<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] Note that $$z^8+16z^4+256$$
$$=z^8+32z^4+256-16z^4$$
$$=(z^4+16)^2-(4z^2)^2$$
$$=(z^4-4z^2+16)(z^4+4z^2+16)$$
$$=(z^4-4z^2+16)(z^2-2z+4)(z^2+2z+4)$$
Now considering $(z^4-4z^2+16)=0$, let $x=z^2$ then we have $$x^2-4x+16=0$$
$$x=\frac{4\pm\sqrt{-48}}{2}=\frac{4\pm 4i\sqrt{3}}{2}=2\pm2i\sqrt{3}$$
So you have $z=\pm\sqrt{2+2i\sqrt{3}}$ and $z=\pm\sqrt{2-2i\sqrt{3}}$ in this case. The other cases are quadratics.
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