Home Mathematics Simple sufficient conditions for starlikeness and convexity for meromorphic functions
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Simple sufficient conditions for starlikeness and convexity for meromorphic functions

  • Pranay Goswami , Teodor Bulboacă and Rubayyi T. Alqahtani EMAIL logo
Published/Copyright: August 12, 2016

Abstract

In this paper we investigate some extensions of sufficient conditions for meromorphic multivalent functions in the open unit disk to be meromorphic multivalent starlike and convex of order α. Our results unify and extend some starlikeness and convexity conditions for meromorphic multivalent functions obtained by Xu et al. [2], and some interesting special cases are given.

MSC 2010: 30C45

1 Introduction and definitions

Let Σp, n denote the class meormorphic multivalent functions of the form

f(z)=zp+k=nakpzkp(p,nN:={1,2,3}),(1)

which are analytic in the punctured unit disk U:=u{0}, where U:={zC:z<1}.

Definition 1.1

  1. A function f ∈ Σp, n is said to be meromorphic starlike functions of order α, if it satisfies the inequality

    Rezf(z)f(z)>α,zU,

    for some real α (0 ≤ α < p), and we denote this subclass byΣSp,n(α).

  2. A function f ∈ Σp, n is said to be meromorphic convex functions of order α, if it satisfies the inequality

    Re1+zf(z)f(z)>α,zU,

    for some real α (0 ≤ α < p), and we denote this subclass by ΣKp, n(α).

Let An denote the class of analytic functions in U of the form

f(z)=z+k=nak+1zk+1,zU(nN).
Definition 1.2

  1. LetSndenote the class of n-starlike functions inU, i.e. fAnand satisfies

    Rezf(z)f(z)>0,zU.
  2. Further, we denote byKnthe class of n-convex functions inU, i.e. fAnand satisfies

    Re1+zf(z)f(z)>0,zU.

In the recent papers of Goyal et al. [1] and Xu et al. [2], the authors obtained some sufficient conditions for multivalent and meromorphic starlikeness and convexity, respectively. In this paper we will derive some extensions of these sufficient conditions for starlikeness and convexity of order α for meromorphic multivalent functions.

2 Main results

In order to find some simple sufficient conditions for the starlikeness and convexity of order α for a function f ∈ Σp, n, we will recall the following lemma due to P. T. Mocanu (see also [3]):

Lemma 2.1

Lemma 2.1 ([4, Corollary 1.2])

IffAnand satisfies the inequality

f(z)1<n+1(n+1)2+1,zU,

thenfSn.

Remark that, for the special case n = 1, this result was previously obtained in [5, Theorem 3].

Theorem 2.2

If f ∈ Σp, n, with f(z) ≠ 0 for allzU, satisfies the inequality

zpf(z)1αpzf(z)f(z)+α+pα<(n+1)(pα)(n+1)2+1,zU,(2)

for some real values of α (0 ≤ α < p), thenfΣSp,n(α). (The power is the principal one).

Proof

For f ∈ Σp, n, with f(z) ≠ 0 for all zU, let us define a function h by

h(z)=zzpf(z)1αp,zU.(3)

Since f ∈ Σp, n and f(z) ≠ 0 for all zU, it follows that the power function

φ(z)=zpf(z)1αp

has an analytic branch in U with φ(0) = 1, so h is analytic in U. If the function f is of the form (1), a simple computation shows that

h(z)=z+anpαpzn+1+,

zU. Thus hAn.

Now, differentiating logarthmically the definition relation (3) we obtain that

h(z)h(z)=1pαf(z)f(z)+αz,zU,(4)

which gives

h(z)1=1pαzpf(z)1αpzf(z)f(z)+α+pα,zU.

From the above relation, by using the assumption (2) of the theorem we get

h(z)1<n+1(n+1)2+1,zU,

hence, according to Lemma 2.1, we deduce that hSn.

