How to find all third roots of $-i$? How to find all third roots of $-i$?
I have no idea.
Could you give me some hint?

If you know de Moivre's law, or representation of complex numbers as exponentials, this problem is simple.
If you don't, here is an algebraic approach.
There exists some complex number $a+bi$ such that $(a+bi)^3 = -i$
Multiplying it out and combining terms.
$a^3 + 3a^2bi + 3a(bi)^2 + (bi)^3 = -i\\
(a^3 - 3ab^2) + (3a^2b - b^3)i = -i$
Set the real parts and the imaginary parts equal to one annother.
$a^3 -3ab^2 = 0\\
a^2b - b^3 = -1$
Working with the first equation.
$a(a^2-3b^2) = 0$
$a = 0$ or $a^2 = 3b^2$
If $a = 0$ substituting into the second equation.
$-b^3 = -1\\
b = -1\\
a+bi = -i$
This is one of our answers.
If $a^2 = 3b^2$
$9b^3 - b^3 = -1\\
8b^3 = -1\\
b^3 = -\frac {1}{8}\\
b = -\frac {1}{2}$
Substituting back into $a^2 = 3b^2$
$a^2 = \frac {3}{4}\\
a = \pm \frac {\sqrt 3}{2}\\
\frac {\sqrt 3}{2} - \frac 12 i, -\frac {\sqrt 3}{2} - \frac 12 i$
are our other 2 solutions.
