Appl. Math. Inf. Sci. 8, No. 5, 2455-2463 (2014)
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Applied Mathematics & Information Sciences An International Journal http://dx.doi.org/10.12785/amis/080541
Higher Order Close-to-Convex Functions related with Conic Domain Khalida Inayat Noor∗ and Muhammad Aslam Noor Mathematics Department, COMSATS Institute of Information Technology, Park Road, Islamabad, Pakistan Received: 13 Sep. 2013, Revised: 11 Dec. 2013, Accepted: 12 Dec. 2013 Published online: 1 Sep. 2014
Abstract: In this paper we define and study a class of analytic functions which map the open unit disk onto same conic regions and are related to Bazilevic and higher order close-to-convex functions. We investigate rate of growth of coefficients, Hankel determinant problem, inclusion result and establish univalence criterion. Some other interesting properties of this class are also studied. Keywords: close-to-convex, univalent, conic regions, bounded boundary rotation, Caratheodary functions, convolution, subordination. 2010 AMS Subject Classification: 30C45, 30C50
1 Introduction Let A be the class of functions analytic in the open unit disc E = {z : |z| < 1} and be given by ∞
f (z) = z + ∑ an z . n
(1)
n=2
Let S ⊂ A be the class of functions which are univalent and also K, S∗ ,C be the well known subclasses of S which, respectively, contain close-to-convex, starlike and convex functions. Let Vm (ρ ), m ≥ 2, 0 ≤ ρ < 1, be the class of functions f analytic and locally univalent in E and satisfying the condition (z f ′ (z))′ Z2π ′ (z) − ρ f ℜ d θ ≤ mπ . (2) 1 − ρ 0
When ρ = 0, we obtain the class Vm (m ≥ 2) of functions with bounded boundary rotation, see [4]. The class Vm (ρ ) was introduced and discussed in some detail in [8]. It can easily be shown that f ∈ Vm (ρ ) if and only if there exists f1 ∈ Vm such that (3) f ′ (z) = ( f1′ (z))1−ρ . The convolution of two functions f (z) given by (1) and ∞
g(z) = z + ∑ bn zn is defined as n=2
∞
( f ∗ g)(z) = (g ∗ f )(z) = z + ∑ an bn zn . n=2
∗ Corresponding
In [5], the domain Ωk , k ∈ [0, ∞) is defined as follows: q Ωk = {u + iv : u > k (u − 1)2 + v2 }. (4)
For fixed k, Ωk represents the conic region bounded, successively, by the imaginary axis (k = 0), the right branch of a hyperbola (0 < k < 1) and a parabola (k = 1) and an ellipse (k > 1). Also, we note that, for no choice of k(k > 1), Ωk reduces to a disc, see [5, 18]. In this paper we will choose k ∈ [0, 1]. For k ∈ [0, 1], the following functions, denoted by pk (z), are univalent in E, continuous as regard to k, have real coefficients and map E onto Ωk such that pk (0) = 1, p′k (0) > 0:
pk (z) =
1+z 1−z ,
1+
1+
(k = 0),
2 π2
√ 2 1+ z log 1−√z ,
2 1−k2
sinh
2
h
2 π
(k = 1),
see[5] . (5)
p i arccos k arctan (z) ,
(0 < k < 1).
Let P denote the class of Caratheodory functions of positive real part. Then the class P(pk ) ⊂ P is defined as follows Definition 1. Let p(z) be analytic in E with p(0) = 1. Then p ∈ P(pk ), if p(z) is subordinate to pk (z) given by (5). We write p ∈ Ppk implies p(z) ≺ pk (z) in E, and p(E) ∈ pk (E). We note that P(p0 ) = P. It is easy to verify that P(pk ) is a convex set and k P(pk ) ⊂ P(ρ ), ρ = , k+1
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where P(ρ ) is the class of functions with real part greater than ρ , see [6]. Also, for p ∈ P(pk ), it is known [22] that π 2 , (k = 0), σ π arctan 1k , (k 6= 0), = (6) | arg p(z)| < 2 π , (k = 1). 4 We extend the class P(pk ) as given below
Definition 2. Let p(z) be analytic in E with p(0) = 1. Then p ∈ Pm (pk ), if and only if, for m ≥ 2, k ∈ [0, 1], we have m 1 m 1 p(z) = p1 (z) − p2 (z), (7) + − 4 2 4 2 p1 , p2 ∈ P(pk ).
Definition 5. Let f ∈ A. Then, for a ≥ 0, 0 ≤ γ < 1, f ∈ k −UTm (a, γ , φ ) if and only if there exists g ∈ k −UTm (φ ) such that z f ′ (z) + a f (z) = (a + 1)z(g′ (z))γ .
