On Booth lemniscate of starlike functions

0.36 MB·

by  R. Kargar

ON BOOTH LEMNISCATE OF STARLIKE FUNCTIONS
RAHIMKARGAR,ALIEBADIANANDJANUSZSOKO´L(cid:32)
Abstract. Assumethat∆istheopenunitdiskinthecomplexplaneandA
istheclassofnormalizedanalyticfunctionsin∆. Inthispaperweintroduce
andstudytheclass
(cid:26) (cid:18)zf(cid:48)(z) (cid:19) z (cid:27)
BS(α):= f ∈A: −1 ≺ ,z∈∆ ,
7 f(z) 1−αz2
1
where0≤α≤1and≺isthesubordinationrelation. Somepropertiesofthis
0
class like differential subordination, coefficients estimates and Fekete-Szeg¨o
2
inequalityassociatedwiththek-throottransformareconsidered.

n
a
J

6 1. Introduction
2
Let A denote the class of functions f(z) of the form:
]
V ∞
(cid:88)
C (1.1) f(z)=z+ anzn,
. n=2
h
which are analytic and normalized in the open unit disk ∆ = {z ∈ C : |z| < 1}.
t
a The subclass of A consisting of all univalent functions f(z) in ∆ is denoted by
m
S. A function f ∈ S is called starlike (with respect to 0), denoted by f ∈ S∗, if
[ tw ∈f(∆)wheneverw ∈f(∆)andt∈[0,1]. Robertsonintroducedin[8],theclass

S∗(γ) of starlike functions of order γ ≤1, which is defined by
1
v (cid:26) (cid:26)zf(cid:48)(z)(cid:27) (cid:27)
9 S∗(γ):= f ∈A: Re >γ, z ∈∆ .
f(z)
7
7 Ifγ ∈[0,1),thenafunctioninS∗(γ)isunivalent. InparticularweputS∗(0)≡S∗.
7
We denote by B the class of analytic functions w(z) in ∆ with w(0) = 0 and
0
|w(z)| < 1, (z ∈ ∆). If f and g are two of the functions in A, we say that
.
1
f is subordinate to g, written f(z) ≺ g(z), if there exists a w ∈ B such that
0
f(z)=g(w(z)), for all z ∈∆.
7
1 Furthermore, if the function g is univalent in ∆, then we have the following
: equivalence:
v
i f(z)≺g(z)⇔(f(0)=g(0) and f(∆)⊂g(∆)).
X
r We now recall from [7], a one-parameter family of functions as follows:
a
∞
z (cid:88)
(1.2) F (z):= =z+ αnz2n+1 (z ∈∆, 0≤α≤1).
α 1−αz2
n=1
ThefunctionF (z)isstarlikeunivalentfor0≤α<1. WehavealsoF (∆)=D(α),
α α
where
D(α)=(cid:26)x+iy ∈C: (cid:0)x2+y2(cid:1)2− x2 − y2 <0(cid:27),
(1−α)2 (1+α)2
2000 Mathematics Subject Classification. 30C45.
Key words and phrases. Boothlemniscate,Subordination,Starlike,StronglyStarlike,Fekete-
Szeg¨oInequality.
1
2 R.KARGAR,A.EBADIANANDJ.SOKO´L(cid:32)
when 0≤α<1 and
D(1)={x+iy ∈C: (∀t∈(−∞,−i/2]∪[i/2,∞))[x+iy (cid:54)=it]}.
Figure 1. The boundary curve of D(1/2)
The Persian curve (cf. [10]) is a plane algebraic curve of order four that is the
line of intersection between the surface of a torus and a plane parallel to its axis.
The equation in rectangular coordinates is
(cid:0)x2+y2+p2+d2−r2(cid:1)2 =4d2(cid:0)x2+p2(cid:1),
where r is the radius of the circle describing the torus, d is the distance from the
origin to its center and p is the distance from the axis of the torus to the plane.
