# Piecewise rotations: limit set for the non -bijective maps

Nicolas Bédaride111 Aix Marseille Université, CNRS, I2M UMR 7373, 13453
Marseille, France. Email<EMAIL_ADDRESS>Jean-François
<EMAIL_ADDRESS>Idrissa Kaboré333Université Nazi Boni,01
PB 1091, Bobo-Dioulasso, Burkina Faso. Email<EMAIL_ADDRESS>
###### Abstract

We consider non-bijective piecewise rotations of the plane. These maps belong
to a family introduced in previous papers by Boshernitzan and Goetz. We derive
in this paper some upper bounds to the size of the limit set. This improves
results of [5].

##  1 Introduction

Consider a line $\mathbf{D}$ in $\mathbb{R}^{2}$, it splits the plane in two
half-planes. Now define a dynamical system $T$ on $\mathbb{R}^{2}$ such that
the restriction to each half-plane of $T$ is given by a euclidean rotation.
The two rotations are of the same angle but with different centers, which may
well lie outside of the corresponding half-plane.

This type of map is called _a piecewise isometry_. A lot of examples have been
studied in the last years, see [10], [2] or [6]. The maps studied here have
the advantage of belonging to a family with a lot of interesting properties,
see [5], [11] and [7] for references.

Throughout this paper we follow the notation and conventions of Boshernitzan-
Goetz [5]. They studied non-bijective maps. They prove that either every orbit
is bounded uniformly by some constant $M$, or else every point outside a ball
of radius $M$, has a divergent orbit.

The bijective case was not treated in this paper; it has been done by Goetz
and Quas for a rational angle in the symmetric case (i.e. the middle of the
segment between the two centers of rotations belongs to the line
$\mathbf{D}$), see [11]. In [4] two authors of this paper give a complete
description of the bijective symmetric map if the angle belongs to a specified
finite set. In [3] they describe the non-symmetric bijective maps, assuming
the angle belongs to a particular finite set.

The aim of this paper is to find, for any non-bijective map in the family, an
explicit bound for $M$, for every given angle. If the angle is rational, the
strategy is to use the proof of [5] and improve it, where this is possible.
For the irrational case, a complete different approach is presented.

###  1.1 Outline of the paper

In Section 2 we recall the definition of our maps, give some notations and
introduce some tools used later. In Section 3 we introduce the notion of limit
sets which allows us to state correctly a previously known result, see Theorem
15. Moreover we give some results on limit sets and periodic islands, see
Proposition 19 and Proposition 20.

In Section 4 we consider the case where the angle is an irrational multiple of
$\pi$ and we prove Theorem 23. Finally in Section 5 we consider the rational
case $p/q$ where $q>2$ is an even number, see Theorem 31.

##  2 Definitions

###  2.1 Definitions of a piecewise rotation

Consider the euclidean space $\mathbb{R}^{2}$ equipped with an orientation. In
the following we will use the classical identification between $\mathbb{C}$
and the euclidean plane. Let us fix a line $\mathbf{D}$. The line cuts the
plane in two half planes, denoted by $\mathbf{P}_{0}$ and $\mathbf{P}_{1}$.
Moreover we assume that $\mathbf{P}_{1}$ is closed, and that $\mathbf{P}_{0}$
is open, in order to have a partition of $\mathbb{C}$.

######  Definition 1.

Now we fix a real number $\alpha$ and two points $C_{0},C_{1}$. We define
$\mathcal{T}$ as the set of maps $T$

$\begin{array}[]{ccc}\mathbb{C}&\rightarrow&\mathbb{C}\\\
z&\mapsto&T(z)=e^{i\alpha}(z-C_{j})+C_{j}\quad\text{ if }\quad
z\in\mathbf{P}_{j},j\in\\{0,1\\}.\end{array}$

######  Remark 2.

The cases with $C_{0}=C_{1}$ or $\alpha=0$ will not be treated in the sequel.
But the dynamic is obvious in these cases.

$\mathbf{P}_{0}$$\mathbf{P}_{1}$$\mathbf{D}$•$e^{i\gamma}$$ie^{i\gamma}$$-ie^{i\gamma}$$C_{0}$$C_{1}$$\beta$
Figure 1:  Example of an application $T\in\mathcal{T}$ with $C_{0}=1+1.5i$,
$C_{1}=4.5+0.25i$, $\gamma=\pi-\arctan(2/3)$.

######  Notations 1.

Let us fix some notation, see Figure 1.

  * •

We parametrize the line $\mathbf{D}$ by a point $z_{0}$ and an angle $\gamma$.
The angle $\gamma$ is chosen so that $z_{0}+ie^{i\gamma}$ belongs to
$\mathbf{P}_{0}$.

  * •

We call $r_{0},r_{1}$ the isometries of the plane which coincide with the
restriction of $T$ to the two half planes: for all $z\in\mathbb{C}$,
$r_{j}(z)=e^{i\alpha}(z-C_{j})+C_{j}$ with $j\in\\{0,1\\}$.

  * •

The two centers have complex coordinates given by $C_{j}=|C_{j}|e^{ic_{j}}$,
and we denote $\beta=\arg(C_{1}-C_{0})$.

  * •

In the following we will need the quantity
$\Delta=-2|C_{1}-C_{0}|\times\sin(\alpha/2)\cos(\gamma+\alpha/2-\beta)$.

These notations shows that the map $T$ is given by some parameters. The
following remark will explain how to reduce the number of such parameters.

######  Remark 3 (Conjugacy).

Let $T\in\mathcal{T}$, and consider the maps $f_{0}:z\mapsto\rho
e^{i\theta}z+b$ and $f_{1}:z\mapsto\rho e^{i\theta}\bar{z}+b$ with
$\rho\in\mathbb{R}_{+}^{*}$, $\theta\in\mathbb{R}$ and $b\in\mathbb{C}$. Then
we denote $T_{j}=f_{j}\circ T\circ f_{j}^{-1}$ for $j\in\\{0,1\\}$. This map
is in $\mathcal{T}$, and we have the following parameters:

$\begin{array}[]{|c||c|c|c|c|c||c|}\hline\cr
T&\alpha&C_{0}&C_{1}&\beta&\gamma&\Delta\\\ \hline\cr\hline\cr
T_{0}&\alpha&f_{0}(C_{0})&f_{0}(C_{1})&\theta+\beta&\theta+\gamma&\rho\Delta\\\
\hline\cr
T_{1}&-\alpha&f_{1}(C_{0})&f_{1}(C_{1})&\theta-\beta&\theta-\gamma&-\rho\Delta\\\
\hline\cr\end{array}$

Thus we can always assume $0\in\mathbf{D}$, and the given data for
$\gamma,\alpha,C_{0},C_{1}$ determine the map $T$. We call these data the
parameters of $T$.

Now we recall a known result for the maps from $\mathcal{T}$, see [5]. For
completeness we include a proof.

######  Lemma 4.

  * •

A map $T\in\mathcal{T}$ is surjective if and only if $\Delta\geq 0$. In this
case the strip $T(\mathbf{P}_{0})\cap T(\mathbf{P}_{1})$ has width $2\Delta$.

  * •

A map $T\in\mathcal{T}$ is injective if and only if $\Delta\leq 0$. In this
case the set $\mathbb{R}^{2}\setminus(T(\mathbf{P}_{0})\cup
T(\mathbf{P}_{1}))$ is a strip of width $-2\Delta$.

###### Proof.

The image of $\mathbf{D}$ by $r_{0},r_{1}$ are two parallel lines. Thus the
images of the two half planes intersect or not. It is enough to compute the
following scalar product to conclude, see Figure 2. We have for all
$z\in\mathbf{D}$:

$\displaystyle\left\langle{r_{1}(z-ie^{i\gamma})-r_{1}(z)},{r_{0}(z)-r_{1}(z)}\right\rangle$
$\displaystyle=\left\langle{-e^{i\alpha}ie^{i\gamma}},{e^{i\alpha}(C_{1}-C_{0})-(C_{1}-C_{0})}\right\rangle$
$\displaystyle=\left\langle{-e^{i\alpha}ie^{i\gamma}},{(e^{i\alpha}-1)(C_{1}-C_{0})}\right\rangle$
$\displaystyle=\left\langle{e^{i\alpha}ie^{i\gamma}},{2i\sin(\alpha/2)e^{i\alpha/2}|C_{1}-C_{0}|e^{i\beta}}\right\rangle$
$\displaystyle=-2|C_{1}-C_{0}|\sin(\alpha/2)\left\langle{e^{i\alpha}ie^{i\gamma}},{ie^{i\alpha/2}e^{i\beta}}\right\rangle=\Delta\qed$

$\mathbf{P}_{0}$$\mathbf{P}_{1}$$C_{0}$$C_{1}$$\longrightarrow$

$T(\mathbf{P}_{0})$$T(\mathbf{P}_{1})$$C_{0}$$C_{1}$$-2\Delta$

Figure 2: Injective example: same parameters as in Figure 1 with an angle
$\alpha=-\pi/10$.

$\mathbf{P}_{0}$$\mathbf{P}_{1}$$C_{0}$$C_{1}$$\longrightarrow$

$T(\mathbf{P}_{0})$$T(\mathbf{P}_{1})$$T(\mathbf{P}_{1})$$C_{0}$$C_{1}$$2\Delta$

Figure 3: Surjective example: same parameters as in Figure 1 with an angle
$\alpha=\pi/10$.

###  2.2 Tools

######  Definition 5.

Consider a map $T:\mathbb{C}\to\mathbb{C}$. We define:

${\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}=\sup\left\\{\big{|}|T(z)|-|z|,z\in\mathbb{C}\right\\}\in\mathbb{R}_{+}\cup\\{+\infty\\}.$

######  Lemma 6.

For all $k\geq 0$ and $z\in\mathbb{C}$ and for every map $T\in\mathcal{T}$ one
has:

$\displaystyle|T^{k}(z)-e^{ik\alpha}z|\leq
k{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
$\displaystyle{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\leq
2\times|\sin(\alpha/2)|\times\max(|C_{0}|,|C_{1}|).$

###### Proof.

For $z\in\mathbb{C}$ we have $T(z)=e^{i\alpha}(z-C_{j})+C_{j}$, and thus:

$\big{|}|T(z)|-|z|\big{|}=\big{|}|T(z)|-|e^{i\alpha}z|\big{|}\leq|T(z)-e^{i\alpha}z|=|C_{j}(1-e^{i\alpha})|\leq
2|\sin(\alpha/2)|\max(|C_{0}|,|C_{1}|).$

Hence we have proved the formula for $k=1$ and
${\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\leq
2|\sin(\alpha/2)|\max(|C_{0}|,|C_{1}|)$.

