If $a$ is invertible in a unital C$ ^*$-algebra, then $a^{-1} \in C^*(a)$ This is Exercise 1.6(iii) from Intro. to K-Theory for $C^*$-Algebras by Rørdam et al.
If $A$ is a unital $C^*$-algebra, and $a \in A$ is invertible, then I want to show that $a^{-1} \in C^*(a)$. What I know is:

*

*$a$ is invertible if and only if $aa^*$ and $a^*a$ are invertible, with $a^{-1} = a^*(aa^*)^{-1} = (a^*a)^{-1}a^*$

*If $b \in A$ is invertible and normal, then there is some $f \in C(\sigma(b))$ such that $f(b) = b^{-1}$ (proven using the continuous functional calculus). Another question here shows that $b^{-1}, 1 \in C^*(b) = C^*(b, 1)$.

*$C^*(a)$ is the closed linear span of $a^ma^{*n}$ and $a^{*s}a^t$, where $m,n,s,t \in \mathbb{N}$.

My issue here is that since $a$ is not a normal element, then I don't know that $C^*(a, 1) = C^*(a)$ (which I  don't think is even true), nor that there is a *-isomorphism between $C(\sigma(a))$ and $C^*(a)$. I suspect that there is something simple that I am not seeing, but I am not sure what it is.
 A: You can use the Polar Decomposition. In principle you have to do this in a von  Neumann algebra, but you always have this if you represent $A\subset B(H)$.
You have $a=v|a|$, with $|a|=(a^*a)^{1/2}\in C^*(a)$. You also have that $|a|$ is invertible and normal, so $|a|^{-1}\in C^*(a)$. Then $v=a|a|^{-1}\in C^*(a)$. This also shows that $v$ is invertible. Now, since $v$ is a partial isometry, being invertible, it is a unitary (since $v^*v$ and $vv^*$ are invertible projections). Then $v^{-1}=v^*\in C^*(a)$. And this gives
$$
a^{-1}=|a|^{-1}v^*\in C^*(a) . 
$$
A: Note that if $a$ is invertible, so is $a^*a$. And $(a^*a)^{-1}\in C^*(a^*a,1)\subset C^*(a, 1)$, and $a^{-1} = (a^*a)^{-1}a^*\in C^*(a, 1)$. This can be used to show that the spectrum of an element doesn't expand when restried to any subalgebra (which is not the case for Banach algebras).
A: Since $a^*a$ is normal, it follows from the linked answer that $C^*(a^*a) = C^*(a^*a,1)$. Therefore we have $1,(a^*a)^{-1} \in C^*(a^*a) \subseteq C^*(a)$, hence also $a^{-1} = (a^*a)^{-1} a^* \in C^*(a)$.
In particular, $C^*(a) = C^*(a,1) = C^*(a,a^{-1})$ whenever $a$ is invertible.
