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Suppose that $\frac{P(z)}{Q(z)}$ is a rational function over $\bf C$ such that $P$ and $Q$ have no common factors and $\text{deg}(P) < \text{deg}(Q)$.

If $\text{deg}(P) + 1 < \text{deg}(Q)$, then the sum of the numerators of the linear rational function terms of the partial fraction decomposition of $\frac{P(z)}{Q(z)}$ is equal to $0$.

If $\text{deg}(P) + 1 = \text{deg}(Q)$, then the sum of the numerators of the linear rational function terms of the partial fraction decomposition of $\frac{P(z)}{Q(z)}$ is equal to $\frac{P_0}{Q_0}$, where $P_0$ and $Q_0$ are the leading coefficients of $P$ and $Q$.

Let $R$ be such that the disk $D:=\{z:|z|<R\}$ contains the roots of $P(z)$ and $Q(z)$, then we have that,

\begin{align} \int_{|z|=R}\frac{P(z)}{Q(z)}dz&=\int_{|w|=R}\frac{P(\frac{R^2}{w})}{Q(\frac{R^2}{w})}\cdot\frac{R^2}{w^2}dw\\ &=\int_{|w|=R}\frac{\frac{R^{2\text{deg}(P)}}{w^{\text{deg}(P)}}\hat{P}(w)}{\frac{R^{2\text{deg}(Q)}}{w^{\text{deg}(Q)}}\hat{Q}(w)}\cdot\frac{R^2}{w^2}dw\\ &=R^{2(\text{deg}(P)-\text{deg}(Q)+1)}\int_{|w|=R}\frac{\hat{P}(w)}{\hat{Q}(w)}\cdot w^{\text{deg}(Q)-\text{deg}(P)-2}dw, \end{align}

with substitution given by $z=\frac{R^2}{w}$. And where the polynomial $\hat{Q}$ now has no roots lying inside $D$.

If $\text{deg}(P) + 1 < \text{deg}(Q)$, then the integrand is an entire function and thus the integral equals $0$.

If $\text{deg}(P) + 1 = \text{deg}(Q)$, then the integrand has a simple pole at zero, and thus by the residue theorem, we have,

\begin{align} R^{2(\text{deg}(P)-\text{deg}(Q)+1)}\int_{|w|=R}\frac{\hat{P}(w)}{\hat{Q}(w)}\cdot w^{\text{deg}(Q)-\text{deg}(P)-2}dw&=\int_{|w|=R}\frac{\hat{P}(w)}{\hat{Q}(w)}\cdot\frac{1}{w}dw\\ &=2\pi i\lim_{w\rightarrow0}w\cdot\frac{\hat{P}(w)}{\hat{Q}(w)}\cdot\frac{1}{w}\\ &=2\pi i\lim_{z\rightarrow\infty}z\frac{P(z)}{Q(z)}\\ &=2\pi i\frac{P_0}{Q_0}. \end{align}

Let $a_1,\ldots,a_r$ be the roots of $Q(z)$, with multiplicities $m_1,\ldots,m_r$.

Then the partial fraction decomposition for $\frac{P(z)}{Q(z)}$ is given by,

$$\frac{P(z)}{Q(z)}=\sum_{i=1}^r\sum_{j=1}^{m_i}\frac{c_{i_j}}{(z-a_i)^j}.$$

Thus to prove the original statement, we simply apply Cauchy's integral formula (for derivatives) to the expression,

\begin{align} \int_{|z|=R}\frac{P(z)}{Q(z)}dz=\int_{|z|=R}\sum_{i=1}^r\sum_{j=1}^{m_i}\frac{c_{i_j}}{(z-a_i)^j}dz&=\sum_{i=1}^r\sum_{j=1}^{m_i}\int_{|z|=R}\frac{c_{i_j}}{(z-a_i)^j}dz\\ &=\sum_{i=1}^r\int_{|z|=R}\frac{c_{i_1}}{z-a_i}dz\\ &=2\pi i\sum_{i=1}^rc_{i_1}. \end{align}

There should thus be an elementary proof of the original statement. However I haven't been able to produce one, can anyone help me find one?

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Just consider $\lim\limits_{z\to\infty}\big(zP(z)/Q(z)\big)$. It gives the claim immediately.

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  • $\begingroup$ this is really slick, +1 $\endgroup$ – Thoth Sep 15 at 20:17

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