Exercise 4 — Candidates who have not chosen the specialty course

The plane is equipped with the direct orthonormal coordinate system $(\mathrm{O}; \vec{u}, \vec{v})$. We are given the complex number $\mathrm{j} = -\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}$. The purpose of this exercise is to study some properties of the number j and to highlight a connection between this number and equilateral triangles.

Part A: properties of the number j

1. a. Solve in the set $\mathbb{C}$ of complex numbers the equation $$z^{2} + z + 1 = 0$$ b. Verify that the complex number j is a solution of this equation.
2. Determine the modulus and an argument of the complex number j, then give its exponential form.
3. Prove the following equalities: a. $j^{3} = 1$; b. $\mathrm{j}^{2} = -1 - \mathrm{j}$.
4. Let $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ be the images respectively of the complex numbers $1, \mathrm{j}$ and $\mathrm{j}^{2}$ in the plane.
   What is the nature of triangle PQR? Justify the answer.

Part B

Let $a, b, c$ be three complex numbers satisfying the equality $a + \mathrm{j}b + \mathrm{j}^{2}c = 0$. Let $\mathrm{A}, \mathrm{B}, \mathrm{C}$ denote the images respectively of the numbers $a, b, c$ in the plane.

1. Using question A-3.b., prove the equality: $a - c = \mathrm{j}(c - b)$.
2. Deduce that $\mathrm{AC} = \mathrm{BC}$.
3. Prove the equality: $a - b = \mathrm{j}^{2}(b - c)$.
4. Deduce that triangle ABC is equilateral.

The plane is equipped with a direct orthonormal coordinate system $(\mathrm{O}; \vec{u}, \vec{v})$.
The purpose of this exercise is to determine the non-zero complex numbers $z$ such that the points with affixes $1$, $z^2$ and $\dfrac{1}{z}$ are collinear. On the graph provided in the appendix, point A has affix 1.
Part A: study of examples
1. A first example
In this question, we set $z = \mathrm{i}$. a. Give the algebraic form of the complex numbers $z^2$ and $\dfrac{1}{z}$. b. Plot the points $N_1$ with affix $z^2$, and $P_1$ with affix $\dfrac{1}{z}$ on the graph given in the appendix. We note that in this case the points $\mathrm{A}$, $N_1$ and $P_1$ are not collinear.
2. An equation
Solve in the set of complex numbers the equation with unknown $z$: $z^2 + z + 1 = 0$.
3. A second example
In this question, we set: $z = -\dfrac{1}{2} + \mathrm{i}\dfrac{\sqrt{3}}{2}$. a. Determine the exponential form of $z$, then those of the complex numbers $z^2$ and $\dfrac{1}{z}$. b. Plot the points $N_2$ with affix $z^2$ and $P_2$, with affix $\dfrac{1}{z}$ on the graph given in the appendix. We note that in this case the points $\mathrm{A}$, $N_2$ and $P_2$ are collinear.
Part B
Let $z$ be a non-zero complex number. We denote by $N$ the point with affix $z^2$ and $P$ the point with affix $\dfrac{1}{z}$.

1. Establish that, for every complex number different from 0, we have: $$z^2 - \frac{1}{z} = \left(z^2 + z + 1\right)\left(1 - \frac{1}{z}\right)$$
2. We recall that if $\vec{U}$ is a non-zero vector and $\vec{V}$ is a vector with affixes respectively $z_{\vec{U}}$ and $z_{\vec{V}}$, the vectors $\vec{U}$ and $\vec{V}$ are collinear if and only if there exists a real number $k$ such that $z_{\vec{V}} = k z_{\vec{U}}$. Deduce that, for $z \neq 0$, the points $\mathrm{A}$, $N$ and $P$ defined above are collinear if and only if $z^2 + z + 1$ is a real number.
3. We set $z = x + \mathrm{i}y$, where $x$ and $y$ denote real numbers. Justify that: $z^2 + z + 1 = x^2 - y^2 + x + 1 + \mathrm{i}(2xy + y)$.
4. a. Determine the set of points $M$ with affix $z \neq 0$ such that the points $\mathrm{A}$, $N$ and $P$ are collinear. b. Trace this set of points on the graph given in the appendix.

