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ARGUMENT ESTIMATES OF CERTAIN MULTIVALENT
FUNCTIONS INVOLVING A LINEAR OPERATOR
NAK EUN CHO, J. PATEL, and G. P. MOHAPATRA
Received 28 December 2001
The purpose of this paper is to derive some argument properties of certain multivalent functions in the open unit disk involving a linear operator. We also investigate their integral preserving property in a sector.
2OOO Mathematics Subject Classification: 30C45.
1. Introduction. LetᏭpdenote the class of functions of the form
f (z)=zp+
∞
n=1
an+pzn+p p∈N= {1,2, . . .} (1.1)
which are analytic in the open unit diskᐁ= {z:|z|<1}. A functionf∈Ꮽpis said to bep-valently starlike of orderαinᐁ, if it satisfies
Re
zf(z) f (z)
> α (0≤α < p; z∈ᐁ). (1.2)
We denote this class by∗p(α). A functionf∈Ꮽpis said to bep-valently convex of orderαinᐁ, if it satisfies
Re
1+zf(z) f(z)
> α (0≤α < p;z∈ᐁ). (1.3)
The class ofp-valently convex functions of orderαis denoted byp(α). It follows from (1.2) and (1.3) that
f∈p(α)⇐⇒ zf
p ∈p(α). (1.4)
Further, a functionf∈Ꮽpis said to bep-valently close-to-convex of orderβand type α, if there exists a functiong∈∗
p(α)such that
Re
zf(z) g(z)
> β (0≤α, β < p;z∈ᐁ). (1.5)
It is well known (see [10]) that everyp-valently close-to-convex function isp-valent inᐁ.
For arbitrary fixed real numbersAandB (−1≤B < A≤1), letᏼ(A, B)denote the class of functions of the form
which are analytic inᐁand satisfies the condition
φ(z)≺11++AzBz (z∈ᐁ), (1.7)
where the symbol≺stands for subordination. The classᏼ(A, B)was introduced and studied by Janowski [8].
We note that a functionφ∈ᏼ(A, B), if and only if
φ(z)−11−−ABB2
< A−B
1−B2 (B≠−1, z∈ᐁ), Reφ(z)>1−A
2 (B= −1, z∈ᐁ).
(1.8)
For a function f ∈Ꮽ, given by (1.1), the generalized Bernardi-Libera-Livingston integral operatorF [1] is defined by
F (z)=γ+p zγ
z
0t
γ−1f (t) dt
=zp+ ∞
n=1 γ+p
γ+p+nan+pz
n+p (γ >−p; z∈ᐁ). (1.9)
It readily follows from (1.9) that
f∈Ꮽp ⇒F∈Ꮽp. (1.10)
Let
φp(a, c;z)= ∞
n=0 (a)n (c)n
zn+p (c≠0,−1,−2, . . .; z∈ᐁ), (1.11)
and we define a linear operatorLp(a, c)onᏭpby
Lp(a, c)f (z)=φp(a, c;z)∗f (z) (z∈ᐁ), (1.12)
where(x)n=Γ(n+x)/Γ(x)and the symbol∗is the Hadamard product or convolu-tion. Clearly,Lp(a, c)mapsᏭp into itself. Further,Lp(a, a) is the identity operator and
Lp(a, c)=Lp(a, b)Lp(b, c)=Lp(b, c)Lp(a, b) (b, c≠0,−1,−2, . . .). (1.13)
Thus, ifa≠0,−1,−2, . . . ,thenLp(a, c)has an inverseLp(c, a). We also observe that forf∈Ꮽp,
Lp(p+1, p)f (z)= zf(z)
p , Lp(µ+p,1)f (z)=D
µ+p−1f (z), (1.14)
L(a, c)introduced by Carlson and Shaffer [3] in their systemic investigation of certain classes of starlike, convex, and prestarlike hypergeometric functions.
In the present paper, we give some argument properties of certain class of analytic functions inᏭp involving the linear operatorLp(a, c). An application of a certain integral operator is also considered. The results obtained here, besides extending the works of Bulboac˘a [2], Chichra [4], Cho et al. [5], Fukui et al. [6], Libera [9], Nunokawa [13], and Sakaguchi [16], it yields a number of new results.
