Find least positive integer such that \(\frac {(2i)^n}{(1-i)^{n-2}}\) is a positive integer

For n = 1, \(\frac {(2i)^n}{(1-i)^{n-2}}\) = \(\frac {(2i)^1}{(1-i)^{1-2}}\)= (2i) (1-i)

= 2i - 2i² = 2i + 2 which is a complex number

For n = 2, \(\frac {(2i)^n}{(1-i)^{n-2}}\) = \(\frac {(2i)^2}{(1-i)^{2-2}}\)= \(\frac {2^2i^2}{1}\)= -4 which is a negative integer

For n = 3, \(\frac {(2i)^n}{(1-i)^{n-2}}\) = \(\frac {(2i)^3}{(1-i)^{3-2}}\)= \(\frac {2^3i^3}{1-i}\)= \(\frac {8i}{1-i}\) 

= \(\frac {8i}{1-i}\) x \(\frac {1+c}{1+c}\) = \(\frac {-8i-8i^2}{1-i^2}\)

= \(\frac {-8i-8}{1+1}\) = -4i + 4 which is a complex number

For n = 4, \(\frac {(2i)^n}{(1-i)^{n-2}}\)= \(\frac {(2i)^4}{(1-i)^{4-2}}\)= \(\frac {2^4i^4}{(1-i)^2} = \frac {16}{1-2i + i^2}\)

= \(\frac {16}{1-2i - 1} = \frac {-8}{i}\)

= \(\frac {-8i}{i^2}\) = 8i which is a complex number

For n = 5, \(\frac {(2i)^n}{(1-i)^{n-2}}\) = \(\frac {(2i)^5}{(1-i)^{3}}\)= \(\frac {2^5i^5}{1-i^3+3i^2=3i}\)

= \(\frac {32\,i}{1+i-3-3\,i} = \frac {32\,i}{-2(1+i)} = \frac {32\,i(1-i)}{-2(1-i^2)}\)

= \(\frac {16\,i + 16\,i^2}{2}\) = -8 i -8 which is a complex number

 For n = 6, \(\frac {(2i)^n}{(1-i)^{n-2}}\) = \(\frac {(2i)^6}{(1-i)^{4}}\)= \(\frac {2^6i^6}{1-4i+6i^2-4i^3+1}\)

= \(\frac {64\,i^4i^2}{2-6-4i+4i} = \frac {-64}{-4}\) = 16 which is a positive integer

Hence, required least positive integer is 6