Gen. Math. Notes, Vol. 17, No. 1, July, 2013, pp.63-73
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2013
www.i-csrs.org
Available free online at http://www.geman.in
On a Class of Meromorphic Univalent Functions
Defined by Hypergeometric Function
Abdul Rahman S. Juma1 and Hazha Zirar2
1
Department of Mathematics
University of Anbar, Ramadi, Iraq
E-mail: dr− juma@hotmail.com
2
Department of Mathematics, College of Science
University of Salahaddin, Erbil, Kurdistan, Iraq
E-mail: hazhazirar@yahoo.com
(Received: 20-4-13 / Accepted: 27-5-13)
Abstract
In this paper, we introduce the class of meromorphic univalent functions
defined by hypergeometric function. We obtain some interesting properties like
coefficient inequality, distortion and growth theorems, Hadamard product(or
convolution), radii of starlikeness and convexity for the functions in the class
H ∗ (β, α, k).
Keywords: Meromorphic univalent function, Hypergeometric functions,
Hadamard product, Starlike function, Convex function.
1
Let
Introduction
P
denote the class of functions of the form
∞
f (z) =
1 X
+
an z n ,
z n=1
which are analytic and meromorphic univalent in the punctured unit disk ∗ =
{z ∈ C : 0 < |z| < 1} = −{0}.
64
Abdul Rahman S. Juma et al.
Let A be a subclass of
P
of functions of the form
∞
f (z) =
1 X
+
an z n , (an ≥ 0, n ∈ N).
z n=1
(1)
A function f ∈ A is meromorphic starlike function of order ρ, (0 ≤ ρ < 1)if
0 zf (z)
> ρ, (z ∈).
−<
f (z)
The class of all such functions is denoted by A∗ (ρ). A function f ∈ A is
meromorphic convex function of order ρ, (0 ≤ ρ < 1) if
zf 00 (z)
> ρ, (z ∈).
−< 1 + 0
f (z)
The Hadamard product (or convolution) of two functions, f given by (1) and
∞
g(z) =
1 X
bn z n , (bn ≥ 0, n ∈ N),
+
z n=1
is defined by
∞
1 X
(f ∗ g)(z) = +
an b n z n ,
z n=1
see [9], let us define the function φ̃(a, c; z) defined by
∞
φ̃(a, c; z) =
1 X (a)n+1
+
|
|an z n ,
z n=0 (c)n+1
for c 6= 0, −1, −2, ..., and a ∈ C − {0}, where (λ)n = λ(λ + 1)n+1 is the
Puchhammer symbol. We note that
φ̃(a, c; z) =
1
z
where
2 F1 (b, a, c; z)
=
2 F1 (1, a, c; z),
∞
X
(b)n (a)n z n
n=0
(c)n
n!
,
is the well-known Gaussian hypergeometric function. Corresponding
to the
P
function φ̃(a, c; z), using the
PHadamard product for f ∈ , we define a new
linear operator L∗ (a, c) on
by
∞
1 X (a)n+1
L (a, c)f (z) = φ̃(a, c; z) ∗ f (z) = +
|
|an z n .
z n=1 (c)n+1
∗
(2)
65
On a Class of Meromorphic Univalent Functions...
The meromorphic functions with the generalized hypergeometric functions
were considered recently by Dziok and Srivastava [2],[3], Liu [5], Liu and Srivastava [6], [7], [8], Cho and Kim [1]. For a function f ∈ L∗ (a, c)f (z) we
define
I 0 (L∗ (a, c)f (z)) = L∗ (a, c)f (z),
and for k = 1, 2, 3, ...,
2
I k (L∗ (a, c)f (z)) = z(I k−1 L∗ (a, c)f (z))0 + ,
z
∞
=
1 X k (a)n+1
+
n |
|an z n ,
z n=0
(c)n+1
(3)
We note that I k (L∗ (a, c)f (z)) studied by Frasin and Darus [4].
