I know that De Moivre's Theorem does not necessarily work for non-integer powers.
The classic counter-example is by considering $\left (\cos \theta + i \sin \theta \right )^n=\cos n\theta + i \sin n \theta$ when $n=\frac{1}{2}$ and setting $\theta=0$ versus $\theta=2\pi$, which yield 1 and -1 respectively.
My question is, if this is the case, why is it that we can use the following to solve say $z^5=1$?
$z^5=\cos (\theta+2k\pi) + i \sin (\theta+2k\pi) $ where $k\in\mathbb{z}$
And then raising both sides to the power of $\frac{1}{5}$ etc?
Best Answer
It does work for $n=\frac{1}{2}$. By putting $\theta =0$ and $n=\frac{1}{2}$ you're working out $1^{1/2} = \pm 1$.