Question #f6317

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Answer 1 · 2016-05-18T04:21:55

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

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

$(cis theta)^n = cis ntheta = cos ntheta + i sin ntheta$

compare $(sqrt 3 - i )$ with $r cis theta$ .

$r cos theta = sqrt 3 and r sin theta = -1$ . Solving,

$r = 2 and theta = -pi/6$ .

So, $(sqrt 3 - i )^3= (2(cis (-pi/6))^3= 2^3 cis (3(-pi/6))=8 cis (-pi/2)=8 (cos(- pi/2)+isin(-pi/2))=8(0-i)=-8i$ .

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