We obtain by triangular inequality

$\displaystyle|T^{k}(z)-e^{ik\alpha}z|$
$\displaystyle\leq\sum_{l=0}^{k-1}|e^{i(k-1-l)\alpha}T^{l+1}(z)-e^{i(k-l)\alpha}T^{l}(z)|$
$\displaystyle\leq\sum_{l=0}^{k-1}|T(T^{l}(z))-e^{i\alpha}T^{l}(z)|\leq\sum_{l=0}^{k-1}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\leq
k{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}.\qed$

######  Lemma 7.

Let $y\in\mathbb{C}$ with $R>|y|$. Then we have:

$\big{|}|R+y|-R-\operatorname{Re}(y)\big{|}\leq\frac{|y|^{2}}{R-|y|}.$

###### Proof.

Remark that $|R+y|^{2}=R^{2}+2R\operatorname{Re}(y)+|y|^{2}$ :

$|R+y|-R-\operatorname{Re}(y)=\frac{|R+y|^{2}-R^{2}}{|R+y|+R}-\operatorname{Re}(y)=\frac{|y|^{2}+2R\operatorname{Re}(y)}{|R+y|+R}-\operatorname{Re}(y)=\frac{|y|^{2}+(R-|R+y|)\operatorname{Re}(y)}{|R+y|+R}.$

If we denote $|R+y|-R-\operatorname{Re}(y)=h$, then we have
$h=\frac{|y|^{2}-(h+\operatorname{Re}(y))\operatorname{Re}(y)}{|R+y|+R}=\frac{\operatorname{Im}^{2}(y)-\operatorname{Re}(y)h}{|R+y|+R}$.
We deduce $|h|\leq\frac{|y|^{2}+|y|\cdot|h|}{R}$ and
$0\leq|y|^{2}+|h|(|y|-R)$, which gives the result.

∎

######  Definition 8.

Let us define the map $g$ :

$\begin{array}[]{ccc}\mathbb{R}&\rightarrow&\mathbb{R}\\\
x&\mapsto&g(x)=\left\\{\begin{array}[]{l}-2|C_{0}|\sin(\alpha/2)\sin(x+\alpha/2-c_{0})\quad\text{if}\quad\gamma<x<\gamma+\pi\mod
2\pi.\\\
-2|C_{1}|\sin(\alpha/2)\sin(x+\alpha/2-c_{1})\quad\text{otherwise}.\end{array}\right.\end{array}$

######  Proposition 9.

Consider a map $T\in\mathcal{T}$. Let $z=Re^{i\theta}\in\mathbb{C}$ such that
$|z|>{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$,
see Definition 5. Then we obtain the following inequalities, where
$\theta_{1}$ is the argument of $T(z)$:

$\displaystyle\big{|}|T(z)|-|z|-g(\theta)\big{|}\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}{|z|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}},$
(1)
$\displaystyle{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}=2|\sin(\alpha/2)|\max(|C_{0}|,|C_{1}|).$
(2) $\displaystyle|\theta_{1}-(\theta+\alpha)\mod
2\pi|\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}.$
(3)

###### Proof.

  * •

_Proof of ( 1) and (2)._ Let $z=Re^{i\theta}\in\mathbb{C}$ such that
$R>2|\sin(\alpha/2)|\max(|C_{0}|,|C_{1}|)$. Since
$0\in\mathbf{D}=e^{i\gamma}\mathbb{R}$, we have

$z\in\mathbf{P}_{0}\iff\gamma<\theta<\gamma+\pi\mod 2\pi.$

Let us denote $j\in\\{0,1\\}$ such that $z\in\mathbf{P}_{j}$. We obtain :

$|T(z)|=|e^{i\alpha}(z-C_{j})+C_{j}|=|Re^{i\theta}-C_{j}+e^{-i\alpha}C_{j}|=|R+C_{j}e^{-i\theta}(e^{-i\alpha}-1)|$

Now we define

$y=C_{j}e^{-i\theta}(e^{-i\alpha}-1)=C_{j}e^{-i(\theta+\alpha/2)}(e^{-i\alpha/2}-e^{i\alpha/2})=2\sin(\alpha/2)C_{j}e^{-i(\theta+\alpha/2+\pi/2)}$

Note that $\operatorname{Re}(y)=g(\theta)$, with $g$ defined in Definition 8.
We apply Lemma 7 with $R=|z|$ and obtain:

$\big{|}|T(z)|-|z|-g(\theta)\big{|}=\big{|}|R+y|-R-\operatorname{Re}(y)\big{|}\leq\frac{|y|^{2}}{|z|-|y|}.$

For $|z|>|y|$ we deduce

$\big{|}|T(z)|-|z|\big{|}\geq|g(\theta)|-\frac{|y|^{2}}{|z|-|y|}.$

We conclude
${\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\geq|y|$
using the fact that $|y|=\sup|g(\theta)|$ and
$\lim_{|z|\to+\infty}\frac{|y|^{2}}{|z|-|y|}=0$. We conclude by means of Lemma
6 to obtain Equation $(2)$. Then we deduce $(1)$.

  * •

_Proof of ( 3)._

$\left|\frac{T(z)e^{-i\alpha}}{z}-1\right|=\left|\frac{T(z)}{z}-e^{i\alpha}\right|=\left|\frac{e^{i\alpha}(z-C_{j})+C_{j}}{z}-e^{i\alpha}\right|=\frac{|C_{j}(1-e^{i\alpha})|}{|z|}\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|}.$

If
$\frac{{\left|\kern-0.60275pt\left|\kern-0.60275pt\left|T\right|\kern-0.60275pt\right|\kern-0.60275pt\right|}}{|z|}<1$
we deduce :

$\displaystyle\left|\operatorname{Im}\left(\frac{T(z)e^{-i\alpha}}{z}\right)\right|=\left|\operatorname{Im}\left(\frac{T(z)e^{-i\alpha}}{z}-1\right)\right|\leq\left|\frac{T(z)e^{-i\alpha}}{z}-1\right|\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|}$
$\displaystyle\left|\operatorname{Re}\left(\frac{T(z)e^{-i\alpha}}{z}\right)\right|=1+\operatorname{Re}\left(\frac{T(z)e^{-i\alpha}}{z}-1\right)\geq
1-\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|}.$

Thus we obtain

$|\tan(\theta_{1}-\alpha-\theta)|=\frac{\left|\operatorname{Im}\left(\frac{T(z)e^{-i\alpha}}{z}\right)\right|}{\left|\operatorname{Re}\left(\frac{T(z)e^{-i\alpha}}{z}\right)\right|}\leq\frac{\frac{{\left|\kern-0.60275pt\left|\kern-0.60275pt\left|T\right|\kern-0.60275pt\right|\kern-0.60275pt\right|}}{|z|}}{1-\frac{{\left|\kern-0.60275pt\left|\kern-0.60275pt\left|T\right|\kern-0.60275pt\right|\kern-0.60275pt\right|}}{|z|}}=\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}.$

Thus we conclude

$0\leq|\theta_{1}-(\theta+\alpha)|\leq\arctan\left(\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}\right)\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}.\qed$

##  3 Limit sets

###  3.1 Definitions

######  Definition 10.

Let $T:\mathbb{C}\to\mathbb{C}$ be an application. For $M>0$, let $B(0,M)$ be
the open ball of radius $M$ centered at $0$. We define the following sets for
any $n\in\mathbb{N}$:

$\mathbf{A}_{M,n}(T)=T^{n}B(0,M)\quad\text{and}\quad\operatorname{\mathbf{B}}_{M,n}(T)=T^{-n}B(0,M).$

$\operatorname{\mathbf{A}}(T)=\bigcup_{M>0}\bigcap_{n\geq
0}\operatorname{\mathbf{A}}_{M,n}(T)\quad\text{and}\quad\operatorname{\mathbf{B}}(T)=\bigcup_{M>0}\bigcap_{n\geq
0}\operatorname{\mathbf{B}}_{M,n}(T).$

These sets will be called _limit sets_ in the rest of the paper.

######  Remark 11.

Let $z\in\mathbb{C}$. The previous sets can be characterized as follows:

  * •

$z\in\operatorname{\mathbf{A}}(T)\iff\exists(y_{n})_{n}\in\mathbb{C}^{\mathbb{N}}$
a bounded sequence such that $\forall n\in\mathbb{N}$, $z=T^{n}(y_{n})$.

  * •

$z\in\operatorname{\mathbf{B}}(T)$ if and only if $(T^{n}(z))_{n}$ is a
bounded sequence.

######  Remark 12.

By construction we obtain the following inclusions:

  * •

$\forall n\in\mathbb{N}$,
$\operatorname{\mathbf{A}}(T)\subset\operatorname{\mathbf{A}}(T^{n})$ and
$\operatorname{\mathbf{B}}(T)\subset\operatorname{\mathbf{B}}(T^{n})$.

  * •

$\operatorname{\mathbf{A}}(T)\subset\bigcap_{n\geq
0}\operatorname{Im}(T^{n})$.

###  3.2 Global attraction and global repulsion

######  Definition 13.

Let $T$ be an application from $\mathbb{C}$ to $\mathbb{C}$.

  * •

The application $T$ is globally attractive if there exists $M\geq 0$ such that
$\forall L\geq 0$, $\limsup\limits_{n\to+\infty}\sup\limits_{|z|\leq
L}|T^{n}(z)|\leq M$.

  * •

The map $T$ is globally repulsive if there exists $M\geq 0$ such that $\forall
L\geq M$, $\liminf\limits_{n\to+\infty}\inf\limits_{|z|\geq
L}|T^{n}(z)|=+\infty$.

We will denote $M(T)$ the infimum of such real numbers $M$.

######  Remark 14.

Let $T$ be an application from $\mathbb{C}$ to $\mathbb{C}$.

  * •

If $T$ is globally attractive, then $\operatorname{\mathbf{A}}(T)$ is bounded
and included in $B(0,M(T))$.

  * •

If $T$ is globally repulsive, then, $\operatorname{\mathbf{B}}(T)$ is bounded
and included in $B(0,M(T))$.

We are interested in the size of the limit sets. Boshernitzan and Goetz have
proved the following formula, reformulated here using our terminology.

######  Theorem 15.

[5]

  * •

Let $T\in\mathcal{T}$ be a surjective and non-injective map, then $T$ is
globally attractive and $\operatorname{\mathbf{A}}(T)$ is a bounded set.

  * •

Let $T\in\mathcal{T}$ be an injective and non-surjective map, then $T$ is
globally repulsive and $\operatorname{\mathbf{B}}(T)$ is a bounded set.

We obtain an upper bound for $M(T)$, in the irrational and the rational case,
by distinct methods presented in Sections 4 and 5.

###  3.3 Periodic islands and limit sets

In this part we want to describe the limit sets $\operatorname{\mathbf{B}}(T)$
and $\operatorname{\mathbf{A}}(T)$. Note that these sets can be quite
complicated, as explained in [9] and [8]. In Proposition 19 we describe some
explicit subsets obtained by periodic islands (see Definition 18). Moreover,
in Proposition 20 and Corollary 22 we explain how these sets change when the
parameters change, see Figure 4.

Let us fix $\alpha$, $C_{0}$, $C_{1}$. Now let
$u=u_{0},\ldots,u_{n-1}\in\\{0,1\\}^{n}$ be a finite word and consider the
maps $r_{k}(z)=e^{i\alpha}(z-C_{k})+C_{k}$. We observe that

$r_{u_{n-1}}\circ\cdots\circ
r_{u_{0}}(z)=e^{in\alpha}z+\sum_{k=0}^{n-1}e^{-(n-k-1)i\alpha}C_{u_{k}}(1-e^{i\alpha}).$
(4)

######  Definition 16.

Consider the word $u=u_{0}\ldots u_{n-1}\in\\{0,1\\}^{n}$ with $n\geq 1$ and
assume $e^{in\alpha}\neq 1$. Then we define _the almost periodic point
associated to $u$_ as the point

$z_{u}=\frac{1-e^{i\alpha}}{1-e^{in\alpha}}\sum_{k=0}^{n-1}e^{-(n-k-1)i\alpha}C_{u_{k}}.$

The following lemma appears already in [11] and [7].

