How do you use DeMoivre's theorem to simplify ${(3 (cos (\frac{5 π}{6}) + i sin (\frac{5 π}{6})))}^{4}$ $(3(cos((5pi)/6)+isin((5pi)/6)))^4$ ?

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Answer 1 · 2016-08-25T12:53:35

$- \frac{81}{2} (1 + i \sqrt{3})$ $-81/2(1+isqrt3)$ .

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

$z = r cos θ + i sin θ, z^{n} = r^{n} (cos n θ + i sin n θ)$ $z=rcostheta+isintheta, z^n=r^n(cos ntheta+isin ntheta)$

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

$= 3^{4} {cos (4 (5 \frac{π}{6})) + i sin (4 (5 \frac{π}{6}))}$ $=3^4{cos(4(5pi/6))+isin(4(5pi/6))}$ .

$= 81 (cos (10 \frac{π}{3}) + i sin (10 \frac{π}{3}))$ $=81(cos(10pi/3)+isin(10pi/3))$ .

$= 81 (cos (3 π + \frac{π}{3}) + i sin (3 π + \frac{π}{3}))$ $=81(cos(3pi+pi/3)+isin(3pi+pi/3))$ .

$= 81 (- cos (\frac{π}{3}) - i sin (\frac{π}{3}))$ $=81(-cos(pi/3)-isin(pi/3))$ .

$= - 81 (\frac{1}{2} + i \frac{\sqrt{3}}{2})$ $=-81(1/2+isqrt3/2)$ .

$= - \frac{81}{2} (1 + i \sqrt{3})$ $=-81/2(1+isqrt3)$ .

How do you use DeMoivre's theorem to simplify (3(cos((5pi)/6)+isin((5pi)/6)))^4(3(cos(5π6)+isin(5π6)))4(3(cos((5pi)/6)+isin((5pi)/6)))^4?