= Solution
First, a <maximal left ideal> $L$ is <closed>. Its <norm> closure is again a <left ideal>, since multiplication is <continuous>. If that closure were all of $A$, some $\ell\in L$ would satisfy $\|1-\ell\|<1$ and would be invertible by the <Neumann series>. But a proper <left ideal> cannot contain an invertible element: $\ell^{-1}\ell=1$ would put $1$ in $L$. Maximality therefore forces $\overline L=L$.
Consequently $X=A/L$ is a nonzero <quotient Banach space>. It is a <simple module> for left multiplication by $A$, because the inverse image of any <submodule> under $A\to A/L$ is a <left ideal> containing $L$, and hence is either $L$ or $A$. Since $La\subseteq L$, right multiplication defines a well-defined bounded <module endomorphism>
$$
T:X\longrightarrow X,\qquad T(u+L)=ua+L,\qquad\|T\|\le\|a\|.
$$
Indeed, changing $u$ by an element of $L$ changes $ua$ by an element of $L$, and taking the infimum over representatives proves the <quotient norm> bound. Right multiplication commutes with the left action by <associativity>.
Choose $\lambda$ in the nonempty <spectrum of an element> of $T$ in the <Banach algebra> $\mathcal B(X)$. Both the <kernel> and range of $T-\lambda I$ are <submodules>. If this <module endomorphism> were nonzero, simplicity would force its <kernel> to be zero and its range to be all of $X$. It would be a bounded bijection of <Banach spaces>, with bounded inverse by the <bounded inverse theorem>, contradicting the choice of $\lambda$. Thus $T=\lambda I$. Applying this identity to $1+L$ yields $a-\lambda1\in L$. If also $a-\mu1\in L$, then $(\lambda-\mu)1\in L$; since $1\notin L$, $\lambda=\mu$. Therefore
$$
\boxed{\text{There is exactly one }\lambda\in\mathbb C\text{ with }a-\lambda1\in L.}
$$
This is the <scalar right action on a maximal-left-ideal quotient>; it does not assume that $L$ is a two-sided ideal or that $A/L$ is an algebra.
Now consider the <center of an associative algebra> $Z$. For each $u\in A$, the map $z\mapsto zu-uz$ is a <continuous> <linear map>, so its <kernel> is <closed>. Their intersection $Z$ is therefore <closed>. The identity belongs to $Z$, as do sums and scalar multiples of its elements. If $z,w\in Z$, then for every $u\in A$,
$$
(zw)u=z(wu)=z(uw)=(zu)w=(uz)w=u(zw).
$$
Thus $zw\in Z$. Also $zw=wz$, since $z$ commutes with every element of $A$, including $w$. Hence \b[the center is a <closed> <commutative> unital <subalgebra>], and is a <Banach algebra> in the inherited <norm>.
For every $z\in Z$, the inclusion $Lz=zL\subseteq L$ allows the first argument to apply. Let $\chi(z)$ be the unique scalar for which $z-\chi(z)1\in L$. Equivalently, the induced right multiplication on $A/L$ is $T_z=\chi(z)I$. Since
$$
T_{z+w}=T_z+T_w,\qquad T_{\alpha z}=\alpha T_z,\qquad T_{zw}=T_wT_z,\qquad T_1=I,
$$
the map $\chi:Z\to\mathbb C$ is a complex-linear unital <character of an algebra>. It is surjective because $\chi(\alpha1)=\alpha$, and its <kernel> is precisely $L\cap Z$. Consequently
$$
\boxed{Z/(L\cap Z)\cong\mathbb C,\qquad L\cap Z\text{ is a maximal ideal of }Z.}
$$
Indeed any ideal properly containing the <kernel> would have a nonzero image in the <field> $\mathbb C$ and thus would be all of $Z$. The <algebra character> is also <continuous>: $|\chi(z)|=\|\chi(z)I\|=\|T_z\|\le\|z\|$, although <continuity> is not needed for maximality.
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