######  Lemma 17.

Let $T\in\mathcal{T}$ and $u=u_{0}\ldots u_{n-1}\in\\{0,1\\}^{n}$ such that
$e^{in\alpha}\neq 1$. Then the almost periodic point $z_{u}$ is a periodic
point of period $n$ for $T$ if and only if for all $k\in\llbracket
0,n-1\rrbracket$, $T^{k}(z_{u})\in\mathbf{P}_{u_{k}}$.

A coding of the map $T$ is a map from $\mathbb{C}$ to $\\{0,1\\}^{\mathbb{N}}$
which sends the point $z$ to the sequence $(u_{n})_{n\in\mathbb{N}}$ where
$T^{n}z\in P_{u_{n}}$.

######  Definition 18.

Let $n$ be such that $e^{in\alpha}\neq 1$ and consider a word $u=u_{0}\ldots
u_{n-1}\in\\{0,1\\}^{n}$. Assume that $z_{u}$ is a periodic word for $T$ with
coding $u^{\omega}$ (a periodic sequence of period $u$). Then we define the
_periodic island_ as the set of complex numbers $z$ such that
$T^{k}(z)\in\mathbf{P}_{u_{(k\mod n)}}$ for all integers $k\geq 0$.

Moreover we define the _weight_ of $z_{u}$ as
$w(z_{u})=\min\big{\\{}d(T^{k}(z),\mathbf{D});k\in{\llbracket
0,n-1\rrbracket}\big{\\}}.$

Note that a periodic island is a convex set obtained as intersection of half-
planes. If $z_{u}$ is a periodic point, then $B(z_{u},w(z_{u}))$ is included
inside the periodic island associated to $z_{u}$, see [11] (this inclusion
becomes an equality if $\alpha$ is irrational, see Section 2 of [11]).

######  Proposition 19.

For any $T\in\mathcal{T}$ we have

$\bigcup_{z_{u}\in\operatorname{Per}}B(z_{u},w(z_{u}))\subset\operatorname{\mathbf{B}}(T)\quad\text{and}\quad\bigcup_{z_{u}\in\operatorname{Per}}B(z_{u},w(z_{u}))\subset\operatorname{\mathbf{A}}(T),$

where $\operatorname{Per}=\\{z_{u}\text{ periodic point of some period $n$
such that }e^{in\alpha}\neq 1\\}$.

###### Proof.

Let $z_{u}$ be a periodic point of period $n$ with $e^{in\alpha}\neq 1$, and
let $z\in B(z_{u},w(z_{u}))$. Notice that the restriction of $T^{n}$ to
$B(z,w(z_{u}))$ is a rotation of angle $n\alpha$ with center $z_{u}$.

  * •

Let $k\in\mathbb{N}$ and consider the euclidean division: $k=qn+r$ with $0\leq
r<n$. The point $T^{qn}(z)$ is in $B(z_{u},w(z_{u}))$. By definition of
${\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
and Lemma 6 we deduce

$|T^{k}(z)|=|T^{r}\circ T^{qn}(z)|\leq
r{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+|T^{qn}(z)|<n{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+|z_{u}|+w(z_{u}).$

Thus we obtain $\sup\\{|T^{k}z|\mid
k\in\mathbb{N}\\}<n{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+|z_{u}|+w(z_{u})<+\infty$
and $z\in\operatorname{\mathbf{B}}(T)$.

  * •

The restriction of $T$ to $\cup_{i=0}^{n-1}T^{i}B(z_{u},w(z_{u}))$ is a
bijection onto its image, which is the same set. Let us denote this map by
$\hat{T}$. Then for all $k\in\mathbb{N}$, if we denote
$y_{k}={\hat{T}}^{-k}(z)$, we have $z=T^{k}(y_{k})$ and we conclude
$z\in\operatorname{\mathbf{A}}(T)$.

∎

######  Proposition 20.

Let us consider an element $T$ of $\mathcal{T}$ with parameters $\alpha$,
$\gamma$, $\mathbf{D}$ and $C_{j}$. For all $\varepsilon>0$, let us denote by
$T_{\varepsilon}$ the element of $\mathcal{T}$ with the same parameters,
except for the discontinuity line
$\mathbf{D}_{\varepsilon}=e^{i\varepsilon}\mathbf{D}$.
Consider a periodic point $z_{u}$ of $T$ of weight $w$ associated to the word
$u$ of length $n$. Then for all $\varepsilon>0$ such that

$w_{\varepsilon}=w-2\big{(}|z_{u}|+n{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\big{)}\times|\sin(\varepsilon/2)|>0,$

the point $z_{u}$ is a periodic point for $T_{\varepsilon}$ with weight bigger
or equal to $w_{\varepsilon}$.

###### Proof.

Let us consider $u=u_{0}\ldots u_{n-1}\in\\{0,1\\}^{n}$ and a periodic point
$z_{u}$ for $T$ of weight $w$, associated to $u$. By definition we have:

$z_{u}=\frac{1-e^{i\alpha}}{1-e^{in\alpha}}\sum_{k=0}^{n-1}e^{-(n-k-1)i\alpha}C_{u_{k}}$

Now we fix $\varepsilon$ small enough to have

$w_{\varepsilon}=w-2\big{(}|z_{u}|+n{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\big{)}\times|\sin(\varepsilon/2)|>0.$

Let us prove that $z_{u}$ is a periodic point for $T_{\varepsilon}$. We denote
by $\mathbf{P}_{0}^{\varepsilon}$ and $\mathbf{P}_{1}^{\varepsilon}$ the half
planes defined by $T_{\varepsilon}$.

  * •

We need to check that $k\in\llbracket 0,n-1\rrbracket$,
$T_{\varepsilon}^{k}(z_{u})\in\mathbf{P}_{u_{k}}^{\varepsilon}$ and
$d(T_{\varepsilon}^{k}(z_{u}),e^{i\varepsilon}\mathbf{D})\geq
w_{\varepsilon}$.
We’ll give a proof by induction over $k$.

  * •

For $k=0$ we have by hypothesis

$d(e^{i\varepsilon}z_{u},z_{u})=|z_{u}e^{-i\varepsilon}-z_{u}|=2|z_{u}|\times|\sin(\varepsilon/2)|.$

We deduce $z_{u}\in\mathbf{P}_{u_{0}}^{\varepsilon}$ and

$\displaystyle d(z_{u},e^{i\varepsilon}\mathbf{D})$ $\displaystyle\geq
d(z_{u}e^{i\varepsilon},e^{i\varepsilon}\mathbf{D})-d(z_{u},e^{i\varepsilon}z)=d(z_{u},\mathbf{D})-d(z_{u},e^{i\varepsilon}z)$
$\displaystyle>w-2|z_{u}|\times|\sin(\varepsilon/2)|\geq w_{\varepsilon}.$

  * •

Assume that the result is true for $k<n-1$. Then $T_{\varepsilon}^{k}(z_{u})$
and $T^{k}(z_{u})$ are in the same half planes
$\mathbf{P}_{u_{k}}\cap\mathbf{P}_{u_{k}}^{\varepsilon}$ and
$T^{k+1}_{\varepsilon}(z_{u})=T^{k+1}(z_{u})$ and then we deduce

$\displaystyle d(T^{k+1}(z_{u}),e^{i\varepsilon}T^{k+1}(z_{u}))$
$\displaystyle=|T^{k+1}(z_{u})-e^{i\varepsilon}T^{k+1}(z_{u})|=|1-e^{i\varepsilon}|\times|T^{k+1}(z_{u})|$
$\displaystyle\leq
2|\sin(\varepsilon/2)|\times(|z_{u}|+(k+1){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}).$

Thus we obtain

$\displaystyle d(T^{k+1}(z_{u}),e^{i\varepsilon}\mathbf{D})$
$\displaystyle\geq
d(e^{i\varepsilon}T^{k+1}(z_{u}),e^{i\varepsilon}\mathbf{D})-d(T^{k+1}(z_{u}),e^{i\varepsilon}T^{k+1}(z_{u}))$
$\displaystyle=d(T^{k+1}(z_{u}),\mathbf{D})-d(T^{k+1}(z_{u}),e^{i\varepsilon}T^{k+1}(z_{u}))$
$\displaystyle\geq
w-2|\sin(\varepsilon/2)|\times(|z_{u}|+(k+1){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|})\geq
w_{\varepsilon}.$

We have proved our claim by induction. ∎

Figure 4: Red part: periodic islands for a bijective case. Blue part: periodic
islands for a perturbation of the parameter $\gamma$, see Proposition 20.

By using the conjugacy by $z\mapsto e^{i\varepsilon}z$ and Remark 3 we can
rephrase the previous proposition as follow:

######  Proposition 21.

Let us consider the element $T$ of $\mathcal{T}$ with parameters $\alpha$,
$\gamma$, $\mathbf{D}$ and $C_{k}$. For all $\varepsilon$, let us denote by
$T_{\varepsilon}$ the element of $\mathcal{T}$ with the same parameters but
with centers $e^{i\varepsilon}C_{k}$.

Consider a periodic point $z_{u}$ of $T$ of weight $w$ associated to the word
$u$ of length $n$. Then for all $\varepsilon$ such that

$w_{\varepsilon}=w-2\big{(}|z_{u}|+n{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\big{)}\times|\sin(\varepsilon/2)|>0$

we have that $e^{i\varepsilon}z_{u}$ is a periodic point for $T_{\varepsilon}$
with weight bigger or equal to $w_{\varepsilon}$.

We deduce:

######  Corollary 22.

For all $M>0$, there exists a non-bijective piecewise rotation
$T\in\mathcal{T}$ such that $\operatorname{\mathbf{B}}$ or
$\operatorname{\mathbf{A}}$ are non empty and contain a ball of radius $M$.

###### Proof.

We apply Proposition 21 with $\beta=\alpha/2+\gamma-\pi/2$. In this case we
have a bijective map, thus it has some periodic islands. Hence for some
$\varepsilon>0$ we can find some non-bijective maps $T_{\varepsilon}$ such
that $\operatorname{\mathbf{B}}$ or $\operatorname{\mathbf{A}}$ are non-empty
and we apply Proposition 19 to conclude. ∎

##  4 Size of the limit set for irrational angle

###  4.1 Statement of the result

Let $T=T(\alpha,C_{0},C_{1},\mathbf{D},\gamma)\in\mathcal{T}$ be a non-
bijective map. Let us denote $\alpha=2\pi a$ and assume that $a$ is an
irrational number. Recall that $M(T)$ is defined in Definition 13.

Let $(p_{\ell})_{\ell\in\mathbb{N}}$ and $(q_{\ell})_{\ell\in\mathbb{N}}$ be
the sequence of convergents of $a$ in the continued fraction expansion. Let
$\ell_{0}$ be such that
$\frac{1}{q_{\ell_{0}}}\left(\left(8+\frac{\pi}{2q_{\ell_{0}}}\right){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+4|\Delta|\right)<\frac{|\Delta|}{\pi}$.
Then we will prove the following:

######  Theorem 23.

Let $T\in\mathcal{T}$ be a non-bijective map with an irrational angle. We
obtain the following bound:

$M(T)\leq
q_{\ell_{0}}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell_{0}}+1\right)$

To prove this theorem we first need some background on the Denjoy-Kosma
inequality and the Three gaps theorem.

