We are learning about singularities in my Complex Analysis course right now. I understand what it means to be each type of singularity however, a question I had was about the singularities of composition functions. In better words:
Let $f$ be holomorphic on $U=D(P,r)/\{P\}$ and $g$ be holomorphic on $f(U)$. Then if f has a singularity (removable, pole, essential, etc.) at $P$, does that imply that $g\circ f$ has one also?
Any help would be great. Thanks!
First of all, $g$ needs to be holomorphic on $f(U)$, not on $U$. Unless $f$ maps $U$ into itself, the behavior of $g$ on $U$ is immaterial. With that correction:
If $f$ has a removable singularity at $P$, and if $g$ is holomorphic at $\lim_{z\to P}f(z)$, then $g\circ f$ has a removable singularity at $P$.
If $f$ has a pole at $P$, then the behavior of $g\circ f$ at $P$ is the same as the behavior of $g(1/z)$ at $z = 0$. For example,
Lastly, if $f$ has an essential singularity and $g$ is not constant, then $g\circ f$ will have an essential singularity. The easiest way to see this is by means of a fairly high-powered theorem: $f(V)$ is dense in $\Bbb C$ for every deleted neighborhood $V$ of $P$ (in fact $f(V)$ includes all of $\Bbb C$ except possibly for two points). From this it follows that $g(\overline{f(V)})$ includes the entire range of $g$, which does not occur for removable singularities or poles (since $g$ is not constant). Therefore $P$ has to be an essential singularity of $g\circ f$.