Fekete-Szegö Coefficient Functional for Certain Subclasses of Close-to-Star Functions

Abstract

We introduce some subclasses of close-to-star functions defined by subordination and obtain sharp upper bounds of the functional |𝑎₃−𝜇𝑎²₂|, 𝜇 real, for an analytic function 𝑓(𝑧)=𝑧+∑^∞_𝑛=2𝑎_𝑛𝑧^𝑛, |𝑧|<1, belonging to these sub-classes.

1. Introduction

Let Ω be the class of bounded functions of the following form: 𝑤(𝑧)=^∞_𝑛=1𝑑_𝑛𝑧^𝑛, (1) which are analytic in the unit disc 𝐸={𝑧∶|𝑧|<1} and satisfy the conditions 𝑤(0)=0 and |𝑤(𝑧)|<1.

It is known (see [1]) that 𝑑₁≤1,𝑑₂≤1−𝑑₁². (2)

Let 𝐴 denote the class of functions of the following form: 𝑓(𝑧)=𝑧+^∞_𝑛=2𝑎_𝑛𝑧^𝑛 (3) which are analytic in 𝐸. Let 𝑆 be the class of functions of the form (3) which are analytic univalent in 𝐸.

We will concentrate on the coefficient problem for the subclasses of 𝐴. In 1916, Bieberbach [2] proved that |𝑎₂|≤2 for 𝑓(𝑧)∈𝑆 as an elementary corollary area theorem. He conjectured that, for each function 𝑓(𝑧)∈𝑆, |𝑎_𝑛|≤𝑛; the equality holds for the Koebe function 𝑘(𝑧)=𝑧/(1−𝑧)², which maps the unit disc 𝐸 onto the entire complex plane minus the slit along the negative real axis from −1/4 to −∞. De Branges [3] solved the Bieberbach conjecture in 1985. The contribution of Löwner [4] in proving that |𝑎₃|≤3 for the class 𝑆 was huge.

With the known estimates |𝑎₂|≤2 and |𝑎₃|≤3, it was natural to seek some relation between 𝑎₃ and 𝑎²₂ for the class 𝑆. This thought prompted Fekete and Szegö [5] and they used Löwner's method to prove the following well-known result for the class 𝑆: if 𝑓(𝑧)∈𝑆, then 𝑎₃−𝜇𝑎²₂≤3−4𝜇if𝜇≤0,1+2exp−2𝜇1−𝜇if0≤𝜇≤1,4𝜇−3if𝜇≥1. (4)

The inequality (4) plays a very important role in determining estimates of higher coefficients for some subclasses of 𝑆 (see Chichra [6], Babalola [7]).

Next, we define some subclasses of 𝐴 and obtain the analogy of (4).

We denote by 𝑆^∗ the class of univalent star-like functions 𝑔(𝑧)=𝑧+∑^∞_𝑛=2𝑏_𝑛𝑧^𝑛∈𝐴 which satisfy the following condition: ℜ𝑧𝑔^′(𝑧)𝑔(𝑧)>0,𝑧∈𝐸. (5)

We denote by 𝐾 the class of convex univalent functions ℎ(𝑧)=𝑧+∑^∞_𝑛=2𝑐_𝑛𝑧^𝑛∈𝐴 which satisfy the condition ℜ𝑧ℎ^′(𝑧)^′ℎ^′(𝑧)>0,𝑧∈𝐸. (6)

Let 𝑓 and 𝑔 be two analytic functions in 𝐸. Then, 𝑓 is said to be subordinate to 𝑔 (symbolically 𝑓≺𝑔) if there exists a bounded function 𝑤(𝑧)∈Ω, such that 𝑓(𝑧)=𝑔(𝑤(𝑧)).

