# Subclass of Multivalentβ-Uniformly Functions with Varying Arguments

Authors: R. M. El-Ashwah; M. K. Aouf; A. A. M. Hassan; A. H. Hassan
Journal: Journal of Mathematics
 (2013)

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## Abstract

We introduce a class of multivalentβ-uniformly functions with varying arguments, give some properties of functions in
this class, and obtain sharp results including coefficient estimates and distortion theorems. Finally, we give the extreme points for this
class.


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## Body

## 1. Introduction

LetA(p) denote the class of functions of the following form:
(1)f(z)=zp+∑k=p+1∞‍akzk(p∈ℕ={1,2,3,…}),
which are analytic and p-valent in the open unit disc U={z∈ℂ:|z|<1}. A function f(z)∈A(p) is said to be p-valent starlike of order α if it satisfies the following inequality:
(2)Re{zf′(z)f(z)}>α(0≤α<p;p∈ℕ;z∈U).
We denote by Sp(α) the class of all p-valent starlike functions of order α. Also a function f(z)∈A(p) is said to be p-valent convex of order α, if it satisfies the following inequality:
(3)Re{1+zf′′(z)f′(z)}>α(0≤α<p;p∈ℕ;z∈U).
We denote by Kp(α) the class of all p-valently convex functions of order α. The classes Sp(α) and Kp(α) were introduced by Patil and Thakare [1] and Owa [2]. Further from (2) and (3), we can see that
(4)f(z)∈Kp(α)⟺zf′(z)p∈Sp(α)(0≤α<p;p∈ℕ).
Also for a function f(z)∈A(p), the differential operator Dpnf(z) was defined by Kamali and Orhan [3], Orhan and Kiziltunç [4], and Aouf and Mostafa [5] as follows:
(5)Dp0f(z)=f(z),Dp1f(z)=Dpf(z)=zpf′(z)=zp+∑k=p+1∞‍(kp)akzk,Dp2f(z)=Dp(Dp1f(z))=zp+∑k=p+1∞‍(kp)2akzk,
and (in general)
(6)Dpnf(z)=zp+∑k=p+1∞‍(kp)nakzk(p∈ℕ;n∈ℕ0=ℕ∪{0}).
We note that putting p=1 in (6), we have the operator D1n=Dn which was introduced by Sălăgean [6].In [7], Eker and Şeker introduced the class Np(m,n;α,β) as follows.Definition 1.
For0≤α<1,β≥0,m∈ℕ,n∈ℕ0,m>n, and z∈U, let Np(m,n;α,β) be the class of functions f(z)∈A(p) which satisfy
(7)Re{Dpmf(z)Dpnf(z)-α}>β|Dpmf(z)Dpnf(z)-1|.
Specializing the parameters   β,m,n, and p we obtain the following subclasses studied by various authors:(i)
