KYUNGPOOK Math. J. 2019; 59(2): 215-224
Coeﬃcient Bounds for Several Subclasses of Analytic and Biunivalent Functions
Samira Rahrovi and Hossein Piri, Janusz SokÓŁ
Department of Mathematics, Basic Science Faculty, University of Bonab, Bonab, Iran
e-mail : s.rahrovi@bonabu.ac.ir and hossein_piri1979@yahoo.com

Faculty of Mathematics and Natural Sciences, University of Rzeszów, Prof. Pigonia street 1, 35-310 Rzeszów, Poland
e-mail : jsokol@ur.edu.pl
* Corresponding Author.
Received: December 17, 2016; Revised: March 20, 2019; Accepted: April 23, 2109; Published online: June 23, 2019.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

In the present paper, some generalizations of analytic functions have been considered and the bounds of the coefficients of these classes of bi-univalent functions have been investigated.

Keywords: bi-univalent function, starlike function of order α, strongly starlike functions, strongly convex functions of order α, subordination.
1. Introduction and Preliminaries

Let denote the class of functions of the form

$f(z)=z+∑n=2∞anzn,$

which are analytic in the open unit disk U = {z ∈ ℂ: |z| < 1}. Also let S denote the subclass of consisting of univalent functions in U. A function f in S is said to be starlike of order α (0 ≤ α < 1) in U if Re(zf′(z)/f(z)) > α. We denote by S*(α), the class of all such functions. A function f in S is said to be convex of order α (0 ≤ α < 1) in U if Re (1 + zf″(z)/f′(z)) > α. Let K(α) denote the class of all those functions which are convex of order α in U. Mocanu [10] studied linear combinations of the representations of convex and starlike functions and defined the class of α-convex functions. In [9], it was shown that if

$Re {(1-α)zf′(z)f(z)+α (1+zf″(z)f′(z))}>0, z∈U$

then f is in the class of starlike functions S*(0) for α real and is in the class of convex functions K(0) for α ≥ 1.

A function f(z) ∈ S is said to be in the class of strongly starlike functions of order α, denoted by $Ss*(α)$, 0 ≤ α < 1, if it satisfies the following inequality

$|arg (zf′(z)f(z))|≤απ2, z∈U,$

and is said to be in the class of strongly convex functions of order α, 0 ≤ α < 1, denoted by Kc(α), if it satisfies the following inequality

$|arg(1+zf″(z)f′(z))|≤απ2, z∈U.$

For two functions f and g, analytic in U, we say that the function f is subordinate to g, and write f(z) ≺ g(z), if there exists an analytic function ϕ(z) defined in U with ϕ(0) = 0 and |ϕ(z)| < 1 such that f(z) = g(ϕ(z)).

For each fS, the Koebe one-quarter theorem [5] ensures the image of U under f contains a disk of radius 1/4. Thus every univalent function fS has an inverse f−1 satisfying

$f-1(f(z))=z, z∈U$

and

$f(f-1(w))=w, ∣w∣

A function is said to be bi-univalent in U if both f and f−1 are univalent in U. If f given by (1.1) is bi-univalent, then a simple computation shows that g = f−1 has the expansion

$g(w)=w-a2w2+(2a22-a3)w3-(5a23-5a2a3+a4)w4+….$

Let ∑ denote the class of all bi-univalent functions defined in the unit disk U.

Lewin [8] considered the class of bi-univalent functions ∑ and obtained the following bound for the second coefficient in the Tylor-Maclaurin expansion as follows

$∣a2∣<1.51,$

for function $f(z)=z+∑n=2∞anzn∈Σ$. After that, Brannan and Clunie [3] conjectured that $∣a2∣≤2$ for f ∈ ∑. Netanyahu [11] showed that max |a2| = 4/3 if f ∈ ∑. Brannan and Taha [4] studied some subclasses of bi-univalent functions, similar to the familiar subclasses of univalent functions consisting of strongly starlike, starlike and convex functions. They introduced the class of bi-starlike functions and bi-convex functions and investigated estimates on the two initial coefficients. Until now, the coefficient estimate problem for each of the following Taylor-Maclaurin coefficients

$∣an∣, n∈ℕ\{1,2},ℕ={1,2,…},$

is presumably still an open problem. The bi-univalent functions was actually revived in the recent years by the pioneering work by Srivastava et al. [15]. In [15], two new subclasses of the function class ∑ has been introduced and the estimates on the coefficients |a2| and |a3| for functions in these new subclasses of the function class ∑ has been considered. After that, many authors investigated bounds for various subclasses of bi-univalent functions [1, 6, 12, 15, 16].

