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If $(x+i y)=\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3$, then the true statement among the following is
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Verified Answer
The correct answer is:
$x>y$
Given $(\mathrm{x}+\mathrm{iy})=\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^3-\left(\frac{1-\mathrm{i}}{1+\mathrm{i}}\right)^3$
$$
\begin{aligned}
& \Rightarrow x-i y=\left\{\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)-\left(\frac{1-i}{1+i} \times \frac{1-i}{1+i}\right)\right\} \\
& =\frac{1}{8}[-8 i-8 i] \\
& =-2 i \\
& x+i y=0+(-2) i \\
& \Rightarrow x>y
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow x-i y=\left\{\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)-\left(\frac{1-i}{1+i} \times \frac{1-i}{1+i}\right)\right\} \\
& =\frac{1}{8}[-8 i-8 i] \\
& =-2 i \\
& x+i y=0+(-2) i \\
& \Rightarrow x>y
\end{aligned}
$$
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