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volume 7, issue 4, article 140, 2006.

Received 05 July, 2005;

accepted 18 October, 2006.

Communicated by:G. Kohr

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Journal of Inequalities in Pure and Applied Mathematics

PARTIAL SUMS OF SOME MEROMORPHIC FUNCTIONS

S. LATHA AND L. SHIVARUDRAPPA

Department of Mathematics Maharaja’s College University of Mysore Mysore - 570005, INDIA.

EMail:drlatha@gmail.com

Department of Mathematics, S.J.C.E Visweshwarayah Technological University Belgam, INDIA.

EMail:shivarudrappa@lycos.com

c

2000Victoria University ISSN (electronic): 1443-5756 203-05

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Partial Sums of Some Meromorphic Functions S. Latha and L. Shivarudrappa

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Abstract

In the present paper we give some results concerning partial sums of certain meromorphic functions.We also consider the partial sums of certain integral operator.

2000 Mathematics Subject Classification:30C45.

Key words: Partial sums, Meromorphic functions, Integral operators, Meromorphic starlike functions, Meromorphic convex functions, Meromorphic close to convex functions.

Contents

1 Introduction. . . 3 2 Some Preliminary Results . . . 5 3 Main Results . . . 7

References

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Partial Sums of Some Meromorphic Functions S. Latha and L. Shivarudrappa

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1. Introduction

LetΣbe the class consisting of functions of the form

(1.1) f(z) = 1

z +

X

k=1

akzk

which are regular in the punctured disc E = {z : 0 < |z| < 1}with a simple pole at the origin and residue1there.

Letfn(z) = 1z +Pn

k=1akzk be thenth partial sum of the series expansion for f(z) ∈ Σ.LetΣ(A, B),ΣK(A, B),Σc(A, B),−1 ≤ A < B ≤ 1be the subclasses of functions inΣsatisfying

(1.2) −

zf0(z) f(z)

≺ 1 +Az

1 +Bz, z ∈ U =E∪ {0}

(1.3) −

zf00(z) f0(z) + 1

≺ 1 +Az

1 +Bz, z ∈ U.

(1.4) −z2f0(z)≺ 1 +Az

1 +Bz, z ∈ U

respectively [5]. The classesΣ(2α−1,1)andΣK(2α−1,1)are respectively the well known subclasses of Σ consisting of functions meromorphic starlike of order α and meromorphic convex of orderα and meromorphically close to convex of orderαdenoted byΣ(α),ΣK(α)andΣc(α)respectively.

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Iff(z) = 1z +P

k=1akzkandg(z) = 1z +P

k=1bkzk,then their Hadamard product (or convolution), denoted byf(z)∗g(z)is the function defined by the power series

f(z)∗g(z) = 1 z +

X

k=1

akbkzk.

In the present paper, we give sufficient conditions forf(z)to be inΣ(A, B), ΣK(A, B)and further investigate the ratio of a function of the form (1.1) to its sequence of partial sums when the coefficients are sufficiently small to satisfy conditions

X

k=1

k{k(1 +B) + (1 +A)}|ak| ≤B−A,

X

k=1

{k(1 +B) + (1 +A)}|ak| ≤B−A.

More precisely, we will determine sharp lower bounds for<nf(z)

fn(z)

o ,<nf

n(z) f(z)

o ,

<n

f0(z) fn0(z)

o

and<n

fn0(z) f0(z)

o

.Further, we give a property for the partial sums of certain integral operators in connection with functions belonging to the class Σc(A, B).

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Partial Sums of Some Meromorphic Functions S. Latha and L. Shivarudrappa

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2. Some Preliminary Results

Theorem 2.1. Letf(z) = 1z +P

k=1akzk, z ∈E.If (2.1)

X

k=1

k{k(1 +B) + (1 +A)}|ak| ≤B −A, then f(z)∈ΣK(A, B).

Proof. It suffices to show that

1 + zff00(z)(z) + 1 B

1 + zff000(z)(z)

+A

<1,

that is,

zf00(z) + 2f0(z) Bzf00(z) + (A+B)f0(z)

<1.