We notice that the class ∑ is not empty. For example, the following functions are members of ∑:

$z, z1-z, -log(1-z), 12log1+z1-z.$

However, the Koebe function is not a member of ∑.

In the next section, motivated essentially by the recent work of Srivastava et al. [15], Frasin and Aouf [6], Xu et al. [16] and other authors, we investigte interesting subclasses of analytic and bi-univalent functions in the open unit dick U. So we estimate the first two Tylor-Maclaurin coefficients |a2| and |a3| for these classes.

2. Coefficient Estimates

In the sequel, it is assumed that ϕ is an analytic function with positive real part in the unit disk U, satisfying ϕ(0) = 1, ϕ′(0) > 0 and ϕ(U) is symmetric with respect to the real axis. Assume also that

$ϕ(z)=1+B1z+B2z+B3z+…, B1>0.$

Suppose that u(z) and v(z) are analytic in the unit disk U with u(0) = v(0) = 0, |u(z)| < 1, |v(z)| < 1 and suppose that

$u(z)=b1z+∑n=2∞bnzn, v(z)=c1z+∑n=2∞cnzn, ∣z∣<1.$

It is well known that

$∣b1∣≤1, ∣b2∣≤1-∣b1∣2, ∣c1∣≤1, ∣c2∣≤1-∣c1∣2.$

By a simple calculation, we have

$ϕ(u(z))=1+B1b1z+(B1b2+B2b12)z2+…,$$ϕ(v(w))=1+B1c1w+(B1c2+B2c12)w2+….$

### Definition 2.1

A function f given by (1.1) is said to be in the class H(α, β, ϕ), 0 ≤ α ≤ 1, β ≥ 0, if f ∈ ∑ and satisfies the following conditions is said to be in the class

$(1-β)[(1-α)zf′(z)f(z)+α (1+zf″(z)f′(z))]+βzf′(z)+αz2f″(z)(1-α)f(z)+αzf′(z)≺ϕ(z),$

and

$(1-β)[(1-α)wg′(w)g(w)+α (1+wg″(w)g′(w))]+βwg′(w)+αw2g″(w)(1-α)g(w)+αwg′(w)≺ϕ(w),$

where g(w) = f−1(w) is given by (1.2).

For special choices for the number α, β and function ϕ, we can obtain several important classes of analytic functions. For example, $HΣ(0,0,1+(1-2α)z1-z)$is the class S*(α), $HΣ(1,0,1+(1-2α)z1-z)$ is the class K(α), $HΣ(0,0,(1+z1-z)α)$ is the class $Ss*(α)$ and $HΣ(1,0,(1+z1-z)α)$ is the class Kc(α). These classes of analytic functions were well investigated by many authors [2, 7, 9, 10, 13, 14].

For proving our results we used the subordination concept which have been used by many authors (for example see [1]).

### Theorem 2.2

If fH(α, β, ϕ) is in, then

$∣a2∣≤B1B1∣(1+α) (B12-(1+α)B2)+B12αβ(1-α)∣+(α+1)2B1,$

and

$∣a3∣≤{B12(1+2α),B1≤(1+α)22(1+2α),∣(1+α) (B12-(1+α)B2)+B12αβ(1-α)∣+2(1+2α)B132(1+2α)[(α+1)2B1+∣(1+α) (B12-(1+α)B2)+B12αβ(1-α)∣],B1≥(1+α)22(1+2α).$
Proof

Let fH(α, β, ϕ) and g = f−1. Then there exist two functions u and v; analytic in U; with |u(z)| < 1, |v(w)| < 1, u(0) = v(0) = 0, z, wU, given by (2.2), such that

