Find a Laurent expansion that converges for \(\abs{z} > 1\) of \[ f(z) \da {1 \over e^{1-z}} .\]
Solution
\[ f(z) = e^{-(1-z)} = e^{z-1} = \inverseof{e} e^z = \inverseof{e}\sum_{k\geq 0} {z^k\over k!} .\] Since \(e^z\) is entire, this converges on \(\CC\).