On Certain Class of Non-Bazilevič Functions of Order 𝛼+𝑖𝛽 Defined by a Differential Subordination

Abstract

We introduce a new subclass 𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵) of Non-Bazilevič functions of order 𝛼+𝑖𝛽. Some subordination relations and inequality properties are discussed. The results obtained generalize the related work of some authors. In addition, some other new results are also obtained.

1. Introduction

Let 𝐴_𝑛 denote the class of the functions 𝑓 of the form 𝑓(𝑧)=𝑧+^∞_𝑘=𝑛+1𝑎_𝑘𝑧^𝑘,(𝑛∈ℕ={1,2,3,…}) (1) which are analytic in the open unit disk 𝕌={𝑧∈ℂ∶|𝑧|<1}. Let 𝑓(𝑧) and 𝐹(𝑧) be analytic in 𝕌. Then we say that the function 𝑓(𝑧) is subordinate to 𝐹(𝑧) in 𝕌 if there exists an analytic function 𝑤(𝑧) in 𝕌 such that |𝑤(𝑧)|≤1 and 𝑓(𝑧)=𝐹(𝑤(𝑧)), denoted 𝑓≺𝐹 or 𝑓(𝑧)≺𝐹(𝑧). If 𝐹(𝑧) is univalent in 𝕌, then the subordination is equivalent to 𝑓(0)=𝐹(0) and 𝑓(𝕌)⊂𝐹(𝕌).

Assume that 0<𝛼<1, a function 𝑓(𝑧)∈𝐴_𝑛, is in 𝑁(𝛼) if and only if ℜ𝑓^′(𝑧)𝑧𝑓(𝑧)^1+𝛼>0,(𝑧∈𝕌). (2) The class 𝑁(𝛼) was introduced by Obradović [1] recently. This class of functions was said to be of Non-Bazilevič type. To this date, this class was studied in a direction of finding necessary conditions over 𝛼 that embeds this class into the class of univalent functions or its subclasses which is still an open problem.

Assume that 0<𝛼<1, 𝜆∈ℂ,  −1≤𝐵≤1, 𝐴≠𝐵, and 𝐴∈ℝ, we consider the following subclass of 𝐴_𝑛: 𝑁(𝜆,𝛼,𝐴,𝐵)=𝑓(𝑧)∈𝐴_𝑛∶()𝑧1+𝜆𝑓(𝑧)^𝛼−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)×𝑧𝑓(𝑧)^𝛼≺1+𝐴𝑧,(),1+𝐵𝑧𝑧∈𝕌 (3) where all the powers are principal values, and we apply this agreement to get the following definition.

Definition 1. Let 𝑁(𝜆,𝛼,𝜇) denote the class of functions in 𝐴_𝑛 satisfying the inequality ℜ)𝑧(1+𝜆𝑓(𝑧)^𝛼−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^𝛼>𝜇, (4) where 0<𝛼<1, 𝜆∈ℂ, 0≤𝜇<1, and 𝑧∈𝕌.

The classes 𝑁(𝜆,𝛼,𝐴,𝐵) and 𝑁(𝜆,𝛼,𝜇) were studied by Wang et al. [2].

In the present paper, similarly we define the following class of analytic functions.

Definition 2. Let 𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵) denote the class of functions in 𝐴_𝑛 satisfying the inequality ()𝑧1+𝜆𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺1+𝐴𝑧,(),1+𝐵𝑧𝑧∈𝕌 (5) where 𝜆∈ℂ, 𝛼≥0, 𝛽∈ℝ, −1≤𝐵≤1, 𝐴≠𝐵, and 𝐴∈ℝ. All the powers in (5) are principal values.

We say that the function 𝑓(𝑧) in this class is Non-Bazilevič functions of type 𝛼+𝑖𝛽.

Definition 3. Let 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,𝛽,𝜇) if and only if 𝑓(𝑧)∈𝐴_𝑛 and it satisfies ℜ)𝑧(1+𝜆𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}>𝜇, (6) where 𝜆∈ℂ,  𝛼≥0,  𝛽∈ℝ,  0≤𝜇<1,  and 𝑧∈𝕌.

In particular, if 𝛽=0, it reduces to the class 𝑁(𝜆,𝛼,𝐴,𝐵) studied in [2].