Consider

zf00(z) + 2f0(z) Bzf00(z) + (A+B)f0(z)

(2.2)

=

P

k=1k(k+ 1)akzk+1 (B−A) +BP

k=1k(k+ 1)akzk+1−(2B−A)P

k=1kakzk+1

≤ Σk(k+ 1)|ak| (B−A)−P

k=1k(kB+A)|ak|. (2.2) is bounded by1if

X

k=1

k(k+ 1)|ak| ≤(B −A)

X

k=1

k(kB+A)|ak|

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Partial Sums of Some Meromorphic Functions S. Latha and L. Shivarudrappa

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which reduces to (2.1)

Similarly we can prove the following theorem.

Theorem 2.2. Letf(z) = 1z +P

k=1akzk, z ∈E.If

(2.3)

X

k=1

{k(1 +B) + (1 +A)}|ak| ≤B−A, then f(z)∈Σ(A, B).

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3. Main Results

Theorem 3.1. Iff(z)of the form (1.1) satisfies (2.3), then

<

f(z) fn(z)

≥ 2(n+ 1 +A)

2n+ 2 +A+B, z ∈ U.

The result is sharp for everyn,with extremal function

(3.1) f(z) = 1

z + B−A

2n+ 2 +A+Bzn+1, n ≥0.

Proof. Consider 2n+ 2 +A+B

B−A

f(z)

fn(z)− 2n+A+B 2n+ 2 +A+B

= 1 +Pn

k=1akzk+1+ 2n+2+A+BB−A P

k=n+1akzk+1 1 +Pn

k=1akzk+1

= 1 +w(z) 1−w(z), where

w(z) =

2n+2+A+B B−A

P

k=n+1akzk+1 2 + 2Pn

k=1akzk+12n+2+A+BB−A P

k=n+1akzk+1 and

|w(z)| ≤

2n+2+A+B B−A

P k=1|ak| 2−2Pn

k=1|ak| −2n+2+A+BB−A P

k=n+1|ak|.

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Partial Sums of Some Meromorphic Functions S. Latha and L. Shivarudrappa

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Now

|w(z)| ≤1 if and only if

2

2n+ 2 +A+B B−A

X

k=n+1

|ak| ≤2−2

n

X

k=1

|ak|,

which is equivalent to (3.2)

n

X

k=1

|ak|+ 2n+ 2 +A+B B −A

X

k=n+1

|ak| ≤1.

It suffices to show that the left hand side of (3.2) bounded above by

X

k=1

2k+A+B (B−A) |ak|, which is equivalent to

n

X

k=1

2(k+A) B−A

|ak|+

X

k=n+1

2(k−n−1) B−A

|ak| ≥0.

To see that the functionf(z)given by (3.1) gives the sharp result, we observe for

z =ren+2πi

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that f(z)

fn(z) = 1+ B−A

2n+ 2 +A+Bzn+2 →1− B−A

2n+ 2 +A+B = 2(n+ 1 +A) 2(n+ 1) +A+B whenr→1.

Therefore we complete the proof of Theorem3.1.

Corollary 3.2. ForA= 2α−1, B = 1,we get Theorem2.1in [3] which states as follows:

Iff(z)of the form (1.1) satisfies condition

X

1

(k+α)|ak| ≤1−α,

then

<

f(z) fn(z)

≥ n+ 2α

n+ 1 +α, z ∈ U. The result is sharp for everyn, with extremal function

f(z) = 1

z + 1−α

n+ 1 +αzn+1, n≥0.

Theorem 3.3. Iff(z)of the form (1.1) satisfies (2.1), then

<

f(z) fn(z)

≥ (n+ 2)(2n+A+B)

(n+ 1)(2n+ 2 +A+B), z∈ U. The result is sharp for everyn,with extremal function

(3.3) f(z) = 1

z + B−A

(n+ 1)(2n+ 2 +A+B)zn+1, n ≥0.

