Find the argument of #(-sqrt3 +i)^8#?

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Answer 1 · 2017-04-02T15:34:15

Argument is #(2pi)/3#

#(-sqrt3 +i)^8#

= #[2(-sqrt3/2+ixx1/2)]^8#

= #[2(cos((5pi)/6)+isin((5pi)/6))]^8#

Now according to DeMoivre's Theorem

#(r(costheta+isintheta))^n=r^n(cosntheta+isinntheta)#

Hence #[2(cos((5pi)/6)+isin((5pi)/6))]^8#

= #[2^8(cos((5pi)/6xx8)+isin((5pi)/6xx8))]#

= #256(cos((20pi)/3)+isin((20pi)/3))#

= #256(cos(6pi+(2pi)/3)+isin(6pi+(2pi)/3))#

= #256(cos((2pi)/3)+isin((2pi)/3))#

Hence while modulus of #(-sqrt3 +i)^8# is #256#, argument is #(2pi)/3#