A) 1
B) - 1
C) 3
D) 0
Correct Answer: D
Solution :
\[{{\omega }^{99}}+{{\omega }^{1\omega }}+{{\omega }^{101}}\] = \[{{\omega }^{99}}[1+\omega +{{\omega }^{2}}]\] [Since \[1+\omega +{{\omega }^{2}}=0,\,{{\omega }^{3}}=1\]] = 0.You need to login to perform this action.
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