Romanian Mathematical Olympiad

If $a = \cos \frac{\pi}{2n} + i \sin \frac{\pi}{2n}$, then $|a| = 1$ and $a^n = i$. We obtain from here that
$$i - 1 = a^n - 1 = (a - 1)(a^{n-1} + a^{n-2} + \dots + 1),$$
and thus $\sqrt{2} \le |a - 1| \cdot n$.
On the other hand, $|a - 1| = |-2 \sin^2 \frac{\pi}{4n} + 2i \sin \frac{\pi}{4n} \cos \frac{\pi}{2n}| = 2 \sin \frac{\pi}{4n}$, which complete the proof.