TITLE: Removable singularity when multiplied by linear factor
QUESTION [5 upvotes]: I was thinking about this situation: If $f(z)$ is holomorphic everywhere except at $a$ (where it has a pole), and has a removable singularity or a pole at $\infty$, does $(z-a)f(z)$ also have a removable singularity or a pole at $\infty$?
In the pole case, $f(z)/z^k$ is holomorphic in some neighborhood of $\infty$. Since $(z-a)$ is holomorphic, $(z-a)f(z)/z^k$ should also be holomorphic in that neighborhood of $\infty$. But can we conclude that? I'm not sure because $z-a$ is not defined at $\infty$.
In the removable singularity case, we can define $f(\infty)$ so that $f(z)$ is holomorphic in a neighborhood of $\infty$. Again, can we conclude anything about $(z-a)f(z)$?

REPLY [2 votes]: Yes, in that situation, $(z-a)f(z)$ has a pole at infinity, or, if $f$ had a zero at infinity, a removable singularity.
When $k$ is such that $f(z)/z^k$ is holomorphic in a neighbourhood of infinity, then
$$\frac{(z-a)f(z)}{z^{k+1}} = \frac{f(z)}{z^k} - a\frac{f(z)}{z^{k+1}}$$
also is holomorphic in a neighbourhood of $\infty$.