2. Main results. To establish our main results, we need the following lemmas.
Lemma2.1[11]. Leth(z)be convex (univalent) inᐁand letψ(z)be analytic inᐁ withRe{ψ(z)} ≥0. Ifφ(z)is analytic inᐁandφ(0)=ψ(0), then
φ(z)+ψ(z)zφ(z)≺h(z) (z∈ᐁ) (2.1)
implies
φ(z)≺h(z) (z∈ᐁ). (2.2)
Lemma2.2[12]. Letφ(z)be analytic inᐁ,φ(0)=1, φ(z)≠0inᐁand suppose that there exists a pointz0∈ᐁsuch that
argφ(z)<π 2η
|z|<z0, argφ
z0= π 2η,
(2.3)
whereη >0. Then
z0φ
z0
φz0
=ikη, (2.4)
where
k≥12
d+1d
whenargφz0
=π2η,
k≤ −12
d+1d
when argφz0
= −π2η,
(2.5)
where
φz01/η= ±id (d >0). (2.6) We now derive the following theorem.
Theorem2.3. Leta >0,−1≤B < A≤1,f∈Ꮽp, and suppose thatg∈Ꮽpsatisfies
Lp(a+1, c)g(z) Lp(a, c)g(z) ≺
1+Az
1+Bz (z∈ᐁ). (2.7)
If
arg
(1−λ)Lp(a, c)f (z) Lp(a, c)g(z)+
λLp(a+1, c)f (z) Lp(a+1, c)g(z)−
β
<π
2δ (λ≥0; 0≤β <1; 0< δ≤1; z∈ᐁ),
then
arg
Lp(a, c)f (z) Lp(a, c)g(z)−
β<π
2η (z∈ᐁ), (2.9)
whereη (0< η≤1)is the solution of the equation
δ=
η+π2tan−1
ληsin(π /2)1−t(A, B)
a(1+A)/(1+B)+ληcos(π /2)1−t(A, B)
, forB≠−1,
η, forB= −1,
(2.10)
when
t(A, B)=π2sin−1 A−B
1−AB
. (2.11)
Proof. Let
Lp(a, c)f (z) Lp(a, c)g(z)=
β+(1−β)φ(z). (2.12)
Thenφ(z)is analytic inᐁwithφ(0)=1. On differentiating both sides of (2.12) and using the identity
zLp(a, c)f (z)=aLp(a+1, c)f (z)−(a−p)Lp(a, c)f (z) (2.13)
in the resulting equation, we deduce that
(1−λ)Lp(a, c)f (z) Lp(a, c)g(z)+
λLp(a+1, c)f (z) Lp(a+1, c)g(z)−
β=(1−β)
φ(z)+λzφ(z) ar (z)
, (2.14)
where
r (z)=Lp(a+1, c)g(z) Lp(a, c)g(z)
. (2.15)
If we let
r (z)=ρe(π θ/2)i, (2.16)
then from (2.7) followed by (1.8), it follows that
1−A 1−B < ρ <
1+A 1+B,
−t(A, B) < θ < t(A, B) forB≠−1,
(2.17)
whent(A, B)is given by (2.11), and
1−A
2 < ρ <∞,
−1< θ <1 forB= −1.
Leth(z)be the function which maps onto the angular domain{w:|arg{w}|< (π /2)δ}
withh(0)=1. ApplyingLemma 2.1for thish(z)withψ(z)=λ/(ar (z)), we see that Reφ(z) >0 inᐁand henceφ(z)≠0 inᐁ.
If there exists a point z0∈ ᐁ such that conditions (2.3) are satisfied, then by
Lemma 2.2we obtain (2.4) under restrictions (2.5) and (2.6).
At first, suppose thatp(z0)1/η=id (d >0). For the caseB≠−1, we obtain
arg
(1−λ)Lp(a, c)f
z0
Lp(a, c)g
z0
+λLp(a+1, c)f
z0
Lp(a+1, c)g
z0 −β
=argφz0+arg
1+ λ arz0
z0φ
z0
φz0
=π2η+arg
1+iηkλe− (π θ/2)i
ρa
=π2η+tan−1
ληksin(π /2)(1−θ) ρa+ληkcos(π /2)(1−θ)
≥π2η+tan−1
ληsin(π /2)1−t(A, B)
a(1+A)/(1+B)+ληcos(π /2)1−t(A, B)
≥π
2δ,
(2.19)
whereδandt(A, B)are given by (2.10) and (2.11), respectively. Similarly, for the case B= −1, we have
arg
(1−λ)Lp(a, c)f
z0 Lp(a, c)gz0+
λLp(a+1, c)f
z0 Lp(a+1, c)gz0−
β
≥π2η. (2.20)
This is a contradiction to the assumption of our theorem.