If 𝛽=0, 𝜆=−1, 𝑛=1, 𝐴=1, and 𝐵=−1, then the class 𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵) reduces to the class of non-Bazilevič‌ functions. If 𝛽=0,𝜆=−1,𝑛=1,𝐴=1−2𝜇, and 𝐵=−1, then the class 𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵) reduces to the class of non-Bazilevič functions of order 𝜇(0≤𝜇<1). Tuneski and Darus studied the Fekete-Szegö problem of the class 𝑁(−1,𝛼,0,1−2𝜇,−1) [3]. Other works related to Bazilevič and non-Bazilevič can be found in ([4–9]).

In the present paper, we will discuss the subordination relations and inequality properties of the class 𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵). The results presented here generalize and improve some known results, and some other new results are obtained.

2. Some Lemmas

Lemma 4 (see [10]). Let 𝐹(𝑧)=1+𝑏_𝑛𝑧^𝑛+𝑏_𝑛+1𝑧^𝑛+1+⋯ be analytic in 𝕌 and ℎ(𝑧) be analytic and convex in 𝕌, ℎ(0)=1. If 𝐹(𝑧)+1𝑐𝑧𝐹^′(𝑧)≺ℎ(𝑧), (7) where 𝑐≠0 and Re𝑐≥0, then 𝐹(𝑧)≺𝑐𝑛𝑧^{−𝑐/𝑛𝑧}₀𝑡^{(𝑐/𝑛)−1}ℎ(𝑡)𝑑𝑡≺ℎ(𝑧), (8) and (𝑐/𝑛)𝑧^−𝑐/𝑛∫^𝑧₀𝑡^{(𝑐/𝑛)−1}ℎ(𝑡)𝑑𝑡 is the best dominant for the differential subordination (7).

Lemma 5 (see [11]). Let −1≤𝐵₁≤𝐵₂<𝐴₂≤𝐴₁≤1; then 1+𝐴₂𝑧1+𝐵₂𝑧≺1+𝐴₁𝑧1+𝐵₁𝑧. (9)

Lemma 6 (see [12]). Let 𝐹(𝑧) be analytic and convex in 𝕌, 𝑓(𝑧)∈𝐴_𝑛, 𝑔(𝑧)∈𝐴_𝑛. If 𝑓(𝑧)≺𝐹(𝑧),𝑔(𝑧)≺𝐹(𝑧),0≤𝜆≤1, (10) then 𝜆𝑓(𝑧)+(1−𝜆)𝑔(𝑧)≺𝐹(𝑧). (11)

Lemma 7 (see [13]). Let 𝑓(𝑧)=∑^∞_𝑘=1𝑎_𝑘𝑧^𝑘 be analytic in 𝕌 and 𝑔(𝑧)=∑^∞_𝑘=1𝑏_𝑘𝑧^𝑘 analytic and convex in 𝕌. If 𝑓(𝑧)≺𝑔(𝑧), then |𝑎_𝑘|≤|𝑏_𝑘|, for 𝑘=1,2,….

Lemma 8. Let 𝜆∈ℂ, 𝛼≥0, 𝛽∈ℝ, 𝛼+𝑖𝛽≠0, −1≤𝐵≤1, 𝐴≠𝐵, and 𝐴∈ℝ. Then 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵) if and only if 𝐹(𝑧)+𝜆𝛼+𝑖𝛽𝑧𝐹^′(𝑧)≺1+𝐴𝑧,1+𝐵𝑧 (12) where 𝐹(𝑧)=𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}. (13)

Proof. Let 𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}=𝐹(𝑧). (14) Then, by taking the derivatives of both sides of (14) and through simple calculation, we have ()𝑧1+𝜆𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}(𝑧)+𝜆=𝐹𝛼+𝑖𝛽𝑧𝐹^′(𝑧); (15) since 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵), we have 𝐹(𝑧)+𝜆𝛼+𝑖𝛽𝑧𝐹^′(𝑧)≺1+𝐴𝑧.1+𝐵𝑧 (16)

3. Main Results

Theorem 9. Let 𝜆∈ℂ, 𝛼≥0, 𝛽∈ℝ, 𝛼+𝑖𝛽≠0, −1≤𝐵≤1, 𝐴≠𝐵, and 𝐴∈ℝ. If 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵), then 𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺𝛼+𝑖𝛽𝜆𝑛¹₀1+𝐴𝑧𝑢𝑢1+𝐵𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑑𝑢≺1+𝐴𝑧.1+𝐵𝑧 (17)

