Abstract

A new generalised derivative operator 𝜇𝜆𝑛,𝑚1,𝜆2 is introduced. This operator generalised many well-known operators studied earlier by many authors. Using the technique of differential subordination, we will study some of the properties of differential subordination. In addition we investigate several interesting properties of the new generalised derivative operator.

1. Introduction and Preliminaries

Let 𝒜 denote the class of functions of the form

𝑓(𝑧)=𝑧+𝑘=2𝑎𝑘𝑧𝑘,where𝑎𝑘iscomplexnumber,(1.1) which are analytic in the open unit disc 𝑈={𝑧|𝑧|<1} on the complex plane . Let 𝑆,𝑆(𝛼),𝐶(𝛼)(0𝛼<1) denote the subclasses of 𝒜 consisting of functions that are univalent, starlike of order 𝛼, and convex of order 𝛼 in 𝑈, respectively. In particular, the classes 𝑆(0)=𝑆and𝐶(0)=𝐶 are the familiar classes of starlike and convex functions in 𝑈, respectively. A function 𝑓𝐶(𝛼) if Re(1+𝑧𝑓/𝑓)>𝛼. Furthermore a function 𝑓 analytic in 𝑈 is said to be convex if it is univalent and 𝑓(𝑈) is convex.

Let (𝑈) be the class of holomorphic function in unit disc 𝑈={𝑧|𝑧|<1}.

Let

𝒜𝑛=𝑓(𝑈)𝑓(𝑧)=𝑧+𝑎𝑛+1𝑧𝑛+1,+,(𝑧𝑈)(1.2) with 𝒜1=𝒜.

For 𝑎 and 𝑛={1,2,3,} we let

[]=𝑎,𝑛𝑓(𝑈)𝑓(𝑧)=𝑧+𝑎𝑛𝑧𝑛+𝑎𝑛+1𝑧𝑛+1.+,(𝑧𝑈)(1.3) Let 𝑓(𝑧)=𝑧+𝑘=2𝑎𝑘𝑧𝑘and𝑔(𝑧)=𝑧+𝑘=2𝑏𝑘𝑧𝑘be analytic in the open unit disc 𝑈={𝑧|𝑧|<1}. Then the Hadamard product (or convolution) 𝑓𝑔 of the two functions 𝑓, 𝑔 is defined by

𝑓(𝑧)𝑔(𝑧)=(𝑓𝑔)(𝑧)=𝑧+𝑘=2𝑎𝑘𝑏𝑘𝑧𝑘.(1.4) Next, we state basic ideas on subordination. If 𝑓 and 𝑔 are analytic in 𝑈, then the function 𝑓 is said to be subordinate to 𝑔, and can be written as

𝑓𝑔,𝑓(𝑧)𝑔(𝑧),(𝑧𝑈),(1.5) if and only if there exists the Schwarz function 𝑤, analytic in 𝑈, with 𝑤(0)=0and|𝑤(𝑧)|<1such that𝑓(𝑧)=𝑔(𝑤(𝑧)), (𝑧𝑈).

Furthermore if 𝑔 is univalent in 𝑈, then 𝑓𝑔 if and only if 𝑓(0)=𝑔(0) and 𝑓(𝑈)𝑔(𝑈) (see [1, page 36]).

Let 𝜓3×𝑈 and let be univalent in 𝑈. If 𝑝 is analytic in 𝑈 and satisfies the (second-order) differential subordination

𝜓𝑝(𝑧),𝑧𝑝(𝑧),𝑧2𝑝(𝑧);𝑧(𝑧),(𝑧𝑈),(1.6) then 𝑝 is called a solution of the differential subordination.

The univalent function 𝑞 is called a dominant of the solutions of the differential subordination, or more simply a dominant, if 𝑝𝑞 for all 𝑝 satisfying (1.6). A dominant ̃𝑞 that satisfies ̃𝑞𝑞 for all dominants 𝑞 of (1.6) is said to be the best dominant of (1.6). (Note that the best dominant is unique up to a rotation of 𝑈.)

