How do you use DeMoivre's theorem to simplify ${(5 (cos (\frac{π}{9}) + i sin (\frac{π}{9})))}^{3}$ $(5(cos(pi/9)+isin(pi/9)))^3$ ?

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Answer 1 · 2016-08-06T08:56:08

$= 125 (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ $=125(1/2+ (sqrt(3))/2i)$

Could also write as $125 e^{\frac{i π}{3}}$ $125e^((ipi)/3)$ using Euler's formula if you so desired.

De Moivre's theorem states that for complex number

$z = r (cos θ + i sin θ)$ $z = r(costheta + isintheta)$

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

So here,

$z = 5 (cos (\frac{π}{9}) + i sin (\frac{π}{9}))$ $z = 5(cos(pi/9) + isin(pi/9))$

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

$= 125 (\frac{1}{2} + \frac{\sqrt{3}}{2} i)$ $=125(1/2+ (sqrt(3))/2i)$

How do you use DeMoivre's theorem to simplify (5(cos(π9)+isin(π9)))3(5(cos(pi/9)+isin(pi/9)))^3?