If (x+iy) = left(frac{1+i}{1-i}right)^3 - lef | vedclass

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(B) Given $(x+iy) = \left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3$. First,simplify the base fractions: $\frac{1+i}{1-i} = \frac{(1+i

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$x > y$

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(B) Given $(x+iy) = \left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3$. First,simplify the base fractions: $\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+2i+i^2}{1-i^2} = \frac{1+2i-1}{1+1} = \frac{2i}{2} = i$. Similarly,$\frac{1-i}{1+i} = \frac{1}{i} = -i$. Substituting these into the expression: $x+iy = (i)^3 - (-i)^3$. $x+iy = -i - (-i^3) = -i - (i) = -2i$. Comparing the real and imaginary parts: $x = 0$ and $y = -2$. Since $0 > -2$,we have $x > y$.

If $(x+iy) = \left(\frac{1+i}{1-i}\right)^3 - \left(\frac{1-i}{1+i}\right)^3$,then the true statement among the following is

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