Theoretical Mathematics & Applications, vol.2, no.3, 2012, 57-73
ISSN: 1792-9687 (print), 1792-9709 (online)
Scienpress Ltd, 2012
Harmonic Meromorphic Functions Involving
Generalized Incomplete Beta Functions
Amit Kumar Yadav1
Abstract
In this Article, a class MH ([α1 ]) of complex valued harmonic meromorphic functions of the form f = h + g ∈ MH is introduced with the
use of inverse function involving generalized incomplete beta function.
A subclass MH ([α1 ]) of MH ([α1 ]) is considered for various properties.
Using coefficient condition for functions belonging to MH ([α1 ]) class,
bounds, extreme points, closure theorems and integral operator for those
functions are also obtained.
Mathematics Subject Classification: 30C45, 30C50
Keywords: Harmonic functions, Meromorphic functions, Incomplete beta
function
1
Introduction
Hengartner and Schober [4] studied and gave the concept of the special
classes of harmonic functions, which are defined on the exterior of the unit
1
Department of Information Technology, Ibra College of Technology, PO Box 327,
Postal Code: 400, Alsharqiyah, Ibra, Sultanate of Oman, e-mail: amitkumar@ict.edu.om
Article Info: Received : August 13, 2012. Revised : October 16, 2012
Published online : December 30, 2012
58
Harmonic Meromorphic Functions ...
e = {z : |z| > 1}. They showed that these functions are complex valued,
disk U
harmonic, sense preserving, univalent mappings f , admits the representation
f (z) = h(z) + g(z) + A log |z| ,
e = {z : |z| > 1}.
where h(z) and g(z) are analytic in U
Let MH denote a class of functions which are harmonic meromorphic in
the unit disk U = {z : |z| < 1} and are of the form:
f (z) = h(z) + g(z),
where
∞
h(z) =
(1)
∞
X
1 X n
+
an z , and g(z) =
bn z n
z n=1
n=1
are meromorphic in U and h(z) has a simple pole at the origin with residue 1
there. The class MH is studied in [2], [5], [6] and [8]. Whereas MH denotes a
subclass of MH consisting of functions f = h + g, with h and g are of the form
h(z) =
∞
∞
X
1 X
+
|an | z n , z ∈ U \ {0} and g(z) = −
|bn | z n , z ∈ U
z n=1
n=1
(2)
and are called respectively meromorphic part and co-meromorphic part. A
?
function f = h + g ∈ MH is said to be in the class M SH
of meromorphically
harmonic starlike functions in U \ {0} if it satisfies the condition
)
(
zh0 (z) − zg 0 (z)
> 0 (z ∈ U ) .
Re −
h(z) + g(z)
?
Jahangiri and Silverman studied the class M SH
in [5]. For positive real
numbers αi and βi , i = 1, 2, 3, ...s, a generalized incomplete beta function,
φ ((αi )1,s , (βi )1,s , z) ≡ φ ([α1 ] , z) is defined as
φ ([α1 ] , z) = z s+1 Fs ((αi )1,s , 1; (βi )1,s ; z)
and its series representation is given as:
φ ([α1 ] , z) =
=
∞
X
(α1 )n ......(αs )n
(β1 )n ......(βs )n
n=0
∞
X
n=o
∇sn ([α1 ]) z n+1 ,
z n+1
(3)
59
Amit Kumar Yadav
where
∇sn ([α1 ]) :=
(α1 )n ......(αs )n
, s ∈ N = {1, 2, .....}
(β1 )n ......(βs )n
(4)
and (a)n is the Pochhammer symbol defined as:
(a)n =
Γ(a + n)
= a (a + 1) · · · · · · (a + n − 1)
Γ(a)
for n ∈ N = {1, 2, ....}.
A differential operator θ [8] on the function φ ([α1 ] , z) given in (3) is defined
as
d
θ (φ ((αi )1,s , (βi )1,s , z)) = z φ ((αi )1,s , (βi )1,s , z) .
dz
The series expansion of θ (φ ((αi )1,s , (βi )1,s , z)) is given as
∞
X
(α1 )n ......(αs )n n+1
θ (φ ([α1 ] , z)) =
(n + 1)
z
(β1 )n ......(βs )n
n=o
(5)
∞
X
=
(n + 1)∇sn ([α1 ]) z n+1 .
n=o
Throughout this paper, the following notations are being used:
θ (φ ([α1 ] , 1)) =
∞
X
(n + 1)∇sn ([α1 ]) =: θ (φ ([α1 ]))
(6)
n=o
and
∞
X
φ ([α1 ] , 1) =
∇sn ([α1 ]) =: φ ([α1 ]) ,
(7)
n=o
provided the corresponding series are absolutely convergent, i.e if
s
s
P
P
(βi − αi ) > 2 respectively.
(βi − αi ) > 1 and
i=1
i=1
Generalized hypergeometric functions are used to define harmonic functions
in various research papers such as [1] and [3]. For harmonic functions
∞
∞
1 X j n X j n
A z +
Bn z , j = 1, 2
fj (z) = +
z n=1 n
n=1
the convolution f1 ˜?f2 is defined by
∞
∞
1 X 1 2 n X 1 2 n
(f1 ˜?f2 ) (z) := +
A A z +
Bn Bn z .
z n=1 n n
n=1
60
Harmonic Meromorphic Functions ...
Corresponding to the function φ ([α1 ] , z), defined in (3), consider
1
φe1 ([α1 ], z) = 2 φ ([α1 ] , z) , z ∈ U \ {0}
z
−1
e
and its inverse function φ1 ([α1 ], z)
defined by
−1
e
e
φ1 ([α1 ], z) ? φ1 ([α1 ], z)
=
1
, z ∈ U \ {0}.
z(1 − z)
The series expansion of this inverse function is given as:
∞
−1
1 X (β1 )n+1 ........(βs )n+1 n
e
+
z
φ1 ([α1 ], z)
=
z n=0 (α1 )n+1 .........(αs )n+1
∞
where
1 X
1
=
+
zn
s
z n=0 ∇n+1 ([α1 ])
(α1 )n+1 .........(αs )n+1
.
(8)
(β1 )n+1 ........(βs )n+1
From the contiguous relation of Pochammer symbol (a)n+1 = a (a + 1)n+1 , it
is noted that
!
s
Y
α
i
∇sn+1 ([α1 ]) =
∇sn ([α1 + 1]) .
(9)
β
i
i=1
∇sn+1 ([α1 ]) =
Let
−1
H(z) :=
φe1 ([α1 ], z)
−
1
s
∇1 ([α1 ])
∞
1 X
1
zn
= +
s
z n=1 ∇n+1 ([α1 ])
(10)
which is meromorphic function in U. Again, corresponding to the function
φ ([α1 ] , z), defined in (3), consider
1
φe2 ([α1 ], z) = φ ([α1 ] , z) , z ∈ U \ {0}
z
−1
and its inverse function φe2 ([α1 ], z)
defined by
−1
e
e
φ2 ([α1 ], z) ? φ2 ([α1 ], z)
=
1
, z ∈ U \ {0}.
z(1 − z)
The series expansion of this inverse function is given as:
∞
−1
X
(β1 )n , ........, (βs )n n
e
φ2 ([α1 ], z)
=
z
(α
)
,
..........,
(α
)
1
n
s
n
n=0
=
∞
X
∇s
n=0 n
1
zn.
([α1 ])
61
Amit Kumar Yadav
Let
G(z) :=
φe2 ([α1 ], z)
−1
1
− s
∇0 ([α1 ])
=
∞
X
∇s
n=1 n
1
zn
([α1 ])
(11)
which is an analytic function, hence meromorphic in U. Now, harmonic meromorphic function defined as
F (z) = H(z) + G(z) ∈ MH ,
(12)
where H and G and are of the form (10) and (11) are called respectively
meromorphic part and co-meromorphic part of F (z). Using convolution “˜?”
of harmonic meromorphic functions F (z) = H(z) + G(z) given by (12) and
f (z) = h(z) + g(z) given by (1) , a linear operator Fs ([α1 ]) f (z) : MH → MH
is defined as:
Fs ([α1 ]) f (z) = F (z)e
?f (z) = H(z) ? h(z) + G(z) ? g(z)
∞
∞
X
1 X
1
1
n
=
+
bn z n .
a
z
+
n
s
s
z n=1 ∇n+1 ([α1 ])
∇ ([α1 ])
n=1 n
(13)
Involving operator Fs ([α1 ]) , a class MH ([α1 ]) is defined as follows:
Definition 1.1. Let MH ([α1 ]) denote the family of harmonic meromorphic
functions f (z) = h(z) + g(z) ∈ MH satisfying
z (Fs ([α1 ]) f (z))0
Re −
>0
(z ∈ U ) ,
(14)
Fs ([α1 ]) f (z)
or

0 
0


 z (H(z) ? h(z)) − z G(z) ? g(z) 
>0
Re −


H(z) ? h(z) + G(z) ? g(z)


(z ∈ U ) .
Denote MH ([α1 ]) = MH ([α1 ]) ∩ MH .
2
Coefficient Conditions
In this section, sufficient coefficient condition for the class MH ([α1 ]) is
established and then it is proved that this coefficient condition is necessary for
its subclass MH ([α1 ]) .
62
Harmonic Meromorphic Functions ...
Theorem 2.1. Let f (z) = h(z) + g(z) be of the form (1) and if
∞
X
n
n=1
1
1
|an | + s
|bn |
s
∇n+1 ([α1 ])
∇n ([α1 ])
≤1
(15)
where ∇sn ([α1 ]) and ∇sn+1 ([α1 ]) are given in (4) and (8) respectively, for poss
P
(βi − αi ) > 1, then f (z)is harmonic,
itive real numbers αi and βi with
i=1
orientation preserving and univalent in U \ {0} and f ∈ MH ([α1 ]).
Proof. Let the function f (z) = h(z)+g(z) given by (1) , satisfying (15). Under
s
P
(βi − αi ) > 1, it follows that 0 < ∇sn+1 ([α1 ]) < ∇sn ([α1 ]) < 1
the condition
i=1
for all n ≥ 1. Then for 0 < |z1 | ≤ |z2 | < 1,
|f (z1 ) − f (z2 )| ≥ |h(z1 ) − h(z2 )| − |g(z1 ) − g(z2 )|
∞
X
|z1 − z2 |
− |z1 − z2 |
(|an | + |bn |) z1n−1 + · · · + z2n−1 ≥
|z1 | |z2 |
n=1
#
"
∞
X
|z1 − z2 |
>
n (|an | + |bn |)
1 − |z2 |2
|z1 | |z2 |
n=1
"
#
∞
X
|z1 − z2 |
1
1
>
|an | + s
|bn |
1−
n
s
|z1 | |z2 |
∇
([α
])
∇
([α
])
1
1
n+1
n
n=1
≥ 0,
if (15) holds, then f is univalent in U \ {0} .
In order to show that f is sense preserving in U \ {0}, it only needs to show
that |h0 (z)| > |g 0 (z)| . For 0 < |z| = r < 1, on using (15) , it follows that
∞
X
1
−
n |an | |z|n−1
|h (z)| ≥
2
|z|
n=1
∞
X
1
n |an | rn−1
> 2−
r
n=1
0
> 1−
∞
X
n=1
n |an |
63
Amit Kumar Yadav
∞
X
≥ 1−
n
1
|an |
s
∇n+1 ([α1 ])
n=1
∞
X
1
|b |
≥
n
s ([α ]) n
∇
1
n
n=1
>
∞
X
n |bn | >
n=1
=
∞
X
∞
X
n |bn | rn−1
n=1
n |bn | |z|n−1 > |g 0 (z)| ,
n=1
which proves that the map f is sense preserving in U \ {0}.
Now, in order to show that f ∈ MH ([α1 ]) , it suffices to show that

0 
0


 z (H(z) ? h(z)) − z G(z) ? g(z) 
Re −
> 0.