Proof. First let 𝐹(𝑧)=(𝑧/𝑓(𝑧))^{𝛼+𝑖𝛽}; then 𝐹(𝑧)=1+𝑏_𝑛𝑧^𝑛+𝑏_𝑛+1𝑧^𝑛+1+⋯ is analytic in 𝕌. Now, suppose that 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵); by Lemma 8, we know that 𝐹(𝑧)+𝜆𝛼+𝑖𝛽𝑧𝐹^′(𝑧)≺1+𝐴𝑧.1+𝐵𝑧 (18) It is obvious that ℎ(𝑧)=(1+𝐴𝑧)/(1+𝐵𝑧) is analytic and convex in 𝕌, ℎ(0)=1. Since 𝛼+𝑖𝛽≠0, 𝛼≥0, 𝜆≠0, and ℜ{(𝛼+𝑖𝛽)/𝜆}≥0; therefore, it follows from Lemma 4 that 𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}(𝑧)≺=𝐹𝛼+𝑖𝛽𝑧𝜆𝑛^{−(𝛼+𝑖𝛽)/𝜆𝑛𝑧}₀𝑡^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}ℎ(𝑡)=𝑑𝑡𝛼+𝑖𝛽𝜆𝑛¹₀1+𝐴𝑧𝑢𝑢1+𝐵𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑑𝑢≺1+𝐴𝑧.1+𝐵𝑧 (19)

Corollary 10. Let 𝜆∈ℂ, 𝛼≥0, 𝛽∈ℝ, 𝛼+𝑖𝛽≠0, and 𝜇≠1. If 𝑓(𝑧)∈𝐴_𝑛 satisfies ()𝑧1+𝜆𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺1+1−2𝜇𝑧(),1−𝑧𝑧∈𝕌 (20) then 𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺𝛼+𝑖𝛽×𝜆𝑛₁₀1+1−2𝜇𝑧𝑢𝑢1−𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}(),𝑑𝑡𝑧∈𝕌 (21) or equivalent to 𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺𝜇+1−𝜇𝛼+𝑖𝛽×𝜆𝑛₁₀1+𝑧𝑢𝑢1−𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑑𝑡(𝑧∈𝕌). (22)

Corollary 11. Let 𝜆∈ℂ, 𝛼≥0, 𝛽∈ℝ, 𝛼+𝑖𝛽≠0, and ℜ{𝜆}≥0; then 𝑁_𝑛𝜆,𝛼,𝛽,𝐴,𝐵⊂𝑁_𝑛0,𝛼,𝛽,𝐴,𝐵. (23)

Theorem 12. Let 0≤𝜆₁≤𝜆₂, 𝛼≥0, 𝛽∈ℝ, 𝛼+𝑖𝛽≠0, and −1≤𝐵₁≤𝐵₂<𝐴₂≤𝐴₁≤1; then 𝑁_𝑛𝜆₂,𝛼,𝛽,𝐴₂,𝐵₂⊂𝑁_𝑛𝜆₁,𝛼,𝛽,𝐴₁,𝐵₁. (24)

Proof. Suppose that 𝑓(𝑧)∈𝑁_𝑛(𝜆₂,𝛼,𝛽,𝐴₂,𝐵₂) we have 𝑓(𝑧)∈𝐴_𝑛, and 1+𝜆₂𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆₂𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺1+𝐴₂𝑧1+𝐵₂𝑧(𝑧∈𝕌). (25) Since −1≤𝐵₁≤𝐵₂<𝐴₂≤𝐴₁≤1, therefore it follows from Lemma 5 that 1+𝜆₂𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆₂𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺1+𝐴₁𝑧1+𝐵₁𝑧(𝑧∈𝕌); (26) that is 𝑓(𝑧)∈𝑁_𝑛(𝜆₂,𝛼,𝛽,𝐴₁,𝐵₁). So Theorem 12 is proved when 𝜆₁=𝜆₂≥0.

When 𝜆₂>𝜆₁≥0, then we can see from Corollary 11 that 𝑓(𝑧)∈𝑁_𝑛(0,𝛼,𝛽,𝐴₁,𝐵₁); then 𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺1+𝐴₁𝑧1+𝐵₁𝑧. (27) But 1+𝜆₁𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆₁𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}=𝜆1−¹𝜆₂𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}+𝜆₁𝜆₂×1+𝜆₂𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆₂𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}. (28) It is obvious that ℎ₁(𝑧)=(1+𝐴₁𝑧)/(1+𝐵₁𝑧) is analytic and convex in 𝕌. So we obtain from Lemma 6 and differential subordinations (26) and (27) that 1+𝜆₁𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆₁𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺1+𝐴₁𝑧1+𝐵₁𝑧; (29) that is, 𝑓(𝑧)∈𝑁_𝑛(𝜆₁,𝛼,𝛽,𝐴₁,𝐵₁). Thus we have 𝑁_𝑛𝜆₂,𝛼,𝛽,𝐴₂,𝐵₂⊂𝑁_𝑛𝜆₁,𝛼,𝛽,𝐴₁,𝐵₁. (30)