Now, (𝑥)𝑘 denotes the Pochhammer symbol (or the shifted factorial) defined by

(𝑥)𝑘=1for𝑘=0,𝑥{0},𝑥(𝑥+1)(𝑥+2)(𝑥+𝑘1)for𝑘={1,2,3,},𝑥.(1.7) To prove our results, we need the following equation throughout the paper:

𝜇𝜆𝑛,𝑚+11,𝜆2𝑓(𝑧)=1𝜆1𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)+𝜆1𝑧𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧),(𝑧𝑈),(1.8) where 𝑛,𝑚0={0,1,2,}, 𝜆2𝜆10, and 𝜙𝜆2(𝑧) is analytic function given by

𝜙𝜆2(𝑧)=𝑧+𝑘=2𝑧𝑘1+𝜆2.(𝑘1)(1.9) Here 𝜇𝜆𝑛,𝑚1,𝜆2 is the generalized derivative operator which we shall introduce later in the paper. Moreover, we need the following lemmas in proving our main results.

Lemma 1.1 (see [2, page 71]). Let be analytic, univalent, and convex in 𝑈, with (0)=𝑎, 𝛾0 and, Re𝛾0. If 𝑝[𝑎,𝑛]and 𝑝(𝑧)+𝑧𝑝(𝑧)𝛾(𝑧),(𝑧𝑈),(1.10) then 𝑝(𝑧)𝑞(𝑧)(𝑧),(𝑧𝑈),(1.11) where 𝑞(𝑧)=(𝛾/𝑛𝑧𝛾/𝑛)𝑧0(𝑡)𝑡(𝛾/𝑛)1𝑑𝑡,(𝑧𝑈).

The function 𝑞 is convex and is the best (𝑎,𝑛)-dominant.

Lemma 1.2 (see [3]). Let 𝑔 be a convex function in 𝑈 and let (𝑧)=𝑔(𝑧)+𝑛𝛼𝑧𝑔(𝑧),(1.12) where 𝛼>0 and 𝑛 is a positive integer.
If
𝑝(𝑧)=𝑔(0)+𝑝𝑛𝑧𝑛+𝑝𝑛+1𝑧𝑛+1+,(𝑧𝑈),(1.13) is holomorphic in 𝑈 and 𝑝(𝑧)+𝛼𝑧𝑝(𝑧)(𝑧),(𝑧𝑈),(1.14) then 𝑝(𝑧)𝑔(𝑧),(1.15) and this result is sharp.

Lemma 1.3 (see [4]). Let 𝑓𝒜, if Re1+𝑧𝑓(𝑧)𝑓1(𝑧)>2,(1.16) then 2𝑧𝑧0𝑓(𝑡)𝑑𝑡,(𝑧𝑈,𝑧0),(1.17) belongs to the class of convex functions.

2. Main Results

In the present paper, we will use the method of differential subordination to derive certain properties of generalised derivative operator 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧). Note that differential subordination has been studied by various authors, and here we follow similar works done by Oros [5] and G. Oros and G. I. Oros [6].

In order to derive our new generalised derivative operator, we define the analytic function

𝐹𝑚𝜆1,𝜆2(𝑧)=𝑧+𝑘=21+𝜆1(𝑘1)𝑚1+𝜆2(𝑘1)𝑚1𝑧𝑘,(2.1) where 𝑚0={0,1,2,} and 𝜆2𝜆10. Now, we introduce the new generalised derivative operator 𝜇𝜆𝑛,𝑚1,𝜆2 as follows.

Definition 2.1. For 𝑓𝒜 the operator 𝜇𝜆𝑛,𝑚1,𝜆2 is defined by 𝜇𝜆𝑛,𝑚1,𝜆2𝒜𝒜𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)=𝐹𝑚𝜆1,𝜆2(𝑧)𝑅𝑛𝑓(𝑧),(𝑧𝑈),(2.2) where 𝑛,𝑚0={0},𝜆2𝜆10,and𝑅𝑛𝑓(𝑧) denotes the Ruscheweyh derivative operator [7], given by 𝑅𝑛𝑓(𝑧)=𝑧+𝑘=2𝑐(𝑛,𝑘)𝑎𝑘𝑧𝑘,𝑛0,,𝑧𝑈(2.3) where 𝑐(𝑛,𝑘)=(𝑛+1)𝑘1/(1)𝑘1.
If 𝑓 is given by (1.1), then we easily find from equality (2.2) that
𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)=𝑧+𝑘=21+𝜆1(𝑘1)𝑚1+𝜆2(𝑘1)𝑚1𝑐(𝑛,𝑘)𝑎𝑘𝑧𝑘,(𝑧𝑈),(2.4) where 𝑛,𝑚0={0,1,2},𝜆2𝜆10,and𝑐(𝑛,𝑘)=(𝑛+𝑘1𝑛)=(𝑛+1)𝑘1/(1)𝑘1.