Np(1,0;α,β)=UST(α,β,p) and Np(2,1;α,β)=UCV(α,β,p)(0≤α<p,β≥0,p∈ℕ) (see Khairnar and More [8]);(ii)
Np(1,0;α,o)=S*(p,α) and Np(2,1;α,0)=K(p,α)(0≤α<p,p∈ℕ) (see Patil and Thakare [1] and Owa [2]);(iii)
N1(m,n;α,β)=Nm,n(α,β)(0≤α<1,β≥0,m∈ℕ,n∈ℕ0,m>n) (see Eker and Owa [9]);(iv)
N1(m,n;α,0)=Km,n(α)(0≤α<1,m∈ℕ,n∈ℕ0,m>n) (see Eker and Owa [10]);(v)
N1(n+1,n;α,β)=S(n,α,β)(0≤α<1,β≥0,n∈ℕ0) (see Rosy and Murugusundaramoorthy [11] and Aouf [12]);(vi)
N1(1,0;α,β)=SD(α,β) and N1(2,1;α,β)=KD(α,β)(0≤α<1,β≥0) (see Shams et al. [13] and Bharati et al. [14]).
Also we note that(8)Np(n+1,n;α,β)=Np(n;α,β)={f∈A(p):Re{Dpn+1f(z)Dpnf(z)-α}>β|Dpn+1f(z)Dpnf(z)-1|(0≤α<1;β≥0;n∈ℕ0;z∈U)Dpn+1f(z)Dpnf(z)}.
Silverman [15] defined the class of univalent functions f(z)=z+∑k=2∞akzk for which arg(ak) prescribed in such a way that f(z) is univalent if and only if f(z) is starlike. In this paper, we introduce the subclass of p-valent functions with varying arguments as follows.Definition 2.
A functionf(z) of the form (1) is said to be in the class Vp(θk) if f∈A(p) and arg(ak)=θk for all k≥p+1. If furthermore there exists a real number δ such that θk+(k-p)δ≡π(mod2π) for all k≥p+1, then f(z) is said to be in the class Vp(θk,δ). The union of Vp(θk,δ) taken over all possible sequences {θk} and all possible real numbers δ is denoted by Vp.
LetVp(m,n;α,β) denote the subclass of Vp consisting of functions f(z)∈Np(m,n;α,β).
We note thatV1(1,0;α,0)=V*(α), which was introduced and studied by Silverman [15].
Specializing the parametersβ,m,n, and p, we obtain the following classes with varying arguments which are associated to the subclasses mentioned before:(i)
Vp(1,0;α,β)=VST(α,β,p) and Vp(2,1;α,β)=VCV(α,β,p)(0≤α<p,β≥0,p∈ℕ);(ii)
Vp(1,0;α,o)=VS*(p,α) and Vp(2,1;α,0)=VK(p,α)(0≤α<p,p∈ℕ);(iii)
V1(m,n;α,β)=Vm,n(α,β)(0≤α<1,β≥0,m∈ℕ,n∈ℕ0,m>n);(iv)
V1(m,n;α,0)=Vm,n(α)(0≤α<1,m∈ℕ,n∈ℕ0,m>n);(v)
V1(n+1,n;α,β)=V(n,α,β)(0≤α<1,β≥0,n∈ℕ0);(vi)
V1(1,0;α,β)=VSD(α,β) and V1(2,1;α,β)=VKD(α,β)(0≤α<1,β≥0);(vii)
Vp(n+1,n;α,β)=Vp(n;α,β).In this paper, we obtain coefficient bounds for functions in the classVp(m,n;α,β), further we obtain distortion theorems and the extreme points for functions in this class.


