The least positive integer n,for which frac{( | vedclass

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(C) Given expression: $\frac{(1+i)^{n}}{(1-i)^{n-2}}$$= \frac{(1+i)^{n}}{(1-i)^{n}} 	imes (1-i)^{2}$$= \left(\frac{1+i}{1-i}ight)^{n} 	imes (1 + i

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$1$

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(C) Given expression: $\frac{(1+i)^{n}}{(1-i)^{n-2}}$$= \frac{(1+i)^{n}}{(1-i)^{n}} 	imes (1-i)^{2}$$= \left(\frac{1+i}{1-i}ight)^{n} 	imes (1 + i^{2} - 2i)$$= \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}ight)^{n} 	imes (1 - 1 - 2i)$$= \left(\frac{1 + i^{2} + 2i}{1 - i^{2}}ight)^{n} 	imes (-2i)$$= \left(\frac{2i}{2}ight)^{n} 	imes (-2i)$$= i^{n} 	imes (-2i) = -2i^{n+1}$For the expression to be a positive real number,$-2i^{n+1}$ must be positive.If $n=1$,$-2i^{1+1} = -2i^{2} = -2(-1) = 2$,which is positive.Thus,the least positive integer $n$ is $1$.

The least positive integer $n$,for which $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ is positive,is

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