Covering Problems for Functions $n$-Fold Symmetric and Convex in the Direction of the Real Axis II

Abstract

Let ${\mathcal {F}}$ denote the class of all functions univalent in the unit disk $\Delta \equiv \{\zeta \in {\mathbb {C}}\,:\,\left| \zeta \right| <1\}$ and convex in the direction of the real axis. The paper deals with the subclass ${\mathcal {F}}^{(n)}$ of these functions $f$ which satisfy the property $f(\varepsilon z)=\varepsilon f(z)$ for all $z\in \Delta$, where $\varepsilon =e^{2\pi i/n}$. The functions of this subclass are called $n$-fold symmetric. For ${\mathcal {F}}^{(n)}$, where $n$ is odd positive integer, the following sets, $\bigcap _{f\in {\mathcal {F}}^{(n)}} f(\Delta )$—the Koebe set and $\bigcup _{f\in {\mathcal {F}}^{(n)}} f(\Delta )$—the covering set, are discussed. As corollaries, we derive the Koebe and the covering constants for ${\mathcal {F}}^{(n)}$.

1 Introduction

Let ${\mathcal {F}}$ denote the class of all functions $f$ which are univalent in $\Delta \equiv \{\zeta \in {\mathbb {C}}\,:\,\left| \zeta \right| <1\}$, convex in the direction of the real axis and normalized by $f(0)={f^{\prime }}(0)-1=0$. Recall that an analytic function $f$ is said to be convex in the direction of the real axis if the intersection of $f(\Delta )$ with each horizontal line is either a connected set or empty.

For a given subclass $A$ of ${\mathcal {F}}$, the following sets: $\bigcap _{f\in A} f(\Delta )$ and $\bigcup _{f\in A} f(\Delta )$ 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 ${\mathcal {F}}$, we consider functions which satisfy the property of $n$-fold symmetry:

$$\begin{aligned} f(\varepsilon z)=\varepsilon f(z)\quad \text {for all}\quad z\in \Delta \ , \end{aligned}$$

where $\varepsilon =e^{2\pi i/n}$. The subclass of ${\mathcal {F}}$ consisting of $n$-fold symmetric functions is denoted by ${\mathcal {F}}^{(n)}$. By the definition, for every $f\in {\mathcal {F}}^{(n)}$ a set $f(\Delta )$ is $n$-fold symmetric, which means that $f(\Delta )=\varepsilon f(\Delta )$. In other words, $f(\Delta )$ may be obtained as the union of rotations about a multiple of $2\pi /n$ from a set $f(\Delta )\cap \left\{ w: \arg w \in \left[ 0,2\pi /n\right] \right\}$. From this reason, the following notation is useful:

$$\begin{aligned} \Lambda _0=\left\{ w: \arg w \in \left[ 0,\frac{2\pi }{n}\right] \right\} ,\quad \Lambda _j=\varepsilon ^j \Lambda _0\ ,\ j=1,2,\ldots ,n-1 \end{aligned}$$

and

$$\begin{aligned} \Lambda ^*=\left\{ w: \arg w \in \left[ \frac{\pi }{2}-\frac{\pi }{n},\frac{\pi }{2}\right] \right\} . \end{aligned}$$

The main aim of the paper is to find the Koebe set and the covering set for the class ${\mathcal {F}}^{(n)}$ 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 ${\mathcal {F}}^{(n)}$.

In [4], we proved that

Theorem 1

The sets $K_{{\mathcal {F}}^{(n)}}$ and $L_{{\mathcal {F}}^{(n)}}$, for $n\in {\mathbb {N}}$, are symmetric with respect to both axes of the coordinate system.

Theorem 2

The sets $K_{{\mathcal {F}}^{(n)}}$ and $L_{{\mathcal {F}}^{(n)}}$, for $n\in {\mathbb {N}}$, are $n$-fold symmetric.

To prove both the theorems, it is enough to consider functions

$$\begin{aligned} g(z)=\overline{f(\overline{z})} \end{aligned}$$

(1)

and

$$\begin{aligned} h(z)=-f(-z). \end{aligned}$$

(2)

Obviously,

$$\begin{aligned} f\in {\mathcal {F}}^{(n)} \Leftrightarrow g,h\in {\mathcal {F}}^{(n)}. \end{aligned}$$

(3)

Moreover, if $D=f(\Delta )$ then $g(\Delta )=\overline{D}$, $h(\Delta )=-D$.

Taking (3) into account, it is clear that the coordinate axes are the lines of symmetry for both the sets $K_{{\mathcal {F}}^{(n)}}$ and $L_{{\mathcal {F}}^{(n)}}$ for all $n\in {\mathbb {N}}$. Furthermore,

Lemma 1

Each straight line $\varepsilon ^{j/4}\cdot \{\zeta = t\ ,\ t\in R\}$, $j=0,1,\dots ,4n-1$ is the line of symmetry of $K_{{\mathcal {F}}^{(n)}}$ and $L_{{\mathcal {F}}^{(n)}}$ for every positive odd integer $n$.

Proof

Let $n$ be a positive odd integer and let $D$ be one of the two sets: $K_{{\mathcal {F}}^{(n)}}$ or $L_{{\mathcal {F}}^{(n)}}$.

Since $D$ is symmetric with respect to the real axis and the positive real half-axis contains one side of the sector $\Lambda _0$, each rotation of the real axis about a multiple of $2\pi /n$ is the line of symmetry of $D$. Because of the equality

$$\begin{aligned} \{\zeta = t\ ,\ t\in R\}\cdot \varepsilon ^{1/2} = \{\zeta = t\ ,\ t\in R\}\cdot \varepsilon ^{(n+1)/2}\ , \end{aligned}$$

our claim is true for all even $j$, $j=0,1,\dots ,4n-1$.

Let $n=4k+1$, $k\ge 1$. The bisector of $\Lambda _{(n-1)/4}$ divides this sector into two subsectors: $\{w: \arg w \in [\pi /2-\pi /2n, \pi /2+\pi /2n]\}$ and $\{w: \arg w \in [\pi /2+\pi /2n, \pi /2+3\pi /2n]\}$. Hence the imaginary axis is the bisector of the former. For this reason, each rotation of the imaginary axis about a multiple of $2\pi /n$ is the line of symmetry of $D$. Moreover,

$$\begin{aligned} \{\zeta = it\ ,\ t\in R\}\cdot \varepsilon ^{1/2} = \{\zeta = it\ ,\ t\in R\}\cdot \varepsilon ^{(n+1)/2}. \end{aligned}$$

Hence our claim is valid also for all odd $j$, $j=0,1,\dots ,4n-1$.

If $n=4k+3$, $k\ge 0$, then the bisector of $\Lambda _{(n-3)/4}$ divides this sector into two subsectors: $\{w: \arg w \in [\pi /2-3\pi /2n, \pi /2-\pi /2n]\}$ and $\{w: \arg w \in [\pi /2-\pi /2n, \pi /2+\pi /2n]\}$. 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 $n$. $\square$

Theorem 3

The sets $K_{{\mathcal {F}}^{(n)}}$ and $L_{{\mathcal {F}}^{(n)}}$, for positive odd integers $n$, are $2n$-fold symmetric.

Proof

Let $D$ be one of the two sets: $K_{{\mathcal {F}}^{(n)}}$ or $L_{{\mathcal {F}}^{(n)}}$. Let $w_0=|w_0|e^{i\varphi _0}$ be an arbitrary point belonging to the boundary of $D$ such that $\arg \varphi _0\in [0,\pi /2n]$.

