Solve for z is Complex number : $z^8 + z^4 - 12 = 0$

Question

As far as I came is that I have two solutions to do now:

$z^4 = 3$ and $z^4 = -4$

We've done something like this with polar form in school but I can't remember how.

Anyone can show me how it's done?

Answer 1

Observe that

$$z^4 = 3 = 3 e^{2 \pi i k}$$

for any integer $k$. Take 4-th roots:

$$ z_{k} = 3^{\frac 1 4} e^{\frac {2\pi i k}{4}}$$

where $k = 1,2,3,4$. plug in these values of $k$ to get 4 roots.

Proceed similarly with $z^4 = -4 = 4e^{\pi i +2 \pi i k}$.

Answer 2

Write $z$ in polar form, that is, $$ z=r(\cos\theta+i\sin\theta) $$ Then we know (by de Moivre's Formula) $$ z^4=r^4(\cos4\theta+i\sin4\theta) $$ Now write, for example, $3$ in polar form (in this case $3\in\mathbb{R}$ so it's easy): $$ 3=3(\cos0+i\sin0) $$ Now, you want $z^4=3$, that is, $$ r^4(\cos4\theta+i\sin4\theta)=3(\cos0+i\sin0) $$ This will be true if, and only if, \begin{cases} r^4=3\\ 4\theta=0+2\pi n\text{ for a certain }n\in\mathbb{Z} \end{cases} The first equality implies $r=\sqrt[4]{3}$ (since $r\in\mathbb{R}_{>0}$) and the second equality implies $$ \theta=\frac{\pi n}{2}\text{ for a certain }n\in\mathbb{Z} $$ Now, we see that $n=0,1,2,3$ give four different values of $\theta$, say $\theta_0,\theta_1,\theta_2,\theta_3$, and that choosing any other integer will give back one of those four angles modulo $2\pi$. All in all, the four complex solutions to $z^4=3$ are $$ z_n:=\sqrt[4]{3}(\cos\theta_n+i\sin\theta_n)\quad(n=0,1,2,3) $$

Answer 3

HINT:

$$z^4=3\Longleftrightarrow$$ $$z^4=3e^{0i}\Longleftrightarrow$$ $$z=\left(3e^{2\pi ki}\right)^{\frac{1}{4}}\Longleftrightarrow$$ $$z=\sqrt[4]{3}e^{\frac{\pi ki}{2}}\Longleftrightarrow$$ $$z=\begin{cases}\pm\sqrt[4]{3}\\ \pm i\sqrt[4]{3} \end{cases}$$

With $k\in\mathbb{Z}$ and $k:0-3$