We fix a natural number $p$ greater than or equal to 2. We denote $\mathcal { V } _ { p } = \left\{ \mathrm { e } ^ { \frac { 2 \mathrm { i } k \pi } { p } } ; k \in \llbracket 1 , p - 1 \rrbracket \right\}$, the set of $p$-th roots of unity different from 1. Let $a$ and $\lambda$ be non-zero complex numbers. Assume that $\frac { a } { \lambda } \notin \mathcal { V } _ { p }$, which means that either $a = \lambda$ or $\frac { a ^ { p } } { \lambda ^ { p } } \neq 1$. Prove that the complex number $\sum _ { j = 0 } ^ { p - 1 } \lambda ^ { p - 1 - j } a ^ { j }$ is non-zero.

Let $\omega$ denote a cube root of unity which is not equal to 1. Then the number of distinct elements in the set $$\left\{ \left( 1 + \omega + \omega ^ { 2 } + \cdots + \omega ^ { n } \right) ^ { m } \quad : \quad m , n = 1, 2, 3, \cdots \right\}$$ is
(A) 4
(B) 5
(C) 7
(D) infinite.

Let $\omega$ denote a cube root of unity which is not equal to 1. Then the number of distinct elements in the set $$\left\{ \left( 1 + \omega + \omega ^ { 2 } + \cdots + \omega ^ { n } \right) ^ { m } \quad : \quad m , n = 1, 2, 3, \cdots \right\}$$ is
(A) 4
(B) 5
(C) 7
(D) infinite

Let $\omega$ denote a cube root of unity which is not equal to 1. Then the number of distinct elements in the set $$\left\{ \left( 1 + \omega + \omega ^ { 2 } + \cdots + \omega ^ { n } \right) ^ { m } \quad : \quad m , n = 1, 2, 3, \cdots \right\}$$ is
(A) 4
(B) 5
(C) 7
(D) infinite

Let $i$ be a root of the equation $x^{2} + 1 = 0$ and let $\omega$ be a root of the equation $x^{2} + x + 1 = 0$. Construct a polynomial
$$f(x) = a_{0} + a_{1}x + \ldots + a_{n}x^{n}$$
where $a_{0}, a_{1}, \ldots, a_{n}$ are all integers such that $f(i + \omega) = 0$.

Let $1 , \omega , \omega ^ { 2 }$ be the cube roots of unity. Then the product $$\left( 1 - \omega + \omega ^ { 2 } \right) \left( 1 - \omega ^ { 2 } + \omega ^ { 2 ^ { 2 } } \right) \left( 1 - \omega ^ { 2 ^ { 2 } } + \omega ^ { 2 ^ { 3 } } \right) \cdots \left( 1 - \omega ^ { 2 ^ { 9 } } + \omega ^ { 2 ^ { 10 } } \right)$$ is equal to:
(A) $2 ^ { 10 }$
(B) $3 ^ { 10 }$
(C) $2 ^ { 10 } \omega$
(D) $3 ^ { 10 } \omega ^ { 2 }$

Let $z_k = \cos\left(\frac{2k\pi}{10}\right) + i\sin\left(\frac{2k\pi}{10}\right)$; $k = 1,2,\ldots,9$.
List I P. For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j = 1$ Q. There exists a $k \in \{1,2,\ldots,9\}$ such that $z_1 \cdot z = z_k$ has no solution $z$ in the set of complex numbers. R. $\frac{|1-z_1||1-z_2|\cdots|1-z_9|}{10}$ equals S. $1 - \sum_{k=1}^{9} \cos\left(\frac{2k\pi}{10}\right)$ equals
List II
1. True
2. False
3. 1
4. 2
P Q R S
(A) 1243
(B) 2134
(C) 1234
(D) 2143

