EPREUVE DE : Mathématiques (Piste 1 → 3 points; Piste 2 → 4 points; Piste 3 → 4points; Piste 4 → 9points )

The given complex numbers are:

$$a = \sqrt{2} \left(\cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right)\right)$$ $$b = \sqrt{2} \left(\cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right)\right)$$ $$c = \sqrt{2} \left(\cos\left(\frac{\pi}{3}\right) + i \sin\left(\frac{\pi}{3}\right)\right)$$

These are in the form $r(\cos\theta + i \sin\theta)$, where $r = \sqrt{2}$ for all three.

De Moivre's Theorem states that $(r(\cos\theta + i \sin\theta))^n = r^n(\cos(n\theta) + i \sin(n\theta))$.
For $a^6$:

$$a^6 = (\sqrt{2})^6 \left(\cos\left(6 \cdot \frac{\pi}{6}\right) + i \sin\left(6 \cdot \frac{\pi}{6}\right)\right)$$ $$a^6 = 2^3 (\cos(\pi) + i \sin(\pi))$$ $$a^6 = 8 (-1 + i \cdot 0)$$ $$a^6 = \boxed{-8}$$

For $b^6$:

$$b^6 = (\sqrt{2})^6 \left(\cos\left(6 \cdot \frac{\pi}{4}\right) + i \sin\left(6 \cdot \frac{\pi}{4}\right)\right)$$ $$b^6 = 2^3 \left(\cos\left(\frac{3\pi}{2}\right) + i \sin\left(\frac{3\pi}{2}\right)\right)$$ $$b^6 = 8 (0 + i \cdot (-1))$$ $$b^6 = \boxed{-8i}$$

For $c^6$:

$$c^6 = (\sqrt{2})^6 \left(\cos\left(6 \cdot \frac{\pi}{3}\right) + i \sin\left(6 \cdot \frac{\pi}{3}\right)\right)$$ $$c^6 = 2^3 (\cos(2\pi) + i \sin(2\pi))$$ $$c^6 = 8 (1 + i \cdot 0)$$ $$c^6 = \boxed{8}$$