Next, suppose thatφ(z0)1/η= −id (d >0). For the caseB≠−1, applying the same method as above, we have
arg
(1−λ)Lp(a, c)f
z0 Lp(a, c)gz0+
λLp(a+1, c)f
z0 Lp(a+1, c)gz0−
β
≤ −π2η−tan−1
ληsin(π /2)1−t(A, B)
a(1+A)/(1+B)+ληcos(π /2)1−t(A, B)
≤ −π
2δ,
(2.21)
whereδandt(A, B)are given by (2.10) and (2.11), respectively and for the caseB= −1, we have
arg
(1−λ)Lp(a, c)f
z0 Lp(a, c)gz0+
λLp(a+1, c)f
z0 Lp(a+1, c)gz0−
β
≤ −π2η (2.22)
Remark2.4. Fora=c=p,A=1,B= −1, andλ=1,Theorem 2.3is the recent result obtained by Nunokawa [13].
Takinga=µ+p (µ >−p), c=1, A=1, andB=0 inTheorem 2.3, we have the following corollary.
Corollary2.5. Iff∈Ꮽpsatisfies
arg
(1−λ)D
µ+p−1f (z) Dµ+p−1g(z)+λ
Dµ+pf (z) Dµ+pg(z)−β
<π
2δ (λ≥0; 0< δ≤1; 0≤β <1;z∈ᐁ)
(2.23)
for someg∈Ꮽpsatisfying the condition
DDµµ++pp−g(z)1g(z)−1
< α (0< α≤1; z∈ᐁ), (2.24)
then
arg
Dµ+p−1f (z) Dµ+p−1g(z)
<π2η (z∈ᐁ), (2.25)
whereη (0< η≤1)is the solution of the equation
δ=η+π2tan−1
ληsinπ /2−sin−1α (µ+p)(1+α)+ληcosπ /2−sin−1α
. (2.26)
LettingB→A (A <1)andg(z)=zpinTheorem 2.3, we get the following corollary.
Corollary2.6. Iff∈Ꮽpsatisfies
arg
(1−λ)Lp(a, c)f (z) zp +λ
Lp(a+1, c)f (z)
zp −β
<π
2δ (a >0; λ≥0; 0≤β <1; 0< δ≤1; z∈ᐁ),
(2.27)
then
arg
Lp(a, c)f (z) zp −β
<
π
2η (z∈ᐁ), (2.28)
whereη (0< η≤1)is the solution of the equation
δ=η+π2tan−1 λη
a
. (2.29)
Corollary2.7. Under the hypothesis ofCorollary 2.6, we have argH(z)−β<π
where the functionH(z)is defined inᐁby
H(z)= z
0
Lp(a, c)f (t)
tp dt (2.31)
andη (0< η≤1)is the solution of (2.29).
Remark2.8. Takinga=c=p,λ=1, andβ=0 inCorollary 2.6, a=c=p and β=0 inCorollary 2.7, we get the corresponding results obtained by Cho et al. [5].
SettingA=1−(2α/p) (0≤α < p),B= −1, andδ=1 inTheorem 2.3, we have the following corollary.
Corollary2.9. Leta >0,f∈Ꮽp, andg∈∗p(α). If
Re
(1−λ)Lp(a, c)f (z) Lp(a, c)g(z)+
λLp(a+1, c)f (z) Lp(a+1, c)g(z)
> β (λ≥0; 0≤β <1;z∈ᐁ), (2.32)
then
Re
Lp(a, c)f (z) Lp(a, c)g(z)
> β (z∈ᐁ). (2.33)
Remark2.10. Fora=c=p=1 andα=0,Corollary 2.9is the result by Bulboac˘a [2]. If we puta=c=p=1,β=0, andg(z)=zinCorollary 2.9, then we have the result due to Chichra [4]. Further, takinga=c=p,λ=1, andα=β=0 inCorollary 2.9, we get the corresponding results of Libera [9] and Sakaguchi [16].