(8)
We note that z k −UTm 0, 1, = k −UTm , 1−z z = Tm . 0 −UTm 0, 1, 1−z Also
0 −UT2 (0, 1, − log(1 − z)) = C∗ ,
the class of quasi-convex functions, see [17].
When k = 0, we obtain the class Pm introduced and studied in [20]. Also P2 (pk ) = P(pk ). We now define the following Definition 3. Let f ∈ A. Then f ∈ k −UVm , k ∈ [0, 1], m ≥ 2 if and only if z( f ′′ (z)) 1+ ∈ Pm (pk ), z ∈ E. f ′ (z)
Throughout this paper, we assume that k ∈ [0, 1], γ ∈ (0, 1], m ≥ 2, ℜ(a) > −1, z ∈ E, unless otherwise specified.
2 Preliminaries Lemma 1([19]). Let q(z), be analytic in E with q(0) = 1. If α ≥ 1, ℜ(c) ≥ 0, 0 ≤ θ1 < θ2 ≤ 2π , z = reiθ , then Zθ2
k − UVm is called the class of functions with k-uniform boundary rotation. For k = 0, 0 −UVm = Vm , see [4, 12, 13, 14].
θ1
α zq′ (z) ℜ q(z) + d θ > −π cα + q(z)
implies
Zθ2
The corresponding class k −URm is defined as k −URm = {F ∈ A : F = z f ′ , f ∈ k −UVm }. We note that: (i) k −UV2 = k −UCV, is the class of uniformly convex functions.
θ1
ℜq(z)d θ > −π .
Lemma 2([16]). Let f ∈ k −URm . Then there exist si ∈ k − ST, i = 1, 2 such that f (z) =
(ii) k − UR2 = k − ST is the class of uniformly starlike functions. For details of these special case, we refer to [22]. Definition 4. Let f ∈ A. Then f ∈ k − UTm if there exists g ∈ k −UVm such that f ′ (z) ∈ P(pk ), z ∈ E. g′ (z) For k = 0, we have the class Tm , introduced and discussed in [10]. Also, for k = 0, m = 2, k − UTm reduces to well known class K of close-to-convex functions, see [7]. Let φ ∈ A. Then f ∈ k −UTm (φ ) if and only if ( f ∗ φ ) ∈ k −UTm for z ∈ E. c 2014 NSP
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(s1 (z))
m+2 4
(s2 (z))
m−2 4
.
Lemma 3([10]). Let g ∈ Vm (ρ ). Then, for 0 ≤ ρ < 1, θ1 < θ2 , (i) g′ (z) = (g′1 (z))1−ρ , g1 ∈ Vm . (ii)
Rθ2
θ1
ℜ
n
(zg′ (z))′ g′ (z) d θ
o
>−
m 2
− 1 (1 − ρ ) π .
3 Main Results Theorem 1. Let G ∈ k −UTm . Then, for θ1 < θ2 , z = reiθ , Zθ2
θ1
ℜ
(zG′ (z))′ dθ G′ (z)
>−
m−2 + σ π. 2(k + 1)
Appl. Math. Inf. Sci. 8, No. 5, 2455-2463 (2014) / www.naturalspublishing.com/Journals.asp
Proof. Since G ∈ k − UTm , there exist G1 ∈ k − UVm and k k −UVm ⊂ Vm (ρ ), ρ = k+1 , such that G′ (z) = hσ (z), G′1 (z)
Zθ2
θ1
(10)
m−2 dθ > − + σ π. 2(k + 1) ⊔ ⊓
ℜ p(z) +
θ1
zp′ (z) d θ > −γ a + p(z)
where p(z) =
(m − 2) + σ π, 2(k + 1)
(zG′ (z))′ + (1 − γ ). G′ (z)
d θ > −π ,
1 2
and in this case f (z) is univalent in E
Theorem 3. For 0 < γ1 < γ2 ≤ 1,
and
with
z ∈ E,
Proof. Let f ∈ k −UT2 (a, γ1 , φ ). Then
z f ′ (z) + a f (z) = (a + 1)z(G′ (z))γ1 , G(z) = (g ∗ φ )(z) ∈ k −UT2 , = (a + 1)z(H ′ (z))γ2 , γ1
H ′ (z) = (G′ (z)) γ2 ,
H = h∗φ.
We now show that H ∈ k − UT2 and this will prove that f ∈ k −UT2 (a, γ2 , φ ). Now γ1
H ′ (z) = (G′ (z)) γ2 ,
G ∈ k −UT2 ,
γ1 < 1. γ2
1
γ1
z f ′ (z) + a f (z) = (a + 1)z(G′ (z))γ , G ∈ k −UTm .
