Removing poles of a meromorphic function by multiplication with the help of a holomorphic function

Question

Let $f:\mathbb{C}\to\mathbb{C}$ be a meromorphic function with a pole of order $k$ at $z_0$. One way of removing the pole at $z_0$ is to consider the function $f(z)(z-z_0)^k$, which is holomorphic at $z_0$. I am wondering if this holds for general holomorphic function $g:\mathbb{C}\to\mathbb{C}$, so is the function $f(z)(g(z)-g(z_0))^k$ holomorphic at $z_0$, and is this thus a way of removing all the singularties from $f$? I think it holds, by simply taking the power series expansion of $g$ around $z_0$.

Answer 1

Clearly, $z_0$ is a zero of $g(z)-g(z_0)$ and therefore the order of $z_0$ as a zero of $\bigl(g(z)-g(z_0)\bigl)^k$ is at least $k$. So, yes, $z_0$ is a removable singularity of $f(z)\bigl(g(z)-g(z_0)\bigl)^k$.