Recall that $\mathrm{Res}(f(z);z_{0}) = \lim_{z \rightarrow z_{0}} \frac{[(z-z_{0})^{k}f(z)]^{k-1}}{(k-1)!}$
If, for example $h(z)/g(z)$ have a pole with grade 1, the formula ends up being $\lim\limits_{z \rightarrow z_{0}} 1/0!= 1$ This is wrong(I think). I've found this formula on my notebook but the general formula is different, it involves derivatives and seems cumbersome. I suspect that the top-right k-1 is wrong, what is the correct way to calculate the residue?
So the correct formula should be $ \mathrm{Res}(f(z);z_{0}) = \lim_{z \rightarrow z_{0}} \frac{[(z-z_{0})^{k}f(z)]^{(k-1)}}{(k-1)!} $ ? If the pole is grade 1 then the result is $\lim\limits_{z \rightarrow z_{0}}(z-z_{0})f(z)$ Which is the formula for a simple pole right?