How to prove the formula for the residue of $f$ at a pole of order $m$?

Question

Let $f$ holomorphic on $z_0$. I saw this awesome formula on a book : the residual of $f$ on $z_0$ is given by $$\text{Res}_{z_0}(f)=\frac{1}{(m-1)!}\frac{\mathrm d^m}{\mathrm dz^{m-1}}(z-z_0)^mf(z)$$ How can I prove it? ($m$ is the order of pole that $f$ is assumed to have at $z_0$).

Answer 1

I think it's more $$Res_{z_0}(f)=\lim_{z\to z_0}\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}(z-z_0)^mf(z)$$

To prove it, since $z_0$ is a pole of order $m$, $$f(z)=\frac{c_{-m}}{(z-z_0)^m}+...+\frac{c_{-1}}{(z-z_0)}+c_0+...$$ and thus $$(z-z_0)^mf(z)= c_{-m}+...+c_{-1}(z-z_0)^{m-1}+c_0(z-z_0)^m+...$$

Then, $$\frac{d^{m-1}}{dz^{m-1}}(z-z_0)^mf(z)=(m-1)!c_{-1}+(z-z_0)(c_0+...).$$

Take the limit when $z\to z_0$ to conclude.