Using again (4), we get

zh(z)h(z)=1pαzf(z)f(z)+α,

and according to the fact that hSn, the above relation implies

Rezf(z)f(z)>α,zU,

that is fΣSp,n(α). □

Theorem 2.3

For f ∈ Σp, n, with f(z) ≠ 0 for allzU, let’s define the function h by

h(z)=zzpf(z)1αp,zU(0α<p).(5)

If h satisfies the inequality

h(z)n+1(n+1)2+1,zU,(6)

thenfΣSp,n(α). (The power is the principal one).

Proof

As in the proof of Theorem 2.2, we have hAn. Moreover, from the assumption (6) we deduce that

h(z)1=0zh(t)dt0zh(ρeiθ)dρ(n+1)z(n+1)2+1<n+1(n+1)2+1,zU.

Therefore, the function h satisfies the condition of Lemma 2.1, and thus hSn. Now, using the same reasons as in the last part of the proof of Theorem 2.2, we finally obtain that fΣSp,n(α). □

Next, we will give some sufficient conditions for a function f ∈ Σp, n to be a convex function of order α.

Theorem 2.4

If f ∈ Σp, n, with f′(z) ≠ 0 for allzU, satisfies the inequality

zp+1f(z)p1αp1+zf(z)f(z)+α+pα<(n+1)(pα)(n+1)2+1,zU,(7)

for some real values of α (0 ≤ α < p), then f ∈ ΣKp, n(α). (The power is the principal one).

Proof

For f ∈ Σp, n with with f′(z) ≠ 0 for all zU, the power function

φ(z)=zp+1f(z)p1αp(8)

has an analytic branch in U with φ(0) = 1, and if the function f is of the form (1), then

φ(z)=1npp(αp)anpzn+,zU.

It follows that the function h defined by

h(z)=0zφ(t)dt=znpp(αp)(n+1)anpzn+1+,zU,(9)

belongs to An. Thus, we deduce that the function g defined by

g(z)=zh(z)=znpp(αp)anpzn+1+,zU,

is in An. From the above definition relation, we get

g(z)=1αpzp+1f(z)p1αp1+zf(z)f(z)+α.

From here and using the assumption (7), we obtain

g(z)1<n+1(n+1)2+1,zU.

Therefore, from Lemma 2.1 it follows that g(z)=zh(z)Sn, which is equivalent to hKn. Noting that

zh(z)h(z)=1αpzf(z)f(z)+p+1,

since hKn, we obtain that

Re1+zh(z)h(z)=Re1αp1+α+zf(z)f(z)>0,zU,

that is

Re1+zf(z)f(z)>α,zU,

hence f ∈ ΣKp, n(α). □

Theorem 2.5

If f ∈ Σp, n, with f′(z) ≠ 0 for allzU, satisfies the inequality

1zzp+1f(z)p1αp1+zf(z)f(z)+p(n+1)(pα)2(n+1)2+1,zU,(10)

for some real values of α (0 ≤ α < p), then f ∈ ΣKp, n(α). (The power is the principal one).

Proof

For f ∈ Σp, n with f′(z) ≠ 0 for allzU, the function φ defined by (8) is in An, therefore the function h given by (9) is in An.

Further, letting g(z) = zh′(z), we obtain that

g(z)=znpp(αp)anpzn+1+An,

and

g(z)1=h(z)+zh(z)1h(z)1+zh(z)=0zh(t)dt+zh(z)0zh(ρeiθ)dρ+zh(z),zU.(11)

Since

h(z)=1z(αp)zp+1f(z)p1αp1+zf(z)f(z)+p,

using the assumption (10) we get

h(ρeiθ)=1pα1zzp+1f(z)p1αp1+zf(z)f(z)+pz=ρeiθn+12(n+1)2+1,

and from (11), using again (10) we deduce that

g(z)1(n+1)z2(n+1)2+1+(n+1)z2(n+1)2+1<n+1(n+1)2+1,zU.