###  4.2 Bounded variation function and three gaps theorem

######  Definition 24.

Consider a periodic function $f:\mathbb{R}\to\mathbb{R}$ of period $T$. If the
following quantity exists we say that $f$ has bounded variation with variation
$\operatorname{Var}(f)$:

$\operatorname{Var}(f)=\sup\left\\{\sum_{k=0}^{P-1}|f(x_{k+1})-f(x_{k})|;\\{x_{0},\ldots,x_{P}\\}\in\mathcal{P}\right\\}.$

where $\mathcal{P}$ is the set of subdivisions $P=\\{x_{0},\ldots,x_{P-1}\\}$
of $[0,T]$ such that

$0=x_{0}\leq x_{1}\leq\cdots\leq x_{P-1}\leq x_{P}=T.$

######  Proposition 25.

Let $f:\mathbb{R}\to\mathbb{R}$ be a $2\pi$-periodic function which is
piecewise $\mathcal{C}^{1}$. Assume that the discontinuity points of $f$ on
$]0,2\pi[$ are denoted by $d_{1},\ldots,d_{m}$, and set $d_{0}=0$. Then $f$
has bounded variation and

$\operatorname{Var}(f)=\int_{0}^{2\pi}|f^{\prime}(t)|dt+\sum_{k=0}^{m}|f(d_{k}^{+})-f(d_{k}^{-})|.$

We associate to any irrational positive real number $0<a<1$ its continued
fraction expansion:

$a=0+\frac{1}{\displaystyle a_{1}+\frac{1}{\displaystyle
a_{2}+\frac{1}{\ddots}}}\mbox{ that we denote by
$a=[0,a_{1},\dots,a_{\ell},\dots]$. }$

We also consider two sequences of integers $(p_{k})_{k\in\mathbb{N}}$ and
$(q_{k}))_{k\in\mathbb{N}}$, such that for any integer $\ell\geq 0$ one has
$[0,a_{1},\dots,a_{\ell}]=\frac{p_{\ell}}{q_{\ell}}$. They are defined by
induction as follows,

$\left\\{\begin{array}[]{ll}p_{-2}=0\\\ q_{-2}=1\end{array}\mbox{ ,
}\begin{array}[]{ll}p_{-1}=1\\\ q_{-1}=0\end{array}\right.\mbox{ and for any
integer }\ell\geq
0:\left\\{\begin{array}[]{ll}p_{\ell}=a_{\ell}p_{\ell-1}+p_{\ell-2}.\\\
q_{\ell}=a_{\ell}q_{\ell-1}+q_{\ell-2}.\end{array}\right.$

We recall Denjoy–Koksma inequality:

######  Theorem 26.

[12] Let $a\in[0,1[$ be an irrational number, and let us denote by
$\frac{p_{\ell}}{q_{\ell}}$ the sequence of partial quotients in the continued
fraction expansion of $a$. Let $f:\mathbb{R}\to\mathbb{R}$ be a
$2\pi$-periodic function, piecewise $\mathcal{C}^{1}$. Then for all
$x\in\mathbb{R}$ and for all $\ell\in\mathbb{N}^{*}$ one has:

$\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}f(x+2\pi
ka)-\frac{1}{2\pi}\int_{0}^{2\pi}f(t)dt\right|\leq\frac{1}{q_{\ell}}\operatorname{Var}(f).$

The following theorem has been proved by V. Sos:

######  Theorem 27 (Three gaps theorem).

[1] Let $0<a<1$ be an irrational number and $\ell$ a positive integer. Let us
denote by $p_{\ell}/q_{\ell}$ a partial quotient. The points
$\\{\\{ka\\},0\leq k\leq q_{\ell}-1\\}$ partition the unit circle into
$q_{\ell}$ intervals. The lengths of the latter take on three values, and the
minimal value is equal to $|q_{\ell-1}a-p_{\ell-1}|\geq\frac{1}{2q_{\ell}}$.

###  4.3 Auxiliary function

######  Lemma 28.

Consider the map $g$ from Definition 8. We have the following properties:

  * •

The map $g$ is of bounded variation and we have $\operatorname{Var}(g)\leq
2|\Delta|+4{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$.

  * •

$\displaystyle\int_{-\pi}^{\pi}g(x)dx=-2\Delta$.

###### Proof.

  * •

The map $g$ has two discontinuities, in $\gamma$ and $\gamma+\pi$, and we have

$\displaystyle|g(\gamma^{+})-g(\gamma^{-})|+|g(\gamma+\pi^{+})-g(\gamma+\pi^{-})|$
$\displaystyle\quad=2|\sin(\alpha/2)|\times\big{|}|C_{0}|\sin(\gamma+\alpha/2-c_{0})-|C_{1}|\sin(\gamma+\alpha/2-c_{1})\big{|}$
$\displaystyle\quad\quad+2|\sin(\alpha/2)|\times\big{|}|C_{0}|\sin(\gamma+\pi+\alpha/2-c_{0})-|C_{1}|\sin(\gamma+\pi+\alpha/2-c_{1})\big{|}$
$\displaystyle\quad=2|\sin(\alpha/2)|\times\big{|}\operatorname{Im}(\bar{C}_{0}e^{i(\gamma+\alpha/2)})-\operatorname{Im}(\bar{C}_{1}e^{i(\gamma+\alpha/2)})\big{|}$
$\displaystyle\quad\quad+2|\sin(\alpha/2)|\times\big{|}\operatorname{Im}(\bar{C}_{0}e^{i(\gamma+\pi+\alpha/2)})-\operatorname{Im}(\bar{C}_{1}e^{i(\gamma+\pi+\alpha/2)})\big{|}$
$\displaystyle\quad=4|\sin(\alpha/2)|\times|C_{0}-C_{1}|\times|\sin(\gamma+\alpha/2-\beta)|$
$\displaystyle\quad=2|\Delta|.$

We obtain for the computation of the integral

$\displaystyle\int_{\gamma-\pi}^{\gamma}|g^{\prime}(x)|dx+\int_{\gamma}^{\gamma+\pi}|g^{\prime}(x)|dx$
$\displaystyle=2|C_{1}||\sin(\alpha/2)|\int_{\gamma-\pi}^{\gamma}|\cos(x+\alpha/2-c_{1})|dx+2|C_{0}||\sin(\alpha/2)|\int_{\gamma}^{\gamma+\pi}|\cos(x+\alpha/2-c_{0})|dx$
$\displaystyle=4|\sin(\alpha/2)|(|C_{0}|+|C_{1}|)\leq
8|\sin(\alpha/2)|\max(|C_{0}|,|C_{1}|).$

We use (1) in Proposition 9 and Proposition 25 to conclude at the first point.

  * •

We obtain:

$\displaystyle\int_{-\pi}^{\pi}g(x)dx$
$\displaystyle=\int_{\gamma-\pi}^{\gamma}g(x)dx+\int_{\gamma}^{\gamma+\pi}g(x)dx$
$\displaystyle=-2|C_{1}|\sin(\alpha/2)\int_{\gamma-\pi}^{\gamma}\sin(x+\alpha/2-c_{1})dx-2|C_{0}|\sin(\alpha/2)\int_{\gamma}^{\gamma+\pi}\sin(x+\alpha/2-c_{0})dx$
$\displaystyle=2|C_{1}|\sin(\alpha/2)\left[\cos(x+\alpha/2-c_{1})\right]_{\gamma-\pi}^{\gamma}+2|C_{0}|\sin(\alpha/2)\left[\cos(x+\alpha/2-c_{0})\right]_{\gamma}^{\gamma+\pi}$
$\displaystyle=4|C_{1}|\sin(\alpha/2)\cos(\gamma+\alpha/2-c_{1})-4|C_{0}|\sin(\alpha/2)\cos(\gamma+\alpha/2-c_{0})$
$\displaystyle=4|C_{1}-C_{0}|\sin(\alpha/2)\cos(\gamma+\alpha/2-\beta)$

where we use the fact that
$\operatorname{Re}(\bar{C}_{1}e^{i(\alpha/2+\gamma)})-\operatorname{Re}(\bar{C}_{0}e^{i(\alpha/2+\gamma)})=\operatorname{Re}\left(\overline{(C_{1}-C_{0})}e^{i(\alpha/2+\gamma)}\right)$.

∎

Now we state the key part in the proof of Theoem 23.

######  Proposition 29.

Consider $\ell\in\mathbb{N},z\in\mathbb{C}$ and assume
$|z|>q_{\ell}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell}+1\right)$.
Then we obtain:

$\left|\frac{|T^{q_{\ell}}z|-|z|}{q_{\ell}}+\frac{\Delta}{\pi}\right|\leq\frac{1}{q_{\ell}}\left((8+\frac{\pi}{2q_{\ell}}){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+4|\Delta|\right).$

###### Proof.

We will need the following notations. If $z\in\mathbb{C}$, then we denote by
$\theta$ its argument, and for all integer $k\geq 0$ we set $z_{k}:=T^{k}(z)$
and $\theta_{k}:=\arg(z_{k})$. Moreover we will denote, by abuse of notations,
$d(\theta,\theta^{\prime})=|\theta-\theta^{\prime}\mod 2\pi|$. We obtain:

$\displaystyle\left|\frac{|T^{q_{\ell}}z|-|z|}{q_{\ell}}+\frac{\Delta}{\pi}\right|$
$\displaystyle=\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}\left(|z_{k+1}|-|z_{k}|\right)+\frac{\Delta}{\pi}\right|$
$\displaystyle\leq\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}\left||z_{k+1}|-|z_{k}|-g(\theta_{k})\right|+\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}g(\theta_{k})+\frac{\Delta}{\pi}\right|.$

By Lemma 6 and equation (2) we obtain, if
$|z|>{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}q_{\ell}$,
that
$|z_{k}|>{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
for $0\leq k\leq q_{\ell}-1$.
Then by equation (1) we have:

$\displaystyle\left|\frac{|T^{q_{\ell}}z|-|z|}{q_{\ell}}+\frac{\Delta}{\pi}\right|\leq\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}{|z_{k}|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}+\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}g(\theta_{k})+\frac{\Delta}{\pi}\right|$
$\displaystyle\qquad\leq\underbrace{\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}{|z_{k}|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}}_{(A)}+\underbrace{\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}g(\theta_{k})-g(\theta_{0}+k\alpha)\right|}_{(B)}+\underbrace{\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}g(\theta_{0}+k\alpha)+\frac{\Delta}{\pi}\right|}_{(C)}.$

We will produce upper bound for each term.

  * $(A)$

First term:

$\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}{|z_{k}|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}\leq\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}{|z|-(k+1){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}{|z|-q_{\ell}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}.$

  * $(B)$

Second term:

    * –

By equation (3) in Proposition 9: if
$|z|>{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$,
we obtain
$d(\theta_{1},\theta_{0}+\alpha)\leq\frac{{\left|\kern-0.60275pt\left|\kern-0.60275pt\left|T\right|\kern-0.60275pt\right|\kern-0.60275pt\right|}}{|z|-{\left|\kern-0.60275pt\left|\kern-0.60275pt\left|T\right|\kern-0.60275pt\right|\kern-0.60275pt\right|}}$.

