Failure of De Moivre's Theorem

Question

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?

Answer 1

It does work for $n=\frac{1}{2}$. By putting $\theta =0$ and $n=\frac{1}{2}$ you're working out $1^{1/2} = \pm 1$.

Answer 2

To solve $z^5=1$, take $z=\cos(t)+i\sin(t)$ (since |z|=1).

So you have (by de Moivre's theorem for a positive integer power) $$z^5 = \cos(5t) + i \sin(5t) = 1 \\ \Rightarrow \cos(5t) + i \ sin(5t) = \cos(2 \pi k) + i \sin (2 \pi k) $$

On equating real and imaginary parts of this equation, you get two trigonometric equations, $\cos(5t)=\cos(2\pi k)$ and $\sin(5t)=\sin(2 \pi k)$.

The common solution is $5t = 2 \pi k + 2 \pi n$, so $t = \dfrac{2 m \pi}{5}$ is the solution. So there is no need to raise anything to fractional powers anywhere.

To convince yourself that there are only five roots, set m={0,1,2,3,4,5}; when m=5 you get the same root as m=0 and so on.