On Certain Subclasses of Analytic Functions Defined by Differential Subordination

Abstract

We introduce and study certain subclasses of analytic functions which are defined by differential subordination. Coefficient inequalities, some properties of neighborhoods, distortion and covering theorems, radius of starlikeness, and convexity for these subclasses are given.

1. Introduction

Let 𝒯(𝑗) be the class of analytic functions 𝑓 of the form𝑓(𝑧)=𝑧−^∞_𝑘=𝑗+1𝑎_𝑘𝑧^𝑘,𝑎_𝑘≥0,𝑗∈ℕ={1,2,…}, (1.1) defined in the open unit disc 𝒰={𝑧∈ℂ∶|𝑧|<1}.

Let Ω be the class of functions 𝜔 analytic in 𝒰 such that 𝜔(0)=0, |𝜔(𝑧)|<1.

For any two functions 𝑓 and 𝑔 in 𝒯(𝑗), 𝑓 is said to be subordinate to 𝑔 that is denoted 𝑓≺𝑔, if there exists an analytic function 𝜔∈Ω such that 𝑓(𝑧)=𝑔(𝜔(𝑧)) [1].

Definition 1.1 (see [2]). For 𝑛∈ℕ and 𝜆≥0, the Al-Oboudi operator 𝐷^𝑛_𝜆∶𝒯(𝑗)→𝒯(𝑗) is defined as 𝐷⁰_𝜆𝑓(𝑧)=𝑓(𝑧), 𝐷¹_𝜆𝑓(𝑧)=(1−𝜆)𝑓(𝑧)+𝜆𝑧𝑓^′(𝑧)=𝐷_𝜆𝑓(𝑧), and 𝐷^𝑛_𝜆𝑓(𝑧)=𝐷_𝜆(𝐷_𝜆^𝑛−1𝑓(𝑧)).

For 𝜆=1, we get Sǎlǎgean differential operator [3].

Further, if 𝑓(𝑧)=𝑧−∑^∞_𝑘=𝑗+1𝑎_𝑘𝑧^𝑘, then𝐷^𝑛_𝜆𝑓(𝑧)=𝑧−^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛𝑎_𝑘𝑧^𝑘𝑎_𝑘≥0. (1.2) For any function 𝑓∈𝒯(𝑗) and 𝛿≥0, the (𝑗,𝛿)-neighborhood of 𝑓 is defined as𝒩_𝑗,𝛿𝑓=𝑔(𝑧)=𝑧−^∞_𝑘=𝑗+1𝑏_𝑘𝑧^𝑘∈𝒯𝑗∶^∞_𝑘=𝑗+1𝑘𝑎_𝑘−𝑏_𝑘≤𝛿. (1.3) In particular, for the identity function 𝑒(𝑧)=𝑧, we see that𝒩_𝑗,𝛿(𝑒)=𝑔(𝑧)=𝑧−^∞_𝑘=𝑗+1𝑏_𝑘𝑧^𝑘∈𝒯𝑗∶^∞_𝑘=𝑗+1𝑘𝑏_𝑘≤𝛿. (1.4) The concept of neighborhoods was first introduced by Goodman [4] and then generalized by Ruscheweyh [5].

Definition 1.2. A function 𝑓∈𝒯(𝑗) is said to be in the class 𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆) if 𝐷_𝜆^𝑛+𝑚𝑓(𝑧)𝐷^𝑛_𝜆𝑓(𝑧)≺1+𝐴𝑧1+𝐵𝑧,𝑧∈𝒰, (1.5) where 𝑛∈ℕ₀, 𝑚∈ℕ, 𝜆≥1, and −1≤𝐵<𝐴≤1.

We observe that 𝒯_𝑗(𝑛,𝑚,1−2𝛼,−1,1)≡𝒯_𝑗(𝑛,𝑚,𝛼)  [6],  𝒯_𝑗(0,1,1−2𝛼,−1,1)≡𝒮^⋆_𝑗(𝛼)  [7], the class of starlike functions of order 𝛼 and 𝒯_𝑗(1,1,1−2𝛼,−1,1)≡𝒞_𝑗(𝛼)  [7], the class of convex functions of order 𝛼.

