Formula for residues of a simple pole

Question

If we have The rational function: $$f(z)=\frac{\phi(z)}{\psi(z)},~~~~ \phi(z_0)\neq 0 , ~\psi(z_0)=0 , ~\psi'(z_0)\neq 0$$ and the point at $z_0$ is a simple pole, then the residue can be calculated by: $$\lim_{z \to z_0}(z-z_0)\frac{\phi(z)}{\psi(z)}=\frac{\phi(z_0)}{\psi'(z_0)}$$ My question is why can't $\psi'(z_0)=0$ along with $\phi(z_0)=0$ such that they cancel out? For example take the function: $$f(z)=\frac{z^2}{e^{2\pi iz^3}-1} \, z_0=0 $$ the formula above gives a result for the residue $\frac{1}{6\pi i}$ where infact the residue is $\frac{1}{2\pi i}$

Answer 1

then the residue can be calculated by: $$\lim_{z \to z_0}(z-z_0)\frac{\phi(z)}{\psi(z)}=\frac{\phi(z_0)}{\psi'(z_0)}\tag{0}$$

Note this formula works under your prerequisite $$\phi(z_0)\neq 0 , ~\psi(z_0)=0 , ~\psi'(z_0)\neq 0\tag{1}$$

But for your new example,

$$f(z)=\frac{z^2}{e^{2\pi iz^3}-1},~~~\phi(z)=z^2,~~\psi(z)= e^{2\pi iz^3}-1,~~z_0=0$$ it doesn't satisfy (1), you can't use (0), otherwise you will get $\frac0 0$. Instead, you can modify (0) and compute the limit by applying L'Hopital's rule,

$$\lim_{z \to z_0}(z-z_0)\frac{\phi(z)}{\psi(z)}=\lim_{z \to z_0}\frac{[(z-z_0)\phi(z)]'}{\psi'(z)}=\dots=\lim_{z \to z_0}\frac{[(z-z_0)\phi(z)]'''}{\psi'''(z)}\tag{0'}$$