for instance $$\dfrac{\sin(\dfrac{1}{z})}{z}$$
$z=0$ is a pole for the denominator but $z=0$ is an essential singularity for the numerator too. So how does it work ? i have two residues ? or it's different some way ?
2026-03-25 03:20:44.1774408844
What happens if I have an essential singularity and a pole for the same $z$?
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$z=0$ is an essential singularity for your function too. with the expansion $sin z=\sum_{n=0}\frac{(-1)^nz^{2n+1}}{(2n+1)!}$ we see $$\frac{1}{z}sin \frac{1}{z}=\sum_{n=0}\frac{(-1)^n}{(2n+1)!z^{2n+2}}=\frac{1}{z^2}-\frac{1}{3!z^4}+\frac{1}{5!z^6}-...$$ Can you find the residue?