The complex plane is equipped with a direct orthonormal coordinate system $( \mathrm { O } , \vec { u } , \vec { v } )$. We denote by i the complex number such that $\mathrm { i } ^ { 2 } = - 1$. We consider the point A with affixe $z _ { \mathrm { A } } = 1$ and the point B with affixe $z _ { \mathrm { B } } = \mathrm { i }$. To any point $M$ with affixe $z _ { M } = x + \mathrm { i } y$, with $x$ and $y$ two real numbers such that $y \neq 0$, we associate the point $M ^ { \prime }$ with affixe $z _ { M ^ { \prime } } = - \mathrm { i } z _ { M }$. We denote by $I$ the midpoint of the segment [AM]. The purpose of the exercise is to show that for any point $M$ not belonging to (OA), the median $( \mathrm { O } I )$ of the triangle $\mathrm { OA} M$ is also a height of the triangle $\mathrm { OB } M ^ { \prime }$ (property 1) and that $\mathrm { B } M ^ { \prime } = 2 \mathrm { O } I$ (property 2).

1. In this question and only in this question, we take $z _ { M } = 2 \mathrm { e } ^ { - \mathrm { i } \frac { \pi } { 3 } }$. a. Determine the algebraic form of $z _ { M }$. b. Show that $z _ { M ^ { \prime } } = - \sqrt { 3 } - \mathrm { i }$. Determine the modulus and an argument of $z _ { M ^ { \prime } }$. c. Place the points $\mathrm { A } , \mathrm { B } , M , M ^ { \prime }$ and $I$ in the coordinate system $( \mathrm { O } , \vec { u } , \vec { v } )$ using 2 cm as the graphical unit. Draw the line $( \mathrm { O } I )$ and quickly verify properties 1 and 2 using the graph.
2. We return to the general case by taking $z _ { M } = x + \mathrm { i } y$ with $y \neq 0$. a. Determine the affixe of point $I$ as a function of $x$ and $y$. b. Determine the affixe of point $M ^ { \prime }$ as a function of $x$ and $y$. c. Write the coordinates of points $I$, B and $M ^ { \prime }$. d. Show that the line $( \mathrm { O } I )$ is a height of the triangle $\mathrm { OB } M ^ { \prime }$. e. Show that $\mathrm { B } M ^ { \prime } = 2 \mathrm { O } I$.

Throughout the problem, the set $\mathbf { C }$ of complex numbers is considered as the Euclidean affine plane equipped with its canonical orthonormal coordinate system $(0,1,i)$ (where $i^2 = -1$). We denote by $K$ the set of triplets $(\alpha, \beta, \gamma)$ of $\mathbf{R}^3$ consisting of three non-negative real numbers such that $\alpha + \beta + \gamma = 1$. If $(a,b,c) \in \mathbf{C}^3$, we denote by $\widehat{abc}$ the filled triangle defined by $\widehat{abc} = \{\alpha a + \beta b + \gamma c / (\alpha,\beta,\gamma) \in K\}$. We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$.
a) Let $a \in \mathbf{C}$ and $\theta \in \mathbf{R}$. Prove that the image $z'$ of the complex number $z$ by the reflection whose axis is the line passing through $a$ and directed by $\mathrm{e}^{\mathrm{i}\theta}$ satisfies the relation: $$z' - a = \mathrm{e}^{2\mathrm{i}\theta} \overline{(z-a)}$$
b) Establish a relation analogous to that of the previous question between a complex number $z$ and its image $z'$ by the homothety with center $a$ and ratio $\rho > 0$.
c) Prove that $\phi_0$ is the composition of a reflection whose axis we will specify and a homothety with strictly positive ratio to be specified and whose center belongs to the axis of the reflection. Prove an analogous property for $\phi_1$. Are these decompositions unique?

Throughout the problem, the set $\mathbf{C}$ of complex numbers is considered as the Euclidean affine plane equipped with its canonical orthonormal coordinate system $(0,1,i)$ (where $i^2=-1$). We denote by $K$ the set of triplets $(\alpha,\beta,\gamma)$ of $\mathbf{R}^3$ consisting of three non-negative real numbers such that $\alpha+\beta+\gamma=1$. If $(a,b,c)\in\mathbf{C}^3$, we denote by $\widehat{abc}$ the filled triangle defined by $\widehat{abc} = \{\alpha a + \beta b + \gamma c / (\alpha,\beta,\gamma)\in K\}$. We denote $\tau = \widehat{(-1)\,1\,\mathrm{i}}$. We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$.
What is the image of a filled triangle $\widehat{abc}$ by $\phi_0$ and by $\phi_1$? Determine $\phi_0(\tau)$ and $\phi_1(\tau)$.