## 2. Coefficient Estimates

Unless otherwise mentioned, we assume in the reminder of this paper that0≤α<1,β≥0,m∈ℕ,n∈ℕ0,m>n, and z∈U.We will need the following lemma.Lemma 3.
The sufficient condition forf(z) given by (1) to be in the class Np(m,n;α,β) is that
(9)∑k=p+1∞‍[(1+β)(kp)m-(α+β)(kp)n]|ak|≤1-α.Proof.
It suffices to show that inequality (7) holds true. Then it suffices to show that
(10)β|Dpmf(z)Dpnf(z)-1|-Re{Dpmf(z)Dpnf(z)-1}≤1-α.
We have
(11)β|Dpmf(z)Dpnf(z)-1|-Re{Dpmf(z)Dpnf(z)-1}≤(1+β)|Dpmf(z)Dpnf(z)-1|,
or, equivalently,
(12)β|Dpmf(z)Dpnf(z)-1|-Re{Dpmf(z)Dpnf(z)-1}≤(1+β)|∑k=p+1∞[(k/p)m-(k/p)n]akzkzp+∑k=p+1∞(k/p)nakzk|≤(1+β)|∑k=p+1∞[(k/p)m-(k/p)n]akzk-p1+∑k=p+1∞(k/p)nakzk-p|,
then we have
(13)β|Dpmf(z)Dpnf(z)-1|-Re{Dpmf(z)Dpnf(z)-1}≤(1+β)∑k=p+1∞[(k/p)m-(k/p)n]|ak||z|k-p1-∑k=p+1∞(k/p)n|ak||z|k-p.
Letting z→1- through real values, then we have
(14)β|Dpmf(z)Dpnf(z)-1|-Re{Dpmf(z)Dpnf(z)-1}≤(1+β)∑k=p+1∞[(k/p)m-(k/p)n]|ak|1-∑k=p+1∞(k/p)n|ak|.
The last expression is bounded above by (1-α) if
(15)∑k=p+1∞‍[(1+β)(kp)m-(α+β)(kp)n]|ak|≤1-α.
And hence the proof of Lemma 3 is completed.Remark 4.
(i) The result obtained from Lemma3 corrects the result obtained by Eker and Şeker [7, Theorem 1].
(ii) Puttingp=1 in Lemma 3, we correct the result obtained by Eker and Owa [9, Theorem 2.1].Corollary 5.
The sufficient condition forf(z) given by (1) to be in the class Np(n;α,β), which is defined by (8) is that
(16)∑k=p+1∞‍(kp)n[kp(1+β)-(α+β)]|ak|≤1-α.Theorem 6.
Letf(z) of the form (1), then f(z)∈Vp(m,n;α,β) if and only if
(17)∑k=p+1∞‍[(1+β)(kp)m-(α+β)(kp)n]|ak|≤1-α.Proof.
In view of Lemma3, we need only to show that function f(z)∈Vp(m,n;α,β) satisfies the coefficient inequality (17). If f(z)∈Vp(m,n;α,β), then from (7), we have
(18)Re{Dpmf(z)Dpnf(z)-α}>β|Dpmf(z)Dpnf(z)-1|,
then we have
(19)Re{(1-α)zp+∑k=p+1∞[(k/p)m-α(k/p)n]akzkzp+∑k=p+1∞(k/p)nakzk}>β|∑k=p+1∞[(k/p)m-(k/p)n]akzkzp+∑k=p+1∞(k/p)nakzk|
or
(20)Re{(1-α)+∑k=p+1∞[(k/p)m-α(k/p)n]akzk-p1+∑k=p+1∞(k/p)nakzk-p}>β|∑k=p+1∞[(k/p)m-(k/p)n]akzk-p1+∑k=p+1∞(k/p)nakzk-p|.
Since f(z)∈V, f(z)lies in the class V(θk,δ) for some sequence {θk} and a real number δ such that
(21)θk+(k-p)δ≡π(mod2π)(k≥p+1).
Set z=reiδ in (20), we get
(22)(1-α)-∑k=p+1∞[(k/p)m-α(k/p)n]|ak|rk-p1-∑k=p+1∞(k/p)n|ak|rk-p>β[∑k=p+1∞[(k/p)m-(k/p)n]|ak|rk-p1-∑k=p+1∞(k/p)n|ak|rk-p].
Letting r→1, then we have the inequality (17).
Hence the proof of Theorem6 is completed.Corollary 7.
Iff(z)∈Vp(m,n;α,β), then
(23)|ak|≤1-α(1+β)(k/p)m-(α+β)(k/p)n(k≥p+1;p∈ℕ).
The result is sharp for the function f(z) given by
(24)f(z)=zp+1-α(1+β)(k/p)m-(α+β)(k/p)neiθkzk(k≥p+1;p∈ℕ;z∈U).