It is sufficient to apply Lemma 1. Firstly, the symmetric point to $w_0$ with respect to the straight line $\varepsilon ^{1/4}\cdot \{\zeta = t\ ,\ t\in R\}$ is $w_1=|w_0|e^{i(\pi /n-\varphi _0)}$. Secondly, the symmetric point to $w_0$ with respect to the real axis is $w_2=|w_0|e^{-i\varphi _0}$. Both points $w_1$, $w_2$ also belong to the boundary of $D$. Consequently,

$$\begin{aligned} w_1=\varepsilon ^{1/2}\cdot w_2\ , \end{aligned}$$

which results in

$$\begin{aligned} D=\varepsilon ^{1/2}\cdot D. \end{aligned}$$

$\square$

Remark 1

On the basis of this lemma, we can observe that in order to find sets $K$ and $L$, it is enough to determine their boundaries only in the sector of the measure $\pi /2n$.

2 Extremal Polygons and Functions

Let $n$ be a fixed positive odd integer, $n\ge 5$, and let $K$ denote the Koebe set for ${\mathcal {F}}^{(n)}$.

We denote by ${\mathcal {W}}$, the family of $n$-fold symmetric polygons $W$ convex in the direction of the real axis and which have $2n$ sides and interior angles $\pi +\pi /n$ and $\pi -3\pi /n$ alternately.

Suppose that $w_*\in \partial K \cap \Lambda ^*$, where $\partial K$ stands for the boundary of $K$. According to Theorem 1, $-\overline{w_*}\in \partial K$. Consider the straight horizontal line containing a segment $I=\{-\lambda \overline{w_*}+(1-\lambda )w_*:\lambda \in (0,1)\}$. There are two possibilities: the intersection of this line with $K$ is either $I$ or the empty set. Assume now that $I\subset K$. We shall see that the second case holds only if ${{\mathrm{Re}}}w_*=0$.

Since $K$ is $n$-fold symmetric, all points $w_*\cdot \varepsilon ^j$, $j=0,1,\ldots , n-1$ belong to $\partial K$. On the one hand, $w_*\in \Lambda ^*$ means that $w_*$ has the greatest imaginary part among points $w_*\cdot \varepsilon ^j$. On the other hand, $w_*\in \partial K$ means that there exists $f_*\in {\mathcal {F}}^{(n)}$ such that $w_*\in \partial f_*(\Delta )$.

From the convexity of $f_*$ in the direction of the real axis, at least one of the two horizontal rays emanating from $w_*$ is disjoint from $f_*(\Delta )$. Since $\{w_*-t: t\ge 0\}\cap K=I$, it is a ray $l=\{w_*+t: t\ge 0\}$ is disjoint from $f_*(\Delta )$. Taking into account the $n$-fold symmetry of $f_*$, all rays $l\cdot \varepsilon ^j$, $j=0,1,\ldots , n-1$ are disjoint from $f_*(\Delta )$.

Observe that the point $w_*\cdot \varepsilon ^{(n+1)/2}$ has the lowest imaginary part among points $w_*\cdot \varepsilon ^j$, $j=0,1,\ldots , n-1$. Only if $\arg w_*=\pi /2$, this point is one of two points with the same imaginary part. The convexity in the direction of the real axis of $f_*$ implies that one of two horizontal rays emanating from $w_*\cdot \varepsilon ^{(n+1)/2}$ is also disjoint from $f_*(\Delta )$. If this ray is of the form $k_1=\{w_*\cdot \varepsilon ^{(n+1)/2}+t: t\ge 0\}$, then $(k_1\cdot \varepsilon ^j) \cap f_*(\Delta ) =\emptyset$ and, conseqeuntly, $f_*(\Delta )$ is included in a polygon of the family ${\mathcal {W}}$. Indeed, the rays $l$ and $k_1\cdot \varepsilon ^{(n-1)/2}$ form a sector with the vertex in $w_*$ and the opening angle $\pi +\pi /n$. From the $n$-fold symmetry of $f_*$, we obtain the polygon mentioned above. The conjugate angle to this opening angle is the vertex angle of the polygon at $w_*$. It is easy to check that the angle of the polygon between $l\cdot \varepsilon$ and $k_1\cdot \varepsilon ^{(n+1)/2}$ has the measure $\pi -3\pi /n$.

But there is another possibility, i.e., $k_2\cap f_*(\Delta ) =\emptyset$, where $k_2=\{w_*\cdot \varepsilon ^{(n+1)/2}-t: t\ge 0\}$.

If $\arg w_*=\pi /2$, the set ${\mathbb {C}}\setminus \{k_2\cdot \varepsilon ^j, j=0,1,\ldots , n-1\}$ consists of two parts: an unbounded part and a bounded one which is a regular $n$-gon, see Fig. 1. A regular polygon is convex and it can be treated as the generalization of a set of the family ${\mathcal {W}}$. Every second side of this generalized polygon has the length 0.

Fig. 1

Extremal $n$-gon for $n=5$

If $\arg w_*\in [\pi /2-\pi /n,\pi /2)$, then the set ${\mathbb {C}}\setminus \{l\cdot \varepsilon ^j, k_2\cdot \varepsilon ^j, j=0,1,\ldots , n-1\}$ is not bounded and it is not convex in the direction of the real axis. Since $w_*\cdot \varepsilon \in \partial f_*(\Delta )$, one of two horizontal rays emanating from this point is also disjoint from $f_*(\Delta )$. If $m_1=\{w_*\cdot \varepsilon +t: t\ge 0\}$ has no common points with $f_*(\Delta )$, then $w_*\notin \partial f_*(\Delta )$, because ${{\mathrm{Im}}}(w_*\varepsilon )<{{\mathrm{Im}}}w_*$ and ${{\mathrm{Re}}}(w_*\varepsilon )<{{\mathrm{Re}}}w_*$, a contradiction. For this reason, $m_2\cap f_*(\Delta ) =\emptyset$, where $m_2=\{w_*\cdot \varepsilon -t: t\ge 0\}$. Hence $(m_2\cdot \varepsilon ^j) \cap f_*(\Delta ) =\emptyset$.

In this way, we obtain $3n$ rays emanating from $n$ points: $w_*\cdot \varepsilon ^j$, $j=0,1,\ldots , n-1$. Let us take three rays starting from $w_*$. These rays are: $l=\{w_*+t: t\ge 0\}$, $k_2\varepsilon ^{-(n+1)/2}=\{w_*-t\cdot \varepsilon ^{-(n+1)/2}: t\ge 0\}$ and $m_2\varepsilon ^{-1}=\{w_*-t\cdot \varepsilon ^{-1}: t\ge 0\}$. The angles between them and the positive real half-axis are equal to: 0, $-\pi /n$, $\pi -2\pi /n$. It means that $l$ lies in the sector with the vertex in $w_*$ and with the sides $k_2\varepsilon ^{-(n+1)/2}$ and $m_2\varepsilon ^{-1}$. The opening angle of this sector is equal to $\pi -3\pi /n$.

Consequently, $f_*(\Delta )$ is included in a polygon generated by $k_2\varepsilon ^{j}$ and $m_2\varepsilon ^{j}$, $j=0,1,\ldots , n-1$. This polygon belongs to the family ${\mathcal {W}}$. Two examples of members of ${\mathcal {W}}$ are shown in Fig. 2.

Fig. 2

Extremal polygons for $n=5$

From now on, we assume that all members of ${\mathcal {W}}$ are open sets.

