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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(3): 477-484

Published online September 30, 2020

### On the Fekete-SzegÖ Problem for Starlike Functions of Complex Order

Hanan Darwish and Abdel-Moniem Lashin, Bashar Al Saeedi*

Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516 Egypt
e-mail : darwish333@yahoo.com and aylashin@mans.edu.eg
Department of Mathematics, Faculty of Science & Engineering, Manchester Metropolitan University, Manchester, M15GD, UK
e-mail : basharfalh@yahoo.com

Received: January 28, 2016; Revised: May 3, 2020; Accepted: May 4, 2020

### Abstract

For a non-zero complex number b and for m and n in let Ψn,m(b) denote the class of normalized univalent functions f satisfying the condition $ℜ [1+1b(Dn+mf(z)Dnf(z)-1)]>0$ in the unit disk U, where Dnf(z) denotes the Salagean operator of f. Sharp bounds for the Fekete-Szegö functional $∣a3-μa22∣$ are obtained.

Keywords: coefficient estimates, Salagean operator, Fekete Szego problem, starlike functions of complex order

### 1. Introduction

Fekete and Szegö proved the remarkable result that the estimate

$|a3-λa22|≤1+2exp(-2λ1-λ)$

holds with 0 ≤ λ ≤ 1 for any normalized univalent function

$f(z)=z+a2z2+a3z3+…$

in the open unit disk U. This inequality is sharp for all λ (see [7]). The coefficient functional

$φ (f(z))=a3-λa22=16(f‴(0)-3λ2[f″(0)]2)$

in the unit disk represents various geometric quantities. For example, when λ = 1, $φ (f(z))=a3-λa22$ becomes Sf (0)/6 where Sf denotes the Schwarzian derivative (f/f′)′ − (f/f′)2/2. Note that, if we consider the nth root transform [f(zn)]1/n = z + cn+1zn+1 + c2n+1z2n+1 + … of f with the power series (1.1), then cn+1 = a2/n and $c2n+1=a3/n+(n-1)a22/2n2$, so that

$a3-λa22=n(c2n+1-μcn+12),$

where μ = λn + (n − 1)/2. Moreover, φ (f(z)) behaves well with respect to the rotation, namely φ ((ef(ez)) = e2φ(f), for θR.

This is quite natural to discuss the behavior of φ (f(z)) for subclasses of normalized univalent functions in the unit disk. This is called the Fekete-Szegö problem. Many authors have considered this problem for typical classes of univalent functions (see, for instance [1, 2]).

We denote by A the set of all functions of the from (1.1) that are normalized analytic and univalent in the unit disk U. Also, for 0 ≤ α < 1, let S*(α) and C(α) denote the classes of starlike and convex univalent functions of order α, respectively, i.e., let

$S*(α)={f(z)∈A:ℜ(zf′(z)f(z))>α,z∈U}$

and

$C(α)={f(z)∈A:ℜ(1+zf″(z)f′(z))>α,z∈U}.$

The notions of α-starlikeness and α-convexity were generalized onto a complex order by Nasr and Aouf [4], Wiatrowski [8], and Nasr and Aouf [3].

Observe that S* (0) = S* and C(0) = C represent standard starlike and convex univalent functions, respectively.

Let $f(z)=z+∑k=2∞akzk$ and $g(z)=z+∑k=2∞bkzk$ be analytic functions in U. The Hadamard product (convolution) of f and g, denoted by f * g is defined by

$(f*g)(z)=z+∑k=2∞akbkzk,z∈U.$

For a function f(z) in A, we define

$D0f(z)=f(z),D1f(z)=Df(z)=zf′(z)$

and

$Dnf(z)=D(Dn-1f(z)), n∈N={0,1,2,…}.$

With the above operator Dn, we say that a function f(z) belonging to A is in the class A(n, m, α) if and only if

$ℜ {Dn+mf(z)Dnf(z)}>α (n,m∈N0=N∪{0})$

for some α (0 ≤ α < 1), and for all zU.

We note that A(0, 1, α) = S*(α) is the class of starlike functions of order α, A(1, 1, α) = C(α) is the class of convex functions of order α, and that A(1, 0, α) = An(α) is the class of functions defined by Salagean [6].

### Definition 1.1

Let b be a nonzero complex number, and let f be an univalent function of the form (1.1), such that Dn+mf(z) ≠ 0 for zU\ {0}. We say that f belongs to Ψn,m(b) if

$ℜ {1+1b (Dn+mf(z)Dnf(z)-1)}>0, z∈U.$

By giving specific values to n,m and b, we obtain the following important subclasses studied by various researchers in earlier works, for instance, Ψ1,0(b) = S*(1 − b) (Nasr and Aouf [4]), Ψ1,1(b) = C(1 − b) (Wiatrowski [8], Nasr and Aouf [3]).

