Question #2e282

1 Answer
Cesareo R.
Mar 16, 2017

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))

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