Find all complex roots for $z^6 + (1 + i)z^3 + i = 0$ Find all complex roots of $z^6 + (1 + i)z^3 + i = 0$.
I've tried by first settings $w = z^3$, so then I get $$w^2 + (1 + i)w + i = 0$$
Now I rewrite it as $$(w+(\frac{1+i}{2}))^2 = -i + (\frac{1+i}{2})^2$$
Set $w+(\frac{1+i}{2}) = a+bi$
$$(a+bi)^2 = -i + 2i=i$$
Which implies that
$$a^2-b^2=0$$
$$a^2+b^2=\sqrt{0^2+1^2} = 1$$
$$2ab=1$$
So I get $w=\frac{1+i}{2}\pm(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)$. Not sure where to go from here. I get answers for $w$, but how do I get the answers for $z$?

It's $$(z^3+i)(z^3+1)=0.$$
We have two cases.


*

*$z^3+i=0$ or $z^3-i^3=0$ or
$$(z-i)(z^2+iz-1)=0,$$ which gives
$$\left\{i,\pm\frac{\sqrt3}{2}-\frac{1}{2}i\right\}$$

*$z^3+1=0$ or
$$(z+1)(z^2-z+1)=0,$$ which gives
$$\left\{-1,\frac{1}{2}\pm\frac{\sqrt3}{2}i\right\}$$
Done!

