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If $f$ is holomorphic on $D\setminus \{0\}$ and takes no values in $(-\infty,0]$ then $0$ is a removable singularity.

I thought to prove this by elimination, but I can't really tell anything about the behavior of $f$ around $0$. How can one translate the information about the definition of $f$ in semi-open interval $(-\infty,0]$.

Meitar
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1 Answers1

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$G := \Bbb C \setminus (-\infty, 0]$ can be mapped conformally onto the unit disk. That is generally true for all simply-connected domains due to the Riemann mapping theorem. In this particular case the mapping can be described explicitly as $$ \varphi(z) = \frac{\sqrt z - 1}{\sqrt z + 1} $$ where $\sqrt z$ is the holomorphic branch mapping $G$ onto the right halfplane.

Then $g := \varphi \circ f$ is holomorphic in $D\setminus \{0\}$ with values in the unit disk, i.e. $g$ is bounded. It follows that $g$ has a removable singularity at $z= 0$, and then the same holds for $f$.

One could also use the "Great Picard Theorem" which states that a holomorphic function takes on all possible complex values, with at most a single exception, infinitely often in a punctured neighbourhood of an essential singularity. But that is an "advanced" result in complex analysis.

Martin R
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  • I am a little confused. What makes $g$ bounded? And what exactly indicates the existence of the removable singularity? – Meitar Jan 12 '16 at 17:52
  • I am so sorry to have neglected this thread. It was unintentional and we moved on with the materials. – Meitar Jan 12 '16 at 17:53
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    @Meitar: $g$ is bounded by $1$ because its values are in the unit disk. Riemann's theorem (see e.g. https://en.wikipedia.org/wiki/Removable_singularity) states that if $f$ is bounded in the neighbourhood of an isolated singularity then this singularity is removable. – Martin R Jan 12 '16 at 17:59
  • @MartinR It has been a while, but is there a simple way to show that if $0$ is removable for $g$ then $0$ is removable for $f$? I ask since I tried to show that $f$ is also bounded around $0$ in this case (which must be true if $0$ is indeed a removable singularity) but didn't manage to do so. – user5721565 Jul 11 '19 at 07:06
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    @dan: If $\tilde g$ is the holomorphic extension of $g$ to $D$ then $\tilde f = \phi^{-1} \circ \tilde g = \left( \frac{1+\tilde g}{1-\tilde g} \right)^2$ is the holomorphic extension of $f$ to $D$. – Martin R Jul 11 '19 at 07:13