It follows from (2) that
z(L(a, c)f (z))0 = aL(a + 1, c)f (z) − (a + 1)L(a, c)f (z).
(4)
Also, from (3) and (4) we get
z(I k L(a, c)f (z))0 = aI k L(a + 1, c)f (z) − (a + 1)I k L(a, c)f (z).
Now we define the following:
Definition 1.1 : A function f ∈ A of the form (1) is in the class H ∗ (β, α, k)
if it satisfies the following condition:
z(I k L∗ (a,c)f (z))00
+2
(I k L∗ (a,c)f (z))0
|
| z(I k L∗ (a,c)f (z))00
+ 2α
(I k L∗ (a,c)f (z))0
< β,
(5)
where λ > −1, 0 ≤ α < 1, 0 < β ≤ 1, k = 1, 2, 3, ....
In this paper we obtain coefficient estimates for the class H ∗ (β, α, k),
Hadamard product, growth and distortion theorems, radii of starlikeness and
convexity.
2
Coefficient Inequality
Theorem 2.1 : A function f defined by (1) is in the class H ∗ (β, α, k), if and
only if
∞
X
n=1
nk+1 |
(a)n+1
|[n(1 + β) + (1 + β(2α − 1))]an ≤ 2β(1 − α).
(c)n+1
(6)
66
Abdul Rahman S. Juma et al.
The result is sharp for the function
1
2β(1 − α)
fn (z) = +
zn, n ≥ 1
z nk+1 | (a)n+1 |[n(1 + β) + (1 + β(2α − 1))]
(7)
(c)n+1
Proof : To proof the sufficient part, let the inequality (6) holds true and
let |z| = 1,
by (5), we have
|
z(I k L∗ (a, c)f (z))00
z(I k L∗ (a, c)f (z))00
+
2|
−
β|
+ 2α| < 0,
(I k L∗ (a, c)f (z))0
(I k L∗ (a, c)f (z))0
|z(I k L∗ (a, c)f (z))00 +2(I k L∗ (a, c)f (z))0 |−β|z(I k L∗ (a, c)f (z))00 +2α((I k L∗ (a, c)f (z))0 )|,
∞
∞
X
2(1 − α) X
(a)n+1
n−1
k+1 (a)n+1
|an z −β|
|an z n−1 |,
+ (n−1+2α)nk+1 |
=|
(n+1)n |
2
(c)
z
(c)
n+1
n+1
n=1
n=1
≤
∞
X
n=1
nk+1 |
(a)n+1
|[n(1 + β) + (1 + β(2α − 1))]an − 2β(1 − α) ≤ 0.
(c)n+1
Thus by the maximam modulus theorem, we have f ∈ H ∗ (β, α, k).
Conversely, suppose that f of the form (1) is in the class H ∗ (β, α, k), then by
(5) we have
z(I k L∗ (a,c)f (z))00
+2
(I k L∗ (a,c)f (z))0
| z(I k L∗ (a,c)f (z))00
| < β,
+
2α
k
∗
0
(I L (a,c)f (z))
P∞
k+1 (a)n+1
| (c)n+1 |an z n−1
n=1 (n + 1)n
| < β.
| 2(1−α) P∞
n+1
+ n=1 (n − 1 + 2α)nk+1 | (a)
|a z n−1
z2
(c)n+1 n
Since |<(z)| ≤ |z| for all z, we get
P∞
k+1 (a)n+1
| (c)n+1 |an z n−1
n=1 (n + 1)n
<{ 2(1−α) P∞
} < β.
k+1 | (a)n+1 |a z n−1
+
(n
−
1
+
2α)n
n
2
n=1
z
(c)n+1
(8)
k
∗
00
L (a,c)f (z))
Now by choosing the value of z on the real axis so that the value of z(I
(I k L∗ (a,c)f (z))0
is real, then clearing the denominator of (8) and letting z → 1− through real
values, we get the inequality (6).
