Abstract

We consider some subclasses of meromorphic multivalent functions and obtain certain simple sufficiency criteria for the functions belonging to these classes. We also study the mapping properties of these classes under an integral operator.

1. Introduction

Let (𝑝,𝑛) denote the class of functions 𝑓(𝑧) of the form 𝑓(𝑧)=𝑧𝑝+𝑘=𝑛𝑎𝑘𝑧𝑘𝑝+1(𝑝),(1.1) which are analytic and 𝑝-valent in the punctured unit disk 𝕌={𝑧0<|𝑧|<1}. Also let 𝜆(𝑝,𝑛,𝛼) and 𝜆𝑐(𝑝,𝑛,𝛼) denote the subclasses of (𝑝,𝑛) consisting of all functions 𝑓(𝑧) which are defined, respectively, by Re𝑒𝑖𝜆𝑧𝑓(𝑧)𝑓(𝑧)>𝛼cos𝜆(𝑧𝕌),Re𝑒𝑖𝜆𝑧𝑓(𝑧)𝑓(𝑧)>𝛼cos𝜆(𝑧𝕌).(1.2)

We note that for 𝜆=0 and 𝑛=1, the above classes reduce to the well-known subclasses of (𝑝) consisting of meromorphic multivalent functions which are, respectively, starlike and convex of order 𝛼(0𝛼<𝑝). For the detail on the subject of meromorphic spiral-like functions and related topics, we refer the work of Liu and Srivastava [1], Goyal and Prajapat [2], Raina and Srivastava [3], Xu and Yang [4], and Spacek [5] and Robertson [6].

Analogous to the subclass of (1,1) for meromorphic univalent functions studied by Wang et al. [7] and Nehari and Netanyahu [8], we define a subclass 𝜆𝑁(𝑝,𝑛,𝛼) of (𝑝,𝑛) consisting of functions 𝑓(𝑧) satisfying Re𝑒𝑖𝜆𝑧𝑓(𝑧)𝑓(𝑧)<𝛼cos𝜆(𝛼>𝑝,𝑧𝕌).(1.3)

For more details of the above classes see also [9, 10].

Motivated from the work of Frasin [11], we introduce the following integral operator of multivalent meromorphic functions (𝑝)𝐻𝑚,𝑝1(𝑧)=𝑧𝑝+1𝑧0𝑚𝑗=1𝑢𝑝𝑓𝑗(𝑢)𝛼𝑗𝑑𝑢.(1.4)

For 𝑝=1, (1.4) reduces to the integral operator introduced and studied by Mohammed and Darus [12, 13]. Similar integral operators for different classes of analytic, univalent, and multivalent functons in the open unit disk are studied by various authors, see [1419].

In this paper, first, we find sufficient conditions for the classes 𝜆(𝑝,𝑛,𝛼) and 𝜆𝑐(𝑝,𝑛,𝛼) and then study some mapping properties of the integral operator given by (1.4).

We will assume throughout our discussion, unless otherwise stated, that 𝜆 is real with |𝜆|<𝜋/2, 0𝛼<𝑝, 𝑝,𝑛, 𝛼𝑗>0 for 𝑗{1,𝑚}.

To obtain our main results, we need the following Lemmas.

Lemma 1.1 (see [20]). If 𝑞(𝑧)(1,𝑛) with 𝑛1 and satisfies the condition ||𝑧2𝑞||<𝑛(𝑧)+1𝑛2+1(𝑧𝕌),(1.5) then 𝑞(𝑧)0(1,𝑛,0).(1.6)

Lemma 1.2 (see [21]). Let Ω be a set in the complex plane and suppose that Ψ is a mapping from 2×𝕌 to which satisfies Ψ(𝑖𝑥,𝑦,𝑧)Ω for 𝑧𝕌, and for all real 𝑥,𝑦 such that 𝑦(𝑛/2)(1+𝑥2). If (𝑧)=1+𝑐𝑛𝑧𝑛+ is analytic in 𝕌 and Ψ((𝑧),𝑧(𝑧),𝑧)Ω for all 𝑧𝕌, then Re(𝑧)>0.

