Residue = 0 if singularity is "analytically continuable"

Question

calculate the residue of $f(z)=\frac{z-\sin z}{z}$

We'll use the Laurent-Series, which is:

$f(z)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{2k + 1} = \frac{1}{3!}z^2 - \frac{1}{5!}z^4 + \frac{1}{7!}z^6 - ...$

now, apparently, we see that $z_1=0$ is a singularity which is "analytically continuable", so we get $Res(f;z_1)=0$

I don't get the argumentation. How do we see that it is "analytically continuable" and why do we get the reisdue 0 from that fact?

Answer 1

If $f$ has a removable singularity at $0$ and if

$f(z)= \sum_{n=- \infty}^{\infty}a_nz^n$ is the Laurent expansion around $0$, then $a_{-n}=0$ for all $n \in \mathbb N$, hence

$Res(f;0)=a_{-1}=0$.