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Proof. Consider

(n+ 1)(2n+ 2 +B+A) B−A

f(z)

fn(z)− (n+ 2)(2n+A+B) (n+ 1)(2n+ 2 +A+B)

= 1 +Pn

k=1akzk+1+(n+1)(2n+2+A+B) B−A

P

k=n+1akzk+1 1 +Pn

k=1akzk+1 := 1 +w(z)

1−w(z),

where

w(z) =

(n+1)(2n+2+A+B) B−A

P

k=n+1akzk+1 2 + 2Pn

k=1akzk+1+(n+1)(2n+2+B+A) B−A

P

k=n+1akzk+1. Now

|w(z)| ≤

(n+1)(2n+2+A+B) B−A

P

k=n+1|ak| 2−2Pn

k=1|ak| − (n+1)(2n+2+A+B) B−A

P

k=n+1|ak| ≤1 if

(3.4)

n

X

k=1

|ak|+ (n+ 1)(2n+ 2 +A+B) B −A

X

k=n+1

|ak| ≤1.

The left hand side of (3.4) is bounded above by

X

k=1

k(2k+A+B) B−A |ak|

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if 1 B−A

( n X

k=1

(k(2k+A+B)−(B−A))|ak|

+

X

k=n+1

(k(2k+A+B)−(n+ 1)(2n+ 2 +A+B))|ak| )

≥0,

and the proof is completed.

Corollary 3.4. ForA= 2α−1, B = 1,we get Theorem2.2in [3] which reads:

Iff(z)of the form (1.1) satisfies condition

X

1

k(k+α)|ak| ≤1−α,

then

<

f(z) fn(z)

≥ (n+ 2)(n+α)

(n+ 1)(n+ 1 +α), z ∈ U. The result is sharp for everyn, with extremal function

f(z) = 1

z + 1−α

(n+ 1)(n+ 1 +α)zn+1, n≥0.

We next determine bounds for<n

fn(z) f(z)

o .

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Partial Sums of Some Meromorphic Functions S. Latha and L. Shivarudrappa

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Theorem 3.5.

(a) Iff(z)of the form (1.1) satisfies the condition (2.3), then

<

fn(z) f(z)

≥ 2n+ 2 +A+B

n+ 2 , z ∈ U.

(b) Iff(z)of the form (1.1) satisfies condition (2.1), then

<

fn(z) f(z)

≥ 2(n+ 1)(2n+ 2 +A+B)

2(n+ 1)(n+ 2)−n(B−A), z ∈ U. Equalities hold in(a) and(b)for the functions given by (3.1) and (3.3) respectively.

Proof. We prove(a).The proof of(b)is similar to(a)and will be omitted.

Consider 2(n+ 2)

B−A

fn(z)

f(z) − 2n+ 2 +A+B 2(n+ 2)

= 1 +Pn

k=1akzk+1+ 2n+2+A+BB−A P

k=n+1akzk+1 1 +Pn

k=1akzk+1 := 1 +w(z) 1−w(z), where

|w(z)| ≤

n+2 B−A

P

k=n+1|ak| 1−Pn

k=1|ak| −n+A+BB−A P

k=n+1|ak| ≤1.

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This last inequality is equivalent to (3.5)

n

X

k=1

|ak|+ 2n+ 2 +A+B B −A

X

k=n+1

|ak| ≤1.

Since the left hand side of (3.5) is bounded above by

X

k=1

2k+A+B B−A |ak|, the proof is completed.

Corollary 3.6. ForA= 2α−1, B = 1,we get Theorem2.3in [3] which reads:

(a) Iff(z)of the form (1.1) satisfies condition

X

1

(k+α)|ak| ≤1−α,

then

<

fn(z) f(z)

≥ (n+ 1 +α)

(n+ 2) , z ∈ U.

(b) Iff(z)of the form (1.1) satisfies condition

X

1

k(k+α)|ak| ≤1−α,

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then

<

fn(z) f(z)

≥ (n+ 1)(n+ 1 +α)

(n+ 1)(n+ 2)−n(1−α), z ∈ U. Equalities hold in (a) and (b) for the functions given by

f(z) = 1

z + 1−α

(n+ 1 +α)zn+1, n ≥0,

f(z) = 1

z + 1−α

(n+ 1)(n+ 1 +α)zn+1, n≥0 respectively.

We turn to ratios involving derivatives. The proof of Theorem 3.7is similar to that in Theorem 3.1 and (a) of Theorem 3.5 and so the details may be omitted.

Theorem 3.7. If f(z) of form (1.1) satisfies condition (2.3) with A = −B, then

<

f0(z) fn0(z)

≥0, z ∈ U, (a)

<

fn0(z) f0(z)

≥ 1

2, z ∈ U. (b)

In both the cases, the extremal function is given by (3.1) withA=−B.