    * –

If
$|z|>2{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$,
then
$|z_{1}|=|T(z)|\geq|z|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}>{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
and we deduce

$\displaystyle d(\theta_{2},\theta_{0}+2\alpha)$ $\displaystyle\leq
d(\theta_{2},\theta_{1}+\alpha)+d(\theta_{1}+\alpha,\theta_{0}+2\alpha)$
$\displaystyle\leq
d(\theta_{2},\theta_{1}+\alpha)+d(\theta_{1},\theta_{0}+\alpha)$
$\displaystyle\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z_{1}|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}+\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}$
$\displaystyle\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-2{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}+\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}$

    * –

Now if
$|z|>n{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$,
we have $\forall k\in\llbracket 0,n-1\rrbracket$,
$|z_{k}|>{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
and

$\displaystyle d(\theta_{n},\theta_{0}+n\alpha)$
$\displaystyle\leq\sum_{k=0}^{n-1}d(\theta_{n-k}+k\alpha,\theta_{n-k-1}+(k+1)\alpha)$
$\displaystyle\leq\sum_{k=0}^{n-1}d(\theta_{n-k},\theta_{n-k-1}+\alpha)$
$\displaystyle\leq\sum_{k=0}^{n-1}\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z_{k}|-{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}$
$\displaystyle\leq\sum_{k=1}^{n}\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-k{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}$
$\displaystyle\leq\frac{n{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-n{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}$

    * –

By Theorem 27 the smallest interval in the subdivision $\\{k\alpha,0\leq k\leq
q_{\ell}-1\\}$ has length at least
$\frac{2\pi}{2q_{\ell}}=\frac{\pi}{q_{\ell}}$.

    * –

we observe the following fact:

$\displaystyle(q_{\ell}-1)\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{|z|-(q_{\ell}-1){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}<\frac{\pi}{2q_{\ell}}$
$\displaystyle\iff\frac{2}{\pi}q_{\ell}(q_{\ell}-1){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}<|z|-(q_{\ell}-1){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
$\displaystyle\iff|z|>(q_{\ell}-1){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell}+1\right).$

Thus if
$|z|>(q_{\ell}-1){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell}+1\right)$
and
$|z|>q_{\ell}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
then for all $0\leq k\leq q_{\ell}-1$, we obtain
$d(\theta_{k},\theta_{0}+k\alpha)<\frac{\pi}{2q_{\ell}}$.

$\blacksquare$$\theta_{0}+0\alpha$$\blacksquare$$\theta_{0}+1\alpha$$\blacksquare$$\theta_{0}+2\alpha$$\blacksquare$$\theta_{0}+3\alpha$$\blacksquare$$\theta_{0}+4\alpha$$\blacksquare$$\theta_{0}+5\alpha$$\blacksquare$$\theta_{0}+6\alpha$$\blacksquare$$\theta_{0}+7\alpha$$\bullet$$\theta_{0}$$\bullet$$\theta_{1}$$\bullet$$\theta_{2}$$\bullet$$\theta_{3}$$\bullet$$\theta_{4}$$\bullet$$\theta_{5}$$\bullet$$\theta_{6}$$\bullet$$\theta_{7}$
Figure 5:  The blue points are the $\\{k\alpha\\}$ with $0\leq k\leq 7$ and
$\alpha=(1+\sqrt{5})/2$. The red points represent the $\theta_{k}$.

    * –

Thus if
$|z|>q_{\ell}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell}+1\right)$,
we obtain for the second term:

$\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}g(\theta_{k})-g(\theta_{0}+k\alpha)\right|\leq\frac{1}{q_{\ell}}\operatorname{Var}(g).$

  * $(C)$

Third term: we use the Denjoy–Koksma inequality (Theorem 26) and Lemma 28 to
deduce:

$\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}g(\theta_{0}+k\alpha)+\frac{\Delta}{\pi}\right|=\left|\frac{1}{q_{\ell}}\sum_{k=0}^{q_{\ell}-1}g(\theta_{0}+k\alpha)-\frac{1}{2\pi}\int_{-\pi}^{\pi}g(x)dx\right|\leq\frac{1}{q_{\ell}}\operatorname{Var}(g).$

Thus if we have
$|z|>q_{\ell}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell}+1\right)$,
we deduce from Lemma 28 that

$\displaystyle\left|\frac{|T^{q_{\ell}}z|-|z|}{q_{\ell}}+\frac{\Delta}{\pi}\right|$
$\displaystyle\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}{|z|-q_{\ell}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}+\frac{2}{q_{\ell}}\operatorname{Var}(g)$
$\displaystyle\leq\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}{q_{\ell}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell}+1\right)-q_{\ell}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}+\frac{2}{q_{\ell}}\left(2|\Delta|+4{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\right)$
$\displaystyle\leq\frac{1}{q_{\ell}}\left(\frac{{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{\frac{2}{\pi}q_{\ell}}+4|\Delta|+8{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|})\right)=\frac{1}{q_{\ell}}\left(\left(8+\frac{\pi}{2q_{\ell}}\right){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+4|\Delta|\right)$

which proves the result. ∎

###  4.4 Conclusion of the proof

Let $\ell_{0}$ such that
$\frac{1}{q_{\ell_{0}}}\left(\left(8+\frac{\pi}{2q_{\ell_{0}}}\right){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+4|\Delta|\right)<\frac{|\Delta|}{\pi}$
and let us denote
$\epsilon=\frac{|\Delta|}{\pi}-\frac{1}{q_{\ell_{0}}}\left(\left(8+\frac{\pi}{2q_{\ell_{0}}}\right){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+4|\Delta|\right).$

  * •

If $\Delta>0$, then by Proposition 29 for
$|z|>q_{\ell_{0}}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell_{0}}+1\right)$
one has:

$|T^{q_{\ell_{0}}}z|\leq|z|-q_{\ell_{0}}\epsilon$

Thus we obtain

$\operatorname{\mathbf{A}}(T^{q_{\ell_{0}}})\subset
B\left(0,q_{\ell_{0}}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell_{0}}+1\right)-q_{\ell_{0}}\epsilon\right)\subset
B\left(0,q_{\ell_{0}}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell_{0}}+1\right)\right).$

By definition of $\operatorname{\mathbf{A}}(T)$, we obtain (see Remark 12):

$\operatorname{\mathbf{A}}(T)\subset\operatorname{\mathbf{A}}(T^{q_{\ell_{0}}})\subset
B\left(0,q_{\ell_{0}}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell_{0}}+1\right)\right).$

  * •

If $\Delta<0$, then by Proposition 29 we obtain for
$|z|>q_{\ell_{0}}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell_{0}}+1\right)$
:

$|T^{q_{\ell_{0}}}z|\geq|z|+q_{\ell_{0}}\epsilon$

Thus for every integer $n$ we deduce

$|T^{nq_{\ell_{0}}}z|\geq|z|+nq_{\ell_{0}}\epsilon$

and $z\notin
B\left(0,q_{\ell_{0}}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell_{0}}+1\right)\right)$.
Then by Remark 12 we conclude:

$\operatorname{\mathbf{B}}(T)\subset
B\left(0,q_{\ell_{0}}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{\ell_{0}}+1\right)\right).$

###  4.5 Example

Consider $T$ with the following parameters :

$\begin{array}[]{|c|c|c|c|c|c||c|c|c|c|}\hline\cr\alpha&a&C_{0}&C_{1}&\gamma&\mathbf{D}&\beta&\Delta&{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\\\
\hline\cr\frac{\sqrt{2}}{2}\pi&\frac{\sqrt{2}}{4}&-e^{1.14i}&1.5e^{1.14i}&\frac{\pi}{2}&i\mathbb{R}&1.14&-0.13>\Delta>-0.14&2.68<|3\sin(\alpha/2)|\leq
2.69\\\ \hline\cr\end{array}$

Since $\Delta<0$ the map $T$ is injective and is not bijective.

Notice that the continued fraction of $a$ is equal to
$a=[0,2,1,4,1,4,1,4,\ldots]$

$\begin{array}[]{|c||c|c|c|c|c|c|c|c|c|c|c|}\hline\cr\ell&0&1&2&3&4&5&6&7&8&9\\\
\hline\cr\hline\cr p_{\ell}&0&1&1&5&6&29&35&169&204&985\\\ \hline\cr
q_{\ell}&1&2&3&14&17&82&99&478&577&2786\\\ \hline\cr\end{array}$

We obtain

$\frac{1}{q_{8}}\left(\left(8+\frac{\pi}{2q_{8}}\right){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+4|\Delta|\right)<0.0383<\frac{|\Delta|}{\pi}<0.0460<\frac{1}{q_{7}}\left(\left(8+\frac{\pi}{2q_{7}}\right){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+4|\Delta|\right).$

and, see Figure 6, we deduce the bound
$M=q_{8}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{8}+1\right)<571283.$
Notice that the real size of the set seems to be much smaller.

In fact we can obtain a better bound due to the following remark:

######  Remark 30.

The bound from $M$ can be improved. In order to avoid technicalities we only
explain this for the above example. Consider some conjugation in order to
change the origin: let us define $p$ as the point of $\mathbf{D}$ which
minimizes both distances to $C_{0}$ and $C_{1}$. This point exists by a
compactness argument. If $p$ is the new origin, then we obtain new coordinates
for the centers of rotations, and
${\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T^{\prime}\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\leq{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$.
In this example we obtain:

The point $p$ is at the intersection of $\mathbf{D}$ with the bisection line
of $[C_{0},C_{1}]$. This point is given by $p=\frac{0.25}{\sin(1.14)}i\simeq
0.275i.$ The new centers of rotations are
$C^{\prime}_{0}=C_{0}-p\quad\text{and}\quad C^{\prime}_{1}=C_{1}-p.$ Then the
new map fulfills
${\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T^{\prime}\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}=2\max(|C^{\prime}_{0}|,|C^{\prime}_{1}|)\sin(\alpha/2)\simeq
2.249.$ We obtain
$\frac{1}{q_{8}}\left(\left(8+\frac{\pi}{2q_{8}}\right){\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T^{\prime}\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}+4|\Delta|\right)<\frac{|\Delta|}{\pi}$.
We deduce the new bound

$M^{\prime}=q_{8}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T^{\prime}\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{2}{\pi}q_{8}+1\right)<536048<M.$

Figure 6: Two pictures of the attractive set of the piecewise isometry of
Example 4.5.

##  5 Size of the limit set for the rational angle

Now we prove the same type of result for a piecewise rotation of rational
angle.

###  5.1 Notations and statement

Consider a non-bijective map
$T=T(\alpha,C_{0},C_{1},\mathbf{D},\gamma)\in\mathcal{T}$. In all of this
section we will use the following notation: $\alpha=\frac{p}{q}2\pi$ with
$q>2$ an even number and $\gcd(p,q)=1$. We will prove:

######  Theorem 31.

Let $T\in\mathcal{T}$ a non-bijective map with a rational angle $\alpha$ such
that $\alpha=\frac{p}{q}2\pi$ for some even number $q>2$ and $\gcd(p,q)=1$.
Then we obtain:

$M(T)\leq
q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{1}{2\tan(|\beta-\alpha/2-\gamma+\pi/2|)}+\frac{1}{2\tan(\pi/q)}+1\right).$

######  Remark 32.

Up to a conjugacy by $z\mapsto e^{-i(\gamma+\pi/2)}$ we can always assume that
$\gamma=\pi/2$ and that the discontinuity line is $\mathbf{D}=i\mathbb{R}$. We
will assume this in the proof and thus we will prove:

$M(T)\leq
q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{1}{2\tan(|\beta-\alpha/2|)}+\frac{1}{2\tan(\pi/q)}+1\right).$

We finish this paragraph with another remark.

######  Remark 33.

If we conjugue by $z\mapsto\bar{z}$ (see Remark 3), then we can assume
$\alpha\in]0,\pi[$. This allows us to assume $\sin(\alpha/2)>0$.

