Find the zeros and poles, their orders and the residue at each pole?

Question

let $$f(z)={1/(z-1)(z-2)}$$

I know the poles are 1 and 2 of order 1 and using the formula for residue for simple poles will give me $-1$ and $1$. How do i find the zeroes? because i am ending up with $1/0$. I am confused.

Answer 1

It has no zeros: $\frac1{(z-1)(z-2)}=0\iff1=0$.

Answer 2

$g=(z-1)(z-2)$ is entire$\implies f=\frac{1}{g}$ has no zeros.