=γ
(z f ′ (z))′ f ′ (z)
Since G ∈ k −UT2 , there exists a function ′ ∈ P(pk ) in E. G1 = (g1 ∗ φ ) ∈ k −UV2 such that GG′ (z) (z)
Proof. Let G(z) = (g ∗ φ )(z). Then, by definition,
′ (z))′ a + (z ff ′ (z) 1 + a z ff (z) ′ (z)
where
z f ′ (z) . f (z)
Differentiating logarithmically, computations, we have
ℜ
k −UT2 (a, γ1 , φ ) ⊂ k −UT2 (a, γ2 , φ ).
Theorem 2. Let f ∈ k −UTm (a, γ , φ ), ℜ(a) ≥ 0, 0 < γ ≤ 1, θ1 < θ2 and z = reiθ . Then
z f ′ (z) d θ > −π , f ∈ k −UTm (a, γ , φ ). f (z)
If k = 1, then σ = for 2 ≤ m ≤ 4.
2r . 1 − r2 Now differentiating (9) logarithmically and using Lemma 3 together with (10), we obtain
Zθ2
and hence f (z) is univalent in E, see [7].
2r 1 − r2
This completes the proof.
θ1
= π − 2 cos−1
θ1
ℜ
Zθ2
h∈P
zp′ (z) (zG′ (z))′ }≥γℜ . a + p(z) G′ (z)
Corollary 2. Let f ∈ k − UTm (0, 1, φ ). Then, for m ≤ 2{(1 − σ )(k + 1) + 1},
Hence Zθ2 i θ ′ iθ r e h (re ) d θ max ℜ h(reiθ ) h∈P θ1 = max arg h(reiθ2 ) − arg h(reiθ1 )
(zG′ (z))′ ℜ G′ (z)
we have
i h Corollary 1. For m ≤ 2(1−γσγ )(k+1) + 2 , we use Lemma 1 to have from Theorem 2,
θ1
z f ′ (z) f (z) ,
Using Theorem 1, we obtain the required result.
and so Zθ2 iθ ′ iθ re h (re ) d θ = arg h(reiθ2 ) − arg h(reiθ1 ). ℜ h(reiθ )
Zθ2
ℜ{p(z) +
(9)
where σ is given by (6) and h ∈ P. Also we observe that, for h ∈ P, o ∂ ∂ n arg h(reiθ ) = ℜ −i ln h(reiθ ) ∂θ ∂θ iθ ′ iθ re h (re ) =ℜ , h(reiθ )
≤ 2 sin−1
That is, with p(z) =
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Let G′∗ (z) = (G′1 (z)) γ2 , γγ12 < 1. It is easy to verify that G∗ ∈ k −UV2 in E. Thus
simple
H ′ (z) = G′∗ (z) since
γ1 γ2
G′ (z) G′1 (z)
γ1 γ2
∈ P(pk ),
< 1. This completes the proof.
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Remark. From definition 5, the following integral representation for the class k − UTm (a, γ , φ ) can easily be obtained. A function f ∈ k −UTm (a, φ , γ ) if and only if there exists a function G ∈ k −UTm (∞, γ , φ ), such that a+1 f (z) = a z
Za
za−1 G(t)dt.
(11)
0
Theorem 4. Let f ∈ 0−Tm (a, 1, φ ) = Tm (a, 1, φ ). Then f is a Bazilevic function and hence univalent in |z| < rm , where rm is given by o p 1n m − m2 − 4 . rm = (12) 2 Proof. We can write, for f ∈ Tm (a, 1, φ ), a+1 f (z) = a z
Zz 0
t a−1 F(t)dt, F ∈ Tm (∞, 1, φ ).
Zz
t c p(t)g(t)t id−1 dt,
(13)
0
where p ∈ P, g ∈ 0 −URm = Rm . We define 1 g(z) c+1 . G(z) = z z Then zG′ (z) 1 zg′ (z) 1 + = 1− . G(z) c+1 c + 1 g(z)
0
.
f1 (z) is Bazilevic function, see [1], and hence univalent in |z| < rm . Therefore f1z(z) 6= 0, |z| < rm . We note that 1 f (z) a+1 , a = c + id. f1 (z) = z z This means that f (z), given by (13), is analytic and for 1 f (z) a+1 , it is possible to select uniform branch which z takes the value one for z = 0 and which is analytic for |z| < rm and also allows us to compute the derivative in |z| < rm . Thus we conclude that f (z) is univalent in |z| < rm , where rm is given by (12). This completes the proof. ⊓ ⊔ c 2014 NSP
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G = (g ∗ φ ) ∈ Tm .
(15)
Logarithmic differentiation of (15) gives us z f ′ (z) γ (zG′ (z))′ = + (1 − γ ). f (z) G′ (z)
and right hand side is positive for |z| < rm1 . This proves the result. ⊓ ⊔ ∞
∞
n=2
n=2
We have:
c+1+id
Gc+1 (t)p(t)t id−1 dt
f (z) = z(G′ (z))γ ,
z + ∑ bn zn , φ (z) = z + ∑ cn zn .
