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.
 A: $z^3+8=0;$
$(z+2)(z^2-2z +2^2)=0;$
$z_1=-2;$
Solve quadratic equation:
$z_{2,3} = \dfrac{2\pm \sqrt{4-(4)2^2}}{2}$;
$z_{2,3}= \dfrac{2\pm i 2√3}{2}.$
A: We know that because of the $^3$ that there will be three roots. You have rightly determined that the modulus of the roots is $2$. The trick when using polar coordinates is to add $\frac{2\pi}{x}$, where $x$ is the number of roots we have, to the argument. So in this case you would add $\frac{2\pi}{3}$ to the argument.
This gives $z = 2e^{i\pi},  2e^{\frac{5\pi}{3}i}, 2e^{\frac{\pi}{3}i}.$
A: Instead of having $-2$ in polar coordinates, use $-8$:
$$-8=8e^{i\pi}$$
Repeatedly adding $2\pi i$ to the exponent and cube rooting gives all the cube roots:
$$2e^{i\pi/3}\qquad 2e^{i\pi}\qquad 2e^{5i\pi/3}$$
These can be converted back into Cartesian coordinates as $-2$ and $1\pm\sqrt3i$.
A: Alright, so we have $z^3=8e^{i(\pi+2\pi k)}$ for each $k\in\mathbb{Z}$. Now to find $z$ we need take the third root from the module and divide the argument by $3$. So we get $z=2e^{i(\frac{\pi}{3}+\frac{2\pi}{3}k)}$. So now compute it for $k\in\{0,1,2\}$, after that the roots will repeat. So for $k=0$ we get:
$z_1=2e^{i\frac{\pi}{3}}=2(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3}))=1+i\sqrt{3}$
Now put $k=1$ and $k=2$ to find the other roots. 
A: Whenever we deal with exponentials, the $re^{i\theta}$ form often is more convenient.
We know that the magnitude of $-8$ is $8$, so the magnitude of $z$ is $\sqrt[3]8=2$.
So, we know that $$(2e^{i\theta})^3=-8\to e^{i\cdot3\theta}=-1$$
We know that $e^{i\pi}=-1$, so we look for all $\theta$ such that $3\theta = 2\pi\cdot n+\pi$. We find the answers to be $$\frac\pi3,\pi,\frac5{3\pi}$$so we know that our answers are $$2e^{i\frac\pi3},2e^{i\pi},2e^{i\frac{5\pi}3}$$
The a+bi form of these complex numbers is $$\color{red}{-2,1\pm\sqrt3i}$$
