Hilbert transform property

Question

I’m stuck with the following problem:I have an operator of Hilbert transform which is $H : f \to (vp) \int\frac{f(t)dt}{(t-x)}$. I need to prove that it's square ($H^2$) is some sort of scalar operator on the space of all proper rational functions with no poles on $\mathbb{R}$ So far if $f$ is proper rational function with no poles on $\mathbb{R}$ point $x$ in the integral is the pole of order 1. So this integral equals $2\pi i \frac{1}{2} res_{x}\frac{f(t)}{(t-x)}$. Since this integral via complex analysis is the integral over bound of upper semicirlce and the integral over arc vanishesh whenever radius tends to $\infty$. So this integral equals $i \pi f(x)$. If we apply Hilbert transformation again we get the same situation and integral is equal to $i \pi 2i\pi \frac{1}{2} res_{y}\frac{f(x)}{(x-y)} = -\pi^2 f(y)$ So far what does it mean to be a scalar operator (is it just the image of operator is scalar?) What should I do next? Is this all right?