## Excel in math and science

### Master concepts by solving fun, challenging problems.

## It's hard to learn from lectures and videos

### Learn more effectively through short, conceptual quizzes.

## Our wiki is made for math and science

###
Master advanced concepts through explanations,

examples, and problems from the community.

## Used and loved by 4 million people

###
Learn from a vibrant community of students and enthusiasts,

including olympiad champions, researchers, and professionals.

## Comments

Sort by:

TopNewestConsider \(f(z) = \mathrm{cosec}(\tfrac{\pi}{z})\). Then \(f(z)\) has a pole at \(z=\tfrac{1}{n}\) for any nonzero integer \(n\), which makes the singularity at \(0\) non-isolated. – Mark Hennings · 3 years, 7 months ago

Log in to reply

– R G · 3 years, 7 months ago

Thnxs dude can you think of one more functionLog in to reply

Generalize my previous example. Let \(f(z)\) be any nonconstant entire function with an infinite number of zeros. Then the zeros are isolated, countable and unbounded. Then \(f(z^{-1})^{-1}\) has a non-isolated singularity at \(z=0\), since \(w^{-1}\) is a pole for any zero \(w\neq0\) of \(f(z)\). For example, \(f(z) = e^z - 1\) would do the trick. – Mark Hennings · 3 years, 7 months ago

Log in to reply