We remark that a curve described by
(cid:0)x2+y2(cid:1)2−(cid:0)n4+2m2(cid:1)x2−(cid:0)n4−2m2(cid:1)y2 =0 (x,y)(cid:54)=(0,0),
is a special case of Persian curve that studied by Booth and is called the Booth
lemniscate [2]. The Booth lemniscate is called elliptic if n4 > 2m2 while, for
n4 <2m2, it is termed hyperbolic. Thus it is clear that the curve
(cid:0)x2+y2(cid:1)2− x2 − y2 =0 (x,y)(cid:54)=(0,0),
(1−α)2 (1+α)2
is the Booth lemniscate of elliptic type (see figure 1). Two other special case of
Persian curve are Cassini oval and Bernoulli lemniscate.
A plane algebraic curve of order four whose equation in Cartesian coordinates
has the form:
(cid:0)x2+y2(cid:1)2−2c2(cid:0)x2−y2(cid:1)=a4−c4.
The Cassini oval is the set of points such that the product of the distances from
each point to two given points p = (−c,0) and p = (c,0) (the foci) is constant.
√ 2 1 √
When a ≥ c 2 the Cassini oval is a convex curve; when c < a < c 2 it is a
curve with ”waists” (concave parts); when a = c it is a Bernoulli lemniscate; and
when a<c it consists of two components. Cassini ovals are related to lemniscates.
CassiniovalswerestudiedbyG.Cassini(17thcentury)inhisattemptstodetermine
the Earth’s orbit.
The Bernoulli lemniscate plane algebraic curve of order four, the equation of
which in orthogonal Cartesian coordinates is
(cid:0)x2+y2(cid:1)2−2a2(cid:0)x2−y2(cid:1)=0,
and in polar coordinates
ρ2 =2a2cos2φ.
The Bernoulli lemniscate is symmetric about the coordinate origin, which is a
nodewithtangentsy =±xandthepointofinflection. Theproductofthedistances
of any point M to the two given points p =(−a,0) and p =(a,0) is equal to the
1 2
ON BOOTH LEMNISCATE OF STARLIKE FUNCTIONS 3
square of the distance between the points p and p . The Bernoulli lemniscate is
1 2
a special case of the Cassini ovals, the lemniscates, and the sinusoidal spirals. The
Bernoulli spiral was named after Jakob Bernoulli, who gave its equation in 1694.
In [4], the authors introduced and studied the class M(δ) as follows:
Definition A. Let π/2 ≤ δ < π. Then the function f ∈ A belongs to the class
M(δ) if f satisfies:
δ−π (cid:26)zf(cid:48)(z)(cid:27) δ
(1.3) 1+ <Re <1+ (z ∈∆).
2sinδ f(z) 2sinδ
By definition of subordination and by (1.3), we have that f ∈M(δ) if and only if
(cid:18)zf(cid:48)(z) (cid:19) 1 (cid:18) 1−z (cid:19)
−1 ≺B (z):= log (z ∈∆),
f(z) α 2isinδ 1−ze−iδ
where π/2≤δ <π. The above function B (z) is convex univalent in ∆ and maps
α
∆ onto Γ ={w :(δ−π)/(2sinδ)<Re{w}<δ/(2sinδ)}, or onto the convex hull
δ
of three points (one of which may be that point at infinity) on the boundary of Γ .
δ
In other words, the image of ∆ may be a vertical strip when π/2≤δ <π, while in
other cases, a half strip, a trapezium, or a triangle.
It was proved in [6], that for α < 1 < β, the following function P : ∆ → C,
α,β
defined by
β−α (cid:32)1−e2πiβ1−−ααz(cid:33)
(1.4) P (z)=1+ ilog (z ∈∆),
α,β π 1−z
maps ∆ onto a convex domain
(1.5) P (∆)={w ∈C:α<Re{w}<β},
α,β
conformally. Therefore, the function P (z) defined by (1.4) is convex univalent
α,β
in ∆ and has the form:
∞
(cid:88)
P (z)=1+ B zn,
α,β n
n=1
where
β−α (cid:16) (cid:17)
(1.6) Bn = nπ i 1−e2nπiβ1−−αα (n=1,2,...).
