# Isolated and non isolated essential singularity at same point?

I need to find the singularities of $$f(z) = \frac{1-e^z}{2+e^z}$$ My effort: Poles of function are given by $$2+e^z=0\implies e^z = -2 \implies z = \log 2+i(2k+1)\pi$$ for k integer.

All these are singularities termed as simple poles. By definition, limit point of these which is $$\infty$$ is a non-isolated singularity.

Further, limit point of zeros is again infinity which is a isolated-essential singularity.

But if both of isolated and non isolated coincides we take it as a non-isolated singularity. Am i correct? These are the only singularities?

## 1 Answer

Every neighborhood of $$\infty$$ contains a pole, implying, as you correctly state, that $$\infty$$ is not an isolated singularity. What makes you believe that it is also an isolated singularity?

• I have read somewhere else that limit point of zeros is such a singularity. – Mittal G Nov 26 '18 at 13:10
• Probably you misremember what you read. – Julián Aguirre Nov 26 '18 at 15:31