How do you use DeMoivre's Theorem to simplify #(2(cos(pi/10)+isin(pi/10)))^5#?

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Answer 1 · 2016-10-21T08:27:59

#(2(cos(pi/10)+isin(pi/10)))^5=32i#

DeMoivre's Theorem states that for a complex number #z=r(costheta+isintheta)#,

#z^n=r^n(cosntheta+irsinntheta)#

Hence #(2(cos(pi/10)+isin(pi/10)))^5#

= #2^5(cosn(5xxpi/10)+isin(5xxpi/10))#

= #32(cosn(pi/2)+isin(pi/2))#

= #32(0+ixx1)#

= #32i#