Sufficient Conditions for a New Class of Polynomial Analytic Functions of Reciprocal Order alpha

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Sufficient Conditions for a New Class of

Polynomial Analytic Functions

of Reciprocal Order

α

Mohammad Taati

?

_{, Sirous Moradi and Shahram Najafzadeh}

Abstract

In this paper, we consider a new class of analytic functions in the unit disk using polynomials of order α. We give some sufficient conditions for functions belonging to this class.

Keywords: Polynomial analytic functions, starlike functions, meromorphic functions.

2010 Mathematics Subject Classification: 30C45, 30C80.

1. Introduction

LetP

denotes the class of all functions of the form

f(z) =1

z +

∞

X k=0

akzk, (1)

which are analytic in the punctured open unit disk

U∗={z: z∈Cand0<|z|<1}=U− {0}. A function f ∈P

is said to be in the class MS∗₍_α₎_{of meromorphic starlike}

functions of orderαif it satisfies the inequality

< zf

0

(z)

f(z) !

<−α, (0≤α <1, z∈_U). (2)

?_{Corresponding author (E-mail: [email protected] & [email protected])} Academic Editor: Gholam Hossein Fath-Tabar

Received 21 November 2016, Accepted 23 January 2017 DOI: 10.22052/mir.2017.67848.1047

c

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We get,MS∗_{(0) =}_MS∗_.

Furthermore, a functionf ∈ MS∗ is said to be in the classN S∗(α) of mero-morphic starlike of reciprocal orderαif and only if

< _f₍_z₎

zf0

(z)

<−α, (0≤α <1 , z∈_U). (3)

In recent years, several authors studied meromorphic starlike functions and starlike functions of reciprocal. We can see the earlier work by Ali and Ravichan-dran [1–3] and Cho et al. [5, 6] and (more recently) by Nunokawa et al. [9, 10], Silverman et al. [11], Srivastava et al. [12], Wang et al. [14–19] and the references therein.

Yong Sun et al. [13] obtained some sufficient conditions for the functions be-longing to the classN S∗₍_α_).

In this paper, we introduce a new class of analytic starlike functions. Also we give some sufficient conditions for functions which belongs to the new class.

2. Preliminaries

LetP denote the class of functions P(z)given by

P(z) = 1 +

∞

X k=1

Pkzk (z∈U), (4)

which are analytic inU.

Lemma 2.1. [[7]]If the functionp∈ Pis given by (4)and satisfied the condition

<(p(z))>0, then |pk| ≤2,k∈N.

Let0≤α <1,P∗(α)denotes the class of all functions p(z)∈ P and satisfies the condition

< zp

0

(z)

p(z) !

<1−α. (5)

We setP∗_{(0) =}_P∗_.

Finally, let0≤α <1. We introduce the notationN P∗₍_α₎_{for the class of all}

mappingsp(z)∈ P∗ _{satisfying the following condition:}

<

_p₍_z₎

zp0

(z)−p(z)

<−α. (6)

Obviously,p(z)∈ N P∗₍_α₎_{if and only if}_f₍_z_{) =}1

zp(z)∈ N S

∗₍_α₎_.

Remark 1. For0< α <1, the functionp∈ Pbelongs to the class N P∗₍_α₎_{if and}

only if

zp0(z)

p(z) + 1−2α

2α

< 1

(3)

In the following, we give several examples of functions of belonging to the class N P∗(α).

Example 2.2. Letp∈ P satisfies the inequality

zp0(z)

p(z)

<1−α (0≤α <1, z∈_U).

Then

zp0(z)

p(z) +

α

2

≤

zp0(z)

p(z)

+α

2 <1−α+

α

2 ≤

α+ 2 2 , thereforep∈ N P∗₍ 1

α+2).

Example 2.3. Let the functionp(z)∈ P be in the form

p(z) =e(1−α)z (0< α <1, z∈_U).

This gives us that

< zp

0

(z)

p(z) !

=<((1−α)z)<1−α.

Therefore,p(z)∈ P∗₍_α_{). Moreover, we have}

p(z)

zp0(z)−p(z) = 1 (1−α)z−1. It follows that

<

p(z)

zp0

(z)−p(z)

=<

1 (1−α)eiθ₋₁

<− 1

2−α (z=e

iθ₎_.

Therefore,p(z)∈ N P∗₍ 1 2−α).

In order to obtain our main results, we need the following lemmas.

Lemma 2.4. (Jack’s lemma [8]) Let ϕbe a non-constant regular function in _U. If |ϕ| attains its maximum value on circle |z|=r <1 atz0, then

z0ϕ

0

(z0) =kϕ(z0),

wherek≥1 is a real number.