2. Some Properties of the Classes 𝜆(𝑝,𝑛,𝛼) and 𝜆𝑐(𝑝,𝑛,𝛼)

Theorem 2.1. If 𝑓(𝑧)(𝑝,𝑛) satisfies ||||(𝑧𝑝𝑓(𝑧))𝑒𝑖𝜆/(𝑝𝛼)cos𝜆𝑒𝑖𝜆𝑧𝑓(𝑧)+||||<𝑛𝑓(𝑧)+𝛼cos𝜆+𝑖𝑝sin𝜆(𝑝𝛼)cos𝜆𝑛2(+1𝑝𝛼)cos𝜆(𝑧𝕌),(2.1) then 𝑓(𝑧)𝜆(𝑝,𝑛,𝛼).

Proof. Let us set a function (𝑧) by 1(𝑧)=𝑧(𝑧𝑝𝑓(𝑧))𝑒𝑖𝜆/(𝑝𝛼)cos𝜆=1𝑧+𝑒𝑖𝜆𝑎𝑛𝑧(𝑝𝛼)cos𝜆𝑛+(2.2) for 𝑓(𝑧)(𝑝,𝑛). Then clearly (2.2) shows that (𝑧)(1,𝑛).
Differentiating (2.2) logarithmically, we have (𝑧)=𝑒(𝑧)𝑖𝜆𝑓(𝑝𝛼)cos𝜆(𝑧)+𝑝𝑓(𝑧)𝑧1𝑧(2.3) which gives ||𝑧2||=||||(𝑧)+1(𝑧𝑝𝑓(𝑧))𝑒𝑖𝜆/(𝑝𝛼)cos𝜆1𝑒(𝑝𝛼)cos𝜆𝑖𝜆𝑧𝑓(𝑧)𝑓||||.(𝑧)+𝛼cos𝜆+𝑖𝑝sin𝜆+1(2.4)
Thus using (2.1), we have ||𝑧2||𝑛(𝑧)+1𝑛2+1(𝑧𝕌).(2.5)
Hence, using Lemma 1.1, we have (𝑧)0(1,𝑛,0).
From (2.3), we can write 𝑧(𝑧)=1(𝑧)(𝑝𝛼)cos𝜆𝑒𝑖𝜆𝑧𝑓(𝑧).𝑓(𝑧)+(𝛼cos𝜆+𝑖𝑝sin𝜆)(2.6)
Since (𝑧)0(1,𝑛,0), it implies that Re(𝑧(𝑧)/(𝑧))>0. Therefore, we get 1(𝑝𝛼)cos𝜆Re𝑒𝑖𝜆𝑧𝑓(𝑧)𝑓(𝑧)𝛼cos𝜆=Re𝑧(𝑧)(𝑧)>0(2.7) or Re𝑒𝑖𝜆𝑧𝑓(𝑧)𝑓(𝑧)>𝛼cos𝜆,(2.8) and this implies that 𝑓(𝑧)𝜆(𝑝,𝑛,𝛼).
If we take 𝜆=0, we obtain the following result.

Corollary 2.2. If 𝑓(𝑧)(𝑝,𝑛) satisfies ||||(𝑧𝑝𝑓(𝑧))1/(𝑝𝛼)𝑒𝑖𝜆𝑧𝑓(𝑧)+||||<1𝑓(𝑧)+𝛼(𝑝𝛼)2(𝑝𝛼)(𝑧𝕌),(2.9) then 𝑓(𝑧)(𝑝,𝑛,𝛼).

Theorem 2.3. If 𝑓(𝑧)(𝑝,𝑛) satisfies |||||𝑧𝑝+1𝑓(𝑧)𝑝𝑒𝑖𝜆/(𝑝𝛼)cos𝜆𝑒𝑖𝜆𝑧𝑓(𝑧)𝑓|||||<(𝑧)+1+𝛼cos𝜆+𝑖𝑝sin𝜆+(𝑝𝛼)cos𝜆(𝑛+1)(𝑝𝛼)cos𝜆(𝑛+1)2(+1𝑧𝕌),(2.10) then 𝑓(𝑧)𝜆𝑐(𝑝,𝑛,𝛼).

