Mathematical and Computer Modelling 57 (2013) 1203–1211
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On a class of starlike functions related to a shell-like curve connected with Fibonacci numbers Jacek Dziok a , Ravinder Krishna Raina b,1 , Janusz Sokół c,∗ a
Department of Mathematics, Institute of Mathematics, University of Rzeszów, ul. Rejtana 16A, 35-310 Rzeszów, Poland
b
M.P. University of Agri. and Technology, Udaipur, Rajasthan, India
c
Department of Mathematics, Rzeszów University of Technology, Al. Powstańców Warszawy 12, 35-959 Rzeszów, Poland
article
info
Article history: Received 14 November 2011 Received in revised form 14 October 2012 Accepted 15 October 2012 Keywords: Univalent functions Convex functions Starlike functions Subordination Fibonacci numbers Conchoid of de Sluze Trisectrix of Maclaurin
abstract In this paper we investigate an interesting subclass SL of analytic univalent functions in the open unit disc on the complex plane. This class was introduced by Sokół [J. Sokół, On starlike functions connected with Fibonacci numbers, Folia Scient. Univ. Tech. Resoviensis 175 (1999) 111–116]. The class SL is strongly related to the class KSL considered earlier by the authors of the present work in their paper [J. Dziok, R. K. Raina, J. Sokół, Certain results for a class of convex functions related to shell-like curve connected with Fibonacci Numbers, Comput. Math. Appl. 61 (2011) 2606–2613]. Apart from furnishing some genuine remarks, we present certain new results for the class SL of functions, and also mention some relevant cases for this function class. © 2012 Elsevier Ltd. All rights reserved.
1. Introduction Let A be the class of all holomorphic functions f in the open unit disc ∆ with the normalization f (0) = 0, f ′ (0) = 1, and let S denote the subset of A which is composed of univalent functions. We say that f is subordinate to F in ∆, written as f ≺ F , if and only if f (z ) = F (ω(z )) for some holomorphic function ω such that ω(0) = 0 and |ω(z )| < 1, for all z ∈ ∆. The class of starlike functions S ∗ can be defined in various ways, and for example, we say that f ∈ A is starlike if it satisfies the condition that zf ′ (z ) f (z )
≺ p(z ) (z ∈ ∆),
(1.1)
where p(z ) = (1 + z )/(1 − z ). Several subclasses of S ∗ have been defined in the literature by choosing appropriately the arbitrary function p(z ) in (1.1). We consider it worthwhile here to mention some useful geometric transformations that arise when the function p(z ) is chosen suitably. Thus, it is easily observed that when 1+(1−2α)z
, α < 1, then under this transformation, the image of the unit circle |z | = 1 is a straight line ℜ(w) = (i) p(z ) = 1−z α , while the image of the unit disc ∆ is the half plane ℜ(w) > α. In this case, a function f ∈ A satisfying (1.1) is called starlike of order α and the family of all such functions is denoted by S ∗ (α).
∗
Corresponding author. E-mail addresses:
[email protected] (J. Dziok),
[email protected] (R.K. Raina),
[email protected],
[email protected] (J. Sokół).
1 Present address: 10/11 Ganpati Vihar, Opposite Sector 5, Udaipur 313002, India. 0895-7177/$ – see front matter © 2012 Elsevier Ltd. All rights reserved. doi:10.1016/j.mcm.2012.10.023
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Az (ii) p(z ) = 11+ , −1 < B < A ≤ 1, then p(∆) is the disc D(C (A, B), R(A, B)) with the centre C (A, B) = (1 + AB)/(1 − B2 ), +Bz and the radius R(A, B) = (A + B)/(1 − B2 ); see [1,2].
β
+z (iii) p(z ) = 11− , 0 < β ≤ 1, then p(∆) is an angle {w ∈ C : Argw < βπ /2}; see [3]. In this case, a function f ∈ A z satisfying (1.1), is called strongly starlike of order β .