Lemma 2.5. (See, [4]) LetΩbe a set in the complex plane Cand suppose that Φ

is a mapping from C2×U toCwhich satisfies φ(ix, y;z)∈/ Ωforz ∈U, and for

all real numbers x, ysuch that y≤ −1+x2

2 . If p(z)∈ P andφ(p(z), zp

0

(z);z)∈Ω

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3. Main Results

We begin this section by presenting the following coefficient sufficient conditions for functions belonging to the classN P∗₍_α_).

Theorem 3.1. If p∈ P satisfies

∞

X k=1

[1 +α(k−1)]|pk| ≤ 1

2(1− |1−2α|). (8)

Thenp∈ N P∗₍_α₎_{, for} ₀_{< α <}₁_.

Proof. Using Remark 1 only need to show that

2αzp0(z)

p(z) + 1−2α

<1 (z∈U). (9) We first observe that

2αzp0(z) + (1−2α)p(z)

p(z)

=

(1−2α) +

∞

P k=1

[1 + 2α(k−1)]pkzk

1 +

∞

P k=1

pkzk

≤

|1−2α|+

∞

P k=1

[1 + 2α(k−1)]|pk||z|k

1−

∞

P k=1

|pk||zk

<

|1−2α|+

∞

P k=1

[1 + 2α(k−1)]|pk|

1−

∞

P k=1

|pk|

.

Now, by using the inequality (8), we have |1−2α|+

∞

P k=1

[1 + 2α(k−1)]|pk|

1 +

∞

P k=1

|pk|

<1, (10)

which, combined with (9) and (10), completes the proof of theorem. Example 3.2. The functionp(z)given by

p(z) = 1 +

∞

X k=1

1− |1−2α|

k(k+ 1)[1 +α(k−1)]z n

(5)

By using Jack’s lemma, we now obtain the following result for the classN P∗₍_α_).

z2_p00₍_z₎

zp0

(z)−p(z)−

zp0(z)

p(z)

<1−α. (11)

Thenp∈ N P∗₍_α₎_{, for} 1

2 ≤α <1.

Proof. Let

ϕ(z) = α 1−α

zp0(z)

p(z)

₁

2 ≤α <1;z∈U

. (12)

Then the function ϕ(z)is analytic in U with ϕ(0) = 0 and it follows from (12) that

z2_p00₍_z₎

zp0

(z)−p(z)−

zp0(z)

p(z) =

(1−α)zϕ0(z)

(1−α)ϕ(z)−α, (13)

therefore

z2_p00₍_z₎

zp0

(z)−p(z)−

zp0(z)

p(z) =

(1−α)zϕ0(z) (1−α)ϕ(z)−α

<1−α.

Next, we claim that |ϕ(z)| <1. Indeed, if not, there exists a point z0 ∈U such that

max

|z|≤|z0|

=|ϕ(z0)|= 1.

Applying Jack’s lemma toϕ(z)at the pointsz0, we have

ϕ(z0) =eiθ, z0ϕ

0

(z0)

ϕ(z0)

=k (k≥1).

This gives us z2 0p 00

(z0)

z0p

0

(z0)−p(z0) −z0p

0

(z0)

p(z0) 2 =

k(1−α) (1−α)−αe−iθ

≥

1−α

(1−α)−αe−iθ

.

This implies that

z2₀p00(z0)

z0p

0

(z0)−p(z0) −z0p

0

(z0)

p(z0) 2

≥ (1−α)

2

(1−α)2₊_α2₋₂_α₍₁₋_α_{) cos}_θ

≥ (1−α)

2

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This contradicts to the condition (11). Therefore, we conclude that |ϕ(z)| < 1 which shows that

|ϕ(z)|= α

1−α zp0(z)

p(z) <1 or

zp0(z)

p(z)

< 1−α α

1

2 ≤α <1;z∈U

.

Thun, we have

zp0(z)

p(z) + 1−2α

2α ≤

zp0(z)

p(z) +

1−2α

2α

<1−α

α −

1−2α

2α =

1 2α,

which completes the proof.

Example 3.4. Let us consider the functionp(z)∈ P given by

p(z) = 1 +p1z (z∈U), with

p1= 1−α

2−α

for some 1₂ ≤α <1, then we see that0< p1≤ 1₃. According to

z2_p00₍_z₎

zp0(z)−p(z)−

zp0(z)

p(z) =

−p1z 1 +p1z

< p1

1−p1

= 1−α,

and

<

_p₍_z₎

zp0

(z)−p(z)

=<(−1−p1z)≤p1−1 = 1

α−2 <−α, we have,p(z)∈ N P∗₍_α₎_for 1

2 ≤α <1. Theorem 3.5. If p∈ P satisfies

R

z2_p00₍_z₎

zp0(z)−p(z)−

zp0(z)

p(z) ! <        α

2(1−α) (0≤α≤ 1 2), 1−α

2α (

1

2 ≤α <1),

(14)

thenp∈ N P∗₍_α₎_{, for} ₀_≤_{α <}₁_.