′
f1 (z) = (c + 1 + id)
Proof. f ∈ Tm (∞, γ , φ ) implies that
We now investigate the rate of growth of coefficients of f ∈ k − UTm (a, γ , φ ). Let f (z) be given by (1) and let g(z) =
(z) ∈ Pm and Pm is a convex set, so G ∈ Rm and, it Now zgg(z) is known [20] that G ∈ Rm is starlike for |z| < rm where rm is given by (12). Further we define f1 (z) as 1
Zz
where m1 = (m + 2)γ and γ1 = (2γ − 1).
Therefore, using a result [11] for G ∈ Tm , we obtain ′ z f (z) (2γ − 1)r2 − γ (m + 2)r + 1 ℜ ≥ , f (z) 1 − r2
Let a = c + id, c > 0. Then we have (c + 1) + id f (z) = zc+id
Theorem 5. Let f ∈ 0 −UTm (∞, γ , φ ) = Tm (∞, γ , φ ). Then the radius rm1 of the disc which f maps onto a starlike domain is given by q o n 1 1 2 − 4γ m − m 1 , γ 6= 2 , 1 1 2γ1 rm1 = , (14) m11 , γ = 21 .
Theorem 6. Let f o ∈ k − UT (a, γ , φ ). n (2−σ γ )(k+1) −2 , m> σ
Then,
for
a + 1 { γ ( m +1)+γσ −1} n k+1 2 |an | ≤ C(m, γ , k) , n → ∞, n+a
where C(m, γ , k) is constant depending only on m, γ and k. Proof. We can write z f ′ (z) + a f (z) = (a + 1)z(G′ (z))γ , where
(16)
G(z) = (g ∗ φ )(z) ∈ k −UTm .
This implies there exists G1 ∈ k −UVm such that G′ (z) = (G′1 (z))(h(z))σ .
Then, with z =
reiθ ,
(17)
we have
(n + a)|an | Z2π −inθ 1 ′ = z f (z) + a f (z) e d θ 2π rn 0 Z2π 1 γ ′ = (a + 1)(G (t)) d θ , G ∈ k −UTm . n−1 2π r 0
(18)
Appl. Math. Inf. Sci. 8, No. 5, 2455-2463 (2014) / www.naturalspublishing.com/Journals.asp
Now, since G ∈ k −UTm , there exists G1 ∈ k −UVm , such that G′ (z) = G′1 (z)hσ (z), h ∈ P,
We note the following special cases. z (i) Let f ∈ 1 −UTm ∞, 1, 1−z = UTm (∞, 1). Then ρ = 21 , γ = 12 and
and σ is given by (6). Using Lemma 2 together with the known result [22] that k k − ST ⊂ S∗ (ρ ), ρ = k+1 , we have
G′1 (z) =
m 1 t1 (z) (1−ρ )( 4 + 2 ) z
,
(1−ρ )( m4 − 12 )
t1 ,t2 ∈ S∗ .
m
an = O(1)n 4 f or m > 4, z = UTm (∞, 1), we get and, for f ∈ 1 − UTm ∞, 1, 1−z m −1 an = O(1).n 4 , m > 4. z (ii) Let k = 0 and f ∈ Tm ∞, γ , 1−z . Then, for m ≥ 2
(19)
(t2 (z)z) o n σγ iθ Define, for m > σ2− (1−ρ ) − 2 , z = re . Iγ (r) =
Z2π 0
|G′ (z)|γ d θ ,
an = O(1)nγ ( 2 +1)+γ −1 = O(1)n m
Iγ (r) =
rγ (1−ρ )
0
m
γ (1−ρ )( m − 1 ) 4 2 4 ≤ γ (1−ρ ) r r
an = O(1)n 2 ,
1
0
m
1
γ (1−ρ )( m − 1 ) 4 2 1 4 Iγ (r) ≤ γ (1−ρ ) r r 2−σ γ ×
0
2
|t1 (z)|
1 ≤ C(m, γ , k) 1−r
γσ 2
1 1−r
1 Iγ (r) = O(1) 1−r
Z2π
m
an = O(1)n 4 −1 ,
0
2
m
an = O(1)n 2 ,
2
|h(z)| d θ 2
2
γ (1−ρ )( m +1)+ γσ −1
⊔ ⊓
With k = 1, γ = 1, we have σ =
γσ
γ (1−ρ )( m +1)+ γσ −1
Now, with r = 1 − 1n , we have from (18), a + 1 γ (m+2) +γσ −1 n 2(k+1) |an | ≤ C(m, γ , k) , n+a This completes the proof.
γ (m+2)
an = O(1)n 2(k+1) 1 2
+γσ −2
.
and so, for m > 4 (n → ∞).