According to Lemma 2.1 we obtain that g(z)=zh(z)Sn, which is equivalent to hKn. Consequently, as in the last part of the proof of Theorem 2.4 it follows that f ∈ ΣKp, n(α). □

Remarks 2.6

  1. If we put n = 1 in Theorem 2.2 and Theorem 2.3, we get the results established by Xu et al. [2].

  2. For the special case n = 1, Theorem 2.4 and Theorem 2.5 represent the results of Xu et al. [2].

For f ∈ Σp, n, with f(z) ≠ 0 for all zU, let’s define the function F by

F(z)=0z[tpf(t)]γdt,zU(γC),

where the power is the principal one. Thus, F(z)=z+γn+1anpzn+1+An, and considering this integral operator we derive the next result:

Theorem 2.7

If f ∈ Σp, n, with f(z) ≠ 0 for allzU, satisfies the inequality

γ[zpf(z)]γzzf(z)f(z)+pn+12(n+1)2+1,zU,(12)

forγ1p, thenfΣSp,np+1γ. (The power is the principal one).

Proof

If f ∈ Σp, n, with f(z) ≠ 0 for all zU, then

F(z)=γ[zpf(z)]γzzf(z)f(z)+p,zU.(13)

Defining the function g(z) = zF′(z), it follows that gAn, and

g(z)1=F(z)+zF(z)1F(z)1+zF(z)=0zF(t)dt+zF(z)0zF(ρeiθ)dρ+zF(z),zU.(14)

From (13), using the assumption (12) we get

F(ρeiθ)=γ[zpf(z)]γzzf(z)f(z)+pz=ρeiθn+12(n+1)2+1,

and the inequality (14) combined again with (12) implies that

g(z)1(n+1)z2(n+1)2+1+(n+1)z2(n+1)2+1<n+1(n+1)2+1,zU.

Consequently, from Lemma 2.1 we obtain that g(z)=zF(z)Sn, which is equivalent to FKn. Using the fact that F′(z) = [zp f(z)]γ, it follows that

1+zF(z)F(z)=γp+zf(z)f(z)+1,

and since FKn we conclude that fΣSp,np+1γ. □

3 Special cases

Let’s consider the function f defined by

f(z)=1zp1λ+λsinzzαp,zU,(15)

where 0 ≤ α < p, the power is the principal one, and assuming that the parameter λ ∈ ℂ is chosen such that

1λ1sinzz,zU.(16)

Using MAPLE™ software, from Figure 1a we may see that

maxz11sinzz<0.18,(17)
Fig. 1
Fig. 1

therefore (16) holds whenever λ509=5.555..., and consequently, if λ ∈ ℂ satisfies this inequality then f ∈ Σp,2.

Using again MAPLE™ software, from Figure 1b we have that

maxz1sin2z2<0.28,(18)

and a simple computation leads to

zpf(z)1αpzf(z)f(z)+α+pα=2(pα)λsin2z2<2(pα)λ0.28,zU.

Thus, according to Theorem 2.2 we obtain the following special case:

Example 3.1

If λ ∈ ℂ andλ751410=1.6941..., then

f(z)=1zp1λ+λsinzzαpΣSp,2(α),

for some real values of α (0 ≤ α < p), where the power is the principal one.

Remark 3.2

For the function f given by (15), the function h defined by (5) is of the form

h(z)=(1λ)z+λsinz,zU.

Therefore, h(z)=λsinz, and using MAPLE™ software, the Figure 2a yields that

maxz1sinz<1.2.

Now, according to Theorem 2.3 we obtain the following special case:

If λ ∈ ℂ andλ5210=0.79057..., then

f(z)=1zp1λ+λsinzαpΣSp,2(α),

for some real values of α (0 ≤ α < p), where the power is the principal one.

If we compare the result given by Example 3.1 with the above one, for this special choice of the function f the Example 3.1 gives a better result.

Fig. 2
Fig. 2

As we already proved, if λ509=5.555... then the relation (16) holds. Therefore, there exists a function f ∈ Σp,2 such that

f(z)=pzp+11λ+λsinzzαp,zU,

where 0 ≤ α < p, and the power is the principal one, assuming that λ ∈ ℂ is chosen such that λ509=5.555...

A simple computation combined with (18) shows that

zp+1f(z)p1ap1+zf(z)f(z)+α+pα=2(pα)λsin2z2<2(pα)λ0.28,zU,

and from Theorem 2.4 we obtain the following special case:

Example 3.3

If λ ∈ ℂ andλ751410=1.6941..., then f ∈ Σp,2with

f(z)=pzp+11λ+λsinzzap,zU,(19)

for some real values of α (0 ≤ α < p), is in ΣKp, n(α). (The power is the principal one).