(iv) p(z ) = 1 + π22
√
1+ z
log 1−√z
w = u + iv ∈ C : u >
2
, then after some elementary calculations, one can find that p(∆) is a parabolic domain
z (u − 1)2 + v 2 . In this case, if a function f ∈ A satisfies (1.1), then the function 0
called a uniformly convex function; see [4–6]. √
1 2 (v) p(z ) = 1−β (arccos β)i log 1−√z 2 cos π 1+ z
C:u>β
−
β2
1−β 2
f (t ) t
dt is
, 0 < β < 1, then p(∆) is an interior of hyperbola w = u + iv ∈
(u − 1)2 + v 2 ; see [7–9]. √ (vi) p(z ) = 1 + k22−1 sin2 2Kπ(t ) F ( z /t , t ) , k > 1, F (1, t ) = K (t ), where w dx , F (w, t ) = √ √ 2 1 − x 1 − t 2 x2 0 is called the Jacobi elliptic integral, and t ∈ (0, 1) is such that
k = cosh
π K ′ (t ) , 2K (t )
(u − 1)2 + v 2 ; see [7–9]. √ √ (vii) p(z ) = 1 + z, where the branch of the square root is chosen in order that 1 = 1, then p(∆) is an interior of the right loop of the Lemniscate of Bernoulli {w ∈ C : ℜ (w) > 0, |w 2 − 1| < 1}; see [10,11]. 1/α +z , a ≥ 1, b ≥ 1/2, where the branch of the square root is chosen in order that p(0) = 1, then (viii) p(z ) = 1+(11− b)/bz p(∆) is a leaf-like domain {w ∈ C : |w α − b| < b, Argw ≤ π /(2α)}; see [12]. then p(∆) is an elliptic domain w = u + iv ∈ C : u > β
In cases (i)–(viii), the function p is a convex univalent function. In [13], Ma and Minda proved some general results for functions f ∈ A satisfying (1.1), where p is assumed to be univalent, p(∆) is assumed to be symmetric with respect to real axis and starlike with respect to p(0) = 1. The problems in the class defined by (1.1) become much more difficult if the function p is not univalent. We will consider such class of functions in the present work. An interesting case, when the function p is convex but is not univalent, was considered in [14]. It would be very interesting to find what extra information can be attained (or gained) by using the defining condition (1.1), instead of the weaker condition that zf ′ (z ) f (z )
∈ p(∆),
for all z ∈ ∆,
(1.2)
given a non-univalent p. 2. Main results We first recall here the following class of functions introduced in [15], in which the estimates of coefficients and other connected results were investigated. The related classes of functions were also studied in [16,17]. Definition 1. The function f ∈ A belongs to the class SL, if it satisfies the condition (1.1) with
p(z ) =
1 + τ 2z2 1 − τ z − τ 2z2
where τ = (1 −
(z ∈ ∆),
(2.1)
√
5)/2 ≈ −0.618.
The function (2.1) is not univalent in ∆, but it is univalent in the disc |z | < (3 − √
p(− 21τ ) = 1 and p(e±i arccos(1/4) ) =
5 , 5
√
5)/2 ≈ 0.38. For example, p(0) =
and it may also be noticed that
|τ | = , |τ | 1 − |τ | 1
which shows that number |τ | divides the iϕ [0, 1] such that it fulfils the golden section of this segment Let us put ℜ p(eiϕ ) = x and ℑ p(e ) = y, ϕ ∈ [0, 2π ) \ {π }; then upon performing simple calculations, we find that
√
x=
5 2(3 − 2 cos ϕ)
,
y=
sin ϕ(4 cos ϕ − 1) 2(3 − 2 cos ϕ)(1 + cos ϕ)
,
ϕ ∈ [0, 2π ) \ {π }.
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We observe that the image of the unit circle |z | = 1 under p is a curve described by
(10x −
√
√
√
5)y2 = ( 5 − 2x)( 5x − 1)2 ,
(2.2)
and we rewrite (2.2) in the following form:
√
5
5
√ √
3 −x
+
3 5 10
√
2
5
−x
5
5
+
5
−x −
√ 5
10
y2 = 0.
