Invertitbility of operator in functional calculus

Question

Let $X$ be a complex banach space and $A$ be a bounded linear operator from $X$ to $X$. Further, let $F$ be an analytic function in a neighborhood of $\sigma(A)$ such that $1/F$ is an analytic function in a neighborhood of $\sigma(A)$ too. Now I want to show that $F(A)=\sum_{n=0}^{\infty}{a_{n}A^{n}}$ is invertible and $F(A)^{-1}=1/F(A)$. Can someone help?

Answer 1

Let $G:=1/F$. Then there is a neighborhood $U$ of $ \sigma(A)$ such that

$F(z)G(z)=1$ for all $z \in U$. One of the properties of the functional calculus says: we have

$$I=F(A)G(A)=G(A)F(A).$$