The present authors (see [3]) introduced the class V(α,β) as follows:
Definition B. Let α<1 and β >1. Then the function f ∈A belongs to the class
V(α,β) if f satisfies:
(cid:40)(cid:18) z (cid:19)2 (cid:41)
(1.7) α<Re f(cid:48)(z) <β (z ∈∆).
f(z)
Therefore, by definition of subordination, we have that f ∈V(α,β) if and only if
(cid:18) z (cid:19)2
f(cid:48)(z)≺P (z) (z ∈∆).
f(z) α,β
MotivatedbyDefinitionA,DefinitionBandusingF ,weintroduceanewclass.
α
Our principal definition is the following.
Definition 1.1. Let f ∈A and 0≤α<1. Then f ∈BS(α) if and only if
(cid:18)zf(cid:48)(z) (cid:19)
(1.8) −1 ≺F (z) (z ∈∆),
f(z) α
where F defined by (1.2).
α
In our investigation, we require the following result.
4 R.KARGAR,A.EBADIANANDJ.SOKO´L(cid:32)
Corollary 1.1. We have that f ∈BS(α) if and only if
(cid:90) z F (w(t))−1
(1.9) f(z)=zexp α dt (z ∈∆),
t
0
for some function w(z), analytic in ∆, with |w(z)|≤|z| in ∆.
Proof. From (1.8) it follows that there exists a function w(z), analytic in ∆, with
|w(z)|≤|z| in ∆, such that
(cid:18)f(cid:48)(z) 1(cid:19)
z − =F (w(z)) (z ∈∆),
f(z) z α
or
(cid:18) f(z)(cid:19)(cid:48)
z log =F (w(z)) (z ∈∆).
z α
This gives (1.9). On the other hand, it is a easy calculation that a function having
the form (1.9) satisfies condition (1.8). (cid:3)
Applying formula (1.9) for w(z)=z gives that
√ √
(cid:18) (cid:19)1/ α
1+ αz
(1.10) f (z)=z √ (z ∈∆),
0 1− αz
is in the class BS(α).
Lemma 1.1. Let F (z) be given by (1.2). Then
α
1 1
(1.11) <Re{F (z)}< (0≤α<1).
α−1 α 1−α
Proof. If α = 0, then we have −1 < Re{F } = Re(z) < 1. For 0 < α < 1, the
α
function {F } does not have any poles in ∆ and is analytic in ∆, thus looking
α
for the min{Re{F (z)} : |z| < 1} it is sufficient to consider it on the boundary
α
∂F (∆)={F (eiϕ):ϕ∈[0,2π]}. A simple calculation give us
α α
Re(cid:8)F (eiϕ)(cid:9)= (1−α)cosϕ (ϕ∈[0,2ϕ]).
α 1+α2−2αcos2ϕ
So we can see that Re{F (z)} is well defined also for ϕ=0 and ϕ=2π. Define
α
(1−α)x
g(x)= (−1≤x≤1),
1+α2−2α(2x2−1)
then for 0<α<1, we have g(cid:48)(x)>0. Thus for −1≤x≤1, we have
1 1
=g(−1)≤g(x)≤g(1)= .
α−1 1−α
This completes the proof. (cid:3)
We note that from Lemma 1.1 and by definition of subordination, the function
f ∈A belongs to the class BS(α), 0≤α<1, if it satisfies the condition
1 (cid:18)zf(cid:48)(z) (cid:19) 1
<Re −1 < (z ∈∆),
α−1 f(z) 1−α
or equivalently
α (cid:18)zf(cid:48)(z)(cid:19) 2−α
(1.12) <Re < (z ∈∆).
α−1 f(z) 1−α
It is clear that BS(0) ≡ S(0,2) ⊂ S∗, where the class S(α,β), α < 1 and β > 1,
was recently considered by K. Kuroki and S. Owa in [6].
