Consider the matrix $A$ as follows\begin{pmatrix} \sqrt{-xy}&x \\ y&-\sqrt{-xy} \end{pmatrix}
where $(x,y)\in\mathbb C^2$
Let $D$ a matrix such that : $D^2=A$
$D$ can be written as follows\begin{pmatrix} a&b \\ c&d \end{pmatrix}
Find $a$, $b$, $c$, $d$ in terms of $x$ and $y$.
Of course if $x = y = 0$, this is trivial.
In the remaining cases, however, the matrix $A$ satisfies $A \neq 0$ and $A^2 = 0$. As such, there are no matrices $D$ for which $D^2 = A$.
To see that this holds, suppose for the purpose of contradiction that $D^2 = A$. Then, $D$ must have $0$ as its only eigenvalue. Thus, the characteristic polynomial is $\det(D - xI) = x^2$. Thus, by the Cayley-Hamilton theorem we have $D^2 = 0$, contradicting our premise.