Domain of holomorphicity of $f(z)=\frac{1}{e^z+1}$

Question

Domain of holomorphicity of $$f(z)=\frac{1}{e^z+1}$$

Is this simply everywhere, because there is no $\bar z$ dependence?

It's holomorphic everywhere it is defined, which is everywhere that $e^z \neq -1$. — Travis Willse, Apr 23 '16 at 22:12

score 3 · Answer 1 · answered Apr 23 '16 at 22:12

3

What happens if $z=(2n-1)\pi i$?

answered Apr 23 '16 at 22:12

Ruzayqat

1 Answers1