$D(P,r)\backslash \{P\}\xrightarrow{f}U\xrightarrow{g}\mathbb{C}$. Both $f$ and $g$ are holomorphic. If $f$ has a removable (or pole, or essential) singularity at $P$, does $g\circ f$ also have one?

I think in any case $g\circ f$ will not have singularities any more. It is just a hunch, and I cannot explain it clearly.

  • 1
    $\begingroup$ You should try to look for potential functions $g$ that cancel out the singularities, and for functions that don't. (Hint: What happens if $g(x)=x$? What if $g$ is constant? And if the singularity of $f$ is removable, what happens if $g$ has a singularity at $f(P)$?) $\endgroup$ Oct 31, 2012 at 18:53
  • $\begingroup$ Thank Lukas and Old John for your hints and details. If $f$ has essential singularity at $P$, by Casorati-Weierstrass theorem, $f(D(P,r)\backslash \{P\})=U$ is dense in $\mathbb{C}$. Does it mean $g$ has no singularity is this case? $\endgroup$
    – Sam
    Oct 31, 2012 at 19:48

1 Answer 1


A couple of things to consider:

If $g$ is the function $g:z \mapsto z$, then the composite will have exactly the same behaviour at $P$ as $f$ does.

At the other extreme, if $g$ is the constant function that maps everything to zero, then the composite function will have at worst a removable singularity, depending on whether you consider $\infty$ to be a point in the co-domain of $g$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .