Finding the n-th root of a complex number I am trying to solve $z^6 = 1$ where $z\in\mathbb{C}$.
So What I have so far is :
$$z^6 = 1 \rightarrow r^6\operatorname{cis}(6\theta) = 1\operatorname{cis}(0 + \pi k)$$
$$r = 1,\ \theta = \frac{\pi k}{6}$$
$$k=0: z=\operatorname{cis}(0)=1$$
$$k=1: z=\operatorname{cis}\left(\frac{\pi}{6}\right)=\frac{\sqrt{3} + i}{2}$$
$$k=2: z=\operatorname{cis}\left(\frac{\pi}{3}\right)=\frac{1 + \sqrt{3}i}{2}$$
$$k=3: z=\operatorname{cis}\left(\frac{\pi}{2}\right)=i$$
According to my book I have a mistake since non of the roots starts with $\frac{\sqrt{3}}{2}$, also even if I continue to $k=6$ I get different (new) results, but I thought that there should be (by the fundamental theorem) only 6 roots.
Can anyone please tell me where my mistake is? Thanks!

Hint: Imagine that there is a unit circle on the Argand Plane. Now the roots will be the  6 equidistant points on that circle each having the argument as multiple of $\frac{2\pi}{6}=\frac{\pi}{3}$
