In the derivation of the formula

\[
\Res_{z=\infty} f(z) = -\Res_{z=0} \left [ \frac{1}{z^2} f \left (
\frac{1}{z} \right) \right ],
\]

I *didn’t* know why the factor $\dfrac{1}{z^2}$ was needed. Luckily,
googling the title of this post, I quickly found two posts on
Mathematics Stack Exchange about the intuition behind.^{1}

In fact, this factor appears when we make the substitution $z \mapsto \dfrac{1}{z}$. For any simple closed contour $C$ and a complex-valued function $f$ which is analytic at the interior points of the region bounded by $C$ except for finitely many points,

\[
\int_C f(z) \ud z = -\int_C \frac{1}{z^2} f \left ( \frac{1}{z}
\right) \ud z.
\]