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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 en.wikipedia.org/wiki/… –  Andrew Oct 1 '12 at 22:11
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In a neighbourhood of the order $1$ pole $z_0,$ a Laurent series expansion of $f$, say$f(z)=\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots$, exists. The formula will be $$\lim_{z\to z_0}\dfrac{(z-z_0)(\frac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+\cdots)}{1}.$$

You can show this is equal to $a_{-1},$ which equals by definition the residue of $f$ at $z_0.$ The general formula works similarly, where your Laurent expansion will begin with $\frac{a_{-n}}{(z-z_0)^n}+\cdots.$

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