The result is sharp for the function
1
(c)n+1
2β(1 − α)
fn (z) = + |
| k+1
zn, n ≥ 1
z
(a)n+1 n [n(1 + β) + (1 + β(2α − 1))]
Corollary 2.1 : If f ∈ H ∗ (β, α, k), then
an ≤ |
(c)n+1
2β(1 − α)
| k+1
,
(a)n+1 n [n(1 + β) + (1 + β(2α − 1))]
where λ > −1, 0 ≤ α < 1, 0 < β ≤ 1, k = 1, 2, 3, ....
On a Class of Meromorphic Univalent Functions...
3
67
Growth and Distortion Theorems
A growth and distortion property for the function f ∈ H ∗ (β, α, k) is given as
follows:
Theorem 3.1 : A function f defined by (1) is in the class H ∗ (β, α, k), then
for 0 < |z| = r < 1, we have
1
(c)2 β(1 − α)
1
(c)2 β(1 − α)
−|
|
r ≤ |f (z)| ≤ + |
|
r,
r
(a)2 (1 + αβ)
r
(a)2 (1 + αβ)
with equality for
f (z) =
1
(c)2 β(1 − α)
+|
|
z.
z
(a)2 (1 + αβ)
Proof : Since f ∈ H ∗ (β, α, k), we have from Theorem 2.1 the inequality
∞
X
nk+1 |
n=1
(a)n+1
|[n(1 + β) + (1 + β(2α − 1))]an ≤ 2β(1 − α),
(c)n+1
Then
∞
X
1
|f (z)| ≤ | | +
an |z|n ,
z
n=1
for 0 < |z| = r < 1, we have
|f (z)| ≤
≤
Also
∞
X
1
+r
an ,
r
n=1
(c)2 β(1 − α)
1
+|
|
r.
r
(a)2 (1 + αβ)
∞
X
1
|f (z)| ≥ | | −
an |z|n ,
z
n=1
1
(c)2 β(1 − α)
−|
|
r, |z| = r.
r
(a)2 (1 + αβ)
Hence the proof is complete.
Theorem 3.2 : A function f defined by (1) is in the class H ∗ (β, α, k),
then for 0 < |z| = r < 1, we have
≥
(c)2 β(1 − α)
1
(c)2 β(1 − α)
1
−|
|
≤ |f 0 (z)| ≤ 2 + |
|
,
2
r
(a)2 (1 + αβ)
r
(a)2 (1 + αβ)
with equality for
f (z) =
1
(c)2 β(1 − α)
+|
|
z.
z
(a)2 (1 + αβ)
68
Abdul Rahman S. Juma et al.
Proof : From Theorem 2.1 we have
∞
X
n=1
Thus
nk+1 |
(a)n+1
|[n(1 + β) + (1 + β(2α − 1))]an ≤ 2β(1 − α).
(c)n+1
∞
X
−1
|f (z)| ≤ | 2 | +
nan |z|n−1 ,
z
n=1
0
for 0 < |z| = r < 1, we get:
∞
X
1
nan ,
|f (z)| ≤ | 2 | +
r
n=1
0
≤
and
1
(c)2 2β(1 − α)
+|
|
,
2
r
(a)2 (1 + αβ)
∞
X
−1
|f (z)| ≥ | 2 | −
nan |z|n−1 ,
z
n=1
0
∞
X
1
nan
≥ | 2| −
r
n=1
≥
1
(c)2 2β(1 − α)
−|
|
.
2
r
(a)2 (1 + αβ)
This completes the proof.
4
Hadamard Product
Theorem 4.1 : Let the functions f, g ∈ H ∗ (β, α, k). Then (f ∗g) ∈ H ∗ (δ, α, k)
for
∞
1 X
f (z) = +
an z n ,
z n=1
∞
g(z) =
and
1 X
+
bn z n ,
z n=1
∞
1 X
(f ∗ g)(z) = +
an b n z n ,
z n=1
On a Class of Meromorphic Univalent Functions...
