Let $A$ be a unital $C^*$-algebra.
Assume that $a\in A$ is a normal and invertible element i.e $aa^*=a^*a$ and $aa^{-1}=a^{-1}a=1$.
let $C^*({a}) $ be the $C^*$-algebra generated by $a$.
I know that $C^*({a}) $ is the closed linear span of $a^{m}a^{*{n}}$ such that $m,n\in N$.
I want to know $1 , a^{-1} \in C^*({a}) $
Q: Is it true?"$1 , a^{-1} \in C^*({a}) $"
How can I prove it?
The spectrum of $a$ is a compact set that does not contain $0$. So there is a disk $D$ around $0$ with $D\cap\sigma(a)=\emptyset$. Thus, on $\sigma(a)$, $f:t\longmapsto 1/t$ is continuous, so $f\in C(\sigma(a))$. Then $f(a)\in C^*(a)$ via the Gelfand transform.
Edit: in view of Josse's comments, here's a clarification. Since $ a $ is invertible, $\sigma (a)\cap\{0\}=\varnothing $, so there exists a continuous function $f $ with $f (0)=0$ and $f (t)=1/t $ on $\sigma (a) $. By using Stone-Weierstrass on a closed disk, we can write $f $ as a uniform limit of polynomials (on $z $ and $\bar z $) with constant term zero. This implies that $a^{-1}\in C^*(a) $, and a fortiori $1\in C^*(a) $.