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 $z$?
 A: Hint:
$$
w^2+w+iw+i=0 \iff w(w+1)+i(w+1)=0\iff (w+1)(w+i)=0
$$
So the solutions in $z$ are the three cubic roots of $-1$ and the three  cubic roots of $-i$.
Your method is really intricate, and start with a mistake because
$$
\left(\frac{1+i}{2} \right)^2=\frac{1}{2}i
$$
A: The first step is good. The roots of $w^2+(1+i)w+i=0$ can be computed with the usual formula:
$$
\frac{-(1+i)\pm\sqrt{(1+i)^2-4i}}{2}
$$
but it's easier finding two numbers having sum $-1-i$ and product $i$: they're $-1$ and $-i$.
Now just find the cube roots of $-1=e^{\pi i}$ and $-i=e^{3\pi i/2}$.

If you want to do it the hard way, with $w=a+bi$ you get
$$
a^2-b^2+2abi+a+bi+ai-b+i=0
$$
so
\begin{cases}
a^2-b^2+a-b=0\\
2ab+a+b+1=0
\end{cases}
The first equation can be rewritten as $(a-b)(a+b+1)=0$. For $a=b$, the second equation is $2a^2+2a+1=0$, which has no solution. With $b=-1-a$ you get
$$
-2a-2a^2=0
$$
so $a=0$ or $a=-1$.
You seem to have done wrongly the completion of the square.
Note that
$$
(1+i)^2=2i
$$
so
$$
-i+\frac{(1+i)^2}{4}=-i+\frac{i}{2}=-\frac{i}{2}
$$
A: 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!

