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 · 2016-11-23T14:03:44

The answer is $= 1$ $=1$

Explanation:

Let $z = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ $z=-1/2+isqrt3/2$

We have to write $z$ $z$ in trigonometric form in order to apply DeMoivre's theorem

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

$r = 1$ $r=1$

$cos θ = - \frac{1}{2}$ $costheta=-1/2$

$sin θ = \frac{\sqrt{3}}{2}$ $sintheta=sqrt3/2$

So, $θ$ $theta$ is located in the 2nd quadrant

$θ = 2 \frac{π}{3}$ $theta=2pi/3$

Therefore, $z = cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3})$ $z=cos((2pi)/3)+isin((2pi)/3)$

We need $z^{3}$ $z^3$

We use DeMoivre 's theorem,
${(cos θ + i sin θ)}^{n} = cos n θ + i sin n θ$ $(costheta+isintheta)^n=cosntheta+isinntheta$

$z^{3} = {(cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3}))}^{3}$ $z^3=(cos((2pi)/3)+isin((2pi)/3))^3$

$= cos (2 \frac{π}{3} \cdot 3) + i sin (2 \frac{π}{3} \cdot 3)$ $=cos(2pi/3*3)+isin(2pi/3*3)$

$= cos 2 π + i sin 2 π$ $=cos2pi+isin2pi$

$= 1$ $=1$

Answer 2 · 2016-11-23T14:05:03

${(- \frac{1}{2} + \frac{\sqrt{3}}{2} i)}^{3} = 1$ $(-1/2+sqrt3/2i)^3=1$

Explanation:

According to DeMoivre's theorem, if $a + i b = r cos θ + i r sin θ$ $a+ib=rcostheta+irsintheta$

then ${(a + i b)}^{n} = r^{n} (cos n θ + i sin n θ)$ $(a+ib)^n=r^n(cosntheta+isinntheta)$

Writing $- \frac{1}{2} + \frac{\sqrt{3}}{2} i$ $-1/2+sqrt3/2i$ in trigonometric form

$- \frac{1}{2} + \frac{\sqrt{3}}{2} i = cos (\frac{2 π}{3}) + i sin (\frac{2 π}{3})$ $-1/2+sqrt3/2i=cos((2pi)/3)+isin((2pi)/3)$

Hence ${(- \frac{1}{2} + \frac{\sqrt{3}}{2} i)}^{3} = 1^{3} (cos (3 \times \frac{2 π}{3}) + i sin (3 \times \frac{2 π}{3}))$ $(-1/2+sqrt3/2i)^3=1^3(cos(3xx(2pi)/3)+isin(3xx(2pi)/3))$

= $cos (2 π) + i sin (2 π)$ $cos(2pi)+isin(2pi)$

= $1 + i 0$ $1+i0$

= $1$ $1$

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