Being $z=x+yi$ how can I factorize the polynomial $z^4+1$ as a product of real quadratic polynomials? Being $z=x+yi$ how can I factorize the polynomial $z^4+1$ as a product of real quadratic polynomials?

I don't really understand what I am asked to do. How can I start with it?
 A: I think, the following is better. $$z^4+1=z^4+2z^2+1-2z^2=(z^2+1)^2-(\sqrt2z)^2=$$
$$=(z^2-\sqrt2z+1)(z^2+\sqrt2z+1).$$
Also, $$z^6+1=(z^2+1)(z^4-z^2+1)=(z^2+1)((z^2+1)^2-3z^2)=$$
$$=(z^2+1)(z^2-\sqrt3z+1)(z^2+\sqrt3z+1).$$
A: I agree with the other answers but prefer an intuitive approach,
which uses the idea that $e^{(i\theta)} = \cos \theta + i\sin \theta$.
You want all values $e^{(i\alpha)}$ such that 
$\left[e^{(i\alpha)}\right]^4 = e^{(i4\alpha)} = -1
= e^{(i\pi)}.$
The easiest way to do that is to pretend that 
$e^{(i\pi)}$ can actually be represented by
the 4 elements $\{e^{(i\pi)}, e^{(i3\pi)}, e^{(i5\pi)}, e^{(i7\pi)}\}.$
Then, with the argument of each of the 4 elements divided by 4, 
you see that the 4 distinct roots are 
$\{e^{(i\pi/4)}, e^{(i3\pi/4)}, e^{(i5\pi/4)}, e^{(i7\pi)/4}\}.$
Having identified the 4 roots, you need to combine them into
conjugate pairs, and then use each pair of roots to form a
quadratic.
This results in $\left[\left(z - e^{(i\pi/4)}\right)
\left(z - e^{(i7\pi/4)}\right)\right] \times
\left[\left(z - e^{(i3\pi/4)}\right)
\left(z - e^{(i5\pi/4)}\right)\right]
$
$=\left[
\left(z - \langle\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\rangle\right)
\left(z - \langle\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\rangle\right)
\right]$
$\times
\left[
\left(z - \langle-\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\rangle\right)
\left(z - \langle-\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}\rangle\right)
\right]
$
$= \left(z^2 -\sqrt{2}z + \frac{1}{2} + \frac{1}{2}\right) \times
\left(z^2 +\sqrt{2}z + \frac{1}{2} + \frac{1}{2}\right)
$
$= \left(z^2 -\sqrt{2}z + 1\right) \times
\left(z^2 +\sqrt{2}z + 1\right)
$
$= (z^4 + 1).$
Addendum
Per OP's request : attacking $(z^6 + 1) = 0.$
As in the original answer, pretend that 
$e^{(i\pi)}$ can actually be represented by
the 6 elements $\{e^{(i\pi)}, e^{(i3\pi)}, e^{(i5\pi)}, e^{(i7\pi)}, e^{(i9\pi)}, e^{(i11\pi)}\}.$
Then, with the argument of each of the 6 elements divided by 6, 
you see that the 6 distinct roots are 
$\{e^{(i\pi/6)}, e^{(i3\pi/6)}, e^{(i5\pi/6)}, e^{(i7\pi)/6}, e^{(i9\pi/6)}, e^{(i11\pi)/6}\}.$
Having identified the 6 roots, you need to combine them into
conjugate pairs, and then use each pair of roots to form a
quadratic.
This results in $\left[\left(z - e^{(i\pi/6)}\right)
\left(z - e^{(i11\pi/6)}\right)\right]$
$\times
\left[\left(z - e^{(i3\pi/6)}\right)
\left(z - e^{(i9\pi/6)}\right)\right]
$
$\times
\left[\left(z - e^{(i5\pi/6)}\right)
\left(z - e^{(i7\pi/6)}\right)\right].
$
The rest of the conversion into real quadratics would follow
the same method as in the original answer, simply multiplying everything out.
What makes this answer convenient is that all 6 roots of 
$\left[e^{i\pi}\right]^{(1/6)}$ are special angles 
each of whose sin and cosine key off of 
$e^{(i\pi/6)}.$
A: So you understood the hint
\begin{eqnarray*}
(z^2+\alpha z +1)(z^2-\alpha z +1)=z^4+\underbrace{(2-\alpha^2)}_{2-\alpha^2=0}z^2+1.
\end{eqnarray*}
To do the next one in your comment ... Factorise $z^6+1=(z^2+1)(z^4-z^2+1)$
\begin{eqnarray*}
(z^2+\alpha z +1)(z^2-\alpha z +1)=z^4+\underbrace{(2-\alpha^2)}_{2-\alpha^2=-1}z^2+1.
\end{eqnarray*}
So
\begin{eqnarray*}
z^6+1=(z^2+1)(z^2+\sqrt{3}z+1)(z^2-\sqrt{3}z+1).
\end{eqnarray*}
A: Direct factorization treating imaginary $ (i^2=-1) $ number algebraically
$$ z^4+1= (z^2-i)(z^2+i)=(z-\sqrt i)(z+\sqrt i)(z-i\sqrt i)(z+i\sqrt i)$$
The arguments in the complex plane are odd multiples of $\pi/4$ because exponent directly multiplies/divides the argument, tips of radius vector are at $ (2k-1) \pi/2$ on unit circle.

If $\sin \pi/4=  \cos \pi/4=\dfrac{1}{\sqrt2}=q,\;$ then the four factors are
$$=(z+(-1-q))(z+(-1+q))(z+(1-q))(z+(1+q)).$$
The equation is encountered as elastic foundation ode for plates.
