Question

Question #07a2a

Answer 1

`cos (14^@) + i sin (14^@)`

Explanation:

De Moivre's theorem
`Z^n = r^n(cos (ntheta) + i sin(ntheta))`

`Z^(1/3) = r^(1/3)(cos (1/3theta) + i sin(1/3theta))`

`Z = cos(42^@) + i sin(42^@)` where `r =1`

`root(3)Z = Z^(1/3) = cos (1/3)(42^@) + i sin (1/3)(42^@)`

`Z^(1/3) = cos (14^@) + i sin (14^@)`