Cesáro partial sums of certain analytic functions

Rabha W Ibrahim; Maslina Darus

doi:10.1186/1029-242X-2013-51

Research
Open Access

Cesáro partial sums of certain analytic functions

Rabha W Ibrahim¹ and
Maslina Darus²Email author

Journal of Inequalities and Applications20132013:51

https://doi.org/10.1186/1029-242X-2013-51

Received: 4 August 2012
Accepted: 25 January 2013
Published: 14 February 2013

Abstract

The aim of the present paper is to consider geometric properties such as starlikeness and convexity of the Cesáro partial sums of certain analytic functions in the open unit disk. By using the Cesáro partial sums, we improve some recent results including the radius of convexity.

AMS Subject Classification:30C45.

Keywords

analytic function
univalent functions
starlike functions
unit disk
Cesáro partial sums
partial sums

1 Introduction

Let

U : = {z : | z | < 1}

be a unit disk in the complex plane ℂ and let ℋ denote the space of all analytic functions on U. Here we suppose that ℋ is a topological vector space endowed with the topology of uniform convergence over compact subsets of U. Also, for

a \in C

and

n \in N

, let

H [a, n]

be the subspace of ℋ consisting of functions of the form

f (z) = a + a_{n} z^{n} + a_{n + 1} z^{n + 1} + \dots

. Further, let

A : = {f \in H : f (0) = f^{'} (0) - 1 = 0}

and

S

denote the class of univalent functions in

A

. A function

f \in A

is called starlike if

f (U)

is a starlike domain with respect to the origin, and the class of univalent starlike functions is denoted by

S^{*}

. It is called convex

C

, if

f (U)

is a convex domain. Each univalent starlike function f is characterized by the analytic condition

ℜ (\frac{z f^{'} (z)}{f (z)}) > 0, z \in U .

Also, it is known that

z f^{'} (z)

is starlike if and only if f is convex, which is characterized by the analytic condition

ℜ (1 + \frac{z f^{″} (z)}{f^{'} (z)}) > 0, z \in U .

For a function

f (z) \in A

, we introduce the partial sum of

f (z)

by

f_{k} (z) = z + \sum_{n = 2}^{k} a_{n} z^{n}, z \in U .

(1)

For the partial sums

f_{n} (z)

of

f (z) \in S^{*}

, Szegö [1] showed that if

f (z) \in S^{*}

, then

f_{n} (z) \in S^{*}

for

| z | < \frac{1}{4}

and

f_{n} (z) \in C

for

| z | < \frac{1}{8}

. Owa [2] considered the starlikeness and convexity of partial sums,

f_{n} (z) = z + a_{n} z^{n},

of certain functions in the unit disk. Moreover, Darus and Ibrahim [3] determined the conditions under which the partial sums of functions of bounded turning are also of bounded turning.

In this paper, we consider the Cesáro partial sums, it is showed that this kind of partial sums preserve the properties of the analytic functions in the unit disk. Robertson [4] showed that if $f (z) \in A$ is univalent, then also all the Cesáro sums are univalent in the unit disk. Moreover, if its ordinary partial sums (1) is univalent in U, then the Cesáro sums are univalent. By employing the concept of the subordination, these results were extended by Ruscheweyha and Salinas [5]. The classical Cesáro means play an important role in geometric function theory (see [6–10]).

From the partial sum

s_{k} (z) = \sum_{n = 1}^{k} a_{n} z^{n}, z \in U,

with

a_{1} = 1

, we construct the Cesáro means

σ_{k} (z)

of

f \in A

by

\begin{array}{rcl} σ_{k} (z) & = & \frac{1}{k} \sum_{n = 1}^{k} s_{n} (z) \\ = & \frac{1}{k} [s_{1} (z) + \dots + s_{k} (z)] \\ = & \frac{1}{k} [z + (z + a_{2} z^{2}) + \dots + (z + \dots + a_{k} z^{k})] \\ = & \frac{1}{k} [k z + (k - 1) a_{2} z^{2} + \dots + a_{k} z^{k}] \\ = & z + \sum_{n = 2}^{k} (\frac{k - n + 1}{k}) a_{n} z^{n} \\ = & f (z) * [z + \sum_{n = 2}^{k} (\frac{k - n + 1}{k}) z^{n}] \\ = & f (z) * g_{k} (z), \end{array}

where

g_{k} = z + \sum_{n = 2}^{k} (\frac{k - n + 1}{k}) z^{n} .

