Can I get help with this question?
z^6 + (i+1)z^3 + i =0
Another approach:
I would factor as follows:
$\displaystyle z^6+(1+i)z^3+i=0$
$\displaystyle (z^6+z^3)+i(z^3+1)=0$
$\displaystyle z^3(z^3+1)+i(z^3+1)=0$
$\displaystyle (z^3+i)(z^3+1)=0$
Can you proceed from here?
yes, the quadratic formula works in ANY field, as long as 1+1 ≠ 0.
so we have:
$\displaystyle w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(i+1) \pm \sqrt{(i+1)^2 - 4i}}{2}$
$\displaystyle = \frac{-(i+1) \pm \sqrt{2i - 4i}}{2} = \frac{-(i+1) \pm \sqrt{-2i}}{2}$
the two square roots of -2i are: -1+i and 1-i (check this by squaring them).
this gives:
$\displaystyle w = \frac{-(i+1) \pm (-1+i)}{2} \text{, so } w = -1 \text{ or } -i$.
since w = z^{3}, we are solving $\displaystyle z^3 = -1 \iff z^3 + 1 = 0 \text{ or }z^3 = -i \iff z^3 + i = 0$
which is exactly where MarkFL2's solution leads you to.