\section*{Exercise 3 -- Candidates who have NOT followed the specialization course}
The complex plane is equipped with a direct orthonormal coordinate system.\\
We consider the equation
$$(E): \quad z^2 - 2z\sqrt{3} + 4 = 0$$

\begin{enumerate}
  \item Solve the equation $(E)$ in the set $\mathbb{C}$ of complex numbers.
  \item We consider the sequence $(M_n)$ of points with affixes $z_n = 2^n \mathrm{e}^{\mathrm{i}(-1)^n \frac{\pi}{6}}$, defined for $n \geqslant 1$.\\
    a. Verify that $z_1$ is a solution of $(E)$.\\
    b. Write $z_2$ and $z_3$ in algebraic form.\\
    c. Plot the points $M_1, M_2, M_3$ and $M_4$ on the figure provided in the appendix and draw, on the figure provided in the appendix, the segments $[M_1, M_2]$, $[M_2, M_3]$ and $[M_3, M_4]$.
  \item Show that, for every integer $n \geqslant 1$, $z_n = 2^n\left(\frac{\sqrt{3}}{2} + \frac{(-1)^n \mathrm{i}}{2}\right)$.
  \item Calculate the lengths $M_1M_2$ and $M_2M_3$.
\end{enumerate}

For the rest of the exercise, we admit that, for every integer $n \geqslant 1$, $M_nM_{n+1} = 2^n\sqrt{3}$.

5. We denote $\ell^n = M_1M_2 + M_2M_3 + \cdots + M_nM_{n+1}$.\\
a. Show that, for every integer $n \geqslant 1$, $\ell^n = 2\sqrt{3}(2^n - 1)$.\\
b. Determine the smallest integer $n$ such that $\ell^n \geqslant 1000$.