Let

$$\begin{aligned} f_\alpha (z)&= \int _0^z (1-\zeta ^n e^{-in\alpha })^{\frac{1}{n}} (1+\zeta ^n e^{-in\alpha /3})^{-\frac{3}{n}}d\zeta \quad ,\quad \alpha \in \left[ -\frac{3\pi }{n},\frac{3\pi }{n}\right] \ ,\end{aligned}$$

(4)

$$\begin{aligned} g_\alpha (z)&= \int _0^z (1-\zeta ^n e^{-in\alpha })^{\frac{1}{n}} (1-\zeta ^n e^{-i(n\alpha /3+5\pi /3)})^{-\frac{3}{n}}d\zeta \quad , \quad \alpha \in \left[ -\frac{2\pi }{n},\frac{4\pi }{n}\right] .\nonumber \\ \end{aligned}$$

(5)

We choose the principal branch of $n$-th root. Since the exponential function is periodic, in the above definitions we restrict the range of variability of $\alpha$ to the intervals of length $6\pi /n$. The choice of these intervals depends on the properties of $f_\alpha$ and $g_\alpha$. Some additional information will be given in Remark 2.

The definition of the family ${\mathcal {W}}$ may be extended for $n=3$. In this case, the sets belonging to ${\mathcal {W}}$ may be treated as generalized polygons. The measure of the angles is equal to $4\pi /3$ and 0 alternately. These sets have the shape of an unbounded three-pointed star, see Fig. 3. Moreover, for $n=3$, the functions $g_\alpha$ map $\Delta$ onto these generalized polygons.

Fig. 3

Extremal polygons for $n=3$

Lemma 2

All functions $f_\alpha$, $\alpha \in \left[ -\frac{3\pi }{n},\frac{3\pi }{n}\right]$, belong to ${\mathcal {F}}^{(n)}$ for $n=4k+1$, $k\ge 1$.

Lemma 3

All functions $g_\alpha$, $\alpha \in \left[ -\frac{2\pi }{n},\frac{4\pi }{n}\right]$, belong to ${\mathcal {F}}^{(n)}$ for $n=4k+3$, $k\ge 0$.

Proof of Lemma 2

At the beginning, we shall show that the functions $f_\alpha$, $\alpha \in \left[ -3\pi /n,3\pi /n\right]$ are univalent. Observe that

$$\begin{aligned} {f_\alpha ^{\prime }} (z)=p(z)\cdot \frac{h(z)}{z}\ , \end{aligned}$$

(6)

where

$$\begin{aligned} p(z)=\left( \frac{1-z^n e^{-ia}}{1+z^n e^{-ib}}\right) ^{1/n} \end{aligned}$$

and

$$\begin{aligned} h(z)=\frac{z}{(1+z^n e^{-ib})^{2/n}} \end{aligned}$$

with $a=n\alpha$, $b=n\alpha /3$.

A Möbius function $p_1(z)=(1-z e^{-ia})/(1+z e^{-ib})\ ,a,b\in {\mathbb {R}}$ satisfies the condition ${{\mathrm{Re}}}e^{i\beta } p_1(z)>0$ with some $\beta \in {\mathbb {R}}$; hence, for all $n\in {\mathbb {N}}$, the inequality ${{\mathrm{Re}}}e^{i\beta } p(z)={{\mathrm{Re}}}e^{i\beta } p_1(z^n)^{1/n}>0$ holds with the same $\beta$. A function $h_1(z)=z/(1+z e^{-ib})^2$, $b\in {\mathbb {R}}$ is starlike, so is $h(z)=\root n \of {h_1(z^n)}$. Combining these two facts with (6), we conclude that

$$\begin{aligned} {{\mathrm{Re}}}e^{i\beta } p(z)={{\mathrm{Re}}}e^{i\beta } \frac{{zf_\alpha ^{\prime }}(z)}{h(z)}>0\ , \end{aligned}$$

which means that $f_\alpha$ is close-to-convex and consequently univalent.

Next, we claim that each polygon $f_\alpha (\Delta )$ is a set which is convex in the direction of the real axis. It is sufficient to discuss the argument of the tangent line to $\partial f_\alpha (\Delta )$. Observe that

$$\begin{aligned} \arg \left( \frac{\partial }{\partial \varphi }\left( f_\alpha (e^{i\varphi })\right) \right)&= \arg \left( {f_\alpha ^{\prime }}(e^{i\varphi })ie^{i\varphi }\right) \nonumber \\ \!&= \!\frac{1}{n}\arg \left( 1+e^{i(\pi +n\varphi -n\alpha )}\right) -\frac{3}{n}\arg \left( 1+e^{i(n\varphi -n\alpha /3)}\right) +\frac{\pi }{2}+\varphi .\nonumber \\ \end{aligned}$$

(7)

Let $\varphi \in [\alpha /3-\pi /n,\alpha )$. Then, $\pi +n\varphi -n\alpha$ as well as $n\varphi -n\alpha /3$ are in $[-\pi ,\pi )$ and from (7) we get

$$\begin{aligned} \arg \left( \frac{\partial }{\partial \varphi }\left( f_\alpha (e^{i\varphi })\right) \right) \!&= \! \frac{1}{2n}\left( \pi +n\varphi -n\alpha \right) \!-\!\frac{3}{2n}\left( n\varphi -n\frac{\alpha }{3}\right) +\frac{\pi }{2}+\varphi = \frac{\pi }{2}+\frac{\pi }{2n}.\nonumber \\ \end{aligned}$$

(8)

The above means that the tangent for $\varphi$ in $(\alpha /3-\pi /n,\alpha )$ has the constant argument $\pi /2+\pi /2n$. Since $f_\alpha (\Delta )$ is a polygon with angles measuring $\pi +\pi /n$ and $\pi -3\pi /n$ alternately, the argument of the tangent line takes values $\pi /2+\pi /2n+2j\pi /n$ and $\pi /2-\pi /2n+2j\pi /n$, $j=1,2,\ldots ,n$ alternately. What is more, putting $j=k$ in $\pi /2+\pi /2n+2j\pi /n$, we obtain the argument equal to $\pi$ and putting $j=3k+1$ in $\pi /2-\pi /2n+2j\pi /n$, we obtain the argument equal to $2\pi$. Hence two of the sides of $f_\alpha (\Delta )$ are horizontal; consequently $f_\alpha \in {\mathcal {F}}^{(n)}$. $\square$

The proof of Lemma 3 is similar.

Remark 2

A polygon $f_\alpha (\Delta )$ has vertices in points $f_\alpha (e^{i\alpha })\cdot \varepsilon ^j$ and $f_\alpha (e^{i(\alpha /3+\pi /n)})\cdot \varepsilon ^j$, $j=0,1,\ldots ,n-1$. These vertices correspond to angles measuring $\pi +\pi /n$ and $\pi -3\pi /n$, respectively. It is worth pointing out some particular cases of polygons belonging to ${\mathcal {W}}$. For $\alpha =-3\pi /2n$ and $\alpha =3\pi /2n$, they become regular $n$-gons and for $\alpha =-3\pi /n$, $\alpha =0$, and $\alpha =3\pi /n$ these sets are $n$-pointed stars symmetric with respect to the real axes. The functions $f_{-3\pi /n}$, $f_0$, and $f_{3\pi /n}$ have real coefficients. In all other cases, coefficients are nonreal. These particular functions are as follows:

$$\begin{aligned}&f_{-3\pi /2n}(z) = \int _0^z (1+i\zeta ^n)^{-\frac{2}{n}} d\zeta \quad ,\quad f_{3\pi /2n}(z) = \int _0^z (1-i\zeta ^n)^{-\frac{2}{n}} d\zeta \ ,\\&f_0(z) = \int _0^z (1-\zeta ^n)^{\frac{1}{n}}(1+\zeta ^n)^{-\frac{3}{n}} d\zeta ,\\&f_{-3\pi /n}(z) = f_{3\pi /n}(z) = \int _0^z (1+\zeta ^n)^{\frac{1}{n}}(1-\zeta ^n)^{-\frac{3}{n}} d\zeta . \end{aligned}$$

Likewise, for $\alpha =-\pi /2n$ and $\alpha =5\pi /2n$ the sets $g_\alpha (\Delta )$ are regular $n$-gons, and for $\alpha =-2\pi /n$, $\alpha =\pi /n$ and $\alpha =4\pi /n$ these sets are $n$-pointed stars. These functions $g_\alpha$ which map $\Delta$ on $n$-pointed stars have real coefficients.

