Calculate $\sum_{i=0}^{\lfloor n / 3\rfloor}\binom{n}{3 i}$.
We are inspired by the proof $2 \sum_{i=0}^{\lfloor n / 2\rfloor}\binom{n}{2 i}=(1+1)^{n}+(1-1)^{n}$, given by the binomial formula:

$$
3 \sum_{i=0}^{\lfloor n / 3\rfloor}\binom{n}{3 i}=(1+1)^{n}+(1+j)^{n}+\left(1+j^{2}\right)^{n}
$$

We have used the fact that $1+j+j^{2}=0$. The desired result is therefore $\frac{2^{n}+2 \cos n \pi / 3}{3}$.

We could continue indefinitely, as complex numbers are important in all areas of mathematics. I will conclude by giving my favorite example of the use of complex numbers: the prime number theorem. The most natural proofs of this difficult theorem all involve complex numbers:

Theorem 8. The proportion of prime numbers less than $n$ is equivalent to $\frac{\log n}{n}$.

## - Proof of the D'Alembert-Gauss Theorem -

We have seen how useful it can be to take the roots of a polynomial, so useful that we will use this technique to give a sketch of the proof of the D'Alembert-Gauss theorem itself. Without going into details, the construction made in the first part, consisting of adding to a set $\mathbb{R}$ the root of an algebraic equation, is in fact perfectly general. Starting from a polynomial $P$ with real coefficients, this construction allows us to construct a set $K$ containing $\mathbb{C}$ (which also has an addition and a multiplication compatible with that of $\mathbb{C}$) on which the polynomial $P$ is split. This set is called the splitting field of the polynomial $P$. We now need a tool to relate these roots, which belong to a set $K$ that we control very poorly, to the coefficients of the polynomial, which are real. This link is provided by the theory of symmetric polynomials.

Definition 9. A polynomial in the variables $\left(X_{1}, \ldots, X_{n}\right)$ is symmetric if it is invariant under permutation of its variables.

Definition 10. Let $1 \leq k \leq n$. We define the $k$-th elementary symmetric polynomial $\sigma_{k}$ in the variables $\left(X_{1}, \ldots, X_{n}\right)$ by

$$
\sigma_{k}\left(X_{1}, \ldots, X_{n}\right)=\sum_{S \in\{1, \ldots, n\},|S|=k} \prod_{i \in S} X_{i}
$$

## Proposition 11.

$$
\prod_{i=1}^{n}\left(X-\lambda_{i}\right)=X^{n}+\sum_{k=1}^{n}(-1)^{n} \sigma_{k}\left(\lambda_{1}, \ldots, \lambda_{n}\right) X^{n-k}
$$

This proposition is almost obvious: it suffices to expand. We can therefore express the coefficients (real) of our polynomial $P$ in terms of its roots (belonging to its splitting field). We even have more:

Theorem 12. Let $Q$ be a symmetric polynomial with real coefficients in $n$ variables $\left(X_{1}, \ldots, X_{n}\right)$. Then there exists a polynomial $P$ with real coefficients and $n$ variables such that:

$$
\left.Q\left(X_{1}, \ldots, X_{n}\right)=P\left(\sigma_{1}\left(X_{1}, \ldots, X_{n}\right)\right), \ldots, \sigma_{n}\left(X_{1}, \ldots, X_{n}\right)\right)
$$

Proof. This theorem has a simple algorithmic proof: we order the monomials of $Q$ lexicographically (i.e., we associate with each monomial the sequence of exponents of the variables $\left(X_{1}, \ldots, X_{n}\right)$, and we classify this with the dictionary order). We then subtract from $Q$ a term in the symmetric polynomials chosen to kill the largest monomial, and we iterate.

Thus, any symmetric polynomial in the roots of $P$ (in the splitting field) can be expressed in terms of the coefficients of $P$ (the reals), and is therefore real. We are now able to prove the D'Alembert-Gauss theorem.

We reason by induction on the 2-adic valuation of the degree of $p$. The base case is that every polynomial of odd degree has a root: we look at its values at $\pm \infty$ (which is an analytic result, it is indeed impossible to give a purely algebraic proof of the D'Alembert-Gauss theorem).

Let us show the result at rank $k$. We write the degree $n$ of $P$ in the form $2^{k} m$, with $m$ odd, and we denote its roots $\left(r_{1}, \ldots, r_{n}\right)$, in $K$. Let $t$ be a real number, we define:

$$
q_{t}(z)=\prod_{1 \leq i, j \leq n}\left(z-r_{i}-r_{j}-t r_{i} r_{j}\right)
$$

The coefficients of this polynomial are symmetric in the $r_{i}$, and are therefore real according to the previous discussion. Since it is of degree $2^{k-1} m(n-1)$, we can apply the induction hypothesis: one of its roots $r_{i}+r_{j}+t r_{i} r_{j}$ is in fact complex. Since this result is true for all $t$, and there are infinitely many real numbers, we can even find a pair $(i, j)$ and two real numbers $t$ and $s$ such that $r_{i}+r_{j}+s r_{i} r_{j}$ and $r_{i}+r_{j}+t r_{i} r_{j}$ are both complex. Thus, $r_{i} r_{j}$ and $r_{i}+r_{j}$ are complex, and $r_{i}$ and $r_{j}$ are therefore the roots of the polynomial with complex coefficients $X^{2}-\left(r_{i}+r_{j}\right) x+r_{i} r_{j}$. By the usual formulas, these roots are complex, which closes the induction.

## - Applications to Plane Geometry -

The idea of using complex numbers in geometry is to associate each point in the plane with its complex coordinate (which we call its affix), and to do the calculations on these coordinates. We will denote the point in uppercase, and its affix in lowercase. This very powerful method has many advantages over classical Cartesian geometry. Essentially:

- we work with one coordinate instead of two,
- trigonometric calculations are violently simplified by the exponential notation,
- we have a new operation, multiplication, which has very strong geometric properties.

The following few examples of results are thus immediate consequences of the geometric interpretation of complex multiplication.

Proposition 13. $\quad-A B \perp C D \Leftrightarrow \frac{d-c}{b-a}$ is purely imaginary

- $A B \| C D$ or $A B C D$ are collinear $\Leftrightarrow \frac{d-c}{b-a}$ is real

More generally, the oriented angle between the lines $A B$ and $C D$ (in this order) is the argument of $\frac{d-c}{b-a}$. The interpretation of multiplication in terms of similarities shows, for example:

Proposition 14. The triangles $A B C$ and $D E F$ are similar (with the same orientation) if and only if

$$
\frac{a-c}{b-c}=\frac{d-f}{e-f}
$$

Finally, the inscribed angle theorem is rewritten as:

Proposition 15. The points $A B C D$ are concyclic or collinear if and only if

$$
\frac{a-c}{b-c}: \frac{a-d}{b-d} \in \mathbb{R}
$$

The symbol : here corresponds to a division. The quantity used in this proposition is called the complex cross-ratio, and is also a key element of complex projective geometry.