How do you use DeMoivre's theorem to simplify #(sqrt(e^(10i)))^6#?

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Answer 1 · 2017-04-04T12:42:21

#(sqrt(e^(10i)))^6=cos30+isin30=sqrt3/2+i1/2#

According to DeMoivre's Theorem if

#z=re^(itheta)=r(costheta+isintheta)#, then for all #ninQ#

#z^n=r^n e^(i ntheta)=r^n(cosntheta+isinntheta)#

Hence, #(sqrt(e^(10i)))^6#

= #((e^(10i))^(1/2))^6#

= #(e^(10i))^(1/2xx6)#

= #(e^(10i))^3#

= #e^(10ixx3)#

= #e^(30i)#

= #cos30+isin30#

= #sqrt3/2+i1/2#