Also, if we take k = 0, γ = 1. Then σ = 1 and so
2
where C(m, γ , k) is a constant depending only on m, γ , k. That is
(n → ∞).
(iv) Let f ∈ k o− UTm (∞, γ , log(1 − z)). Then, for n 2−γσ m > 2 γ (1−ρ ) − 1 ,
|t1 (z)|γ (1−ρ )( 4 + 2 ) |h(z)|γσ d θ ,
γ (1−ρ )( m4 + 21 )
,
This result is proved in [11]. See also [15].
where we have used well-known distortion result for the starlike function t2 (z). We now apply Holder’s inequality, use subordination for starlike functions and a result due to Pommerenke [21] for h ∈ P to have, for σ (1 − ρ )(m + 2) > 2 − σ γ ,
Z2π
γm 2 +2γ −2
and with γ = 1, we obtain, for m ≥ 2
2
×
an = O(1)n 2 +1 . z . Then, for m ≥ 2 0, γ , 1−z an = O(1)n
m 1 |t1 (z)|γ (1−ρ )( 4 + 2 ) γσ m 1 |h(z)| d θ |t (z)|γ (1−ρ )( 4 − 2 )
Z2π
, (n → ∞).
m
(iii) Let f ∈ Tm Z2π
γm 2 +2γ −1
When we take γ = 1, then
G ∈ k −UTm .
Then, from (19) 1
2459
,
Theorem 7. Let f ∈ k − UTm (0, 1, φ ). Denote by L(r, f ), the length of the image of the circle |z| = r under f , by A(r, f ), the area of f (|z| < r) and M(r, f ) = max | f (reiθ )|. θ
Then
L(r, f ) = O(1)M(r, f ) log where O(1) is a constant.
2
,
(n → ∞).
(n → ∞).
1 , 1−r
Proof. Since f ∈ k −UTm (0, 1, φ ), we have
z f ′ (z) = zG′ (z), G(z) = (g ∗ φ )(z) ∈ k −UTm .
This implies that f ∈ k − UTm . So there exists G1 ∈ k − UVm ⊂ Vm (ρ ) such that G′ = p ∈ P(pk ) ⊂ P(ρ ). G′1 c 2014 NSP
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Now, with z = reiθ , L(r, f ) =
=
=
≤
≤
Z2π 0 Z2π 0 Z2π
Also, since A(r, f ) ≤ π M 2 (r, f ), we have
|z f ′ (z)|d θ
I1 (r) ≤
Zr Z2π 0 0
G1Vm (ρ ), p ∈ P(ρ )
| f ′ (z)H(z)|d θ d ξ +
0 0
I1 (r) =
0 0
Z2π 0
1 + zeit d µ (t), 1 − zeit
1−ρ p (z) = π ′
|zp′ (z)G′1 (z)|d θ d ξ
= I1 (r) + I2 (r). Zr Z2π
1−ρ I2 (r) ≤ π
(20)
Zr Z2πZ2π
ℜp(z) =
∞ (zG′1 (z))′ dn zn , = 1 + H(z) = ∑ G′1 (z) n=1 k m f (z) given by (1), |dn | ≤ m 1 − k+1 , and for n ≥ 1, = k+1 we have
I1 (r) 1 1 2 2 Zr Z2π Z2π ′ 2 2 ≤ | f (z)| d θ |H(z)| d θ dξ = 2π
Zr 0
∑ n |an | ξ 2
2 2n−2
n=1
√ ≤ 2
m π k+1
√ 2
m π k+1
≤
!1
2
∞
∑ |αn | ξ
n=0
∞
n2 ∑ 2n − 1 |an |2 r2n−1 n=1 !1 ∞
∑ n|an |2 r2n−1
n=1
!1
!1 2
2
log
1+r log 1−r 1
1
2
0 0 0
1−ρ π
.
∞
n=1
|z| < r by w = f (z). Therefore 1 1 √ 1+r 2 m A(r, f ) 2 log π . I1 (r) ≤ 2 k+1 πr 1−r c 2014 NSP
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|zG′1 (z)| d µ (t)d θ d ξ . |1 − zeit |2
Z2π 0
1−ξ2 d µ (t), |1 − zeit |2
I2 (r) ≤ 2(1 − ρ )
Zr Z2π
0 0 Z2π 0
|zG′ (z)|ℜH(z)d θ ′
dξ 1−ξ2
ℜ{zG′ (z)e−i arg zG1 }d θ
dξ . 1−ξ2
Zr 0
M(r, f ) dξ . 1−ξ2
(22)
From (20), (21) and (22), we obtain the desired result. ⊓ ⊔ We study arc-length problem with a different technique as follows.
