Exercise · ZW-ISOLATED-SINGULARITY-006 · ★★☆ Intermediate
If \(|f|\) is bounded on a punctured disc \(0<|z-z_0|<r\), what kind of isolated singularity does \(f\) have at \(z_0\)?
Choose one option.
Show answer
- D. Removable
Explanation
This is Riemann's theorem: a function bounded near an isolated singularity has a finite limit there, so the singularity is removable.
Review the concept: Isolated singularity
An isolated singularity of a function \(f\) is a point \(z_0\) where \(f\) is not holomorphic but is holomorphic on a punctured disc \(0<|z-z_0|<r\) around it.
More practice problems: isolated singularity
- ZW-ISOLATED-SINGULARITY-001 What kind of singularity does (sin z)/(z) have at z=0?
- ZW-ISOLATED-SINGULARITY-002 What kind of singularity does e^1/z have at z=0?
- ZW-ISOLATED-SINGULARITY-003 What kind of singularity does 1z^2+1 have at z=i?
- ZW-ISOLATED-SINGULARITY-004 The singularity of (cos z-1)/(z^2) at z=0 is removable. What value must f(0) take to remove it?
- ZW-ISOLATED-SINGULARITY-005 What is the order of the pole of 1z^2(z-1) at z=0?
- ZW-ISOLATED-SINGULARITY-007 Which of these functions has at z=0 a singular point that is NOT an isolated singularity?
- ZW-ISOLATED-SINGULARITY-008 Classify the singularity of (z)/(sin z) at z=π.
- ZW-ISOLATED-SINGULARITY-009 Classify the singularity of zsin 1z at z=0.