H(z) ? h(z) + G(z) ? g(z)


(16)
It is known that Re (p(z)) > 0, if and only if p(z)−1
< 1 for an analytic
p(z)+1 function p(z) = 1 + p1 z + p2 z + ....
Let
0
(17)
A(z) := −z (H(z) ? h(z))0 + z G(z) ? g(z)
and
B(z) := H(z) ? h(z) + G(z) ? g(z).
(18)
It is observe that (16) holds if
|A(z) + B(z)| − |A(z) − B(z)| > 0.
(19)
Now from (17) and (18), it follows that
|A(z) + B(z)|
0
0
= −z (H(z) ? h(z)) + z G(z) ? g(z) + H(z) ? h(z) + G(z) ? g(z)
∞
∞
2 X
X
1
1
bn z n an z n +
(n − 1) s
(n + 1) s
= −
z
∇n+1 ([α1 ])
∇n ([α1 ])
n=1
∞
X
n=1
∞
X
1
1
2
n
−
|an | |z| −
|bn | |z|n
(n − 1) s
(n + 1) s
≥
|z| n=1
∇n+1 ([α1 ])
∇n ([α1 ])
n=1
64
Harmonic Meromorphic Functions ...
and
|A(z) − B(z)|
0
0
= −z (H(z) ? h(z)) + z G(z) ? g(z) − H(z) ? h(z) − G(z) ? g(z)
∞
∞
X
X
1
1
n
n
(n + 1) s
= bn z an z −
(n − 1) s
n=1
∇n+1 ([α1 ])
∇n ([α1 ])
n=1
≤
∞
X
(n + 1)
n=1
∞
X
1
1
n
|a
|
|z|
+
|bn | |z|n .
(n − 1) s
n
s
∇n+1 ([α1 ])
∇
([α
])
1
n+1
n=1
Thus, from (15)
≥
≥
≥
≥
|A(z) + B(z)| − |A(z) − B(z)|
∞
∞
X
X
1
1
2
|bn | |z|n
−2 n s
|an | |z|n − 2 n s
|z|
∇n+1 ([α1 ])
∇n ([α1 ])
n=1
( n=1∞
)
∞
X
X
2
1
1
|bn | |z|n+1
|an | |z|n+1 −
1−
n s
n s
|z|
∇n+1 ([α1 ])
∇n ([α1 ])
n=1
n=1
(
)
∞
X
1
1
2 1−
n s
|bn |
|an | + n s
∇
∇
([α
])
([α
])
1
1
n+1
n
n=1
(
)
∞
X
1
1
2 1−
n
|an | + s
|bn |
s
∇
([α
])
∇
([α
])
1
1
n+1
n
n=1
≥ 0.
This proves the result.
Now, we prove that the condition (15) is necessary for functions f ∈MH ([α1 ]) .
Theorem 2.2. Let f (z) = h(z) + g(z) ∈ MH with h and g are of the form
(2) , then f ∈ MH ([α1 ]) if and only if
∞
X
n
n=1
1
1
|an | + s
|bn |
s
∇n+1 ([α1 ])
∇n ([α1 ])
≤ 1,
where ∇sn ([α1 ]) and ∇sn+1 ([α1 ]) are given in (4) and (8) respectively.
(20)
65
Amit Kumar Yadav
Proof. In view of Theorem 2.1, it only needs to prove the “only if”’ part of the
Theorem. Since MH ([α1 ]) ⊂ MH ([α1 ]), it suffices to show that f ∈
/ MH ([α1 ])
if the condition (20) does not hold. If f ∈ MH ([α1 ]) then