69
where
δ=
2β 2 (1 − α)(n + 1)
(c)n+1
|nk+1 [n(1 + β) + (1 + β(2α − 1))]2
2β 2 (1 − α)(n + 2α − 1) − | (a)
n+1
Proof : Since f, g ∈ H ∗ (β, α, k), then by Theorem 2.1 we have:
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))]
|
|
an ≤ 1,
(c)n+1
2β(1 − α)
n=1
and
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))]
|
|
bn ≤ 1.
(c)
2β(1
−
α)
n+1
n=1
We must find the largest δ such that
∞
X
(a)n+1 nk+1 [n(1 + δ) + (1 + δ(2α − 1))]
|
|
an bn ≤ 1.
(c)
2δ(1
−
α)
n+1
n=1
By Cauchy- Schwarz inequality, we get
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))] p
|
|
an bn ≤ 1,
(c)
2β(1
−
α)
n+1
n=1
(9)
To prove the theorem it is enough to show that
(a)n+1 nk+1 [n(1 + δ) + (1 + δ(2α − 1))]
|
an b n
(c)n+1
2δ(1 − α)
≤|
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))] p
|
an b n ,
(c)n+1
2β(1 − α)
which is equivalent to
p
an bn ≤
δ[n(1 + β) + (1 + β(2α − 1))]
.
β[n(1 + δ) + (1 + δ(2α − 1))]
From (9), we have
p
(c)n+1
2β(1 − α)
an b n ≤ |
| k+1
.
(a)n+1 n [n(1 + β) + (1 + β(2α − 1))]
We must show that
|
(c)n+1
2β(1 − α)
δ[n(1 + β) + (1 + β(2α − 1))]
| k+1
≤
,
(a)n+1 n [n(1 + β) + (1 + β(2α − 1))]
β[n(1 + δ) + (1 + δ(2α − 1))]
70
Abdul Rahman S. Juma et al.
which gives
δ≤
2β 2 (α − 1)(n + 1)
(c)n+1
2β 2 (1 − α)(n + 2α − 1) − | (a)
|nk+1 [n(1 + β) + (1 + β(2α − 1))]2
n+1
.
Hence the proof is complete.
Theorem 4.2 : Let the functions fi (i = 1, 2) defined by
∞
1 X
an,i z n , (an,i ≥ 0, i = 1, 2)
fi (z) = +
z n=1
be in the class H ∗ (β, α, k). Then the function g defined by
∞
1 X 2
(an,1 + a2n,2 )z n ,
g(z) = +
z n=1
is in the class H ∗ (η, α, k), where
η=
4β 2 (α − 1)(n + 1)
n+1
4β 2 (1 − α)(n + 2α − 1) − | (a)
|nk+1 [n(1 + β) + (1 + β(2α − 1))]2
(c)n+1
Proof : Since fi ∈ H ∗ (β, α, k), (i = 1, 2),then by Theorem 2.1 we have:
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))]
|
|
an,i ≤ 1, i = 1, 2.
(c)n+1
2β(1 − α)
n=1
Hence
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))] 2 2
|
) an,i
(|
(c)
2β(1
−
α)
n+1
n=1
≤(
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))]
|
|
an,i )2 ≤ 1, (i = 1, 2.)
(c)
2β(1
−
α)
n+1
n=1
Thus
∞
X
1 (a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))] 2 2
(|
|
) (an,1 + a2n,2 ) ≤ 1,
2
(c)
2β(1
−
α)
n+1
n=1
To prove the theorem, we must find the largest η such that
n+1 k+1
| (a)
n [n(1 + β) + (1 + β(2α − 1))]2
n(1 + η) + (1 + η(2α − 1))]
(c)n+1
≤
, n ≥ 1,
η
4β 2 (1 − α)
so that
η≤
4β 2 (α − 1)(n + 1)
n+1
4β 2 (1 − α)(n + 2α − 1) − | (a)
|nk+1 [n(1 + β) + (1 + β(2α − 1))]2
(c)n+1
.
