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?
 A: As $i^3=-i,$
If $x^3=-i,$
$$x^3-i^3=(x-i)(x^2+ix+i^2)=0$$
Can you solve the quadratic equation?
A: If you don't know polar form you can do $(a+bi)^3 = a^3 + 3a^2bi + 3ab^2i^2 + b^3i^3 = a^3 + 3a^2bi -3ab^2 -b^3i = (a^3-3ab^2) +(3a^2b -b^3)i = -i$.  So $a^3-3ab^2 = 0; 3a^2b-b^3 = -1$.
Solve for real $a,b$.
If you do know polar form then $-i = e^{\frac {3\pi}2i}$. so $(-i)^{\frac 13} = e^{\frac {\frac {3\pi}2+2\pi}3 i}$.
Also $i^3 = (i^2)i = -i$ so $i$ is one of the cube roots.
A: 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.
