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Question

The smallest positive integral value of n for which (1+3i)n/2 is real is

A
3
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B
6
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C
12
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D
0
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Solution

The correct option is A 6

Given, (1+3i)n2


=2n2[1+3i2]n2


=2n2(eiπ3n2)


=2n2.einπ6

Hence we get a real value for nϵ6k where k=0,1,2...

Hence the smallest integral value of n is 6.



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