If a $2×2$ matrix has determinant $0$, why is its rank $1$ if $\vec 0$ is not in the domain?

Question

Let $A=\mathbf{J}\left(\frac{x^2-y^2}{x^2+y^2},\frac{xy}{x^2+y^2}\right)$, then the $\text{det}A=0$ for every point in $\mathbb{R}^2$ except $\vec 0$ where the function is undefined. Its rank can only be $0$ or $1$. At first I thought its rank was $0$ since it sent the basis vectors to $\vec 0$, but I suspect this may be wrong since $A$ is not linear (given that it changes with its arguments). The book says its rank is $1$ because its domain does not include the zero vector, but why does not including a zero vector imply the rank of the transformation is $1$?