###  5.2 Vectors of translations

First of all, we consider the following partition of the complement of
$B(0,q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|})$,
see Figure 7:

  * •

For $k=0,\ldots,q-1$, we define the cone
$\mathbf{C}_{k}=\left\\{z\mid\frac{\pi}{2}+2\pi\frac{k}{q}<\arg(z)<\frac{\pi}{2}+2\pi\frac{(k+1)}{q}\right\\}$.

  * •

For $k=0,\ldots,q-1$, let us consider $\mathbf{E}_{k}$ which is the translate
of $\mathbf{C}_{k}$ such that its vertex is the point

$O_{k}=\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{\sin(\pi/q)}\exp\left(i\left(\frac{\pi}{q}+2\pi\frac{k}{q}+\frac{\pi}{2}\right)\right).$

  * •

For $k=0,\ldots,q-1$, let us denote $\mathbf{G}_{k}$ the strip which is the
intersection of the set of points of modulus bigger than
$q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
and of the closed
$q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$-neighborhood
of $\mathbf{C}_{k-1}\cap\mathbf{C}_{k}$ where
$\mathbf{C}_{-1}=\mathbf{C}_{q-1}$.

$\mathbf{E}_{0}$$\mathbf{E}_{1}$$\mathbf{E}_{2}$$\mathbf{E}_{3}$$\mathbf{E}_{4}$$\mathbf{E}_{5}$$\mathbf{E}_{6}$$\mathbf{E}_{7}$$\mathbf{E}_{8}$$\mathbf{E}_{9}$$O_{0}$$O_{1}$$O_{2}$$O_{3}$$O_{4}$$O_{5}$$O_{6}$$O_{7}$$O_{8}$$O_{9}$$\mathbf{G}_{0}$$\mathbf{G}_{1}$$\mathbf{G}_{2}$$\mathbf{G}_{3}$$\mathbf{G}_{4}$$\mathbf{G}_{5}$$\mathbf{G}_{6}$$\mathbf{G}_{7}$$\mathbf{G}_{8}$$\mathbf{G}_{9}$

$\mathbf{E}^{\prime}_{0}$$\mathbf{E}^{\prime}_{1}$$\mathbf{E}^{\prime}_{2}$$\mathbf{E}^{\prime}_{3}$$\mathbf{E}^{\prime}_{4}$$\mathbf{E}^{\prime}_{5}$$\mathbf{E}^{\prime}_{6}$$\mathbf{E}^{\prime}_{7}$$\mathbf{E}^{\prime}_{8}$$\mathbf{E}^{\prime}_{9}$$\mathbf{G}^{\prime}_{0}$$\mathbf{G}^{\prime}_{1}$$\mathbf{G}^{\prime}_{2}$$\mathbf{G}^{\prime}_{3}$$\mathbf{G}^{\prime}_{4}$$\mathbf{G}^{\prime}_{5}$$\mathbf{G}^{\prime}_{6}$$\mathbf{G}^{\prime}_{7}$$\mathbf{G}^{\prime}_{8}$$\mathbf{G}^{\prime}_{9}$

Figure 7:  The different zones $\mathbf{E}_{k}$, $\mathbf{G}_{k}$,
$\mathbf{E}^{\prime}_{k}$ and $\mathbf{G}_{k}^{\prime}$ (cf Lemma 34).

The map $T^{q}$ is clearly a piecewise translation. In the next lemma we
compute the vectors of translation which define this map. We identify complex
numbers with coordinates of points, in order to simplify.

######  Lemma 34.

Let us denote: $v=2\frac{\sin(\alpha/2)}{\sin(\pi/q)}|C_{1}-C_{0}|$ and
$w=2\frac{\sin(\alpha/2)}{\tan(\pi/q)}|C_{1}-C_{0}|.$ For $k=0,\dots,q-1$, let
us consider the vectors:

$v_{k}=ve^{i\left(\beta-\frac{\alpha}{2}+k\frac{2\pi}{q}+\frac{\pi}{q}\right)},\quad\text{and}\quad
w_{k}=we^{i\left(\beta-\frac{\alpha}{2}+k\frac{2\pi}{q}\right)}.$

Then the map $T^{q}$ is a piecewise translation with vectors $t_{i}$ with
$i\in I$ for some finite set $I$. We have for each $k\in\\{0,\dots,q-1\\}$.

  * •

The set $\mathbf{E}^{\prime}_{k}$, where the vector of translation of $T^{q}$
is $v_{k}$, is a subcone of $\mathbf{C}_{k}$ which contains $\mathbf{E}_{k}$,
see Figure 7.

  * •

The restriction of $T^{q}$ to $\mathbf{G}_{k}$ is a translation by some vector
$t_{k}$ which is equal to $v_{k-1},w_{k}$ or $v_{k}$.

More precisely,

    * –

consider the subset $\mathbf{G}^{\prime}_{k}$ of $\mathbf{G}_{k}$ where
$T^{q}$ is a translation by $w_{k}$. Then $\mathbf{G}^{\prime}_{k}$ is a
substrip of $\mathbf{G}_{k}$.

    * –

There exists a subset of $\mathbf{G}_{k}$ inside $\mathbf{C}_{k-1}$ where
$t_{k}=v_{k-1}$.

Recall that we have assumed that $\alpha$ satisfies $\sin(\alpha/2)>0$ (cf
Remark 33). Thus we have $v=|v_{k}|$ and $w=|w_{k}|$.

######  Remark 35.

A statement similar to Lemma 34 above appears already in [Bosh.Goet.03]. Their
statement, however, was not quite correct: they claimed that the vectors of
translation for $T^{q}$ restricted to $\mathbf{G}_{k}$ all have the same norm.
Moreover the description on the sets
$\mathbf{G}^{\prime}_{k},\mathbf{C}^{\prime}_{k}$ was not in their paper.

###### Proof of Lemma 34.

Let us fix $k=0,\dots,q-1$. We consider several different cases.

  * •

Let $z\in\mathbf{E}_{k}$. We have $e^{i\ell\alpha}O_{k}=O_{k+p\ell}$ by
definition of $O_{k}$ for all $\ell\in\mathbb{N}$. This gives us
$e^{i\ell\alpha}z\in\mathbf{E}_{k+p\ell}$. Then by Lemma 6 we obtain for
$\ell\leq q-1$,
$d(T^{\ell}(z),\mathbf{E}_{k+p\ell})<q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$.
We deduce

$T^{\ell}(z)\in\mathbf{C}_{k+p\ell}\quad 0\leq\ell\leq q-1.$

Thus the first $q-1$ iterates of $z$ under the action of $T$ meet all the
cones $\mathbf{C}_{i}$ and in particular no two subsequent iterates lie in the
same cone.

Thus the restriction of $T^{q}$ to $\mathbf{E}_{k}$ is an honest translation
and not a piecewise translation. Now we will compute the vector of translation
of $T^{q}$. We use the notation

$\eta(i)=\begin{cases}0\quad i\leq q/2,\\\ 1\quad i>q/2.\end{cases}$

We compute the vector $T^{q}(z)$ for $z\in\mathbf{E}_{k}$ by the following
formula, see Equation (4)

$T^{q}(z)=z+(1-e^{i\alpha})\sum_{n=0}^{q-1}C_{\eta(k+np)}e^{i\alpha(q-n-1)}=z+(1-e^{i\alpha})\sum_{n=0}^{q-1}C_{\eta(k+np)}e^{i\alpha(-n-1)}.$

Modulo $q$ the map $n\mapsto k+np$ is a bijection since $\gcd(p,q)=1$

$T^{q}(z)=z+(1-e^{i\alpha})\sum_{m=0}^{q-1}C_{\eta(m)}e^{i\alpha(-1-\frac{m-k}{p})}=z+(1-e^{i\alpha})\sum_{m=0}^{q-1}C_{\eta(m)}e^{2i\pi\frac{-p-m+k}{q}}.$

Thus we obtain :

$T^{q}(z)=z+(1-e^{i\alpha})C_{0}\sum_{m=0}^{q/2-1}e^{2i\pi\frac{-p-m+k}{q}}+(1-e^{i\alpha})C_{1}\sum_{m=q/2}^{q-1}e^{2i\pi\frac{-p-m+k}{q}}$
(5)

By means of well known trigonometric formula we obtain

$\displaystyle T^{q}(z)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{-p+k}{q}}\left(C_{0}\sum_{m=0}^{q/2-1}e^{-2i\pi\frac{m}{q}}+C_{1}\sum_{m=q/2}^{q-1}e^{-2i\pi\frac{m}{q}}\right)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}(C_{0}-C_{1})\frac{2}{1-e^{-2i\pi/q}}$
$\displaystyle=z+2\frac{\sin(\alpha/2)}{\sin(\pi/q)}|C_{1}-C_{0}|\exp\left(i\left(\beta+2\pi\frac{k-p}{q}+\frac{\alpha}{2}+\frac{\pi}{q}\right)\right)$
$\displaystyle=z+2\frac{\sin(\alpha/2)}{\sin(\pi/q)}|C_{1}-C_{0}|e^{i\left(\beta-\frac{\alpha}{2}+\frac{\pi}{q}+2\pi\frac{k}{q}\right)}$
$\displaystyle=z+v_{k}.$

  * •

Now consider $z\in\mathbf{G}_{k}$. We will denote by $R$ the rotation with
center $0$ and angle $\alpha$. If there exists some $n\geq 0$ such that
$e^{in\alpha}z$ and $T^{n}(z)$ are in the two different half planes defined by
$\operatorname{Re}(z)\geq 0$ and $\operatorname{Re}(z)\leq 0$, then this
implies that the two points are both in $\mathbf{G}_{0}$ or $\mathbf{G}_{s}$.
Indeed by Equation (6) they are at distance at most
$q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
of each other. Thus there are several choices for the different codings :
either the codings of the orbits of $T^{n}z,R^{n}z$ for $z\in\mathbf{G}_{k}$
differ in one position $j_{1}$, or else they differ at two positions
$j_{1},j_{2}$. Note that $j_{1},j_{2}$ are defined by $k+j_{1}p=q/2\mod q$ or
$k+j_{2}p=0\mod q$, according to the points in $\mathbf{G}_{0}$ or
$\mathbf{G}_{s}$. In other words: $R^{j_{1}}(z)$ and $T^{j_{1}}(z)$ are
contained in different half spaces.

Now we want to do the same computation as in the case $z\in\mathbf{E}_{k}$. We
can resume the situation as indicated in the lines below: the first one
concerns the coding of a point in $\mathbf{E}_{k}$, and the next three lines
correspond to one change in the coding if the change is at position $j_{1}$ or
$j_{2}$, or else to two changes in the coding.

$\begin{cases}u_{0}\dots u_{j_{1}}\dots u_{j_{2}}\dots\\\ u_{0}\dots
u_{j_{1}}\dots(1-u_{j_{2}})\dots\\\ u_{0}\dots(1-u_{j_{1}})\dots
u_{j_{2}}\dots\\\ u_{0}\dots(1-u_{j_{1}})\dots(1-u_{j_{2}})\dots\\\
\end{cases}$

We consider the two possibilities, starting from Equation (5).

    1. 1.

