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Let $f: D(P, r) \setminus \{P \} \rightarrow \mathbb{C}$ be holomorphic. and define $U = f(D(P,r) \setminus \{P \})$ be open.

and define $g: U \rightarrow \mathbb{C}$ be holomorphic.

If $f$ has a removable singularity at $P$ does $g \circ f$ have one also?


My guess is that the properties of $g\circ f$ depends on $g$,

\begin{align} \lim_{z \rightarrow P} g\left( f(z) \right) = \end{align}

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  • $\begingroup$ You mean $g : f(D(P,r)\setminus\{P\})\to\mathbb C$? $\endgroup$ – Yiorgos S. Smyrlis Oct 29 '16 at 10:07
  • $\begingroup$ Why are you looking at $g(\infty)$? This seems more appropriate if $f$ has a pole, not a removable singularity. This question may depend on if $\lim_{z\rightarrow P}f(z)\in U$. $\endgroup$ – Michael Burr Oct 29 '16 at 10:16
  • $\begingroup$ @ YiorgosS.Smyrlis, yes. @ Michael Burr, I got your point. $\endgroup$ – phy_math Oct 29 '16 at 13:07

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