# Critical confusion with the spectrum and resolvent of a compact operator

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...

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.
• The Churchill complex variables and application book has an exercise 11 in p.247 that if $R$ is a region inside and on a simple closed contour and $f$ is analytic on $R$ except for poles in the interior of $R$, then $f$ must have at most finite poles. Oct 28 '19 at 9:17
• @Keith I don't know about "more severe", but yes, if $0$ is in the spectrum then $0$ is certainly not a pole of the resolvent, since it's not a isolated singularity. Oct 28 '19 at 12:30