A) 2
B) zero
C) \[-1\]
D) 1
Correct Answer: C
Solution :
Key Idea: \[\omega =\frac{-1+i\sqrt{3}}{2}\]and\[{{\omega }^{2}}=\frac{-1-i\sqrt{3}}{2}\]where\[\omega ,{{\omega }^{2}}\]are cube roots of unity such that \[1+\omega +{{\omega }^{2}}=0\] Given \[{{\left( \frac{-1+\sqrt{-3}}{2} \right)}^{100}}+{{\left( \frac{-1-\sqrt{-3}}{2} \right)}^{100}}\] \[={{\left( \frac{-1+i\sqrt{3}}{2} \right)}^{100}}+{{\left( \frac{-1-i\sqrt{3}}{2} \right)}^{100}}\] \[={{(\omega )}^{100}}+{{({{\omega }^{2}})}^{100}}=\omega +{{\omega }^{2}}=-1\]
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