Find all complex roots for $z^6 + (1 + i)z^3 + i = 0$

Question

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 $z$?

Answer 1

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 $-i=e^{3\pi i/2}$.

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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 $$ -i+\frac{(1+i)^2}{4}=-i+\frac{i}{2}=-\frac{i}{2} $$

Answer 2

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

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

Answer 3

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 $$ \left(\frac{1+i}{2} \right)^2=\frac{1}{2}i $$