Let $n=4k+1, k\ge 1$ be fixed. Let us denote by $W_\alpha$, a set $f_\alpha (\Delta )$ for a fixed $\alpha \in \left[ -3\pi /n,3\pi /n\right]$.

Observe that the following equalities hold:

$$\begin{aligned} f_{-\alpha }^{\prime }(z) = \overline{f_{\alpha }^{\prime }(\overline{z})}\quad \text {for}\; \alpha \in \left[ -\frac{3\pi }{n},\frac{3\pi }{n}\right] \end{aligned}$$

and

$$\begin{aligned} {f_{\frac{3\pi }{2n}-\gamma }^{\prime }}(-z) = \overline{{f_{\frac{3\pi }{2n}+\gamma }^{\prime }}(\overline{z})}\quad \text {for}\; \gamma \in \left[ 0,\frac{3\pi }{2n}\right] \ , \end{aligned}$$

which means that

$$\begin{aligned} f_{-\alpha }(z) = \overline{f_{\alpha }(\overline{z})}\quad \text {for}\quad \alpha \in \left[ -\frac{3\pi }{n},\frac{3\pi }{n}\right] \end{aligned}$$

and

$$\begin{aligned} -f_{\frac{3\pi }{2n}-\gamma }(-z) = \overline{f_{\frac{3\pi }{2n}+\gamma }(\overline{z})}\quad \text {for}\quad \gamma \in \left[ 0,\frac{3\pi }{2n}\right] . \end{aligned}$$

Consequently,

$$\begin{aligned} W_{-\alpha }=\overline{W_{\alpha }}\quad \text {and}\quad -W_{\frac{3\pi }{2n}-\gamma }=\overline{W_{\frac{3\pi }{2n}+\gamma }}\ . \end{aligned}$$

For this reason, a polygon $W$ as well as $\overline{W}$ and $-\overline{W}$ belong to the family ${\mathcal {W}}$. From the geometric construction of polygons in ${\mathcal {W}}$, it follows that the ratio of lengths of any two adjacent sides of a polygon varies from 0 to infinity as $\alpha$ is changing in $[-3\pi /n,3\pi /n]$; in each case a polygon is convex in the direction of the real axis. Multiplying sets $W_{\alpha }$, $\alpha \in \left[ -3\pi /n,3\pi /n\right]$ by $\lambda >0$ we obtain all members of the set ${\mathcal {W}}$.

We have proved one part of the following lemma (the second one can be proved analogously)

Lemma 4

• 1. ${\mathcal {W}}=\left\{ \lambda \cdot f_\alpha (\Delta ), \lambda >0, \alpha \in \left[ -3\pi /n,3\pi /n\right] \right\} \quad \text {for}\quad n=4k+1, k\ge 1$ ,

• 2. ${\mathcal {W}}=\left\{ \lambda \cdot g_\alpha (\Delta ), \lambda >0, \alpha \in \left[ -2\pi /n,4\pi /n\right] \right\} \quad \text {for}\quad n=4k+3, k\ge 0$.

3 Koebe Sets for ${\mathcal {F}}^{(n)}$

Let us define

$$\begin{aligned} F(\alpha )\equiv f_\alpha (e^{i\alpha }),\quad \alpha \in \left[ -\frac{3\pi }{n},\frac{3\pi }{n}\right] \end{aligned}$$

and

$$\begin{aligned} G(\alpha )\equiv g_\alpha (e^{i\alpha }),\quad \alpha \in \left[ -\frac{2\pi }{n},\frac{4\pi }{n}\right] \ . \end{aligned}$$

From (4) and (5)

$$\begin{aligned} F(\alpha )=e^{i\alpha }\int _0^1 (1-t^n)^{\frac{1}{n}} (1+t^n e^{2in\alpha /3})^{-\frac{3}{n}}dt \end{aligned}$$

and

$$\begin{aligned} G(\alpha )=e^{i\alpha }\int _0^1 (1-t^n)^{\frac{1}{n}} (1-t^n e^{i(2n\alpha /3-5\pi /3)})^{-\frac{3}{n}}dt. \end{aligned}$$

It can be easily checked that

$$\begin{aligned} \arg F(\alpha )&= \alpha \quad \text {for}\; \alpha \in \left\{ -\frac{3\pi }{n},-\frac{3\pi }{2n},0,\frac{3\pi }{2n},\frac{3\pi }{n}\right\} \ , \end{aligned}$$

(9)

$$\begin{aligned} \arg G(\alpha )&= \alpha \quad \text {for}\; \alpha \in \left\{ -\frac{2\pi }{n},-\frac{\pi }{2n},\frac{\pi }{n},\frac{5\pi }{2n},\frac{4\pi }{n}\right\} \ . \end{aligned}$$

(10)

Theorem 4

The Koebe set $K_{{\mathcal {F}}^{(n)}}$, for a fixed $n=4k+1$, $k\in {\mathbb {N}}$, is a bounded and $2n$-fold symmetric domain such that

$$\begin{aligned} \partial K_{{\mathcal {F}}^{(n)}}\cap \left\{ w:\arg w\in \left[ -\frac{\pi }{2n},\frac{\pi }{2n}\right] \right\} = F\left( \left[ -\alpha _F,\alpha _F\right] \right) \ , \end{aligned}$$

(11)

where $\alpha _F$ is the only solution of the equation

$$\begin{aligned} \arg F\left( \alpha \right) =\frac{\pi }{2n} \end{aligned}$$

(12)

in $\left[ 0,3\pi /2n\right]$.

Proof

Let $K$ denote the Koebe set for ${\mathcal {F}}^{(n)}$.

Let us consider a polygon $V_{\alpha }=f_\alpha (\Delta )$ belonging to ${\mathcal {W}}$, such that one of its vertices, let say $v_*$, lies in $\Lambda ^*$ (its argument is in $[\pi /2-\pi /n,\pi /2]$) and the interior angle at $v_*$ has the measure $\pi (1+1/n)$. Suppose additionaly that $w_*$ is a point of the boundary of $K$, such that $\arg w_*= \arg v_*$ and $|w_*| < |v_*|$. We denote the quotient $w_*/v_*= |w_*|/|v_*|$ by $\lambda$. Hence $\lambda <1$.

Since $w_*\in \partial K$, there exists $f_*\in {\mathcal {F}}^{(n)}$ such that

$$\begin{aligned} f_*(\Delta )\subset \lambda V_\alpha \subsetneq V_\alpha =f_\alpha (\Delta ). \end{aligned}$$

(13)

Therefore, $f_*\prec f_\alpha$ and $1={f_*^{\prime }}(0)\le {f_\alpha ^{\prime }}(0)=1$. Consequently $f_*= f_\alpha$, which contradicts (13). It means that $v_*=w_*$, or in other words, $w_*$ coincides with some vertex of $f_\alpha (\Delta )$. Hence $w_*$ is equal to $f_\alpha (e^{i\alpha })$ rotated about a multiple of $2\pi /n$, namely about $2\pi /n\cdot (n-1)/4$. However, it is true only for those $\alpha$, for which $w_*=F(\alpha )\cdot \varepsilon ^{(n-1)/4}$ is in $\Lambda ^*$.

