# On Subclasses of Analytic Functions with respect to Symmetrical Points

Authors: Muhammad Arif; Khalida Inayat Noor; Rafiullah Khan
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

In our present investigation, motivated from Noor work, we define the classℛks(b) of functions of bounded radius rotation of complex order b with respect to symmetrical points and learn some of its basic properties. We also apply this concept to define the class ℋks(α,b,δ). We study some interesting results, including arc length, coefficient difference, and univalence sufficient condition for this class.


---

## Body

## 1. Introduction

Let𝒜 denote the class of analytic function satisfying the condition f(0)=0, f′(0)-1=0 in the open unit disc 𝒰={z:|z|<1} and in more simple form:(1.1)f(z)=z+∑n=2∞anzn(z∈U).
By 𝒮, 𝒞, and 𝒮*, we means the well-known subclasses of 𝒜 which consists of univalent, convex, and starlike functions, respectively. In [1], Sakaguchi introduced the class 𝒮s* of starlike functions with respect to symmetrical points and is defined as follows: a function f(z) given by (1.1) belongs to the class 𝒮s*, if and only if(1.2)Re{zf′(z)f(z)-f(-z)}>0(z∈U).
Motivated from Sakaguchi work, Das and Singh [2] extend the concepts of 𝒮s* to other class in 𝒰, namely, convex functions with respect to symmetrical points. Let 𝒞s denote the class of convex functions with respect to symmetrical points and satisfying the following condition:(1.3)Re{(zf′(z))′f′(z)-f′(-z)}>0(z∈U).
Let 𝒫k(δ),  0≤δ<1, be the class of functions p(z) analytic in 𝒰 with p(0)=1 and(1.4)∫02π|Rep(z)-δ1-δ|dθ≤kπ,z=reiθ,k≥2.
This class was introduced in [3]. For δ=0, we obtain the class 𝒫k defined by Pinchuk [4], and for k=2, the class 𝒫k reduces to the class 𝒫 of functions with positive real part.Now, with the help of the aforementioned concepts, we define the classℛks(b) of functions of bounded radius rotation of complex order b with respect to symmetrical points as follows.Definition 1.1.
Letf(z)∈𝒜 in 𝒰. Then f(z)∈ℛks(b), if and only if
(1.5)1+1b{2zf′(z)f(z)-f(-z)-1}∈Pk(z∈U),
where k≥2 and b∈ℂ-{0}.
Using the classℛks(b), we define the class ℋks(α,b,δ) as follows.Definition 1.2.
Letf(z)∈𝒜 in 𝒰. Then f(z)∈ℋks(α,b,δ), if and only if there exists g(z)∈ℛks(b) such that
(1.6)zf′(z)f(z)(2f(z)g(z)-g(-z))α∈P(δ),
where α>0, 0≤δ<1, and b∈ℂ-{0}.
It is noticed that, by giving specific values toα,  b,  δ, and k in ℛks(b) and ℋks(α,b,δ), we obtain many well-known as well as new subclasses of analytic and univalent functions; for details see [5–11].
Throughout this paper, we will assume, unless otherwise stated, thatk≥2, α>0, 0≤δ<1, and b∈ℂ-{0}.Lemma 1.3.
Letp(z) be analytic in 𝒰 where p(0)=1 belongs to P(δ).Then
(1.7)12π∫02π|p(z)|2dθ≤1+(4(1-δ)2-1)r21-r2
(see [8, 12]).Lemma 1.4.
Lets1(z) be univalent function in 𝒰. Then there exists ξ with |ξ|=r such that for all z, |z|=r,
(1.8)|z-ξ||s1(z)|≤2r21-r2
(see [13]).


