A) 1
B) 2
C) \[-i\]
D) \[-1\]
Correct Answer: B
Solution :
\[z=\frac{\sqrt{3}+i}{-2}\] \[\Rightarrow \] \[z=-\frac{\sqrt{3}}{2}-\frac{i}{2}\] \[\Rightarrow \]\[z=i\left( \frac{i\sqrt{3}}{2}-\frac{1}{2} \right)=i\left( -\frac{1}{2}+i\frac{\sqrt{3}}{2} \right)=i\omega \] \[\therefore \]\[{{z}^{69}}={{(i\omega )}^{69}}={{({{i}^{4}})}^{17}}.i{{({{\omega }^{3}})}^{23}}\] \[=i[\because {{\omega }^{3}}=1]\]
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