Answer
$z_0=2(cos45^\circ+i\ sin45^\circ)=\sqrt 2+\sqrt 2\ i$;
$z_1=2(cos135^\circ+i\ sin135^\circ)=-\sqrt 2+\sqrt 2\ i$;
$z_2=2(cos225^\circ+i\ sin225^\circ)=-\sqrt 2-\sqrt 2\ i$;
$z_3=2(cos315^\circ+i\ sin315^\circ)=\sqrt 2-\sqrt 2\ i$.
Work Step by Step
Let $z=-16=16(-1+0i)=16(cos180^\circ+i\ sin180^\circ)$, we have
$z_k=z^{1/4}=16^{1/4}(cos(\frac{360k+180}{4})^\circ+i\ sins(\frac{360k+180}{4})^\circ)=2(cos(90k+45)^\circ+i\ sins(90k+45)^\circ)$ where $k=0,1,2,3$.
Thus, $z_0=2(cos45^\circ+i\ sin45^\circ)=\sqrt 2+\sqrt 2\ i$;
$z_1=2(cos135^\circ+i\ sin135^\circ)=-\sqrt 2+\sqrt 2\ i$;
$z_2=2(cos225^\circ+i\ sin225^\circ)=-\sqrt 2-\sqrt 2\ i$; and
$z_3=2(cos315^\circ+i\ sin315^\circ)=\sqrt 2-\sqrt 2\ i$.