We denote by $\mathcal{E}$ the set of continuous maps $g$ from $[0,1]$ to $\mathbf{C}$ such that $g(0)=-1$ and $g(1)=1$.
Determine the unique element $f_0$ of $\mathcal{E}$ which is affine.

We denote by $\mathcal{E}$ the set of continuous maps $g$ from $[0,1]$ to $\mathbf{C}$ such that $g(0)=-1$ and $g(1)=1$. We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. If $g\in\mathcal{E}$, we denote by $Tg$ the map from $[0,1]$ to $\mathbf{C}$ defined by: $$Tg(x) = \phi_0(g(2x)) \text{ if } x\in\left[0,\frac{1}{2}\right] \text{ and } Tg(x) = \phi_1(g(2x-1)) \text{ if } x\in\left]\frac{1}{2},1\right]$$
Show that $Tg\in\mathcal{E}$ for all $g\in\mathcal{E}$.

We denote by $\mathcal{E}$ the set of continuous maps $g$ from $[0,1]$ to $\mathbf{C}$ such that $g(0)=-1$ and $g(1)=1$. The norm of uniform convergence on the $\mathbf{C}$-vector space of continuous maps from $[0,1]$ to $\mathbf{C}$ is denoted $\|\cdot\|_\infty$. We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. If $g\in\mathcal{E}$, we denote by $Tg$ the map from $[0,1]$ to $\mathbf{C}$ defined by: $$Tg(x) = \phi_0(g(2x)) \text{ if } x\in\left[0,\frac{1}{2}\right] \text{ and } Tg(x) = \phi_1(g(2x-1)) \text{ if } x\in\left]\frac{1}{2},1\right]$$
Let $g_1$ and $g_2$ be two elements of $\mathcal{E}$. Prove that: $$\|Tg_2 - Tg_1\|_\infty = \frac{1}{\sqrt{2}}\|g_2 - g_1\|_\infty$$

We denote $\tau = \widehat{(-1)\,1\,\mathrm{i}}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$ and $f([0,1])\subset\mathbf{C}$. We denote by $\mathbf{Z}\left[\frac{1}{2}\right]$ the set of rationals of the form $\frac{k}{2^n}$ where $k\in\mathbf{Z}$ and $n\in\mathbf{N}$.
a) Show that $f\left([0,1]\cap\mathbf{Z}\left[\frac{1}{2}\right]\right)\subset\tau$.
b) Show that $f([0,1])\subset\tau$.

We denote $\tau = \widehat{(-1)\,1\,\mathrm{i}}$, $\tau_0 = \widehat{0\,(-1)\,(-\mathrm{i})}$, $\tau_1 = \widehat{0\,1\,\mathrm{i}}$. We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$ and $f([0,1])\subset\tau$.
Conversely, let $z\in\tau$.
a) Show that we can define two sequences $(z_n)_{n\geq 0}$ and $(r_n)_{n\geq 1}$ in the following way:

• $z_0 = z$ and, if $n\geq 1$:
• if $z_{n-1}\in\tau_0$ then $r_n = 0$ and $z_n = (\phi_0)^{-1}(z_{n-1})$
• otherwise $r_n = 1$ and $z_n = (\phi_1)^{-1}(z_{n-1})$.

Prove that, for every integer $n\in\mathbb{N}$, $z_n$ belongs to $\tau$.
b) Prove that $f\left(\sum_{n=1}^{\infty}\frac{r_n}{2^n}\right) = z$ (one may express $z$ in terms of $z_n$ and the $\phi_{r_i}$).
c) Write a function that takes as argument a complex number $z$ (which we will assume is in $\tau$) and a real number $\epsilon$ and which returns an approximate value to within $\epsilon$ of a preimage of $z$.

The map $f\in\mathcal{E}$ satisfies $Tf = f$, $f([0,1]) = \tau$ where $\tau = \widehat{(-1)\,1\,\mathrm{i}}$, and $f(x) = -\overline{f(1-x)}$ for all $x\in[0,1]$.
a) Prove that $f$ is not injective (one may use the relation $f(1-x) = -\overline{f(x)}$).
b) More generally show that there exists no continuous bijection from $[0,1]$ onto $\tau$ (one may use an argument of arc-connectedness).