## 3. Distortion Theorems

Theorem 8.
Let the functionf(z) defined by (1) be in the class Vp(m,n;α,β). Then
(25)|z|p-1-αΦ(p+1)|z|p+1≤|f(z)|≤|z|p+1-αΦ(p+1)|z|p+1,
where
(26)Φ(p+1)=(1+β)(p+1p)m-(α+β)(p+1p)n.
The result is sharp.Proof.
We employ the same technique as used by Silverman [15]. In view of Theorem 6, since
(27)Φ(k)=(1+β)(kp)m-(α+β)(kp)n,
is an increasing function of k(k≥p+1;p∈ℕ), we have
(28)Φ(p+1)∑k=p+1∞‍|ak|≤∑k=p+1∞‍Φ(k)|ak|≤1-α,
that is
(29)∑k=p+1∞‍|ak|≤1-αΦ(p+1).
Thus, we have
(30)|f(z)|≤|z|p+|z|p+1∑k=p+1∞‍|ak|,|f(z)|≤|z|p+1-α(1+β)((p+1)/p)m-(α+β)((p+1)/p)n|z|p+1.
Similarly, we get
(31)|f(z)|≥|z|p-|z|p+1∑k=p+1∞‍|ak|≥|z|p-1-α(1+β)((p+1)/p)m-(α+β)((p+1)/p)n×|z|p+1.
This completes the proof of Theorem8. Finally the result is sharp for the following function:
(32)f(z)=zp+1-α(1+β)((p+1)/p)m-(α+β)((p+1)/p)n×eiθp+1zp+1,
at z=±|z|e-iθp+1.Corollary 9.
Under the hypotheses of Theorem8, f(z) is included in a disc with center at the origin and radius r1 given by
(33)r1=1+1-α(1+β)((p+1)/p)m-(α+β)((p+1)/p)n.Theorem 10.
Let the functionf(z) defined by (1) be in the class Vp(m,n;α,β). Then
(34){p-(p+1)(1-α)Φ(p+1)|z|}|z|p-1≤|f′(z)|≤{p+(p+1)(1-α)Φ(p+1)|z|}|z|p-1,
where Φ(p+1) is defined by (26). Also the result of Theorem 10 is sharp.Proof.
Similarly forΦ(k) defined by (27) it is clear that Φ(k)/k is an increasing function of k(k≥p+1;p∈ℕ), in view of Theorem 6, we have
(35)Φ(p+1)p+1∑k=p+1∞‍k|ak|≤∑k=p+1∞‍Φ(k)kk|ak|=∑k=p+1∞‍Φ(k)|ak|≤1-α,
that is
(36)∑k=p+1∞‍k|ak|≤(p+1)(1-α)Φ(p+1).
Thus, we have
(37)|f′(z)|≤p|z|p-1+|z|p∑k=p+1∞‍k|ak|≤{p+|z|∑k=p+1∞‍k|ak|}|z|p-1≤{p+(p+1)(1-α)Φ(p+1)|z|}|z|p-1.
Similarly,
(38)|f′(z)|≥p|z|p-1-|z|p∑k=p+1∞‍k|ak|≥{p-|z|∑k=p+1∞‍k|ak|}|z|p-1≤{p-(p+1)(1-α)Φ(p+1)|z|}|z|p-1.
Finally, we can see that the assertions of Theorem 10 are sharp for the function f(z) defined by (32). This completes the proof of Theorem 10.Corollary 11.
Under the hypotheses of Theorem10, f′(z) is included in a disc with center at the origin and radius r2 given by
(39)r2=p+(p+1)(1-α)Φ(p+1).


## 4. Extreme Points

Theorem 12.
Let the functionf(z) defined by (1) be in the class Vp(m,n;α,β), with arg(ak)=θk, where θk+(k-p)δ≡π(mod2π)(k≥p+1). Define
(40)fp(z)=zp,fk(z)=zp+1-α(1+β)(k/p)m-(α+β)(k/p)neiθkzk(k≥p+1;z∈U).
Then f(z)∈Vp(m,n;α,β) if and only if f(z) can be expressed in the form f(z)=∑k=p∞μkfk(z), where μk≥0 and ∑k=p∞μk=1.Proof.
Iff(z)=∑k=p∞μkfk(z) with μk≥0 and ∑k=p∞μk=1, then
(41)∑k=p+1∞‍[(1+β)(kp)m-(α+β)(kp)n]×1-α(1+β)(k/p)m-(α+β)(k/p)nμk=∑k=p+1∞‍(1-α)μk=(1-α)(1-μp)≤(1-α).
Hence, f(z)∈Vp(m,n;α,β).
Conversely, let the functionf(z) defined by (1) be in the class Vp(m,n;α,β). Define
(42)μk=[(1+β)(k/p)m-(α+β)(k/p)n]1-α|ak|(k≥p+1;p∈ℕ),μp=1-∑k=p+1∞‍μk.
From Theorem 6, ∑k=p+1∞μk≤1 and so μp≥0. Since μkfk(z)=μkzp+akzk, then
(43)∑k=p∞‍μkfk(z)=zp+∑k=p+1∞‍akzk=f(z).
This completes the proof of  Theorem 12.Remark 13.
(i) Puttingβ=n=0 and p=m=1 in all the above results, we obtain the corresponding results obtained by Silverman [15].
(ii) Putting (i)β=n=0and m=1 and (ii) β=0,n=1and m=2 in all the above results, we obtain the corresponding results obtained by Aouf et al. [16].
(iii) Puttingm=n+1(n∈ℕ0) in all the above results, we obtain the corresponding results of the classes Vp(n;α,β), defined by (8).



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