On Certain Subclasses of Meromorphic Univalent Funtions with Positive Coefficients

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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

Theorem-1 : 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 Theorem-1 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 Theorem-3.

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

  1. f z is meromoriphically starlike of order 0 1 in

    Theorem-3 : 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

  2. 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 Theorem-2 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 )

Theorem-1 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 +

[ ta

1 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

    Theorem-9: 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:

  1. For the choice of

    n q 0

    in Theorem-1, 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].

  2. If in Theorem-2 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

  3. For

    n q 0

    and

    m 0in Theorem-3, 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

    Corollary-1. 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

  4. 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)

    Corollary-2. 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).

  5. If in Theorem-8 we take n q 0 then it

    reduces to the know corollary due to Uralegaddi and Ganigi [6].

  6. For n q 0 in Theorem-9 we have the following corollary:

Corollary-3. Let the function f j z j 1,2defined by (5.2.4) be in the class * . Then

*

*

f1 f 2 z ( ) , Where

REFERENCES

  1. AOUF, M. K., On a certain class of meromorphic univalent functions with positive coefficients, Rend. Mat. Appl., 7 (11), No. 2, (1991), 209-219.

  2. ATSHAN, W, G., Subclass of meromorphic functions with positive coefficients defined by Ruscheweyh derivative, II. Surv. Math. Appl., 3, (2008), 67-77.

  3. 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.

  4. MOGRA, M. L., REDDY, T. R. and JUNEJA, O. P., Meromorphic univalent functions with positive coefficients, Bull. Austral. Math. Soc., 32, No. 2, (1985), 161-176.

  5. 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

  6. URALEGADDI, B. A. and GANIGI, M. D., A certain class of meromorphically starlike functions with positive coefficients, Pure Appl. Math. Sci., 26, (1987), 75-81.

f z = 1 + 1 zk

(5.3.4)

j

(k N )

z k

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