# Hi guys, find all complex numbers satisfying `z^6-1=0.` I have seen someone do the working out as follows:     `z^6-1=(z^3-1)(z^3+1) = (z-1)(z^2+z+1) (z+1)(z^2-z+1).`   However, I am not sure what the next step is. thanks daniel

Because of the factors `z-1` and `z+1,` we know that `+-1` are solutions.

Now we can use the quadratic formula for the remaining factors. For

`z^2+z+1,` we get

`z=(-1+-sqrt(1^2-4(1)(1)))/2=(-1+-i sqrt(3))/2`

and for `z^2-z+1` we get

`z=(1+-i sqrt(3))/2` .

The six solutions are `+-1,(-1+-i sqrt(3))/2,(1+-i sqrt(3))/2.`

Note that if `z` is a solution, then so is ``the conjugate of `z.` This is true for any solution of `z^n=1` for any `n.` Also, if `z` is a solution, so is `-z.` This is the case for the solutions of `z^n=1` only when `n` is even (which is why it works for this example).