(2.3)
Recall here that a shell-like curve also called as the conchoid of de Sluze (René François Baron de Sluze 1622–1685), can be described as follows. A ray OA is drawn from the point O(0, 0) and it cuts the directrix x = a, where a > 0 at the point A(a, b). From the point A(a, b), segments AB and AC are laid off in either direction along the ray such that
|OA| · |AB| = k2 and |OA| · |AC | = k2 , where k > 0 is given. As the ray revolves, the point M describes a curve (called the conchoid of de Sluze) whose equation is given by a(x − a)(x2 + y2 ) + k2 x2 = 0,
(2.4)
while the point N describes a curve which may be termed as being conjugate with the conchoid of de Sluze, whose equation is given by a(x − a)(x2 + y2 ) − k2 x2 = 0,
(2.5)
and it may be noted that b = ay/x. For k = 2a, the conchoid of de Sluze (2.4) becomes the trisectrix of Maclaurin (Colin Maclaurin 1698–1746) whose equation is x3 + 3ax2 + (x − a)y2 = 0,
(2.6)
while the conjugate with the conchoid (2.5) becomes the conjugate with the trisectrix of Maclaurin whose equation is of the form: x3 − 5ax2 + (x − a)y2 = 0.
(2.7)
Therefore, it follows that the image of the unit circle under the p, described in (2.3), is translated and revolved √ function 5 / 10; see Fig. 1 below. The curve p(reit ) is a closed curve trisectrix of Maclaurin given in (2.6) with a = ( 1 − 2 τ )/( 10 ) = √ with a loop for 1 > r > (3 − 5)/2 ≈ 0.38. For r = 1, it has a vertical asymptote. It is easy to observe that
ℜ[ p(z )] → a and |ℑ[ p(z )]| → ∞,
when z → −1+ .
Thus, if f ∈ SL, then
ℜ
zf ′ (z ) f (z )
1 − 2τ
> a for z ∈ ∆, a =
10
≈ 0.2236,
which leads to the following corollary. Corollary 1. Let SL and S ∗ (a) be defined as above; then
SL ⊂ S ∗ (a) = f ∈ A : ℜ
zf ′ (z )
f (z )
> a, z ∈ ∆ ,
√ where a =
5/10 ≈ 0.2236, which means that if f ∈ SL, then it is starlike of order a and so it is univalent in the unit disc ∆.
From (2.1), we infer that
p(1) = p(τ 4 ) = 5a,
p e± arccos(1/4) = 2a,
1 = 1. p(0) = p − 2τ
With the particular assigned values (as described above), the graphic representations of the functions arising from Definition 1 are given in Fig. 1. We now show the relevant connection of the function defined by (2.1) with the Fibonacci numbers. Let {un } be the sequence of Fibonacci numbers u0 = 0, u1 = 1, un+2 = un + un+1 (n = 0, 1, 2, 3, . . .); then we have
(1 − τ )n − τ n un = , √ 5
√ τ=
1− 2
5
(n = 0, 1, 2, 3, . . .).
(2.8)
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Fig. 1. p(eiϕ ).
On putting τ z = t, we get
p(z ) =
1 + τ 2z2
1
= t+ t
1 − τ z − τ 2z2 1
= √
t+
5
=
t+
=
t+
1 t 1 t
1
t 1 − t − t2
1
−
t
1 − (1 − τ )t
∞
(1 − τ ) − τ n t √ n
1 1 − τt
n
5
n =1
∞
un t n = 1 +
n =1
∞ (un−1 + un+1 )τ n z n .
(2.9)
n =1
Using now (2.8) and (2.9), we get
p(z ) = 1 +
∞
pn z n = 1 + (u0 + u2 )τ z + (u1 + u3 )τ 2 z 2
(2.10)
n =1
+
∞ (un−3 + un−2 + un−1 + un )τ n z n n =3
= 1 + τ z + 3τ 2 z 2 + 4τ 3 z 3 + 7τ 4 z 4 + 11τ 5 z 5 + · · · ,
(2.11)
with the coefficients pn satisfying p1 = τ , p2 = 3τ , 2
pn = (un−1 + un+1 )τ n = (un−3 + un−2 + un−1 + un )τ n = τ pn−1 + τ 2 pn−2
(n = 3, 4, 5, . . .).