Remark 2.2. Special cases of this operator include the Ruscheweyh derivative operator in two cases 𝜇𝑛,10,𝜆2𝑅𝑛 and 𝜇𝜆𝑛,01,0𝑅𝑛 [7], the Salagean derivative operator 𝜇0,𝑚1,0𝑆𝑛 [8], the generalised Ruscheweyh derivative operator in two cases 𝜇𝜆𝑛,11,𝜆2𝑅𝑛𝜆 and 𝜇𝜆𝑛,01,𝜆2𝑅𝑛𝜆 [9], the generalised Salagean derivative operator 𝜇𝜆0,𝑚1,0𝑆𝑛𝛽 introduced by Al-Oboudi [10], and the generalised Al-Shaqsi and Darus derivative operator 𝜇𝜆𝑛,𝑚1,0𝐷𝑛𝜆,𝛽 that can be found in [11].

Now, we remind the well-known Carlson-Shaffer operator 𝐿(𝑎,𝑐) [12] associated with the incomplete beta function 𝜙(𝑎,𝑐;𝑧), defined by

𝐿𝐿(𝑎,𝑐)𝒜𝒜,(𝑎,𝑐)𝑓(𝑧)=𝜙(𝑎,𝑐;𝑧)𝑓(𝑧),(𝑧𝑈),(2.5) where

𝜙(𝑎,𝑐;𝑧)=𝑧+𝑘=2(𝑎)𝑘1(𝑐)𝑘1𝑧𝑘,(2.6)𝑎 is any real number, and 𝑐𝑧0;𝑧0={0,1,2,}.

It is easily seen that

𝜇𝜆0,01,0𝑓(𝑧)=𝜇0,𝑚0,0𝑓(𝑧)=𝜇0,10,𝜆2𝑓(𝑧)=𝜇1,20,1𝜇𝑓(z)=𝐿(𝑎,𝑎)𝑓(𝑧)=𝑓(𝑧),𝜆1,01,0𝑓(𝑧)=𝜇1,𝑚0,0𝑓(𝑧)=𝜇1,10,𝜆2𝑓(𝑧)=𝜇𝜆0,01,1𝑓(𝑧)=𝐿(2,1)𝑓(𝑧)=𝑧𝑓(𝑧),(2.7) and also

𝜇𝜆𝑎1,01,0𝑓(𝑧)=𝜇𝑎1,10,𝜆2𝑓(𝑧)=𝜇𝑎1,𝑚0,0𝑓(𝑧)=𝐿(𝑎,1)𝑓(𝑧),(2.8) where 𝑎=1,2,3,.

Next, we give the following.

Definition 2.3. For 𝑛,𝑚0,𝜆2𝜆10, and 0𝛼<1, let 𝑅𝜆𝑛,𝑚1,𝜆2(𝛼) denote the class of functions 𝑓𝒜 which satisfy the condition 𝜇Re𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)>𝛼,(𝑧𝑈).(2.9) Also let 𝐾𝜆𝑛,𝑚1,𝜆2(𝛿) denote the class of functions 𝑓𝒜 which satisfy the condition 𝜇Re𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)>𝛿,(𝑧𝑈).(2.10)

Remark 2.4. It is clear that 𝑅𝜆0,11,0(𝛼)𝑅(𝜆1,𝛼), and the class of functions 𝑓𝒜 satisfying 𝜆Re1𝑧𝑓(𝑧)+𝑓(𝑧)>𝛼,(𝑧𝑈)(2.11) is studied by Ponnusamy [13] and others.