A function 𝑓(𝑧)∈𝐴 is said to be close to star if there exists a function 𝑔(𝑧)∈𝑆^∗ such that ℜ𝑓(𝑧)𝑔(𝑧)>0,𝑧∈𝐸. (7)

The class of close to star functions was introduced by Reade [8] and is denoted by 𝐶^∗. A close to star function need not be univalent. The immediate shoot of 𝐶^∗ is the following subclass: 𝐶^∗₁=𝑓(𝑧)𝑓(𝑧)∈𝐴∶ℜℎ(𝑧)>0,ℎ(𝑧)∈𝐾,𝑧∈𝐸. (8)

In this paper, we consider the following subclasses of close-to-star functions: 𝐶^∗()=𝑓(𝑧)𝑓(𝑧)𝐴,𝐵∈𝐴∶𝑔(𝑧)≺1+𝐴𝑧,𝑔(𝑧)1+𝐵𝑧∈𝑆^∗,−1≤𝐵<𝐴≤1,𝑧∈𝐸. (9)

The class 𝐶^∗(𝐴,𝐵) was introduced and studied by Mehrok et al. [9]. In particular, 𝐶^∗(1,−1)≡𝐶^∗, 𝐶^∗₁()=𝑓(𝑧)𝑓(𝑧)𝐴,𝐵∈𝐴∶ℎ(𝑧)≺1+𝐴𝑧,ℎ(𝑧),1+𝐵𝑧∈𝐾,−1≤𝐵<𝐴≤1,𝑧∈𝐸 (10) is the class introduced and studied by Mehrok et al. [10]. Obviously, 𝐶^∗₁(1,−1)≡𝐶^∗₁.

Hummel [11] proved the conjecture of 𝑉. Singh that |𝑐₃−𝑐²₂|≤1/3 for the class 𝐾. Keogh and Merkes [12] obtained the estimates (4) for the classes 𝑆^∗ and 𝐾. We obtain sharp upper bounds of the functional |𝑎₃−𝜇𝑎²₂| for functions belonging to the classes 𝐶^∗(𝐴,𝐵) and 𝐶^∗₁(𝐴,𝐵).

2. Preliminary Lemmas

Lemma 1. Let 𝑔(𝑧)∈𝑆^∗. Then, 𝑏₃−𝜇𝑏²₂≤13−4𝜇𝑖𝑓𝜇≤2,11𝑖𝑓2≤𝜇≤1,4𝜇−3𝑖𝑓𝜇≥1. (11)

This lemma is due to Keogh and Merkes [12].

Lemma 2. Let ℎ(𝑧)∈𝐾. Then, 𝑐₃−𝜇𝑐²₂≤21−𝜇𝑖𝑓𝜇≤3,132𝑖𝑓34≤𝜇≤3,4𝜇−1𝑖𝑓𝜇≥3. (12)

This result was proved by Keogh and Merkes [12].

3. Main Results

Theorem 3. Let 𝑓(𝑧)∈𝐶^∗(𝐴,𝐵). Then, 𝑎₃−𝜇𝑎²₂≤[3+(𝐴−𝐵)(2−𝐵)]−[2+(𝐴−𝐵)]²𝜇𝑖𝑓𝜇≤−𝐵()()2+𝐴−𝐵𝐵<0,𝐴+𝐵<0𝑜𝑟𝜇≤−𝐵()(),()+()𝐴−𝐵𝐵>03+𝐴−𝐵−4𝜇𝐴−𝐵1−2𝜇²(1+𝐵)+𝜇(𝐴−𝐵)𝑖𝑓−𝐵()12+𝐴−𝐵≤𝜇≤2𝑜𝑟−𝐵2+(𝐴−𝐵)≤𝜇≤−𝐵()(),()+()𝐴−𝐵𝐵<01+𝐴−𝐵𝐴−𝐵2𝜇−1²()()11−𝐵−𝜇𝐴−𝐵𝑖𝑓2()≤𝜇≤2−𝐵(),())]2+𝐴−𝐵𝐴−𝐵)(4+𝐴−𝐵𝜇−[(𝐴−𝐵)(2−𝐵−1𝑖𝑓2−𝐵2+(𝐴−𝐵)≤𝜇≤1−𝐵,[]𝐴−𝐵2+(𝐴−𝐵)²𝜇−[3+(𝐴−𝐵)(2−𝐵)]𝑖𝑓𝜇≥1−𝐵.𝐴−𝐵 (13)

These results are sharp.

