The expression frac{(1+i)^{n}}{(1-i)^{n-2}} e | vedclass

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Question

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

Vedclass · Accepted Answer

$-2i^{n+1}$

Vedclass · Answer

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

The expression $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ equals

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