JEE MAIN - Mathematics (2021 - 26th August Evening Shift - No. 23)
The least positive integer n such that $${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}},i = \sqrt { - 1} $$ is a positive integer, is ___________.
Answer
6
Explanation
$${{{{(2i)}^n}} \over {{{(1 - i)}^{n - 2}}}} = {{{{(2i)}^n}} \over {{{( - 2i)}^{{{n - 2} \over 2}}}}}$$
$$ = {{{{(2i)}^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}} = {{{2^{{{n + 2} \over 2}}};{i^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}}$$
This is positive integer for n = 6
$$ = {{{{(2i)}^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}} = {{{2^{{{n + 2} \over 2}}};{i^{{{n + 2} \over 2}}}} \over {{{( - 1)}^{{{n - 2} \over 2}}}}}$$
This is positive integer for n = 6
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