Question #eece4

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Answer 1 · 2017-12-02T00:09:22

$-128+128sqrt3*i$

$-1-sqrt3*i=2*(-1/2-sqrt3/2*i)=2cis((4pi)/3)$

Hence,

$(-1-sqrt3*i)^8$

= $2^8*Cis((4pi)/3*8)$

= $2^8*Cis((32pi)/3)$

= $2^8*Cis((2pi)/3)$

= $2^8*(cos((2pi)/3)+i*sin((2pi)/3))$

= $2^8*(-1/2+sqrt3/2*i)$

= $-128+128sqrt3*i$