left(frac{1-i}{1+i}right)^{2022}+left(frac{1+ | vedclass

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(D) First,simplify the base expressions: $\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

Vedclass · Accepted Answer

$i-1$

Vedclass · Answer

(D) First,simplify the base expressions: $\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1-2i+i^2}{1-i^2} = \frac{1-2i-1}{1+1} = \frac{-2i}{2} = -i$ $\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{1+2i-1}{1+1} = \frac{2i}{2} = i$ Now substitute these into the expression: $(-i)^{2022} + (i)^{2021}$ $= (i)^{2022} + (i)^{2021}$ $= (i^{4})^{505} \cdot i^2 + (i^{4})^{505} \cdot i^1$ $= (1)^{505} \cdot (-1) + (1)^{505} \cdot i$ $= -1 + i$

$\left(\frac{1-i}{1+i}\right)^{2022}+\left(\frac{1+i}{1-i}\right)^{2021}=$

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