Instead of answering the question you posted, I will try to explain why it is not quite the question you linked to, and how the question you linked to is useful for the problem.
The question you asked is "How do we integrate, and what happens when we take the limit as $R\to \infty$?" which most people would answer with "Use the residue theorem", because that is usually how most contour integrals are most easily done (at least if you want an exact answer and expect that explicit integration will be messy, if not impossible).
However, the question you linked to had the hint of showing that the integral you listed approached zero for large $R$ (and because the only poles of the function are within a circle with small radius, the integral will hence be exactly $0$ for all sufficiently large $R$). This is most easily done by parameterizing a path on the circle, converting the contour integral into a regular integral on, say, $[0,2\pi]$, and then using the fact that $\left|\int_I f dx\right|\leq \int_I \left| f \right| dx$ to bound your integral by something easy to integrate. Because of this, the computation of the hint requires no calculation of residues.
However, the hint implies that the sum of all the residues is equal to zero, and hence (by part 3) $-f_n$ is equal to the sum of the residues not at $z=0$. Because all of the roots of the denominator except the one at $z=0$ are simple, the only truly difficult-to-calculate residue is no longer needed, and the task becomes simplifying the sum of the other residues. This simplification is an exercise in algebra left to the reader.