### 2. Main Results

We denote by ℘ class of the analytic functions in U with p(0) = 1 and ℜ(p(z)) > 0. We shall require the following:

### Lemma 2.1.([5], p.166)

Let p ∈ ℘ with p(z) = 1+c1z + c2z2 + …, then

$∣cn∣ ≤2, for n≥1.$

If |c1| = 2 then p(z) ≡ p1(z) = (1+γ1z)/(1−γ1z) with γ1 = c1/2. Conversely, if p(z) ≡ p1(z) for some 1| = 1, then c1 = 2γ1 and |c1| = 2. Furthermore we have

$|c2=c122|≤2-∣c1∣2.$

If |c1| < 2 and $|c2-c122|=2-∣c1∣2$, then p(z) ≡ p2(z), where

$p2(z)=1+γ2z+γ11+γ¯1+γ2z1-γ2z+γ11+γ¯1+γ2z,$

and γ1 = c1/2, $γ2=2c2-c124-∣c1∣2$. Conversely if p(z) ≡ p2(z) for some 1| < 1 and 2| = 1, then γ1 = c1/2, $γ2=2c2-c124-∣c1∣2$ and $|c2-c122|=2-∣c1∣2$.

### Theorem 2.2

Let n,m ≥ 0 and let b be non-zero complex number. If f of the form (1.1) is in Ψn,m(b), then

$∣a2∣ ≤2∣b∣2n(2m-1),$$∣a3∣ ≤2∣b∣3n(3m-1)max {1,∣(2m-1)+2b∣(2m-1)},$

and

$|a3-22n(2m-1)3n(3m-1)a22|≤2∣b∣3n(3m-1).$

Equality in (2.1) holds if Dn+mf(z)/Dnf(z) = 1+b [p1(z) − 1] , and in (2.2) if Dn+mf(z)/Dnf(z) = 1+b [p2(z) − 1] , where p1, p2are given in Lemma 2.1.

Proof

Denote F(z) = Dnf(z) = z + A2z2 + A3z3 + …,DmF(z) = Dn+mf(z) = z + 2mA2z2 + 3mA3z3 + …, where A2 = 2na2 and A3 = 3na3, …., then

$2mA2=2n+ma2, 3mA3=3n+ma3.$

By the definition of the class Ψn,m(b) there exists p ∈ ℘ such that $DmF(z)F(z)=1-b+bp(z)$, so that

$z+2mA2z2+3mA3z3+…z+A2z2+A3z3+…=1-b+b(1+c1z+c2z2+…),$

which implies the equality

$z+2mA2z2+3mA3z3+…=z+(A2+bc1)z2+(A3+bc1A2+bc2)z3+…$

Equating the coefficients of both sides we have

$A2=b2m-1c1, A3=b3m-1 [c2+bc122m-1],$

so that, on account of Dn+mf(z)

$a2=bc12n(2m-1), a3=b3n(3m-1) [c2+b(2m-1)c12].$

Taking into account (2.5) and Lemma 2.1, we obtain

$∣a2∣=|b2n(2m-1)c1|≤2∣b∣2n(2m-1),$

and

$∣a3∣=|b3n(3m-1) [c2-c122+(2m-1)+2b2 (2m-1)c12]|≤∣b∣3n(3m-1) [2-∣c12∣2+∣(2m-1)+2b∣2 (2m-1)∣c12∣]=∣b∣3n(3m-1) [2+∣(2m-1)+2b∣-(2m-1)2 (2m-1)∣c12∣]≤2∣b∣3n(3m-1)max {1,1+∣(2m-1)+2b∣-(2m-1)(2m-1)}.$

Thus

$∣a3∣ ≤2∣b∣3n(3m-1)max {1,∣(2m-1)+2b∣(2m-1)}.$

Moreover

$|a3-22n (2m-1)3n (3m-1)a22|=|b3n (3m-1) [c2+b(2m-1)c12]-b2c12[2n(2m-1)]222n (2m-1)3n(3m-1)|=|bc23n(3m-1)|≤2∣b∣3n(3m-1),$

as asserted.

### Remark 2.3

Putting b = 1−α in the above theorem we get the following corollary.

### Corollary 2.4

Let n,m ≥ 0 . If f of the form (1.1) is in A(n, m, α), then

$∣a2∣ ≤2 (1-α)2n(2m-1),∣a3∣ ≤2 (1-α)3n(3m-1)max {1,(2m+1-2α)(2m-1)},$

and

$|a3-22n (2m-1)3n(3m-1)a22|≤2 (1-α)3n(3m-1).$

### Remark 2.5

In the above Theorem 2.2 and Corollary 2.4 a special case of Fekete- Szegö problem e.g. for real μ = 22n (2m − 1) /3n(3m − 1) occurred very naturally and simple estimate was obtained.

Now, we consider functional $∣a3-μa22∣$ for complex μ.