Now we begin with the first result as follows.

Theorem 2.5. Let (𝑧)=1+(2𝛼1)𝑧1+𝑧,(𝑧𝑈),(2.12) be convex in 𝑈, with h(0)=1 and 0𝛼<1. If 𝑛,𝑚0,𝜆2𝜆10, and the differential subordination 𝜇𝑛,𝜆𝑚+11,𝜆2𝑓(𝑧)(𝑧),(𝑧𝑈),(2.13) holds, then 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)𝑞(𝑧)=2𝛼1+2(1𝛼)𝜆1𝑧1/𝜆1𝜎1𝜆1,(2.14) where 𝜎 is given by 𝜎(𝑥)=𝑧0𝑡𝑥11+𝑡𝑑𝑡,(𝑧𝑈).(2.15) The function 𝑞 is convex and is the best dominant.

Proof. By differentiating (1.8), with respect to 𝑧, we obtain 𝜇𝜆𝑛,𝑚+11,𝜆2𝑓(𝑧)=𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)+𝜆1𝑧𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧).(2.16) Using (2.16) in (2.13), differential subordination (2.13) becomes 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)+𝜆1𝑧𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)(𝑧)=1+(2𝛼1)𝑧1+𝑧.(2.17) Let 𝜇𝑝(𝑧)=𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)=𝑧+𝑘=21+𝜆1(𝑘1)𝑚1+𝜆2(𝑘1)𝑚𝑐(𝑛,𝑘)𝑎𝑘𝑧𝑘=1+𝑝1𝑧+𝑝2𝑧2[]+,(𝑝1,1,𝑧𝑈).(2.18) Using (2.18) in (2.17), the differential subordination becomes 𝑝(𝑧)+𝜆1𝑧𝑝(𝑧)(𝑧)=1+(2𝛼1)𝑧.1+𝑧(2.19) By using Lemma 1.1, we have 1𝑝(𝑧)𝑞(𝑧)=𝜆1𝑧1/𝜆1𝑧0(𝑡)𝑡(1/𝜆1)1=1𝑑𝑡𝜆1𝑧1/𝜆1𝑧01+(2𝛼1)𝑡𝑡1+𝑡(1/𝜆1)1𝑑𝑡=2𝛼1+2(1𝛼)𝜆1𝑧1/𝜆1𝜎1𝜆1,(2.20) where 𝜎 is given by (2.15), so we get 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)𝑞(𝑧)=2𝛼1+2(1𝛼)𝜆1𝑧1/𝜆1𝜎1𝜆1.(2.21) The function 𝑞 is convex and is the best dominant. The proof is complete.

Theorem 2.6. If 𝑛,𝑚0,𝜆2𝜆10, and 0𝛼<1, then one has 𝑅𝜆𝑛,𝑚+11,𝜆2(𝛼)𝐾𝜆𝑛,𝑚1,𝜆2(𝛿),(2.22) where 𝛿=2𝛼1+2(1𝛼)𝜆1𝜎1𝜆1,(2.23) and 𝜎 is given by (2.15).

Proof. Let 𝑓𝑅𝜆𝑛,𝑚+11,𝜆2(𝛼), then from (2.9) we have 𝜇Re𝜆𝑛,𝑚+11,𝜆2𝑓(𝑧)>𝛼,(𝑧𝑈),(2.24) which is equivalent to 𝜇𝜆𝑛,𝑚+11,𝜆2𝑓(𝑧)(𝑧)=1+(2𝛼1)𝑧.1+𝑧(2.25) Using Theorem 2.5, we have 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)𝑞(𝑧)=2𝛼1+2(1𝛼)𝜆1𝑧1/𝜆1𝜎1𝜆1.(2.26) Since 𝑞 is convex and 𝑞(𝑈) is symmetric with respect to the real axis, we deduce that 𝜇Re𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)>Re𝑞(1)=𝛿=𝛿𝛼,𝜆1=2𝛼1+2(1𝛼)𝜆1𝜎1𝜆1,(2.27) from which we deduce 𝑅𝜆𝑛,𝑚+11,𝜆2(𝛼)𝐾𝜆𝑛,𝑚1,𝜆2(𝛿). This completes the proof of Theorem 2.6.

