Question:
If $z=\frac{\sqrt{3}}{2}+\frac{i}{2}(i=\sqrt{-1})$, then $\left(1+i z+z^{5}+i z^{8}\right)^{9}$ is equal
to:
Correct Option: , 4
Solution:
$\frac{\sqrt{3}}{2}+\frac{i}{2}=-i\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=-i \omega$
where $\omega$ is imaginary cube root of unity.
Now, $\left(1+i z+z^{5}+i z^{8}\right)^{9}$
$=\left(1+\omega-i \omega^{2}+i \omega^{2}\right)^{9}=(1+\omega)^{9}$
$=\left(-\omega^{2}\right)^{9}=-\omega^{18}=-1 \quad\left(\because 1+\omega+\omega^{2}=0\right)$
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