Find all roots of $x^{6} + 1$ I'm studying for my linear algebra exam and I came across this exercise that I can't solve.
Find all roots of polynomial $x^{6} + 1$. Hint: use De Moivre's formula.
I guessed that two roots are $i$ and $-i$, since: 
$i^{6} = (i^{2})^{3} = (-1)^{3} = -1 $
therefore, $i$ is root and his complex conjugate $-i$ has to be root too. However that was just guessing. I have no idea how can I use De Moivre's formula here.
Can you help me solve this?
 A: Hint: if $x^6=-1$, then $|x|^6=1$ and you can write $x=\cos\theta + i\sin\theta$.
details: 
Then the equation is, thanks to De Moivre theorem and 
$\cos^2 + \sin^2 =1$, equivalent to
$$
\cos 6\theta =-1\\
6\theta = \pi\mod 2\pi\\
\theta\in \frac \pi 6+\left\{0, \frac\pi 3, \frac{2\pi}3,\pi,\frac{4\pi} 3,
\frac{5\pi}3
\right\}.
$$
A: I'd like to add that this is possible to solve using well known formulas for $x^n\pm a^n$:
$$\begin{align}x^6+1&=(x^3)^2-i^2=(x^3+i)(x^3-i)=(x^3-i^3)(x^3+i^3)\\&=(x-i)(x^2+ix+i^2)\cdot(x+i)(x^2-ix+i^2)\end{align}$$
And now you can use the formula for quadratic roots to get them all in an explicit form.
A: If $z^6 + 1= 0$ then $z^6 = -1$. We can write $-1 = \mathrm{e}^{\mathrm{i}(\pi+2\pi n)}$ where $n \in \mathbb{Z}$. It follows that
$$(-1)^{1/6} = \{\mathrm{e}^{\mathrm{i}(\pi/6+\pi n/3)} : n \in \mathbb{Z}\}$$
Putting $n=0,1,2,3,4,5$ will give you all of the solutions you need. For example, when $n=1$:
$$z = \mathrm{e}^{\mathrm{i}(\pi/6+\pi /3)} = \mathrm{e}^{\mathrm{i}(\pi/2)} = \cos\tfrac{\pi}{2}+\mathrm{i}\sin\tfrac{\pi}{2} = \mathrm{i}$$
A: The easiest way to do these problems is using the polar form of a complex number. 
$$
x^6 + 1 = 0 \rightarrow x^6 = -1
$$
Write $-1$ as a complex number in polar form with the $+2\pi n$ (the most general form), then write $x$ as a general complex number:
$$
\left(re^{i\theta}\right)^6 = 1e^{\pi i + 2\pi n i} \\
r^6 e^{6\theta i} = 1 e^{\left(\pi + 2\pi n\right)i}
$$
From here, $r = 1$ (because $x^6 = 1$ has only one real root: $x = \sqrt[6]{1} = 1$).  Then you just set the angles equal:
$$
6\theta = \pi + 2\pi n \\
\theta = \frac{\pi + 2\pi n}{6} = \frac{\pi}{6} + \frac{\pi}{3}n
$$
Now just keep incrementing $n$ until you start looping (which will happen after $6$ consecutive values of $n$):
$$
\theta = \frac{\pi}{6}, \frac{\pi}{6} + \frac{\pi}{3}, \frac{\pi}{6} + \frac{2\pi}{3}, ...
$$
If you want to write them as $a +bi$ then you have to convert each of those to $r\cos(\theta) + ri\sin(\theta)$.