2
But A(r, f ) = π ∑ n|an |2 r2n is the area of the image of
eit d µ (t). (1 − zeit )2
I2 (r) ≤ [2m(1 − ρ ) + 2ρ ]π
dξ
1+r 1−r
0
Integrating by parts gives us
2
2 2n
d µ (t) = 2π .
and hence
= 2(1 − ρ )
0
∞
0
Also
| f ′ (z)H(z)|d θ d ξ ,
0
Z2π
Z2π
Therefore
where
0
(21)
implies that we can write
So
(zG′1 (z))′ H(z) = G′1 (z)
Zr Z2π
k k+1 ,
1−ρ p(z) = 2π
|G′1 (z)p(z)|H(z) + G′1 (z)(zp′ (z))d ξ d θ ,
Now
1 1 1+r 2 m M(r, f ) log . k+1 r 1−r
p ∈ P(ρ ), ρ =
|zG′1 (z)p(z)|d θ ,
0 0
We now estimate I2 (r).
|zG′ (z)|d θ
0 Z2πZr
√ 2
Theorem 8. Let f o∈ k − UTm (0, γ , φ ). Then, for n m > (2−σ γγ)(k+1) − 2 ,
1 L(r, f ) = O(1) 1−r
γ k+1
( m2 +1)+σ γ −1
, (r → 1),
where σ is given by (6) and O(1) is a constant.
Appl. Math. Inf. Sci. 8, No. 5, 2455-2463 (2014) / www.naturalspublishing.com/Journals.asp
Proof. We can write z f ′ (z) = z(G′ (z))γ , G = (g ∗ φ ) ∈ k −UTm = z(G′ (z)hσ (z))γ , G = (g ∗ φ ) ∈ k −UTm , h ∈ P γ ( m + 1 ) 4 2 z s1z(z) σγ (23) = γ ( m − 1 ) h (z), s1 , s2 ∈ k − ST, s2 (z) z
4
2
by using Lemma 2. Also si ∈ k − ST implies that si ∈ S∗ (ρ ), ρ = Therefore, for z = reiθ L(r, f ) =
Z2π 0
|z f (z)|d θ = ′
Z2π 0
k k+1 , i =
1, 2.
γ m 1 |s1 (z)| k+1 ( 4 + 2 ) σγ dθ . γ m 1 |h(z)| |s (z)| k+1 ( 4 − 2 )
0
Holder’s inequality together with subordination for starlike functions, we have L(r, f ) σγ 2 γ ( m − 1 ) Z2π 4 2 k+1 4 |h(z)|2 d θ ≤ r 0
Z2π
0
r |1 − reiθ |
1 ≤ O(1) 1−r for
γ (m+2) k+1
γ k+1
2γ k+1
Let f ∈ A and be given by (1). Suppose that the qth Hankel determinant of f is defined for q ≥ 1, n ≥ 1 by an an+1 . . . an+q−1 .. an+1 an+2 . . . . Hq (n) = . (24) .. .. .. . . an+q−1 . . . . . . an+2q−2
Theorem 9. Let f ∈ UTm (0, γ , φ ) and let the qth Hankel determinant of o q ≥ 1, n ≥ 1, be defined by (24). n f (z) for 8q The, for m ≥ γ − 2 , mγ 2 Hq (n) = O(1)n( 4 +γ −1)q−q , (n → ∞),
2
Since s2 (z) is starlike and hence univalent, so we have γ ( m − 1 ) Z2π γ m+1 4 k+1 4 2 L(r, f ) ≤ |s1 (z)| k+1 ( 4 2 ) |h(z)|σ γ d θ . r
( m4 + 21 ) 2−2σ γ
( m2 +1)+σ γ −1
2−σ γ
dθ
⊓ ⊔
(i) For γ = 1, m = 2 and k = 1, which gives us σ = 21 . This gives us 1
Lemma 4. Let f ∈ A and be given by (1) and let the qth Hankel determinant of f be defined by (24). Then, writing ∆ j (n) = ∆ j (n, z1 , f ). We have Hq (n) ∆2q−1 (n) ∆2q−3 (n + 1) . . . ∆q−1 (n + q − 1) ∆2q−3 (n + 1) ∆2q−4 (n + 2) . . . ∆q−2 (n + q) = , .. .. .. . . . ∆ (n + q − 1) (n + 2q − 2) ∆ . . . . . . q q−1
2
.
(ii) We take k = 0 and γ = 1. Then σ = 1 and f ∈ Tm . This gives us L(r, f ) = O(1)
1 1−γ
m +1
(25)
(26)
, v ≥ 0 and integer
∆ j (n + v, v, x, z f ′ (z)) j l j y (v − (l − 1)n) ∆ j−l (n + v, v + l, y, f ). =∑ l (n + 1)l l=0 Proof(Theorem 9). Since f ∈ UTm (0, γ , φ ), we can write z f ′ (z) = z(G′ (z))γ , g ∈ UTm , G = (g ∗ φ ).
