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

I'm trying to find the roots of $x^4+1$. I've already found in this site solutions for polynomials like this $x^n+a$, where $a$ is a negative term. I don't remember how to solve an equation when $a$ is a positive term as the equation above.

Thanks

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 $\pm\left(\frac{1+i}{\sqrt 2}\right),\pm\left(\frac{-1+i}{\sqrt 2}\right)$

$$ x^4=-1 $$ $$ x^2=\pm i = \pm\left(\cos\frac\pi2 + i\sin\frac\pi2\right) $$ $$ \text{If }x^2 = \cos\frac\pi2 + i\sin\frac\pi2 \text{ then } x = \pm\left(\cos\frac\pi4+i\sin\frac\pi4\right). $$ $$ \text{If }x^2 = -\left(\cos\frac\pi2 + i\sin\frac\pi2\right) = \cos\frac\pi2 - i\sin\frac\pi2\text{ (since $\cos\frac\pi2=0$)} $$ $$ \text{then }x=\pm\left(\cos\frac\pi4-i\sin\frac\pi4\right). $$

Because $x^8-1=(x^4+1)(x^4-1)$, the roots of $x^4+1$ are the roots of $x^8-1$ that are not roots of $x^4-1$.

The roots of $x^n-1$ are easy: $e^{\frac{2k\pi i}{n}}$ for $k=0,\dots,n-1$.

Therefore, roots of $x^4+1$ are $e^{\frac{2k\pi i}{8}}$ for $k$ odd.