(C) Let $z = \frac{1+i}{1-i}-\frac{1-i}{1+i}$.Simplifying the expression:$z = \frac{(1+i)^2 - (1-i)^2}{(1-i)(1+i)}$Using the identities $(1+i)^2 = 1 +

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(C) Let $z = \frac{1+i}{1-i}-\frac{1-i}{1+i}$.Simplifying the expression:$z = \frac{(1+i)^2 - (1-i)^2}{(1-i)(1+i)}$Using the identities $(1+i)^2 = 1 + i^2 + 2i = 2i$ and $(1-i)^2 = 1 + i^2 - 2i = -2i$:$z = \frac{2i - (-2i)}{1^2 - i^2} = \frac{4i}{1 - (-1)} = \frac{4i}{2} = 2i$.The modulus is $|z| = |2i| = \sqrt{0^2 + 2^2} = 2$.