Let $U \subseteq \mathbb C$ be open, $A$ a (unital, associative) complex Banach algebra and $f : U \to A$ holomorphic and invertible in a punctured neighborhood of $0 \in U$, so that $0$ is an isolated singularity of $1/f$.
Must $1/f$ be meromorphic at $0$? I.e. has the Laurent expansion at $0$ a finite singular part?
In general, I think the answer is no, because if $f(0)$ is nonzero and not a zero divisor, then looking at the Laurent expansion shows that if $1/f$ is meromorphic at $0$, it must be holomorphic, which gives a contradiction if we suppose in addition that $f(0)$ is not invertible.
I'm interested in the case of the algebra of operators on a complex Banach algebra, and $f(0)$ of the form $1+$ a compact operator, so that this counterexample cannot occur (by the Fredholm alternative).
Are there counterexamples with $f(0) = 1 + K$ with $K$ compact ?
I managed to prove that if $K$ is compact, normal and trace class, and $f : U \to \mathbb C$ is holomorphic, then even when $(1+f(s)K)^{-1}$ is not meromorphic it does send meromorphic functions to meromorphic functions. The proof analyzes the Laurent-expansion and uses the spectral theorem to eventually reduce to the finite-dimensional case.