Is complex residue related to the word residue?

Question

I know little formal math terminology and don't understand much of anything about complex analysis. Also, if this isn't a good starting point for complex integration feel free to say (I'm learning about it partly for Cauchy's residue theorem).

My first and intuitive idea of residue has to do with remainder, subtraction, division, etc., But more generally something that's leftover, extra, or unused by an operation or something. Do these ideas tie together?

Answer 1

The residue (latin residuere - remain) is named that way because $\frac{1}{2\pi i}\int_{|w-z_0|=r}f(w)dw=\sum_{n=-\infty}^{\infty} a_n\int_{|w-z_0|=r}(w-z_0)^n\,dw= a_{-1}$ is what remains after integration.

Answer 2

If $a\in\mathbb C$, $r\in(0,\infty)$, $f\colon B_r(a)\setminus\{a\}\longrightarrow\mathbb C$ is an analytic function, and $\gamma\colon[a,b]\longrightarrow B_r(a)\setminus\{a\}$ is a simple loop around $a$, then$$\frac1{2\pi i}\int_\gamma f(z)\,\mathrm dz=\operatorname{res}_{z=a}\bigl(f(z)\bigr).$$So, the residue is what's leftover after integrating along such a loop.

Answer 3

Yes, it is. The residue of the Laurent series of a function in an annulus is the coefficient $c_{-1}$ of this series, that is what is left after integration of the function in a loop on the annulus (that is, after the integration of the Laurent series that represent the function in this annulus).

Note that

$$\oint z^{-k}\, dz=\begin{cases}0,& k\in\Bbb Z\setminus\{1\}\\2\pi i,& k=1\end{cases}$$

Hence (assuming a Laurent series around a pole in the origin for simplicity) we have that

$$\oint f(z)\, dz=\oint\sum_{k\in\Bbb Z}c_k z^k\, dz=\sum_{k\in\Bbb Z}c_k \oint z^k\, dz=c_{-1}2\pi i$$