- #1

krindik

- 65

- 1

The function [itex]\frac{1}{\sin(\frac{\pi}{z})}[/itex] has i

**solated singularities**at z=+-1, +-1/2, ....

However, it is said that it has an

**non-isolated singularity**at z=0.

A

**non-isolated singularity**has to be a point where its neigborhood too is also singular.

But, for some [itex]\epsilon > 0 ,\, \frac{1}{\sin(\frac{\pi}{\epsilon})}[/itex] is not singular eg. [itex]\epsilon = 0.000001 [/itex]

Can u pls explain.

Thanks