frac{(1+i)^{2011}}{(1-i)^{2009}} is equal to | vedclass

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(D) We know that $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$. Given expression: $E = \frac{(1+i)^{2011}}{(1-i)^{2009}} = \frac{(1

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-$2$

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(D) We know that $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$. Given expression: $E = \frac{(1+i)^{2011}}{(1-i)^{2009}} = \frac{(1+i)^{2009} \cdot (1+i)^2}{(1-i)^{2009}}$. $E = \left(\frac{1+i}{1-i}\right)^{2009} \cdot (1+i)^2$. Simplify $\frac{1+i}{1-i}$: $\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{1+i^2+2i}{1-i^2} = \frac{2i}{2} = i$. Substitute back: $E = (i)^{2009} \cdot (2i) = 2 \cdot i^{2010}$. Since $i^4 = 1$,$i^{2010} = (i^4)^{502} \cdot i^2 = 1^{502} \cdot (-1) = -1$. Therefore,$E = 2 \cdot (-1) = -2$.

$\frac{(1+i)^{2011}}{(1-i)^{2009}}$ is equal to

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