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Koebe domains for certain subclasses of starlike functions

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U N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K A L U B L I N – P O L O N I A

VOL. LXI, 2007 SECTIO A 129–135

MAGDALENA SOBCZAK-KNEĆ

Koebe domains for certain subclasses of starlike functions

Abstract. The Koebe domain’s problem in the class of starlike functions with real coefficients was considered by M. T. McGregor [3]. In this paper we determined the Koebe domain for the class of starlike functions with real coefficients and the fixed second coefficient.

1. Introduction. Let S denote the class of analytic and univalent func- tions f in the unit disk ∆ = {z ∈ C : |z| < 1} such that f (0) = f0(0) − 1 = 0 and

Rezf0(z)

f (z) > 0, z ∈ ∆.

The class S is called the class of starlike functions.

In this paper we will study a subclass of the class S, i.e. the class SR which contains the starlike functions with real coefficients. In 1964 M. T.

McGregor [3] found the setT

f ∈SRf (∆), which is called the Koebe domain for the class SR.

Theorem 1 ([3]). The Koebe domain for the class SR is symmetric with respect to the real axis and the boundary of this domain in the upper half

2000 Mathematics Subject Classification. Primary 30C45; Secondary 30C80.

Key words and phrases. Real coefficients, Koebe domain, starlike functions, fixed sec- ond coefficient.

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plane is given by the polar equation w = ρ(θ)e, where

(1) ρ(θ) = 1

4

 θ π

θ

π 1 − θ

π

πθ−1

, θ ∈ [0, π].

The extremal functions are of the form

Fθ(z) = z

(1 − z)π(1 + z)2(1−πθ), z ∈ ∆, θ ∈ [0, π].

2. Main Results.

Theorem 2. If f ∈ SR and ρe ∈ f (∆), then f ≺ M ·F/ θ, where M = ρ(θ)ρ , θ ∈ [0, π] and ρ(θ) is given by (1).

Proof. Let f ∈ SR and ρe ∈ f (∆). Since f ∈ S/ R, it means that f does not admit values, which are on the rays l and l, where

l : {ζ ∈ C : ζ = ρet, t ≥ 1}, l : {ζ : ζ ∈ l}.

The function

ρ ρ(θ)Fθ

maps the unit disk ∆ onto the plane C without the rays l and l. Moreover, f ∈ SR, so

f (∆) ⊂ ρ

ρ(θ)Fθ(∆).

From the above as well as from the univalence of Fθ we conclude that

f ≺ M · Fθ, where M = ρ(θ)ρ , θ ∈ [0, π]. 

Remark 1. Theorem 1 results from Theorem 2. We have f ≺ M · Fθ.

Hence

1 = f0(0) ≤ M · Fθ0(0).

This condition is equivalent to M ≥ 1.

Let f = z + a2z2 + · · · ∈ SR and ρe ∈ f (∆). In the next theorem/ we determine the region of values (ρ, a2) for a fixed θ ∈ [0, 2π]. In this research we can discuss only θ ∈ [0, π], because the region of values (ρ, a2) is symmetric with respect to the real axis.

Theorem 3. If f = z + a2z2+ · · · ∈ SR and ρe ∈ f (∆), then for a fixed/ θ ∈ [0, π], the region of values (ρ, a2) is of the form

Aρ,a2 :=



(ρ, a2) : 1 ρ

θ − 1θπ−1

− 2 ≤ a2 ≤ 2 −1 ρ

θ − 1πθ .

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Proof. Let f ∈ SR and ρe ∈ f (∆). From Theorem 2 and [2] we have/ f (z) = M · Fθ h(z)

M

 ,

where M = ρ(θ)ρ ≥ 1. The function h(z) is univalent, with real coefficients, bounded by M and such that

M · Fθ h(z) M



∈ S. Denoting

f (z) = z + a2z2+ . . . Fθ(z) = z + b2(θ)z2+ . . .

h(z) = z + c2z2+ . . . . we have

a2 = c2+ 1

Mb2(θ) and b2(θ) = 2 2θ π − 1

 . For the function h(z), the following inequalities are true [1]:

−2

 1 − 1

M



≤ c2≤ 2

 1 − 1

M

 . Hence

a2≤ 2

 1 − 1

M

 + 2

M

 2θ π − 1

 , and consequently

a2 ≤ 2 −1 ρ

π θ − 1πθ

. Moreover,

a2 ≥ −2

 1 − 1

M

 + 2

M

 2θ π − 1

 , and

a2 ≥ −2 +1 ρ

θ − 1πθ−1

. Then we have

1 ρ

θ − 1θπ−1

− 2 ≤ a2 ≤ 2 −1 ρ

π θ − 1πθ

.

We shall prove, that for the fixed θ ∈ [0, π] and ρ > ρ(θ) there are functions f ∈ SR, ρe ∈ f (∆) such that/ f002!(0) assumes all values from the range

 1 ρ

θ − 1θπ−1

− 2, 2 − 1 ρ

θ − 1θπ .

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We consider the univalent functions

w = fM,t(z), fM,t(z) = z + c2(t)z2+ . . . for which the following equation is satisfied

z

1 − 2tz + z2 = w 1 − 2tMw +Mw22

.

These functions map the unit disk ∆ on the disk |w| < M with one or two slits on the real axis. Their coefficients c2(t) = 2t(1 − M1), t ∈ [−1, 1], assume all values from the range [−2(1 −M1 ), 2(1 −M1 )]. Since the functions

f (z) = M · Fθ h(z) M



= z + a2(t)z2+ . . . , where h(z) = fM,t(z), are starlike, ρe ∈ f (∆), therefore a/ 2(t) assumes all values from the range h1

ρ π

θ − 1θπ−1

− 2, 2 − 1ρ πθ − 1θπi

. 

On figure 1 there is the set Aρ,a2 for fixed θ.

Figure 1. The set Aρ,a2 for θ = 23π.

Definition 1. We say, that the function f is in Saif f ∈ Sand 12f00(0) = a, a ≥ 0 i.e.

Sa = {f ∈ S : f (z) = z + az2+ . . . }.

Rogosinski in paper [4] determined the Koebe domain for the class Sa. Theorem 4. The Koebe domain for the class Sa, a ∈ [0, 2), is symmetric with respect to the real axis and the boundary of this domain in the upper half plane is given by the polar equation w = ρ(θ)e, where

ρ(θ) = 2 + a cos θ

4 − a2 , a ≥ 0, θ ∈ [0, π].

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We determine the Koebe domain for the class SaR consisting of the func- tions from the class Sa which have real coefficients. From Theorem 3 we conclude the following theorem for the class SaR.

Theorem 5. The Koebe domain for the class SaR is symmetric with respect to the real axis and the boundary of this domain in the upper half plane is given by the polar equation w = ρa(θ)e, where

(2) ρa(θ) =

1 2−a

π θ − 1πθ

, θ ∈h

0,(2+a)π4 i ,

1 2+a

π

θ − 1πθ−1

, θ ∈(2+a)π

4 , πi . Proof. Let a2 = a. From Theorem 3 we have

ρ ≥ 1 2 − a

π θ − 1πθ

, where θ ∈



0,(2 + a)π 4



and

ρ ≥ 1 2 + a

π θ − 1θ

π−1

, where θ ∈ (2 + a)π 4 , π



. 

On figures 2, 3, 4 there are the Koebe domains for the class SaR for some fixed a2 = a.

Figure 2. The Koebe domain for the class SaR, a = 0.

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Figure 3. The Koebe domain for the class SaR, a = 1.

Figure 4. The Koebe domain for the class SaR, a = −1.

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Definition 2. We say that the function f (z) is n-symmetric function in ∆, if for fixed z ∈ ∆ the following condition is satisfied

f

 e2πin z



= e2πin f (z).

We say that the set D is n-symmetric, if the set satisfies the condition e2πin D = D. The set λD is understood as {λz : z ∈ D}.

We denote by SRnthe class of starlike and n-symmetric functions with real coefficients. From Theorem 5 we have

Corollary 1. The Koebe domain for the class SRn with fixed an+1 = b, n ≥ 2 is n-symmetric, symmetric with respect to the real axis and the line ζ = eπint and the boundary of this domain in the setζ ∈ C : 0 ≤ arg ζ ≤ πn is given by the polar equation w = ρb,n(θ)e where

ρb,n(θ) = pn

ρa(nθ), a = bn, 0 ≤ θ ≤ π n.

Proof. For the function f ∈ SRn the following condition is satisfied (3) f ∈ SaR ⇐⇒ g ∈ SRn, g(n+1)(0)

(n + 1)! = a2 n,

where g(z) = pf(zn n). Let b = an2. We determine the set of the form T

SRng(∆). From Theorem 5 we know that the boundary of the Koebe domain in the class SR is of the form w = ρa(θ)e where ρa(θ) is given by (2). From (3) we have

n

w = pn

ρa(t)eitn, t ∈ [0, π], and consequently for a = bn, θ = nt ∈ [0,πn] we have

n

w = pn

ρbn(nθ)e. 

References

[1] Goodman, A. W., Univalent Functions, Mariner Pub. Co., Tampa, 1983.

[2] Koczan, L., On classes generated by bounded functions, Ann. Univ. Mariae Curie- Skłodowska, Sect. A 52, No. 2 (1998), 95–101.

[3] McGregor, M. T., On three classes of univalent functions with real coefficients, J.

London Math. Soc. 39 (1964), 43–50.

[4] Rogosinski, W. W., On subordinate functions, Proc. Camb. Phil. Soc. 35 (1939), 1–26.

Magdalena Sobczak-Kneć

Department of Applied Mathematics Lublin University of Technology ul. Nadbystrzycka 38D

20-618 Lublin, Poland

e-mail: m.sobczak-knec@pollub.pl Received May 24, 2007

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