What is complex number z if $z^8+16z^4+256=0$?

Question

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 $z$?

Answer 1

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.

Answer 2

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.

Answer 3

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?

Answer 4

$$z^8+16z^4+256=0$$ $$(\frac{z}{2})^8+(\frac{z}{2})^4+1=0$$ 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

Answer 5

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:
$-1-i\sqrt{3}$
$-1+i\sqrt{3}$
$1-i\sqrt{3}$
$1+i\sqrt{3}$
$-\sqrt{2-2i\sqrt{3}}$
$\sqrt{2-2i\sqrt{3}}$
$-\sqrt{2+2i\sqrt{3}}$
$\sqrt{2+2i\sqrt{3}}$

Answer 6

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 $$z^4=r^4\mathrm e^{4i\theta}=2^4\mathrm e^{\pm\tfrac{2i\pi}3}\iff\begin{cases}r=2,\\4\theta\equiv\pm\frac{2\pi}3\mod 2\pi\iff \theta\pm\frac\pi 6\mod\frac\pi 2.\end{cases} $$