We denote by $\phi_0$ and $\phi_1$ the maps from $\mathbf{C}$ to $\mathbf{C}$ defined by: $\phi_0(z) = \frac{1+\mathrm{i}}{2}\bar{z} + \frac{-1+\mathrm{i}}{2}$ and $\phi_1(z) = \frac{1-\mathrm{i}}{2}\bar{z} + \frac{1+\mathrm{i}}{2}$. The map $f\in\mathcal{E}$ satisfies $Tf = f$ and $f([0,1]) = \tau$ where $\tau = \widehat{(-1)\,1\,\mathrm{i}}$.
a) For $(i,j)\in\{0,1\}^2$, determine the complex expression of $\phi_i\circ\phi_j$, recognize it, specify its fixed point and the image of $\tau$. Make a drawing.
b) Let $r_1, r_2, \ldots, r_p$ be elements of $\{0,1\}$. Prove that $\phi = \phi_{r_1}\circ\phi_{r_2}\circ\cdots\circ\phi_{r_p}$ has a unique fixed point which we will not necessarily try to express simply.
c) Exhibit, using the map $f$, a fixed point of $\phi$.
d) Show that the set $X$ of complex numbers $z$ which are fixed points of the composition of a finite number of maps $\phi_0$ and $\phi_1$ is dense in $\tau$.

The map $f\in\mathcal{E}$ satisfies $Tf = f$ and $f([0,1]) = \tau$. Suppose that $f$ is differentiable on $[0,1]$.
Let $x\in[0,1]$, $(\alpha_n)_{n\geq 1}$ and $(\beta_n)_{n\geq 1}$ be two sequences of elements of $[0,1]$, convergent to $x$ and such that $\alpha_n \leq x \leq \beta_n$ and $\alpha_n < \beta_n$ for all $n$.
Show that the sequence with general term $\frac{f(\beta_n) - f(\alpha_n)}{\beta_n - \alpha_n}$ converges to $f'(x)$.

The integer part of the real number $x$ is denoted $[x]$. For every real $x$ and every non-zero natural number $n$: $r_n(x) = [2^n x] - 2[2^{n-1}x]$. The map $f\in\mathcal{E}$ satisfies $Tf = f$ and $f([0,1]) = \tau$.
Let $x\in[0,1]$.
a) If $x\in[0,1[$, by choosing: $$\alpha_n = \frac{r_1(x)}{2} + \cdots + \frac{r_n(x)}{2^n} \text{ and } \beta_n = \alpha_n + \sum_{k=n+1}^{\infty}\frac{1}{2^k},$$ prove that $f$ is not differentiable at $x$.
b) Prove that $f$ is not differentiable at $1$.

A particle $P$ starts from the point $z _ { 0 } = 1 + 2 i$, where $i = \sqrt { - 1 }$. It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point $z _ { 1 }$. From $z _ { 1 }$ the particle moves $\sqrt { 2 }$ units in the direction of the vector $\hat { i } + \hat { j }$ and then it moves through an angle $\frac { \pi } { 2 }$ in anticlockwise direction on a circle with centre at origin, to reach a point $z _ { 2 }$. The point $z _ { 2 }$ is given by
(A) $6 + 7 i$
(B) $- 7 + 6 i$
(C) $7 + 6 i$
(D) $- 6 + 7 i$

We have a triangle ABC on the complex plane whose vertices are the three points $\mathrm { A } ( \alpha )$, $\mathrm { B } ( \beta )$ and $\mathrm { C } ( \gamma )$ that satisfy
$$\frac { \gamma - \alpha } { \beta - \alpha } = 1 - i$$
(In the following, the range of an argument $\theta$ is $0 \leqq \theta < 2 \pi$.)
(1) When we express the complex number $\frac { \gamma - \alpha } { \beta - \alpha }$ in polar form, we have
$$\frac { \gamma - \alpha } { \beta - \alpha } = \sqrt { \mathbf { N } } \left( \cos \frac { \mathbf { O } } { \mathbf { P } } \pi + i \sin \frac { \mathbf { O } } { \mathbf { P } } \pi \right) .$$
Hence we see that point C is the point resulting from rotating point B by $\frac { \square \mathbf { Q } } { \mathbf{R} } \pi$ around point A and then changing its distance from point A to its distance multiplied by $\sqrt { \mathbf { S } }$. From this we also see that the absolute value and the argument of the complex number $w = \frac { \gamma - \beta } { \alpha - \beta }$ are
$$| w | = \mathbf { T } \quad \text { and } \quad \arg w = \frac { \mathbf { U } } { \mathbf { 4 } } \pi .$$
(2) If $\alpha + \beta + \gamma = 0$, then we have that
$$| \alpha | : | \beta | : | \gamma | = \sqrt { \mathbf { W } } : \sqrt { \mathbf { X } } : \sqrt { \mathbf { Y } } .$$