2. Neighborhoods for the Class 𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆)

Theorem 2.1. A function 𝑓∈𝒯(𝑗) belongs to the class 𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆) if and only if ^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)𝑎_𝑘≤𝐴−𝐵 (2.1) for 𝑛∈ℕ₀, 𝑚∈ℕ, 𝜆≥1, and −1≤𝐵<𝐴≤1.

Proof. Let 𝑓∈𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆). Then, 𝐷_𝜆^𝑛+𝑚𝑓(𝑧)𝐷^𝑛_𝜆𝑓(𝑧)≺1+𝐴𝑧1+𝐵𝑧,𝑧∈𝒰. (2.2) Therefore, 𝜔(𝑧)=𝐷^𝑛_𝜆𝑓(𝑧)−𝐷_𝜆^𝑛+𝑚𝑓(𝑧)𝐵𝐷_𝜆^𝑛+𝑚𝑓(𝑧)−𝐴𝐷^𝑛_𝜆𝑓(𝑧). (2.3) Hence, |𝜔(𝑧=𝐷)|^𝑛𝜆𝑓(𝑧)−𝐷_𝜆^𝑛+𝑚𝑓(𝑧)𝐵𝐷_𝜆^𝑛+𝑚𝑓(𝑧)−𝐴𝐷^𝑛_𝜆𝑓(𝑧)=∑^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛[1+(𝑘−1)𝜆]^𝑚−1𝑎_𝑘𝑧^𝑘(𝐴−𝐵)𝑧+∑^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛𝐵[1+(𝑘−1)𝜆]^𝑚−𝐴𝑎_𝑘𝑧^𝑘<1. (2.4) Thus, ℜ∑^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛[1+(𝑘−1)𝜆]^𝑚−1𝑎_𝑘𝑧^𝑘(𝐴−𝐵)𝑧+∑^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛𝐵[1+(𝑘−1)𝜆]^𝑚−𝐴𝑎_𝑘𝑧^𝑘<1. (2.5) Taking |𝑧|=𝑟, for sufficiently small 𝑟 with 0<𝑟<1, the denominator of (2.5) is positive and so it is positive for all 𝑟 with 0<𝑟<1, since 𝜔(𝑧) is analytic for |𝑧|<1. Then, inequality (2.5) yields ^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛[1+(𝑘−1)𝜆]^𝑚−1𝑎_𝑘𝑟^𝑘<(𝐴−𝐵)𝑟+𝐵^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛+𝑚𝑎_𝑘𝑟^𝑘−𝐴^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛𝑎_𝑘𝑟^𝑘. (2.6) Equivalently, ^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)𝑎_𝑘𝑟^𝑘≤(𝐴−𝐵)𝑟, (2.7) and (2.1) follows upon letting 𝑟→1.
Conversely, for |𝑧|=𝑟, 0<𝑟<1, we have 𝑟^𝑘<𝑟. That is, ^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)𝑎_𝑘𝑟^𝑘≤^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)𝑎_𝑘𝑟≤(𝐴−𝐵)𝑟. (2.8) From (2.1), we have ^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛[1+(𝑘−1)𝜆]^𝑚−1𝑎_𝑘𝑧^𝑘≤^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛[1+(𝑘−1)𝜆]^𝑚−1𝑎_𝑘𝑟^𝑘<(𝐴−𝐵)𝑟+^∞_𝑘=𝑗+1𝐵[1+(𝑘−1)𝜆]^𝑚−𝐴[1+(𝑘−1)𝜆]^𝑛𝑎_𝑘𝑟^𝑘<(𝐴−𝐵)𝑧+^∞_𝑘=𝑗+1𝐵[1+(𝑘−1)𝜆]^𝑚−𝐴[1+(𝑘−1)𝜆]^𝑛𝑎_𝑘𝑧^𝑘. (2.9) This proves that 𝐷_𝜆^𝑛+𝑚𝑓(𝑧)𝐷^𝑛_𝜆𝑓(𝑧)≺1+𝐴𝑧1+𝐵𝑧,𝑧∈𝒰, (2.10) and hence 𝑓∈𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆).