Theorem2.11. Iff∈Ꮽpsatisfies
arg
Lp(a, c)f (z) zp −β
<π2δ (0≤β <1; 0< δ≤1; z∈ᐁ), (2.34)
then arg
(γ+p)0ztγ−1Lp(a, c)f (t) dt
zγ+p −β
<
π
2η (0< γ+p; z∈ᐁ), (2.35)
whereη (0< η≤1)is the solution of the equation
δ=η+π2tan−1
η γ+p
. (2.36)
Proof. Consider the functionφ(z)defined inᐁby
(γ+p)0ztγ−1Lp(a, c)f (t) dt
zγ+p =β+(1−β)φ(z). (2.37)
Thenφ(z) is analytic inᐁ with φ(0)=1. Differentiating both sides of (2.37) and simplifying, we get
Lp(a, c)f (z)
zp −β=(1−β)
φ(z)+zφ (z)
γ+p
. (2.38)
Takinga=p+1,c=p,β=ρ/p, andδ=1 inTheorem 2.11, we have the following corollary.
Corollary2.12. Iff∈Ꮽpsatisfies
Re f(z)
zp−1
> ρ (0≤ρ < p; z∈ᐁ), (2.39)
then
arg
(γ+p)0ztγ−1f(t) dt
zγ+p −ρ
<
π
2η (z∈ᐁ), (2.40)
whereη (0< η≤1)is the solution of the equation
η+2 πtan
−1 η γ+p
=1. (2.41)
Theorem2.13. Iff∈Ꮽpsatisfies
arg
Lp(a+1, c)f (z) Lp(a, c)f (z) −
a−p−γ a
<π2δ (a >0; p+γ >0; 0< δ≤1;z∈ᐁ), (2.42)
then
arg
zγL
p(a, c)f (z) z
0tγ−1Lp(a, c)f (t) dt
<
π
2η (z∈ᐁ), (2.43)
whereη (0< η≤1)is the solution of (2.36).
Proof. Our proof ofTheorem 2.13is much akin to that ofTheorem 2.3. Indeed, in place of (2.37), we define the functionφ(z)by
φ(z)= zγLp(a, c)f (z) (γ+p)0ztγ−1Lp(a, c)f (t) dt
(z∈ᐁ), (2.44)
and applyLemma 2.1(withψ(z)=1/(γ+p)) as before. We choose to skip the details involved.
Settinga=c=pandδ=1 inTheorem 2.13, we obtain the following corollary.
Corollary2.14. Iff∈Ꮽpsatisfies
Re
zf(z) f (z)
>−γ (γ+p >0; z∈ᐁ), (2.45)
then
arg
zγf (z) z
0tγ−1f (t) dt
<π2η (z∈ᐁ), (2.46)
whereη (0< η≤1)is the solution of (2.41).
Corollary2.15. Iff∈Ꮽpsatisfies
Re
1+zff(z)(z)
>−γ (γ+p >0;z∈ᐁ), (2.47)
then
arg
zf(z) f (z)−(γ/zγ)z
0tγ−1f (t) dt
<
π
2η (z∈ᐁ), (2.48)
whereη (0< η≤1)is the solution of (2.41).
By settingγ=0 inCorollary 2.15, we have the following corollary.
Corollary2.16. Iff∈p(0), then
arg
zf(z)
f (z)
<π2η (z∈ᐁ), (2.49)
whereη (0< η≤1)is the solution of the equation:
η+π2tan−1 η
p
=1. (2.50)
Similarly, we have the following theorem.
Theorem2.17. Iff∈Ꮽpsatisfies
arg
Lp(a+1, c)f (z) Lp(a, c)f (z) −
β
<
π
2δ (a >0; 0≤β <1; 0< δ≤1;z∈ᐁ), (2.51)
then
arg
Lp(a, c)f (z) zp
<
π
2η (z∈ᐁ), (2.52)
whereη (0< η≤1)is the solution of the equation
δ= 2 πtan
−1 η (1−β)a
. (2.53)
Theorem2.18. Letf∈Ꮽpand suppose that
B < A≤B+p(1−B)
a (a >0; −1≤B < A≤1). (2.54)
If arg
(1−λ)Lp(a+1, c)f (z) Lp(a, c)g(z) +
λ
Lp(a+1, c)f (z)
Lp(a, c)g(z) −β
<π
2δ (λ≥0; 0≤β <1; 0< δ≤1; z∈ᐁ),
for someg∈Ꮽpsatisfying
Lp(a+1, c)g(z) Lp(a, c)g(z) ≺
1+Az
1+Bz (z∈ᐁ), (2.56)
then
arg
Lp(a+1, c)f (z) Lp(a, c)g(z) −
β
<π2η (z∈ᐁ), (2.57)
whereη (0< η≤1)is the solution of the equation
δ=
η+2 πtan
−1 ληsin(π /2)
1−t(A, B)
p(1+B)+a(A−B)/(1+B)+ληcos(π /2)1−t(A, B)
, forB≠−1,
η, forB=−1,
(2.58)
when
t(A, B)=π2sin−1
a(A−B) p1−B2−aB(A−B)
. (2.59)
Proof. Let
Lp(a+1, c)f (z) Lp(a, c)g(z) =
β+(1−β)φ(z), r (z)=Lp(a+1, c)g(z) Lp(a, c)g(z)
, (2.60)
we have
(1−λ)Lp(a+1, c)f (z) Lp(a, c)g(z) +
λ
Lp(a+1, c)f (z)
Lp(a+1, c)g(z)
−β=(1−β)
φ(z)+ λzφ (z)
ar (z)+p−a
.