Our aim is to consider geometric properties such as starlikeness and convexity of the Cesáro partial sums of certain analytic functions in the open unit disk.

2 Main results

We define the function

S_{k}

which is a partial sum of

f \in A

by

S_{k} (z) = z + (\frac{a_{k}}{k}) z^{k}, k \geq 2, a_{k} \neq 0 .

(2)

Theorem 1 The function

S_{k} (z)

satisfies

\frac{1 - | a_{k} | r^{k - 1}}{1 - \frac{| a_{k} |}{k} r^{k - 1}} \leq ℜ (\frac{z S_{k}^{'} (z)}{S_{k} (z)}) \leq \frac{1 + | a_{k} | r^{k - 1}}{1 + \frac{| a_{k} |}{k} r^{k - 1}}

(3)

for

0 \leq r < \sqrt[k - 1]{\frac{k}{| a_{k} |}} \leq 1, | a_{k} | \neq 0 .

Furthermore,

S_{k} (z) \in S^{*} (α)

for

0 \leq r < \sqrt[k - 1]{\frac{1 - α}{(1 - α / k) | a_{k} |}} \leq 1, | a_{k} | \neq 0 .

Proof

Noting that

\frac{z S_{k}^{'} (z)}{S_{k} (z)} = k - \frac{(k - 1)}{1 + \frac{a_{k}}{k} z^{k - 1}},

it follows that for

cos θ \to 1

, we obtain

\begin{array}{rcl} ℜ (\frac{z S_{k}^{'} (z)}{S_{k} (z)}) & = & k - (k - 1) \frac{1 + \frac{| a_{k} |}{k} cos θ r^{k - 1}}{1 + 2 \frac{| a_{k} |}{k} r^{k - 1} cos θ + {(\frac{| a_{k} |}{k})}^{2} r^{2 (k - 1)}} \\ \leq & \frac{1 + | a_{k} | r^{k - 1}}{1 + \frac{| a_{k} |}{k} r^{k - 1}} . \end{array}

Moreover, we also observe that

ℜ (\frac{z S_{k}^{'} (z)}{S_{k} (z)}) \geq \frac{1 - | a_{k} | r^{k - 1}}{1 - \frac{| a_{k} |}{k} r^{k - 1}} .

Now assume that

\frac{1 - | a_{k} | r^{k - 1}}{1 - \frac{| a_{k} |}{k} r^{k - 1}} > α

for

0 \leq r < \sqrt[k - 1]{\frac{1 - α}{(1 - α / k) | a_{k} |}} \leq 1, | a_{k} | \neq 0 .

This completes the proof. □

Remark 2 For example, the values $α = 0.5$ , $k = 2$ and $| a_{k} | = 1$ imply the radius of starlikeness of $S_{k} (z)$ is $r = 0.8164965 \dots$ , and for the same values, the radius of starlikeness of the ordinary partial sums $f_{k} (z) = z + a_{k} z^{k}$ is $r = 0.577350 \dots$ (see [2]).

Next, we derive the radius of convexity.

Theorem 3 The function

S_{k} (z)

satisfies

\frac{1 - k | a_{k} | r^{k - 1}}{1 - | a_{k} | r^{k - 1}} \leq ℜ (1 + \frac{z S_{k}^{''} (z)}{S_{k}^{'} (z)}) \leq \frac{1 + k | a_{k} | r^{k - 1}}{1 + | a_{k} | r^{k - 1}}

(4)

for

0 \leq r < \sqrt[k - 1]{\frac{1}{| a_{k} |}} \leq 1, | a_{k} | \neq 0 .

Furthermore,

S_{k} (z) \in C (α)

for

0 \leq r < \sqrt[k - 1]{\frac{1 - α}{(k - α) | a_{k} |}} \leq 1, | a_{k} | \neq 0 .

Proof

A computation gives

1 + \frac{z S_{k}^{''} (z)}{S_{k}^{'} (z)} = k - \frac{(k - 1)}{1 + a_{k} z^{k - 1}} .

