Given two sequences of complex numbers $\{a_1,\dots,a_r\}$ and $\{b_1,\dots,b_r\}$. Consider the function $$f(z)=z(z-1)\prod_{i=1}^r \left(1-\frac{2}{z-b_i} \right)\left(1+\frac{2}{z-1+a_i} \right).$$ What is the coefficient of $z^{-1}$ in the Laurent expansion of $f(z)$ around infinity?
This is an Exercise of Fulton-Harris "Representation theory: a first course", where $a_i, b_j$ are assumed to be integers; but it seems that this is non-relevant. The answer should be $$2\sum_{i=1}^r(a_i(a_i+1)-b_i(b_i+1)).$$ WolframAlpha can also confirm this result for small $r$. It seems that a direct calculation is very complicate. I am wondering if there is a simple way to see this. Thanks in advance.
Write $f(z)= z(z-1) P(z)$ where $P=\Pi_j g_j$ and each $g_j$ is a product of two linear-fractional functions of $z$.
The special Laurent coefficient sought is also the value of the path integral $\oint_C f(z) dz$ taken over a large contour $C$. Thus it is clear that translation of the function $f(z)$ has no affect on the value of the Laurent coefficient.
The direct calculation of that one special Laurent coefficient involves finding the third-order expansion of $P(z)$ at infinity. The calculation can be simplified somewhat by noting that the substitution $z=2\zeta+1$ converts $z(z-1)$ to the more symmetric form $4(\zeta^2-1)$, which is an even function of $\zeta$.
This symmetrization simplifies the direct computation of the Laurent expansion of $(4\zeta^2-1) P$ because we only need to know the odd terms in the expansion of $P$ in the variable $\zeta$,and only up to order three. That cuts the book-keeping burden in half!
Writing $w=\zeta^{-1}$, and expanding $P$ as a Taylor series about $w=0$ we see that each $g_j= (1 + c_j w^2 + d_j w^3)$ lacks first powers of $w$ in its Taylor expansion. Thus up to third order, the odd terms in the Taylor expansion of $ \Pi_j g_j$ is simply $(\sum_j d_j ) w^3$.
More explicitly each factor expands to third order as $g=(1-\frac{ w}{1-rw})(1+ \frac{w}{1+sw}) = (1- w - r w^2 - r^2 w^3)(1+ w -sw^2 +s^2 w^3)$
$ =1+ \ldots + [(s^2-r^2) +(s-r)]w^3$.