If \left(\frac{1-i}{1+i}\right)^{2m}\left(\frac{1+i}{1-i}\right)^{2n}=1 where i=\sqrt{-1} then what is the <strong>SMALLEST</strong> positive value of (m-n)?

Explanation

Simplify the inner fractions: \frac{1-i}{1+i} = \frac{(1-i)^2}{1^2 - i^2} = -i and \frac{1+i}{1-i} = i. The equation becomes (-i)^{2m}(i)^{2n} = 1. This simplifies to ((-i)^2)^m ((i)^2)^n = 1 \implies (-1)^m (-1)^n = 1, which gives (-1)^{m+n} = 1. For this to hold, m+n must be an even integer. If m+n is even, m and n share the same parity, so their difference m-n must also be even. The smallest positive even integer is 2.

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