Let $\mathcal{A}$ be a unital Banach algebra and $a\in \mathcal{A}$. If $f: G\rightarrow \mathbb{C}$ is analytic, $\sigma(a)\subset G$ and $f(z)\neq 0$, then the inverse of $f(a)$ is $f(a)^{-1}$, which is the analytic functional calculus by $\frac{1}{f(z)}$.
My question is: Does there exist a analytic functional $g(z)$ such that the analytic functional calculus $g(a)$ is the left inverse of $f(a)$, i.e. $g(a)f(a)=1$?
If $f,g: G\rightarrow \mathbb{C}$ are analytic and $ \sigma(a) \subset G$, then
$\quad f(a)$ and $g(a)$ commute.
Hence if $g(a)f(a)=1$, then $g(a)f(a)=1=f(a)g(a).$