# If $f$has a removable singularity $1/f$ has a removable singularity or a pole

Let $f$ be analytic in $D_r(z_0)$ \ {$z_0$}. I am trying to prove that if $f$ has a removable singularity at $z_0$, then $1/f$ has either a removable singularity or a pole at $z_0$.

My proof: Since $f$ has a removable singularity at $z_0$ it can be redefined at $z_0$ so that $f$ is analytic in $D_r(z_0).$ Suppose $f(z_0)=w$.

Case I: $w=0$

Then $f$ has a zero of some finite order at $z_0$. Thus $1/f$ has a pole of the same order at $z_0$

Case II: $w\neq0$

Then $f(z)\neq0$ on some disk $D_R(z_0)$ where $0<R\leq r$. Thus $1/f$ is analytic in $D_R(z_0)$. So $\lim_{z\rightarrow z_0}(1/f(z))$ exits. Thus, $1/f$ has a removable singularity at $z_0$.

This is my proof. If there are any mistakes please point them out I would much appreciate it. Thanks

Alternate way of proving there is a pole

Since $f(z_0)=0$ $$\lim_{z\rightarrow z_0}|1/f(z)|\rightarrow \infty$$. So $z_0$ is a pole of $1/f$

• In case I, you have not explained how you know that $f$ has a zero of some finite order at $z_0$.
– 5xum
Dec 2 '14 at 14:26
• @5xum Thanks for pointing it out I never thought of that. Dec 2 '14 at 14:30
• @5xum Can you explain how it can be done? thanks Dec 2 '14 at 14:32
• No problem. The other parts of your proof are OK, so you only need to examine the case when $f$ has a zero of infinite order at $z_0$. I don't know how to do that, I'm afraid.
– 5xum
Dec 2 '14 at 14:33
• @5xum Check my alternate proof I think that should be correct Dec 2 '14 at 14:44

If $$f$$ has a removable singularity at $$z_0$$ then in some punctured disk it can be represented uniquely in a Laurent series with all $$b_k=0$$. Say, $$f=a_0+a_1(z-z_0)+O((z-z_0)^2)$$. If $$a_0\neq 0$$ then $$1/f=1/a_0(1+a_1/a_0 (z-z_0)+ O((z-z_0)^2)$$ via a geometric expansion. If $$a_0=0$$ then $$1/f= 1/a_j(z-z_0)^j (1+ a_{j+1}/a_j (z-z_0) +O((z-z_0)^2))$$ so $$1/f$$ has a pole of order $$j$$ where $$a_j$$ is the first nonzero term in the Laurent expansion of $$f$$.