How to find all third roots of $-i$?

Question

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

Answer 1

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?

Answer 2

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.

Answer 3

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.