 Open Access
 Total Downloads : 25
 Authors : B. B. Jaimini, Jyoti Gupta
 Paper ID : IJERTCONV3IS31009
 Volume & Issue : ATCSMT – 2015 (Volume 3 – Issue 31)
 Published (First Online): 24042018
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
On Certain Subclasses of Meromorphic Univalent Funtions with Positive Coefficients
B. B. Jaimini 1
1 Department of Mathematics, Govt. P.G. College, Kota 324001(India)
Jyoti Gupta 2
2 Department of Mathematics, Maharishi Arvind International Institute of
Technology, Kota
Abstract: In this paper we have introduced a subclass
An, q,, of meromorphic univalent functions with
MAIN RESULTS
positive coefficients in the punctured unit disk
We establish the following ten properties for a function
f z
U z C : 0 z 1. Coefficient estimate, distortion theorem, radii of starlikeness and convexity, closure theorems and Hadamard product of functions belonging to this class are
obtained. Further properties using integral operators are also
belonging to in the class condition given in (5.1.4)
Coefficient Estimate
An, q,, defined under
obtained for the same class.
Let denote the class of meromorphic functions in the punctured
Theorem1 : Let the function f z defined by (5.1.1) be in the class . Then the function f z belong to the class
unit disk U * z C : 0 z 1 of the form
An, q,, if and only if
k! k qn k qa
f z = 1 + a zk
(k q)! k
z
z
k
k 1
k 1
q
n
a 0 ; k N {1,2,..}
(5.1.1)
1
q! q 1 1 q
(5.2.1)
k
For each
f z we define the following differential
( 0 1 q ; 0 ; n, q N0 )
operator
Proof: Let us suppose that the inequality (5.2.1) hold true. Then in
D
D
n f q z = D [ Dn1
f q z]
view of condition given in (5.1.4) and z 1we have
q q 1n
k!
n k q
n (q)
= 1
q! zq1
+ (k q)! k q
ak z
z D f
(z)
(1 q)
k 1
(5.1.2)
Dn f (q) (z)
q, n N where f q z is the qth derivative of
f z
k! ( k q)n (k 1)a
k 1
k 1
0
defined in (5.1.1) and
(k q)! k
z f z'
(1)q ( q 1)n q! k! ( k q)n a
D[ f z]
z 1
(5.1.3)
(k q)! k
k 1
k 1
(1 q )
With help of the differential operator Dn , we say that a
Therefore the values of the functions
function only if
f z belongs to is in the class
An, q,, if and
z
zDn f (q) (z)
Dn f (q) (z)
(5.2.2)
zDn f (q) (z)
Re
Dn f (q) (z)
(5.1.4)
lie in a circle which is a centered at 1 q and
0
0
where z U * ; 0 ; 0 1 q ; n, q N
and n
whose radius is 1 q .
D
D
Hence the function satisfies the condition given in (5.1.4).
is defined in (5.1.2).
Now conversely, assume that the function
f z is in the class
(m)
m! 1q q!( q 1)n (1 q )
z m
An, q,, . Then, we have f (1)m z
z
z
zDn f (q) (z)
(1 q)n (1 q)
(5.2.8)
Re
=
Dn f (q) (z)
z U;0 (1 q) ; q, n, m N0; 0)
1q q 1n q!q 1 k! k qn k qa z k 1
The result is sharp for the function
f z given by
(k q)!
k!
k
1 1 q!( q 1) (1 q )(1 q)!
Re
k 1
1q q 1n q!
k qn a z k 1
q n
f (z) z
k 1 (k q)!
k
(5.2.3)
z (1 q)n (1 q)
(5.2.9)
Proof: The function f z is in the class An, q,, then in
for some ( 0 1 q ; 0 ; n, q N0 ) and z U * choose value of z on the real axis so that z given by (5.2.2) is real. Upon clearing the denominator in (5.2.3) and
view of the assertion (5.2.1) of Theorem1 we can see that
(1 q)n (1 q) k!a
letting z 1
through the real values we can see that inequality in
(1 q)!
k
k 1
(5.2.3) lead to inequality (5.2.1). It completes the proof of
k! ( k q)n (k q)a
Theorem1.
k 1
(k q)! k
Theorem 2 : Let the function
f z defined by (5.1.1) be in the
1q q!( q 1)n (1 q )
class An, q,, . Then
which evidently yields
(1)q q!( q 1)n (1 q )
k!
k!
ak
( k q)n (k q)
(k q)!
…(5.2.5)
k!ak
k 1
1q q!( q 1)n (1 q )(1 q)! (1 q)n (1 q)
(5.2.10)
Now on differentiating both sides of (5.2.1) m times , we have
( k 1; q, n N0 ; 0 )
(m)
m m! k!
k
k
The result is sharp for the function
f z given by
f (z) = (1)
m1
z
z
k 1
(k q)!
a zk m
1 (1)q q!( q 1)n (1 q ) k
(5.2.6)
(5.2.11)
f (z)
z k!
(k q)!
z
( k q)n (k q)
Now taking the modulus of both sides of (5.2.11) and using (5.2.10) we at once arrive at the desired results in (5.2.7) and
k 1; q, n N0 ; 0
(5.2.8).
This completes the proof of Theorem3.
Proof: As f z An, q,, therefore in (5.2.1) k th term will be less then equal to the sum on L.H.S. of (5.2.1). Therefore (5.2.5) is true for the function defined in (5.2.6).
Distortion Theorem
Radii of Starlikeness and Convexity
Theorem 4: Let the function defined by (5.1.1) be in the class
An, q,, . Then

