= Solution
A <character of an algebra> is a nonzero multiplicative complex <linear functional> $\phi:A\to\mathbb C$. Nonzeroness and multiplicativity force $\phi(1)=1$, so it is onto $\mathbb C$. Its kernel is an <ideal>, and the quotient is the <field> $\mathbb C$, making the kernel maximal.
<Continuity> is automatic. The element $a-\phi(a)1$ cannot be invertible, since applying $\phi$ to an inverse equation would give $0=1$. On the other hand $a-\lambda1$ is invertible whenever $|\lambda|>\|a\|$, by the <Neumann series>. Hence $|\phi(a)|\leq\|a\|$, the <automatic continuity of characters> bound.
Conversely let $M$ be a <maximal ideal>. Its closure is an <ideal>, because multiplication is continuous. That closure is proper: otherwise there would be $m\in M$ with $\|1-m\|<1$, making $m$ invertible by the <Neumann series> and forcing $M=A$. Maximality gives $\overline M=M$. The quotient $A/M$ is consequently a <Banach algebra>. It is a division <algebra>: for any $a\notin M$, the larger <ideal> $M+Aa$ equals $A$, so the class of $a$ has an inverse. By the <Gelfand-Mazur theorem>, $A/M\cong\mathbb C$, and its quotient map gives a continuous <character of an algebra> with kernel $M$.
If two <algebra characters> have the same kernel, then $a-\phi(a)1$ lies in the second kernel, giving $\psi(a)=\phi(a)$ for every $a$. Thus
$$
\boxed{\Phi_A\longrightarrow\mathcal M_A,\qquad\phi\longmapsto\ker\phi}
$$
is a bijection. This proves that <maximal ideals of a commutative complex unital Banach algebra are character kernels>.
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