Corollary 13. Let 0≤𝜆₁≤𝜆₂,0≤𝜇₁≤𝜇₂<1,𝛼≥0,𝛽∈ℝ, and 𝛼+𝑖𝛽≠0; then 𝑁_𝑛𝜆₂,𝛼,𝛽,𝜇₂⊂𝑁_𝑛𝜆₁,𝛼,𝛽,𝜇₁. (31)

Theorem 14. Let 𝜆∈ℂ,𝛼≥0,𝛽∈ℝ,𝛼+𝑖𝛽≠0,−1≤𝐵≤1,𝐴≠𝐵, and 𝐴∈ℝ. If 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,𝛽,𝐴,𝐵), then inf_𝑧∈𝕌ℜ𝛼+𝑖𝛽𝜆𝑛¹₀1+𝐴𝑧𝑢𝑢1+𝐵𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^{𝛼+𝑖𝛽}<sup_𝑧∈𝕌ℜ𝛼+𝑖𝛽×𝜆𝑛₁₀1+𝐴𝑧𝑢𝑢1+𝐵𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑑𝑢. (32)

Proof. Suppose that 𝑓(𝑧)∈𝑁_n(𝜆,𝛼,𝛽,𝐴,𝐵); then from Theorem 9 we know that 𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}≺𝛼+𝑖𝛽𝜆𝑛¹₀1+𝐴𝑧𝑢𝑢1+𝐵𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑑𝑢. (33) Therefore, from the definition of the subordination, we have ℜ𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}>inf_𝑧∈𝕌ℜ𝛼+𝑖𝛽𝜆𝑛¹₀1+𝐴𝑧𝑢𝑢1+𝐵𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1},ℜ𝑧𝑑𝑢𝑓(𝑧)^{𝛼+𝑖𝛽}<sup_𝑧∈𝕌ℜ𝛼+𝑖𝛽𝜆𝑛¹₀1+𝐴𝑧𝑢𝑢1+𝐵𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑑𝑢. (34)

Corollary 15. Let 𝜆∈ℂ,𝛼≥0,𝛽∈ℝ,𝛼+𝑖𝛽≠0, and 𝜇<1. If 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,𝛽,1−2𝜇,−1), then 𝜇+1−𝜇inf_𝑧∈𝕌ℜ𝛼+𝑖𝛽𝜆𝑛¹₀1+𝑧𝑢𝑢1−𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^{𝛼+𝑖𝛽}<𝜇+1−𝜇sup_𝑧∈𝕌ℜ𝛼+𝑖𝛽×𝜆𝑛₁₀1+𝑧𝑢𝑢1−𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑑𝑢. (35)

Corollary 16. Let 𝜆∈ℂ,𝛼≥0,𝛽∈ℝ,𝛼+𝑖𝛽≠0, and 𝜇>1. If 𝑓(𝑧)∈𝐴_𝑛; then ℜ()𝑧1+𝜆𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}(),<𝜇𝑧∈𝕌 (36) then 𝜇+1−𝜇sup_𝑧∈𝕌ℜ𝛼+𝑖𝛽𝜆𝑛¹₀1+𝑧𝑢𝑢1−𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^{𝛼+𝑖𝛽}<𝜇+1−𝜇inf_𝑧∈𝕌ℜ𝛼+𝑖𝛽×𝜆𝑛₁₀1+𝑧𝑢𝑢1−𝑧𝑢^{((𝛼+𝑖𝛽)/𝜆𝑛)−1}𝑑𝑢. (37)

Corollary 17. Let 𝜆∈ℂ,𝛼≥0, and −1≤𝐵<𝐴≤1. If 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,0,𝐴,𝐵), then 𝛼𝜆𝑛¹₀1−𝐴𝑢𝑢1−𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^𝛼<𝛼𝜆𝑛¹₀1+𝐴𝑢𝑢1+𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢(𝑧∈𝕌), (38) and inequality (38) is sharp, with the extremal function defined by 𝑓_{𝜆,𝛼,𝐵,𝐴}(𝑧)𝛼=𝑧𝜆𝑛¹₀1+𝐴𝑧^𝑛𝑢1+𝐵𝑧^𝑛𝑢𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢^−(1/𝛼). (39)