(27) 1
Now, for G ∈ UTm , there exists G1 ∈ UVm ⊂ Vm 2 such ′ that GG′ ∈ P(p1 ). Also, for p ∈ P(p1 ), we have | arg p(z)| < π 4
1
which gives us σ = 12 . Thus we can write (27) as f ′ (z) = [(G′2 (z)) 2 p 2 (z)]γ 1
n n+1 y
Lemma 5. With x =
As special cases, we note the following
1 L(r, f ) = O(1) 1−r
To prove Theorem 9, we need the following results and for these we refer to [9].
∆ j (n, z, f ) = ∆ j−1 (n, z1 , f ) − z1 ∆ j−1 (n + 1, z1 , f ).
,
where O(1) is a constant depending upon γ , m and q only.
where, with ∆0 (n, z1 , f ) = an , we define for j ≥ 1,
2
> (2 − σ γ ). This completes the proof.
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2
,
(r → 1).
We shall estimate the growth rate of Hq (n) for the functions in the class UTm (0, γ , φ ). This is the main motivation of next result.
1
γ
= (G′1 (z)p(z)) 2 , G2 ∈ Vm , p ∈ P γ m 1 s1 (z) 2 ( 4 + 2 ) 2 γ (p(z)) 2 , = γ m 1 − ( ) 2 4 2 s2 (z) 2
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K. I. Noor, M. A. Noor: Higher Order Close-to-Convex Functions related...
where s1 , s2 ∈ S∗ , where we have used a result due to Brannan [2]. Also we can choose a z1 with |z1 | = r such that for any univalent functions s(z) max |(z − z1 )s(z)| ≤ |z|=r
2r2 , 1 − r2
(28)
see [3]. Now, for j ≥ 0, z1 any nonzero complex number, consider
∆ j (n, z1 , f ′ (z)) γ m 1 (z − z ) j s1 (z) 2 ( 4 + 2 ) 1 γ z 1 (p(z)) 2 d θ . = γ m 1 n+ j − 2π r s2 (z) 2 ( 4 2 ) z Z2π
|z − z1 | j
≤
rn+ j−1
1 × 2π
2r2 1 − r2 Z2π 0
0
4−γ 4
m 1 4 |s1 (z)| ( 4 + 2 )− j 4−γ
j γ 2r2 1 + 3r2 4 ≤ C(m, γ , j) 1 − r2 1 − r2 n o 1 m [γ ( + −2 j)] 4 −1 4−γ 4 2 4−γ 4 1 × 1−r mγ − j+γ −1 4 1 = O(1) , 1−r
O(1) is a constant and we have used (28), distortion results for starlike functions, Holder’s inequality and a result for h ∈ P, see [21]. Choosing r = 1 − 1n , we have, for γ (m + 2) ≥ 8( j + 1),
∆ j (n, z1 , f ′ ) = O(1).n
mγ 4 +γ − j−1
,
and using Lemma 5, we obtain
∆ j (n, eiθn , f ) = O(1)n
mγ 4 +γ − j−2
,
(n → ∞).
(29)
We use Lemma 4 and follow the similar argument given in [9], to have mγ 2 Hq (n) = O(1)n( 4 +γ −1)q−q ,
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(n → ∞)
m ≥ 14.
References
γ 4 j γ (m−1) Z2π 4 2 4 2 1 2 |p(z)| d θ r 2π γ 2
For this case we note that
The authors are grateful to Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Pakistan for providing excellent research and academic environment.
0
1
m 2 Hq (n) = O(1)n( 4 )q−q , n → ∞.
Acknowledgement
γ m 1 |s1 (z)| 2 ( 4 + 2 ) γ m 1 γ 2π rn+ j−1 |s2 (z)| 2 ( 4 − 2 ) |p(z)| 2 d θ 0 j γ (m−1) 1 2r2 4 2 4 2 ≤ n+ j−1 r 1 − r2 r 2 π Z γ m 1 γ 1 × |s1 (z)| 2 ( 4 + 2 )− j |p(z)| 2 d θ 2π
Special Case. m When γ = 1, m ≥ 6, we have an = O(1)n 4 −1 and
m
∆ j (n, z1 , f ′ ) 1
⊓ ⊔
H2 (n) = O(1)n 2 −4 ,
Thus, for γ (m + 2) ≥ 8( j + 1),
≤
for γ (m + 2) ≥ 8q. This completes the proof.
