<s>[INST] How to find all values of $z$ such that $z^3=−8$ The question is
Solve the equation
$z^3=-8$
My attempt
I attempt to write it out in polar co-ordinates
since $z = r(\sin(x) + i\sin(x)) \\ z^3 = r^3(\cos(3x) + i\sin(3x))$
then
$r^3\sin(3x) = -8$ and $r^3\cos(3x) = 0$
but from here I'm not really sure where to go , I've searched up the solution to this before and people have written $r^3 = 8$ so $\cos(3x) = -1$ and $\sin(3x) = 0$ .
 [/INST] $$z^3=-8$$
$$z^3 = r^3\cdot[\cos(3\cdot\theta) + i\sin(3\cdot\theta)]$$
Let's substitute $R = r^3$, $\rho= 3\cdot\theta$ so that:
$$R\cdot\sin(\rho) = 0$$
$$R\cdot\cos(\rho) = -8$$
Since $|z^3| = R = |-8| = 8$ you can get $\rho$ by doing:
$$8\cdot\cos(\rho) = -8$$
$$\cos(\rho) = -1$$
$$\rho = (2k-1)\cdot\pi$$
where $k$ is an arbitrary integer $k \in \mathbb{Z}$.
Let's make the anti-substitution and we obtain:
$$r = \sqrt[3]{R}=2$$
$$\theta = \frac{2k-1}{3}\cdot\pi$$
So,
$$z = 2\cdot\left[\cos\left(\frac{2k-1}{3}\cdot\pi\right) + i\sin\left(\frac{2k-1}{3}\cdot\pi\right)\right]$$
Although it seems that there are infinite solutions for $z(k)$, this is not true, because $\cos()$ and $\sin()$ are periodic. Because of the fundamental theorem of algebra $z^3+8=0$ will have 3 solutions, which also matches with the $\cos()$ and $\sin()$ periodicity for $z$. Thus, possible solutions are defined this way:
For $k=0$, $z = 2\cdot\left[\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right] = 1-\sqrt{3}\cdot i$.
For $k=1$, $z = 2\cdot\left[\cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right)\right] = 1+\sqrt{3}\cdot i$.
For $k=2$, $z = 2\cdot[\cos(\pi) + i\sin(\pi)] = -2$
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