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#