# Holomorphic function and removable singularity

If a function is holomorphic on $\Bbb D$\ ${0}$, does that mean 0 is a removable singularity of the function if the residue at 0 is 0?

I think this is false because I thought I did some practice problems and sometimes the residue at poles can be 0 too. But can someone please offer me a proof? Thanks.

• Consider $f(z) = 1/z^2$. – Ethan Alwaise Dec 14 '16 at 9:06
• That's a counterexample. Could you tell me how to prove it? I could think of some counterexamples, but don't know how to prove it.. Thanks! – J.doe Dec 14 '16 at 9:07
• Do you mean prove that it is a counterexample? – Ethan Alwaise Dec 14 '16 at 9:08
• @EthanAlwaise No I meant is there a way to prove it directly, instead of providing counterexamples. – J.doe Dec 14 '16 at 9:09
• The statement is false, so providing a counterexample is a proof. – Ethan Alwaise Dec 14 '16 at 9:14

## 1 Answer

All kinds of singularities are possible:

$f(z)=\frac{\sin z}{z}$ has a removable singularity at $0$

$f(z)=\frac{1}{z^{123456}}$ has a pole at $0$

$f(z)=e^{1/z}$ has an essential singularity at $0$

• you meant $e^{1/z^2}$ or $e^{1/z} - 1/z$ such that $Res(f(z),0) = 0$ – reuns Dec 14 '16 at 15:04
• yes, this is a better example – Fred Dec 14 '16 at 15:18