f z is meromoriphically starlike of order 0 1 in
Theorem3 : If the function
An, q,, then
f z defined by (5.1.1) is in the class
z r1 , where
1
k 1
k 1
r =
k! k q n k q
(1 )
k 1
f (m) (z)
1 Inf k q!1q q 1n q!1 q
(k 2 )
(1)
m m! z
1q q!( q 1)n (1 q )
(1 q)n (1 q)
m
z
z
z
(5.2.7)
( k 1; q, n N0 ; 0 )
(5.2.12)
and

f z is meromoriphically convex of order 0 1 in
z r2 , where
r2 = Inf
k! k qn k q
1
(1 ) k 1
f z = 1 + a zk
for z U
(5.2.18)
q n
j
k , j
k 1 k q!1 q 1 q!1 q
k(k 2 )
(5.2.13)
z k 1
( k 1; q, n N0 ; 0 )
Each of these results is sharp for the function
f z given
be in the class An, q,, for every Then the function F z defined by
j 1,2,…, m .
by (5.2.6)
Proof: Let
f z An, q,, . Then by Theorem2 we have
F z = 1
z
k
k
+ bk z
k1
belongs to the class
An, q,, where
1q q 1n 1 q q!
(5.2.14) 1
m
m
ak k!
k q!

k q
n

k q
bk = ak , j k 1
(k N )
To obtain the radius of starlike function (5.1.1) given in (5.2.12) it is sufficient to show that
Proof: Since
f j (z) An, q,, , it follows from
zf '(z) 1
f (z)
(1 )
(5.2.15 )
Theorem1 that
k!
k , j
k , j
q n
The L.H.S. of (5.2.15):
zf '(z)
k 1a zk
k 1ak
z k 1
k 1
(k q)!
( k q)n (k q)a
(1) q!( q 1) (1 q )
k
k
(5.2.19)
1 =
k 1
k 1
for every
j 1,2,.., m . Hence
f (z)
1
z k 1
a zk
1 ak
k 1
z k 1
k 1
k! k qn k q bk
(k q)!
k
k
Then in view of (5.2.15) this will be bounded by ( 1 ) therefore
1 m
k! k qn k qa
m
m
on
k 2 ak
z k 1
j 1
(k q)!
(1)q q!( q 1)n (1 q )
k , j
k 1
k 1
k 1 1 (1 )
(5.2.16)
By Theorem 1, it follows that F z An, q,, .
is true if
In view of (5.2.14) it follows that the inequality in (5.2.16)
Theorem 6: The class
An, q,, is closed under
(k 2 ) z k 1
(1 )
convex linear combination.
Proof: Let the functions f j (z) j 1,2 defined by
k! k qn k q
(k q)!1q q 1n q!1 q
(5.2.17)
(5.2.18) be in the class An, q,,
the function
. It is sufficient to show that
H z = t
f1z + 1 t f z
0 t 1
2
2
Setting z r1 in (5.2.17) we get desired result in (5.2.12).
is also belongs to the class An, q,, .
Similarly, to prove that f z is meromorphically convex of
Since 0 t 1
order it is sufficient to show that
H z= 1 +
[ ta1 ta
] zk
zf "(z) 2
f ' (z)
(1 )
z
We observe that
k 1
k ,1
k ,2
for radius
( k 1; q, n N0 ; 0 )
z r2 given in (5.2.13).
k 1
k! (k q)!
k qn k q{
tak ,1
1 tak ,2 }
Closure Theorems
Theorem 5 : Let the function by
k ,1
k ,1
f j (z)
( j 1,2,….) defined
= t
k 1
k!
(k q)!
k qn k q a
