Answer
Ans. -√2(1+i), √2(1+i), 2i
Work Step by Step
Given:-z=-8i
$ z^{1/3}=(-8i)^{1/3}=-2i^{1/3}$
As we can write i=$cosπ/2+isinπ/2$
Also
$ i=[cos{(\frac{π}{2}+{2πk})}+isin({\frac{π}{2}+{2πk})}]$
$ -2i^{1/3}=-2[cos(4k+1)π/6+isin(4k+1)π/6]$
After putting k=0, 1,2
We get
= -2$[cos\frac{π}{6}+isin\frac{π}{6}]$,-2$[cos\frac{5π}{6}+isin\frac{5π}{6}]$, -2$[cos\frac{3π}{2}+isin\frac{3π}{2}]$
=-√2(1+i), √2(1+i), 2i
Ans.
