= Resolvent spectral mapping identity
{title2=$\sigma((A-z)^{-1})\setminus\{0\}=\{(\lambda-z)^{-1}:\lambda\in\sigma(A)\}$}
For a closed <linear operator> and $z$ in its resolvent, the displayed identity follows by factoring $(A-z)^{-1}-wI$ for $w\ne0$. 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.
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