Proof. Let us set (𝑧)=𝑧01𝑡2𝑡𝑝+1𝑓(𝑡)𝑝𝑒𝑖𝜆/(𝑝𝛼)cos𝜆1𝑑𝑡=𝑧+𝑛𝑝+1𝑒𝑛𝑝𝑖𝜆𝑎𝑛𝑧(𝑝𝛼)cos𝜆𝑛+.(2.11)
Also let 𝑔(𝑧)=𝑧1(𝑧)=𝑧𝑧𝑝+1𝑓(𝑧)𝑝𝑒𝑖𝜆/(𝑝𝛼)cos𝜆=1𝑧+𝑝𝑛1𝑝𝑒𝑖𝜆𝑎𝑛𝑧(𝑝𝛼)cos𝜆𝑛+.(2.12)
Then clearly (𝑧) and 𝑔(𝑧)(1,𝑛). Now 1𝑔(𝑧)=𝑧𝑧𝑝+1𝑓(𝑧)𝑝𝑒𝑖𝜆/(𝑝𝛼)cos𝜆.(2.13)
Differentiating logarithmically and then simple computation gives us ||𝑧2𝑔||=|||||𝑧(𝑧)+1𝑝+1𝑓(𝑧)𝑝𝑒𝑖𝜆/(𝑝𝛼)cos𝜆1𝑒(𝑝𝛼)cos𝜆𝑖𝜆𝑧𝑓(𝑧)𝑓|||||<𝑛(𝑧)+1+𝛼cos𝜆+𝑖𝑝sin𝜆+1𝑛2.+1(2.14)
Therefore, by using Lemma 1.1, we have 𝑔(𝑧)=𝑧(𝑧)0(1,𝑛,0)(2.15) which implies that (𝑧)0𝑐(1,𝑛,0). Since 1+𝑧(𝑧)=𝑒(𝑧)𝑖𝜆(𝑝𝛼)cos𝜆𝑧𝑓(𝑧)𝑓(𝑧)+(𝑝+1)1,(2.16) therefore Re1+𝑧(𝑧)=1(𝑧)𝑒(𝑝𝛼)cos𝜆Re𝑖𝜆1+𝑧𝑓(𝑧)𝑓(𝑧)+𝑝𝑒𝑖𝜆=1(𝑝𝛼)cos𝜆(𝑝𝛼)cos𝜆Re𝑒𝑖𝜆1+𝑧𝑓(𝑧)𝑓.(𝑧)+𝛼cos𝜆(2.17) Since (𝑧)0𝑐(1,𝑛,0), so 1(𝑝𝛼)cos𝜆Re𝑒𝑖𝜆1+𝑧𝑓(𝑧)𝑓(𝑧)+𝛼cos𝜆>0,(2.18) or Re𝑒𝑖𝜆1+𝑧𝑓(𝑧)𝑓(𝑧)>𝛼cos𝜆.(2.19)
It follows that 𝑓(𝑧)𝜆𝑐(𝑝,𝑛,𝛼).

Theorem 2.4. If 𝑓(𝑧)(𝑝,𝑛) satisfies 𝑒Re𝑖𝜆𝑧𝑓(𝑧)𝛼𝑓(𝑧)𝑧𝑓(𝑧)𝑓>𝑀(𝑧)124𝐿+𝑁(𝑧𝕌),(2.20)
then 𝑓(𝑧)𝜆(𝑝,𝑛,𝛽), where 0𝛼1, 0𝛽<𝑝 and 𝑛𝐿=𝛼(𝑝𝛽)2𝛽+(𝛽𝑝)cos2𝜆cos𝜆,𝑀=𝛼(𝛽𝑝)(1𝛽cos𝜆)sin2𝜆cos𝜆,𝑁=𝛼2cos2𝜆+sin2𝑛𝜆2(𝛽𝑝)cos𝜆+(1+𝛼)𝛽cos𝜆.(2.21)