Remark 2.7. Special case of Theorem 2.6 with 𝜆2=0 was given earlier in [11].

Theorem 2.8. Let 𝑞 be a convex function in 𝑈, with 𝑞(0)=1, and let (𝑧)=𝑞(𝑧)+𝜆1𝑧𝑞(𝑧),(𝑧𝑈).(2.28) If 𝑛,𝑚0,𝜆2𝜆10,and 𝑓𝒜 and satisfies the differential subordination 𝜇𝜆𝑛,𝑚+11,𝜆2𝑓(𝑧)(𝑧),(𝑧𝑈),(2.29) then 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)𝑞(𝑧),(𝑧𝑈),(2.30) and this result is sharp.

Proof. Using (2.18) in (2.16), differential subordination (2.29) becomes 𝑝(𝑧)+𝜆1𝑧𝑝(𝑧)(𝑧)=𝑞(𝑧)+𝜆1𝑧𝑞(𝑧),(𝑧𝑈).(2.31) Using Lemma 1.2, we obtain 𝑝(𝑧)𝑞(𝑧),(𝑧𝑈).(2.32) Hence 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)𝑞(𝑧),(𝑧𝑈).(2.33) And the result is sharp. This completes the proof of the theorem.

We give a simple application for Theorem 2.8.

Example 2.9. For 𝑛=0,𝑚=1,𝜆2𝜆10,𝑞(𝑧)=(1+𝑧)/(1𝑧),𝑓𝒜, and 𝑧𝑈 and applying Theorem 2.8, we have (𝑧)=1+𝑧1𝑧+𝜆1𝑧1+𝑧1𝑧=1+2𝜆1𝑧𝑧2(1𝑧)2.(2.34) By using (1.8) we find that 𝜇𝜆0,11,𝜆2𝑓(𝑧)=1𝜆1𝑓(𝑧)+𝜆1𝑧𝑓(𝑧).(2.35) Now, 𝜇𝜆0,11,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)=1𝜆1𝑧+𝑘=2𝑎𝑘𝑧𝑘1+𝜆2(𝑘1)+𝜆1𝑧+𝑘=2𝑎𝑘𝑘𝑧𝑘1+𝜆2.(𝑘1)(2.36) A straightforward calculation gives the following: 𝜇𝜆0,11,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)=1+𝑘=21+𝜆1(𝑘1)𝑘𝑎𝑘1+𝜆2𝑧(𝑘1)𝑘1=𝑧+𝑘=2𝑘𝑎𝑘1+𝜆1/(𝑘1)1+𝜆2𝑧(𝑘1)𝑘𝑧=𝑓(𝑧)𝜙𝜆2(𝑧)𝑧+𝑘=2𝑘1+𝜆1𝑧(𝑘1)𝑘𝑧.(2.37) Similarly, using (1.8), we find that 𝜇𝜆0,21,𝜆2𝑓(𝑧)=1𝜆1𝜇𝜆0,11,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)+𝜆1𝑧𝜇𝜆0,11,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧),(2.38) then 𝜇𝜆0,21,𝜆2𝑓(𝑧)=𝜇𝜆0,11,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)+𝜆1𝑧𝜇𝜆0,11,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧).(2.39) By using (2.37) we obtain 𝜇𝜆0,11,𝜆2𝑓(𝑧)𝜙𝜆2(𝑧)=𝑘=2𝑘(𝑘1)𝑎𝑘1+𝜆1(𝑘1)1+𝜆2𝑧(𝑘1)𝑘2.(2.40) We get 𝜇𝜆0,21,𝜆2𝑓(𝑧)=1+𝑘=2𝑘𝑎𝑘1+𝜆1(𝑘1)1+𝜆2𝑧(𝑘1)𝑘1+𝜆1𝑘=2𝑘(𝑘1)𝑎𝑘1+𝜆1(𝑘1)1+𝜆2𝑧(𝑘1)𝑘1=1+𝑘=2𝑘𝑎𝑘1+𝜆1(𝑘1)21+𝜆2𝑧(𝑘1)𝑘1=𝑓(𝑧)𝜙𝜆2(𝑧)𝑧+𝑘=2𝑘1+𝜆1(𝑘1)2𝑧𝑘𝑧.(2.41) From Theorem 2.8 we deduce that 𝑓(𝑧)𝜙𝜆2(𝑧)𝑧+𝑘=2𝑘1+𝜆1(𝑘1)2𝑧𝑘𝑧1+2𝜆1𝑧𝑧2(1𝑧)2,(2.42) implies that 𝑓(𝑧)𝜙𝜆2(𝑧)𝑧+𝑘=2𝑘1+𝜆1𝑧(𝑘1)𝑘𝑧1+𝑧,1𝑧(𝑧𝑈).(2.43)

