THE 68th 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 $$i-1=a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots +1),$$ 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.