# bounded real parts and removable singularity

I was reading this proof here: holomorphic function with bounded real part on punctured neighborhood $\dot{D}_{\epsilon}(z_0)$, which was proving that

Let $$f$$ be holomorphic on a punctured disk $$D_\epsilon(z_0)/\{z_0\}$$ and $$Re(f(z)) < M$$ for all $$z \in D_\epsilon(z_0)/\{z_0\}$$. Then it implies that $$f$$ has a removable singularity at $$z_0$$.

In the answer to the original post, they supposed that $$z_0$$ is a pole for the sake of contradiction, but why $$g = \frac{1}{f}$$ holomorphic in a "possible smaller" punctured neighbourhood $$D_\delta(z_0)/\{z_0\}$$? Also, why is $$g(D_\delta(z_0)/\{z_0\})$$ an open neighbourhood of $$0$$? The open mapping theorem might have been used here but why is it a neighbourhood of $$0$$? Last, why is $$\frac{1}{D_r(0)\setminus \{0\}} = \mathbb{C}\setminus \overline{D_{\frac{1}{r}}(0)}$$?

Thank you .

## 3 Answers

1) Because $$f$$ can have a zero in $$D_\varepsilon(z_0) \setminus \{z_0\}$$ so $$1/f$$ can be non holomorphic on $$D_\varepsilon(z_0) \setminus \{z_0\}$$. One have to consider a smaller disk: if $$f$$ has a pole at $$z = z_0$$, then by continuity, $$f$$ does not vanish around $$z = z_0$$ that is to say in $$D_\delta(z_0) \setminus \{z_0\}$$ for $$\delta \leq \varepsilon$$, so $$g=1/f$$ is holomorphic on $$D_\delta(z_0) \setminus \{z_0\}$$.

2) By the open mapping theorem, $$g(D_\delta(z_0) \setminus \{z_0\})$$ is open because $$D_\delta(z_0) \setminus \{z_0\}$$ is open. It is a neighbourhood of 0 because if one takes a sequence $$(z_n)_n$$ in $$D_\delta(z_0) \setminus \{z_0\}$$ such that $$z_n \to z_0$$, then $$g(z_n) = 1/f(z_n) \to 0$$, so $$0 \in \overline{g(D_\delta(z_0) \setminus \{z_0\})}$$.

3) $$z \in 1/\dot{D}_r(0)$$ iff ($$\vert 1/z \vert < r$$ and $$z \neq 0$$) iff $$\vert z \vert > r$$ iff $$z \in \mathbb C \setminus \overline{D_{1/r}(0)}$$.

• Your proof of 2) has some problems. $g(D(z_0)\setminus\{z_0\})$ is not an open neighborhood of $0$ because $0$ is not in that set. And I'm not sure what you're doing with $z_n\to z_0$ etc. – zhw. Oct 13 '19 at 16:24

The proof in the link given does not say $$g(D(z_0)\setminus\{z_0\})$$ is an open neighborhood of $$0.$$ That doesn't even make sense: $$0\notin g(D(z_0)\setminus\{z_0\})!$$ What is stated is that $$g$$ has a removable singularity at $$z_0.$$ Still using the notation $$g$$ for the extension, we have $$g(z_0)=0.$$ It is the extended $$g$$ that maps $$D(z_0)$$ onto a neighborhood of $$0.$$

First question: if $$f$$ has a pole at $$z_0$$ then $$|f(z)| \to \infty$$ as $$z \to z_0$$ so $$f$$ does not vanish in some deleted neighborhood of $$z_0$$. Hence $$\frac 1 f$$ is analytic there.

Second question: it is already stated in the link that $$g$$ is defined to be $$0$$ at $$z_0$$ (to get rid of removable singularity). Now open mapping theorem shows that $$g(D_{\delta} (z_0))\setminus \{z_0\}$$ is an open set containing $$0$$.

Third question: $$0<|z| iff $$z\neq 0$$ and $$|\frac 1 z| >\frac 1 r$$.

• thank you, for the first question i was asking why is the punctured disk "possible smaller"? – beigecamellia Oct 13 '19 at 0:22
• If $f$ has a zero in the original disk then $g$ is not defined there. So you have to take a smaller disk to make sure that $f$ has no zeros there and this is possible because $|f(z)| \to \infty$. @beigecamellia – Kavi Rama Murthy Oct 13 '19 at 0:29