Observe that for $\alpha \in \left[ -3\pi /n,3\pi /n\right]$, we have

$$\begin{aligned} F\left( -\alpha \right) = \overline{F\left( \alpha \right) }\ , \end{aligned}$$

that is,

$$\begin{aligned} \arg F\left( -\alpha \right) = -\arg F\left( \alpha \right) . \end{aligned}$$

From this and (9), (12) it follows that

$$\begin{aligned} \left\{ F(\alpha )\cdot \varepsilon ^{(n-1)/4}: \alpha \in \left[ -\alpha _F,\alpha _F\right] \right\} \end{aligned}$$

is the boundary of the Koebe set for ${\mathcal {F}}^{(n)}$ in $\Lambda ^*$. Combining this with Theorem 3, the equality (11) follows.

Finally, we claim that $\alpha _F$ is the only solution of (12) in $\left[ 0,3\pi /2n\right]$. On the contrary, assume that there exist two different numbers $\alpha _1, \alpha _2\in [0,3\pi /2n]$ such that

$$\begin{aligned} \arg F(\alpha _1) = \arg F(\alpha _2)\ , \end{aligned}$$

or equivalently,

$$\begin{aligned} \arg f_{\alpha _1}(e^{i\alpha _1}) = \arg f_{\alpha _2}(e^{i\alpha _2}). \end{aligned}$$

The sets $W_{\alpha _1}=f_{\alpha _1}(\Delta )$, $W_{\alpha _2}=f_{\alpha _2}(\Delta )$ are polygons of the family ${\mathcal {W}}$. This and the definition of ${\mathcal {W}}$ (or Lemma 4) result in

$$\begin{aligned} W_{\alpha _1}\subset W_{\alpha _2}\quad \text {or}\quad W_{\alpha _2}\subset W_{\alpha _1}. \end{aligned}$$

The normalization of $f_{\alpha _1}$ and $f_{\alpha _2}$ leads to $W_{\alpha _1}= W_{\alpha _2}$. Hence $\alpha _1=\alpha _2$, a contradiction. This means that (12) has only one solution in the set $[0,3\pi /2n]$. $\square$

The above proof gives more. Namely, $F$ is starlike for $\alpha \in [-3\pi /2n,3\pi /2n]$. Moreover,

$$\begin{aligned} \arg F\left( \alpha +\frac{3\pi }{n}\right) = \arg F(\alpha )+\frac{3\pi }{n}. \end{aligned}$$

It implies that $F$ is starlike for $\alpha \in [-\pi ,\pi ]$.

Furthermore, it is not difficult to see that there do not exist two different points in the set $\partial K_{{\mathcal {F}}^{(n)}}\cap \Lambda ^*$ with the same imaginary part. For contrary suppose that it is not the case, i.e., there exist $v_1$ and $v_2$ such that $v_1\ne v_2$, $v_1, v_2\in \partial K_{{\mathcal {F}}^{(n)}}\cap \Lambda ^*$ and ${{\mathrm{Im}}}v_1={{\mathrm{Im}}}v_2$.

With use of an argument similar to those in the proof of previous theorem, we can see that there exist two polygons $V_1, V_2\in {\mathcal {W}}$ with vertices $v_1$, $v_2$, respectively. The angles at these vertices have the same measure. Hence the sides of these polygons are pairwise parallel and if ${{\mathrm{Re}}}v_1<{{\mathrm{Re}}}v_2$, then $V_1\subset V_2$. This means that there exist $f_1, f_2 \in {\mathcal {F}}^{(n)}$ that $f_1(\Delta )=V_1$, $f_2(\Delta )=V_2$, and $f_1\prec f_2$. But the normalization of $f_1$ and $f_2$ is the same, hence $f_1=f_2$; a contradiction.

Theorem 5

The Koebe set $K_{{\mathcal {F}}^{(n)}}$, for a fixed $n=4k+3$, $k\ge 0$ is a bounded and $2n$-fold symmetric domain such that

$$\begin{aligned} \partial K_{{\mathcal {F}}^{(n)}}\cap \left\{ w:\arg w\in \left[ \frac{\pi }{2n},\frac{3\pi }{2n}\right] \right\} = G\left( \left[ \alpha _G,\frac{2\pi }{n}-\alpha _G\right] \right) \ , \end{aligned}$$

(14)

where $\alpha _G$ is the only solution of the equation

$$\begin{aligned} \arg G\left( \alpha \right) =\frac{\pi }{2n} \end{aligned}$$

(15)

in $\left[ -\pi /2n,\pi /n\right]$.

Proof

A consideration similar to the above shows that $\alpha _G$ is the only solution of the Eq. (15) in $\left[ -\pi /2n,\pi /n\right]$.

Suppose that $w_*\in \partial K\cap \Lambda ^*$. The analogous argument to this in the proof of Theorem 4 yields that $K$ is contained in some polygon $W$ of the family ${\mathcal {W}}$.

Let $g_*$ be a function from ${\mathcal {F}}^{(n)}$ for which $w_*\in \partial g_*(\Delta )$. We have $g_*(\Delta )\subset W=g_\alpha (\Delta )$ for some $\alpha \in [-\pi /2n,5\pi /n]$. For this reason $g_*\prec g_\alpha$, but taking into account the normalization of both functions, we obtain $g_*= g_\alpha$. Hence $w_*=g_\alpha (e^{i\alpha })\cdot \varepsilon ^{(n-3)/4}$, but only if $w_*\in \Lambda ^*$.

For $\alpha \in [-\pi /2n,\pi /n]$,

$$\begin{aligned} G\left( \frac{2\pi }{n}-\alpha \right) = \varepsilon \overline{G\left( \alpha \right) }\ , \end{aligned}$$

and so

$$\begin{aligned} \arg G\left( \frac{2\pi }{n}-\alpha \right) = \frac{2\pi }{n}-\arg G\left( \alpha \right) . \end{aligned}$$

From this, (10) and (15), we conclude that

$$\begin{aligned} \left\{ G(\alpha )\cdot \varepsilon ^{(n-3)/4}: \alpha \in \left[ \alpha _G,\frac{2\pi }{n}-\alpha _G\right] \right\} \end{aligned}$$

is the boundary of the Koebe set for ${\mathcal {F}}^{(n)}$ in $\Lambda ^*$. Theorem 3 concludes the proof of our theorem. $\square$

Now, we can derive the Koebe constant for ${\mathcal {F}}^{(n)}$.

Theorem 6

For a fixed positive odd integer $n$, $n\ge 3$ and for every function $f\in {\mathcal {F}}^{(n)}$ the disk $\Delta _{r_n}$, where $r_n=B(1/n,1/2n+1/2)/n\root n \of {4}$, is included in $f(\Delta )$. The number $r_n$ cannot be increased.

The symbol $B$ stands for the Beta and $\Delta _r$, $r>0$ means $\Delta _r =\{\zeta \in {\mathbb {C}}\,:\,\left| \zeta \right| <r\}$.