(2.61)
The remaining part of the proof ofTheorem 2.18is similar to that ofTheorem 2.3. So we omit the details.
Puta=c=p,λ=1,A=α/p, andB=0 inTheorem 2.18, we have the following corollary.
Corollary2.19. Iff∈Ꮽpsatisfies
arg
zf(z) g(z) −β
<
π
2δ (0≤β < p; 0< δ≤1; z∈ᐁ), (2.62)
for someg∈Ꮽpsatisfying the condition
zg(z)
g(z) −p
then
arg
zf(z)
g(z) −β
<π2η (z∈ᐁ), (2.64)
whereη (0< η≤1)is the solution of the equation
δ=η+π2tan−1
ηsinπ /2−sin−1(α/p) p+α+ηcosπ /2−sin−1(α/p)
. (2.65)
Lemma2.20. Let
α=ξ+ ξ γ+p+aξ
0≤(a−1)/a < ξ < α <1 (2.66)
and the functionG(z)be defined by
G(z)=γz+γp z
0
tγ−1g(t) dt g∈Ꮽp
(2.67)
forγ > (aξ2+(p+1−a)ξ−p)/(1−ξ). Ifg∈Ꮽ
psatisfies
Lp(a+1, c)g(z) Lp(a, c)g(z) −
1< α (z∈ᐁ), (2.68)
then
Lp(a+1, c)G(z) Lp(a, c)G(z) −
1< ξ (z∈ᐁ). (2.69)
Proof. Defining the functionw(z)by
Lp(a+1, c)G(z) Lp(a, c)G(z) =
1+ξw(z), (2.70)
we see thatw(z)is analytic inᐁwithw(0)=0. Now, using the identities
zLp(a, c)G(z)=aLp(a+1, c)G(z)−(a−p)Lp(a, c)G(z), (2.71)
zLp(a, c)G(z)
=(γ+p)L
p(a, c)g(z)−γLp(a, c)G(z) (2.72)
in (2.70), we get
Lp(a, c)G(z) Lp(a, c)g(z)=
γ+p
γ+p+aξw(z). (2.73)
Making use of the logarithmic differentiation of both sides of (2.73) and using identity (2.71) for bothg(z)andf (z)in the resulting equation, we deduce that
Lp(a+1, c)g(z) Lp(a, c)g(z) −
1=ξw(z)+ zw (z)
γ+p+aξw(z)
We assume that there exists a pointz0∈ᐁsuch that max|z|<|z0||w(z)| = |w(z0)| =1. Then by Jack’s lemma [7], we havez0w(z0)=kw(z0) (k≥1). Letw(z0)=eiθ, and apply this result tow(z)atz0∈ᐁ, we get
Lp(a+1, c)gz0 Lp(a, c)g
z0
−1=ξ1+ k γ+p+aξeiθ
=ξ
(γ+p+k)2+2aξ(γ+p+k)cosθ+(aξ)2 (γ+p)2+2aξ(γ+p)cosθ+(aξ)2
1/2 .
(2.75)
Since the right side of (2.75) is decreasing for 0≤θ <2πandγ >{aξ2+(p+1−a)ξ− p}/(1−ξ), we obtain
Lp(a+1, c)gz0 Lp(a, c)gz0 −
1≤ξ(γ+p+1+aξ)
γ+p+aξ , (2.76)
which contradicts our hypothesis and hence we get
w(z)=ξ1LpL(a+1, c)G(z)
p(a, c)G(z) −
1<1 (z∈ᐁ). (2.77)
This completes the proof ofLemma 2.20.
