Pure imaginary numbers : $ ( j^{2n} - j^n ) \in i\Bbb R $? Considering the complex number $j$ such that $$ j = \frac{-1}{2} + i\frac{\sqrt3}{2} $$
Prove that $ \forall n \in \Bbb Z : $
$$ ( j^{2n} - j^n ) \in i\Bbb R $$
( $i\Bbb R$ being the set of pure imaginary numbers)


$$j=e^{2\pi i/3}$$
$$j^m=e^{2\pi i m/3}=\cos\dfrac{2m\pi }3+i\sin\dfrac{2m\pi }3$$ using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i*\sin(\varphi)$?
Now the real part $j^{2n}-j^n$
$$=\cos\dfrac{4n\pi }3-\cos\dfrac{2n\pi}3$$ $$=-2\sin2n\pi\sin\dfrac{2n\pi}3=?$$