Proof

According to Theorems 4 and 5, the Koebe constant is equal to

$$\begin{aligned} \min \left\{ |F(\alpha )|: \alpha \in \left[ -\alpha _F,\alpha _F\right] \right\} \quad \text {for} \quad n=4k+1\ , \end{aligned}$$

or

$$\begin{aligned} \min \left\{ |G(\alpha )|: \alpha \in \left[ \alpha _G,\frac{2\pi }{n}-\alpha _G\right] \right\} \quad \text {for} \quad n=4k+3. \end{aligned}$$

But

$$\begin{aligned} |F(\alpha )|^2\ge \left( \int _0^1 (1-t^n)^{\frac{1}{n}} {{\mathrm{Re}}}(1+t^n e^{2in\alpha /3})^{-\frac{3}{n}}dt\right) ^2 \end{aligned}$$

and the integrand in this expression is nonnegative; thus

$$\begin{aligned} |F(\alpha )|\ge \int _0^1 (1-t^n)^{\frac{1}{n}} {{\mathrm{Re}}}q_F(\alpha ,t) dt,\quad q_F(\alpha ,t)=(1+t^n e^{2in\alpha /3})^{-\frac{3}{n}}. \end{aligned}$$

Likewise,

$$\begin{aligned} |G(\alpha )|\ge \int _0^1 (1-t^n)^{\frac{1}{n}} {{\mathrm{Re}}}q_G(\alpha ,t) dt,\quad q_G(\alpha ,t)=(1-t^n e^{i(2n\alpha /3-5\pi /3)})^{-\frac{3}{n}}. \end{aligned}$$

It is easy to check that for $n\ge 3$ the functions $p(z)=(1\pm t^n z)^{-3/n}$ are convex in $\Delta$ and they have real coefficients. This means that

$$\begin{aligned} {{\mathrm{Re}}}(1\pm t^n z)^{-\frac{3}{n}}\ge (1+t^n)^{-\frac{3}{n}}. \end{aligned}$$

(16)

Applying (16) for both $q_F$ and $q_G$, we get

$$\begin{aligned} |F(\alpha )|\ge q_0\quad \text {and}\quad |G(\alpha )|\ge q_0\ , \end{aligned}$$

where

$$\begin{aligned} q_0 = \int _0^1 (1-t^n)^{\frac{1}{n}} (1+t^n)^{-\frac{3}{n}} dt. \end{aligned}$$

(17)

This results in

$$\begin{aligned} \min \left\{ |F(\alpha )|: \alpha \in \left[ -\alpha _F,\alpha _F\right] \right\} = \left| F\left( 0\right) \right| \end{aligned}$$

and

$$\begin{aligned} \min \left\{ |G(\alpha )|: \alpha \in \left[ \alpha _G,\frac{2\pi }{n}-\alpha _G\right] \right\} = \left| G\left( \frac{\pi }{n}\right) \right| . \end{aligned}$$

Moreover, substituting $t^n=\tan ^2(x/2)$ in (17), we get

$$\begin{aligned} q_0=\frac{2}{n\root n \of {4}}\int _0^{\pi /2}(\sin x)^{\frac{2}{n}-1}(\cos x)^\frac{1}{n}\ dx =\frac{1}{n\root n \of {4}}B\left( \frac{1}{n},\frac{1}{2n}+\frac{1}{2}\right) \ . \end{aligned}$$

$\square$

Corollary 1

For a fixed positive odd integer $n$, $n\ge 3$, the Koebe constant for ${\mathcal {F}}^{(n)}$ is equal to $r_n=B(1/n,1/2n+1/2)/n\root n \of {4}$.

The Koebe sets and the Koebe disks for $n=3$ and $n=5$ are shown in Fig. 4.

Fig. 4

Koebe domain (solid line) and Koebe disk (dashed line) for ${\mathcal {F}}^{(3)}$ and ${\mathcal {F}}^{(5)}$

Remark 3

The results established in Theorem 4 and in Corollary 1 are actually valid also for $n=1$. They were obtained by Złotkiewicz and Reade in [6].

One can check that for $n=1$, the function $F$ takes the form

$$\begin{aligned} F(\alpha )=\,e^{i\alpha }\int _0^1 \frac{1-t}{(1+t e^{2i\alpha /3})^3}dt= \frac{e^{2i\alpha /3}}{4\cos (\alpha /3)}. \end{aligned}$$

The Eq. (12) gives $\alpha _F=3\pi /4$; thus the boundary of the Koebe set in the upper half-plane can be written as follows

$$\begin{aligned} u=\frac{\cos (2\alpha /3)}{4\cos (\alpha /3)}\ ,\quad v=\frac{1}{2}\sin (\alpha /3)\ ,\quad \alpha \in \left[ -\frac{3\pi }{4},\frac{3\pi }{4}\right] . \end{aligned}$$

This fact, we can rewrite in a different way

$$\begin{aligned} K_{{\mathcal {F}}}=\left\{ w\in {\mathbb {C}}:8|w|\left( |w|+|{{\mathrm{Re}}}w|\right) <1\right\} . \end{aligned}$$

The extremal functions $f_\alpha$ given by (4) are of the form

$$\begin{aligned} f_\alpha (z)=\frac{z+Bz^2}{(1+ze^{-i\alpha /3})^2}\ ,\ B=i\sin (\alpha /3)e^{-2i\alpha /3} \end{aligned}$$

and

$$\begin{aligned} g_\alpha (z)=-f_\alpha (-z)\ , \end{aligned}$$

where $\alpha \in \left[ -3\pi /4,3\pi /4\right]$. The image set $f_\alpha (\Delta )$ for a fixed $\alpha \in (-3\pi /4,3\pi /4)$ coincides with the plane with a horizontal ray excluded. For $\alpha =-3\pi /4,3\pi /4$, the sets $f_\alpha (\Delta )$ are half-planes.

Moreover, $r_1=B(1,1)/4=1/4$.

4 Covering Domains for ${\mathcal {F}}^{(n)}$

Theorem 7

The covering set $L_{{\mathcal {F}}^{(n)}}$ for odd $n\ge 5$ is a bounded and $2n$-fold symmetric domain such that

$$\begin{aligned} \partial L_{{\mathcal {F}}^{(n)}}\cap \left\{ w:\arg w\in \left[ 0,\frac{\pi }{2n}\right] \right\} = H\left( \left[ 0,\frac{\pi }{2n}\right] \right) . \end{aligned}$$

(18)

In the proof of this theorem, we need the following lemma.

Lemma 5

Let $n\ge 5$ be a fixed odd integer. If $f\in {\mathcal {F}}^{(n)}$ and $w\in f(\Delta )\cap \Lambda ^*$, then $f(\Delta )$ contains a polygon $W\in {\mathcal {W}}$ such that $W$ has one of its vertices at $w$ and the interior angle at $w$ has the measure $\pi -3\pi /n$.

Proof

Let $f\in {\mathcal {F}}^{(n)}$ and $w\in f(\Delta )\cap \Lambda ^*$, i.e., $\arg w\in [\pi /2-\pi /n,\pi /2]$. Because of the $n$-fold symmetry of $f$ every point $w\cdot \varepsilon ^j$, $j=0,1,\ldots , n-1$ belongs to $f(\Delta )$.

