Question #0c21c

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Answer 1 · 2017-04-11T16:33:57

$- 128 (1 + i \sqrt{3}) .$ $-128(1+isqrt3).$

Let $z = 1 - i \sqrt{3} = r (cos θ + i sin θ) = r c i s θ, r > 0 .$ $z=1-isqrt3=r(costheta+isintheta)=rcistheta, r >0.$

$:. rcostheta=1, rsintheta=-sqrt3.$

Squaring and adding, $r^2(cos^2theta+sin^2theta)=1+3.$

$rArr r=2.$

$:. costheta=1/r=1/2, and, sintheta=-sqrt3/2.$

$:. theta=-pi/3.$

$rArr z=rcistheta=2cis(-pi/3).$

By De' Moivre's Theorem, then,

$(1-isqrt3)^8=z^8={2cis(-pi/3)}^8=2^8cis{8(-pi/3)}.$

$=2^8{cos(-8pi/3)+isin(-8pi/3)},$

$=2^8{cos(8pi/3)-isin(8pi/3)},$

$=2^8{cos(3pi-pi/3)-isin(3pi-pi/3)},$

$=2^8{-cos(pi/3)-isin(pi/3)},$

$=-2^8(1/2+isqrt3/2),$

$=-2^7(1+isqrt3).$

Hence, the Reqd. Value= $-128(1+isqrt3).$

Enjoy Maths.!