Prove that $A^2 + B^2 = O_2$ given conditions Show that, given $A,B$ second order matrices with real entries, such that $AB = BA$, $\det{(A + iB)} = 0$ and $4 \det{A} >( \text{tr}{A} )^2$, then $A^2 + B^2 = O_2$.
My progress:
Considering the polynomial $\det(A + xB)$, since $i$ is a root, $-i$ is also a root, and thus $\det(A + xB) = x^2 + 1$. From this we deduce that $\det(A) = \det(B) = 1$. 
Since $AB = BA$, we have that $\det(A^2 + B^2) = \det(A + iB) \det(A - iB)  = 0$.
Since we know that $\det(A^2 - B^2) = 4$, we can also deduce the form of the polynomial $\det(A^2 + xB^2)$, but that doesn't seem to help.
I know that the condition $4 \det{A} >( \text{tr}{A} )^2$ implies that $A$ has two distinct complex eigenvalues, but I don't know how to use that.
Any ideas?

$det(A+iB)=0$ implies that there exists a  non zero complex vector $u$ such that $(A+iB)(u)=0$, this implies that $(A^2+B^2)(u)=(A-iB)(A+iB)(u)=0$ since $AB=BA$.
$det(A-iB)=0$ implies that there exists a complex vector $v$ such that $(A-iB)(v)=0$, we deduce that $(A^2+B^2)(v)=(A+iB)(A-iB)=0$.
Suppose that $v=cu$, $(A+iB)(v)=(A+iB)(cu)=c(A+iB)(u)=0$ implies that $cA(u)+ciB(u)=A(v)+iB(v)=0$. We have $A(v)-iB(v)=0$, we deduce that $A(v)=iB(v)=-iB(v)$. This implies that $A(v)=0$.
The characteristic polynomial $P_A$ of $A$ is $X^2-tr(A)X+det(A)$, the condition $tr(A)^2<4det(A)$ implies that $P_A$ does not has zero as a root, so we cannot have $A(v)=0$ thus $u,v$ are linearly independent.
Since $(A^2+B^2)(u)=(A^2+B^2)(v)=0$, we deduce that $A^2+B^2=0$.
