Let $S \in B(l^2_{\mathbb{Z}_{+}})$ be a right shift operator ($e_n \mapsto e_{n+1}$). Take the C*-subalgebra $A$ in $B(l^2)$generated by $S$ and all the compact operators $K(l^2)$. I want to show that $A/K(l^2)$ is isomorphic to $C(\mathbb{T})$.
I know that the Gelfand spectrum of this algebra is homeomorphic to $\sigma_e(S)$ and I know that $\sigma_e(S)$ is in $\mathbb{T}$. All the authors say that it is obvious that $\sigma_e(S)$ is rotationally invariant (if $\lambda$ is in the spectrum than $z \lambda$ is also in the spectrum fot all $\lambda \in \sigma_e(S)$ and all $z \in \mathbb{T}$). I don’t see how is it obvious, though.
Note that $C^*(S)$ already contains the compact operators (it's very easy to see that it contains some compacts; then one uses that it is irreducible and so it has to contain them all).
There might be an obvious argument, but my guess is that the authors (I don't know who you are talking about, though) say that $\sigma_e(S)$ is rotationally invariant because they already know that the image on the Calkin is isomorphic to $C(\mathbb T)$. The canonical way to do that is to identify $S$ with the Toeplitz operator $T_z$ of multiplication by the identity function.