Determine all triplets of complex numbers $(x, y, z)$ satisfying the following system:

$$
\begin{aligned}
x+y+z & =1 \\
x y z & =1 \\
|x|=|y|=|z| & =1
\end{aligned}
$$
In the system, we find several elementary symmetric functions, which can encourage us to consider the unitary polynomial of degree 3 whose roots are $x, y, z$. This is the polynomial $t^{3}-t^{2}+a t-1$ (where we have used Viète's formulas) with $a=x y+y z+z x$. If we manage to determine $a$, it will remain to find the roots of the polynomial. Now:

$$
\begin{aligned}
a & =x y+y z+z x \\
& =\frac{1}{z}+\frac{1}{y}+\frac{1}{x} \\
& =\bar{z}+\bar{y}+\bar{x} \\
& =\overline{z+y+x} \\
& =1
\end{aligned}
$$

$x, y, z$ are therefore the roots of the polynomial $t^{3}-t^{2}+t-1=\left(t^{2}+1\right)(t-1)=(t+i)(t-i)(t-1)$, that is, $\{x, y, z\}=\{-i, 1, i\}$.

## Complex Numbers in Geometry

Who could believe that an object as abstract as a complex number could have so many applications in a domain as "visual" as geometry? And yet, one might guess this since we recall that to every complex number we can associate a pair $(a, b) \in \mathbb{R}^{\not t}$, which reminds us of the coordinates of a point.

Thus was born the complex plane: a plane equipped with an orthonormal coordinate system, with a real axis corresponding to the set of points with coordinates $(x, 0)$ where $x$ is a real number, and an imaginary axis corresponding to the set of points with coordinates $(0, y)$ where $y$ is a real number.

The advantage of the complex plane over the usual plane is that points in the complex plane have only one coordinate $z=a+i b$ called the affix. Here we find the interest of the modulus of a complex number. This modulus strangely resembles a distance in the Euclidean plane. Indeed, we can easily convince ourselves that the modulus of a complex number $z$ corresponds to the distance between the point in the complex plane with affix $z$ and the origin of the coordinate system. It is also allowed to calculate the distance between two complex numbers $z$ and $z^{\prime}$, which is the number $\left|z-z^{\prime}\right|$.

To identify the point $A$ with affix $z$, we can use its algebraic form or we can use its modulus and the angle formed by the real axis and the line $(O A)$. This angle $\theta$ is called the argument of a complex number and is denoted $\arg z$. The question then arises: how do we determine the argument of a complex number?

First, we need to calculate its modulus and note that

$$
z=|z|\left(\frac{x}{|z|}+i \frac{y}{|z|}\right)
$$

We then notice that the point with affix $\frac{x}{|z|}+i \frac{y}{|z|}$ lies on the unit circle, so there exists $\theta \in]-\pi, \pi]$ such that $\frac{x}{|z|}=\cos \theta$ and $\frac{y}{|z|}=\sin \theta$. This $\theta$ is the argument we were looking for. We then obtain what is called the trigonometric form of our complex number $z$:

$$
z=|z|(\cos \theta+i \sin \theta)
$$

We can then look for some properties of the argument: For example, what is the argument of $\bar{z}$? We can easily find that $\arg \bar{z}=-\arg z$. Or again, we might want to express the argument of a product $\arg z z^{\prime}$ in terms of $\arg z$ and $\arg z^{\prime}$. For this, we simply start by calculating the product of the two complex numbers in question, noting $\arg z=\theta$ and $\arg z^{\prime}=\theta^{\prime}$:

$$
\begin{aligned}
z z^{\prime} & =|z|(\cos \theta+i \sin \theta) \cdot\left|z^{\prime}\right|\left(\cos \theta^{\prime}+i \sin \theta^{\prime}\right) \\
& =\left|z z^{\prime}\right|\left(\cos \theta \cos \theta^{\prime}-\sin \theta \sin \theta^{\prime}+i\left(\cos \theta \sin \theta^{\prime}+\cos \theta^{\prime} \sin \theta\right)\right) \\
& =\left|z z^{\prime}\right|\left(\cos \left(\theta+\theta^{\prime}\right)+i \sin \left(\theta+\theta^{\prime}\right)\right)
\end{aligned}
$$

We thus obtain the following relation:

$$
\arg z z^{\prime}=\arg z+\arg z^{\prime}
$$

which reminds us of a property of the natural logarithm function and thus also of the exponential function. This is why we introduce the following notation, called the exponential form of a complex number $z$ with argument $\theta$:

$$
z=|z| e^{i \theta}
$$

This form proves very useful and allows us, in particular, to easily calculate the powers of complex numbers.

Now that we have several ways to denote a complex number, doing geometry in the complex plane should be easier. We recall that any point in the complex plane has an affix. This is also the case for vectors in the complex plane. The translation in the complex plane of a vector $\vec{u}$ therefore associates with any point in the plane with affix $z$ the point in the plane with affix $z+a$ where $a$ is the affix of the vector $\vec{u}$. Similarly, a similarity of ratio $k$, angle $\theta$, and center $A$ with affix $a$ associates with any point with affix $z$ the point in the plane with affix $k e^{i \theta}(z-a)+a$.

We can also define the dot product of two vectors $\vec{u}$ and $\vec{v}$ as the quantity $\vec{u} \cdot \vec{v}=\frac{1}{2}\left(z \overline{z^{\prime}}+\bar{z} z^{\prime}\right)$ and we recover the equivalence with the dot product in the real plane.

This product is zero if and only if the vectors $\vec{u}$ and $\vec{v}$ are orthogonal, indeed, the oriented angle between $\vec{u}$ and $\vec{v}$ corresponds to $\arg \frac{z^{\prime}}{z}$ and thus if $\vec{u} \cdot \vec{v}=0$ with a little

In the same way, the points with respective affixes $z_{1}, z_{2}, z_{3}$ and $z_{4}$ are concyclic or collinear if and only if

$$
\arg \frac{z_{4}-z_{1}}{z_{3}-z_{1}}=\arg \frac{z_{4}-z_{2}}{z_{3}-z_{2}}
$$

or if and only if

$$
\arg \frac{z_{4}-z_{1}}{z_{3}-z_{1}} \cdot \frac{z_{3}-z_{2}}{z_{4}-z_{2}}=0
$$

and thus finally if and only if the quantity $\frac{z_{4}-z_{1}}{z_{3}-z_{1}} \cdot \frac{z_{3}-z_{2}}{z_{4}-z_{2}}$ called the cross-ratio of $z_{1}, z_{2}, z_{3}$ and $z_{4}$ is real.