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So we have a function $f$ which has an isolated singularity at $z_0$.

We define $$\text{Res}(f, z_0) = \frac{1}{2\pi i} \int_{B_{\varepsilon}(z_0)} f(\eta) d\eta $$

for $\varepsilon > 0$ sufficiently small.

Now I want to show this is equal to $$\text{Res}(f, z_0) = \lim_{z\rightarrow z_0} \frac{1}{(n-1)!)} \frac{d^n}{dz^n}((z-z_0)^n f(z_0)) $$

Bit unsure of how to do this.

I know that the residue is the coefficient of $a_{-1}$ of the Laurent series.

And I can write $g(z) = (z-z_0)^n f(z)$ which is holomorphic at $z_0$. But I'm struggling to make the connection to the final result...

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  • $\begingroup$ Note that this not for any isolated singularity. It must be a pole of order (at most) $n$. $\endgroup$ – logarithm May 15 '19 at 21:54
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Since $z_0$ is a pole of order n, then write:

$f(z)=\frac{c_n}{(z-z_0)^n}+...+\frac{c_{-1}}{(z-z_0)}+c_0+O(z-z_0)$.

Multiplying by $(z-z_0)^n$ you get:

$(z-z_0)^nf(z)= c_n+...+c_{-1}(z-z_0)^{n-1}+c_0(z-z_0)^n+O((z-z_0)^{n+1})$.

Now differentiate $(n-1)$ times to get:

$\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^nf(z)=(n-1)!c_{-1}+O(z-z_0).$

Now in the limit $z \to z_0$, and rearranging for $c_{-1}$ you get your result.

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