$$x = \frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{2i}{2} = i \implies \color{red}{x = i \implies y = 2i}$$ Assim$$(x+y)^2 = (i+2i)^2 = (3i)^2 = -9 \implies \boxed{(x+y)^2 = -9}$$ $$\text{Alternativa } \mathbb{(C)}$$