Proof. By definition of 𝐶^∗(𝐴,𝐵), 𝑓(𝑧)(𝑧)(𝑧)=𝑔1+𝐴𝑤1+𝐵𝑤(𝑧),𝑔∈𝑆^∗,𝑤(𝑧)∈Ω. (14)
Identifying terms in the above expansion, 𝑎₂=𝑏₂+(𝐴−𝐵)𝑑₁,𝑎₃=𝑏₃+(𝐴−𝐵)𝑏₂𝑑₁+(𝐴−𝐵)𝑑₂−𝐵𝑑²₁. (15)
From (15) and by using (2), it is easily established that 𝑎₃−𝜇𝑎²₂≤(𝐴−𝐵)+𝑏₃−𝜇𝑏²₂+(+()(𝑥𝐴−𝐵)1−2𝜇𝑥𝑦𝐴−𝐵𝐵+𝜇𝐴−𝐵)−1², (16) where 𝑥=|𝑑₁|≤1 and 𝑦=|𝑏₂|≤2.
Two cases arise: (a) 𝜇<−𝐵/(𝐴−𝐵) and (b) 𝜇≥−𝐵/(𝐴−𝐵).
Consider the case 𝜇<−𝐵/(𝐴−𝐵).
Case I. Suppose that 𝜇≤1/2 so that 𝐴+𝐵≤0.
By Lemma 1, (16) can be written as 𝑎₃−𝜇𝑎²₂≤[(()×()(𝑥3+𝐴−𝐵)]−4𝜇+2𝐴−𝐵1−2𝜇𝑥−𝐴−𝐵)(1+𝐵+𝜇𝐴−𝐵)²=𝐻₁(𝑥)𝐻,say,_′1(𝑥)(()×)(𝐻=2𝐴−𝐵)1−2𝜇−2𝐴−𝐵(1+𝐵+𝜇𝐴−𝐵)𝑥,_1′′(𝑥)=−2(𝐴−𝐵)(1+𝐵)+𝜇(𝐴−𝐵), (17)𝐻^′₁(𝑥)=0 implies that 𝑥=(1−2𝜇)/((1+𝐵)+𝜇(𝐴−𝐵))=𝑥₁.
Subcase I(i). For 𝜇<0, 𝐻^′₁(𝑥)>0 provided 𝜇≤−(1+𝐵)/(𝐴−𝐵)<−𝐵/(𝐴−𝐵).
Where  𝐻₁(𝑥) is an increasing function in [0,1] and max𝐻₁(𝑥)=𝐻₁(1).
Subcase I(ii). Suppose 𝜇≥0 and −(1+𝐵)/(𝐴−𝐵)≤𝜇≤−𝐵/(2+(𝐴−𝐵)).
If 𝜇<−𝐵/(2+(𝐴−𝐵)),then𝐻^′₁(𝑥)>0 and max𝐻₁(𝑥)=𝐻₁(1).
Combining the above two subcases, we obtain the first result of (13). For 𝑥₁>1,𝐻₁^′′(𝑥)<0 which holds for 𝜇<−(1+𝐵)/(𝐴−𝐵).
For −(1+𝐵)/(𝐴−𝐵)≤𝜇≤−𝐵/(2+(𝐴−𝐵))(𝐵<0) or −𝐵/(2+(𝐴−𝐵))≤𝜇≤1/2 (for all 𝐴 and 𝐵), max𝐻₁(𝑥)=𝐻₁(𝑥₁) from which the second result of (13) follows.
We now consider the case 𝜇≥−𝐵/(𝐴−𝐵).
Case II. Suppose that 1/2≤𝜇≤1. Then by Lemma 1, (16) takes the following form: 𝑎₃−𝜇𝑎²₂()()𝑥−()(𝑥≤1+𝐴−𝐵+2𝐴−𝐵2𝜇−1𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)²=𝐻₂(𝑥), (18)
Subcase II(i). For 1/2≤𝜇≤(2−𝐵)/(2+(𝐴−𝐵)), 𝐻^′₂(𝑥)()()(𝐻=2𝐴−𝐵2𝜇−1−2𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)𝑥,_2′′(𝑥)=−2(𝐴−𝐵)(1−𝐵)−𝜇(𝐴−𝐵), (19) where 𝐻^′₂(𝑥)=0 implies that 𝑥=(2𝜇−1)/((1−𝐵)−𝜇(𝐴−𝐵))=𝑥₂.
