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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?

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    The exponent $k-1$ is wrong. This should be $(k-1),$ the iterated derivative. See https://en.wikipedia.org/wiki/Residue_%28complex_analysis%29#Limit_formula_for_higher_order_poles2012-10-01

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