0 
0


 z (H(z) ? h(z)) − z G(z) ? g(z) 
>0
Re −


H(z) ? h(z) + G(z) ? g(z)


is equivalent to


∞
∞
P
P

2
1
1
n
n

 z − 2 n ∇sn+1 ([α1 ]) |an | z − 2 n ∇sn ([α1 ]) |bn | z 

n=1
n=1
Re
> 0.
∞
∞
P
P


1
1
1
n
n


+
|a
|
z
−
|b
|
z
 z

n
∇s ([α1 ])
∇sn ([α1 ]) n
n=1
Since
n+1
(21)
n=1
ξ(z) ≥ Re ξ(z) ≥ 0,
η(z) η(z)
hence the condition (21) holds if
1−
∞
P
1
n+1 ([α1 ])
|an | |z|n+1 −
1
∇sn+1 ([α1 ])
|an | |z|n+1 +
n ∇s
n=1
∞
P
1+
n=1
∞
P
1
|bn | |z|n+1
n ∇s ([α
1 ])
n=1
∞
P
n=1
n
≥ 0.
1
∇sn ([α1 ])
(22)
|bn | |z|n+1
Now, if the condition (20) does not holds, then the numerator of above
equation is negative for z sufficiently close to 1. Which contradicts the required
condition for f ∈ MH ([α1 ]) and this proves the required result.
Corollary 2.3. If f ∈ MH ([α1 ]) then
!
s
∞
Y
X
αi
|an | ≤
(ζ(2)) [{θ (φ ([α1 + 1]))} − {φ ([α1 + 1])}]
β
n=1
i=1 i
and
∞
X
|bn | ≤ (ζ(2)) [{θ (φ ([α1 ]))} − {φ ([α1 ])}] ,
n=1
where ζ(2) =
∞
P
n=1
1
n2
is called a Zeta function and θ (φ ([α1 ])), φ ([α1 ]) are given
in (6) and (7) provided
s
P
i=1
(βi − αi ) > 2.
66
Harmonic Meromorphic Functions ...
Proof. On using relation given in (9), (6) and (7), from Theorem 2.2 it follows
that
∞
X
|an | ≤
n=1
∞
X
∇sn+1 ([α1 ])
n=1
∞ X
n
!
Y
s
1
αi
1
∇sn ([α1 + 1])
(n + 1) − 2
=
2
n
n
β
n=1
i=1 i
#
!
"
s
∞
∞
Y
X
X
αi
1
1 s
s
=
(n + 1)∇n ([α1 + 1]) −
∇
([α1 + 1])
β
n2
n2 n+1
n=1
n=1
i=1 i
!
s
Y
αi
(ζ(2)) [{θ (φ ([α1 + 1]))} − {φ ([α1 + 1])}]
≤
β
i=1 i
and
∞
X
∞
X
∇sn ([α1 ])
n
n=1
∞
X 1
1
=
(n + 1) − 2 ∇sn ([α1 ])
2
n
n
n=1
|bn | ≤
n=1
∞
∞
X
X
1
1 s
s
≤
(n + 1)∇n ([α1 ]) −
∇ ([α1 ])
2
n
n2 n
n=1
n=1
≤ (ζ(2)) [{θ (φ ([α1 ]))} − {φ ([α1 ])}] .
3
Bounds
In this section, bounds for functions belonging to the class MH ([α1 ]) are
determined with the use of Theorem 2.2 and Corollary 2.3.
Theorem 3.1. Let f ∈ MH ([α1 ]) , then for
r<1
s
Y
1
αi
−r
r
β
i=1 i
!
s
P
(βi − αi ) > 1 and 0 < |z| =
i=1
s
Y
1
αi
≤ |f (z)| ≤ + r
r
β
i=1 i
!
.
Amit Kumar Yadav
67
Proof. Let f ∈ MH ([α1 ]), taking the absolute value of f defined in (2) and
using Theorem 2.2, it follows that
∞
∞
1 X
X
n
n
|an | z −
|bn | z |f (z)| = +
z n=1
n=1
≤
∞
∞
X
1 X
+
|an | rn +
|bn | rn
r n=1
n=1
∞
X
1
+ r (|an | + |bn |)
≤
r
n=1
∞
X
1
1
s
(|an | + |bn |)
+ r∇1 ([α1 ])
n s
≤
r
∇n ([α1 ])
n=1
∞
X
1
1
1
s
≤
+ r∇1 ([α1 ])
|bn |
|an | + s
n
s
r
∇
∇
([α
])
([α
])
1
1
n+1
n
!n=1
s
Y
αi
1
+r
≤
r
β
i=1 i
and
∞
∞
1 X
X
n
n
|bn | z |an | z −
|f (z)| = +
z n=1
n=1
∞
∞
X
1 X
n
−
|an | r −
|bn | rn
≥
r n=1
n=1
∞
X
1
− r (|an | + |bn |)
≥
r
n=1
∞
X
1
1
− r∇s1 ([α1 ])
(|an | + |bn |)
n s
r
∇
([α
])
1
n
n=1
∞
X
1
1
1
s
≥
− r∇1 ([α1 ])
|an | + s
|bn |
n
s
r
∇
([α
])
∇
([α
])
1
1
n+1
n
!n=1
s
Y
1
αi
.
≥
−r
r
β
i
i=1
≥
This proves the required result.
Using inequalities obtained in Corollary 2.3, following functional bounds
are estimated.
68
Harmonic Meromorphic Functions ...
Theorem 3.2. If f ∈ MH ([α1 ]) then for 0 < |z| = r < 1

