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Find the modulus of $\frac{1+i}{1-i}-\frac{1-i}{1+i}$
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Verified Answer
$\frac{1+i}{1-i}=\frac{1+i}{1-i} \times \frac{1+i}{1+i}=\frac{(1+i)^2}{1-i^2}=\frac{1+i^2+2 i}{1+1}=\frac{2 i}{2}=i$
$\frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}=\frac{(1-i)^2}{1-i^2}=\frac{1+i^2-2 i}{1+1}=\frac{-2 i}{2}=-i$
We have, $\frac{1+i}{1-i}-\frac{1-i}{1+i}=i-(-i)=2 i$
$\therefore\left|\frac{1+i}{1-i}-\frac{1-i}{1+i}\right|=2$
$\frac{1-i}{1+i}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}=\frac{(1-i)^2}{1-i^2}=\frac{1+i^2-2 i}{1+1}=\frac{-2 i}{2}=-i$
We have, $\frac{1+i}{1-i}-\frac{1-i}{1+i}=i-(-i)=2 i$
$\therefore\left|\frac{1+i}{1-i}-\frac{1-i}{1+i}\right|=2$
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