Proof. Suppose that

q(z) =

− p(z)

zp0(z)−p(z)−α

(7)

thenqis analytic in_U. It follows from (15) that

zp0(z)

p(z) −

z2_p0₍_z₎

zp0

(z)−p(z)=

(1−α)zq0(z)

α+ (1−α)q(z) =φ(q(z), zq

0

(z);z),

where

φ(r, s;z) = (1−α)s

α+ (1−α)r.

For all real numbersxandy satisfyingy≤ −1+x2

2 , we have <(φ(ix, y;z)) = (1−α)αy

α2_{+ (1}₋_α₎2_x ≤ −(1−α)α

2 ·

1 +x2 α2_{+ (1}₋_α₎2_x2

≤         

−(1−α)α

2 ·

1

(1−α)2 =−

α

2(1−α) (0≤α≤ 1 2),

−(1−α)α

2 ·

1

α2 =− 1−α

2α (

1

2 ≤α <1). We now put

Ω =   

z:<(z)>

  

− α

2(1−α) (0≤α≤ 1 2) −1−α

2α ( 1

2 ≤α <1)   

,

thenφ(ix, y;z)∈/ Ω for allx, y such thaty ≤ −1+x2

2 . Moreover, in view of (14), we getφ(q(z), zq0(z);z). Thus, by Lemma 2.5, we deduce that

<(q(z))>0 (z∈U), which implies thatp∈ N P∗₍_α_).

< p(z)

zp0(z)−p(z) 1 +β

z2_p00₍_z₎

zp0(z)−p(z) !!

<1

2β(α+ 3)−α, (16)

thenp∈ N P∗₍_α₎_{, for} ₀_≤_{α <}₁ _and_β_≥₀_.

Proof. Suppose that

q(z) =

− p(z)

zp0(z)−p(z)−α

(8)

Thenφis analytic in_U. It follows from (17) that

1 +β z

2_p00₍_z₎

zp0(z)−p(z) =

β[(1−α)zq0(z)−1]

(1−α)q(z) +α + 1−β. (18)

Combining with (17) and (18), we get

p(z)

zp0

(z)−p(z) 1 +β

z2p00(z)

p0

(z)−p(z) !

=β(1−α)zp0(z) + (1−β)(1−α)p(z) + (1−β)α−β

=φ(q(z), zq0(z);z),

where

φ(r, s;z) =β(1−α)s+ (1−β)(1−α)r+ (1−β)α−β.

For all real numbersxandy satisfyingy≤ −1+x2

2 , we have <(φ(ix, y;z)) =β(1−α)y+ (1−β)α−β

≤ −β(1−α) 2 (1 +x

2_{) + (1}₋_β₎_α₋_β

≤ −β(1−α)

2 + (1−β)α−β =α−1

2β(α+ 3) (0≤α <1). If we set

Ω ={z:<(z)> α−1

2β(α+ 3)}, then, by Lemma 2.5, we conclude that

<(q(z))>0 (z∈_U),

which implies the assertion of theorem holds. Theorem 3.7. If p∈ P satisfies

1−2α+2αzp

0

(z)

p(z) !0

≤β|z|γ, (19)

thenp∈ N P∗₍_α₎_{, for} ₀_{< α <}₁ _,₀_{< β}_≤_γ_{+ 1} _and_γ_≥₀_.

Proof. Forp∈ P, we set

q(z) =z 1 + 2α+2αzp

0

(z)

p(z) !

(9)

thenq(z)is regular in_Uandq(0) = 0. The condition of the theorem gives us

1−2α+2αzp

0

(z)

p(z) !0

=

q(z)

z

0

≤β|z|γ ₍_z_∈ U).

It follow that

_q₍_z₎

z

0

=

Z z 0

_q₍_u₎

u

0

du

≤ Z |z|

0

β|u|γ_d_|_u_|₌ β

γ+ 1|z| γ+1_.

This implies that

_q₍_z₎

z

0

≤ β

γ+ 1|z|

γ+1_<₁ ₍₁_{< β < γ}_{+ 1}_{, γ}_≥₀₎_.

Therefore, by definition ofq(z), we have

1−2α+2αzp

0

(z)

p(z)

<1

or

zp0(z)

p(z) + 1−2α

2α

< 1

2α.

This implies thatp∈ N P∗₍_α₎_.

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Mohammad Taati

Department of Mathematics, Payame Noor University,

P.O.Box 19395-3697, Tehran, Iran

E-mail: [email protected] & [email protected] Sirous Moradi

Department of Mathematics, Faculty of Science,

Arak University,

Arak 38156-8-8349, Iran

E-mail: [email protected] & [email protected] Shahram Najafzadeh

Department of Mathematics, Payame Noor University,