Proof. Let us set 𝑒𝑖𝜆𝑧𝑓(𝑧)=[]𝑓(𝑧)(𝛽𝑝)(𝑧)𝛽cos𝜆𝑖sin𝜆.(2.22)
Then (𝑧) is analytic in 𝕌 with 𝑝(0)=1.
Taking logarithmic differentiation of (2.22) and then by simple computation, we obtain 𝑒𝑖𝜆𝑧𝑓(𝑧)𝛼𝑓(𝑧)𝑧𝑓(𝑧)𝑓(𝑧)1=𝐴𝑧(𝑧)+𝐵2(𝑧)+𝐶(𝑧)+𝐷=Ψ(𝑧),𝑧(𝑧),𝑧(2.23) with 𝐴=(𝛽𝑝)𝛼cos𝜆,𝐵=𝛼𝑒𝑖𝜆(𝛽𝑝)2cos2𝜆,𝐶=(𝑝𝛽)𝛼𝑒𝑖𝜆2𝛽cos2𝜆+𝑖sin2𝜆+(1+𝛼)𝑒𝑖𝜆,cos𝜆𝐷=𝛼𝑒𝑖𝜆𝛽2cos2𝜆sin2𝜆+𝑖𝛽sin2𝜆+(1+𝛼)(𝛽cos𝜆𝑖sin𝜆).(2.24)
Now for all real 𝑥 and 𝑦 satisfying 𝑦(𝑛/2)(1+𝑥2), we have Ψ(𝑖𝑥,𝑦,𝑧)=𝐴𝑦𝐵𝑥2+𝐶(𝑖𝑥)+𝐷.(2.25)
Reputing the values of 𝐴, 𝐵, 𝐶, and 𝐷 and then taking real part, we obtain ReΨ(𝑖𝑥,𝑦,𝑧)𝐿𝑥2+𝑀𝑥+𝑁=𝑀𝐿𝑥2𝐿2+𝑀2<𝑀4𝐿+𝑁24𝐿+𝑁,(2.26) where 𝐿, 𝑀, and 𝑁 are given in (2.21).
Let Ω={𝑤Re𝑤>(𝑀2/4𝐿)+𝑁}. Then Ψ((𝑧),𝑧(𝑧),𝑧)Ω and Ψ(𝑖𝑥,𝑦,𝑧)Ω, for all real 𝑥 and 𝑦 satisfying 𝑦(𝑛/2)(1+𝑥2), 𝑧𝕌. By using Lemma 1.2, we have Re(𝑧)>0, that is 𝑓(𝑧)𝜆(𝑝,𝑛,𝛽).
If we put 𝜆=0, we obtain the following result.

Corollary 2.5. If 𝑓(𝑧)(𝑝,𝑛) satisfies Re𝑧𝑓(𝑧)𝛼𝑓(𝑧)𝑧𝑓(𝑧)𝑓𝛽(𝑧)1>𝛼2𝑛2(𝛽𝑝)+(1+𝛼)𝛽(𝑧𝕌),(2.27) then 𝑓(𝑧)(𝑝,𝑛,𝛽), where 0𝛼1, 0𝛽<𝑝.

Theorem 2.6. For 𝑗{1,𝑚}, let 𝑓𝑗(𝑧)(𝑝,𝑛) and satisfy (2.9). If 𝑚𝑗=1𝛼𝑗<𝑝+1,𝑝𝛽(2.28) then 𝐻𝑚,𝑝(𝑧)𝑁(𝑝,𝑛,𝜁), where 𝜁>𝑝.

