# Find the modulus of $\;\large\frac{1+i}{1-i} - \large\frac{1-i}{1+i}$

$(a)\;2\qquad(b)\;3\qquad(c)\;4\qquad(d)\;0$

Answer : $\;2$
Explanation :
$\;\large\frac{1+i}{1-i} - \large\frac{1-i}{1+i} = \large\frac{(1+i)^{2}-(1-i)^{2}}{(1-i)(1+i)}$
$=\large\frac{1^{2}+i^{2}+2\;.1\;.i - 1-i+2i^{2}}{1^{2}-i^{2}}$
$= \large\frac{4i}{2} =2i$
$|\large\frac{1+i}{1-i} - \large\frac{1-i}{1+i}| = |2i| = \sqrt{2^{2}} = 2$