Assume that there is only one change. For example if one $0$ becomes a $1$, we
need to change in Equation (5) the $C_{0}\sum_{0}^{q/2-1}$ into
$C_{0}\sum_{1}^{q/2-1}$ and $C_{1}\sum_{q/2}^{q-1}$ into
$C_{1}\sum_{q/2}^{q}$. We obtain the following:

$\displaystyle T^{q}(z)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}\left(C_{0}\sum_{m=1}^{q/2-1}e^{2i\pi\frac{-m}{q}}+C_{1}\sum_{m=q/2}^{q}e^{2i\pi\frac{-m}{q}}\right)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}\left(C_{0}\sum_{m=0}^{q/2-1}e^{2i\pi\frac{-m}{q}}-C_{0}+C_{1}\sum_{m=s}^{2s-1}e^{2i\pi\frac{-m}{q}}+C_{1}\right)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}\left((C_{0}-C_{1})\sum_{m=0}^{q/2-1}e^{2\frac{-m}{q}}-(C_{0}-C_{1})\right)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}\left((C_{0}-C_{1})\frac{2}{1-e^{-2i\pi/q}}-(C_{0}-C_{1})\right).$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}(C_{0}-C_{1})\left(\frac{1+e^{-2i\pi/q}}{1-e^{-2i\pi/q}}\right).$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}(C_{0}-C_{1})\frac{2\cos(\pi/q)e^{-i\pi/q}}{1-e^{-2i\pi/q}}.$
$\displaystyle=z+\cos\left({\pi}/{q}\right)e^{-i\frac{\pi}{q}}v_{k}$
$\displaystyle=z+w_{k}.$

    2. 2.

The case where one $1$ becomes a $0$ is similar: we replace
$C_{0}\sum_{0}^{q/2-1}$ by $C_{0}\sum_{1}^{q/2}$ and $C_{1}\sum_{q/2}^{q-1}$
by $C_{1}\sum_{q/2+1}^{q}$. We obtain the following:

$\displaystyle T^{q}(z)$
$\displaystyle=z+v_{k}+(1-e^{i\alpha})C_{0}e^{2i\pi\frac{-p-q/2+k}{q}}-(1-e^{i\alpha})C_{1}e^{2i\pi\frac{-p-q/2+k}{q}}$
$\displaystyle=z+v_{k}-(1-e^{i\alpha})C_{0}e^{2i\pi\frac{-p+k}{q}}-(1-e^{i\alpha})C_{1}e^{2i\pi\frac{-p+k}{q}}$
$\displaystyle=z+\cos(\pi/q)e^{-i\frac{\pi}{q}}v_{k}$ $\displaystyle=z+w_{k}.$

    3. 3.

In the last case (where we exchange one $0$ and one $1$), we need to change
the $C_{0}\sum_{0}^{q/2-1}$ into $C_{0}\sum_{1}^{q/2}$ and
$C_{1}\sum_{q/2}^{q-1}$ into $C_{1}\sum_{q/2+1}^{q}$.

$\displaystyle T^{q}(z)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}\left(C_{0}\sum_{m=1}^{q/2}e^{2i\pi\frac{-m}{q}}+C_{1}\sum_{m=q/2+1}^{q}e^{2i\pi\frac{-m}{q}}\right)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}\left(C_{0}\sum_{m=0}^{q/2-1}e^{2i\pi\frac{-m}{q}}-2C_{0}+C_{1}\sum_{m=q/2}^{q-1}e^{2i\pi\frac{-m}{q}}+2C_{1}\right)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}\left(C_{0}\sum_{m=0}^{q/2-1}e^{2i\pi\frac{-m}{q}}+C_{1}\sum_{m=q/2}^{q-1}e^{2i\pi\frac{-m}{q}}+2(C_{1}-C_{0})\right).$

The same computation shows that

$\displaystyle T^{q}(z)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}(C_{0}-C_{1})\left(\frac{2}{1-e^{-2i\pi/q}}-2\right)$
$\displaystyle=z+(1-e^{i\alpha})e^{2i\pi\frac{k-p}{q}}(C_{0}-C_{1})\frac{2e^{-2i\pi/q}}{1-e^{-2i\pi/q}}$
$\displaystyle=z+v_{k}e^{-2i\pi/q}=z+v_{k-1}$

  * •

Let $z\in\mathbf{G}_{k}$, consider $t>0$. We claim that the codings of $z$ and
$z^{\prime}=z+te^{-i\pi/q}O_{k}$ have the same prefix of size $q+1$ (i.e
$T^{l}(z)$ and $T^{l}(tz)$ are in the same half planes for
$l\in\\{0,\ldots,q\\}$). This proves that $\mathbf{G}^{\prime}_{k}$ is a
substrip of $\mathbf{G}_{k}$, see Figure 7, since for each point
$z\in\mathbf{G}^{\prime}_{k}$ a half-line starting from $z$ will be inside
$\mathbf{G}^{\prime}_{k}$.

$\displaystyle T^{j_{1}}(z^{\prime})-T^{j_{1}}(z)$
$\displaystyle=e^{2i\pi\frac{p}{q}j_{1}}z^{\prime}+\text{Cst}-e^{2i\pi\frac{p}{q}j_{1}}z-\text{Cst}=e^{2i\pi\frac{p}{q}j_{1}}(z^{\prime}-z)=e^{2i\pi\frac{p}{q}j_{1}}te^{-i\frac{\pi}{q}}O_{j}$
$\displaystyle=e^{2i\pi\frac{p}{q}j_{1}}te^{-i\frac{\pi}{q}}\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{\sin(\pi/q)}e^{i\frac{\pi}{q}+2i\pi\frac{j}{q}+i\frac{\pi}{2}}=\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}t}{\sin(\pi/q)}ie^{2i\pi\frac{p}{q}j_{1}}e^{2i\pi\frac{j}{q}}$
$\displaystyle=\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}t}{\sin(\pi/q)}ie^{2i\pi\frac{pj_{1}+j}{q}}=\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}t}{\sin(\pi/q)}ie^{2i\pi\frac{1}{2}}=\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}t(-1)^{k}}{\sin(\pi/q)}i\in
i\mathbb{R}$
$\displaystyle\implies\operatorname{Re}(T^{j_{1}}(z))=\operatorname{Re}(T^{j_{1}}(z^{\prime})).$

Since by now all cases are treated, the claim is proved.

∎

######  Notations 2.

Let us denote $a_{k}$ (or $b_{k}$ respectively) the width of the half strip
$\mathbf{G}_{k}^{\prime}\cap\mathbf{C}_{k}$ (or
$\mathbf{G}_{k}^{\prime}\cap\mathbf{C}_{k-1}$ respectively), see Figure 8 and
Figure 9. Note that we have $0\leq a_{k},b_{k}\leq
q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$.

$v_{k}$$e^{i\left(\frac{\pi}{q}+2\pi\frac{k}{q}+\frac{\pi}{2}\right)}$$x$$x$$a_{k}$$b_{k+1}$$\bullet$$A_{k}$$A^{\prime}_{k}$$\bullet$$B_{k}$$B^{\prime}_{k}$

$w_{k}$$A_{k}$$\bullet$$A^{\prime}_{k}=B^{\prime}_{k-1}$$\bullet$$B_{k-1}$$\bullet$$e^{i\left(2\pi\frac{k}{q}+\frac{\pi}{2}\right)}$$\bullet$$0$$a_{k}$$b_{k}$$x$

Figure 8: Translation vectors $v_{k},w_{k}$ in $\mathbf{E}^{\prime}_{k}$ (red
part) and $\mathbf{G}_{k}^{\prime}$ (yellow) (cf Lemma 34).

###  5.3 Construction of polygons

####  5.3.1  The case $\Delta<0$

In this part we assume $\Delta<0$. By Definition of $\Delta$ and Remark 32, we
deduce that $\beta-\alpha/2$ belongs to $(0,\pi/2)$. The goal is to construct
for each positive number $x$, big enough, some polygon $\mathbf{P}_{x}$. These
polygons will have the following properties:

  * •

the disjoint union of all $\mathbf{P}_{x}$ cover the plane outside the ball
$B(0,q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|})$,

  * •

the map $x\mapsto P_{x}$ is increasing, with respect to the partial order on
subsets of the plane given by the inclusion,

  * •

if $z$ lies outside of outside $\mathbf{P}_{x}$, then the piecewise
translation $T^{q}$ maps $z$ to a point outside $\mathbf{P}_{x^{\prime}}$ for
some $x^{\prime}>x$, see Proposition 40.

This will be the key point to finish the proof of Theorem 31.

######  Definition 36.

Consider
$x>q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$,
then we define a polygon $\mathbf{P}_{x}$ with vertices
$A_{0},B_{0},...,A_{q-1},B_{q-1}$ where (see Figure 8):

  * •

the point $A_{k}$ is defined in
$\mathbf{E}^{\prime}_{k}\cap\mathbf{G}^{\prime}_{k}$ as the point such that
the orthogonal projection $A^{\prime}_{k}$ of $A_{k}$ on $[0,ie^{2i\pi k/q})$
has modulus $x$.

  * •

Similarly we define
$B_{k}\in\mathbf{E}^{\prime}_{k}\cap\mathbf{G}^{\prime}_{k+1}$ such that the
orthogonal projection $B^{\prime}_{k}$ of $B_{k}$ on $[0,ie^{2i\pi(k+1)/q})$
has modulus $x$.

Moreover we define $\delta_{k}(x)$ as the angle (taken in $[0,\pi/2]$) between
the lines $(A_{k}B_{k})$ and $(A^{\prime}_{k}B^{\prime}_{k})$.

######  Lemma 37.

For all
$x>q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
and for all $z$ on the boundary of $\mathbf{P}_{x}$ we obtain

$x\cos(\pi/q)\leq|z|\leq\displaystyle\sqrt{x^{2}+q^{2}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}.$

The proof of this lemma is left to the reader with the help of Figure 8.

######  Lemma 38.

Let us consider
$\bar{x}=q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{1}{2\tan(\beta-\alpha/2)}+\frac{1}{2\tan(\pi/q)}\right)$.
Then one has:

  * •

the map $x\mapsto\delta_{k}(x)$ is decreasing

  * •

$\lim_{x\to+\infty}\delta_{k}(x)=0$

  * •

For $x>\bar{x}$, we have $0\leq\delta_{k}(x)<\beta-\alpha/2$.

###### Proof.

We refer to Figure 9 . There is a polygon with vertices
$A^{\prime}_{k},A_{k},B_{k},B^{\prime}_{k}$ and we define points $C_{k},L_{k}$
such that:

  * •

If $a_{k}\geq b_{k+1}$, then $C_{k}$ belongs to $[A^{\prime}_{k},A_{k}]$ and
$|A^{\prime}_{k}C_{k}|=b_{k+1}$. The point $L_{k}$ is the orthogonal
projection of $A_{k}$ on $[B_{k},C_{k}]$.

  * •

If $b_{k+1}>a_{k}$, $C_{k}$ belongs to $[B^{\prime}k,Bk]$ and
$|B^{\prime}_{k}C_{k}|=a_{k}$. The point $L_{k}$ is defined as the orthogonal
projection of $A_{k}$ on $[A_{k},C_{k}]$.

Let us denote $r_{k}=\max(a_{k},b_{k+1})-\min(a_{k},b_{k+1})$ and
$s_{k}=\min(a_{k},b_{k+1})$. Note that $0\leq r_{k}\leq
q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$.

