Resolvent spectral mapping identity ($\sigma ((A-z{)}^{-1})\setminus \{0\}=\{(\lambda -z{)}^{-1}:\lambda \in \sigma (A)\}$)

For a closed linear operator and $z$ in its resolvent, the displayed identity follows by factoring $(A-z{)}^{-1}-wI$ for $w\ne 0$. Its inverse exists exactly when $A-(z+1/w)I$ has a bounded inverse. Products involving $A$ are bounded after composing with a resolvent, since $A(A-\lambda {)}^{-1}=I+\lambda (A-\lambda {)}^{-1}$. Zero may be in the bounded resolvent's spectrum when $A$ is unbounded.