Contour integral over function $P(x)/Q(x)$: $P(x) = 1$ and $Q(x)$ can be broken into linear factors

Question

a. Let $z_1,z_2,...,z_n$ be distinct complex numbers $(n \geq 2)$. Show that in the partial fractions decomposition \begin{equation} \frac{1}{(z-z_1)(z-z_2)\cdots(z-z_n)} = \frac{A_1}{z-z_1}+\frac{A_2}{z-z_2}+\cdots+\frac{A_n}{z-z_n} \end{equation} we must have $A_1+A_2+...+A_n=0$.

b. Suppose that C is a simple closed path that contains the points $z_1,z_2,...,z_n$ in its interior. Use the result in part a to prove that \begin{equation} \int_C \frac{1}{(z-z_1)(z-z_2)\cdots(z-z_n)} dz = 0. \end{equation}

Assuming part a to be true, we can include epsilon balls around each of the discontinuities $z_1,...,z_n$ and the integral over the region is the sum of the integrals of the epsilon balls. But for each epsilon ball, the integral over all but one of the summands will vanish, and for the non-vanishing one, we get $A_i2\pi i$ (assuming positive orientation on all curves).

Then the answer is $2\pi i(A_1+...+A_n) = 0.$

My question is how to show part a. I've thought about it for a while, now. I know what each $A_i$ is explicitly, by Heaviside's cover-up trick/rule. But there's gotta be an easier way than writing and summing all of those up, right?

Answer 1

Multiply to eliminate the denominators and check the coefficients of $z^{n-1}$ on both sides.

Answer 2

Hint Multiplying both sides of the decomposition by $(z - z_1) \cdots (z - z_n)$ gives $$1 = \sum_k A_k (z - z_1) \cdots \widehat{(z - z_k)} \cdots (z - z_n) ,$$ where the wide hat denotes omission of that factor. Compare the coefficients in $z$ of both sides of the equation.

Remark The same argument shows that if $\deg p < n$, then the decomposition $$\frac{p(z)}{(z - z_1) \cdots (z - z_n)} = \frac{B_1}{z - z_1} + \cdots + \frac{B_n}{z - z_n}$$ satisfies $B_1 + \cdots + B_n = 0$ iff $\deg p < n - 1$.