ
=N(z) D(z),
whereh(z) is analytic withh() = . Now
N(z) D(z) =
–ρ
zf(z) f(z)
f(z) g(z)
α
f(z) z
iβ
–ρ
=h(z)
+ D(z) zD(z)·
zh(z) h(z) ,
where
N(z) = –ρ
α+iβ
(c+α+iβ)zcfα+iβ(z) –czcFα+iβ(z)–ρzc+iβ–Gα(z)
,
D(z) = (c+α+iβ)
z
tc+iβ–gα(t)dt.
SinceG∈S∗[A,B], therefore, we can write
zD(z)
D(z) =c+iβ+α zG(z)
G(z) =re
iπ θ,
where
⎧ ⎨ ⎩
|c+iβ|+α––AB<r<|c+iβ|+α++AB, B= –,
similarly
⎧ ⎨ ⎩
|c+iβ|+α–A<r<∞, B= –,
– <θ< , B= –. (.)
Suppose thath(z)= inE, there exists a pointz∈Esuch that|argh(z)|<πη(|z|<|z|) and|argh(z)|=πη. Now, using Lemma ., we have zh
(z
)
h(z) =ikη. At first suppose that
h(z) = (ix)η(x> ) for the caseB= –, we obtain
arg
zf(z) f(z)
f(z) g(z)
α
f(z) z
iβ
–ρ
=argh(z) +arg
+
c+iβ+αzG(z) G(z)
·zh(z) h(z)
=π η+arg
+rei
π
θ–ikη
=π η+tan
–
kηsinπ( –θ)/ r+kηcosπ( –θ)/
≥π η+tan
–
ηsinπ( –t(α,β,c,A,B))/
|c+iβ|+α++AB+ηcosπ( –t(α,β,c,A,B))/
=π υ,
whereυandt(α,β,c,A,B) are given by (.) and (.) respectively. ForB= –, we have
arg
zf(z) f(z)
f(z) g(z)
α
f(z) z
iβ
–ρ
≥π η,
which is a contradiction to the assumption of our theorem. Now we suppose thath(z) = (–ix)η. For the caseB= – using a similar method, we obtain
arg
zf(z) f(z)
f(z) g(z)
α
f(z) z
iβ
–ρ
≤–π υ,
and forB= –, we have
arg
zf(z) f(z)
f(z) g(z)
α
f(z) z
iβ
–ρ
≤–π η,
which is a contradiction to the assumption of our theorem. Hence, we have the proof.
This kind of problem is also considered in []. Forβ= , we have the following result proved in [].
Corollary . Letα> ,c≥–α––AB and let f ∈A.If
arg
zf(z) f(z)
f(z) g(z)
α
–ρ<π
then
arg
zF(z) F(z)
F(z) G(z)
α
–ρ<π
η ( <η≤),
where
F(z) =
c+α
zc
z
tc–fα(t)dt
α
,
with
υ=
⎧ ⎨ ⎩
η+πtan–( (+B)ηsin(π(–t(α,β,c,A,B))/)
c(+B)+α(+A)+η(+B)cos(π(–t(α,β,c,A,B))/)), B= –,
η, B= –,
and
t(α,β,c,A,B) =
πsin
– α(A–B) c( –B) +α( –AB).
Remark . By using the suitable choices of parametersc,α,AandB, we can find many results proved in the literature.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
MR, SNM and KIN jointly discussed and presented the ideas of this article. MR made the text file and all the communications regarding the manuscript. All authors read and approved the final manuscript.
Author details
1Department of Mathematics, GC University Faisalabad, Faisalabad, Pakistan.2Department of Mathematics, COMSATS Institute of Information Technology, Islamabad, Pakistan.
Acknowledgements
The third author would like to express gratitude to Dr. S.M. Junaid Zaidi, Rector CIIT for his support and for providing excellent research facilities and to the Higher Education Commission of Pakistan for financial assistance.
Received: 11 March 2012 Accepted: 15 October 2012 Published: 29 October 2012 References
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