$\sin(1/(z-a))$ can be expressed in a Laurent series around $a$, I would like to know how I can shift the centre back to 0.
2026-04-01 15:55:56.1775058956
What is the radius of convergence of Laurent Series around 0 of $\sin(1/(z-a))$?
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You don't.
Remember the radius of convergence extends to the nearest pole or essential singularity. (Unfortunately this theorem does not have a standard name.)