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 $z$?
 A: 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.
A: You have the (double) equation:
$$z^4=-8\pm \sqrt{192}i=-8(1 \pm \sqrt{3}i)\tag{1}$$
Consider the "-" case. Divide (1) by $16$, giving:
$$(\frac{z}{2})^4=\cos (2 \pi/3) +\sin (2 \pi/3) i = \cos 4a+\sin 4a i=(e^{ia})^4$$
where $a:=\pi/6$.
Which is equivalent to:
$$\left(\dfrac{z}{2e^{ia}}\right)^4=1 \ \iff \ Z^4=1\tag{2}$$
by setting $Z=2\dfrac{z}{e^{ia}}.$
Therefore $Z$ is one of the fourth roots of unity, i.e., $Z=1,i,-1,-i$ giving finally four values of $z$:
$$z=2e^{ia}, \ 2ie^{ia}, \ -2e^{ia}, \ -2ie^{ia}$$
Up to you for the "+" case and its associated 4 other solutions.
A: 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?
A: $$z^8+16z^4+256=0$$
$$(\frac{z}{2})^8+(\frac{z}{2})^4+1=0$$
let $x=\frac{z}{2}$
$$x^8+x^4+1=0$$
$$(x^4+1)^2-x^4=0$$
$$(x^4-x^2+1)(x^4+x^2+1)=0$$
$$((x^2+1)^2-3x^2)((x^2+1)^2-x^2)=0$$
so
$$(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)(x^2-x+1)(x^2+x+1)=0$$
then you can use the quadratic formula  to find all the complex roots
A: When you factorize the polynomial you'll get the following products:
$$(x^2 - 2 x + 4) (x^2 + 2 x + 4) (x^4 - 4 x^2 + 16) = 0 $$
And therefore you get the following roots by computing quadratic and quartic equations:
$-1-i\sqrt{3}$
$-1+i\sqrt{3}$
$1-i\sqrt{3}$
$1+i\sqrt{3}$
$-\sqrt{2-2i\sqrt{3}}$
$\sqrt{2-2i\sqrt{3}}$
$-\sqrt{2+2i\sqrt{3}}$
$\sqrt{2+2i\sqrt{3}}$
A: I would make it a bit shorter, with the appropriate tools:
Write $z=r\mathrm e^{i\theta}$ ($r>0$, $\theta\in\mathbf R/2\pi \mathbf Z$) and $\:y=8(-1\pm\sqrt 3)=16\bigl(-\frac12\pm\frac{\sqrt 3}2i\bigl)=2^4\mathrm e^{\pm\tfrac{2i\pi}3}$. We have to solve
$$z^4=r^4\mathrm e^{4i\theta}=2^4\mathrm e^{\pm\tfrac{2i\pi}3}\iff\begin{cases}r=2,\\4\theta\equiv\pm\frac{2\pi}3\mod 2\pi\iff \theta\pm\frac\pi 6\mod\frac\pi 2.\end{cases} $$
