<s>[INST] Finding all $z \in \Bbb C$, expressed in the form $z = a + ib$, with $a, b \in \Bbb R$, satisfying equation $z^3 = -i$ I'm trying to find all $z \in \Bbb C$, expressed in the form $z = a + ib$, with $a, b \in \Bbb R$, satisfying equation $z^3 = -i$.
I've figured out that
$$(a + ib)^3 = -i \Longleftrightarrow a + ib = \sqrt[3]{-i} \Longleftrightarrow a + ib = -i$$ 
and also that
$$(a+ib)^3 = a^3 + 3a^2bi + 3ab^2i^2 +i^3b^3$$
but I'm not sure how to proceed from here.
I think I'm supposed to use synthetic division to find the other roots, but not sure what factor to use. I tried $a + ib + i$ and that didn't work. 
Any ideas?
 [/INST] Let $z=a+ib$. You have
\begin{align*}
(a+ib)^3 & = -i\\
a^3 + 3a^2bi - 3ab^2 -ib^3 & =-i\\
(a^3-3ab^2)+i(3a^2b-b^3) & =-i
\end{align*}
Equate the real and imaginary parts on both sides to get
\begin{align*}
a^3-3ab^2 =a(a^2-3b^2)& =0\\
3a^2b-b^3 =b(3a^2-b^2)& =-1
\end{align*}
From the first of these equations, we get either $a=0$ or $a^2=3b^2$. 
With $a=0$, from the second equation, we get $b^3=1$. Since $b \in \mathbb{R}$, so $b=1$. This gives $\color{red}{z=i}$ as a solution.
With $a^2=3b^2$, from the second equation, we get $8b^3=-1$. Since $b \in \mathbb{R}$, so $b=-\frac{1}{2}$. Consequently we get $a=\pm\frac{\sqrt{3}}{2}$ This gives $\color{red}{z=\frac{\sqrt{3}}{2}+\frac{i}{2}}$ and $\color{red}{z=\frac{-\sqrt{3}}{2}+\frac{i}{2}}$ as other two solutions.
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