Originally Posted by

**shawsend** How about for fun then? To be honest I didn't even know this until you brought it up. Here goes: If a function is analytic outside some domain, then we can consider the residue at infinity. So outside the hull, $\displaystyle H$, of zeros for $\displaystyle Q_m(z)$, the quotient is analytic so that we can say:

$\displaystyle \mathop\oint\limits_{|z|>H} \frac{P_n(z)}{Q_m(z)}dz=2\pi i \mathop\text{Res}_{z=0} \left\{\frac{1}{z^2} \frac{P(1/z)}{Q(1/z)}\right\}$

And with some work we get:

$\displaystyle =2\pi i\mathop\text{Res}_{z=0}\left\{\frac{1}{z^2} \frac{z^k(a_0 z^n+a_1 z^{n-1}+\cdots+a_n}{b_0 z^m+b_1 z^{m-1}+\cdots+b_m}\right\}$

but $\displaystyle k\geq 2$ so that:

$\displaystyle =2\pi i\mathop\text{Res}_{z=0}\left\{\frac{z^c(a_0 z^n+\cdots+a_n}{b_0 z^m+\cdots+b_m}\right\}$

with $\displaystyle c\geq 1$ and therefore that's zero.