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Q.
If $z = \frac{\sqrt{3}}{2}+\frac{i}{2}\left(i=\sqrt{-1}\right),$ then $\left(1+iz+z^{5}+iz^{8}\right)^{9}$ is equal to
JEE MainJEE Main 2019Complex Numbers and Quadratic Equations
Solution:
$z = \frac{\sqrt{3}}{2}+\frac{i}{2}=cos \frac{\pi}{6}+i sin \frac{\pi}{6} $
$\Rightarrow z^{5}= cos \frac{5\pi}{6}+i sin \frac{5\pi}{6}=\frac{-\sqrt{3}+i}{2}$
and $ Z^{8}= cos \frac{4\pi}{3}+i sin \frac{4\pi}{3}=-\left(\frac{1+i\sqrt{3}}{2}\right)$
$ \Rightarrow \left(1+iz+z^{5}+iz^{8}\right)^{9}=\left(1+\frac{i\sqrt{3}}{2}-\frac{1}{2}-\frac{\sqrt{3}}{2}+\frac{i}{2}+\frac{i}{2}+\frac{\sqrt{3}}{2}\right)^{9}$
$=\left(\frac{1+i\sqrt{3}}{2}\right)^{9} = cos 3\pi +i sin 3\pi =-1$