 s Q αi
[{θ (φ ([α1 + 1]))} − {φ ([α1 + 1])}]
1

|f (z)| ≤ + rζ(2)  i=1 βi
r
+ [{θ (φ ([α1 ]))} − {φ ([α1 ])}]
and
 1
|f (z)| ≥ − rζ(2) 
r
where ζ(2) =
∞
P
n=1
1
n2
s
Q
αi
i=1
βi
[{θ (φ ([α1 + 1]))} − {φ ([α1 + 1])}]
+ [{θ (φ ([α1 ]))} − {φ ([α1 ])}]


is called a Zeta function and θ (φ ([α1 ])), φ ([α1 ]) are given
in (6) and (7) provided
s
P
(βi − αi ) > 2.
i=1
Proof. Let f ∈ MH ([α1 ]), using Corollary 2.3 and taking the absolute value
of f of the form (2), it follows that
∞
∞
1 X
X
n
n
|f (z)| = +
|an | z −
|bn | z z n=1
n=1
≤
and
∞
∞
X
1 X
+
|an | rn +
|bn | rn
r n=1
n=1
∞
X
1
+ r (|an | + |bn |)
≤
r
n=1
 s 
Q αi
[{θ (φ ([α1 + 1]))} − {φ ([α1 + 1])}]
1

+ rζ(2)  i=1 βi
≤
r
+ [{θ (φ ([α1 ]))} − {φ ([α1 ])}]
∞
∞
1 X
X
n
n
|a | z −
|bn | z |f (z)| = +
z n=1 n
n=1
∞
∞
X
1 X
n
−
|an | r −
|bn | rn
≥
r n=1
n=1
∞
X
1
− r (|an | + |bn |)
≥
r
n=1

 s Q αi
[{θ (φ ([α1 + 1]))} − {φ ([α1 + 1])}]
1

≥
− rζ(2)  i=1 βi
r
+ [{θ (φ ([α1 ]))} − {φ ([α1 ])}]
This proves the required result.
69
Amit Kumar Yadav
Remark: The functional bounds obtained in Theorem 3.2 are best possible
for f ∈ MH ([α1 ]) .
In view of above remark, the following result is obtained which is sharp:
Corollary 3.3. Let f ∈ MH ([α1 ]) then