Theorem 2.2. If ()𝛿=𝐴−𝐵1+𝜆𝑗^𝑛−1(1−𝐵)1+𝜆𝑗^𝑚−(,1−𝐴) (2.11) then 𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆)⊂𝑁_𝑗,𝛿(𝑒).

Proof. It follows from (2.1) that if 𝑓∈𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆), then 1+𝜆𝑗^𝑛−1(1−𝐵)1+𝜆𝑗^𝑚−(1−𝐴)^∞_𝑘=𝑗+1𝑘𝑎_𝑘≤(𝐴−𝐵), (2.12) which implies ^∞_𝑘=𝑗+1𝑘𝑎_𝑘≤(𝐴−𝐵)1+𝜆𝑗^𝑛−1(1−𝐵)1+𝜆𝑗^𝑚−(1−𝐴)=𝛿. (2.13) Using (1.4), we get the result.

3. Neighborhoods for the Classes ℛ_𝑗(𝑛,𝐴,𝐵,𝜆) and 𝒫_𝑗(𝑛,𝐴,𝐵,𝜆)

Definition 3.1. A function 𝑓∈𝒯(𝑗) is said to be in the class ℛ_𝑗(𝑛,𝐴,𝐵,𝜆) if it satisfies 𝐷^𝑛_𝜆𝑓(𝑧)^′≺1+𝐴𝑧(),1+𝐵𝑧𝑧∈𝒰 (3.1) where −1≤𝐵<𝐴≤1, 𝜆≥1 and 𝑛∈𝒩₀.

Definition 3.2. A function 𝑓∈𝒯(𝑗) is said to be in the class 𝒫_𝑗(𝑛,𝐴,𝐵,𝜆) if it satisfies 𝐷^𝑛_𝜆𝑓(𝑧)𝑧≺1+𝐴𝑧(),1+𝐵𝑧𝑧∈𝒰 (3.2) where −1≤𝐵<𝐴≤1, 𝜆≥1 and 𝑛∈𝒩₀.

Lemma 3.3. A function 𝑓∈𝒯(𝑗) belongs to the class ℛ_𝑗(𝑛,𝐴,𝐵,𝜆) if and only if ^∞_𝑘=𝑗+1(1−𝐵)[1+(𝑘−1)𝜆]^𝑛+1𝑎_𝑘≤𝐴−𝐵. (3.3)

Lemma 3.4. A function 𝑓∈𝒯(𝑗) belongs to the class 𝒫_𝑗(𝑛,𝐴,𝐵,𝜆) if and only if ^∞_𝑘=𝑗+1(1−𝐵)[1+(𝑘−1)𝜆]^𝑛𝑎_𝑘≤𝐴−𝐵. (3.4)

Theorem 3.5. ℛ_𝑗(𝑛,𝐴,𝐵,𝜆)⊂𝒩_𝑗,𝛿(𝑒), where ()𝛿=𝐴−𝐵1+𝜆𝑗^𝑛().1−𝐵 (3.5)

Proof. If 𝑓∈ℛ_𝑗(𝑛,𝐴,𝐵,𝜆), we have 1+𝜆𝑗^𝑛∞_𝑘=𝑗+1(1−𝐵)𝑘𝑎_𝑘≤𝐴−𝐵, (3.6) which implies ^∞_𝑘=𝑗+1𝑘𝑎_𝑘≤(𝐴−𝐵)1+𝜆𝑗^𝑛(1−𝐵)=𝛿. (3.7)

