Use the Laurent series to show that the residue at a pole of the order $m$ can be evaluated by:
$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^mf(z))$$
I have no clue how to approach this problem. First, it does look extremely related to Cauchy's Integral Theorem, or at least some generalization of it. However, I'm not exactly sure what I should be expanding and manipulating to show this.
If it's just straight forward plug in, then how do I simplify?
The definition I've been given is
$$f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k+\sum_{k=1}^{\infty}a_{-k}(z-z_0)^{-k}$$
which I know can just be written as:
$$f(z)=\sum_{k=-\infty}^\infty a_k(z-z_0)^k$$
Using this directly we get
$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}((z-z_0)^m\sum_{k=-\infty}^\infty a_k(z-z_0)^k)\implies$$
$$\operatorname{Res}_{z=z_0}f(z)=\frac{1}{(m-1)!}\sum_{k=-\infty}^\infty\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}(a_k(z-z_0)^{k+m})$$
But I don't know how to proceed beyond this point…
Best Answer
It's a basic result of complex analysis, here an approach.
Suppose that $f$ is analytic insde and on a simple closed curve $\gamma$, so indeed $f$ can be written by Laurent series as $$f(z)=\sum_{k=-\infty}^{+\infty}a_{k}(z-z_{0})^{k}$$ But we know that $f$ has a pole $z_{0}$ of order $m$ (we need the pole inside $\gamma$), so $f$ can be written as $$f(z)=\sum_{k=0}^{+\infty}a_{k}(z-z_{0})^{k}+\sum_{k=1}^{m}a_{-k}(z-z_{0})^{-k}$$ $$\implies (z-z_{0})^{m}f(z)=a_{-m}+a_{-m+1}(z-z_{0})+\cdots+a_{-1}(z-z_{0})^{m-1}+a_{0}(z-z_{0})^{m}+\cdots$$ $$\implies \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=(m-1)!a_{-1}+m(m-1)\cdots 2a_{0}(z-z_{0})+\cdots $$ $$\implies \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=\lim_{z\to z_{0}}(m-1)!a_{-1}+m(m-1)\cdots 2a_{0}(z-z_{0})+\cdots $$ $$\implies \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)=(m-1)!a_{-1}$$ $$\implies a_{-1}:={\rm Res}(f,z_{0})= \lim_{z\to z_{0}} \frac{1}{(m-1)!} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right)$$ Therefore, $$\boxed{{\rm Res}(f,z_{0})= \frac{1}{(m-1)!} \lim_{z\to z_{0}} \frac{{\rm d}^{m-1}}{{\rm d}z^{m-1}}\left( (z-z_{0})^{m}f(z)\right) }$$