# Stucked with some complex number arithmetic

I have the following problem from an algebra class.

$z^{12} + z^6 + 1 = 0$

I need to find all $z \in \Bbb C$ for that equation.

I did this:

$z^6 (z^6 + 1) = -1$

Now $z=0$ can't be a solution.

$z^6 + 1 = -1 \div z^6$

$z^6 + 1 = z^{-6}$

I don't know what property should I use to go on from here. Any suggestions? Thanks.

• Use $t=z^6$ then solve $t^2+t+1=0$ (as you would solve any quadratic equation). – barak manos Nov 5 '16 at 18:53

$w=z^6$
$w^2+w+1=0$
which has two unit roots (the two roots of the $w^3=1$ which are not $1$)
Then solve $z^6=w$ for the two cases.
Another idea: multiply the whole thing by $z^6-1$ to get $z^{18}-1=0$ and see which solutions fit the initial equation.