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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

1 Answers 1

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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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    Correct,@User69127.2014-05-06