How do you use DeMoivre's theorem to simplify ${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3}$ $(-1/2-sqrt3/2i)^3$ ?

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How do you use DeMoivre's theorem to simplify ${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3}$ $(-1/2-sqrt3/2i)^3$ ?

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Answer 1 · 2016-09-29T22:58:35

${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3} = 1$ $(-1/2-sqrt(3)/2i)^3 = 1$

Explanation:

de Moivre's theorem tells us that:

${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(cos theta + i sin theta)^n = cos n theta + i sin n theta$

So we find:

${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3} = {(cos (- \frac{2 π}{3}) + i sin (- \frac{2 π}{3}))}^{3}$ $(-1/2-sqrt(3)/2i)^3 = (cos (-(2pi)/3) + i sin (-(2pi)/3))^3$

${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3} = cos (- 2 π) + i sin (- 2 π)$ $color(white)((-1/2-sqrt(3)/2i)^3) = cos (-2pi) + i sin (-2pi)$

${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3} = 1 + i \cdot 0$ $color(white)((-1/2-sqrt(3)/2i)^3) = 1 + i * 0$

${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3} = 1$ $color(white)((-1/2-sqrt(3)/2i)^3) = 1$

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How do you use DeMoivre's theorem to simplify ${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3}$ $(-1/2-sqrt3/2i)^3$ ?

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How do you use DeMoivre's theorem to simplify ${(- \frac{1}{2} - \frac{\sqrt{3}}{2} i)}^{3}$ $(-1/2-sqrt3/2i)^3$ ?