 s 
Q αi

[{θ (φ ([α1 + 1]))} − {φ ([α1 + 1])}] 
βi


w : |w| < 1 − ζ(2)
i=1


+ [{θ (φ ([α1 ]))} − {φ ([α1 ])}]
⊆ C \ f (U \ {0}) .
4
Extreme Points
In this section, extreme points for the class MH ([α1 ]) are provided.
Theorem 4.1. Let f = h + g, where h and g are of the form (2) then
f ∈ MH ([α1 ]) , if and only if f can be expressed as
∞
X
f (z) =
(xn hn (z) + yn gn (z)) ,
(23)
n=0
where z ∈ U \ {0} and
1
h0 (z) = ,
z
hn (z) =
1 ∇sn+1 ([α1 ]) n
+
z , n = 1, 2...
z
n
(24)
1 ∇sn ([α1 ]) n
z , n = 1, 2...
−
z
n
(25)
for
1
g0 (z) = ,
z
and
∞
X
gn (z) =
(xn + yn ) = 1,
xn ≥ 0 and yn ≥ 0.
(26)
n=0
Proof. Let
∞
X
f (z) =
(xn hn (z) + yn gn (z))
n=0
1 ∇sn+1 ([α1 ]) n
1 ∇sn ([α1 ]) n
= x0 h 0 + y0 g 0 +
xn
z
+
z + yn
−
z
n
z
n
n=1
s
∞ s
X
∇n+1 ([α1 ])
1
∇n ([α1 ])
=
+
xn −
yn
zn.
z n=1
n
n
∞
X
70
Harmonic Meromorphic Functions ...
Thus by Theorem 2.2, it follows that f ∈ MH ([α1 ]), since
s
s
∞ X
∇n+1 ([α1 ])
1
∇n ([α1 ])
1
xn − n s
yn
n s
∇
([α
])
n
∇
([α
])
n
1
1
n+1
n
n=1
=
∞
X
(xn + yn ) = (1 − x0 − y0 ) ≤ 1.
n=1
Conversely, suppose that f ∈ MH ([α1 ]) . Set
1
|an | ,
∇sn+1 ([α1 ])
1
= n s
|bn |
∇n ([α1 ])
xn = n
yn
which satisfy (26), thus
∞
∞
X
1 X
n
+
|an | z −
|bn | z n
f (z) =
z n=1
n=1
∞
∞
X
∇sn ([α1 ])
1 X ∇sn+1 ([α1 ])
+
xn z n −
yn z n
z n=1
n
n
n=1
∞ ∞ X
X
1
1
1
xn +
yn
gn −
+
hn −
=
z n=1
z
z
n=1
#
"
∞
∞
∞
∞
X
X
X
X
1
=
h n xn +
1−
g n yn
xn −
yn +
z
n=1
n=1
n=1
n=1
=
∞
∞
X
X
= x0 h 0 + y0 g 0 +
h n xn +
g n yn
n=1
=
∞
X
n=1
(xn hn + yn gn ) .
n=0
This proves the Theorem.
Remark: The extreme points for the class MH ([α1 ]) are given by (24)
and (25).
5
Closure Theorems
In this section, convolution of the class MH ([α1 ]) and convex linear combination of its members are defined and studied.
71
Amit Kumar Yadav
Theorem 5.1. Let f ∈ MH ([α1 ]) and F ∈ MH ([α1 ]), then the convolution
function
∞
∞
X
1 X
n
˜
|an An | z −
|bn Bn | z n ∈ MH ([α1 ]) .
f ?F = +
z n=1
n=1
Proof. Since F ∈ MH ([α1 ]), then by Theorem 2.2, |An | ≤ 1 and |Bn | ≤ 1,
hence
∞
X
1
1
|bn Bn |
|an An | + s
n
∇sn+1 ([α1 ])
∇n ([α1 ])
n=1
∞
X
1
1
≤
n
|bn | ≤ 1
|an | + s
∇sn+1 ([α1 ])
∇n ([α1 ])
n=1
by Theorem 2.2 as f ∈ MH ([α1 ]). Thus, again by Theorem 2.2, f ˜?F ∈
MH ([α1 ]).
Theorem 5.2. Let the functions fj (z) defined as
∞
∞
X
1 X
n
|an,j | z −
|bn,j | z n
fj (z) = +
z n=1
n=1
(27)
be in the class MH ([α1 ]) for every j = 1, 2, 3......., then the function