Proof. Suppose that 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,0,𝐴,𝐵); from Theorem 9 we know 𝑧𝑓(𝑧)^𝛼≺𝛼𝜆𝑛¹₀1+𝐴𝑢𝑧𝑢1+𝐵𝑢𝑧^{(𝛼/𝜆𝑛)−1}𝑑𝑢. (40) Therefore, from the definition of the subordination and 𝐴>𝐵, we have that ℜ𝑧𝑓(𝑧)^𝛼<sup_𝑧∈𝕌ℜ𝛼𝜆𝑛¹₀1+𝐴𝑢𝑧𝑢1+𝐵𝑢𝑧^{(𝛼/𝜆𝑛)−1}≤𝛼𝑑𝑢𝜆𝑛¹₀sup_𝑧∈𝕌1+𝐴𝑢𝑧𝑢1+𝐵𝑢𝑧^{(𝛼/𝜆𝑛)−1}<𝛼𝑑𝑢𝜆𝑛¹₀1+𝐴𝑢𝑢1+𝐵𝑢^{(𝛼/𝜆𝑛)−1}ℜ𝑧𝑑𝑢,𝑓(𝑧)^𝛼>inf_𝑧∈𝕌ℜ𝛼𝜆𝑛¹₀1+𝐴𝑢𝑧𝑢1+𝐵𝑢𝑧^{(𝛼/𝜆𝑛)−1}≥𝛼𝑑𝑢𝜆𝑛¹₀inf_𝑧∈𝕌1+𝐴𝑢𝑧𝑢1+𝐵𝑢𝑧^{(𝛼/𝜆𝑛)−1}>𝛼𝑑𝑢𝜆𝑛¹₀1−𝐴𝑢𝑢1−𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢. (41) It is obvious that inequality (38) is sharp, with the extremal function given by (39).

Corollary 18. Let 𝜆∈ℂ,𝛼≥0, and 𝜇<1. If 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,0,1−2𝜇,−1), then 𝛼𝜆𝑛¹₀1−1−2𝜇𝑢𝑢1+𝑢^{(𝛼/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^𝛼<𝛼𝜆𝑛¹₀1+1−2𝜇𝑢𝑢1−𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢(𝑧∈𝕌), (42) and inequality (42) is equivalent to 𝛼𝜇+1−𝜇𝜆𝑛¹₀1−𝑢𝑢1+𝑢^{(𝛼/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^𝛼𝛼<𝜇+1−𝜇𝜆𝑛¹₀1+𝑢𝑢1−𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢(𝑧∈𝕌). (43) The inequality (42) is sharp, with the extremal function defined by 𝑓_{𝜆,𝛼,𝛽}(𝑧)𝛼=𝑧𝜆𝑛¹₀1+(1−2𝛽)𝑧^𝑛𝑢1−𝑧^𝑛𝑢𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢^−1/𝛼𝑑𝑧. (44)

Corollary 19. Let 𝜆∈ℂ, 𝛼≥0, and −1≤𝐴<𝐵≤1. If 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,0,𝐴,𝐵), then 𝛼𝜆𝑛¹₀1+𝐴𝑢𝑢1+𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^𝛼<𝛼𝜆𝑛¹₀1−𝐴𝑢𝑢1−𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢(𝑧∈𝕌), (45) and inequality (45) is sharp, with the extremal function given by (39).

Proof. Applying similar method as in Corollary 17, we get the result.

Corollary 20. Let 𝜆∈ℂ, 𝛼≥0, and𝜇>1. If 𝑓(𝑧)∈𝐴_𝑛 satisfies ℜ()𝑧1+𝜆𝑓(𝑧)^{𝛼+𝑖𝛽}−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^{𝛼+𝑖𝛽}(),<𝜇𝑧∈𝕌 (46) then 𝛼𝜆𝑛¹₀1+1−2𝜇𝑢𝑢1−𝑢^{(𝛼/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^𝛼<𝛼𝜆𝑛¹₀1−1−2𝜇𝑢𝑢1+𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢, (47) and inequality (47) is equivalent to 𝛼𝜇+1−𝜇𝜆𝑛¹₀1+𝑢𝑢1−𝑢^{(𝛼/𝜆𝑛)−1}𝑧𝑑𝑢<ℜ𝑓(𝑧)^𝛼𝛼<𝜇+1−𝜇𝜆𝑛¹₀1−𝑢𝑢1+𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢(𝑧∈𝕌), (48) and inequality (47) is sharp, with the extremal function defined by equality (44).

