Question #f6317

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Question #f6317

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Answer 1 · 2016-05-18T04:21:55

I have corrected the question. ${(\sqrt{3} - i)}^{3} = - 8 i$ $(sqrt 3 - i)^3=-8i$

Explanation:

De Moivre's Theorem; If $c i s θ = cos θ + i sin θ$ $cis theta = cos theta + i sin theta$ , then

${(c i s θ)}^{n} = c i s n θ = cos n θ + i sin n θ$ $(cis theta)^n = cis ntheta = cos ntheta + i sin ntheta$

compare $(\sqrt{3} - i)$ $(sqrt 3 - i )$ with $r c i s θ$ $r cis theta$ .

$r cos θ = \sqrt{3} and r sin θ = - 1$ $r cos theta = sqrt 3 and r sin theta = -1$ . Solving,

$r = 2 and θ = - \frac{π}{6}$ $r = 2 and theta = -pi/6$ .

So, ${(\sqrt{3} - i)}^{3} = (2 {(c i s (- \frac{π}{6}))}^{3} = 2^{3} c i s (3 (- \frac{π}{6})) = 8 c i s (- \frac{π}{2}) = 8 (cos (- \frac{π}{2}) + i sin (- \frac{π}{2})) = 8 (0 - i) = - 8 i$ $(sqrt 3 - i )^3= (2(cis (-pi/6))^3= 2^3 cis (3(-pi/6))=8 cis (-pi/2)=8 (cos(- pi/2)+isin(-pi/2))=8(0-i)=-8i$ .

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Question #f6317

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