For 𝜇≤(2−𝐵)/(2+(𝐴−𝐵)), we have 𝑥₂<1 and 𝐻^′₂(𝑥)>0. So max𝐻₂(𝑥)=𝐻₂(𝑥₂), which gives the third result of (13).
Subcase II(ii). For (2−𝐵)/(2+(𝐴−𝐵))≤𝜇≤(1−𝐵)/(𝐴−𝐵). For 𝑥₂≥1,(2−𝐵)/(2+(𝐴−𝐵))≤𝜇≤(1−𝐵)/(𝐴−𝐵),𝐻^′₂(𝑥)>0, and max𝐻₂(𝑥)=𝐻₂(1), which leads us to the fourth result of (13).
Case III. For 𝜇≥(1−𝐵)/(𝐴−𝐵)≥1. By Lemma 1, (16) can be put in the following form: 𝑎₃−𝜇𝑎²₂≤()+()𝑥−()(𝑥𝐴−𝐵4𝜇−3+2𝐴−𝐵2𝜇−1𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)²=𝐻₃(𝑥)𝐻,say,_′3(𝑥)(()×)(𝐻=2𝐴−𝐵)2𝜇−1−2𝐴−𝐵(1−𝐵−𝜇𝐴−𝐵)𝑥,_3′′(𝑥)=−2(𝐴−𝐵)(1−𝐵)−𝜇(𝐴−𝐵), (20) where 𝐻^′₃(𝑥)=0 shows that 𝑥=(2𝜇−1)/((1−𝐵)−𝜇(𝐴−𝐵))=𝑥₃.
If 𝑥₃>1,  then (2−𝐵)/(2+(𝐴−𝐵))<𝜇, which is true.
𝐻^′₃(𝑥)>0 for 𝜇≥(1−𝐵)/(𝐴−𝐵)≥1. So max𝐻₃(𝑥)=𝐻₃(1), which takes us straight to the fifth result of (13).
The equality holds in the first and fifth results of (13) for the function 𝑓₁(𝑧) defined by 𝑓₁(𝑧)=𝑧(1−𝑧)²1+𝐴𝑧.1+𝐵𝑧 (21)
The equality holds in the second result of (13) for the function 𝑓₂(𝑧) defined by 𝑓₂(𝑧)=𝑧(1−𝑧)²1+𝐴+𝑥₁𝑧+𝐴𝑧²1+𝐵+𝑥₁𝑧+𝐵𝑧², (22) where 𝑥₁=(1−2𝜇)/((1+𝐵)+𝜇(𝐴−𝐵)).
The equality holds in the third result of (13) for the function 𝑓₃(𝑧) defined by 𝑓₃(𝑧)=𝑧(1−𝑧)²1+𝐴+𝑥₂𝑧+𝐴𝑧²1+𝐵+𝑥₂𝑧+𝐵𝑧², (23) where 𝑥₂=(2𝜇−1)/((1−𝐵)−𝜇(𝐴−𝐵)).
The equality holds in the fourth result of (13) for the function 𝑓₄(𝑧) defined by 𝑓₄(𝑧)=^𝑧₀𝑀1+²−2𝐿𝑀𝑧^{𝑀2/(𝑀2−2𝐿)}𝑑𝑧, (24) where 𝐿=3[(𝐴−𝐵)(2−𝐵)−1] and 𝑀=2[(𝐴−𝐵)(4+𝐴−𝐵)]^1/2 for |(𝑀²−2𝐿)/𝑀|≤1.
Proof of Theorem 3 is completed.

