Answer
Ans. 1, $\frac{-1}{2}+\frac{√3i}{2}$, $\frac{-1}{2}+\frac{-√3i}{2}$
Work Step by Step
Given:- z=1
$ z^{1/3}=1^{1/3}$
As we know that
$ z^{1/3}=r^{1/3}[cos{(\frac{θ}{n}+\frac{2πk}{n})}+isin({\frac{θ}{n}+\frac{2πk}{n})}]$
Put θ=π/2
We get
=$cos\frac{2kπ}{3}+isin\frac{2kπ}{3}$
After putting k=0, 1,2
We get
=$cos0+isin0$, $cos\frac{2π}{3}+isin\frac{2π}{3}$, $cos\frac{4π}{3}+isin\frac{4π}{3}$
=1, $\frac{-1}{2}+\frac{√3i}{2}$, $\frac{-1}{2}+\frac{-√3i}{2}$
Ans.