$(1-β)[(1-α)zf′(z)f(z)+α (1+zf″(z)f′(z))]+βzf′(z)+αz2f″(z)(1-α)f(z)+αzf′(z)=ϕ(u(z)),$

and

$(1-β)[(1-α)wg′(w)g(w)+α (1+wg″(w)g′(w))]+βwg′(w)+αw2g″(w)(1-α)g(w)+αwg′(w)=ϕ(v(w)).$

Since

$(1-β)[(1-α)zf′(z)f(z)+α (1+zf″(z)f′(z))]+βzf′(z)+αz2f″(z)(1-α)f(z)+αzf′(z)=1+(1+α)a2z+[2(1+2α)a3-(1+3α-αβ+α2β)a22]z2+…$

and

$(1-β)[(1-α)wg′(w)g(w)+α (1+wg″(w)g′(w))]+βwg′(w)+αw2g″(w)(1-α)g(w)+αwg′(w)=1-(1+α)a2w-[(1+3α-αβ+α2β)a22-2(1+2α) (2a22-a3)]w2+…,$

it follows from (2.4), (2.5), (2.8) and (2.9) that

$(α+1)a2=B1b1,$$2(1+2α)a3-(1+3α-αβ+α2β)a22=B1b2+B2b12,$$-(α+1)a2=B1c1,$$2(1+2α) (2a22-a3)-(1+3α-αβ+α2β)a22=B1c2+B2c12.$

From (2.10) and (2.12) we get

$c1=-b1,$

and

$2(α+1)2a22=B12(b12+c12).$

By adding the relation (2.11) with the relation (2.13) and use (2.15) we can get

$2[(1+α) (B12-(1+α)B2)+B12αβ(1-α)]a22=B13(b2+c2)$

Now, in view of (2.3), (2.10) and (2.14) we have

$2|(1+α) (B12-(1+α)B2)+B12αβ(1-α)| ∣a2∣2≤B13(∣b2∣+∣c2∣)≤B13((1-∣b1∣2)+(1-∣c1∣2))≤2B13(1-∣b1∣2)=2B13-2(α+1)2B1∣a2∣2.$

Then, we get

$∣a2∣≤B1B1∣(1+α) (B12-(1+α)B2)+B12αβ(1-α)∣+(α+1)2B1.$

Next, from (2.11) and (2.13), we have

$4(1+2α)a3=B1(b2-c2)+4(1+2α)a22.$

Then in view of (2.3) and (2.14), we have

$4(1+2α)∣a3∣≤4(1+2α)∣a2∣2+B1(∣b2∣+∣c2∣)≤4(1+2α)∣a2∣2+2B1(1-∣b1∣2).$

It follows from (2.10) that

$2B1(1+2α)∣a3∣≤[2B1(1+2λ)-(1+α)2]∣a2∣2+B12.$

Now from (2.16) we have

$∣a3∣≤{B12(1+2α),B1≤(1+α)22(1+2α),∣(1+α) (B12-(1+α)B2)+B12αβ(1-α)∣+2(1+2α)B132(1+2α)[(α+1)2B1+∣(1+α) (B12-(1+α)B2)+B12αβ(1-α)∣],B1≥(1+α)22(1+2α).$

This completes the proof.

### Remark 2.3

Let

$ϕ(z)=(1+z1-z)γ=1+2γz+2γ2z2+…, 0<γ≤1$

then inequalities (2.6) and (2.7) become

$∣a2∣≤2γ(1-α) (1+α+2αβ),$

and

$∣a3∣≤{γ1+2α,γ≤(1+α)24(1+2α),(1-α) (1+α+2αβ)+16γ(1+2α)γ(1+2α) (1-α) (1+α+2αβ),γ≥(1+α)24(1+2α).$

### Definition 2.4

A function f given by (1.1) is said to be in the class C(λ, ϕ), 0 ≤ λ < 1, if f ∈ ∑ and satisfies the following conditions

$λz3f″′(z)+(2λ+1)z2f″(z)+zf′(z)λz2f″(z)+zf′(z)≺ϕ(z),$

and

$λw3g″′(w)+(2λ+1)w2g″(w)+wg′(z)λw2g″(w)+wg′(w)≺ϕ(w),$

where g(w) = f−1(w) is given by (1.2).

