So let us take the commutative Banach algebra $B=\mathscr{l}^1(\mathbb{Z}_n)$ over $\mathbb{R}$ with convolution as multiplication $(f\ast g)(x)=\underset{y\in \mathbb{Z}_n}{\sum} f(y)g(x-y)$. I know that there are two multiplicative linear functionals when $n$ is even. Namely, since we can embed the group $\mathbb{Z}_n$ into our algebra by mapping $m\mapsto f_{m}$, where $f_m$ is zero everywhere except in the m-th coordinate, where it is equal to 1, and mapping $f_1$ either to 1 or -1.
My problem is that it seems the quotient rings formed by quotienting out the kernels of these functionals is always the field $\mathbb{R}$. I have been told that one of the quotient rings should be isomorphic to $\mathbb{C}$ but I don't see it, and my calculations seems to show that they're always isomorphic to $\mathbb{R}$.
Let $B$ be your algebra, which has the set $\{f_i:0\leq i<n\}$ as a basis. The identity element is $f_0$, which we can write simply $1$, and the element $f_1$ generates $B$ as an algebra, as $f_i=f_1^i$ for each $0\leq i<n$. Moreover, $f_1^n=1$, and one can easily see that $B$ is generated as an algebra by $f_1$, which I will writ more simply just $x$, subject to the only relation $x^n=1$. In other words, we have $B\cong\mathbb R[x]/(x^n-1)$.
You are looking $\mathbb R$-algebra morphisms $\phi:B\to\mathbb C$. As $x$ generates $B$ as an algebra, $\phi$ is completely determined by the image of $x$. As $x^n=1$, we must have $\phi(x)^n=1$, and whenever we pick an element $u\in\mathbb C$ such that $u^n=1$ there is a unique morphism of algebras $\phi_u:B\to\mathbb C$ such that $\phi_u(x)=u$.
This means that there are exactly $n$ $\mathbb R$-algebra morphisms $B\to\mathbb C$. You can easily check that if $n$ is even, $n-2$ of them are surjective, and if $n$ is odd then $n-1$ of them are surjective.