# Residue at Infinity

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