How to find the roots of $x^4 +1$

Another way is to use some creative rewriting:

$$x^4 + 1 = (x^2 + 1)^2 - 2x^2 = (x^2 + 1 - x\sqrt 2)(x^2 + 1 + x\sqrt2).$$

Then just solve the two quadratic equations.

Using this, $x^4=-1=e^{i\pi}=e^{(2n+1)\pi i}$ as $e^{2m\pi i}=1$ where $m,n$ are integers.

So, $x=e^{\frac{(2n+1)\pi i}4}=\cos\frac{(2n+1)\pi}4 +i\sin \frac{(2n+1)\pi}4 $ where $n$ has any $4$ in-congruent values $\pmod 4$, the most simple set of values of $n$ can be $\{0,1,2,3\}$.

If $x_m=e^{\frac{(2m+1)\pi i}4},x_{m+2}=e^{\frac{(2m+3)\pi i}4}=e^{\frac{(2m+1)\pi i}4}\cdot e^{\frac {i\pi}2}=-x_m$

Also, observe that if $y$ is a solution of $x^4=-1$, so is $-y$

$x_0=\cos\frac{\pi}4 +i\sin \frac{\pi}4=\frac{1+i}{\sqrt 2}$

$x_1=\cos\frac{3\pi}4 +i\sin \frac{3\pi}4=\frac{-1+i}{\sqrt 2}$

$x_2=-x_0$

$x_3=-x_1$

So, the values of $x$ are $\pm\left(\frac{1+i}{\sqrt 2}\right),\pm\left(\frac{-1+i}{\sqrt 2}\right)$

Finding the roots of $X^n=z$ in $\mathbb{C}$ is a simple problem if you use the exponential notation for $z$.

If $z=\rho e^{i\theta}$ then the $n$ roots of this polynomial are: $\{\rho^{\frac{1}{n}}e^{i\frac{\theta+2k\pi}{n}},k\in [\![ 0;n-1 ]\!] \}$.

In this case, $n=4$, $\rho=1$ and $\theta=\pi$, so the roots are $e^{i\pi\frac{2k+1}{4}}$ for $k\in[\![ 0;3 ]\!]$,

which can also be written $e^{i\frac{\pi}{4}}i^k$ for the same values of $k$.