Let \(n\) be a natural number. Show that
\[ [1 / 2(-1+\sqrt{3} i)]^{n}+[1 / 2(-1-\sqrt{3} i)]^{n} \]
is equal to 2 if \(n\) is a multiple of 3 , and it is equal to \(-1\) otherwise.
Solution
The two numbers are \[ \frac{-1+\sqrt3\,i}{2}=e^{2\pi i/3}, \qquad \frac{-1-\sqrt3\,i}{2}=e^{-2\pi i/3}. \] Therefore their \(n\)th powers sum to \[ e^{2\pi i n/3}+e^{-2\pi i n/3} =2\cos\frac{2\pi n}{3}. \] If \(3\mid n\), this equals \(2\). If \(3\nmid n\), then \(n\equiv1\) or \(2\pmod3\), so the cosine is \(-1/2\) and the sum is \(-1\). Thus the stated dichotomy holds.