Most efficient way to solve $z^4 + 1 = 0$? I would like to know  the most efficient way to solve (in $\mathbb{C}$ of course):
$$z^4+1 = 0$$
I already tried:
$$\Longleftrightarrow z^2 = \pm \sqrt {-1} = \pm i$$
$$\Longleftrightarrow z = \pm \sqrt{\pm i}$$
which is quite confusing. I also tried:
$$\rho \exp{(4i\theta)} = -1$$
$$\Longleftrightarrow 4i\theta = \ln \left(\frac{-1}{\rho}\right)$$
which is also confusing, as $\ln(z\le 0)$ does not exist...
 A: You should understand what modulus is:
$z=re^{i\theta}$
then $|z|=\sqrt{z\bar{z}}=\sqrt{re^{i\theta}re^{-i\theta}}=r$
now $z^4=-1\implies r^4e^{4i\theta}=-1$
so the modulus is $r=1$ and we are left with
$e^{4i\theta}=-1\implies 4\theta=\pi+2n\pi\implies\\ \theta=\frac{\pi}{4}+\frac{n}{2}\pi$
A: One can rewrite $z^4$ as $(z^2)^2$, then, $(z^2)^2 = -1,$ and $z^2 = \pm i$
Now recall that since i corresponds with $\frac{\pi}{2}$ on the unit circle, the square root must correspond with $\frac{\pi}{4}$ so the solutions must be:
$$
z=\pm\frac{1+i}{\sqrt 2},\ \ z=\pm\frac{1-i}{\sqrt 2}
$$
And here you have your four roots.
A: We don't need to appeal to the use of the complex exponential--the polynomial can be factored directly by "completing the square" in a clever way. Observe:
$$
z^4 + 1 = (z^4 + 2z^2 + 1) - 2z^2 = (z^2 + 1)^2 - 2z^2 = [(z^2+1)-\sqrt{2}z]\cdot[(z^2+1) + \sqrt{2}z].
$$
In $\mathbb{R}$, this would be as far as we can go in factoring, because the resultant quadratics has discriminant $\Delta = (\pm\sqrt{2})^2 - 4\cdot1\cdot1 = 2-4 = -2 < 0$. But in $\mathbb{C}$, we can just apply the quadratic formula:
$$
z^2 \pm \sqrt{2}z + 1 =0 \implies z = \frac{\mp\sqrt{2} \pm \sqrt{-2}}{2} = \frac{1}{\sqrt{2}}(\mp1 \pm i).
$$
Hence the four roots are
$$
z_1 = \frac{1+i}{\sqrt{2}}, z_2 = -\frac{1+i}{\sqrt{2}}, z_3 = \frac{1-i}{\sqrt{2}}, z_4 = -\frac{1-i}{\sqrt{2}}.
$$
This factoring technique can be used to solve any quartic in $\mathbb{C}$ where only even powers show up.
Edit: This technique was given in the linked answer here--more discussion on the same polynomial is given there.
A: I just thought about a trick: we know that $1 = \exp{(2ik\pi})$ and $-1 = \exp{(2ik\pi+\pi})$ with $k \in \mathbb{Z}$, so why not writing:
$$\rho\exp{(4i\theta)} = \exp{(2ki\pi+\pi)}$$
$$\Longleftrightarrow  \begin{cases}\theta = k\pi/2+\pi/4\\ \rho=1 \end{cases}$$
So the solutions would be:
$$z_k = \exp(ik\pi/2 + i\pi/4)$$