ON BOOTH LEMNISCATE OF STARLIKE FUNCTIONS 5
Corollary 1.2. If f ∈BS(α), then
zf(cid:48)(z)
(1.13) ≺P (z) (z ∈∆),
f(z) α
where
(cid:32) (cid:33)
2 1−eπi(1−α)2z
(1.14) P (z)=1+ ilog (z ∈∆),
α π(1−α) 1−z
is convex univalent in ∆.
Proof. If f ∈ BS(α), then it satisfies (1.12) or zf(cid:48)(z)/f(z) lies in a strip of the
form (1.5). Then applying the definition of subordination and the function (1.4),
we obtain (1.13) and (1.14). (cid:3)
For the proof of our main results, we need the following Lemma.
Lemma 1.2. (See [9]) Let q(z)=(cid:80)∞ C zn be analytic and univalent in ∆, and
n=1 n
suppose that q(z) maps ∆ onto a convex domain. If p(z)=(cid:80)∞ A zn is analytic
n=1 n
in ∆ and satisfies the following subordination
p(z)≺q(z) (z ∈∆),
then
|A |≤|C | n≥1.
n 1
2. Main Results
The first main result is the following theorem.
Theorem 2.1. Let f ∈A and 0≤α<1. If f ∈BS(α) then
f(z) (cid:90) z P (t)−1
(2.1) log ≺ α dt (z ∈∆),
z t
0
where
(cid:32) (cid:33)
2 1−eπi(1−α)2z
P (z)−1= ilog (z ∈∆)
α π(1−α) 1−z
and
(cid:90) z P (t)−1
P(cid:101)α(z)= α t dt (z ∈∆),
0
are convex univalent in ∆.
Proof. If f ∈BS(α), then by (1.13) it satisfies
(cid:26) f(z)(cid:27)(cid:48)
(2.2) z log ≺P (z)−1.
z α
It is known that if F(z) is convex univalent in ∆, then
(cid:20)(cid:90) z f(t) (cid:90) z F(t) (cid:21)
(2.3) [f(z)≺F(z)]⇒ dt≺ dt
t t
0 0
and
(cid:90) z F(t)
F(cid:101)(z)= dt,
t
0
is convex univalent in ∆. By Corollary 1.2, we know that P (z) − 1 is convex
α
univalent in ∆. Therefore, applying (2.3) in (2.2) gives (2.1) with convex univalent
P(cid:101)α(z). (cid:3)
6 R.KARGAR,A.EBADIANANDJ.SOKO´L(cid:32)
Corollary 2.1. If f ∈BS(α) and |z|=r <1, then
(2.4) |mz|=inr(cid:12)(cid:12)(cid:12)expP(cid:101)α(z)(cid:12)(cid:12)(cid:12)≤(cid:12)(cid:12)(cid:12)(cid:12)f(zz)(cid:12)(cid:12)(cid:12)(cid:12)≤|mz|a=xr(cid:12)(cid:12)(cid:12)expP(cid:101)α(z)(cid:12)(cid:12)(cid:12).
Proof. Subordination implies
f(z)
(2.5) z ≺expP(cid:101)α(z)
and expP(cid:101)α(z) is convex univalent. Then (2.5) implies (2.4). (cid:3)
We now obtain coefficients estimates for functions belonging to the class BS(α).
Theorem 2.2. Assume that the function f of the form (1.1) belongs to the class
√
BS(α) where 0≤α≤3−2 2. then |a |≤1 and
2
n−1(cid:18) (cid:19)
1 (cid:89) k
(2.6) |a |≤ (n=3,4,...).
n n−1 k−1
k=2
Proof. Assume that f ∈BS(α). Then from Definition 1.1 we have
(2.7) p(z)≺1+F (z)=1+z+αz3+··· (z ∈∆),
α
where
(2.8) zf(cid:48)(z)=p(z)f(z).
