Intermediate Algebra (12th Edition)

By the definition of $i$ shown on Page 485, we know that $\sqrt -1=i$ because $i^{2}=-1$. Therefore, $(-i)^{2}=(-i\times-i)=(i\times i\times-1\times-1)=(i^{2}\times-1\times-1)=(-1\times-1\times-1)=-1$.