Is complex residue related to the word residue?

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.

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.

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

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