Answer
Ans. 2,(1+√3i), (-1+√3i), -2, (-1-√3i), (1-√3i)
Work Step by Step
Given:-z=64
$z^{1/6}=64^{1/6}$
$z^{1/n}=64^{1/n}[cos(\frac{θ}{n}+\frac{2πk}{n})+isin(\frac{θ}{n}+\frac{2πk}{n})]$
After putting n=6
$z^{1/6}=2[cos(\frac{πk}{3})+isin(\frac{πk}{3})]$
Putting the value of k =0, 1,2,3,4,5
We get
$z^{1/6}=2[cos(0)+isin(0)]$, $2[cos(π/3)+isin(π/3)]$, $2[cos(2π/3)+isin(2π/3)]$, $2[cos(π)+isin(π)]$, $2[cos(4π/3)+isin(4π/3)]$, $2[cos(5π/3)+isin(5π/3)]$
We get after solving
2,(1+√3i), (-1+√3i), -2, (-1-√3i), (1-√3i)
Ans.
