<s>[INST] How to find all values of $z$ such that $z^3=-8i$ If I am asked to find all values of $z$ such that $z^3=-8i$, what is the best method to go about that?
I have the following formula:
$$z^{\frac{1}{n}}=r^\frac{1}{n}\left[\cos\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)+i\sin\left(\frac{\theta}{n}+\frac{2\pi k}{n}\right)\right]$$
for $k=0,\pm1, \pm2,...$
Applying this formula, I find the cubed root of $8$, which is $2$.  And then when I apply it to the formula, I get the following:
$$z = 2\left[\cos\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{3}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$
I am confused, because the given solution is as follows:
$$z = 2\left[\cos\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)+i\sin\left(\frac{\pi}{2}+\frac{2\pi k}{3}\right)\right]$$ for $k=0,\pm1, \pm2,...$
Where did I go wrong?  How would my approach changed if I was asked to find all values for $-8$, or $8i$?
 [/INST] Short way write it as $$z^{ 3 }=-8i\\ z^{ 3 }=8{ e }^{ i\frac { 3\pi  }{ 2 }  }\\ z=\sqrt [ 3 ]{ 8 } { e }^{ i\left( \frac { \frac { 3\pi  }{ 2 } +2\pi k }{ 3 }  \right)  }=2{ e }^{ i\left( \frac { 3\pi +4\pi k }{ 6 }  \right)  },k=0,1,2\quad $$
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