If ℜ𝑤≥0, then (ℜ𝑤)^1/2≤ℜ𝑤^1/2≤ℜ𝑧^1/2 (see [2, 12]). So we have the following.

Corollary 21. Let 𝜆∈ℂ, 𝛼≥0, and−1≤𝐵<𝐴≤1. If 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,0,𝐴,𝐵), then 𝛼𝜆𝑛¹₀1−𝐴𝑢𝑢1−𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢^1/2𝑧<ℜ𝑓(𝑧)^𝛼1/2<𝛼𝜆𝑛¹₀1+𝐴𝑢𝑢1+𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢^1/2(𝑧∈𝕌), (49) and inequality (49) is sharp, with the extremal function defined by equality (39).

Proof. From Theorem 9 we have 𝑧𝑓(𝑧)^𝛼≺1+𝐴𝑧.1+𝐵𝑧 (50) Since −1≤𝐴<𝐵≤1, we have 0≤1−𝐴𝑧1−𝐵<ℜ𝑓(𝑧)^𝛼<1+𝐴.1+𝐵 (51) Thus, from inequality (38), we can get inequality (49). It is obvious that inequality (49) is sharp, with the extremal function defined by equality (39).

Corollary 22. Let 𝜆∈ℂ,𝛼≥0, and −1≤𝐴<𝐵≤1. If 𝑓(𝑧)∈𝑁_𝑛(𝜆,𝛼,0,𝐴,𝐵), then 𝛼𝜆𝑛¹₀1+𝐴𝑢𝑢1+𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢^1/2𝑧<ℜ𝑓(𝑧)^𝛼1/2<𝛼𝜆𝑛¹₀1−𝐴𝑢𝑢1−𝐵𝑢^{(𝛼/𝜆𝑛)−1}𝑑𝑢^1/2(𝑧∈𝕌), (52) and inequality (52) is sharp, with the extremal function defined by equality (39).

Proof. Applying similar method as in Corollary 21, we get the required result.

Remark 23. From Corollaries 21 and 22, we can generalize the corresponding results and some other special classes of analytic functions.

Corollary 24. Let 𝜆∈ℂ, 𝛼≥0, −1≤𝐵<𝐴≤1, and 𝐴∈ℝ; if 𝑓(𝑧)=𝑧+∑^∞_𝑘=𝑛+1𝑎_𝑘𝑧^𝑘∈𝑁_𝑛(𝜆,𝛼,0,𝐴,𝐵), then one has 𝑎_𝑛+1≤|𝐴−𝐵||𝑛𝜆+𝛼| (53) and inequality (53) is sharp, with the extremal function defined by equality (39).

Proof. Suppose that 𝑓(𝑧)=𝑧+∑^∞_𝑘=𝑛+1𝑎_𝑘𝑧^𝑘∈𝑁_𝑛(𝜆,𝛼,0,𝐴,𝐵); then we have ()𝑧1+𝜆𝑓(𝑧)^𝛼−𝜆𝑧𝑓^′(𝑧)𝑓(𝑧)𝑧𝑓(𝑧)^𝛼=1+(−𝑛𝜆−𝛼)𝑎_𝑛+1𝑧^𝑛+⋯≺1+𝐴𝑧.1+𝐵𝑧 (54) It follows from Lemma 7 that 𝑎_𝑛+1≤|𝐴−𝐵|||.𝑛𝜆+𝛼 (55) Thus, we can get (53). Notice that 𝑓(𝑧)=𝑧+𝐴−𝐵𝑧𝑛𝜆+𝛼^𝑛+1+⋯∈𝑁_𝑛(𝜆,𝛼,0,𝐴,𝐵); (56) we obtain that the inequality (53) is sharp.

Remark 25. Setting 𝜆=𝑛=𝐴=1, and 𝐵=−1 in Corollary 24 we get the results obtained by [14].

Conflict of Interests

The authors declare that they have no conflict of interests.

Authors’ Contribution

Both authors read and approved the final paper.

Acknowledgments

The authors would like to acknowledge and appreciate the financial support received from Universiti Kebangsaan Malaysia under the Grant AP-2013-009. The authors also would like to thank the referees for the comments and suggestions to improve the paper.