If a=3+i2 then the value of 1+a3+a6+a9+a12+a15 is

a=3+i2=12×2cos⁡π6+isin⁡π6a3=cos⁡π2+isin⁡π2=ia6=−1a9=−ia12=1a15=i ∴1+a3+a6+a9+a12+a15=1+i−1−i+1+i=1+i

If a=3+i2 then the value of 1+a3+a6+a9+a12+a15 is

a=3+i2=12×2cos⁡π6+isin⁡π6a3=cos⁡π2+isin⁡π2=ia6=−1a9=−ia12=1a15=i ∴1+a3+a6+a9+a12+a15=1+i−1−i+1+i=1+i

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If $a = \frac{\sqrt{3} + i}{2}$ then the value of $1 + a^{3} + a^{6} + a^{9} + a^{12} + a^{15}$ is

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$1 + i$

$\frac{\sqrt{3} + i}{2}$

answer is C.

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Detailed Solution

$\begin{aligned} a = \frac{\sqrt{3} + i}{2} \\ = \frac{1}{2} \times 2 (\cos \frac{π}{6} + i \sin \frac{π}{6}) \\ a^{3} = \cos \frac{π}{2} + i \sin \frac{π}{2} = i \\ a^{6} = - 1 \\ a^{9} = - i \\ a^{12} = 1 \\ a^{15} = i ∴ 1 + a^{3} + a^{6} + a^{9} + a^{12} + a^{15} \\ = 1 + i - 1 - i + 1 + i = 1 + i \end{aligned}$