How do you use DeMoivre's Theorem to simplify ${(2 (cos (\frac{π}{10}) + i sin (\frac{π}{10})))}^{5}$ $(2(cos(pi/10)+isin(pi/10)))^5$ ?

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Answer 1 · 2016-10-21T08:27:59

${(2 (cos (\frac{π}{10}) + i sin (\frac{π}{10})))}^{5} = 32 i$ $(2(cos(pi/10)+isin(pi/10)))^5=32i$

DeMoivre's Theorem states that for a complex number $z = r (cos θ + i sin θ)$ $z=r(costheta+isintheta)$ ,

$z^{n} = r^{n} (cos n θ + i r sin n θ)$ $z^n=r^n(cosntheta+irsinntheta)$

Hence ${(2 (cos (\frac{π}{10}) + i sin (\frac{π}{10})))}^{5}$ $(2(cos(pi/10)+isin(pi/10)))^5$

= $2^{5} (cos n (5 \times \frac{π}{10}) + i sin (5 \times \frac{π}{10}))$ $2^5(cosn(5xxpi/10)+isin(5xxpi/10))$

= $32 (cos n (\frac{π}{2}) + i sin (\frac{π}{2}))$ $32(cosn(pi/2)+isin(pi/2))$

= $32 (0 + i \times 1)$ $32(0+ixx1)$

= $32 i$ $32i$

How do you use DeMoivre's Theorem to simplify (2(cos(pi/10)+isin(pi/10)))^5(2(cos(π10)+isin(π10)))5(2(cos(pi/10)+isin(pi/10)))^5?