Skip to content
Counterexample

Is this true?

|f| bounded near an isolated singularity ⇒ pole

No — it is false.

The counterexample

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

Why it fails

Bounded ⇒ removable (Riemann). Poles have |f| →∞\to \infty.

Save for spaced revision (free account)
Keep practicing

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

See real exam options built on this mistakeDrill statements like this
complexsingularities

Go deeper· Explained in full in Singularities — three types, three tests· Asked in CSIR NET December 2025· From Singularities and Residues › Laurent series, classification of singularities, Casorati–Weierstrass

Published . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram