Covering Problems for Functions \(n\) -Fold Symmetric and Convex in the Direction of the Real Axis II
Bull. Malays. Math. Sci. Soc.
Covering Problems for Functions n-Fold Symmetric and Convex in the Direction of the Real Axis II
Leopold Koczan 0
Paweł Zaprawa 0
Mathematics Subject Classification 0
0 Primary 30C45
Let F denote the class of all functions univalent in the unit disk ≡ {ζ ∈ C : |ζ | < 1} and convex in the direction of the real axis. The paper deals with the subclass F (n) of these functions f which satisfy the property f (εz) = ε f (z) for all z ∈ , where ε = e2πi/n . The functions of this subclass are called n-fold symmetric. For F (n), where n is odd positive integer, the following sets, f ∈F (n) f ( )-the Koebe set and f ∈F (n) f ( )-the covering set, are discussed. As corollaries, we derive the Koebe and the covering constants for F (n). Communicated by Saminathan Ponusammy.
Covering domain; Koebe domain; Convexity in one direction; n-fold symmetry
1 Introduction
Let F denote the class of all functions f which are univalent in ≡ {ζ ∈ C : |ζ | <
1}, convex in the direction of the real axis and normalized by f (0) = f (0) − 1 = 0.
Recall that an analytic function f is said to be convex in the direction of the real axis
if the intersection of f ( ) with each horizontal line is either a connected set or empty.
For a given subclass A of F , the following sets: f ∈A f ( ) and f ∈A f ( ) are
called the Koebe set for A and the covering set for A. We denote them by K A and L A,
respectively. The radius of the largest disk with center at the origin contained in K A
is called the Koebe constant for A. Analogously, the radius of the smallest disk with
center at the origin that contains L A is called the covering constant for A.
In the class F , we consider functions which satisfy the property of n-fold symmetry:
f (εz) = ε f (z) for all z ∈
,
where ε = e2πi/n . The subclass of F consisting of n-fold symmetric functions is
denoted by F (n). By the definition, for every f ∈ F (n) a set f ( ) is n-fold symmetric,
which means that f ( ) = ε f ( ). In other words, f ( ) may be obtained as the union
of rotations about a multiple of 2π/n from a set f ( ) ∩ {w : arg w ∈ [0, 2π/n]}.
From this reason, the following notation is useful:
2π
n
0 =
,
j = ε
j 0 , j = 1, 2, . . . , n − 1
∗ =
.
The main aim of the paper is to find the Koebe set and the covering set for the
class F (n) when n is an odd positive integer. Similar problems in related classes were
discussed, for instance, in [1,2,5] and in the papers of the authors [3,4].
At the beginning, let us consider the general properties of the Koebe sets and the
covering sets for F (n).
In [4], we proved that
Theorem 1 The sets K (n) and LF(n) , for n ∈ N, are symmetric with respect to both
F
axes of the coordinate system.
Theorem 2 The sets KF(n) and LF(n) , for n ∈ N, are n-fold symmetric.
To prove both the theorems, it is enough to consider functions
and
and
Obviously,
g(z) = f (z)
h(z) = − f (−z).
f ∈ F
(n) ⇔ g, h ∈ F (n).
(
1
)
(
2
)
(
3
)
Moreover, if D = f ( ) then g( ) = D, h( ) = −D.
Taking (
3
) into account, it is clear that the coordinate axes are the lines of symmetry
for both the sets KF(n) and LF(n) for all n ∈ N. Furthermore,
Lemma 1 Each straight line ε j/4 · {ζ = t , t ∈ R}, j = 0, 1, . . . , 4n − 1 is the line
of symmetry of KF(n) and LF(n) for every positive odd integer n.
Proof Let n be a positive odd integer and let D be one of the two sets: K (n) or L (n) .
F F
Since D is symmetric with respect to the real axis and the positive real half-axis
contains one side of the sector 0, each rotation of the real axis about a multiple of
2π/n is the line of symmetry of D. Because of the equality
{ζ = t , t ∈ R} · ε1/2 = {ζ = t , t ∈ R} · ε(n+1)/2 ,
our claim is true for all even j , j = 0, 1, . . . , 4n − 1.
Let n = 4k + 1, k ≥ 1. The bisector of (n−1)/4 divides this sector into two
subsectors: {w : arg w ∈ [π/2 − π/2n, π/2 + π/2n]} and {w : arg w ∈ [π/2 +
π/2n, π/2 + 3π/2n]}. Hence the imaginary axis is the bisector of the former. For this
reason, each rotation of the imaginary axis about a multiple of 2π/n is the line of
symmetry of D. Moreover,
{ζ = i t , t ∈ R} · ε1/2 = {ζ = i t , t ∈ R} · ε(n+1)/2.
Hence our claim is valid also for all odd j , j = 0, 1, . . . , 4n − 1.
If n = 4k + 3, k ≥ 0, then the bisector of (n−3)/4 divides this sector into two
subsectors: {w : arg w ∈ [π/2 − 3π/2n, π/2 − π/2n]} and {w : arg w ∈ [π/2 −
π/2n, π/2 + π/2n]}. The imaginary axis is the bisector of the latter. Similar argument
to the one for n = 4k + 1 completes the proof for this choice of n.
Theorem 3 The sets KF(n) and LF(n) , for positive odd integers n, are 2n-fold
symmetric.
pPoroinotfbLeleotnDginbge otonethoef bthoeuntwdaorsyeotsf: DKFsu(nc) horthLaFta(nr)g. ϕL0et∈w[00,=π|/w20n|]e.iϕ0 be an arbitrary
It is sufficient to apply Lemma 1. Firstly, the symmetric point to w0 with respect
to the straight line ε1/4 · {ζ = t , t ∈ R} is w1 = |w0|ei(π/n−ϕ0). Secondly, the
symmetric point to w0 with respect to the real axis is w2 = |w0|e−i (...truncated)