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B
10
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C
9
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D
12
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Solution
The correct option is B12 (√3−i)n=2n (2.ei−π6)n=2n 2ne−inπ6=2n 2n(e−inπ6−1)=0 Since 2n≠0, Hence e−inπ6−1=0 e−inπ6=1 e−inπ6=ei2kπ where k=±1,±2,±3.. −nπ6=2kπ n=−12k Hence n is a multiples of 12.