1. ## de moivre's thm

froom my textbook it writes:
de moivre's thm
$\displaystyle (cosx + i sin x )^{\frac p q } = cos (\frac p q x + \frac {2k\pi} q ) + i sin (\frac p q x + \frac {2k\pi} q )$
for$\displaystyle k = 0, 1, 2, ... , q-1$
i don't understand why k cannot be greater than q-1. thanks

2. it can be, just if you do the solution you get will be exactly the same as one you have found before say for example if k=q
we have
$\displaystyle \cos (\frac{p}{q}x+2\pi) +i \sin (\frac{p}{q}x+2\pi)$

this is identical to $\displaystyle \cos (\frac{p}{q}x) + i \sin (\frac{p}{q}x)$ which is the result with k=0
(because sine and cosine are periodic functions with period 2pi)

so k could be q,q+1...etc just you would just repeat all of the solutions