Therefore, for

cos θ \to 1

, we obtain

\begin{array}{rcl} ℜ (1 + \frac{z S_{k}^{''} (z)}{S_{k}^{'} (z)}) & = & k - (k - 1) \frac{1 + | a_{k} | cos θ r^{k - 1}}{1 + 2 | a_{k} | r^{k - 1} cos θ + {| a_{k} |}^{2} r^{2 (k - 1)}} \\ \leq & \frac{1 + k | a_{k} | r^{k - 1}}{1 + | a_{k} | r^{k - 1}} . \end{array}

Moreover, we impose

ℜ (1 + \frac{z S_{k}^{''} (z)}{S_{k}^{'} (z)}) \geq \frac{1 - k | a_{k} | r^{k - 1}}{1 - | a_{k} | r^{k - 1}} .

Now, consider that

\frac{1 - k | a_{k} | r^{k - 1}}{1 - | a_{k} | r^{k - 1}} > α

for

0 \leq r < \sqrt[k - 1]{\frac{1 - α}{(k - α) | a_{k} |}} \leq 1, | a_{k} | \neq 0 .

This completes the proof. □

Remark 4 In view of Theorem 3, for example, the values $α = 0.5$ , $k = 2$ and $| a_{k} | = 1$ pose the radius of convexity of $S_{k} (z)$ is $r = 0.577350 \dots$ and for the same values, the radius of convexity of the ordinary partial sums $f_{k} (z) = z + a_{k} z^{k}$ is $r = 0.4082 \dots$ (see [2]).

Next, we assume special ordinary partial sums depending so that their coefficients satisfy the relation $| a_{n} | \leq (\frac{k - n + 1}{k})$ .

Theorem 5 Assume the partial sum

f_{3} (z) = z + \frac{k - 1}{k} z^{2} + \frac{k - 2}{k} z^{3}, k \geq 2 .

Then the function $f_{3} (z) \in S^{*} (\frac{1}{2})$ .

Proof We consider α such that

ℜ (\frac{z f_{3}^{'} (z)}{f_{3} (z)}) = ℜ (3 - \frac{2 + \frac{k - 1}{k} z}{1 + \frac{k - 1}{k} z + \frac{k - 2}{k} z^{2}}) > α .

This implies that

ℜ (\frac{2 + \frac{k - 1}{k} z}{1 + \frac{k - 1}{k} z + \frac{k - 2}{k} z^{2}}) < 3 - α,

that is,

ℜ (\frac{1 - \frac{k - 1}{k} z^{2}}{1 + \frac{k - 1}{k} z + \frac{k - 2}{k} z^{2}}) = \frac{1 - \frac{k - 1}{k} r^{2} (2 {cos}^{2} θ - 1)}{1 + \frac{k - 1}{k} r cos θ + \frac{k - 2}{k} r^{2} (2 {cos}^{2} θ - 1)} < 2 - α .

By letting

t = cos θ

, we define the function

g (t)

as follows:

g (t) = \frac{1 - \frac{k - 1}{k} r^{2} (2 t^{2} - 1)}{1 + \frac{k - 1}{k} r t + \frac{k - 2}{k} r^{2} (2 t^{2} - 1)} .

Logarithmic derivative of

g (t)

yields

\begin{array}{rcl} \frac{g^{'} (t)}{g (t)} & = & - {\frac{(4 t \frac{k - 1}{k} r^{2}) [1 + \frac{k - 1}{k} r t + \frac{k - 2}{k} r^{2} (2 t^{2} - 1)] + (\frac{k - 1}{k} r + 4 \frac{k - 2}{k} r^{2} t) [1 - \frac{k - 1}{k} r^{2} (2 t^{2} - 1)]}{[1 - \frac{k - 1}{k} r^{2} (2 t^{2} - 1)] [1 + \frac{k - 1}{k} r t + \frac{k - 2}{k} r^{2} (2 t^{2} - 1)]}} \\ : = & - {\frac{h (t)}{[1 - \frac{k - 1}{k} r^{2} (2 t^{2} - 1)] [1 + \frac{k - 1}{k} r t + \frac{k - 2}{k} r^{2} (2 t^{2} - 1)]}} \\ = & - \frac{A t^{2} + B t + C}{[1 - \frac{k - 1}{k} r^{2} (2 t^{2} - 1)] [1 + \frac{k - 1}{k} r t + \frac{k - 2}{k} r^{2} (2 t^{2} - 1)]}, \end{array}

where

\begin{matrix} A = 2 r^{3} {(\frac{k - 1}{k})}^{2}, \\ B = 4 r^{2} [\frac{k - 1}{k} + \frac{k - 2}{k}], \\ C = \frac{k - 1}{k} r [1 + \frac{k - 1}{k} r^{2}] . \end{matrix}