We present the proof in the case $a_{k}\geq b_{k+1}$; the other case is
similar. We have
$\widehat{B_{k}C_{k}A_{k}}=\widehat{B_{k}A^{\prime}_{k}A_{k}}=\pi/q$ and

$\tan(\delta_{k}(x))=\tan(\widehat{C_{k}B_{k}A_{k}})=\frac{|A_{k}L|}{|B_{k}L_{k}|}=\frac{|A_{k}L_{k}|}{|B_{k}C_{k}|-|C_{k}L_{k}|}=\frac{r_{k}\sin(\pi/q)}{|B_{k}C_{k}|-r_{k}\cos(\pi/q)}.$

We obtain from a classical geometry argument that
$|B_{k}C_{k}|=\frac{x-s_{k}}{x}|B^{\prime}_{k}A^{\prime}_{k}|=2(x-s_{k})\sin(\pi/q)$.
We deduce

$\tan(\delta_{k}(x))=\frac{r_{k}\sin(\pi/q)}{2(x-s_{k})\sin(\pi/q)-r_{k}\cos(\pi/q)}\implies\delta_{k}(x)=\arctan\left(\frac{r_{k}\sin(\pi/q)}{2(x-s_{k})\sin(\pi/q)-r_{k}\cos(\pi/q)}\right).$

Thus we obtain that the map $x\mapsto\delta_{k}(x)$ is decreasing and tends to
$0$ when $x$ approaches $+\infty$.

The angle $\delta_{k}(x)$ is maximal if
$r_{k}=q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}$
and $s_{k}=0$:

$\tan(\delta_{k}(x))\leq\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\sin(\pi/q)}{2x\sin(\pi/q)-q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\cos(\pi/q)}.$

Thus we have :

$\displaystyle\delta_{k}(x)<\beta-\alpha/2$
$\displaystyle\iff\tan(\delta_{k}(x))<\tan(\beta-\alpha/2)$
$\displaystyle\impliedby\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\sin(\pi/q)}{2x\sin(\pi/q)-q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\cos(\pi/q)}<\tan(\beta-\alpha/2)$
$\displaystyle\impliedby\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\sin(\pi/q)}{\tan(\beta-\alpha/2)}<2x\sin(\pi/q)-q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\cos(\pi/q)$
$\displaystyle\impliedby\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\sin(\pi/q)}{\tan(\beta-\alpha/2)}+q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\cos(\pi/q)<2x\sin(\pi/q)$
$\displaystyle\impliedby\frac{q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}{2\tan(\beta-\alpha/2)}+q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\frac{1}{2\tan(\pi/q)}<x$
$\displaystyle\impliedby\bar{x}<x.$

∎

######  Remark 39.

We can also show
$x\leq|A_{k}|=\sqrt{x^{2}+a_{k}^{2}}\leq\sqrt{x^{2}+q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}$
and
$x\leq|B_{k}|=\sqrt{x^{2}+b_{k}^{2}}\leq\sqrt{x^{2}+q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}}$.
Using Proposition 38, we deduce that $\frac{1}{x}P_{x}$ converges in the
Hausdorff topology to the regular polygon centered at $0$ with $q$ sides, and
with $i$ as a vertex.

$a_{k}$$r_{k}$$s_{k}=b_{k+1}$$2x\sin(\pi/q)$$A^{\prime}_{k}$$\bullet$$A_{k}$$\bullet$$\bullet$$B^{\prime}_{k}$$\bullet$$B_{k}$$\bullet$$\bullet$$C_{k}$$\bullet$$L_{k}$$\pi/q$$\pi/q$$\delta_{k}(x)$
Figure 9: The quantity $\delta_{k}(x)$ used in Lemma 38.

######  Proposition 40.

For any $x>\bar{x}$ let us define
$K_{x}=\min\big{(}v\sin(\beta-\alpha/2-\delta_{k}(x)),w\sin(\beta-\alpha/2)\big{)}$,
see Lemma 34 for notations. Then, if $z\in\mathbf{P}_{x}$, we have
$T^{q}(z)\in\mathbf{P}_{x^{\prime}}$ with $x^{\prime}\geq x+K_{x}$. Moreover,
the map $x\mapsto K_{x}$ is increasing.

###### Proof.

First of all, note that by Lemma 38 we have $\beta-\alpha/2-|\delta_{x}|>0$.
From now on we will use the notation from Figure 9.

We split the proof into four cases, due to Lemma 34. In each case we consider
$z\in P_{x}$ and compute $x^{\prime}$ such that $T^{q}z\in P_{x^{\prime}}$.

  * •

Assume $z\in\mathbf{E}_{k}^{\prime}$ and $T^{q}(z)\in\mathbf{E}_{k}^{\prime}$.
The angle between $v_{k}$ and the blue segment is equal to
$\beta-\alpha/2-\delta_{k}(x)$. Indeed, $\beta-\alpha/2$ is the angle between
$A^{\prime}_{k}B^{\prime}_{k}$ and $v_{k}$, see Figure 8. Thus we deduce
$x^{\prime}\geq x+v_{k}\sin(\beta-\alpha/2-\delta_{k}(x))$.

  * •

If $z\in\mathbf{E}_{k}^{\prime}$ and $T^{q}(z)\in\mathbf{G}_{k}^{\prime}$,
then $x^{\prime}-x$ is bigger than previous value (see green arrow in Figure
8).

  * •

Assume $z\in\mathbf{G}_{k}^{\prime}$ and $T^{q}(z)\in\mathbf{G}_{k}^{\prime}$

Then we obtain $x^{\prime}=x+w\sin(\beta-\alpha/2)$.

  * •

Assume $z\in\mathbf{G}_{k}^{\prime}$ and
$T^{q}(z)\in\mathbf{E}_{k-1}^{\prime}$. In this case we can fall back onto the
previous case $x^{\prime}\geq x+w\sin(\beta-\alpha/2)$.

Since $x\mapsto\delta_{k}(x)$ is decreasing, we deduce that $x\mapsto K_{x}$
is increasing. ∎

####  5.3.2 The case $\Delta>0$

The same construction as in the case $\Delta<0$ can be done with a familly of
polygons $\mathbf{Q}_{x}$:

  * •

As in Lemma 37 we can prove:

$\forall z\in\mathbb{C},\quad z\in Q_{x}\implies
x\cos(\pi/q)\leq|z|\leq\sqrt{x^{2}+q^{2}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}$

  * •

As in Proposition 40, for all $x>\bar{x}$ (with $\bar{x}$ defined as in
Proposition 38), there exists $L_{x}>0$ such that for $z\in\mathbf{Q}_{x}$ we
have $T^{q}(z)\in\mathbf{P}_{x^{\prime}}$ with $x^{\prime}\leq x-L_{x}$.
Moreover, the map $x\to L_{x}$ is increasing and tends to $0$ if
$x\to\bar{x}$.

###  5.4 Proof of Theorem 31

We consider two cases:

  * •

First case: $\Delta<0$.

Consider $z$ such that
$|z|>\sqrt{{\bar{x}}^{2}+q^{2}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}$.
By Lemma 37 we deduce $z\in\mathbf{P}_{x_{0}}$, with

$x_{0}\geq\sqrt{|z|^{2}-q^{2}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}>\sqrt{{\bar{x}}^{2}}=\bar{x}.$

Thus we have $x_{0}>\bar{x}$ and by Proposition 40, for all $n\in\mathbb{N}$,
we obtain $T^{nq}z\in\mathbf{P}_{x_{n}}$ with

$x_{n}\geq x_{0}+K_{x_{0}}+\cdots+K_{x_{n-1}}\geq x_{0}+nK_{x_{0}}\text{ since
$x\mapsto K_{x}$ is increasing}.$

Now by Lemma 37 we have

$|T^{nq}z|\geq x_{n}\cos(\pi/q)\geq(x_{0}+K_{x_{0}}n)\cos(\pi/q)\to+\infty.$

Thus we deduce
$M\leq\sqrt{{\bar{x}}^{2}+q^{2}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}$.
We obtain finally

$\displaystyle\sqrt{{\bar{x}}^{2}+q^{2}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}$
$\displaystyle=q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\sqrt{\left(\frac{1}{2\tan(|\beta-\alpha/2|)}+\frac{1}{2\tan(\pi/q)}\right)^{2}+1}$
$\displaystyle\leq
q{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}\left(\frac{1}{2\tan(|\beta-\alpha/2|)}+\frac{1}{2\tan(\pi/q)}+1\right).$

  * •

Second case: $\Delta>0$. Let us denote by $\operatorname{conv}(Q_{x})$ the
convex hull of $\mathbf{Q}_{x}$.

We fix $M>0$, and we deduce $B(0,M)\subset\operatorname{conv}(Q_{x_{0}})$ with
$x_{0}=M/\cos(\pi/q)$.

If $x_{0}\leq\bar{x}$ then the proof is finished.

Otherwise, consider a sequence $(x_{n})_{n}$ of real numbers, defined by
induction according to the following instructions: If $x_{n-1}>\bar{x}$, then
there exists $x_{n}>0$ such that
$T^{qn}B(0,M)\subset\operatorname{conv}(Q_{x_{n}})$ and
$x_{n}<x_{n-1}<\cdots<x_{0}$.

If there exists $n_{1}$ such that $x_{n_{1}}<\bar{x}$, then the proof is
finished. Otherwise the sequence $(x_{n})_{n}$ is a decreasing bounded
sequence, thus it converges to some value $x_{\infty}$. If
$x_{\infty}>\bar{x}$, then $L_{x}$ has limit value $0$ as $x$ tends to
$x_{\infty}$, and by a fact directly analogous to what has been shown in
Proposition 40 we deduce $L_{x_{\infty}}=0$, which is a contradiction. We
conclude $x_{\infty}=\bar{x}$.

The convex polygons $(\operatorname{conv}(Q_{x}))_{x}$ form a decreasing
family (again with respect to the inclusion) and from the definition of
$\operatorname{\mathbf{A}}(T)$ we deduce

$\operatorname{\mathbf{A}}(T)\subset\bigcap_{n\geq
0}T^{qn}B(0,M)\subset\bigcap_{n\geq
0}\operatorname{conv}(x_{n})=\operatorname{conv}(Q_{\bar{x}}).$

We conclude the proof with the observation that
$\operatorname{conv}(Q_{\bar{x}})\subset
B(0,\sqrt{{\bar{x}}^{2}+q^{2}{\left|\kern-0.86108pt\left|\kern-0.86108pt\left|T\right|\kern-0.86108pt\right|\kern-0.86108pt\right|}^{2}}).$

######  Remark 41.

The above derived result can be improved slightly: indeed, it is possible to
show $\operatorname{\mathbf{B}}(T)\subset\operatorname{conv}(P_{\bar{x}})$ and
$\operatorname{\mathbf{A}}(T)\subset\operatorname{conv}(Q_{\bar{x}})$.

###  5.5 Rational example similar to Section 4.5

We consider the irrational angle of Section 4.5 and truncate it with the
convergents of the continued fraction. In other words, we consider the
following:

$\begin{array}[]{|c|c|c|c|c|c||c|c|c|c|}\hline\cr\alpha&a&C_{0}&C_{1}&\gamma&\mathbf{D}&\beta&\Delta\\\
\hline\cr\frac{169}{478}2\pi&\frac{p_{7}}{q_{7}}=\frac{169}{478}&-e^{1.14i}&1.5e^{1.14i}&\frac{\pi}{2}&i\mathbb{R}&1.14&-0.13112>\Delta>-0.13111\\\
\hline\cr\end{array}$

Here we obtain $M\simeq 120\,968$. For the irrational angle, we have obtained
$M\simeq 571283$.

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