(2.12)
Let us set
f (z ) =
1
z
1
1
= √ − . 1 − τ 2z τ 5 1+z iϕ iϕ If we put (as before) ℜ f (e ) = x and ℑ f (e ) = y, ϕ ∈ [0, 2π ) \ {π }, then after some calculations, we get x=
1 − τ z − τ 2z2
−1 , 2τ (3 − 2 cos ϕ)
y=
(1 + τ 2 ) sin ϕ , 2τ (3 − 2 cos ϕ)(1 + cos ϕ) 2
(2.13)
ϕ ∈ [0, 2π ) \ {π }.
Consequently, the function (2.13) maps the unit circle onto a curve which forms a conjugate with the trisectrix of Maclaurin (see (2.7) and its equation is
(x − b)y2 = (5b − x)x2 , where b = (1 − τ )/10 = (1 +
√
5)/20 ≈ 0.1618; see Fig. 2.
J. Dziok et al. / Mathematical and Computer Modelling 57 (2013) 1203–1211
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Fig. 2. f (ei ϕ ).
The functions p and f are connected by the relation z f ′ (z )
f (z )
= p(z ) (z ∈ ∆).
Thus, we have f ∈ SL ⇔
zf ′ (z ) f (z )
z f ′ (z )
≺
(z ∈ ∆).
f (z )
Therefore, f ∈ SL and f plays the role of an extremal function in this class. If we let τ z = t, then
f (z ) =
z 1 − τ z − τ 2z2
1
=
1
τ
1
= √
t 1 − t − t2
−
1
1 − τt 5τ 1 − (1 − τ )t ∞ n n 1 (1 − τ ) − τ n = t √ τ n=1 5
=
∞ 1
τ
∞
un t n =
n=1
un τ n − 1 z n
n=1
= z + τ z + 2τ z + 3τ 3 z 4 + 5τ 4 z 5 + 8τ 5 z 6 + · · · . 2
2 3
(2.14)
Moreover, if f ∈ SL and f (z ) = z +
∞
(z ∈ ∆),
an z n
(2.15)
n =2
then
|an | ≤ |τ |n−1 un ,
(2.16)
for each integer n ≥ 1. This estimation and the next lemma were established in [15]. Lemma 1. A function f belongs to the class SL if and only if there exists an analytic function q, q ≺ p, such that f (z ) = z exp
0
z
q(t ) − 1 t
dt .
(2.17)
Let S ∗ (α) denote the class of starlike functions of order α defined in (i), and let S ∗ (A, B) be the subclass of S ∗ defined in (ii). We now formulate the following theorem for these classes of functions.
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Theorem 1. If a function f belongs to the class SL, then there exist a function g ∈ S ∗ (0, −τ 2 ) and a function h (starlike of order 1/2) such that zf (z ) = g (z )h(z )
(z ∈ ∆).
(2.18)
Proof. Let f ∈ SL. Then by Lemma 1, there exists a holomorphic function ω(z ) with ω(0) = 0 and |ω(z )| < 1, for z ∈ ∆ such that f (z ) = z exp
z
p (ω(t )) − 1 t
0
dt .
