<s>[INST] What is complex number z if $z^8+16z^4+256=0$? So far, I have set y to equal to $z^4$ and used the quadratic equation to solve $y = -8+8\sqrt{3}i$ or $-8-8\sqrt{3}i$. How do I determine the 8 different values of $z$?
 [/INST] As you have mentioned, the correct approach is to let $y=z^4$. Then we get
$$
y^2+16y+256=0
$$
Using the quadratic formula, the solutions are
$$
y=-8\pm8i\sqrt{3}
$$
All we need to do know is solve the equations $z^4=-8+i\sqrt{3}$ and $z^4=-8-i\sqrt{3}$. We can do this by writing the two complex numbers in the polar-coordinate form. As a reminder, I will write down how to derive the polar-coordinate form here:
$$
a+bi=r\text{cis}\theta=r(\cos\theta+i\sin\theta)=re^{i\theta}
$$
The magnitude of $-8\pm8i\sqrt{3}$ is $\sqrt{8^2+(8\sqrt{3})^2}=\sqrt{64+64(3)}=\sqrt{256}=16$. Therefore, $r=16$. The angle $\theta$ depends on whether we are considering $-8+8i\sqrt{3}$ or $-8-8i\sqrt{3}$. In the former case, $\theta=120$ degrees. Then, by symmetry, we can find out $\theta$ in the latter case. Can you take it from here?
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