Let $A$ be a Banach algebra, let $a\in A$ and suppose $f(a)=0$, where $f$ is an analytic function defined on an open set $U$ containing $\sigma(a)$. Prove that $a$ is algebraic in the sense that $p(a)=0$ for some polynomial $p$.
PS: I have just answered an identical question a few minutes ago but then, almost immediately, the questioner deleted the question and with it my nice answer which I enjoyed very much writing. I am therefore asking it again and I'll soon post my answer here.
Should anyone point me to some Stack Exchange guideline I'm disrespecting by doing so, I'll gladly delete everything again.
As stated this result does not hold. First of all $f$ could be zero in which case "$f(x)=0$" gives no information whastoever. Even if $f$ is nonzero, it still could be zero on the connected component of its domain containing $\sigma (x)$, and then "$f(x)=0$" again says nothing.
In order to prove the result it is important to assume that $f$ is not identically zero on any connected component of its domain.
This said, by the Spectral Mapping Theorem, $$ \{0\} = \sigma (f(x)) = f(\sigma (x)), $$ which says that $f$ vanishes on the spectrum of $x$. Because $\sigma (x)$ is compact, and because the set of zeros of an analytic function satisfying our hypothesis cannot have accumulation points, if follows that $\sigma (x)$ is a finite set.
Writing $$ \sigma (x) = \{a_1, a_2, \ldots , a_n\}, $$ for every $i=1,2,\ldots ,n$ we let $n_i$ be the (finite) order of $a_i$ as a zero of $f$, so that $$ f(z)=g(z)\prod_{i=1}^n(z-a_i)^{n_i}, $$ for all $z$, where $g$ is some analytic function defined on the domain of $f$, not vanishing on any $a_i$. It follows that $$ 0=f(x)=g(x)\prod_{i=1}^n(x-a_i)^{n_i}, $$ but since $g(x)$ is invertible (again by the Spectral Mapping Theorem), we must have that $\prod_{i=1}^n(x-a_i)^{n_i}=0$, as desired.