### Theorem 2.6

Let b be a non-zero complex number and let f ∈ Ψn,m(b). Then for μ ∈ ℂ

$∣a3-μa22∣ ≤2∣b∣3n(3m-1)max {1,|1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2|}.$

For each μ there is a function in Ψn,m(b) such that equality holds.

Proof

Applying (2.5) we have

$a3-μa22=b3n(3m-1) [c2+b(2m-1)c12]-μb2c12[2n(2m-1)]2=b3n(3m-1) [c2+b(2m-1)c12-μb3n(3m-1)[2n(2m-1)]2c12]=b3n(3m-1) [c2-c122+c122 (1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2)].$

Then, with the aid of Lemma 2.1, we obtain

$∣a3-μa22∣ ≤∣b∣3n(3m-1) [2-∣c1∣22+∣c1∣22|1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2|]=∣b∣3n(3m-1) [2+∣c1∣22 (|1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2|-1)]≤2∣b∣3n(3m-1)max {1,|1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2|}.$

An examination of the proof shows that equality is attained for the first case, when c1 = 0, c2 = 2, then the functions in Ψn,m(b) is given by

$Dn+mf(z)Dnf(z)=1+(2b-1)z1-z,$

and, for the second case, when c1 = c2 = 2, so that

$Dn+mf(z)Dnf(z)=1+(2b-1)z21-z,$

respectively.

### Remark 2.7

Putting b = 1 − α in the above theorem we get the following corollary.

### Corollary 2.8

Let fA(n, m, α) . Then for μ ∈ ℂ

$∣a3-μa22∣ ≤2 (1-α)3n(3m-1)max {1,|1+2(1-α)(2m-1)-2μ(1-α)3n(3m-1)[2n(2m-1)]2|}.$

For each μ there is a function in A(n, m, α) such that equality holds.

We next consider the case, when μ and b are real. Then we have:

### Theorem 2.9

Let b > 0 and let f ∈ Ψn,m(b) . Then for μR we have

$∣a3-μa22∣ ≤{2b3n(3m-1) [1+2b(2m-1)(1-μ3n(3m-1)22n(2m-1))]if μ≤22n(2m-1)3n(3m-1),b3n(3m-1)if 22n(2m-1)3n(3m-1)≤μ22n(2m-1)[(2m-1)+2b]2b[3n(3m-1)],2b3n(3m-1) [2μb3n(3m-1)[2n(2m-1)]2-1-2b(2m-1)]if μ≥22n(2m-1)[(2m-1)+2b]2b[3n(3m-1)].$

For each there is a function in Ψn,m(b) such that equality holds.

Proof

First, let $μ≤22n(2m-1)3n(3m-1)$ In this case, (2.5) and Lemma 2.1 give

$∣a3-μa22∣ ≤b3n(3m-1) [2-∣c1∣22+∣c1∣22 (1+2b(2m-1)-2μb3n(3m-1)[2n(2m-1)]2)]≤2b3n(3m-1) [1+2b(2m-1) (1-μ3n(3m-1)22n(2m-1))].$

Let, now $22n(2m-1)3n(3m-1)≤μ≤22n(2m-1)[(2m-1)+2b]2b[3n(3m-1)]$. Then, using the above calculations, we obtain

$∣a3-μa22∣ ≤b3n(3m-1).$

Finally, if $μ≥22n(2m-1)[(2m-1)+2b]2b[3n(3m-1)]$, then

$∣a3-μa22∣ ≤b3n(3m-1) [2-∣c1∣22+∣c1∣22 (2μb3n(3m-1)[2n(2m-1)]2-1-2b(2m-1))]=b3n(3m-1) [2+∣c1∣22 (2μb3n(3m-1)[2n(2m-1)]2-2-2b(2m-1))]≤2b3n(3m-1) [2μb3n(3m-1)[2n(2m-1)]2-1-2b(2m-1)].$

Equality is attained for the second case on choosing c1 = 0, c2 = 2 in (2.7) and in (2.8) by choosing c1 = 2, c2 = 2 and c1 = 2i, c2 = −2 for the first and third case, respectively. Thus the proof is complete.

### Remark 2.10

Put b = 1 − α in the above theorem we get the following corollary.

### Corollary 2.11

Let fA(n, m, α) . Then for μR we have

$∣a3-μa22∣ ≤{2(1-α)3n(3m-1) [1+2(1-α)(2m-1)(1-μ3n(3m-1)22n(2m-1))]if μ≤22n(2m-1)3n(3m-1),(1-α)3n(3m-1)if 22n(2m-1)3n(3m-1)≤μ≤22n(2m-1)[2m+1-2α]2(1-α)[3n(3m-1)],2(1-α)3n(3m-1) [2μ (1-α)3n(3m-1)[2n(2m-1)]2-1-2(1-α)(2m-1)]if μ≥22n(2m-1)[2m+1-2α]2(1-α)[3n(3m-1)].$

For each there is a function in A(n, m, α) such that equality holds.

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