# Residues - proving residue form using Laurent Series

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...

• Note that this not for any isolated singularity. It must be a pole of order (at most) $n$. – logarithm May 15 '19 at 21:54

## 1 Answer

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.