### Theorem 2.5

If fC(λ, ϕ) is in, then

$∣a2∣≤B1B1∣2(2λ+1-2λ2)B12-4(λ+1)2B2∣+4(λ+1)2B1,$

and

$∣a3∣≤{B16(1+2λ),B1≤2(1+λ)23(1+2λ),∣(2λ+1-2λ2)B12-2(λ+1)2B2∣+3(1+2λ)B133(1+2λ)[∣2(2λ+1-2λ2)B12-4(λ+1)2B2∣+4(λ+1)2B1],B1≥2(1+λ)23(1+2λ).$
Proof

Let fC(λ, ϕ) and g = f−1. Then there are analytic functions u, v: UU given by (2.2) such that

$λz3f″′(z)+(2λ+1)z2f″(z)+zf′(z)λz2f″(z)+zf′(z)=ϕ(u(z)),$

and

$λw3g″′(w)+(2λ+1)w2g″(w)+wg′(z)λw2g″(w)+wg′(w)=ϕ(v(w)).$

Since

$λz3f″′(z)+(2λ+1)z2f″(z)+zf′(z)λz2f″(z)+zf′(z)=1+2(λ+1)a2z+2((1+2λ) (3a3-2a22)-2λ2a22)z2+…$

and

$λw3g″′(w)+(2λ+1)w2g″(w)+wg′(z)λw2g″(w)+wg′(w)=1-2(λ+1)a2w+(2(1+2λ) (4a22-3a3)-4λ2a22)w2+…,$

it follows from (2.4), (2.5), (2.19) and (2.20) that

$2(λ+1)a2=B1b1,$$2(1+2λ) (3a3-2a22)-4λ2a22=B1b2+B2b12,$$-2(λ+1)a2=B1c1,$$2(1+2λ) (4a22-3a3)-4λ2a22=B1c2+B2c12.$

From (2.21) and (2.23) we get

$c1=-b1,$

and

$8(λ+1)2a22=B12(b12+c12).$

By adding the relation (2.22) with the relation (2.24) and use (2.26) we have

$2(B12(4λ-4λ2+2)-4(λ+1)2B2)a22=B13(b2+c2).$

Now, in view of (2.3), (2.21) and (2.25) we have

$2|B12(4λ-4λ2+2)-4(λ+1)2B2|∣a2∣2≤B13(∣b2∣+∣c2∣)≤B13((1-∣b1∣2)+(1-∣c1∣2))≤2B13(1-∣b1∣2)=2B13-8(λ+1)2B1∣a2∣2.$

Then, we get

$∣a2∣≤B1B1∣2(2λ+1-2λ2)B12-4(λ+1)2B2∣+4(λ+1)2B1.$

Next, from (2.22) and (2.24), we have

$12(1+2λ)a3=B1(b2-c2)+12(1+2λ)a22.$

Then in view of (2.3) and (2.25), we have

$12(1+2λ)∣a3∣≤12(1+2λ)∣a2∣2+B1(∣b2∣+∣c2∣)≤12(1+2λ)∣a2∣2+2B1(1-∣b1∣2).$

It follows from (2.21) that

$6B1(1+2λ)∣a3∣≤[6B1(1+2λ)-4(1+λ)2]∣a2∣2+B12.$

Now from (2.27) we have

$∣a3∣≤{B16(1+2λ),B1≤2(1+λ)23(1+2λ),∣(2λ+1-2λ2)B12-2(λ+1)2B2∣+3(1+2λ)B133(1+2λ)[∣2(2λ+1-2λ2)B12-4(λ+1)2B2∣+4(λ+1)2B1],B1≥2(1+λ)23(1+2λ).$

This completes the proof.

### Remark 2.6

If let

$ϕ(z)=(1+z1-z)α=1+2αz+2α2z2+…, 0<γ≤1$

in theorem 2.5, then inequalities (2.17) and (2.18) become

$∣a2∣≤αλ(2+λ+3αλ),$

and

$∣a3∣≤{α3(1+2λ),α≤(1+λ)23(1+2λ),α (2α+λ(3α+λ))2λ(1+2λ) (1-α) (1+λ+3αλ),γ≥(1+α)24(1+2α).$
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