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 $z_0$).

  • $\begingroup$ Did this awesome book also say what $m$ is? $\endgroup$ – mickep Oct 14 '15 at 20:05
  • $\begingroup$ Sorry, $m$ is the order of the singularity. $\endgroup$ – Rick Oct 14 '15 at 20:06
  • $\begingroup$ You forgot the exponent $m$ in $(z-z_0)^m f(z)$, and that formula is valid only if $z_0$ is a pole of order $m$ of $f$. It then follows immediately from the Laurent expansion of $f$ about $z_0$. It is rarely a good way to compute the residue, in my experience. The Laurent expansion is usually easier. $\endgroup$ – Daniel Fischer Oct 14 '15 at 20:06
  • $\begingroup$ ... which means also, that the first (somewhat cryptic) sentence is not valid. I suggest that you reformulate the question given the comments you have so far. $\endgroup$ – mickep Oct 14 '15 at 20:07
  • $\begingroup$ Yes, I corrected it. $\endgroup$ – Rick Oct 14 '15 at 20:15

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.


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