finding real roots by way of complex I was given 
$$x^4 + 1$$
and was told to find its real factors. I found the $((x^2 + i)((x^2 - i))$ complex factors but am lost as to how the problem should be approached.
My teacher first found 4 complex roots ( different than mine)
$$( x - \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)( x - \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)$$
$$( x + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)( x + \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} i)$$
He then mltiplied both pairs,resulting in:
$$(x^2 - \sqrt{2}x + 1) and (x^2 + \sqrt{2}x + 1)$$
How do I get the first four imaginary points, or the two distinct because I realize that each has a conjugate.
 A: You can for exemple complete the square $$x^4+1 = x^4 + 2x^2 + 1 - 2x^2 = (x^2+1)^2-2x^2 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1).$$
Now you can use the traditionnal method for each factor.
An other possibility is the used the polar form.
A: $$x^4+1=0\Longleftrightarrow$$
$$x^4=-1\Longleftrightarrow$$
$$x^4=|-1|e^{\arg(-1)i}\Longleftrightarrow$$
$$x^4=e^{\pi i}\Longleftrightarrow$$
$$x=\left(e^{\left(2\pi k+\pi\right)i}\right)^{\frac{1}{4}}\Longleftrightarrow$$
$$x=e^{\frac{1}{4}\left(2\pi k+\pi\right)i}$$
With $k\in\mathbb{Z}$ and $k:0-3$

So, the solutions are:
$$x_0=e^{\frac{1}{4}\left(2\pi\cdot0+\pi\right)i}=e^{\frac{\pi i}{4}}$$
$$x_1=e^{\frac{1}{4}\left(2\pi\cdot1+\pi\right)i}=e^{\frac{3\pi i}{4}}$$
$$x_2=e^{\frac{1}{4}\left(2\pi\cdot2+\pi\right)i}=e^{-\frac{3\pi i}{4}}$$
$$x_3=e^{\frac{1}{4}\left(2\pi\cdot3+\pi\right)i}=e^{-\frac{\pi i}{4}}$$
A: change equation to
$$x^4 = -1 = e^{\pi i}$$
$$x = e^{(\frac{\pi}{4} + k\frac{\pi}{2})i}$$
where $k = 0, 1, 2, 3$. All four roots are evenly disitributed on the unit circle in complex plane.
A: From $$x^4 + 1 = (x^2+i)(x^2-i)$$ we need to find the roots of each factor. Remember that $i = e^{i\pi/2}$, so $x=e^{i\pi/4}$ is a root of $x^2-i$. Hence, so is its conjugate $e^{-i\pi/4}$. By Euler's formula,
$$
e^{i\pi/4} = \cos(\pi/4) + i\sin(\pi/4) = \frac{\sqrt{2}}{2} + i\frac{\sqrt 2}{2}.
$$
I will leave you to do the others.