(2.19)
We note that
p (ω(t )) =
1 1 − τ ω(t ) 2
+
1 1 + ω(t )
− 1,
(2.20)
and hence, we can rewrite (2.19) in the form zf (z ) = z 2 exp
z
1/(1 − τ 2 ω(t )) − 1 t
0
= z exp
t
0
1/(1 + ω(t )) − 1 t
0
1/(1 − τ 2 ω(t )) − 1
z
z
dt +
z
dt × z exp
dt
1/(1 + ω(t )) − 1
0
=: g (z )h(z ),
t
dt (2.21)
where × denotes the ordinary multiplication. Using the structural formulae for the classes S ∗ (A, B) (see [2, p. 315]) and S ∗ (α) (see [18, p. 172]), we find that the functions g , h defined in (2.21) satisfy g ∈ S ∗ (0, −τ 2 ) and h ∈ S ∗ (1/2), which proves Theorem 1. In the case, when the functions g ∈ S ∗ (0, −τ 2 ) and h ∈ S ∗ (1/2) are generated by their structural formulae with the same function ω, then we can prove the converse by using (2.20) and (2.21) to show that the function f is such that zf (z ) = g (z )h(z )
(z ∈ ∆)
is in the class SL. Our next result gives the distortion inequalities for a function f belonging to the class SL which is contained in the following theorem. Theorem 2. If f ∈ SL and |z | = r, 0 ≤ r < 1, then r 1 + (1 + τ 2 )r + τ 2 r 2
≤ |f (z )| ≤ | f (−r )| =
r 1 + τ r − τ 2r 2
(2.22)
and 1 + (2a − 1)r
1 + τ 2r 2
≤ |f ′ (z )| ≤ f ′ (−r ) =
2 , (1 + r )3−2a 1 + τ r − τ 2r 2 √ √ where a = 5/10 and τ = (1 − 5)/2. The upper bounds are sharp.
(2.23)
Proof. Let f ∈ SL be given by (2.15), and let
f (z ) = z +
∞
bn z n
(z ∈ ∆).
n =2
Then, upon using (2.16), we conclude that |an | ≤ |bn | = |τ |n−1 un , for each integer n ≥ 0. It may be observed that the even coefficients bn are negative, while the odd are positive, and if z = reiθ , then from the coefficient inequality |an | ≤ |bn |, we get
∞ ∞ f (z ) n−1 = 1 + a z ≤ 1 + |an ||z |n−1 n z n =2 n=2 ≤ 1+
∞ n =2
|bn ||z |n−1 = 1 +
∞
|bn |r n−1
n =2
= 1 − b2 r + b3 r 2 − b4 r 3 + · · · =
f (−r ) , −r
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which gives the upper bound of (2.22). Analogously, we obtain
∞ ∞ n−1 |f (z )| = 1 + nan z n|an ||z |n−1 ≤1+ n =2 n=2 ′
≤ 1+
∞
n|bn ||z |n−1 = 1 +
n =2
∞
n|bn |r n−1
n =2
= 1 − 2b2 r + 3b3 r 2 − 4b4 r 3 + · · · = f ′ (−r ), and we get the upper bound of the inequality (2.23). It is easy to see that the upper bounds are sharp, being attained by the function f ∈ SL at the point z = −r. To find the left-hand side of the inequality (2.22), let us recall (see [2, pp. 315–317]), that if g ∈ S ∗ (0, −τ 2 ), then (for |z | = r, 0 ≤ r < 1) r
≤ |g (z )|.
1 + τ 2r
(2.24)
Moreover, if h ∈ S ∗ (1/2), then (for |z | = r, 0 ≤ r < 1) r
≤ |h(z )|.
1+r
(2.25)
By Theorem 1, we infer that |zf (z )| = |g (z )||h(z )| with g ∈ S ∗ (0, −τ 2 ) and h ∈ S ∗ (1/2), and multiplication of (2.24) with (2.25) leads to the left-hand side of (2.22). To prove the left-hand side of the inequality (2.23), we note that by Corollary 1, if f ∈ SL, then f is starlike of order a. The desired inequality will now follow from the well known inequality for |f ′ (z )| with f ∈ S ∗ (a); see for example [18, pp. 139–140]. Theorem 3. If f (z ) = z + a2 z 2 + a3 z 3 + · · ·, belongs to SL, then
√
5 ≈ 0.764
| an | ≤ b 3 = 3 −
(2.26)
and
√ 5+
lim sup |an | ≤
5
10
n→∞
≈ 0.72,
(2.27)
with equality for the coefficients of f. Proof. If f (z ) = z + b2 z 2 + b3 z 3 + · · ·, then by (2.8) and (2.13), we get
bn = τ
n −1
un =
1−
√ n−1 5
2
1
√
1+
5
√ n
2
5
−
1−
√ n 5
2
,
√
√
bn = bn−1 τ + bn−2 τ 2 , (n ≥ 2). Also, |b2 | ≤ |bn | ≤ b3 , b2 = (1 − 5 − 1)/2 ≈ −0.618, and b3 = 3 − 5 ≈ 0.764. By (2.16), we have |an | ≤ |bn | which implies (2.26). The coefficient inequality |an | ≤ |bn | and the limiting case that
√ lim |bn | =
5+5
n→∞
10
obviously leads to (2.27).