For 𝐴=1 and 𝐵=−1, Theorem 3 gives the following result.

Corollary 4. If 𝑓(𝑧)∈𝐶^∗, then, 𝑎₃−𝜇𝑎²₂≤19−16𝜇𝑖𝑓𝜇≤4,11+𝜇1𝑖𝑓41≤𝜇≤2,3+2𝜇−1²11−𝜇𝑖𝑓23≤𝜇≤4,312𝜇−5𝑖𝑓4≤𝜇≤1,16𝜇−9𝑖𝑓𝜇≥1. (25)

Theorem 5. Let 𝑓(𝑧)∈𝐶^∗₁(𝐴,𝐵). Then,𝑎₃−𝜇𝑎²₂≤[1+(𝐴−𝐵)(1−𝐵)]−[1+𝐴−𝐵]²−()𝜇𝑖𝑓𝜇≤1+2𝐵2()(),()+()1+𝐴−𝐵𝐴+𝐵>01+𝐴−𝐵−𝜇𝐴−𝐵1−2𝜇²4)(−()(1+𝐵+𝜇𝐴−𝐵)𝑖𝑓1+2𝐵2(1+𝐴−𝐵)≤𝜇≤−𝐵−()𝐴−𝐵𝑜𝑟1+2𝐵2()11+𝐴−𝐵≤𝜇≤2,()+()1+𝐴−𝐵−𝜇𝐴−𝐵2𝜇−1²4(1−𝐵)−𝜇(𝐴−𝐵)𝑖𝑓−𝐵2𝐴−𝐵≤𝜇≤3,13[(+()1+3𝐴−𝐵)]𝐴−𝐵2𝜇−1²4)(2(1−𝐵−𝜇𝐴−𝐵)𝑖𝑓3≤𝜇≤3−2𝐵2(),11+𝐴−𝐵3−(𝐴−𝐵)(1−𝐵)+(𝐴−𝐵)(2+𝐴−𝐵)𝜇𝑖𝑓3−2𝐵2()41+𝐴−𝐵≤𝜇≤3,[1+𝐴−𝐵]²[(4𝜇−1+𝐴−𝐵)(1−𝐵)]𝑖𝑓𝜇≥3. (26)

The results are sharp.

