Factor $z^4 +1$ into linear factors $z$ is a complex number, how do I factor $z^4 +1$ into linear factors?
Do I write z in terms of $x+yi$ so that $z^4+1=(x+yi)^4+1?$

If $z^4+1=0$ then $z^4=-1$ so $z^2=\pm i$.  If $z^2=i=\cos90^\circ+i\sin90^\circ$ then $z=\pm(\cos45^\circ+i\sin45^\circ) = \pm\left(\frac1{\sqrt2} + i\frac 1 {\sqrt2}\right)$, and similarly for $-i=\cos(-90^\circ)+i\sin(-90^\circ)$.
So
$$
z^4+1 = \left(z-\frac{1+i}{\sqrt2}\right)\left(z-\frac{-1-i}{\sqrt2}\right)\left(z-\frac{1-i}{\sqrt2}\right)\left(z-\frac{-1+i}{\sqrt2}\right).
$$
Let us note parenthetically that if we group conjugates together, thus
$$
z^4+1 = \underbrace{\left(z-\frac{1+i}{\sqrt2}\right)\left(z-\frac{1-i}{\sqrt2}\right)}_\text{conjugates}\quad  \underbrace{\left(z-\frac{-1+i}{\sqrt2}\right)\left(z-\frac{-1-i}{\sqrt2}\right)}_\text{conjugates}
$$
then when conjugates get multiplied, the imaginary parts cancel, so in this case we get:
$$
z^4+1 = \big(z^2 - z\sqrt2 + 1\big)\big(z^2+z\sqrt2+1\big).
$$
