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?
 A: 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.
A: Without using polar representation:
Note that 
$i^3 = i^2 i = (-1)i = -i; \tag 1$
thus $i$ is a solution of
$z^3 = -i, \tag 2$
which may be re-written as
$z^3 + i = 0; \tag 3$
we may seek the other zeroes of (2)-(3) via synthetic division; we seek
$z - i \overline{)z^3 + i}; \tag 4$
we have:
$z \; \text{into} \; z^3 \; \text{yields} \; z^2; \tag 5$
$z^2 \times (z - i) = z^3 - iz^2; \tag 6$
$z^3 + i - (z^3 - iz^2) = iz^2 +  i = i(z^2 + 1) = i(z + i)(z - i) ; \tag 7$
now we can skip to the chase by noticing that 
$(z - i) \overline{)i(z + i)(z - i)} = i(z + i) = iz - 1; \tag 8$
therefore the quotient should be
$(z - i) \overline{)z^3 + i} = z^2 + iz - 1; \tag 9$
we check:
$(z - i)(z^2 + iz - 1) = z^3 + iz^2 - z - iz^2 + z + i = z^3 + i; \tag{10}$
the quotient (4) is thus the quadratic $z^2 + iz - 1$; we use the quadratic formula:
$z = \dfrac{1}{2} (-i \pm \sqrt{i^2 - 4(1)(-1)}) = \dfrac{1}{2}(-i \pm \sqrt 3), \tag{11}$
which are easily checked to satisfy (3); I leave that simple task to my readers.  As well as the tasks of converting our roots to totally proper $a + bi$ form.
A: To find $z$ such that $z=(-i)^\frac{1}{3}$. So write $-i=\cos(\frac{\pi}{2})+i \sin (-\frac{\pi}{2})$
$z^3=-i$ implies 
$z=(-i)^\frac{1}{3}=\Big(\cos(\frac{\pi}{2})+i \sin (-\frac{\pi}{2})\Big)^\frac{1}{3}=\Big(\cos(2k\pi+\frac{\pi}{2})+i \sin (2k\pi-\frac{\pi}{2})\Big)^\frac{1}{3}$
which is same as $$\cos\Big(2k+\frac{1}{2}\Big)\frac{\pi}{3}+i \sin \Big(2k-\frac{1}{2}\Big)\frac{\pi}{3}$$ where $k=0,1,2$