Now, for all

k \geq 2

and

r \to 1

, the function

h (t)

has unique real negative zeros in the interval

(- \frac{1}{2}, 0)

. This leads to the fact that

g^{'} (t)

has unique positive real zeros for all

k \geq 2

distributed in the interval

(0, \frac{1}{2})

. Therefore, we will calculate α in

t \in [\frac{1}{2}, 1)

. It is easy to check that

g (t)

is decreasing for

r \to 1

in the interval

[\frac{1}{2}, 1)

. Moreover, we have

lim_{k \to \infty} g (\frac{1}{2}) = lim_{k \to \infty} \frac{1 + \frac{k - 1}{2 k}}{1 + \frac{k - 1}{2 k} - \frac{k - 2}{2 k}} = \frac{3}{2} .

We conclude that for

t \in [\frac{1}{2}, 1)

,

g (t) < g (\frac{1}{2}) \leq \frac{3}{2} = 2 - α,

thus $α = \frac{1}{2}$ . This completes the proof. □

By letting $k = 3$ in Theorem 5, we have the following result.

Corollary 6 The Cesáro partial sums

σ_{3} (z) = z + \frac{2}{3} z^{2} + \frac{1}{3} z^{3}, z \in U,

of the function $f (z) = \frac{z}{1 - z}$ are starlike of order $α = \frac{1}{5}$ .

Theorem 7 Assume the partial sum $f_{3} (z)$ as in Theorem 5. Then the function $f_{3} (z) \in C (\frac{1}{5})$ .

Proof We consider α such that

ℜ (1 + \frac{z f_{3}^{''} (z)}{f_{3}^{'} (z)}) = ℜ (3 - \frac{2 (\frac{k - 1}{k} z + 1)}{1 + 2 \frac{k - 1}{k} z + 3 \frac{k - 2}{k} z^{2}}) > α .

This implies that

ℜ (\frac{\frac{k - 1}{k} z + 1}{1 + 2 \frac{k - 1}{k} z + 3 \frac{k - 2}{k} z^{2}}) < \frac{3 - α}{2},

therefore, a computation gives

ℜ (\frac{\frac{k - 1}{k} z + 1}{1 + 2 \frac{k - 1}{k} z + 3 \frac{k - 2}{k} z^{2}}) = \frac{1}{2} + ℜ (\frac{\frac{1}{2} (1 - 3 \frac{k - 2}{k} z^{2})}{1 + 2 (\frac{k - 1}{k}) z + 3 (\frac{k - 2}{k}) z^{2}}),

thus

\frac{\frac{1}{2} (1 - 3 \frac{k - 2}{k} r^{2} (2 {cos}^{2} θ - 1))}{1 + 2 (\frac{k - 1}{k}) r cos θ + 3 (\frac{k - 2}{k}) (2 {cos}^{2} θ - 1)} < \frac{2 - α}{2} .

By putting

t = cos θ

, we define the function

G (t)

as follows:

G (t) = \frac{\frac{1}{2} (1 - 3 \frac{k - 2}{k} r^{2} (2 t^{2} - 1))}{1 + 2 (\frac{k - 1}{k}) r t + 3 r^{2} (\frac{k - 2}{k}) (2 t^{2} - 1)} .

Logarithmic derivative of

G (t)

