Pure imaginary numbers : $ ( j^{2n} - j^n ) \in i\Bbb R $?

Question

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)

Answer 1

Observing that $$j^2=\left(\frac{-1}{2} + i\frac{\sqrt3}{2}\right)^2=-\frac{1}{2} - i\frac{\sqrt3}{2}=\overline{j}$$

then since $\Im(z) =\frac{1}{2i}(z-\overline{z})$ we obviously get $$j^{2n}-j^n =\bar{j}^{n}-j^n =\overline{j^n} - j^n = \color{blue}{-2i\Im(j^n) \in i\Bbb R} $$

Answer 2

$$j=e^{2\pi i/3}$$

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

Now the real part $j^{2n}-j^n$

$$=\cos\dfrac{4n\pi }3-\cos\dfrac{2n\pi}3$$ $$=-2\sin2n\pi\sin\dfrac{2n\pi}3=?$$

Answer 3

Hint:

Note that using Euler's formula, we can write $$ j = \cos \frac{2\pi}3 + i\sin \frac{2\pi}3 = e^{\dfrac{2\pi i} 3}$$ which is a complex root of unity.