Theorem 3.6. 𝒫_𝑗(𝑛,𝐴,𝐵,𝜆)⊂𝒩_𝑗,𝛿(𝑒), where ()𝛿=𝐴−𝐵1+𝜆𝑗^𝑛−1().1−𝐵 (3.8)

Proof. If 𝑓∈𝒫_𝑗(𝑛,𝐴,𝐵,𝜆), we have 1+𝜆𝑗^∞𝑛−1_𝑘=𝑗+1(1−𝐵)𝑘𝑎_𝑘≤𝐴−𝐵, (3.9) which implies ^∞_𝑘=𝑗+1𝑘𝑎_𝑘≤(𝐴−𝐵)1+𝜆𝑗^𝑛−1(1−𝐵)=𝛿. (3.10) Thus, in view of condition (1.4), we get the required result of Theorem 3.6.

4. Neighborhood of the Class 𝒦^𝜆_𝑗(𝑛,𝑚,𝐴,𝐵,𝐶,𝐷)

Definition 4.1. A function 𝑓∈𝒯(𝑗) is said to be in the class 𝒦^𝜆_𝑗(𝑛,𝑚,𝐴,𝐵,𝐶,𝐷) if it satisfies 𝑓(𝑧)𝑔(𝑧)−1<𝐴−𝐵1−𝐵,𝑧∈𝒰, (4.1) for −1≤𝐵<𝐴≤1, −1≤𝐷<𝐶≤1, 𝜆≥1 and 𝑔∈𝒯_𝑗(𝑛,𝑚,𝐶,𝐷,𝜆).

Theorem 4.2. For 𝑔∈𝒯_𝑗(𝑛,𝑚,𝐶,𝐷,𝜆), one has 𝒩_𝑗,𝛿(𝑔)⊂𝒦^𝜆_𝑗(𝑛,𝑚,𝐴,𝐵,𝐶,𝐷) and 1−𝐴1−𝐵=1−1+𝜆𝑗^𝑛−1(1−𝐷)1+𝜆𝑗^𝑚−(1−𝐶)𝛿1+𝜆𝑗^𝑛(1−𝐷)1+𝜆𝑗^𝑚−(−(),1−𝐶)𝐶−𝐷 (4.2) where 𝛿≤(1−𝐷)1+𝜆𝑗−(𝐶−𝐷)1+𝜆𝑗^1−𝑛(1−𝐷)1+𝜆𝑗^𝑚−(1−𝐶)⁻¹. (4.3)

Proof. Let 𝑓∈𝒩_𝑗,𝛿(𝑔) for 𝑔∈𝒯_𝑗(𝑛,𝑚,𝐶,𝐷,𝜆). Then, ^∞_𝑘=𝑗+1𝑘𝑎_𝑘−𝑏_𝑘≤𝛿,^∞_𝑘=𝑗+1𝑏_𝑘≤𝐶−𝐷1+𝜆𝑗^𝑛(1−𝐷)1+𝜆𝑗^𝑚−(.1−𝐶) (4.4) Consider 𝑓(𝑧)𝑔(𝑧)≤∑−1^{∞𝑘=𝑗+1}𝑎_𝑘−𝑏_𝑘1−∑^∞_𝑘=𝑗+1𝑏_𝑘≤𝛿1+𝜆𝑗1+𝜆𝑗^𝑛(1−𝐷)1+𝜆𝑗^𝑚−(1−𝐶)1+𝜆𝑗^𝑛(1−𝐷)1+𝜆𝑗^𝑚−(−()=1−𝐶)𝐶−𝐷1+𝜆𝑗^𝑛−1(1−𝐷)1+𝜆𝑗^𝑚−(1−𝐶)𝛿1+𝜆𝑗^𝑛(1−𝐷)1+𝜆𝑗^𝑚−(−()=1−𝐶)𝐶−𝐷𝐴−𝐵.1−𝐵 (4.5) This implies that 𝑓∈𝒦^𝜆_𝑗(𝑛,𝑚,𝐴,𝐵,𝐶,𝐷).