## 2. Some Properties of the Classesℛks(b)andℋks(α,b,δ)

Theorem 2.1.
Letf(z)∈ℛks(b). Then the odd function
(2.1)ϕ(z)=12[f(z)-f(-z)]
belongs to ℛk(b) in 𝒰.Proof.
Letf(z)∈ℛks(b) and consider
(2.2)ϕ(z)=12[f(z)-f(-z)].
From logarithmic differentiation of the previous relation, we have
(2.3)ϕ′(z)ϕ(z)=f′(z)-f′(-z)f(z)-f(-z),
or, equivalently,
(2.4)zϕ′(z)ϕ(z)=12[p1(z)+p2(z)]
with
(2.5)p1(z)=2zf′(z)f(z)-f(-z),p2(z)=2(-z)f′(-z)f(-z)-f(z)
belongs to 𝒫k(b). Since 𝒫k(b) is a convex set, we have
(2.6)zϕ′(z)ϕ(z)∈Pk(b)(z∈U),
and hence ϕ(z)∈ℛk(b).Theorem 2.2.
Letf(z)∈ℛks(b). Then
(2.7)f′(z)=12[b(p(z)-1)+1]exp⁡{b2∫0z1ξ(p(ξ)-p(-ξ)-2b)dξ}.Proof.
Letf(z)∈ℛks(b). Then by definition we have
(2.8)1+1b[2zf′(z)f(z)-f(-z)-1]=p(z),p(z)∈Pk.
Simple computation yields us
(2.9)f(z)-f(-z)z=exp⁡{b2∫0z1ξ[p(ξ)-p(-ξ)-2b]dξ}.
Using (2.8) in (2.9), we can easily obtain (2.7).
If we putb=1 and k=2 in Theorem 2.1, we obtain the integral representation for 𝒮s* given by Stankiewiez in [14].Theorem 2.3.
Letf(z)∈ℛks(b). Then
(2.10)|a2|≤k|b|2.
The function f0(z)∈ℛks(b) defined by
(2.11)f0′(z)=(1+z2)(k-2)/4(1-z2)(k+2)/4[b{(k+24)(1-z1+z)-(k-24)(1+z1-z)}+(1-b)]
shows that this bound is sharp.Proof .
Sincef(z)∈ℛks(b), there exists an odd function ϕ(z)∈ℛk(b) with
(2.12)ϕ(z)=12[f(z)-f(-z)],
such that
(2.13)zf′(z)=ϕ(z)p(z),
with p(z)∈𝒫k(b). Let
(2.14)ϕ(z)=z+∑n=2∞b2n-1z2n-1,p(z)=1+∑n=1∞cnzn.
Then (2.13) implies that
(2.15)z+∑n=2∞nanzn=[z+∑n=2∞b2n-1z2n-1][1+∑n=1∞cnzn].
Equating the coefficients of z2, we have 2a2=c1, and so
(2.16)|a2|≤k|b|2,
where we have used the coefficient bounds |c1|≤k|b| for the class 𝒫k(b).Corollary 2.4.
The range of every univalent functionf(z)∈ℛks(b) contains the disc
(2.17)|w|<24+k|b|.Proof .
The Koebe one-quarter theorem states that each omitted valuew of the univalent function f(z) of the form (1.1) satisfies
(2.18)|w|>12+|a2|.
Using (2.18) and Theorem 2.3, we obtain the required result.By using the same method as in [1], we obtain the following result.Theorem 2.5.
Letf(z)∈ℛks(b). Then,  for z=reiθ and 0≤θ1<θ2≤2π,
(2.19)∫θ1θ2Re(zf′(z))′f′(z)dθ>-(k-1)|b|π.Theorem 2.6.
Letf(z)∈ℋks(α,b,0). Then, for z=reiθ,
(2.20)∫θ1θ2ReJ(α,f(z))dθ>-(α|b|(k-1)+1)π,
where 0≤θ1<θ2≤2π and