Remark2.21. We note that fora=c=p=1,Lemma 2.20yields the correspond-ing result obtained by Fukui et al. [6].
Theorem 2.22. Letα be as given in (2.66) andγ∗ >max{(aξ2+(p+1−a)ξ− p)/(1−ξ), aξ−p}. Iff∈Ꮽpsatisfies
arg
Lp(a+1, c)f (z) Lp(a, c)g(z) −
β
<π2δ (0≤β <1; 0< δ≤1; z∈ᐁ), (2.78)
for somef∈Ꮽpsatisfying condition (2.68), then
arg
Lp(a+1, c)F (z) Lp(a, c)G(z) −
β
<π2η (z∈ᐁ), (2.79)
where the functionF (z)andG(z)are defined forγ∗by (1.9) and (2.67), respectively andη (0< η≤1)is the solution of the equation
δ=η+2 πtan
−1
ηsinπ /2−sin−1aξ/γ∗+p γ∗+p+aξ+ηcosπ /2−sin−1aξ/γ∗+p
. (2.80)
Proof. Consider the functionφ(z)defined inᐁby
Lp(a+1, c)F (z) Lp(a, c)G(z) =
Thenφ(z)is analytic inᐁwithφ(0)=1. Taking logarithmic differentiation on both sides of (2.81) and using identity (2.71) in the resulting equation, we get
zLp(a+1, c)F (z) Lp(a+1, c)F (z) =
p−a+aLp(a+1, c)G(z) Lp(a, c)G(z) +
(1−β) zφ(z)
β+(1−β)φ(z). (2.82)
From the definition ofF (z), we have
γ∗+pLp(a, c)f (z)=a
Lp(a+1, c)F (z) +γ∗L
p(a+1, c)F (z). (2.83)
Again, from (2.71) and (2.72), it follows that
γ∗+pLp(a+1, c)g(z)=zLp(a+1, c)G(z)+
p+γ∗−aLp(a, c)G(z). (2.84)
Thus, by using (2.83) and (2.84) followed by (2.82), we obtain
Lp(a+1, c)f (z) Lp(a, c)g(z) −
β=(1−β)
φ(z)+ zφ(z) ar (z)+γ∗+p−a
, (2.85)
wherer (z)=Lp(a+1, c)G(z)/Lp(a, c)G(z). By usingLemma 2.20, we have
r (z)≺1+ξz (z∈ᐁ), (2.86)
whereξis given by (2.66). Letting
ar (z)+γ∗+p−a=ρeiπ θ/2 (2.87)
and using the techniques ofTheorem 2.3, the remaining part of the proof ofTheorem 2.22follows.
Remark2.23. We easily find the following:
γ >
aξ−p, ifa−1
a < ξ < 2a−1
2a ,
2(a−p)−1
2 , ifξ=
2a−1 2a ,
aξ2+(p+1−a)ξ−p 1−ξ , if
2a−1
2a < ξ <1.
(2.88)
Takinga=c=pinTheorem 2.22, we get the following corollary.
Corollary2.24. Let
whereγ∗>max{(pξ2+ξ−p)/(1−ξ), p(ξ−1)}. Iff∈Ꮽ
psatisfies
arg
zf(z)
g(z) −β
<π2δ (0≤β < p; 0< δ≤1; z∈ᐁ) (2.90)
for someg∈Ꮽpsatisfying the condition
zgg(z)(z)−p< pα (z∈ᐁ), (2.91)
then
arg
zF(z)
G(z) −β
<
π
2 (z∈ᐁ), (2.92)
whereη (0< η≤1)is the solution of the equation
δ=η+ 2 πtan
−1
ηsinπ /2−sin−1pξ/γ∗+p γ∗+p(1+ξ)+ηcosπ /2−sin−1pξ/γ∗+p
. (2.93)
Acknowledgment. This work was supported by Korea Research Foundation Grant KRF-2001-015-DP0013.
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Nak Eun Cho: Division of Mathematical Sciences, Pukyong National University,
Pusan608-737, Korea
E-mail address:[email protected]
J. Patel: Department of Mathematics, Utkal University, Vani Vihar, Bhubaneswar 751 004, India
E-mail address:[email protected]
G. P. Mohapatra: Silicon Institute of Technology, Near Infocity, Patia,
Bhubaneswar751035, India