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Question #2e282

1 Answer

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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