Answer
$z_0=\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$
$z_1=-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$
$z_2=-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$
$z_3=\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$
Work Step by Step
$z^4+1=0$
$z^4=-1$
$|-1|=1$. Also $-1$ lies in the negative real axis.
Polar form: $1(cos~\pi+i~sin~\pi)$
$z_k=\sqrt[4] 1[cos(\frac{\pi+2k\pi}{4})+i~sin(\frac{\pi+2k\pi}{4})]$
$z_0=1(cos~\frac{\pi}{4}+i~sin~\frac{\pi}{4})=\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$
$z_1=1(cos~\frac{3\pi}{4}+i~sin~\frac{3\pi}{4})=-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$
$z_2=1(cos~\frac{5\pi}{4}+i~sin~\frac{5\pi}{4})=-\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$
$z_3=1(cos~\frac{7\pi}{4}+i~sin~\frac{7\pi}{4})=\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$