Theorem 2.10. Let 𝑞 be a convex function in 𝑈, with 𝑞(0)=1 and let (𝑧)=𝑞(𝑧)+𝑧𝑞(𝑧),(𝑧𝑈).(2.44) If 𝑛,𝑚0,𝜆2𝜆10, and 𝑓𝒜 and satisfies the differential subordination 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)(𝑧),(2.45) then 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧𝑞(𝑧),(𝑧𝑈).(2.46) And the result is sharp.

Proof. Let 𝜇𝑝(𝑧)=𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧=𝑧+𝑘=21+𝜆1(𝑘1)𝑚/1+𝜆2(𝑘1)𝑚1𝑐(𝑛,𝑘)𝑎𝑘𝑧𝑘𝑧=1+𝑝1𝑧+𝑝2𝑧2[]+,(𝑝1,1,𝑧𝑈).(2.47) Differentiating (2.47), with respect to 𝑧, we obtain 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)=𝑝(𝑧)+𝑧𝑝(𝑧),(𝑧𝑈).(2.48) Using (2.48), (2.45) becomes 𝑝(𝑧)+𝑧𝑝(𝑧)(𝑧)=𝑞(𝑧)+𝑧𝑞(𝑧).(2.49) Using Lemma 1.2, we deduce that 𝑝(𝑧)𝑞(𝑧),(𝑧𝑈),(2.50) and using (2.47), we have 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧𝑞(𝑧),(𝑧𝑈).(2.51) This proves Theorem 2.10.

We give a simple application for Theorem 2.10.

Example 2.11. For 𝑛=0,𝑚=1,𝜆2𝜆10,𝑞(𝑧)=1/(1𝑧),𝑓𝒜, and 𝑧𝑈 and applying Theorem 2.10, we have 1(𝑧)=11𝑧+𝑧1𝑧=1(1𝑧)2.(2.52) From Example 2.9, we have 𝜇𝜆0,11,𝜆2𝑓(𝑧)=1𝜆1𝑓(𝑧)+𝜆1𝑧𝑓(𝑧),(2.53) so 𝜇𝜆0,11,𝜆2𝑓(𝑧)=𝑓(𝑧)+𝜆1𝑧𝑓(𝑧).(2.54) Now, from Theorem 2.10 we deduce that 𝑓(𝑧)+𝜆1𝑧𝑓1(𝑧)(1𝑧)2(2.55) implies that 1𝜆1𝑓(𝑧)+𝜆1𝑧𝑓(𝑧)𝑧1.1𝑧(2.56)

Theorem 2.12. Let (𝑧)=1+(2𝛼1)𝑧1+𝑧,(𝑧𝑈),(2.57) be convex in 𝑈, with (0)=1 and 0𝛼<1. If 𝑛,𝑚0,𝜆2𝜆10,𝑓𝒜, and the differential subordination holds as 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)(𝑧),(2.58) then 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧𝑞(𝑧)=2𝛼1+2(1𝛼)ln(1+𝑧)𝑧.(2.59) The function 𝑞 is convex and is the best dominant.

