Verified Solution • 2025–26 Session moderate

QJEEM21Ma57dd: JEE Main 2021 Mathematics - Class 11 JEEM21Ma57dd Solution: The least positive integer n such that is a positive integer,…

The least positive integer n such that (2i)n(1−i)n−2,i=−1{{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}},i = \sqrt { - 1} is a positive integer, is ___________.

Step-by-Step Verified Solution

(2i)n(1−i)n−2=(2i)n(−2i)n−22{{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}} = {{{{(2i)}^n}} \over {{{( - 2i)}^{{{n - 2} \over 2}}}}}

=(2i)n+22(−1)n−22=2n+22;in+22(−1)n−22= {{{{(2i)}^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}} = {{{2^{{{n + 2} \over 2}}};{i^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}}

This is positive integer for n = 6

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