Let $\mathcal{A}:=\{f:\mathbb{Z}\rightarrow C:\|f\|:=\sum_{n\in Z}|f(n)|2^{|n|}<\infty\}$ under the usual function addition and scalar multiplication .Then with function multiplication :
$f\ast g(n):=\sum_{k\in \mathbb{Z}}f(n-k)g(k)$, $\mathcal{A}$ is a commutative
Banach algebra.
$\mathcal{K}:=\{z\in C:1/2\leq |z|\leq2\}, $
$\mathcal{M}:=the\ set\ of\ all\ the\ maximal \ ideals\ in\ Banach \ algera\ \mathcal{A}$
How to prove there is one to one corressponse between $\mathcal{K}\ and\ \mathcal{M}$ and show the Gelfand reprensentation of $\mathcal{A}$ is Laurent series that is absolute convergent over $\mathcal{K}$
Define $\hat{f}:\mathcal{K}\to \mathbb{C}, z\mapsto \sum_{n\in \mathbb{Z}} f(n)z^{n}$ and show that $F:f\mapsto\hat{f}$ is a unital isomorphism.
The following map gives a one to one correspond between $\mathcal{K} $ and $\mathcal{M}$: $$\Phi:\mathcal{K}\to \mathcal{M}, z\mapsto \ker ev_z,$$ where $ev_z:\mathcal{A}\to \mathbb{C}, f\mapsto \hat{f}(z).$
We will show that $\Phi$ is a surjection. Let $M$ be a maxiamal ideal of ${\mathcal{A}}$ and denote all the zero points in $\mathcal K$ of $\hat{f}\in \hat{\mathcal{A}}:=F(\mathcal{A})$ by $Z(\hat{f})$. For any finite functions $f_i\in M, i=1,2,\cdots, n$, $$\sum_{i=1}^n\bar{\hat{f_i}}\hat{f_i}\in \hat{M}:=F(M).$$ If $Z(\sum_{i=1}^n\bar{\hat{f_i}}\hat{f_i})=\emptyset$, then $\frac{1}{\sum_{i=1}^n\bar{\hat{f_i}}\hat{f_i}}\in\hat{\mathcal A}$ and thus $1\in \hat{M}$, a contradiction with $\hat{M}$ being a proper ideal of $\hat{\mathcal{A}}$. Hence $$\bigcap_{i=1}^n Z(\hat{f})\supset Z(\sum_{i=1}^n\bar{\hat{f_i}}\hat{f_i})\neq\emptyset.$$ Therefore $$\bigcap_{f\in M}Z(\hat{f})\neq\emptyset.$$ Note that $Z(f)$ is closed for every continuous function and that $\mathcal{K}$ is compact.
Pick some $z\in \bigcap_{f\in M}Z(\hat f)$, then $\ker ev_z=\{f\in M|\hat f(z)=0\}\supset M$, so $\ker ev_z=M$. Therefore, $\Phi$ is a surjection.
See here as a reference.