= Solution
Take a nonzero complex unital <Banach algebra>, with identity $1$. The <spectrum of an element> is
$$
\boxed{\sigma_A(a)=\{\lambda\in\mathbb C:\lambda1-a\text{ is not invertible in }A\}.}
$$
The standard <Banach algebra spectrum> theorem states that it is a nonempty compact subset of $\mathbb C$, contained in the disk $|\lambda|\leq\|a\|$. Conversely, every nonempty compact set $K\subseteq\mathbb C$ occurs: in the <Banach algebra> $C(K)$, the coordinate function $a(z)=z$ has <Banach algebra spectrum> exactly $K$, since $\lambda-a$ has a continuous reciprocal precisely when $\lambda\notin K$. Thus \b[the possible <Banach algebra spectra> are exactly the nonempty compact subsets of the complex plane].
For a nonconstant <polynomial> $p$, factor $p(z)-\lambda=c\prod_{j=1}^n(z-z_j)$ over $\mathbb C$. Substitution gives
$$
p(a)-\lambda1=c\prod_{j=1}^n(a-z_j1).
$$
These factors commute. A finite commuting product is invertible exactly when every factor is invertible: if $P=\prod_jx_j$ is invertible, its inverse commutes with each $x_j$, and $x_j^{-1}=(\prod_{k\ne j}x_k)P^{-1}$ is a two-sided inverse. Hence $\lambda$ belongs to the <Banach algebra spectrum> of $p(a)$ exactly when some root $z_j$ belongs to $\sigma_A(a)$, which is equivalent to $\lambda\in p(\sigma_A(a))$. Constant <polynomials> give the same result directly, since $\sigma_A(c1)=\{c\}$. This proves the <polynomial> <spectral mapping theorem>:
$$
\boxed{\sigma_A(p(a))=p(\sigma_A(a)).}
$$
For a closed unital <Banach subalgebra> $B\subseteq A$, use the common identity $1_A\in B$. An inverse in $B$ is an inverse in $A$, so $\boxed{\sigma_A(b)\subseteq\sigma_B(b)}$. Let $\lambda\in\partial\sigma_B(b)$. Choose $\lambda_n\notin\sigma_B(b)$ tending to $\lambda$. Then $\lambda_n\notin\sigma_A(b)$, and $(\lambda_n1-b)^{-1}\in B$. If $\lambda$ were outside $\sigma_A(b)$, continuity of inversion in $A$ would give
$$
(\lambda_n1-b)^{-1}\longrightarrow(\lambda1-b)^{-1}\in B,
$$
where membership follows from closedness. This would make $\lambda\notin\sigma_B(b)$, contrary to the closed <Banach algebra spectrum> having its boundary inside it. Therefore $\lambda\in\sigma_A(b)$ and is approached by points outside that <Banach algebra spectrum>, proving
$$
\boxed{\partial\sigma_B(b)\subseteq\partial\sigma_A(b).}
$$
Continuity of inversion here follows, for example, by writing a nearby element as an invertible element times $1+$ a <norm>-small perturbation and using the <Neumann series>. This is <spectrum in a closed unital subalgebra>.
To retain the original <Banach algebra spectrum> in a commutative subalgebra, take the <resolvent-generated commutative algebra>
$$
C=\overline{\operatorname{alg}\bigl(1,a,(\lambda1-a)^{-1}:\lambda\notin\sigma_A(a)\bigr)}^{\|\cdot\|}.
$$
Every generator commutes with $a$ and with every other generator: commuting invertible elements have commuting inverses, and their inverses commute with all elements commuting with them. <commutativity> persists on the algebra they generate and on its <norm> closure. For every $\lambda\notin\sigma_A(a)$, the required inverse is in $C$ by construction, so $\sigma_C(a)\subseteq\sigma_A(a)$. The inclusion for unital subalgebras gives the reverse containment. Hence $\boxed{\sigma_C(a)=\sigma_A(a)}$.
Finally, the <Banach subalgebra generated by one element> is $D=\overline{\{p(a):p\text{ a polynomial}\}}$, and it is commutative. Put $K=\sigma_D(a)$ and suppose its complement has a bounded connected component $U$. The boundary of $U$ lies in $K$. For $\lambda\in U$, the <maximum modulus principle on a bounded domain> applied to any <polynomial> gives
$$
|p(\lambda)|\leq\max_{z\in\partial U}|p(z)|\leq\max_{z\in K}|p(z)|=r_D(p(a))\leq\|p(a)\|.
$$
Here $r_D$ is the <spectral radius>, and the equality uses the already proved <polynomial> <spectral mapping theorem>. Define $\chi(p(a))=p(\lambda)$. The displayed bound makes this well defined, even when two <polynomials> represent the same algebra element, and bounded on the dense <polynomial> algebra. Its continuous extension to $D$ remains linear and multiplicative, with $\chi(1)=1$ and $\chi(a)=\lambda$, so it is a <character of an algebra>. If $\lambda1-a$ were invertible in $D$, applying $\chi$ to its inverse equation would give $0=1$. Thus $\lambda\in K$, contradicting $\lambda\in U$.
There are therefore no bounded complementary components. A compact subset of the plane has exactly one unbounded complementary component: everything outside a sufficiently large disk is connected, and every unbounded component meets it. \b[Consequently $\boxed{\mathbb C\setminus\sigma_D(a)\text{ is connected}.}$]
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