# Prove that $\mathrm{Res}=\frac{f(z_0)}{g'(z_0)}$

I need to prove that if $f$ and $g$ are analytic in $D_r(z_0)$ and $g$ has a simple zero at $z_0$, then $$\mathrm{Res}[f/g,z_0]=\frac{f(z_0)}{g'(z_0)}$$

When $f(z_0)\neq 0$ and since $1/g$ has a simple pole at $z_0$ $$\mathrm{Res}[f/g,z_0]=\lim_{z\to z_0}(z-z_0)\frac{f(z)}{g(z)}=\frac{f(z_0)}{g'(z_0)}$$
But what do I need to do when $f(z_0)=0$? Any help will be appreciated thanks

#### Solutions Collecting From Web of "Prove that $\mathrm{Res}=\frac{f(z_0)}{g'(z_0)}$"

When $f(z_0)=0$ the formula is trivially true because the singularity of $f/g$ is removable and the residue is zero.