I am asked to prove the following theorem:
Let $A$ be a unital (complex) Banach algebra generated by $\{\mathbf 1,x\}$ for some $x\in A$. Show that $\sigma_A(x)$ has no holes (i.e. $\mathbb C\setminus\sigma_A(x)$ is connected).
Now, I am familiar with the spectral permanence theorem, as well as the identification $\sigma_A(x)=\{\hat{x}(p)\ |\ p\in\operatorname{sp}(A)\}$ where $\operatorname{sp}(A)$ denotes the Gelfand spectrum of $A$, but neither of these seem sufficient to prove the theorem.
I only have very crude ideas of how to solve this problem. The first is to categorize all possible forms of $A$ and solve each case. Now, if not for a norm, I could simply treat $A$ as $\Bbb C[x]/\langle p(x)\rangle$ for some irreducible (or $0$) polynomial $p(x)$, and in fact, when $p\ne 0$, by finite dimensionality we have that $A=\Bbb C[x]/\langle p(x)\rangle$ with some appropriate norm.
Case 1: $A=\Bbb C[x]/\langle p\rangle$. In this case, we can use the Chinese Remainder theorem to write $$A\cong\Bbb C[x]/\langle(x-\alpha_1)^{d_1}\rangle\times\cdots\times\Bbb C[x]/\langle(x-\alpha_r)^{d_r}\rangle$$ as a $\Bbb C$-algebra with some choice of norm (all are equivalent). Then we can once again write $$A\cong\Bbb C[\epsilon_{d_1}]\times\cdots\times\Bbb C[\epsilon_{d_r}]$$ with some choice of norm, where $$\Bbb C[\epsilon_d]:=\Bbb C[x]/\langle x^{d+1}\rangle.$$ Now, note that for two Banach algebras $A,B$, we can make $A\times B$ into a Banach algebra by setting $\lVert (x,y)\rVert=\max(\lVert x\rVert_A,\lVert y\rVert_B)$, and we can give $\Bbb C[\epsilon_d]$ a canonical Banach algebra structure with the norm $$ \left\lVert\sum_{i=0}^d a_i\epsilon_d^i\right\rVert:=\sum_{i=0}^d|a_i| $$ which makes $\Bbb C[\epsilon_d]$ into a Banach algebra as can easily be checked. Furthermore, it can be checked that the original element $x\in A$ is mapped to the element $$y:=(\epsilon_{d_1}+\alpha_1,\ldots,\epsilon_{d_r}+\alpha_r)\in B:=\Bbb C[\epsilon_{d_1}]\times\cdots\times\Bbb C[\epsilon_{d_r}]$$ by these isomorphisms, so that in fact $\sigma_A(x)=\sigma_B(y)$. Now, note that for $0\ne\alpha\in\Bbb C$, any element $$\epsilon_d+\alpha\in\Bbb C[\epsilon_d]$$ has an inverse of the form $$\sum_{i=0}^d(-1)^i\frac{\epsilon_d^i}{\alpha^{i+1}}$$ while $\epsilon_d$ is noninvertible given that it is nilpotent, so that $$\sigma_A(x)=\sigma_B(x)=\{\alpha_1,\ldots,\alpha_r\}\subset\Bbb C$$ which has connected complement.
Case 2: $A$ is the (strict) completion of $\Bbb C[x]$ under some norm. It's this case that's causing me trouble. I can solve this only for specific examples, such as $C(\overline{D})\cap\mathcal{H}(D)$, the space of continuous functions on the closed unit disc $\overline{D}=\overline{B(0,1)}\subset\Bbb C$ holomorphic on $D$ with $\sup$ norm, in which $\sigma(x)=\overline{D}$. In general, however, I don't see why the elements of $A$ should be representable as power series.
It's possible I'm missing an entirely simpler way to view this problem, however.