√
We note that F is convex function for 0≤α≤3−2 2 (see [7, Corollary 3.3]). If
α
we define p(z)=1+(cid:80)∞ p zn, then from Lemma 1.2, we have
n=1 n
(2.9) |p |≤1.
n
Equating the coefficients of zn on both sides of (2.8), we find the following relation
between the coefficients:
(2.10) na =p +a p +···+a p +a .
n n−1 2 n−2 n−1 1 n
Making use of (2.9) and (2.10), we get
n−1
1 (cid:88)
(2.11) |a |≤ |a | |a |=1.
n n−1 k 1
k=1
Obvious that, from (2.11) we have |a |≤1. We now need show that
2
n−1 n−1(cid:18) (cid:19)
1 (cid:88) 1 (cid:89) 1
(2.12) |a |≤ 1+ (n=3,4,...).
n−1 k n−1 k−1
k=1 k=2
Weuseinductiontoprove(2.12). Ifwetaken=3intheinequality(2.12),wehave
|a |≤1, therefore the case n=3 is clear. A simple calculation gives us
2
m (cid:32)m−1 (cid:33)
1 (cid:88) 1 (cid:88)
|a |≤ |a |= |a |+|a |
m+1 m k m k m
k=1 k=1
m−1(cid:18) (cid:19) m−1(cid:18) (cid:19)
1 (cid:89) 1 1 1 (cid:89) 1
≤ 1+ + × 1+
m k−1 m m−1 k−1
k=2 k=2
m (cid:18) (cid:19)
1 (cid:89) 1
= 1+ ,
m k−1
k=2
which implies that the inequality (2.12) holds for n=m+1. From now (2.11) and
(2.12), the desired estimate for |a |(n=3,4,...) follows, as asserted in (2.6). This
n
completes the proof. (cid:3)
ON BOOTH LEMNISCATE OF STARLIKE FUNCTIONS 7
Theproblemoffindingsharpupperboundsforthecoefficientfunctional|a −µa2|
3 2
for different subclasses of the normalized analytic function class A is known as the
Fekete-Szeg¨o problem.
Recently, Ali et al. [1] considered the Fekete-Szeg¨o functional associated with
the kth root transform for several subclasses of univalent functions. We recall here
that, for a univalent function f(z) of the form (1.1), the kth root transform is
defined by
∞
(cid:88)
(2.13) F(z)=[f(zk)]1/k =z+ b zkn+1 (z ∈∆).
kn+1
n=1
Following, we consider the problem of finding sharp upper bounds for the Fekete-
Szeg¨o coefficient functional associated with the kth root transform for functions in
the class BS(α).
In order to prove next result, we need the following lemma due to Keogh and
Merkes [5]. Further we denote by P the well-known class of analytic functions p(z)
with p(0)=1 and Re(p(z))>0, z ∈∆.
Lemma 2.1. Let the function g(z) given by
g(z)=1+c z+c z2+··· ,
1 2
be in the class P. Then, for any complex number µ
|c −µc2|≤2max{1,|2µ−1|}.
2 1
The result is sharp.
Theorem 2.3. Let 0 ≤ α < 1, f ∈ BS(α) and F is the kth root transform of f
defined by (2.13). Then, for any complex number µ,
(cid:26) (cid:12) (cid:12)(cid:27)
(2.14) (cid:12)(cid:12)b2k+1−µb2k+1(cid:12)(cid:12)≤ 21k max 1,(cid:12)(cid:12)(cid:12)2(µk−1) +1(cid:12)(cid:12)(cid:12) .
The result is sharp.
Proof. Using (1.2), we first put
∞
(cid:88)
(2.15) 1+F (z)=1+ B zn,
α n
n=1
where B =1, B =0, B =α, and etc. Since f ∈BS(α), from Definition 1.1 and
1 2 3
definition of subordination, there exists w ∈B such that
(2.16) zf(cid:48)(z)/f(z)=1+F (w(z)).
α
We now define
1+w(z)
(2.17) p(z)= =1+p z+p z2+··· .
