How do you use DeMoivre's theorem to simplify #(sqrt2e^(15i))^8#?

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Answer 1 · 2016-12-06T15:25:07

The answer is #=-8+i8sqrt3#

We use

#e^(itheta)=costheta+isintheta#

And DeMoivre's theorem

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

#sqrt2e^(15i)=sqrt2(cos15+isin15)#

#(sqrt2e^(15i))^8=(sqrt2(cos15+isin15))^8#

#=(sqrt2)^8(cos(8*15)+isin (8*15))#

#=16(cos120+isin120)#

#=16(-1/2+isqrt3/2)#

#=8(-1+isqrt3)#

Answer 2 · 2016-12-06T15:32:52

#=13.0 + i 9.40#, nearly.

#(sqrt2 e^(15i))8#

#=(sqrt2)^8((e^(15i))^8#

#=16e^(120i)#

#=16cis120#

#=16cis((38.2pi)#

#=16cis(.2pi)#

#=16(cos(36^o)+i sin (36^o))#

#=12.9 + i 9.40#, nearly.