Remark 3.4

Using MAPLE™ software, from Figure 2b we have that

maxz1zcoszsinzz2<0.38.(20)

From a simple computation combined with (20) we get

1zzp+1f(z)p1αp1+zf(z)f(z)+p=(pα)λzcoszsinzz2<(pα)λ0.38,zU,

and using Theorem 2.5 we obtain the next special case:

If λ ∈ ℂ andλ751910=1.2483..., then f ∈ Σp,2that satisfies (19) has, moreover, the property that f ∈ ΣKp, n(α), for some real values of α (0 ≤; α < p), where the power is the principal one.

Comparing the result given by Example 3.3 with the above one, for this special choice of the function f the Example 3.3 gives a better result.

As we proved at the beginning of this section, the function f ∈ Σp,2, where

f(z)=1zp1λ+λsinzz1γ,zU,(21)

with λ ∈ ℂ, λ509=5.555..., and the power is the principal one. Using the inequality (20), we deduce that

γ[zpf(z)]γzzf(z)f(z)+p=λzcoszsinzz2<λ0.38,zU,

and from Theorem 2.7 we obtain the next special case:

Example 3.5

If λ ∈ ℂ andλ751910=1.2483..., then the function f given by (21) is inΣSp,2p+1γ. forγ1p. (The power is the principal one).

Using MAPLE™ software, we could check that the next inequalities hold (see Figures 3a, 3b, 4a, and 4b):

maxz11+z2ez1z<0.22,(22)
maxz1ez1z<0.73,(23)
maxz1ez1<1.73,(24)
maxz1zezez+1z212<0.51.(25)
Fig. 3
Fig. 3
Fig. 4
Fig. 4

From (22) and (23), using Theorem 2.2 we may easily obtain the following special case:

Example 3.6

If λ ∈ ℂ andλ3007310=1.2996..., then

f(z)=1zp1+λez1z1z2αpΣSp,2(α),

for some real values of α (0 ≤ α < p), where the power is the principal one.

Remark 3.7

From (22) and (24), according to Theorem 2.3 we could similarly obtain the next special case:

If λ ∈ ℂ andλ30017310=0.54837..., then

f(z)=1zp1+λez1z1z2αpΣSp,2(α),

for some real values of α (0 ≤ α < p), where the power is the principal one.

Thus, for this special choice of the function f the Example 3.6 gives a better result.

From (22) and (24), using Theorem 2.4 we easily get the next special case:

Example 3.8

If λ ∈ ℂ andλ3007310=1.2996..., then f ∈ Σp,2with

f(z)=pzp+11+λez1z1z2αp,zU,(26)

for some real values of α (0 ≤ α < p), is in ΣKp, n(α). (The power is the principal one).

Remark 3.9

From (22) and (25), according to Theorem 2.5 we could similarly obtain the next special case:

If λ ∈ ℂ andλ1505110=0.93008..., then f ∈ Σp,2that satisfies (26) has, moreover, the property that f ∈ ΣKp, n(α), for some real values of α (0 ≤ α < p), where the power is the principal one.

Consequently, for this special choice of the function f the Example 3.8 gives a better result.

Finally, from the inequalities (22) and (25), using Theorem 2.7 we obtain the next special case:

Example 3.10

If λ ∈ ℂ andλ1505110=0.93008..., then

f(z)=1zp1+λez1z1z21γ,ΣSp,2p+1γ,

forγ1p. (The power is the principal one).

We will omit the detailed proofs of the last three examples, since these are similar with the previous ones.


Rubayyi T. Alqahtani: Department of Mathematics and Statistics, College of Science, Al-Imam Mohammad Ibn Saud Islamic University (IMSIU), P.O.Box 65892, Riyadh 11566, Saudi Arabia

References

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Received: 2016-1-31
Accepted: 2016-6-10
Published Online: 2016-8-12
Published in Print: 2016-1-1

© 2016 Goswami et al., published by De Gruyter Open

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

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