5. Distortion and Covering Theorems

Theorem 5.1. If 𝑓∈𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆), then 𝑟−𝐴−𝐵1+𝑗𝜆^𝑛(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗+1≤𝑓(𝑧)≤𝑟+𝐴−𝐵1+𝑗𝜆^𝑛(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗+1(0<|𝑧|=𝑟<1), (5.1) with equality for 𝑓(𝑧)=𝑧−𝐴−𝐵1+𝑗𝜆^𝑛(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗+1(𝑧=±𝑟). (5.2)

Proof. In view of Theorem 2.1, we have 1+𝑗𝜆^𝑛(1−𝐵)1+𝑗𝜆^𝑚−(1−𝐴)^∞_𝑘=𝑗+1𝑎_𝑘≤^∞_𝑘=𝑗+1[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)𝑎_𝑘≤𝐴−𝐵. (5.3) Hence, 𝑓(𝑧)≤𝑟+^∞_𝑘=𝑗+1𝑎_𝑘𝑟^𝑘≤𝑟+𝑟^∞𝑗+1_𝑘=𝑗+1𝑎_𝑘≤𝑟+𝐴−𝐵1+𝑗𝜆^𝑛(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗+1,𝑓(𝑧)≥𝑟−^∞_𝑘=𝑗+1𝑎_𝑘𝑟^𝑘≥𝑟−𝑟^∞𝑗+1_𝑘=𝑗+1𝑎_𝑘≥𝑟−𝐴−𝐵1+𝑗𝜆^𝑛(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗+1. (5.4) This completes the proof.

Theorem 5.2. Any function 𝑓∈𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆) maps the disk |𝑧|<1 onto a domain that contains the disk |𝑤|<1−𝐴−𝐵1+𝑗𝜆^𝑛(1−𝐵)1+𝑗𝜆^𝑚−(.1−𝐴) (5.5)

Proof. The proof follows upon letting 𝑟→1 in Theorem 5.1.

Theorem 5.3. If 𝑓∈𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆), then ()1−𝐴−𝐵1+𝑗𝜆^𝑛−1(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗≤𝑓^′(𝑧)≤1+𝐴−𝐵1+𝑗𝜆^𝑛−1(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗(0<|𝑧|=𝑟<1), (5.6) with equality for 𝑓(𝑧)=𝑧−𝐴−𝐵1+𝑗𝜆^𝑛−1(1−𝐵)1+𝑗𝜆^𝑚−(𝑧1−𝐴)^𝑗+1(𝑧=±𝑟). (5.7)

Proof. We have 𝑓^′(𝑧)≤1+^∞_𝑘=𝑗+1𝑘𝑎_𝑘|𝑧|^𝑘−1≤1+𝑟^𝑗∞_𝑘=𝑗+1𝑘𝑎_𝑘. (5.8) In view of Theorem 2.1, ^∞_𝑘=𝑗+1𝑘𝑎_𝑘≤𝐴−𝐵1+𝑗𝜆^𝑛−1(1−𝐵)1+𝑗𝜆^𝑚−(.1−𝐴) (5.9) Thus, 𝑓^′(𝑧)≤1+𝐴−𝐵1+𝑗𝜆^𝑛−1(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗. (5.10) On the other hand, 𝑓^′(𝑧)≥1−^∞_𝑘=𝑗+1𝑘𝑎_𝑘|𝑧|^𝑘−1≥1−𝑟^𝑗∞_𝑘=𝑗+1𝑘𝑎_𝑘≥1−𝐴−𝐵1+𝑗𝜆^𝑛−1(1−𝐵)1+𝑗𝜆^𝑚−(𝑟1−𝐴)^𝑗. (5.11) This completes the proof.

6. Radii of Starlikeness and Convexity

In this section, we find the radius of starlikeness of order 𝜌 and the radius of convexity of order 𝜌 for functions in the class 𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆).

