In our present investigation, motivated from Noor work, we define the class ℛks(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.
1. Introduction
Let 𝒜 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: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 ifRe{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: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∫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 class ℛks(b) of functions of bounded radius rotation of complex order b with respect to symmetrical points as follows.
Definition 1.1.
Let f(z)∈𝒜 in 𝒰. Then f(z)∈ℛks(b), if and only if
1+1b{2zf′(z)f(z)-f(-z)-1}∈Pk(z∈U),
where k≥2 and b∈ℂ-{0}.
Using the class ℛks(b), we define the class ℋks(α,b,δ) as follows.
Definition 1.2.
Let f(z)∈𝒜 in 𝒰. Then f(z)∈ℋks(α,b,δ), if and only if there exists g(z)∈ℛks(b) such that
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 to α, 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, that k≥2, α>0, 0≤δ<1, and b∈ℂ-{0}.
Lemma 1.3.
Let p(z) be analytic in 𝒰 where p(0)=1 belongs to P(δ).Then
12π∫02π|p(z)|2dθ≤1+(4(1-δ)2-1)r21-r2
(see [8, 12]).
Lemma 1.4.
Let s1(z) be univalent function in 𝒰. Then there exists ξ with |ξ|=r such that for all z, |z|=r,
|z-ξ||s1(z)|≤2r21-r2
(see [13]).
2. Some Properties of the Classes ℛks(b)andℋks(α,b,δ)Theorem 2.1.
Let f(z)∈ℛks(b). Then the odd function
ϕ(z)=12[f(z)-f(-z)]
belongs to ℛk(b) in 𝒰.
Proof.
Let f(z)∈ℛks(b) and consider
ϕ(z)=12[f(z)-f(-z)].
From logarithmic differentiation of the previous relation, we have
ϕ′(z)ϕ(z)=f′(z)-f′(-z)f(z)-f(-z),
or, equivalently,
zϕ′(z)ϕ(z)=12[p1(z)+p2(z)]
with
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
zϕ′(z)ϕ(z)∈Pk(b)(z∈U),
and hence ϕ(z)∈ℛk(b).
Theorem 2.2.
Let f(z)∈ℛks(b). Then
f′(z)=12[b(p(z)-1)+1]exp{b2∫0z1ξ(p(ξ)-p(-ξ)-2b)dξ}.
Proof.
Let f(z)∈ℛks(b). Then by definition we have
1+1b[2zf′(z)f(z)-f(-z)-1]=p(z),p(z)∈Pk.
Simple computation yields us
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 put b=1 and k=2 in Theorem 2.1, we obtain the integral representation for 𝒮s* given by Stankiewiez in [14].
Theorem 2.3.
Let f(z)∈ℛks(b). Then
|a2|≤k|b|2.
The function f0(z)∈ℛks(b) defined by
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 .
Since f(z)∈ℛks(b), there exists an odd function ϕ(z)∈ℛk(b) with
ϕ(z)=12[f(z)-f(-z)],
such that
zf′(z)=ϕ(z)p(z),
with p(z)∈𝒫k(b). Let
ϕ(z)=z+∑n=2∞b2n-1z2n-1,p(z)=1+∑n=1∞cnzn.
Then (2.13) implies that
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
|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 function f(z)∈ℛks(b) contains the disc
|w|<24+k|b|.
Proof .
The Koebe one-quarter theorem states that each omitted value w of the univalent function f(z) of the form (1.1) satisfies
|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.
Let f(z)∈ℛks(b). Then, for z=reiθ and 0≤θ1<θ2≤2π,
∫θ1θ2Re(zf′(z))′f′(z)dθ>-(k-1)|b|π.
Theorem 2.6.
Let f(z)∈ℋks(α,b,0). Then, for z=reiθ,
∫θ1θ2ReJ(α,f(z))dθ>-(α|b|(k-1)+1)π,
where 0≤θ1<θ2≤2π and
J(α,f(z))=(1+zf′′(z)f′(z))+(α-1)zf′(z)f(z).
Proof.
We can define, for z=reiθ, r<1, θ real, the following:
S(r,θ)=arg[zf′(z)fα-1(z)],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
|S(r,θ)-V(r,θ)|≤π2.
Now we have from (2.22)
V(r,θ2)-V(r,θ1)=α∫θ1θ2Rezϕ′(z)ϕ(z)dθ,
where ϕ(z) is an odd function of the following form:
ϕ(z)=12[g(z)-g(-z)].
Since g(z)∈ℛks(b), therefore by using Theorem 2.5, we have
∫θ1θ2Rezϕ′(z)ϕ(z)dθ>-(k-1)|b|π.
From (2.22), (2.23), (2.24), and (2.27), we have
|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)
ddθS(r,θ)=Re[(1+zf′′(z)f′(z))+(α-1)zf′(z)f(z)].
Therefore
∫θ1θ2ReJ(α,f(z))dθ>-(α|b|(k-1)+1)π.
Theorem 2.7.
Let f(z)∈ℋks(α,b,δ).Then for α(k/2+1)Reb>1,
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
Lrf(z)=∫02π|zf′(z)|dθ,z=reiθ,0<r<1.
Since f(z)∈ℋks(α,b,δ), therefore
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
zf′(z)=(f(z))1-α(ϕ(z))αp(z).
Therefore, we have
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*,
≤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)Imb.≤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
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
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
Lrf(z)≤C(α,b,δ,k)mα-1(r)(11-r)α(k/2+1)Reb.
Theorem 2.8.
Let f(z)∈ℋks(α,b,δ). Then for α(k/2+1)Reb>1|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, with z=reiθ Cauchy theorem gives
nan=12πrn∫02πzf′(z)e-inθdθ,
therefore
n|an|≤12πrnLrf(z).
Now using Theorem 2.7 for 0<α≤1,we have
n|an|≤12πrnC(α,b,δ,k)M1-α(r)(11-r)α(k/2+1)Reb.
Putting r=1-1/n,we have
|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.
Let f(z)∈ℋks(α,b,δ). Then for α(k/2+1)Reb>3,
||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 for ξ∈𝒰 and n≥1,
|(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
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
zf′(z)=(f(z))1-α(ϕ(z))αp(z).
Thus, for ξ∈𝒰 and n≥1, we have
|(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*,
|(n+1)ξan+1-nan|≤2α(k/2-1)RebeImb(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
≤2α(k/2-1)RebeImb(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
|(n+1)ξan+1-nan|≤2α(k/2-1)RebeImb(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
|(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
||an+1|-|an||≤C2(α,b,δ,k)M1-α(r)(n)α(k/2+1)Reb-2.
Similarly for α>1,we have
||an+1|-|an||≤C2(α,b,δ,k)mα-1(r)(n)α(k/2+1)Reb-2.
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