Proof. Proceeding as in Theorem 3, we have 𝑎₃−𝜇𝑎²₂≤(𝐴−𝐵)+𝑐₃−𝜇𝑐²₂+(+()(𝑥𝐴−𝐵)1−2𝜇𝑥𝑦𝐴−𝐵𝐵+𝜇𝐴−𝐵)−1², (27) where 𝑥=|𝑑₁|≤1 and 𝑦=|𝑐₂|≤1.
Two cases arise: (a) 𝜇<−𝐵/(𝐴−𝐵),  (b) 𝜇≥−𝐵/(𝐴−𝐵).
Consider the case 𝜇<−𝐵/(𝐴−𝐵).
By Lemma 2, (27) can be written as 𝑎₃−𝜇𝑎²₂≤[((𝑥−()(𝑥1+𝐴−𝐵)]−𝜇+𝐴−𝐵)1−2𝜇𝐴−𝐵)(1+𝐵+𝜇𝐴−𝐵)². (28)
 Subcase I(i). For 𝜇<0, 𝑎₃−𝜇𝑎²₂≤[(+()()×)(𝑥1+𝐴−𝐵)]−𝜇𝐴−𝐵1−2𝜇𝑥−𝐴−𝐵(1+𝐵+𝜇𝐴−𝐵)²=𝐻₁(𝑥),𝐻^′₁(𝑥)=()()(𝐻𝐴−𝐵1−2𝜇−2𝐴−𝐵)(1+𝐵+𝜇𝐴−𝐵)𝑥,_1′′(𝑥)=−2(𝐴−𝐵)(1+𝐵)+𝜇(𝐴−𝐵). (29)
For 𝜇<0, 𝐻^′₁(𝑥)>0 which shows that 𝐻₁(𝑥) is an increasing function in [0,1] and max𝐻₁(𝑥)=𝐻₁(1).
Subcase I(ii). Suppose 𝜇≥0.
𝐻^′₁(𝑥)=0 implies that 𝑥=(1−2𝜇)/(2[(1+𝐵)+𝜇(𝐴−𝐵)])=𝑥₁.
For 𝑥₁≥1,−(1+𝐵)/(𝐴−𝐵)≤𝜇≤−(1+2𝐵)/(2(1+(𝐴−𝐵))). So 𝐻^′₁(𝑥)>0 and max𝐻₁(𝑥)=𝐻₁(1).
Combining the above two subcases, we obtain the first result of (26).
For 𝑥₁>1,𝐻₁^′′(𝑥)<0, which holds for −(1+2𝐵)/(2(1+(𝐴−𝐵)))≤𝜇≤−𝐵/(𝐴−𝐵).
So max𝐻₁(𝑥)=𝐻₁(𝑥₁), which leads us to the second result of (26).
We now consider the case 𝜇≥−𝐵/(𝐴−𝐵).
Subcase I(iii). For 1/2≤−𝐵/(𝐴−𝐵)≤𝜇≤(2/3)(𝐵<0,2𝐴+𝐵>0), (27) takes the following form: 𝑎₃−𝜇𝑎²₂()()()×)(𝑥≤1+𝐴−𝐵−𝜇+𝐴−𝐵2𝜇−1𝑥−𝐴−𝐵(1−𝐵−𝜇𝐴−𝐵)²=𝐻₂(𝑥),𝐻^′₂(𝑥)=()()(𝐻𝐴−𝐵2𝜇−1−2𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)𝑥,_2′′(𝑥)=−2(𝐴−𝐵)(1−𝐵)−𝜇(𝐴−𝐵), (30) where 𝐻^′₂(𝑥)=0 implies that 𝑥=(2𝜇−1)/(2[(1−𝐵)−𝜇(𝐴−𝐵)])=𝑥₂.
It is easy to verify that 𝐻^′₂(𝑥)<0. So max𝐻₂(𝑥)=𝐻₂(𝑥₂), which gives the third result of (26).
Case II. For 2/3≤𝜇≤4/3.
 Subcase II(i). For 2/3≤𝜇≤(3−2𝐵)/(2(1+𝐴−𝐵)), (27) reduces to 𝑎₃−𝜇𝑎²₂≤13+()+(𝑥−()(𝑥𝐴−𝐵𝐴−𝐵)2𝜇−1𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)²=𝐻₃(𝑥),𝐻^′₃(𝑥)=(()(𝐻𝐴−𝐵)2𝜇−1−2𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)𝑥,_3′′(𝑥)=−2(𝐴−𝐵)(1−𝐵)−𝜇(𝐴−𝐵), (31) where 𝐻^′₃(𝑥)=0 implies that 𝑥=(2𝜇−1)/(2[(1−𝐵)−𝜇(𝐴−𝐵)])=𝑥₃.