Theorem 6.1. If 𝑓∈𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆), then 𝑓 is starlike of order 𝜌, (0≤𝜌<1) in |𝑧|<𝑟₁(𝑛,𝑚,𝐴,𝐵,𝜆,𝜌), where 𝑟₁𝑛,𝑚,𝐴,𝐵,𝜆,𝜌=inf_𝑘[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)1−𝜌𝑘−𝜌(𝐴−𝐵)^1/(𝑘−1). (6.1)

Proof. It is sufficient to show that |𝑧(𝑓^′(𝑧)/𝑓(𝑧))−1|≤1−𝜌(0≤𝜌<1) for |𝑧|<𝑟₁(𝑛,𝑚,𝐴,𝐵,𝜆,𝜌).
We have 𝑧𝑓^′(𝑧)𝑓(𝑧)≤∑−1^{∞𝑘=𝑗+1}(𝑘−1)𝑎_𝑘|𝑧|^𝑘−11−∑^∞_𝑘=𝑗+1𝑎_𝑘|𝑧|^𝑘−1. (6.2) Thus, |𝑧(𝑓^′(𝑧)/𝑓(𝑧))−1|≤1−𝜌 if 𝑘−𝜌𝑎^∞_{𝑘=𝑗+1𝑘}|𝑧|^𝑘−11−𝜌≤1. (6.3) Hence, by Theorem 2.1, (6.3) will be true if 𝑘−𝜌|𝑧|^𝑘−1≤[()𝜆]1−𝜌1+𝑘−1^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)(𝐴−𝐵) (6.4) or if |𝑧|≤[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)1−𝜌𝑘−𝜌(𝐴−𝐵)^1/(𝑘−1)𝑘≥𝑗+1. (6.5) This completes the proof.

Theorem 6.2. If 𝑓∈𝒯_𝑗(𝑛,𝑚,𝐴,𝐵,𝜆), then 𝑓 is convex of order 𝜌, (0≤𝜌<1) in |𝑧|<𝑟₂(𝑛,𝑚,𝐴,𝐵,𝜆,𝜌), where 𝑟₂𝑛,𝑚,𝐴,𝐵,𝜆,𝜌=inf_𝑘[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)1−𝜌𝑘−𝜌(𝐴−𝐵)^1/(𝑘−1). (6.6)

Proof. It is sufficient to show that |𝑧(𝑓^′′(𝑧)/𝑓^′(𝑧))|≤1−𝜌(0≤𝜌<1) for |𝑧|<𝑟₁(𝑛,𝑚,𝐴,𝐵,𝜆,𝜌).
We have 𝑧𝑓^′′(𝑧)𝑓^′(𝑧)≤∑^∞_𝑘=𝑗+1𝑘(𝑘−1)𝑎_𝑘|𝑧|^𝑘−11−∑^∞_𝑘=𝑗+1𝑘𝑎_𝑘|𝑧|^𝑘−1. (6.7) Thus, |𝑧(𝑓^′′(𝑧)/𝑓^′(𝑧))|≤1−𝜌 if ^∞_𝑘=𝑗+1𝑘𝑘−𝜌𝑎_𝑘|𝑧|^𝑘−11−𝜌≤1. (6.8) Hence, by Theorem 2.1, (6.8) will be true if 𝑘𝑘−𝜌|𝑧|^𝑘−1≤[()𝜆]1−𝜌1+𝑘−1^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)(𝐴−𝐵) (6.9) or if |𝑧|≤[1+(𝑘−1)𝜆]^𝑛(1−𝐵)[1+(𝑘−1)𝜆]^𝑚−(1−𝐴)1−𝜌𝑘𝑘−𝜌(𝐴−𝐵)^1/(𝑘−1)𝑘≥𝑗+1. (6.10) This completes the proof.

Acknowledgment

The author wish to thank the referee for his valuable suggestions.