Suppose we are given the simple expression
$$ F(k) = \frac{1}{E^2-E(k)^2} $$
which has a pole when $E^2 = E(k)^2$ and where $E, E(k)$ are real numbers. When working with this expression (e.g. inside an integral) we can add some infinitesimal constant $\epsilon >0$ to take care of the poles and let this $\epsilon$ go to zero in the end.
What I want to show now is that the imaginary part of the square of $F$ is proportional to a delta function $\delta\left( E^2 - E(k)^2 \right)$. What I tried was simply to calculate
$$ F^2 = \lim_{\epsilon\to 0}\frac{1}{(E^2+E(k)^2+i \epsilon)^2} = \lim_{\eta\to 0} \frac{1}{(E^2-E(k)^2)^2 + i\eta(E^2-E(k)^2)} $$
where $\eta = 2\epsilon$ and where I have set $\epsilon^2 = 0$. My idea was to use the representation of the delta function
$$ \pi\delta(x) = \lim_{\eta \to 0} \frac{\eta}{\eta^2+x^2} $$
but I failed to bring $F^2$ to this form. Is there any way to do this anyway? For example by introducing an infinitesimal parameter somewhere else in the fraction? Or maybe by doing something completely different?