1−w(z) 1 2
Since w ∈B, it follows that p∈P. From (2.15) and (2.17) we have:
(cid:18) (cid:18) (cid:19)(cid:19)
1 1 1 1
(2.18) 1+F (w(z))=1+ B p z+ B p2+ B p − p2 z2+··· ,
α 2 1 1 4 2 1 2 1 2 2 1
where B = 1 and B = 0. Equating the coefficients of z and z2 on both sides of
1 2
(2.16) and substituting B =1 and B =0, we get
1 2
1
(2.19) a = p ,
2 2 1
and
(cid:18) (cid:19)
1 1 1
(2.20) a = p2+ p − p2 .
3 8 1 4 2 2 1
8 R.KARGAR,A.EBADIANANDJ.SOKO´L(cid:32)
A computation shows that, for f given by (1.1),
(cid:18) (cid:19)
1 1 1k−1
(2.21) F(z)=[f(z1/k)]1/k =z+ a zk+1+ a − a2 z2k+1+··· .
k 2 k 3 2 k2 2
From equations (2.13) and (2.21), we have
1 1 1k−1
(2.22) b = a and b = a − a2.
k+1 k 2 2k+1 k 3 2 k2 2
Substituting from (2.19) and (2.20) into (2.22), we obtain
1
b = p ,
k+1 2k 1
and
(cid:18) (cid:19)
1 k−1
b = p − p2 ,
2k+1 4k 2 k 1
so that
(cid:20) (cid:18) (cid:19) (cid:21)
1 1 2(µ−1)
(2.23) b −µb2 = p − +2 p2 .
2k+1 k+1 4k 2 2 k 1
Letting
(cid:18) (cid:19)
1 2(µ−1)
µ(cid:48) = +2 ,
2 k
theinequality(2.14)nowfollowsasanapplicationofLemma2.1. Itiseasytocheck
that the result is sharp for the kth root transforms of the function
(cid:18)(cid:90) z F (w(t)) (cid:19)
(2.24) f(z)=zexp α dt .
t
0
(cid:3)
Putting k =1 in Theorem 2.3, we have:
Corollary 2.2. (Fekete-Szeg¨o inequality) Suppose that f ∈BS(α) and 0≤α<1.
Then, for any complex number µ,
(2.25) (cid:12)(cid:12)a3−µa22(cid:12)(cid:12)≤ 21max{1,|2µ−1|}.
The result is sharp.
It is well known that every function f ∈ S has an inverse f−1, defined by
f−1(f(z))=z, z ∈∆ and
f(f−1(w))=w (|w|<r ; r <1/4),
0 0
where
(2.26) f−1(w)=w−a w2+(2a2−a )w3−(5a3−5a a +a )w4+··· .
2 2 3 2 2 3 4
Corollary 2.3. Let the function f, given by (1.1), be in the class BS(α) where
0≤α<1. Also let the function f−1(w)=w+(cid:80)∞ b wn be inverse of f. Then
n=2 n
(2.27) |b |≤1,
2
and
3
(2.28) |b |≤ .
3 2
ON BOOTH LEMNISCATE OF STARLIKE FUNCTIONS 9
Proof. Relation (2.26) give us
b =−a and b =2a2−a .
2 2 3 2 3
Thus, we can get the estimate for |b | by
2
|b |=|a |≤1.
2 2
For estimate of |b |, it suffices in Corollary 2.2, we put µ = 2. Hence the proof of
3
Corollary 2.3 is completed. (cid:3)
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Department of Mathematics, Payame Noor University, Tehran, Iran
E-mail address: [email protected] (RahimKargar)
Department of Mathematics, Payame Noor University, Tehran, Iran
E-mail address: [email protected] (AliEbadian)
University of Rzeszo´w, Faculty of Mathematics and Natural Sciences, ul. Prof.
Pigonia 1, 35-310 Rzeszo´w, Poland
E-mail address: [email protected] (JanuszSoko´(cid:32)l)