Let $g$ be a holomorphic function on a disk $\mathbb{D}(z_0, r)$ centered at $z_0$ of radius $r>0$. Suppose $f$ has a simple pole at $z_0$. Show that $Res(fg, z_0)=g(z_0)Res(f, z_0)$.
Proof:
Since $g$ is holomorphic on $\mathbb{D}$, we can represent it as $$ g(z) = \sum\limits_{n=0}^\infty d_n(z-z_0)$$ where $d_n=\frac{g^{(n)}(z_0)}{n!}$. Since $f$ has a pole at $z_0$, it is holomorphic on some annulus $\mathbb{A}(r_1, r_2, z_0)$ centered at $z_0$ with $r>r_2>r_1>0$. We can represent $f$ as follows:
$$ f(z) = \frac{h(z)}{z-z_0} = \sum\limits_{n=-1}^\infty c_n(z-z_0)^n$$
where $h(z)$ is a function holomorphic on $\mathbb{D}(z_0, r_2)$ such that $h(z_0)\ne 0$.
Now, $fg(z)=\frac{d_0c_{-1}}{z-z_0}+O(1)$, so that $Res(fg,z_0)=d_0 c_{-1}$.
Also, $g(z_0)Res(f,z_0)=d_0c_{-1}$.
Hence, $Res(fg, z_0)=g(z_0)Res(f, z_0)$.
Do you think my proof is completely correct? I would appreciate your opinion.