How do you use DeMoivre's theorem to simplify $(3(cos((5pi)/6)+isin((5pi)/6)))^4$ ?

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Answer 1 · 2016-08-25T12:53:35

$-81/2(1+isqrt3)$ .

DeMoivre's Theorem states that, for a complex no.

$z=rcostheta+isintheta, z^n=r^n(cos ntheta+isin ntheta)$

Hence, {3(cos(5pi/6)+isin(5pi/6)}^4

$=3^4{cos(4(5pi/6))+isin(4(5pi/6))}$ .

$=81(cos(10pi/3)+isin(10pi/3))$ .

$=81(cos(3pi+pi/3)+isin(3pi+pi/3))$ .

$=81(-cos(pi/3)-isin(pi/3))$ .

$=-81(1/2+isqrt3/2)$ .

$=-81/2(1+isqrt3)$ .

How do you use DeMoivre's theorem to simplify (3(cos((5pi)/6)+isin((5pi)/6)))^4(3(cos((5pi)/6)+isin((5pi)/6)))^4?