If $\mathrm{m}, \mathrm{n}$ are respectively the least positive and greatest negative integer values of $k$…
If $\mathrm{m}, \mathrm{n}$ are respectively the least positive and greatest negative integer values of $k$ such that $\left(\frac{1-i}{1+i}\right)^k=-i$, then $m-n=$
4
0
6
2
Solution
$\text { Given }\left(\frac{1-i}{1+i}\right)^k=-i \Rightarrow\left\{\frac{(1-i)^2}{2}\right\}^k=-i \Rightarrow(-i)^k=-i$
If $k=1$ (least positive integer) $\Rightarrow(-i)^1=-i$
If $k=-3$ (greatest negative integer) $\Rightarrow(-i)^{-3}=-i$
So, $m=1$ and $n=-3 \Rightarrow m-n=4$.