The term $$(1-i)^2/(1+i)^2$$ where $$i=sqrt(-1)$$ is most nearly: 

To solve this problem, we can first begin to simplify both the numerator and denominator:

$\frac{{(1-i)}^2}{{(1+i)}^2}=\frac{(1-i)(1-i)}{(1+i)(1+i)}=\frac{1-2i+i^2}{1+2i+i^2}$ 

Since we know $i = \sqrt{-1}$, we also know $i^2=-1$.

Plugging this back into our expression:

$\frac{1-2i+i^2}{1+2i+i^2}=\frac{1-2i+(-1)}{1+2i+(-1)}=\frac{-2i}{2i}=-1$

Therefore, our answer is -1.