Factorising a complex polynomial over $\mathbb{C}$

Question

I'm given $f(z)=z^6-1$ to factorise over $\mathbb{C}$. My working is as follows up to the point I don't understand:

$f(-1)=0$ and $f(1)=0$
So $(z+1)$ and $(z-1)$ are factors
$(z+1)(z-1)=z^2-1$
$(z^2-1)(z^4+pz^3+qz^2+rz+s)=z^6-1$
$z^6+pz^5+qz^4-z^4+rz^3-pz^3+sz^2-qz^2-rz-s=z^6-1$
$s=1$ because $-s=-1$
$q=1$ because $qz^4-z^4=0$
$p=0$ and $r=0$ because $pz^5-rz=0$
$(z^2-1)(z^4+z^2+1)=z^6-1$
If $z^4+z^2+1=0$
Let $y=z^2$
$y^2+y+1=0$
$y=$$-1\pm\sqrt3i\over2$
$z^2=$$-1\pm\sqrt3i\over2$

Now here's where I get stuck. In the answers section of my textbook the remaining 4 factors are listed as:

$z+{\frac{1}{2}\pm\frac{\sqrt3}{2}i},$
$z-{\frac{1}{2}\pm\frac{\sqrt3}{2}i}$

but I have:

$z=\pm\sqrt{-\frac12+\frac{\sqrt3}{2}i},$
$z=\pm\sqrt{-\frac12-\frac{\sqrt3}{2}i}$

and I don't know if I've made a mistake somewhere, or if I just don't know how to link what I have so far to the answers given by the textbook.

Answer 1

You're answer is right; what you're missing is that you can actually simplify those square roots by actually computing their real and imaginary parts.

Answer 2

This solution refers to the original post: $z^6 + 1$. We have: $z^3 + i$ and $z^3 - i$ are factors. And $f(z) = z^3 + i$ has $f(i) = i^3 + i = -i + i = 0$. So $z - i$ is a factor, and you can easily find that: $z^3 + i = (z - i)(z^2 + iz - 1)$. Let $g(z) = z^3 - i$. Then $g(-i) = 0$. So $z + i$ is a factor of $g(z)$, and you can also find that: $z^3 - i = (z + i)(z^2 - iz - 1)$. And from this you can use quadratic formula to solve for the remaining zeroes, and find all factors and roots.

For the case: $z^6 - 1$. We have: $z^6 - 1 = (z^3 - 1)(z^3 + 1) = (z - 1)(z^2 + z + 1)(z + 1)(z^2 - z + 1)$. And use quadratic formula for the two quadratic factors to find the zeroes and then all the factors and roots.