roots - Find all complex numbers $z$ that satisfy equation $z^3=-8$

Problem

Find all complex numbers $z$ that satisfy equation $z^3=-8$

Attempt to solve

The real solution is quite easily computable or more specifically complex solution where imaginary part is zero.

$$z^3=-8 \iff z_1 = \sqrt[3]{-8}=-2$$

Now WolframAlpha suggests that other complex solutions would be :

$$z_2 = 1 - i\sqrt{3}$$
$$z_3 = 1 + i\sqrt{3}$$

Only problem is i don't have clue on how to derive these. I heard something about using polar form of complex number and then increasing argument by $2\pi$ so that we have all roots. I lack intuition on how this would work.

I could try to represent our complex number $-2$ in polar

$$re^{i \theta}$$

Computing radius via pythagoras theorem

$$r=\sqrt{(-2)^2+(0)^2} = \sqrt{4}=2$$

which is quite intuitive even without pythagoras theorem since our imaginary part is $0$ so "radius" has to be same as real part, just without the $-$ sign.

our angle would be $\pi$ radians since our complex number was $-2+i \cdot 0$.

We get:

$$2e^{i\pi}$$

Now increasing by every $2\pi$

$$2e^{i3\pi},2e^{i5\pi},2e^{i7\pi},\dots$$

which doesn't make any sense since we end up in the same spot over and over again since $2\pi$ in radians is full circle by definition.

Answer

$z^3+8=0;$

$(z+2)(z^2-2z +2^2)=0;$

$z_1=-2;$

Solve quadratic equation:

$z_{2,3} = \dfrac{2\pm \sqrt{4-(4)2^2}}{2}$;

$z_{2,3}= \dfrac{2\pm i 2√3}{2}.$