Critical confusion with the spectrum and resolvent of a compact operator

Question

I know that the spectrum of a compact operator $C$ consists at most of countable set of complex number that accumulates only at $0$. I also know that the resolvent $(\lambda-C)^{-1}$ has a pole at each nonzero element of the spectrum.

Here I meet a serious contradiction. The complex analysis says an analytic function cannot have an infinite (even countable) number of poles in a bounded domain. However, is it possible that a compact operator $C$ has an infinite spectrum so that the poles of the resolvent are infinite in number? I am extremely confused...

Answer 1

We can have an infinite number of poles in a finite region. $\frac 1 {\sin (\frac 1 z)}$ has a pole at each of the points $\frac 1 {n\pi}$.

Note: $0$ is not a pole of the resolvent and there is no disc around it on which the resolvent is analytic except at $0$. So having poles tending to $0$ is not a contradiction to the result you are quoting.