Answer
$z_0=cos~\frac{\pi}{12}+i~sin~\frac{\pi}{12}$
$z_1=cos~\frac{5\pi}{12}+i~sin~\frac{5\pi}{12}$
$z_2=cos~\frac{3\pi}{4}+i~sin~\frac{3\pi}{4}=-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$
$z_3=cos~\frac{13\pi}{12}+i~sin~\frac{7\pi}{12}$
$z_4=cos~\frac{17\pi}{12}+i~sin~\frac{9\pi}{12}$
$z_5=cos~\frac{7\pi}{4}+i~sin~\frac{7\pi}{4}=\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$
Work Step by Step
$z^6-i=0$
$z^6=i$
$r=|i|=1$. Also $i$ lies in the positive real axis.
Polar form of $i$:
$1(cos~\frac{\pi}{2}+i~sin~\frac{\pi}{2})$
$z_k=\sqrt[6]{1}[cos(\frac{\frac{\pi}{2}+2k\pi}{6})+i~sin(\frac{\frac{\pi}{2}+2k\pi}{6})]$
$z_0=1(cos~\frac{\pi}{12}+i~sin~\frac{\pi}{12})$
$z_1=1(cos~\frac{5\pi}{12}+i~sin~\frac{5\pi}{12})$
$z_2=1(cos~\frac{3\pi}{4}+i~sin~\frac{3\pi}{4})=-\frac{\sqrt 2}{2}+\frac{\sqrt 2}{2}i$
$z_3=1(cos~\frac{13\pi}{12}+i~sin~\frac{7\pi}{12})$
$z_4=1(cos~\frac{17\pi}{12}+i~sin~\frac{9\pi}{12})$
$z_5=1(cos~\frac{7\pi}{4}+i~sin~\frac{7\pi}{4})=\frac{\sqrt 2}{2}-\frac{\sqrt 2}{2}i$