(2.21)J(α,f(z))=(1+zf′′(z)f′(z))+(α-1)zf′(z)f(z).Proof.
We can define, forz=reiθ, r<1, θ real, the following:
(2.22)S(r,θ)=arg[zf′(z)fα-1(z)],(2.23)V(r,θ)=arg[g(z)-g(-z)2]α.
The functions S(z)andV(z) are periodic and continuous with period 2π. Since f(z)∈ℋks(α,b,0), therefore from (2.22), it follows that we can choose the branches of argument of S(z) and V(z) as
(2.24)|S(r,θ)-V(r,θ)|≤π2.
Now we have from (2.22)
(2.25)V(r,θ2)-V(r,θ1)=α∫θ1θ2Rezϕ′(z)ϕ(z)dθ,
where ϕ(z) is an odd function of the following form:
(2.26)ϕ(z)=12[g(z)-g(-z)].
Since g(z)∈ℛks(b), therefore by using Theorem 2.5, we have
(2.27)∫θ1θ2Rezϕ′(z)ϕ(z)dθ>-(k-1)|b|π.
From (2.22), (2.23), (2.24), and (2.27), we have
(2.28)|S(r,θ2)-S(r,θ1)|=|S(r,θ2)-V(r,θ2)-(S(r,θ1)-V(r,θ1))+(V(r,θ2)-V(r,θ1))|<π2+π2+α(k-1)|b|π=(α|b|(k-1)+1)π.
Moreover, from (2.22)
(2.29)ddθS(r,θ)=Re[(1+zf′′(z)f′(z))+(α-1)zf′(z)f(z)].
Therefore
(2.30)∫θ1θ2ReJ(α,f(z))dθ>-(α|b|(k-1)+1)π.Theorem 2.7.
Letf(z)∈ℋks(α,b,δ).Then for α(k/2+1)Reb>1,
(2.31)Lrf(z)≤{C(α,b,δ,k)M1-α(r)(11-r)α(k/2+1)Reb,0<α≤1,C(α,b,δ,k)mα-1(r)(11-r)α(k/2+1)Reb,α>1,
where m(r)=min|z|=r|f(z)|,M(r)=max|z|=r|f(z)|,and C(α,b,δ,k) is a constant depending upon α,b,δ, and konly.Proof.
We know that(2.32)Lrf(z)=∫02π|zf′(z)|dθ,z=reiθ,0<r<1.
Since f(z)∈ℋks(α,b,δ), therefore
(2.33)zf′(z)f(z)(2f(z)g(z)-g(-z))α=p(z),p(z)∈P(δ).
By Theorem 2.1, we have, for g(z)∈ℛks(b), the odd function ϕ(z)=(1/2)[g(z)-g(-z)]∈ℛk(b). This implies that
(2.34)zf′(z)=(f(z))1-α(ϕ(z))αp(z).
Therefore, we have
(2.35)Lrf(z)≤∫02π|f(z)|1-α|ϕ(z)|α|p(z)|dθ,≤M1-α(r)∫02π|ϕ(z)|α|p(z)|dθ.
Since ϕ(z)∈ℛk(b), therefore we have for odd functions s1(z),  s1(z)∈𝒮*,
(2.36)≤M1-α(r)∫02π|(s1(z))(k/4+1/2)b(s2(z))(k/4-1/2)b|α|p(z)|dθ,≤cM1-α(r)∫02π|(s1(z))|α(k/4+1/2)Reb|(s2(z))|α(k/4-1/2)Reb|p(z)|dθ,c=e(π/2)Im⁡b.≤cM1-α(r)2α(k/2-1)Rebr-α(k/4-1/2)Reb∫02π|(s1(z))|α(k/4+1/2)Reb|p(z)|dθ.
Now using Cauchy Schwarz inequality, we have
(2.37)Lrf(z)≤cM1-α(r)2α(k/2-1)Rebr-α(k/4-1/2)Reb(12π∫02π|p(z)|2dθ)1/2×(12π∫02π|(s1(z))|2α(k/4+1/2)Rebdθ)1/2.
By Lemma 1.3 and distortion results for the class 𝒮* with a subordination result, we obtain
(2.38)Lrf(z)≤cM1-α(r)2α(k/2-1)RebrαReb(1(1-r)4α(k/4+1/2)Reb-1)1/2(1+(4(1-δ)2-1)r21-r2)1/2=C(α,b,δ,k)M1-α(r)(11-r)α(k/2+1)Reb.