+ 1 t
k!
(k q)!
k qn k q a
0 1 k
k 1
k ,2
k ,2
(1)q q!( q 1)n (1 q )
Now, it follows that
f z = f z
By Theorem 1, it follows that H z An, q,, .
k k
k 0
Theorem 7: Let
f z = 1 and
0 z
This completes the proof of Theorem7.
Integral Operator for function
1 1q q 1n 1 q q!
Theorem 8: Let the function
f z given by (5.1.1) be in
fk (z)
z k!
zk
k qn k q
An, q,, . Then the integral operator
(k q)!
for k 1,2,…. Then
f z An, q,, if and
F z
1
= cuc f
0
uzdu
(5.2.21)
only if f z can be expressed in the form
0 u 1;0 c
f z =
k 0
k fk
z
is in An, q,, where
= 1 q1 qc 2 1 q1 q c
(5.2.22)
where k 0 and
k 1
k 0
c1 q c 21 q
Proof: Let
f z f z,
The result is sharp for the function
f z given by
Then in view of
k 0
k k
we have
f (z)
1 1q q 1n q!1 q

1 z
(5.2.23)
f z 1
k 1
k 0
1q q 1n 1 q q!
z 1 qn 1 q
(1 q)!
z
z
k 1
zk k k! k qn k q
(k q)!
Proof : Let f z An, q,, . Then for
1
Now we have
F z cuc f uzdu
0
1q q 1n 1 q q!
k! ( k q)n ( k q)n
(k q)!
we have
F z 1
c a zk
k 1
k k! k qn k q
(1)q ( q 1)n (1 q )q!
k
z c k 1
(k q)!
k 1
Since
Since
f
f
z
z
A n, q,,
A n, q,,
. We have from Theorem.1
. We have from Theorem.1
= k = 1 0 1 (5.2.20)
k! k qn k qa
k
k
k 1
Therefore from equation (5.2.20) we conclude in view of
i.e.
k 1 (k q)! 1
q n
Theorem1 that f z An, q,, .
1 q 1 q! 1 q
Conversely : As f z An, q,, then in view of Theorem2 we have
Now for function F z in view of Theorem 1, It is sufficient to show that the largest satisfies
k! k qn k q c a
1q q 1n 1 q q!
(k q)!
c k 1 k
ak
k 1
q n
1
k! k qn k q
(k q)!
1 q 1 q! 1 q
(5.2.24)
(k N )
Therefore, the value of
satisfies the range
On setting
k q c k q
(5.2.25)
k! k qn k q
1 q c k 1 1 q
(k q)! a ,
k
k
k
1q q 1n 1 q q!
for each k N . From (5.2.25), we obtain
We have k 1
k 0
k q1 qc k 1 ck q1 q = H k
therefore it is true that
c1 q c k 1k q
We see that H k is increasing therefore H k H (1)
Now in the light of inequality (5.2.30) we have
it means H (1) H k .
Now the result in (5.2.21) follows immediately.
1q q 1n q!1 q
k! k qn k q k q !
k q1 q
k q1 q
..(5.2.31)
Modified Hadamard Product
Theorem9: Let the function
f j z j
1,2
defined by (5.2.18)
On simplifying the inequality (5.2.31) we obtain
G(k)
where the function Gk is
be in the class An, q,, . Then the modified Hadamard
G(k) 1 q
product (or convolution) of f1z and f2 z by
1q q 1n q!1 q 2 (k 1)
(5.2.32)
1 k!
k
n
k
n
2 q n 2
f1 * f2 (z)