It can be easily checked that

$$\begin{aligned} \max \left\{ {{\mathrm{Im}}}\left( w\cdot \varepsilon ^j\right) , j=0,1,\ldots , n-1\right\} = {{\mathrm{Im}}}\left( w\right) \end{aligned}$$

and

$$\begin{aligned} \min \left\{ {{\mathrm{Im}}}\left( w\cdot \varepsilon ^j\right) , j=0,1,\ldots , n-1\right\} = {{\mathrm{Im}}}\left( w\cdot \varepsilon ^{2k+1}\right) . \end{aligned}$$

Let $w_1 = w\cdot \varepsilon$ and $w_2 = w\cdot \varepsilon ^{2k}$. The point $w_1$ has the second biggest imaginary part among points $w,w\cdot \varepsilon ,\ldots , w\cdot \varepsilon ^{n-1}$. Likewise, $w_2$ has the second lowest imaginary part among those points.

Let, moreover, $l_1$ and $l_2$ stand for two horizontal rays emanating from $w_1$ and $w_2$: $l_1=\{w_1+t: t\ge 0\}$, $l_2=\{w_2+t: t\ge 0\}$, respectively.

From the inequality ${{\mathrm{Im}}}w_2>{{\mathrm{Im}}}w\cdot \varepsilon ^{2k+1}$, we conclude that the point $w\cdot \varepsilon ^{2k+1}$ lies on the opposite side of the straight line which contains $l_2$ with respect to the origin. As a consequence, $w_1$ lies on the other side of the straight line including $l_2\cdot \varepsilon ^{-2k}$ with respect to the origin. Hence, two rays $l_1$ and $l_2\cdot \varepsilon ^{-2k}$ have a common point, let say $w_0$.

We shall show that $w_0$ also belongs to $f(\Delta )$. Suppose, contrary to our claim, that $w_0\notin f(\Delta )$. The points $w_0,w_1$ lie on the ray $l_1$ and $w_1\in f(\Delta )$. Therefore, taking into account the convexity in the direction of the real axis of $f$, a ray $m_1=\{w_0+t: t\ge 0\}$ is disjoint from $f(\Delta )$.

Since $w_0, w$ belong to $l_2\cdot \varepsilon ^{-2k}$, the points $w_0\cdot \varepsilon ^{2k}, w\cdot \varepsilon ^{2k}$ belong to $l_2$. Moreover, $w_0\cdot \varepsilon ^{2k}\notin f(\Delta )$ and $w_2\in f(\Delta )$. Consequently, $m_2=\{w_0\cdot \varepsilon ^{2k}+t: t\ge 0\}$ is disjoint from $f(\Delta )$, and, generally, $m_2\varepsilon ^j \cap f(\Delta )=\emptyset$, $j=0,1,\ldots , n-1$.

We have proved that the rays $m_1$ and $m_2\varepsilon ^{-2k}$ with the common vertex $w_0$ are disjoint from $f(\Delta )$. It means that the reflex sector with the vertex in $w_0$ and these two rays as the sides has no common points with $f(\Delta )$. But $w_1$ lies in this reflex sector; hence $w_1\notin f(\Delta )$, a contradiction.

From the argument given above all points $w\varepsilon ^j$, $w_0\varepsilon ^j$, $j=0,1,\ldots , n-1$ belong to $f(\Delta )$. Applying $n$-fold symmetry and the convexity of $f$ in the direction of the real axis, we can see that a polygon $W$ with succeeding vertices at points $w, w_0, w\varepsilon , w_0\varepsilon ,\ldots , w\varepsilon ^{n-1}, w_0\varepsilon ^{n-1}$ is contained in $f(\Delta )$. It is easy to check that this polygon has the interior angles $\pi -3\pi /n$ and $\pi +\pi /n$ alternately. For this reason, $W$ is in ${\mathcal {W}}$. $\square$

According to Lemmas 2 and 3, every function in ${\mathcal {F}}^{(n)}$ mapping $\Delta$ onto a polygon of the family ${\mathcal {W}}$ has the form (4) and (5) with appropriately taken $\alpha$. These functions may be written in the form

$$\begin{aligned} f_\beta (z)&= \int _0^z (1+\zeta ^n e^{-3in\beta })^{\frac{1}{n}} (1-\zeta ^n e^{-in\beta })^{-\frac{3}{n}}d\zeta ,\quad \beta \in \left[ 0,\frac{2\pi }{n}\right] \ ,\end{aligned}$$

(19)

$$\begin{aligned} g_\beta (z)&= \int _0^z (1+\zeta ^n e^{-3in\beta })^{\frac{1}{n}} (1-\zeta ^n e^{-in\beta })^{-\frac{3}{n}}d\zeta ,\quad \beta \in \left[ \frac{\pi }{n},\frac{3\pi }{n}\right] \ , \end{aligned}$$

(20)

equivalent to (4) and (5).

In fact, the functions defined by (4) and (19) are connected by the relation $\beta =\alpha /3+\pi /n$ and the functions in (5) and (20) are connected by $\beta =\alpha /3+5\pi /3n$.

Let us define

$$\begin{aligned} H(\beta )= f_\beta (e^{i\beta })\quad \text {for}\; \beta \in \left[ 0,\frac{2\pi }{n}\right] \ , \end{aligned}$$

and

$$\begin{aligned} H(\beta )= g_\beta (e^{i\beta })\quad \text {for}\; \beta \in \left[ \frac{\pi }{n},\frac{3\pi }{n}\right] . \end{aligned}$$

Hence

$$\begin{aligned} H(\beta )\equiv e^{i\beta }\int _0^1 (1+t^ne^{-2in\beta })^{\frac{1}{n}} (1-t^n)^{-\frac{3}{n}}dt\ ,\quad \beta \in {\mathbb {R}}. \end{aligned}$$

Observe that

$$\begin{aligned} \arg H(\beta ) = \beta \quad \text {for}\quad \beta =\frac{\pi }{2n}\cdot j\ ,\quad j=0,1,\ldots ,4n-1. \end{aligned}$$

(21)

Furthermore,

$$\begin{aligned} H\left( \beta +\frac{\pi }{n}\right) = e^{i\frac{\pi }{n}}H(\beta ). \end{aligned}$$

(22)

Now, we can prove Theorem 7.

Proof of Theorem 7

Let $L$ denote the covering set for ${\mathcal {F}}^{(n)}$. We additionaly assume that $n=4k+1, k\ge 1$. The proof for the case $n=4k+3, k\ge 0$ is almost similar.

Let us consider a polygon $W_{\beta }=f_\beta (\Delta )$ belonging to ${\mathcal {W}}$, such that one of its vertices, let say $w^*$, lies in $\Lambda ^*$ and the interior angle at $w^*$ has the measure $\pi (1-3/n)$. Suppose additionaly that $v^*$ is a point of the boundary of $L$ such that $\arg v^*= \arg w^*$ and $|v^*| > |w^*|$. We denote the quotient $v^*/w^*= |v^*|/|w^*|$ by $\mu$. Hence $\mu >1$.

Since $v^*\in \partial L$, there exists $f^*\in {\mathcal {F}}^{(n)}$ such that $v^*$ is a boundary point of $f^*(\Delta )$. From Lemma 5

$$\begin{aligned} f^*(\Delta )\supset \mu W_\beta \supsetneq W_\beta =f_\beta (\Delta ). \end{aligned}$$

(23)

Therefore, $f_\beta \prec f^*$ and $1={f_\beta ^{\prime }}(0)\le {f^{{*}^{\prime }}}(0)=1$. Consequently $f_\beta = f^*$, which contradicts (23). It means that $w^*=v^*$, or in other words, $v^*$ coincides with some vertex of $f_\beta (\Delta )$. Hence $v^*$ is equal to $f_\beta (e^{i\beta })$ rotated about a multiple of $2\pi /n$, namely about $2\pi /n\cdot (n-1)/4$. It is enough to take such $\beta$ that $v^*=H(\beta )\cdot \varepsilon ^{(n-1)/4}$ is in $\Lambda ^*$. From this, we conclude that $\beta \in [0,\pi /2n]$. $\square$

Theorem 8

For a fixed odd integer $n\ge 5$ and for every function $f\in {\mathcal {F}}^{(n)}$, the set $f(\Delta )$ is included in $\Delta _{R_n}$, where $R_n=B(1/n,1/2-3/2n)/n\root n \of {4}$. The number $R_n$ cannot be decreased.

