Answer
$\pm1,\pm i$.
See graph.
Work Step by Step
Based on the given conditions, we have:
$1=cos0^\circ+i\ sin0^\circ)$, $(i)^{1/4}=cos(\frac{360k+0}{4})^\circ+i\ sin(\frac{360k+0}{4})^\circ)$, $k=0, z_0= cos0^\circ+i\ sin0^\circ=1$, $k=1, z_1= cos90^\circ+i\ sin90^\circ=i$, $k=2, z_2= cos180^\circ+i\ sin180^\circ=-1$, $k=3, z_3= cos270^\circ+i\ sin270^\circ=-i$.
See graph.
