Contour integration of a reciprocal of a polynomial with an arc sufficiently large to contain all its residues.

Question

I am not aware of this being a genuine result so I was wondering if this is in general true. Let $P(z)$ be some polynomial with complex coefficients and suppose we take an arc large enough that it contains all of the poles of $\frac{1}{P(z)}$. We further assume that all the poles are simple.

This then becomes and exercise in just calculating the residues.

My question is should this always be $0$?

For example, for degree two, $$2\pi i\left[Res\left[\frac{1}{(z-a)(z-b)},z=a\right]+Res\left[\frac{1}{(z-a)(z-b)},z=a\right]\right]=2\pi i\left[\frac{1}{a-b}+\frac{1}{b-a}\right]=0$$ Similarly, for degree $3$, $$\frac{1}{(a-b)(a-c)}+\frac{1}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)}=\frac{c-b+a-c+b-a}{(a-b)(b-c)(c-a)}=0$$ So the integral is is indeed 0.

Is there any result from the theory of polynomials, symmetric functions or alike that generalises the result to higher dimension? Or does it fail in higher dimension?

Answer 1

Here is a proof for the quadratic case which is easily extended to higher degree polynomials with distinct zeroes.

Suppose $P(z) = (z-z_0)(z-z_1)$ where $z_0 \neq z_1$. By the theory of partial fractions, $$\frac{1}{P(z)} = \frac{1}{(z-z_0)(z-z_1)} = \frac{a_0}{z-z_0} + \frac{a_1}{z-z_1}$$ where $a_0$ is the residue of $1/P(z)$ at $z_0$ and $a_1$ is the residue at $z_1$. Then $$1 = a_0 (z-z_1) + a_1 (z-z_0)$$ The coefficent of $z$ on the LHS is $0$; the coefficent of $z$ on the RHS is $a_0 + a_1$. So $a_0 + a_1 = 0$.

[EDIT]

I see that the case n=2 is not sufficiently general to satisfy the OP. So here's the general case, although in my opinion it's just the case n=2 plus a lot of notation. Suppose $P(z)$ is a polynomial of degree $n \geq 2$ with distinct zeroes, so $$P(z) = \prod_{j=1}^n (z-z_j)$$ where $z_j \neq z_k$ if $j \neq k$. Then by the theory of partial fractions, $$\frac{1}{P(z)} = \prod_{j-1}^n \frac{1}{z-z_j} = \sum_{k=1}^n \frac{a_k}{z-z_k}$$ where $a_1, a_2, \dots ,a_n$ are the residues of $1/P(z)$ at $z_1, z_2, \dots ,z_n$, respectively. So $$(*)\qquad1 = \sum_{k=1}^n \prod_{j=1}^n (z-z_j) \cdot \frac{a_k}{z-z_k}$$ $\prod_{j=1}^n (z-z_j) \cdot \frac{1}{z-z_k}$ is, for each $k$, a monic polynomial of degree $n-1$ because $1/(z-z_k)$ cancels exactly one of the factors in the product $\prod_{j=1}^n (z-z_j) $. Consequently, the coefficient of $z^{n-1}$ on the RHS of (*) is $\sum_{k=1}^n a_k$, while the coefficient of $z^{n-1}$ on the LHS is $0$. So $\sum_{k=1}^n a_k = 0$.

Answer 2

1. If the polynomial degree $g$ is greater than $1$ ( $\underline{one}$ ), you can take the radius $R$ of the arc/circle to go to $\infty$.
2. The integral magnitud behaves as $\displaystyle{\left.{1 \over R^{\,g - 1}}\right\vert_{\ g\ \geq\ 2} \to 0\ \mbox{as}\ R \to \infty}$.