NETMaths

Is this true?

|f| bounded near an isolated singularity ⇒ pole

No — it is false.

The counterexample

f(z) = sin(z)/z at 0

Bounded ⇒ removable (Riemann). Poles have |f| .

The kind of mistake this is

Execution slip

The idea was right. The computation was not.

Drill statements like this

Others that fail the same way

From Singularities and ResiduesLaurent series, classification of singularities, Casorati–Weierstrass

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