Let $i$ be a unit imaginary part , we have for $\theta=\frac{\pi}{2}$: $\left(\cos \theta + i \sin \theta \right)^n = (0+i\sin(\frac{\pi}{2} ))^n=i\sin(n\frac{\pi}{2})=0$ (Using Moiver formual) , I asked why this is work however ${i}^{n}$ gives only $1$ or $-1$ ?
Edite01 : I edited the question because i meant why I can't let $\cos\theta =0$ before using Moiiver formula
Edite02:I edited the question to give some explanations for the recent
Note : I think I have a bad uses for the Moiiver formula !!
Thank you for any help
It is DeMoivre's rule and you stated it wrong. $$ \left(\cos \theta + i \sin \theta \right)^n = \cos n \theta + i \sin n \theta.$$ If $n\theta$ is a multiple of $\pi$ the end result is $\pm 1$, not $0$.