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.

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