The least positive integer n such that (2i)^n / (1 - i)^n - 2,i = sqrt - 1 is a positive integer, is ___________.

(2i)^n / (1 - i)^n - 2 = (2i)^n / ( - 2i)^n - 2 / 2 = (2i)^n + 2 / 2 / ( - 1)^n - 2 / 2 = 2^n + 2 / 2;i^n + 2 / 2 / ( - 1)^n - 2 / 2This is positive integer for n = 6

The least positive integer n such that 2in 1 in 2i 1 i... | null (Mathematics)

{{{{(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

Explanation

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