Find the nth roots of a complex number. I am being asked to find all fourth roots of $\zeta_3$.
My book teaches that $\zeta_3$=$cos\frac{2\pi}{3}$+$isin\frac{2\pi}{3}$.
From there I did the following but I'm not sure I am correct.
$\zeta_3$=$cos\frac{2\pi}{3}$+$i sin\frac{2\pi}{3}$=($cos\frac{2\pi}{12}$+$isin\frac{2\pi}{12})^4$=$(\zeta_{12})^4$
Am I on track?
All of my roots are $\zeta_{12}$, $(\zeta_{12})^4$, $(\zeta_{12})^7$,$(\zeta_{12})^{10}$.
 A: Let $$\DeclareMathOperator{\cis}{cis}\cis(\theta)=\cos(\theta)+i\sin(\theta).$$
You are aiming to solve $$z^4=\zeta_{3}$$ for $z$;
since $\zeta_3$ is a root of unity, its magnitude is $1$, and therefore all solutions $z$ also have magnitude $1$ (why?);
therefore, you can rewrite (which you did) both sides using $\cis$:
$$\cis(\theta)^4=\cis(2\pi/3)$$
for some $\theta$;
now observe, that since both $\sin$ and $\cos$ are $2\pi$-periodic, $\cis$ is also $2\pi$-periodic, and we can conclude that the right-hand side stays the same for each integer $c$:
$$\cis(\theta)^4=\cis(2\pi/3+2\pi c);$$
now, from your post I can conclude you know of de Moivre's identity; this step is crucial and involves just that, so please observe the following application of the identity which does not change the left-hand side:
$$\cis(4\theta)=\cis(2\pi/3+2\pi c);$$
now conclude (this works and we do not lose solutions, because $\cis$ is $2\pi$-periodic and $c$ enumerates and exhausts every argument of $\cis$ that gives a fixed value):
$$\theta=\frac{2\pi/3+2\pi c}4.$$
A: Just to be perverse, let's re-frame this as an entirely group-theoretic question.
What we have in mind is leveraging an implicit group-homomorphism:
$\phi:\Bbb Z_n \to T$, where $T = \{z \in \Bbb C: |z| = 1\}$, afforded by:
$\phi([k]_n) = \cos(\frac{2\pi k}{n}) + i\sin(\frac{2\pi k}{n})$.
Now since $\zeta_3$ is a third root of $1$, it follows that a fourth root $w$ of $\zeta_3$ would satisfy:
$w^{12} = (w^4)^3 = (\zeta_3)^3 = 1$, that is, $w$ is a $12$-th root of unity.
This suggests we might start with looking at $\Bbb Z_{12}$, a cyclic group of order $12$.
So we are identifying $\zeta_3$ with a particular element of $\phi(\Bbb Z_{12})$, namely the one corresponding to:
$\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}) = \cos(\frac{8\pi}{12}) + i\sin(\frac{8\pi}{12}) = \phi([4]_{12})$.
So what we are really asking is: which $[k]_{12}$ satisfy $4[k]_{12} = [4]_{12}?$
Clearly, one solution is: $k = 1$. This corresponds to your $\zeta_{12}$.
Since the group operation in $\Bbb Z_{12}$ is addition (modulo $12$), the powers of $\zeta_{12}$ correspond to (integral) multiples of $[1]_{12}$.
Now, note that:
$4[3t + 1]_{12} = 4[3t]_{12} + 4[1]_{12} = [12t]_{12} + [1]_{12} = t[12]_{12} + [1]_{12} = t[0]_{12} + [1]_{12} = [t0]_{12} + [1]_{12} = [0]_{12} + [1]_{12} = [1]_{12}.$
Since we are only interested in $0 \leq  3t + 1 < 12$, this quickly gives us another three solutions: $k = 4,7,10$, which correspond to the four powers you found.
Could there be others? Well, we could eliminate the other values of $0 \leq k < 12$ by trial-and-error, but here is a better way:
If $4[k]_{12} = [4k]_{12} = [4]_{12}$, then:
$4k = 4 + 12m$, for some integer $m$. Thus $k = 1 + 3m$ for some integer $m$.
Since $ 0 \leq k < 12$, we must have $0 \leq m < 4$, that is $m = 0,1,2$ or $3$, and we already found those.
Or, knowing that $\Bbb C$ is a field, we have found $4$ solutions to:
$x^4 - \zeta_3 = 0$, which is a quartic polynomial in $\Bbb C[x]$, and thus can have (at most) four solutions.
Note: the "other answers" in dbanets' answer can be obtained from using other integers $k$ outside of $\{0,1,2,\dots,11\}$, by the periodicity of $2\pi$ of the trigonometric functions.