yields

\begin{array}{rcl} \frac{G^{'} (t)}{G (t)} & = & - {\frac{[12 r^{2} \frac{k - 2}{k} t] [1 + 2 (\frac{k - 1}{k}) r t + 3 r^{2} (\frac{k - 2}{k}) (2 t^{2} - 1)]}{[1 - 3 \frac{k - 2}{k} r^{2} (2 t^{2} - 1)] [1 + 2 (\frac{k - 1}{k}) r t + 3 r^{2} (\frac{k - 2}{k}) (2 t^{2} - 1)]} \\ + \frac{[12 r^{2} \frac{k - 2}{k} t + 2 r \frac{k - 1}{k}] [1 - 3 \frac{k - 2}{k} r^{2} (2 t^{2} - 1)]}{[1 - 3 \frac{k - 2}{k} r^{2} (2 t^{2} - 1)] [1 + 2 (\frac{k - 1}{k}) r t + 3 r^{2} (\frac{k - 2}{k}) (2 t^{2} - 1)]}} \\ : = & - {\frac{H (t)}{[1 - 3 \frac{k - 2}{k} r^{2} (2 t^{2} - 1)] [1 + 2 (\frac{k - 1}{k}) r t + 3 r^{2} (\frac{k - 2}{k}) (2 t^{2} - 1)]}} \\ = & - \frac{A t^{2} + B t + C}{[1 - 3 \frac{k - 2}{k} r^{2} (2 t^{2} - 1)] [1 + 2 (\frac{k - 1}{k}) r t + 3 r^{2} (\frac{k - 2}{k}) (2 t^{2} - 1)]}, \end{array}

where

\begin{matrix} A = 12 r^{3} \frac{(k - 1) (k - 2)}{k^{2}}, \\ B = 12 r^{2} \frac{k - 2}{k} [2 + 3 r^{2} (\frac{k - 2}{k})], \\ C = 2 r \frac{k - 1}{k} [1 + 3 r^{2} (\frac{k - 2}{k})] . \end{matrix}

Now, for all

k \geq 3

and

r \to 1

, the function

H (t)

has unique real negative zeros in the interval

[- \frac{1}{2}, 0)

. This leads to the fact that

G^{'} (t)

has unique positive real zeros for all

k \geq 3

in the interval

(0, \frac{1}{2}]

. So, we calculate α in the interval

t \in (\frac{1}{2}, 1)

. A computation yields

G (t)

is decreasing for

r \to 1

in the interval

t \in (\frac{1}{2}, 1)

. Thus, we have

lim_{k \to \infty} G (t) < \frac{9}{10} = \frac{2 - α}{2}, 1 > t > 0.5,

which implies $α = \frac{1}{5}$ . This completes the proof. □

Theorem 8 Assume the Cesáro partial sum

σ_{3} (z) = z + \frac{4}{3} z^{2} + z^{3}

of the function

f (z) = \frac{z}{{(1 - z)}^{2}} = z + 2 z^{2} + 3 z^{3} + \dots .

Then the function $σ_{3} (z) \in C (\frac{1}{2})$ for all $0.2 < r < 0.5$ .

Proof We consider α such that

ℜ (1 + \frac{z σ_{3}^{''} (z)}{σ_{3}^{'} (z)}) = ℜ (3 - \frac{2 (\frac{4}{3} z + 1)}{1 + \frac{8}{3} z + 3 z^{2}}) > α .

This implies that

ℜ (\frac{\frac{4}{3} z + 1}{1 + \frac{8}{3} z + 3 z^{2}}) < \frac{3 - α}{2},

therefore, a computation gives

ℜ (\frac{\frac{4}{3} z + 1}{1 + \frac{8}{3} z + 3 z^{2}}) = \frac{1}{2} + ℜ (\frac{\frac{1}{2} (1 - 3 z^{2})}{1 + \frac{8}{3} z + 3 z^{2}}),

thus

\frac{1 - 3 r^{2} (2 {cos}^{2} θ - 1)}{1 + \frac{8}{3} r cos θ + 3 (2 {cos}^{2} θ - 1)} < 2 - α .

By putting

t = cos θ

, we define the function

ȷ (t)

as follows:

ȷ (t) = \frac{1 - 3 r^{2} (2 t^{2} - 1)}{1 + \frac{8}{3} r t + 3 r^{2} (2 t^{2} - 1)} .

Logarithmic derivative of

ȷ (t)

yields

\begin{array}{rcl} \frac{ȷ^{'} (t)}{ȷ (t)} & : = & - {\frac{ħ (t)}{[1 - 3 r^{2} (2 t^{2} - 1)] [1 + \frac{8}{3} r t + 3 r^{2} (2 t^{2} - 1)]}} \\ = & - \frac{16 r^{3} t^{2} + 24 r^{2} t + \frac{8}{3} r (1 + 3 r^{2})}{[1 - 3 r^{2} (2 t^{2} - 1)] [1 + \frac{8}{3} r t + 3 r^{2} (2 t^{2} - 1)]} . \end{array}

The function

ħ (t)

has a unique real negative zero in the interval

t \in (- 1, 0)

for all

0.2 < r < 0.5

which is around

t ≃ - \frac{1}{2}

. This leads to the fact that

ȷ^{'} (t)

has a unique positive real zero in the interval

(0, 1)

around

t ≃ \frac{1}{2}

. A computation yields

ȷ (t)

is decreasing in the interval

t \in (\frac{1}{2}, 1)

and assuming its maximums at

t = 0.5

and

r = 0.5

. Thus, we have

lim_{r \to 0.5, t \to 0.5} ȷ (t) < \frac{3}{2} = 2 - α,

which implies $α = \frac{1}{2}$ . This completes the proof. □

Note that some other results related to partial sums can be found in [11–15].

Declarations

Acknowledgements

The work was fully supported by UKM-DLP-2011-050 and LRGS/TD/2011/UKM/ICT/03/02. The authors also would like to thank the referees for giving some suggestions for improving the work.

Authors’ Affiliations

(1)

Institute of Mathematical Sciences, Universiti Malaya, Kuala Lumpur, 50603, Malaysia

(2)

School of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, Bangi, Selangor Darul Ehsan, 43600, Malaysia

References

Szego G: Zur theorie der schlichten abbilungen. Math. Ann. 1928, 100: 188–211. 10.1007/BF01448843MathSciNetView ArticleGoogle Scholar
Owa S: Partial sums of certain analytic functions. Int. J. Math. Math. Sci. 2001, 25(12):771–775. 10.1155/S0161171201005099MATHMathSciNetView ArticleGoogle Scholar
Darus M, Ibrahim RW: Partial sums of analytic functions of bounded turning with applications. Comput. Appl. Math. 2010, 29(1):81–88.MATHMathSciNetGoogle Scholar
Robertson MS: On the univalency of Cesáro sums of univalent functions. Bull. Am. Math. Soc. 1936, 42: 241–243. 10.1090/S0002-9904-1936-06279-8View ArticleGoogle Scholar
Ruscheweyha S, Salinas LC: Subordination by Cesáro means. Complex Var. Elliptic Equ. 1993, 3(21):279–285.View ArticleGoogle Scholar
Splina LT: On certain applications of the Hadamard product. Appl. Math. Comput. 2008, 199: 653–662. 10.1016/j.amc.2007.10.031MathSciNetView ArticleGoogle Scholar
Darus M, Ibrahim RW: On Cesáro means for Fox-Wright functions. J. Math. Stat. 2008, 4(3):156–160.MATHMathSciNetView ArticleGoogle Scholar
Darus M, Ibrahim RW: On some properties of differential operator. Acta Didact. Napoc. 2009, 2(2):1–6.MATHGoogle Scholar
Darus M, Ibrahim RW: Coefficient inequalities for concave Cesáro operator of non-concave analytic functions. Eur. J. Pure Appl. Math. 2010, 3(6):1086–1092.MATHMathSciNetGoogle Scholar
Srivastava HM, Darus M, Ibrahim RW: Classes of analytic functions with fractional powers defined by means of a certain linear operator. Integral Transforms Spec. Funct. 2011, 1(22):17–28.MathSciNetView ArticleGoogle Scholar
Frasin BA: Generalization of partial sums of certain analytic and univalent functions. Appl. Math. Lett. 2008, 21: 735–741. 10.1016/j.aml.2007.08.002MATHMathSciNetView ArticleGoogle Scholar
Wang Z-G, Liu Z-H, Catas A: On neighborhoods and partial sums of certain meromorphic multivalent functions. Appl. Math. Lett. 2011, 24: 864–868. 10.1016/j.aml.2010.12.033MATHMathSciNetView ArticleGoogle Scholar
Murugusundaramoorthy G, Uma K, Darus M: Partial sums of generalized class of analytic functions involving Hurwitz-Lerch zeta function. Abstr. Appl. Anal. 2011., 2011: Article ID 849250. doi:10.1155/2011/849250Google Scholar
Ghanim F, Darus M: Partial sums of certain new subclasses for meromorphic functions. Far East J. Math. Sci. 2011, 55(2):181–195.MATHMathSciNetGoogle Scholar
Ibrahim RW, Darus M: Partial sums for certain classes of meromorphic functions. Tamkang J. Math. 2010, 41(1):39–49.MATHMathSciNetGoogle Scholar

Copyright

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.