For 𝑥₃<1, (2/3)≤𝜇≤(3−2𝐵)/(2(1+𝐴−𝐵)), 𝐻₃^′′(𝑥)<0, and max𝐻₃(𝑥)=𝐻₃(𝑥₃), which leads us to the fourth result of (26).
 Subcase II(ii). For (3−2𝐵)/(2(1+𝐴−𝐵))≤𝜇≤1, (27) reduces to the following form: 𝑎₃−𝜇𝑎²₂≤13+()+(𝑥−()(𝑥𝐴−𝐵𝐴−𝐵)2𝜇−1𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)²=𝐻₄(𝑥),𝐻^′₄(𝑥)=(()(𝐴−𝐵)2𝜇−1−2𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)𝑥; (32)𝐻^′₄(𝑥)=0 implies that 𝑥=(2𝜇−1)/(2[(1−𝐵)−𝜇(𝐴−𝐵)]). max𝐻₄(𝑥)=𝐻₄(1).
Combining the above two subcases, max𝐻₄(𝑥)=𝐻₄(1), which take us straight to the fifth result of (26).
Case III. For 𝜇≥4/3. By Lemma 1, (27) can be put in the form 𝑎₃−𝜇𝑎²₂≤()()𝑥−()(𝑥𝐴−𝐵−1+𝜇+𝐴−𝐵2𝜇−1𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)²=𝐻₅(𝑥),𝐻^′₅(𝑥)=(()(𝐴−𝐵)2𝜇−1−2𝐴−𝐵)(1−𝐵−𝜇𝐴−𝐵)𝑥, (33)(34) where 𝐻^′₅(𝑥)=0 shows that 𝑥=(2𝜇−1)/(2[(1−𝐵)−𝜇(𝐴−𝐵)])=𝑥₅.
𝑥₅>1 and max𝐻₅(𝑥)=𝐻₅(1), which yield the sixth result of (26).
The equality holds in the first and sixth results of (26) for the function 𝑓₁(𝑧) defined by 𝑓₁(𝑧)=𝑧1−𝑧1+𝐴𝑧.1+𝐵𝑧 (35)
The equality holds in the second result of (26) for the function 𝑓₂(𝑧) defined by 𝑓₂(𝑧)=𝑧1−𝑧1+𝐴+𝑥₁𝑧+𝐴𝑧²1+𝐵+𝑥₁𝑧+𝐵𝑧², (36) where 𝑥₁=(1−2𝜇)/(2[(1+𝐵)+𝜇(𝐴−𝐵)]).
The equality holds in the third and fourth results of (26) for the function 𝑓₃(𝑧) defined by 𝑓₃(𝑧)=𝑧1−𝑧1+𝐴+𝑥₂𝑧+𝐴𝑧²1+𝐵+𝑥₂𝑧+𝐵𝑧², (37) where 𝑥₂=(2𝜇−1)/(2[(1−𝐵)−𝜇(𝐴−𝐵)]).
The equality holds in the fifth result of (26) for the function 𝑓₄(𝑧) defined by 𝑓₄(𝑧)=^𝑧₀𝑀1+²−2𝐿𝑀𝑧^{𝑀2/(𝑀2−2𝐿)}𝑑𝑧, (38) where 𝐿=1−3(𝐴−𝐵)(1−𝐵) and 𝑀=2[(𝐴−𝐵)(2+𝐴−𝐵)]^1/2 for |(𝑀²−2𝐿)/𝑀|≤1.
Proof of Theorem 5 is completed.

For 𝐴=1,𝐵=−1, Theorem 5 gives the following result.

Corollary 6. If 𝑓(𝑧)∈𝐶^∗₁, then 𝑎₃−𝜇𝑎²₂≤15−9𝜇𝑖𝑓𝜇≤6,12+14𝜇𝑖𝑓61≤𝜇≤2,3−𝜇+2𝜇−1²411−𝜇𝑖𝑓22≤𝜇≤3,73+2𝜇−1²421−𝜇𝑖𝑓35≤𝜇≤6,8𝜇−1135𝑖𝑓64≤𝜇≤3,49𝜇−5𝑖𝑓𝜇≥3. (39)

The results of Corollaries 4 and 6 were proved by the authors of [13].