(A) Given $\left(\frac{1-i}{1+i}\right)^k = -i$. First,simplify the base: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} =

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(A) Given $\left(\frac{1-i}{1+i}\right)^k = -i$. First,simplify the base: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$. So,the equation becomes $(-i)^k = -i$. For $(-i)^k = -i$,$k$ must satisfy $k \equiv 1 \pmod 4$. The least positive integer $m$ is $1$. The greatest negative integer $n$ is $1 - 4 = -3$. Thus,$m - n = 1 - (-3) = 1 + 3 = 4$.

Class 11 Mathematics 4-1.Complex numbers Integral power of iota, Algebraic operations and Equality of complex numbers

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If $m$ and $n$ are respectively the least positive and greatest negative integer values of $k$ such that $\left(\frac{1-i}{1+i}\right)^k = -i$,then $m-n =$

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$6$
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$2$

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Integral power of iota, Algebraic operations and Equality of complex numbers

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