How do you use DeMoivre's theorem to simplify ${(\sqrt{2} e^{15 i})}^{8}$ $(sqrt2e^(15i))^8$ ?

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Answer 1 · 2016-12-06T15:25:07

The answer is $= - 8 + i 8 \sqrt{3}$ $=-8+i8sqrt3$

We use

$e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$

And DeMoivre's theorem

${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(costheta+isintheta)^n=cosntheta+isinntheta$

$\sqrt{2} e^{15 i} = \sqrt{2} (cos 15 + i sin 15)$ $sqrt2e^(15i)=sqrt2(cos15+isin15)$

${(\sqrt{2} e^{15 i})}^{8} = {(\sqrt{2} (cos 15 + i sin 15))}^{8}$ $(sqrt2e^(15i))^8=(sqrt2(cos15+isin15))^8$

$= {(\sqrt{2})}^{8} (cos (8 \cdot 15) + i sin (8 \cdot 15))$ $=(sqrt2)^8(cos(8*15)+isin (8*15))$

$= 16 (cos 120 + i sin 120)$ $=16(cos120+isin120)$

$= 16 (- \frac{1}{2} + i \frac{\sqrt{3}}{2})$ $=16(-1/2+isqrt3/2)$

$= 8 (- 1 + i \sqrt{3})$ $=8(-1+isqrt3)$

Answer 2 · 2016-12-06T15:32:52

$= 13.0 + i 9.40$ $=13.0 + i 9.40$ , nearly.

$(\sqrt{2} e^{15 i}) 8$ $(sqrt2 e^(15i))8$

$= {(\sqrt{2})}^{8} ({(e^{15 i})}^{8}$ $=(sqrt2)^8((e^(15i))^8$

$= 16 e^{120 i}$ $=16e^(120i)$

$= 16 c i s 120$ $=16cis120$

$= 16 c i s ((38.2 π)$ $=16cis((38.2pi)$

$= 16 c i s (.2 π)$ $=16cis(.2pi)$

$= 16 (cos (36^{o}) + i sin (36^{o}))$ $=16(cos(36^o)+i sin (36^o))$

$= 12.9 + i 9.40$ $=12.9 + i 9.40$ , nearly.

How do you use DeMoivre's theorem to simplify (√2e15i)8(sqrt2e^(15i))^8?