<s>[INST] 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?
 [/INST] 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.
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