On a Class of Meromorphic Univalent Functions...
71
Hence the proof is complete.
P∞
1
n
Theorem
4.3
:
If
f
(z)
=
+
∈ H ∗ (β, α, k), and g(z) =
n=1 an z
z
P∞
1
n
∗
+ n=1 bn z with |bn | ≤ 1 is in the class H (β, α, k), then f (z) ∗ g(z) ∈
z
H ∗ (β, α, k).
Proof : From Theorem 2.1 we have
∞
X
(a)n+1 k+1
|
|n [n(1 + β) + (1 + β(2α − 1))]an ≤ 2β(1 − α).
(c)n+1
n=1
Since
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))]
|
|an bn |,
|
(c)n+1
2β(1 − α)
n=1
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))]
|
an |bn |,
|
=
(c)
2β(1
−
α)
n+1
n=1
≤
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))]
|
an ≤ 1.
|
(c)
2β(1
−
α)
n+1
n=1
Thus f (z) ∗ g(z) ∈ H ∗ (β, α, k).
Hence the proof is complete.
P
n
Corollary
4.1 : If f (z) = z1 + ∞
∈ H ∗ (β, α, k), and g(z) =
n=1 an z
P
∞
1
n
∗
+ n=1 bn z with 0 ≤ bn ≤ 1 is in the class H (β, α, k), then f (z) ∗ g(z) ∈
z
H ∗ (β, α, k).
5
Radii of Starlikness and Convexity
Theorem 5.1 : Let the function f (z) defined by (1) be in the class H ∗ (β, α, k).
Then f is meromorphically starlike of order δ(0 ≤ δ < 1) in the disk |z| <
r1 (β, α, k, δ), where
r1 (β, α, k, δ) = inf {|
n
1
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))](1 − δ) n+1
|
}
(c)n+1
2β(n + 2 − δ)(1 − α)
The result is sharp for the function given by (7).
Proof : It is enough to show that
|
zf 0 (z)
+ 1| ≤ 1 − δ,
f (z)
P∞
P∞
+ 1)an z n
(n + 1)an |z|n+1
zf 0 (z)
n=1P
n=1 (n
P
|
+ 1| = | −1
|≤
.
n
n+1
f (z)
z + ∞
1− ∞
n=1 an z
n=1 an |z|
72
Abdul Rahman S. Juma et al.
This will be bounded by 1 − δ,
P∞
(n + 1)an |z|n+1
n=1P
≤ 1 − δ,
n+1
1− ∞
n=1 an |z|
∞
X
(n + 2 − δ)an |z|n+1 ≤ 1 − δ,
n=1
by Theorem 2.1, we have
∞
X
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))]
|
an ≤ 1.
|
(c)
2β(1
−
α)
n+1
n=1
Hence
|z|n+1 ≤ |
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))](1 − δ)
|
,
(c)n+1
2β(n + 2 − δ)(1 − α)
1
(a)n+1 nk+1 [n(1 + β) + (1 + β(2α − 1))](1 − δ) n+1
|z| ≤ {|
|
} .
(c)n+1
2β(n + 2 − δ)(1 − α)
This completes the proof of the theorem.
Theorem 5.2 : Let the function f (z) defined by (1) be in the class H ∗ (β, α, k).
Then f is meromorphically convex of order γ(0 ≤ γ < 1) in the disk |z| <
r2 (β, α, k, γ), where
r2 (β, α, k, γ) = inf {|
n
1
(a)n+1 nk (1 − γ)[n(1 + β) + (1 + β(2α − 1))] n+1
|
}
(c)n+1
2β(n + 2 − γ)(1 − α)
The result is sharp for the function given by (7).
Proof : By using the same technique in the proof of Theorem 5.1 we can show
that
zf 00 (z)
+ 2| ≤ 1 − γ, ((0 ≤ γ < 1).
| 0
f (z)
for |z| < r2 with the aid of Theorem 2.1, we have the assertion of Theorem
5.2.
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