How do you use DeMoivre's theorem to simplify ${(5 e^{15 i})}^{3}$ $(5e^(15i))^3$ ?

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How do you use DeMoivre's theorem to simplify ${(5 e^{15 i})}^{3}$ $(5e^(15i))^3$ ?

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Answer 1 · 2016-11-07T19:29:08

The answer is $= 125 (cos 45 + i sin 45)$ $=125(cos45+isin45)$

Explanation:

DeMoivre theorem states that
${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(costheta+isintheta)^n=cosntheta+isinntheta$
and $e^{i θ} = cos θ + i sin θ$ $e^(itheta)=costheta+isintheta$
Here we have $5 e^{15 i} = 5 (cos 15 + i sin 15)$ $5e^(15i)=5(cos15+isin15)$
$:.(5e^(15i))^3=5^3(cos15+isin15)^3$
$=125(cos45+isin45)$

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How do you use DeMoivre's theorem to simplify ${(5 e^{15 i})}^{3}$ $(5e^(15i))^3$ ?

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How do you use DeMoivre's theorem to simplify ${(5 e^{15 i})}^{3}$ $(5e^(15i))^3$ ?