A) -1
B) 1
C) 0
D) \[{{(-1+2i)}^{9}}\]
Correct Answer: A
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\[{{z}^{8}}=\cos \frac{4\pi }{3}+i\,\sin \frac{4\pi }{3}=-\left( \frac{1+i\sqrt{3}}{2} \right)\] \[\Rightarrow \]\[{{\left( 1+iz+{{z}^{5}}+i{{z}^{8}} \right)}^{9}}\] \[={{\left( 1+\frac{i\sqrt{3}}{2}-\frac{1}{2}-\frac{\sqrt{3}}{2}+\frac{i}{2}-\frac{i}{2}+\frac{\sqrt{3}}{2} \right)}^{9}}\] \[={{\left( \frac{1+i\sqrt{3}}{2} \right)}^{9}}=\cos 3\pi +i\sin 3\pi =-1\]
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