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Let $p(z)$ and $q(z)$ be relatively prime polynomials with complex co-efficients so that $deg(q(z))\ge deg(p(z))+2$ and let $f(z)=p(z)/q(z)$. We need to show that the sum of residues of $f(z)$ over all poles is $0$

Well, I tried like this:

by Residue theorem: If $f$ is analytic in a domain except for isolated singularities at $a_1,\dots a_k$ then for any closed contour $\gamma\in D$ on which none of the points $a_k$ lie, we have $$\frac{1}{2\pi i}\int_{\gamma}f(z)dz=\sum_{1}^{k}n(\gamma;a_k)Res[f(z);a_k]$$

as $p$ and $q$ are relatively prime to each other we have $r,s$ such that $p(z)r(z)+q(z)s(z)=1$

$$\frac{1}{2\pi i}\int_{\gamma}f(z)dz=\sum_{1}^{k}Res[f(z);a_k ]$$

$$\frac{1}{2\pi i}\int_{\gamma}\frac{p(z)}{q(z)}dz=\sum_{1}^{k}n(\gamma;a_k)Res[f(z);a_k]$$

Now, I am confused where to use the given facts, should I replacing $p(z)$ from the relatively prime condition? and how to implement the given degree condition? thank you for help

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up vote 4 down vote accepted

Hint: Let the contour be a circle of radius $R$ (large enough to contain all the poles). Let $R\to\infty$. What happens to the integral?

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Indeed. The “relatively prime” condition is a red herring. – Harald Hanche-Olsen Jul 7 '12 at 19:56
An alternative way would be to "invert" that large circle, and show that the degree condition implies that the substitution $w=1/z$ won't introduce a pole at $w=0$. – Jyrki Lahtonen Jul 7 '12 at 19:59
Ok let informally, $p(z)=z$ and $q(z)=z^3+1$, then $\int_{|z|=R}\frac{z}{z^3+1}= \int_{|\omega|=1/R}\frac{\omega^2}{\omega^3+1}$ and you want to say as it has no poles at $0$ so the integration must be $0$ as it is analytic with in $|\omega|=1/R$ as $r\rightarrow\infty$? – Un Chien Andalou Jul 7 '12 at 20:11
sorry not $r$, $R$ – Un Chien Andalou Jul 7 '12 at 20:16
With $z=1/w,\,dz=-(1/w^2)dw$ I got (observe that the orientation of the circle gets flipped - absorb that with the minus sign) $$\int_{C(0,R)}\frac{z}{z^3+1}\,dz=\int_{C(0,1/R)}\frac{1}{w^3+1}\,dw.$$ If $R>1$ there are no poles inside, so yeah. But notice the difference if you try to do the same with $p(z)=z^2$, $q(z)=z^3+1$! – Jyrki Lahtonen Jul 7 '12 at 20:21

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