Proof. From (1.4), we obtain 𝑧𝑝+1𝐻𝑚,𝑝(𝑧)+(𝑝+1)𝑧𝑝𝐻𝑚,𝑝(𝑧)=𝑚𝑗=1𝑧𝑝𝑓𝑗(𝑧)𝛼𝑗.(2.29)
Differentiating again logarithmically and then by simple computation, we get 𝑧𝐻𝑚,𝑝(𝑧)𝐻𝑚,𝑝𝑝(𝑧)+1+2𝐻1𝑚,𝑝(𝑧)𝑧𝐻𝑚,𝑝𝐻(𝑧)+2𝑝=1+(𝑝+1)𝑚,𝑝(𝑧)𝑧𝐻𝑚,𝑝(𝑧)𝑚𝑗=1𝛼𝑗𝑧𝑓𝑗(𝑧)𝑓𝑗,(𝑧)+𝑝1(2.30) or, equivalently we can write 𝑧𝐻𝑚,𝑝(𝑧)𝐻𝑚,𝑝=𝐻(𝑧)+1𝑚,𝑝(𝑧)𝑧𝐻𝑚,𝑝(𝑧)(𝑝+1)𝑚𝑗=1𝛼𝑗𝑧𝑓𝑗(𝑧)𝑓𝑗+𝑝(𝑧)𝑝+12+1𝑚𝑗=1𝛼𝑗𝑧𝑓𝑗(𝑧)𝑓𝑗(𝑧)𝑝+(1+2𝑝).(2.31) Now taking real part on both sides, we obtain Re𝑧𝐻𝑚,𝑝(𝑧)𝐻𝑚,𝑝𝐻(𝑧)+1=Re𝑚,𝑝(𝑧)𝑧𝐻𝑚,𝑝(𝑧)(𝑝+1)𝑚𝑗=1𝛼𝑗𝑧𝑓𝑗(𝑧)𝑓𝑗+𝑝(𝑧)𝑝+12+1𝑚𝑗=1𝛼𝑗Re𝑧𝑓𝑗(𝑧)𝑓𝑗(𝑧)𝑝+(1+2𝑝).(2.32) This further implies that Re𝑧𝐻𝑚,𝑝(𝑧)𝐻𝑚,𝑝|||||𝐻(𝑧)+1𝑚,𝑝(𝑧)𝑧𝐻𝑚,𝑝(𝑧)(𝑝+1)𝑚𝑗=1𝛼𝑗𝑧𝑓𝑗(𝑧)𝑓𝑗+𝑝(𝑧)𝑝+12|||||+1𝑚𝑗=1𝛼𝑗Re𝑧𝑓𝑗(𝑧)𝑓𝑗(𝑧)𝑝+(1+2𝑝).(2.33)
Let |||||𝐻𝜁=𝑚,𝑝(𝑧)𝑧𝐻𝑚,𝑝(𝑧)(𝑝+1)𝑚𝑗=1𝛼𝑗𝑧𝑓𝑗(𝑧)𝑓𝑗+𝑝(𝑧)𝑝+12|||||+1𝑚𝑗=1𝛼𝑗Re𝑧𝑓𝑗(𝑧)𝑓𝑗.(𝑧)𝑝+(1+2𝑝)(2.34) Clearly we have 𝜁>𝑚𝑗=1𝛼𝑗Re𝑧𝑓𝑗(𝑧)𝑓𝑗.(𝑧)𝑝+(1+2𝑝)(2.35) Then by using (2.28) and Corollary 2.2, we obtain 𝜁>𝑚𝑗=1𝛼𝑗(𝛽𝑝)+(1+2𝑝)>𝑝.(2.36) Therefore 𝐻𝑚,𝑝(𝑧)𝑁(𝑝,𝑛,𝜁) with 𝜁>𝑝.
Making use of (2.27) and Corollary 2.5, one can prove the following result.

Theorem 2.7. For 𝑗{1,𝑚}, let 𝑓𝑗(𝑧)(𝑝,𝑛) and satisfy (2.27). If 𝑚𝑗=1𝛼𝑗<𝑝+1,𝑝𝛽(2.37) then 𝐻𝑚,𝑝(𝑧)𝑁(𝑝,𝑛,𝜁), where 𝜁>𝑝.

Acknowledgment

The author would like to thank Prof. Dr. Ihsan Ali, Vice Chancellor Abdul Wali Khan University Mardan for providing excellent research facilities and financial support.