Proof. Let 𝜇𝑝(𝑧)=𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧=𝑧+𝑘=21+𝜆1(𝑘1)𝑚/1+𝜆2(𝑘1)𝑚1𝑐(𝑛,𝑘)𝑎𝑘𝑧𝑘𝑧=1+𝑝1𝑧+𝑝2𝑧2[]+,(𝑝1,1,𝑧𝑈).(2.60) Differentiating (2.60), with respect to 𝑧, we obtain 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)=𝑝(𝑧)+𝑧𝑝(𝑧),(𝑧𝑈).(2.61) Using (2.61), the differential subordination (2.58) becomes 𝑝(𝑧)+𝑧𝑝(𝑧)(𝑧)=1+(2𝛼1)𝑧1+𝑧,(𝑧𝑈).(2.62) From Lemma 1.1, we deduce that 1𝑝(𝑧)𝑞(𝑧)=𝑧𝑧0=1(𝑡)𝑑𝑡𝑧𝑧01+(2𝛼1)𝑡=11+𝑡𝑑𝑡𝑧𝑧011+𝑡𝑑𝑡+(2𝛼1)𝑧0𝑡1+𝑡𝑑𝑡=2𝛼1+2(1𝛼)ln(1+𝑧)𝑧.(2.63) Using (2.60), we have 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧𝑞(𝑧)=2𝛼1+2(1𝛼)ln(1+𝑧)𝑧.(2.64) The proof is complete.

From Theorem 2.12, we deduce the following corollary.

Corollary 2.13. If 𝑓𝑅𝜆𝑛,𝑚1,𝜆2(𝛼), then 𝜇Re𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧>(2𝛼1)+2(1𝛼)ln2,(𝑧𝑈).(2.65)

Proof. Since 𝑓𝑅𝜆𝑛,𝑚1,𝜆2(𝛼), from Definition 2.3 we have 𝜇Re𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)>𝛼,(𝑧𝑈),(2.66) which is equivalent to 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)(𝑧)=1+(2𝛼1)𝑧.1+𝑧(2.67) Using Theorem 2.12, we have 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧𝑞(𝑧)=(2𝛼1)+2(1𝛼)ln(1+𝑧)𝑧.(2.68) Since 𝑞 is convex and 𝑞(𝑈) is symmetric with respect to the real axis, we deduce that 𝜇Re𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧>Re𝑞(1)=(2𝛼1)+2(1𝛼)ln2,(𝑧𝑈).(2.69)

Theorem 2.14. Let (𝑈), with (0)=1, (0)0 which satisfy the inequality Re1+𝑧(𝑧)(1𝑧)>2,(𝑧𝑈).(2.70) If 𝑛,𝑚0,𝜆2𝜆10,and 𝑓𝒜 and satisfies the differential subordination 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)(𝑧),(𝑧𝑈),(2.71) then 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧1𝑞(𝑧)=𝑧𝑧0(𝑡)𝑑𝑡.(2.72)

Proof. Let 𝜇𝑝(𝑧)=𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧=𝑧+𝑘=21+𝜆1(𝑘1)𝑚/1+𝜆2(𝑘1)𝑚1𝑐(𝑛,𝑘)𝑎𝑘𝑧𝑘𝑧=1+𝑝1𝑧+𝑝2𝑧2[]+,(𝑝1,1,𝑧𝑈).(2.73) Differentiating (2.73), with respect to 𝑧, we have 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)=𝑝(𝑧)+𝑧𝑝(𝑧),(𝑧𝑈).(2.74) Using (2.74), the differential subordination (2.71) becomes 𝑝(𝑧)+𝑧𝑝(𝑧)(𝑧),(𝑧𝑈).(2.75) From Lemma 1.1, we deduce that 1𝑝(𝑧)𝑞(𝑧)=𝑧𝑧0(𝑡)𝑑𝑡,(2.76) and using (2.73), we obtain 𝜇𝜆𝑛,𝑚1,𝜆2𝑓(𝑧)𝑧1𝑞(𝑧)=𝑧𝑧0(𝑡)𝑑𝑡.(2.77) From Lemma 1.3, we have that the function 𝑞 is convex, and from Lemma 1.1, 𝑞 is the best dominant for subordination (2.71). This completes the proof of Theorem 2.14.

3. Conclusion

We remark that several subclasses of analytic univalent functions can be derived and studied using the operator 𝜇𝜆𝑛,𝑚1,𝜆2.

Acknowledgment

This work is fully supported by UKM-ST-06-FRGS0107-2009, MOHE, Malaysia.