How do I find all possible complex roots of a polynomial with a degree of 4?

Question

The problem: find all possible complex roots of

$$P(x)=x^4 + 1$$

and write it down in a form of $a+ib$.

Any hints on how I should start?

Answer 1

$x^4+1=0~=>~\Big(x^2\Big)^2=-1~=>~x^2=\pm i~=>~x=\pm\dfrac{1\pm i}{\sqrt2}$

Answer 2

Rewite as $x^4 = -1$. Now find the 4 roots of unity for $-1$. Rewrite $-1$ as $e^{i\pi}$ for starters.

Answer 3

$\displaystyle x^4 + 1 = (x^2)^2 - (i)^2 = (x^2 + i)(x^2 - i) = [(x)^2 - i^2\cdot (i^{\frac 12})^2]\cdot[(x)^2 - (i^{\frac 12})^2] = (x-i^{\frac 32})(x+i^{\frac32})(x-i^{\frac 12})(x+i^{\frac12})$

Now $\displaystyle i = e^{i\frac{\pi}{2}}$ giving $\displaystyle i^{\frac12} = e^{i\frac{\pi}{4}} = \frac{\sqrt 2}{2}(1+i)$ and $\displaystyle i^{\frac 32} = e^{i\frac{3\pi}{4}} = \frac{\sqrt 2}{2}(-1+i)$

Hence the final factorisation is:

$\displaystyle x^4 + 1 = [x-\frac{\sqrt 2}{2}(-1+i)][x-\frac{\sqrt 2}{2}(1-i)][x-\frac{\sqrt 2}{2}(1+i)][x-\frac{\sqrt 2}{2}(-1-i)]$

From this form, the solutions to $\displaystyle P(x) = 0$ should be obvious.