∞
X
ψ(z) =
cj fj (z)
n=1
is also in the class MH ([α1 ]), where
Proof. It is noted that
∞
P
cj = 1 ,cj ≥ 0 (j = 1, 2, 3.......).
n=1
!
!
∞
∞
∞
∞
X
X
1 X X
cj |an,j | z n −
cj |bn,j | z n .
ψ(z) = +
z n=1 j=1
n=1
j=1
Since fj (z) ∈ MH ([α1 ]) for every j = 1, 2, 3......., then by Theorem 2.2, it
follows that
!#
!
"
∞
∞
∞
X
X
X
1
1
cj |bn,j |
cj |an,j | + n s
n s
∇n+1 ([α1 ]) j=1
∇n ([α1 ]) j=1
n=1
!
∞ ∞
X
X
1
1
|bn,j |
|an,j | + n s
n s
=
cj
∇
∇
([α
])
([α
])
1
1
n+1
n
n=1
j=1
≤
∞
X
j=1
cj ≤ 1
72
Harmonic Meromorphic Functions ...
hence, ψ(z) ∈ MH ([α1 ]), which is the desired result.
6
Integral Operator
In this section, it is shown that the class MH ([α1 ]) is closed under Integral
operator.
Definition 6.1. An integral operator I : MH → MH is defined as:
If (z) =
c
z c+1
Z
z
c
c
t h(t)dt +
0
z c+1
Z
z
tc g(t)dt, for c > 0, z ∈ U \ {0} . (28)
0
Theorem 6.2. Let f ∈ MH ([α1 ]) and If (z) be defined in (28), then If (z) ∈
MH ([α1 ]) .
Proof. From the series representation of I, it follows that
∞
∞
X
1 X
c
c
n
If (z) = +
|an | z −
|bn | z n .
z n=1 n + c + 1
n
+
c
+
1
n=1
Since, f ∈ MH ([α1 ]) , by Theorem 2.2, If (z) ∈ MH ([α1 ]) , since
∞
X
c
n
n
+
c
+
1
n=1
∞
X
≤
n
n=1
1
1
|bn |
|an | + s
s
∇n+1 ([α1 ])
∇n ([α1 ])
1
1
|bn |
|an | + s
s
∇n+1 ([α1 ])
∇n ([α1 ])
≤ 1.
Amit Kumar Yadav
73
References
[1] H.A. Al-Kharsani and R.A. Al-Khal, Univalent Harmonic Functions,
Journal of Inequa. in pu. and App. Math., 8, (Art59), (2007), 8 pp.
[2] H. Bostanci and M. Ozturk, New classes of salagean type meromorphic
harmonic functions, Inter. jour. of math. scie., 2, (2008), 52-57.
[3] J. Dziok and H.M. Srivastava, Classes of analytic functions associated
with the generalized hypergeometric functions, Appl. Math. Comp., 103,
(1999), 1-13.
[4] W. Hengartner and G. Schober, Univalent Harmonic Functions, Trans.
Amer. Math. Soc., 299, (1987), 1-31.
[5] J.M. Jahangiri and H. Silverman, Harmonic Meromorphic univalent Functions with Negative Coefficients, Bull. Korean Math. Soc., 36, (1999),
291-301.
[6] J.M. Jahangiri, Harmonic Meromorphic Starlike Functions, Bull. Korean
Math. Soc., 37, (2001), 763-770.
[7] E.D. Rainvillie, Special Functions, Chelsea Publishing Company, New
York, 1960.
[8] K.Al. Shaqsi and M. Darus, On meromorphic harmonic functions with
respact to k - symmetric points, Jour. of Inequ. and Apps., Art. 259205
(2008).