Similarly for α>1,we have
(2.39)Lrf(z)≤C(α,b,δ,k)mα-1(r)(11-r)α(k/2+1)Reb.Theorem 2.8.
Letf(z)∈ℋks(α,b,δ). Then for α(k/2+1)Reb>1(2.40)|an|≤{C1(α,b,δ,k)M1-α(n)(n)α(k/2+1)Reb-1,0<α≤1,C1(α,b,δ,k)mα-1(n)(n)α(k/2+1)Reb-1,α>1,
where m andMare the same as in Theorem 2.7 and C1(α,b,δ,k) is a constant depending upon α,b,δ, and konly.Proof.
Since, withz=reiθ Cauchy theorem gives
(2.41)nan=12πrn∫02πzf′(z)e-inθdθ,
therefore
(2.42)n|an|≤12πrnLrf(z).
Now using Theorem 2.7 for 0<α≤1,we have
(2.43)n|an|≤12πrnC(α,b,δ,k)M1-α(r)(11-r)α(k/2+1)Reb.
Putting r=1-1/n,we have
(2.44)|an|≤C1(α,b,δ,k)M1-α(r)(n)α(k/2+1)Reb-1.
Similarly we obtain the required result for α>1.Theorem 2.9.
Letf(z)∈ℋks(α,b,δ).  Then for α(k/2+1)Reb>3,
(2.45)||an+1|-|an||≤{C2(α,b,δ,k)M1-α(r)(n)α(k/2+1)Reb-2,0<α≤1,C2(α,b,δ,k)mα-1(r)(n)α(k/2+1)Reb-2,α>1.Proof.
We know that forξ∈𝒰 and n≥1,
(2.46)|(n+1)ξan+1-nan|≤∫02π|z-ξ||zf′(z)|dθ,z=reiθ,0<r<1,0≤θ≤2π.
Since f(z)∈ℋks(α,b,δ), therefore
(2.47)zf′(z)f(z)(2f(z)g(z)-g(-z))α=p(z),p(z)∈P(δ).
By Theorem 2.1, we have, for g(z)∈ℛks(b), the odd function ϕ(z)=(1/2)[g(z)-g(-z)]∈ℛk(b). This implies that
(2.48)zf′(z)=(f(z))1-α(ϕ(z))αp(z).
Thus, for ξ∈𝒰 and n≥1, we have
(2.49)|(n+1)ξan+1-nan|≤M1-α(r)∫02π|z-ξ||ϕ(z)|α|p(z)|dθ.
Since ϕ(z)∈ℛk(b), therefore we have for odd functions s1(z),  s1(z)∈𝒮*,
(2.50)|(n+1)ξan+1-nan|≤2α(k/2-1)RebeIm⁡b(π/2)M1-α(r)2πrn+1-α(k/2-1)Reb∫02π|z-ξ||(s1(z))|α(k/4+1/2)Reb|p(z)|dθ.
By using Lemma 1.4, we have
(2.51)≤2α(k/2-1)RebeIm⁡b(π/2)M1-α(r)2πrn-1-α(k/2-1)Reb(1-r)∫02π|(s1(z))|α(k/4+1/2)Reb-1|p(z)|dθ.
Now using Cauchy Schwarz inequality, we have
(2.52)|(n+1)ξan+1-nan|≤2α(k/2-1)RebeIm⁡b(π/2)M1-α(r)2πrn-1-α(k/2-1)Reb(1-r)(12π∫02π|(s1(z))|2α(k/4+1/2)Reb-2dθ)1/2×(12π∫02π|p(z)|2dθ)1/2.
By Lemma 1.3 and distortion result for the class S* with a subordination result, we obtain
(2.53)|(n+1)ξan+1-nan|≤cM1-α(r)2α(k/4)-RebrαReb-n+1(1(1-r))α(k/2+1)Reb-5/2+1×(1+(4(1-δ)2-1)r21-r2)1/2.
Now putting |ξ|=r=n/(n+1), we obtain
(2.54)||an+1|-|an||≤C2(α,b,δ,k)M1-α(r)(n)α(k/2+1)Reb-2.
Similarly for α>1,we have
(2.55)||an+1|-|an||≤C2(α,b,δ,k)mα-1(r)(n)α(k/2+1)Reb-2.



---

*Source: 790689-2012-03-15.xml*