Question

Question #2e282

Answer 1

`1/2(-1+isqrt(3))`

Explanation:

We have that `z=-1/2(1+isqrt(3))=abs ze^(i "arg"( z))`

where `"arg"(z)=arctan((-sqrt(3))/(-1))`

but `abs z=1` and `"arg"(z)=-(2pi)/3`

so `z^(83)=(e^(-i(2pi)/3))^83 = e^(-i(166pi)/3) = e^(i(2pi)/3) = 1/2(-1+isqrt(3))`