lf a=cos4π3+isin4π3 then 1+a23n=

a=cos4π3+i sin4π3=1+a=1+cos4π3+i sin4π3 =1-12-i32 =12-i321+a=-w(1+a)3n=-(w3)n 1+a23n=-123n

lf a=cos4π3+isin4π3 then 1+a23n=

a=cos4π3+i sin4π3=1+a=1+cos4π3+i sin4π3 =1-12-i32 =12-i321+a=-w(1+a)3n=-(w3)n 1+a23n=-123n

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$lf a = \cos \frac{4 π}{3} + isin \frac{4 π}{3} then {(\frac{1 + a}{2})}^{3 n} =$

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${(\frac{1}{2})}^{3 n}$

${(- \frac{1}{2})}^{3 n}$

${(- \frac{1}{2})}^{n}$

${(\frac{1}{2})}^{n}$

answer is B.

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Detailed Solution

$\begin{array}{rcl} a & = & \cos \frac{4 π}{3} + i \sin \frac{4 π}{3} = \\ 1 + a & = & 1 + \cos \frac{4 π}{3} + i \sin \frac{4 π}{3} \\ = & 1 - \frac{1}{2} - \frac{i \sqrt{3}}{2} \\ = & \frac{1}{2} - \frac{i \sqrt{3}}{2} \\ 1 + a & = & - w \\ {(1 + a)}^{3 n} & = & - {(w^{3})}^{n} \end{array}$