ak ,1ak ,2 z
z
z
k 1
(5.2.26)
k
q! k q k q 1 q 1 q!1 q
f1 f 2 z An, q,, where
(1 q)
We see that G(k) is an increasing function of k , therefore G1 G(k)
21q q 1n q!1 q 2
(5.2.27)
i.e. G(1)
which means in view of (5.2.32) at k 1 gives (5.2.27).
1
1
1 q! 1
qn 1 q2 1q q 1n
q!1 q
2
5.3 APPLICATIONS
The result is sharp for the functions
f j z j 1,2
As application of the theorems established in this section contain
given by
certain known and new results for the known class
* ( ) of
f (z)
1 1q q 1n q!1 q
k
k
z
(5.2.28)
univalent meromorphic functions with positive coefficients.
j z k!
n
We illustrate some results deduced from our main theorems as
(k N )
(k q)! k q k q
follows:

For the choice of
n q 0
in Theorem1, we get
Proof: In order to prove Theorem 9, we have to find the largest
in view of Theorem 2 and (5.2.26) i.e.
known corollary due to Kavitha et al [3, p.111].

If in Theorem2 we put n q 0 then it reduces to the
k!
(k q)!
k qn k q
a a 1
(5.2.29)
known corollary due to Kavitha et al [3, p.111].
1q q 1n q!1 q
k ,1 k ,2

For
n q 0
and
m 0in Theorem3, we get the
Since f1z and f2 z are in An, q,, then for
f1z
following corollary:
and f2 z we have the following inequalities
Corollary1. If f z ( ) then
k! k qn k q
k 1 (k q)! a 1
1 – 1
f (z)
1 + 1
(5.3.1)
1q q 1n q!1 q
k ,1
z 1 z
1
and
k! k qn k q
k 1 (k q)! a 1
( z U * ; 0 1)
The result is sharp for the function
f z
given in
1q q 1n q!1 q k ,2
(5.4.5).
Now by Cauchy Schwartz inequality and then in view of Theorem – 2, we have

At n q 0 in Theorem4 provides the following corollary:
k! k qn k q
(k q)!
1q q 1n q!1 q
ak ,1
.ak ,2 1
(5.2.30)
Corollary2. Let the function
be in the class
be in the class
.
.
*
f z defined by (5.2.1)
Then for the convolution f1z and f2 z
in class
Then f z is meromorphically starlike of order
An, q,, in view of (5.2.29) and (5.2.30) we have
k q1 q
0 1 in
z r1 , where
a .a 1
k ,1
k ,2
k q1 q
r1 = Inf
(k ) (1 ) k 1
(5.3.2)
k 1
(1 ) (k 2 )
The result in (5.4.7) is sharp for the function f z given by (5.4.3).

If in Theorem8 we take n q 0 then it
reduces to the know corollary due to Uralegaddi and Ganigi [6].

For n q 0 in Theorem9 we have the following corollary:
Corollary3. Let the function f j z j 1,2defined by (5.2.4) be in the class * . Then
*
*
f1 f 2 z ( ) , Where
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KAVITHA, S., SIVASUBRAMANIAN, S. and MUTHUNAGAI, K., A new class of meromorphic function with positive coefficients, Bulletin of mathematical analysis and applications, 3, (2010), 109 121.

MOGRA, M. L., REDDY, T. R. and JUNEJA, O. P., Meromorphic univalent functions with positive coefficients, Bull. Austral. Math. Soc., 32, No. 2, (1985), 161176.

SRIVASTAVA, H. M. and OWA, S., Current Topics in Analytic Function Theory, World Scientific Publishing Co., New Jersey, (1992).
= 21 2
1 1 2 1 2
The result is sharp for the functions given by
(5.3.3)
f j z j 1,2

URALEGADDI, B. A. and GANIGI, M. D., A certain class of meromorphically starlike functions with positive coefficients, Pure Appl. Math. Sci., 26, (1987), 7581.
f z = 1 + 1 zk
(5.3.4)
j
(k N )
z k