,
Theorem 4. If
√ 5−1
|c | > √
n 5−1
,
(2.28)
then the function g (z ) = z + cz n does not belong to the class SL. Proof. Let us write G(z ) :=
zg ′ (z ) g (z )
=
1 + ncz n−1 1 + cz n−1
(z ∈ ∆).
We prove that if (2.28) is satisfied, then G(z ) ̸≺ p(z ). It suffices to show that G(∆) ̸⊂ p(∆). The set p(∆) is on the right of 1−n|c | 1+n|c | the curve in Fig. 1. The set G(∆) is a disc with the diameter from x1 = 1−|c | to x2 = 1+|c | . If (2.28) is satisfied, then the one of xi satisfies xi < 2a =
√
5/5, and then G(∆) ̸⊂ p(∆). This proves our theorem.
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Let z = reiϕ , ϕ ∈ [0, 2π ); then we obtain
p reiϕ = =
1 + τ 2 r 2 e2iϕ 1 − τ reiϕ − τ 2 r 2 e2iϕ
(1 + τ 2 r 2 )(1 − τ 2 r 2 − τ r cos ϕ) iτ r (1 − τ 2 r 2 + 4τ r cos ϕ) sin ϕ + , |1 − τ reiϕ − τ 2 r 2 e2iϕ |2 |1 − τ reiϕ − τ 2 r 2 e2iϕ |2
which yields
iϕ ℑ[ τ r (1 − τ 2 r 2 + 4τ r cos ϕ) sin ϕ p re ] = ℜ[ (1 + τ 2 r 2 )(1 − τ 2 r 2 − τ r cos ϕ) p reiϕ ] −τ r (1 − τ 2 r 2 − 4τ r ) := ψ(r ), (1 − τ 2 r 2 + τ r )(1 + τ 2 r 2 )
≤
√
whenever r < r0 = 3−2 r0 }. Therefore
zf ′ (z ) f (z )
5
≺ p(z ),
(2.29)
. For such r, the curve p(reit ), t ∈ [0, 2π ) \ {π }, has no loops and p is univalent in ∆r0 = {z : |z |
0 for all |z | < r , and for all f ∈ SL .
It seems that this radius is the radius of convexity of the function f . For f , we infer that
2τ z + 6τ 2 z 2 + 2τ 4 z 4 = ℜ(5τ 2 i) = 0, ℜ 1+ =ℜ 1+ 2 2 2 2 (1 − τ z − τ z )(1 + τ z ) z =τ i f ′ (z ) z =τ i z f ′′ (z )
and therefore, we state the following conjecture. Conjecture. If f ∈ SL, then
√ Rc (SL) = |τ | =
5−1 2
≈ 0.618.
As we have indicated in Section 1, it would be very interesting to describe what further results would emerge by using the defining condition (1.1), instead of the weaker condition that zf ′ (z ) f (z )
∈ p(∆),
for all z ∈ ∆.
(2.31)
It seems to be a hard problem. For example, the function g (z ) = z + cz n satisfies the condition (2.31), if and only if
√ 5−1
|c | ≤ √
n 5−1
.
(2.32)
This result can be proved by using simple geometric arguments and, moreover, it is equivalent to Theorem 4. Even for such an elementary function, it is hard to find an explicit condition related to (2.32) for the class SL. These difficulties could, however, be avoided and surpassed, if we consider the subordination in the disc |z | < r0 , as in Theorem 5, in which the function p is univalent.
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