Let $z = \frac{-1 + \sqrt{3}\,i}{2}$, where $i = \sqrt{-1}$, and $r, s \in \{1,2,3\}$. Let $P = \left[\begin{array}{cc} (-z)^r & z^{2s} \\ z^{2s} & z^r \end{array}\right]$ and $I$ be the identity matrix of order 2. Then the total number of ordered pairs $(r,s)$ for which $P^2 = -I$ is

Let $\omega \neq 1$ be a cube root of unity. Then the minimum of the set $$\left\{ \left| a + b \omega + c \omega ^ { 2 } \right| ^ { 2 } : a , b , c \text { distinct non-zero integers } \right\}$$ equals $\_\_\_\_$

If $\omega(\neq 1)$ is a cube root of unity, and $(1+\omega)^{7}=A+B\omega$. Then $(A,B)$ equals
(1) $(1,1)$
(2) $(1,0)$
(3) $(-1,1)$
(4) $(0,1)$

Let $\omega$ be a complex number such that $2\omega + 1 = z$ where $z = \sqrt{-3}$. If
$$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -\omega^2 - 1 & \omega^2 \\ 1 & \omega^2 & \omega^7 \end{vmatrix} = 3k$$
Then $k$ can be equal to:
(1) $-z$
(2) $\frac{1}{z}$
(3) $-1$
(4) $1$

If $\alpha , \beta \in C$ are the distinct roots of the equation $x ^ { 2 } - x + 1 = 0$, then $\alpha ^ { 101 } + \beta ^ { 107 }$ is equal to
(1) 2
(2) - 1
(3) 0
(4) 1

Let $z_0$ be a root of quadratic equation, $x^2 + x + 1 = 0$. If $z = 3 + 6iz_0^{81} - 3iz_0^{93}$, then $\arg(z)$ is equal to:
(1) 0
(2) $\frac{\pi}{4}$
(3) $\frac{\pi}{6}$
(4) $\frac{\pi}{3}$

Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$. If $a = (1 + \alpha)\sum_{k=0}^{100}\alpha^{2k}$ and $b = \sum_{k=0}^{100}\alpha^{3k}$, then $a$ and $b$ are the roots of the quadratic equation.
(1) $x^{2} + 101x + 100 = 0$
(2) $x^{2} - 102x + 101 = 0$
(3) $x^{2} - 101x + 100 = 0$
(4) $x^{2} + 102x + 101 = 0$

If $a$ and $b$ are real numbers such that $( 2 + \alpha ) ^ { 4 } = a + b \alpha$, where $\alpha = \frac { - 1 + i \sqrt { 3 } } { 2 }$, then $a + b$ is equal to:
(1) 9
(2) 24
(3) 33
(4) 57

Let integers $\mathrm { a } , \mathrm { b } \in [ - 3,3 ]$ be such that $\mathrm { a } + \mathrm { b } \neq 0$. Then the number of all possible ordered pairs $( \mathrm { a } , \mathrm { b } )$, for which $\left| \frac { z - \mathrm { a } } { z + \mathrm { b } } \right| = 1$ and $\left| \begin{array} { c c c } z + 1 & \omega & \omega ^ { 2 } \\ \omega & z + \omega ^ { 2 } & 1 \\ \omega ^ { 2 } & 1 & z + \omega \end{array} \right| = 1 , z \in \mathrm { C }$, where $\omega$ and $\omega ^ { 2 }$ are the roots of $x ^ { 2 } + x + 1 = 0$, is equal to $\_\_\_\_$ .

$z$ is a complex number, $\operatorname { Im } ( z ) \neq 0$ and $z ^ { 3 } = - 1$. Given this,
$$( z - 1 ) ^ { 10 }$$
Which of the following is this expression equal to?
A) $z + 1$
B) $z - 1$
C) $z$
D) $- z$
E) $- z - 1$