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Theorem 3.8. Iff(z)of form (1.1) satisfies condition (2.1) then,

<

f0(z) fn0(z)

≥ 2(n+A+B)

2n+ 2 +A+B, z ∈ U, (a)

<

fn0(z) f0(z)

≥ 2n+ 2 +A+B

2(n+ 2) , z ∈ U. (b)

In both the cases, the extremal function is given by (3.3)

Proof. It is well known that f(z) ∈ ΣK(A, B) if and only if −zf0(z) ∈ Σ(A, B). In particular, f(z) satisfies condition (2.1)if and only if −zf0(z) satisfies condition (2.3). Thus (a) is an immediate consequence of Theorem 3.1and(b)follows directly from(a)of Theorem 3.5.

For a functionf(z)∈Σ,we define the integral operatorF(z)as follows F(z) = 1

z2 Z z

0

tf(t)dt = 1 z +

X

k=1

1

k+ 2akzk, z ∈E.

Thenth partial sumFn(z)of the integral operatorF(z)is given by Fn(z) = 1

z +

n

X

k=1

1

k+ 2akzk, z ∈E.

The following lemmas will be required for the proof of Theorem3.11below.

Lemma 3.9. For0≤θ ≤π, 12 +Pm k=1

cos(kθ) k+1 ≥0

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Lemma 3.10. LetP be analytic inU withP(0) = 1 and<{P(z)} > 12 in U. For any functionQanalytic inU the functionP ∗Qtakes values in the convex hull of the image onU underQ.

Lemma 3.9is due to Rogosinski and Szego [4] and Lemma 3.10is a well known result ([2] and [6]) that can be derived from the Herglotz representation forP.Finally we derive

Theorem 3.11. Iff(z)∈Σc(A, B),thenFn(z)∈Σc(A, B).

Proof. Letf(z)be the form (1.1) and belong to the classΣc(A, B).

We have,

(3.6) <

(

1− 1 B−A

X

k=1

kakzk+1 )

> 1

2, z ∈ U.

Applying the convolution properties of power series toFn0(z)we may write

−z2Fn0(z) (3.7)

= 1−

n

X

k=1

k

k+ 2akzk+1

= 1− 1 B−A

X

k=1

kakzk+1

!

∗ 1 + (B−A)

X

k=n+1

1 k+ 1zk

! .

Puttingz =re, 0≤r <1, 0≤ |θ| ≤π,and making use of the minimum

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principle for harmonic functions along with Lemma 3.9, we obtain

<

(

1 + (B−A)

n+1

X

k=1

1 k+ 1zk

)

= 1 + (B−A)

n+1

X

k=1

rkcos(kθ) k+ 1 (3.8)

>1 + (B−A)

n+1

X

k=1

coskθ k+ 1

1−

B−A 2

.

In view of (3.6), (3.7), (3.8) and Lemma3.10we deduce that

−<{z2Fn0(z)}>

1−

B −A 2

, 0≤A+B <2, z ∈ U,

which completes the proof of Theorem 3.11

Corollary 3.12. ForA = 2α−1, B = 1,we obtain Theorem2.8in [3] which reads:

Iff(z)∈Σc(α),thenFn(z)∈Σc(α).

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References

[1] S. RUSCHWEYH, Convolutions in Geometric Function Theory, Les Presses de l’Universite de Montreal, 1982.

[2] A.W. GOODMAN, Univalent Functions, Vol. I, Mariner Publ. Co., Tampa, Fl., (1983).

[3] NAK EUN CHO AND S. OWA, On partial sums of certain meromorphic functions, J. Inequal. in Pure and Appl. Math., 5(2) (2004), Art. 30. [ON- LINE:http://jipam.vu.edu.au/article.php?sid=377].

[4] W. ROGOSINSKIANDG. SZEGO, Uber die abschimlte von potenzreihen die in ernein kreise be schranket bleiben, Math. Z., 28 (1928), 73–94.

[5] S. LATHA, Some studies in the theory of univalent and multivalent func- tions, PhD Thesis, (1994).

[6] R. SINGH AND S. SINGH, Convolution properties of a class of starlike fuctions, Proc. Amer. Math. Soc., 106 (1989), 145–152.

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