Proof

We have

$$\begin{aligned} \left| H(\beta ) \right| \le \int _0^1 \left| (1+t^ne^{-2in\beta })^{\frac{1}{n}} (1-t^n)^{-\frac{3}{n}} \right| dt \le \int _0^1 \frac{(1+t^n)^{\frac{1}{n}}}{(1-t^n)^{\frac{3}{n}}} dt = \left| H(0) \right| . \end{aligned}$$

It can be shown that $H(0)=B(1/n,1/2-3/2n)/n\root n \of {4}$. $\square$

Corollary 2

For a fixed odd integer $n\ge 5$, the covering constant for ${\mathcal {F}}^{(n)}$ is equal to $R_n=B(1/n,1/2-3/2n)/n\root n \of {4}$.

The results presented above are valid for positive odd integers greater than or equal to 5. In the last part of this section, we turn to the case $n=3$.

As it was said in Section 2 (see also Fig. 3) for $n=3$ and $\beta \in [\pi /3,\pi ]\setminus \{\pi /2,5\pi /6\}$, the functions given by (20) map $\Delta$ onto the polygons with the interior angles $4\pi /3$ and 0 alternately, and the vertices in points $a\cdot \varepsilon ^j$, $\infty \cdot a\cdot \varepsilon ^j$, $j=0,1,2$ alternately, where $a=g_\beta (e^{i\beta })=H(\beta )$. Both sides adjacent to every vertex in infinity are parallel. Hence $g_\beta (\Delta )$ are star-shaped sets with three unbounded strips. The strips have the direction $\pi /3$, $\pi$, $5\pi /3$ if $\beta \in [\pi /3,\pi /2)\cup (5\pi /6,\pi ]$ and 0, $2\pi /3$, $4\pi /3$ if $\beta \in (\pi /2,5\pi /6)$. The thickness of the strips is changing as $\beta$ varies in $\beta \in [\pi /3,\pi ]\setminus \{\pi /2,5\pi /6\}$, but when $\beta$ tends to $\pi /2$ or $5\pi /6$, the thickness of the strips tends to 0.

For $\beta =\pi /2$ and $\beta =5\pi /6$, the functions

$$\begin{aligned} g_\frac{\pi }{2}(z) = \int _0^z \frac{1}{(1-i\zeta ^3)^{2/3}}d\zeta \end{aligned}$$

and

$$\begin{aligned} g_\frac{5\pi }{6}(z) = \int _0^z \frac{1}{(1+i\zeta ^3)^{2/3}}d\zeta \end{aligned}$$

map $\Delta$ onto the equilateral triangles symmetric with respect to the imaginary axis. The first triangle has one of its vertices in the point $ic$, the second one - in the point $-ic$, where

$$\begin{aligned} c=\frac{B(\frac{1}{3},\frac{1}{6})}{3\root 3 \of {4}}=1.76\ldots . \end{aligned}$$

Let

$$\begin{aligned} \Omega _0=\left\{ w:{{\mathrm{Re}}}w\ge 0, |{{\mathrm{Im}}}w|< \frac{1}{2} c\right\} . \end{aligned}$$

Theorem 9

The covering domain $L_{{\mathcal {F}}^{(3)}}$ is an unbounded and $6$-fold symmetric domain

$$\begin{aligned} L_{{\mathcal {F}}^{(3)}} = \bigcup _{j=0}^5 e^{j\frac{\pi }{3}i} \cdot \Omega _0. \end{aligned}$$

Proof

Let $L$ denote the covering set for ${\mathcal {F}}^{(n)}$ and let $L^*$ stand for $\bigcup _{j=0}^5 e^{j\pi i/3} \cdot \Omega _0$.

At the beginning, we can see that $L$ includes six-pointed star obtained as a union of $g_{\pi /2}(\Delta )$ and $g_{5\pi /6}(\Delta )$. We know that for $\beta \in (\pi /2,5\pi /6)$ each set $g_\beta (\Delta )$ contains a part of a horizontal strip between two rays emanating from $a/\varepsilon$ and $a$, where $a=H(\beta )$. From (21), it follows that the arguments of these points vary continuously from $-\pi /6$ to $\pi /6$ for the point $a/\varepsilon$ and from $\pi /2$ to $5\pi /6$ for the point $a$. This and the symmetry of $L$ with respect to the imaginary axis result in $L^*\subset L$.

Now, we shall prove that $L\subset L^*$. On the contrary, assume that $w_0\notin L^*$ but $w_0\in L$. It means that there exists a function $f_0\in {\mathcal {F}}^{(3)}$ such that $w_0\in f_0(\Delta )$. Without loss of generality, we can assume that $\arg w_0\in (0,\pi /6)$ because of Lemma 1 and Remark 1.

From the 3-fold symmetry of $f_0$, we know that $w_0\varepsilon , w_0\varepsilon ^2\in f_0(\Delta )$. Moreover,

$$\begin{aligned} {{\mathrm{Im}}}w_0 = |w_0|\sin \varphi _0 < |w_0|\sin \left( \varphi _0+\frac{2\pi }{3}\right) = {{\mathrm{Im}}}(w_0\varepsilon )\ , \end{aligned}$$

because $\varphi _0=\arg w_0\in (0,\pi /6)$.

Observe that the point $w_1=\{w_0-t: t\ge 0\}\cap (\varepsilon \cdot \{w_0-t: t\ge 0\})$ also belongs to $f_0(\Delta )$. If it were not the case, the points $w_1\varepsilon , w_1\varepsilon ^2$ would not be in $f_0(\Delta )$ either. But $w_1, w_1\varepsilon ^2\in \{w_0-t: t\ge 0\}$. Combining $w_1, w_1\varepsilon ^2\notin f_0(\Delta )$ with $w_0\in f_0(\Delta )$ yields that the segment connecting $w_1$ and $w_1\varepsilon ^2$ has no common points with $f_0(\Delta )$. From this and the 3-fold symmetry, all three segments connecting $w_1, w_1\varepsilon , w_1\varepsilon ^2$ and, as a consequence, the equilateral triangle $T$ with vertices in these points, would be disjoint with $f_0(\Delta )$, a contradiction. This means that $w_1, w_1\varepsilon , w_1\varepsilon ^2\in f_0(\Delta )$, which results in

$$\begin{aligned} T\subset f_0(\Delta ). \end{aligned}$$

(24)

But

$$\begin{aligned} g_{\frac{5\pi }{6}}(\Delta )\subset T\quad \text {and}\quad g_{\frac{5\pi }{6}}(\Delta )\ne T. \end{aligned}$$

(25)

From (24) and (25), $g_{5\pi /6}$ is subordinated to $f$, but $g_{5\pi /6}$ and $f$ have the same normalization, a contradiction. It means that if $w_0\in L$, then $w_0\in L^*$, which completes the proof. $\square$

Fig. 5

Covering domains for ${\mathcal {F}}^{(3)}$ and ${\mathcal {F}}^{(5)}$

The covering domains for ${\mathcal {F}}^{(3)}$ and ${\mathcal {F}}^{(5)}$ are shown in Fig. 5.