[1] I. E. Bazilevic, On a class of integrability in quadratures of the Loewner-Kufarev equation, Math. Sb., 37, 471-476 (1955). [2] D. A Brannan, On functions of bounded boundary rotations, Proc. Edin. Math. Soc., 2, 339-347 (1968-1969). [3] G. Golusin, On distortion theorems and coefficients of univalent functions, Mat. Sb., 19, 183-203 (1946). [4] A. W. Goodman, Univalent Functions, Vol. I,II, Polygonal Publishing House, Washington, New Jersey, (1983). [5] S. Kanas, Techniques of differential subordination for domains bounded by conic sections, Internat. J. Math. Math. Sci., 38, 2389-2400 (2003). [6] S. Kanas, A. Wisniowska, Conic regions and k-uniform convexity, J. Comput. Appl. Math., 105, 327-336 (1999). [7] W. Kaplan, Close-to-convex Schlicht functions, Michigan Math. J., 1, 169-185 (1952). [8] E. J. Moulis, Generalizations of the Robertson functions, Pacific J. Math., 81, 169-174 (1979). [9] J. W. Noonan, D. K. Thomas, On Hankel determinant of areally mean p-valent functions, Proc. Lond. Math. Soc., 25, 503-524 (1972). [10] K. I. Noor, On subclasses of close-to-convex functions of higher order, Internat. J. Math. Math. Sci., 6, 327-334 (1983). [11] K. I. Noor, On generalizations of close-to-convexity, Mathematica, 23, 217-219 (1981). [12] K. I. Noor, Some recdent developments in the theory of functions with bounded boundary rotation, Inst. Math. Comput. Sci., 2, 25-37 (2007). [13] K. I. Noor, W. Ul-Haq, M. Arif, S. Mustafa, On bounded boundary and bounded radius rotation, J. Inequal. Appl., 2009, Article ID 813687, 12 pages (2009). [14] K. I. Noor, Some properties of analytic functions with bounded radius rotation, Complex Var. Elliptic, Equat., 54, 865-877 (2009). [15] K. I. Noor, Higher order close-to-convex functions, Math. Japonica, 37, 1-8 (1992).
Appl. Math. Inf. Sci. 8, No. 5, 2455-2463 (2014) / www.naturalspublishing.com/Journals.asp
[16] K. I. Noor, On a generalization of uniformaly convex and related functions, Comput. Math. Appl., 61, 117-125 (2011). [17] K. I. Noor, On quasi-convex univalent functions and related topics, Internat. J. Math. Sci., 2, 241-258 (1987). [18] K. I. Noor, R. Fayyaz, M. A. Noor Some classes of kuniformly functions with bounded radius rotation, Appl. Math. Inf. Sci., 8, No. 2, 527-533 (2014). [19] R. Parvatham, S. Radha, On certain classes of analytic functions, Ann. Polon. Math., 49, 31-34 (1980). [20] B. Pinchuk, Functions of bounded boundary rotation, Isr. J. Math., 10, 7-16 (1971). [21] Ch. Pommerenke, On starlike and close-to-convex functions, Proc. London Math. Soc., 3, 290-304 (1963). [22] F. Ronning, Uniform convex functions and a corresponding class of starlike functions, Proc. Amer. Math. Soc., 118, 189-196 (1993).
Khalida Inayat Noor is a leading world-known figure in mathematics. She obtained her PhD from University of Wales (Swansea)(UK). She has a vast experience of teaching and research at university levels in various countries including Iran, Pakistan, Saudi Arabia, Canada and United Arab Emirates. She was awarded HEC best research paper award in 2009 and CIIT Medal for innovation in 2009. She has been awarded by the President of Pakistan: President’s Award for pride of performance on August 14, 2010 for her outstanding contributions in mathematical sciences and other fields. Her field of interest and specialization is Complex analysis, Geometric function theory, Functional and Convex analysis. She introduced a new technique, now called as Noor Integral Operator which proved to be an innovation in the field of geometric function theory and has brought new dimensions in the realm of research in this area. She has been personally instrumental in establishing PhD/MS programs at COMSATS Institute of Information Technology, Pakistan. Dr. Khalida Inayat Noor has supervised successfully several Ph.D and MS/M.Phil students. She has been an invited speaker of number of conferences and has published more than 400 (Four hundred ) research articles in reputed international journals of mathematical and engineering sciences. She is member of editorial boards of several international journals of mathematical and engineering sciences.
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M. Aslam Noor earned his PhD degree from Brunel University, London, UK (1975) in the field of Applied Mathematics (Numerical Analysis and Optimization). He has vast experience of teaching and research at university levels in various countries including Pakistan, Iran, Canada, Saudi Arabia and United Arab Emirates. His field of interest and specialization is versatile in nature. It covers many areas of Mathematical and Engineering sciences such as Variational Inequalities, Operations Research and Numerical Analysis. He has been awarded by the President of Pakistan: President’s Award for pride of performance on August 14, 2008, in recognition of his contributions in the field of Mathematical Sciences. He was awarded HEC Best Research paper award in 2009. He has supervised successfully several Ph.D and MS/M.Phil students. He is currently member of the Editorial Board of several reputed international journals of Mathematics and Engineering sciences. He has more than 750 research papers to his credit which were published in leading world class journals.
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