Question #0c21c

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Answer 1 · 2017-04-11T16:33:57

#-128(1+isqrt3).#

Let #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.!