# Prove that an expression is zero for all sets of distinct $a_1, \dotsc, a_n\in\mathbb{C}$

A while ago one of my professors gave the class a problem “to think about when lying on the beach.”

Well, I’ve been on the beach several times since then to no avail and my curiosity has finally outweighed my desire to solve this personally. The problem is this:

Let $a_1, \dotsc, a_n\in\mathbb{C}$ be distinct. Prove that:

\sum_{i=1}^n\prod_{j\neq i}\frac{1}{a_i – a_j} = 0

It’s pretty easy, if tedious, to show this for a given $n$ but I’m unsure about how to generalise the result.

#### Solutions Collecting From Web of "Prove that an expression is zero for all sets of distinct $a_1, \dotsc, a_n\in\mathbb{C}$"

First, assume $\prod a_i \ne 0$

Now consider the $(n-1)^{th}$ degree polynomial

$$P(z) = \sum_{i=1}^{n} a_i \prod_{j \neq i} \frac{z-a_j}{a_i – a_j }$$

We see that $P(a_i) = a_i$ for each $i$

Thus $P(z) – z$ has at least $n$ roots, and thus must be identically $0$.

$$z \equiv \sum_{i=1}^{n} a_i \prod_{j \neq i} \frac{z-a_j}{a_i – a_j }$$

Now put $z= 0$ in the above divide by $(-1)^n\prod a_i$ to get your identity.

If $\prod a_i = 0$, then wlog, assume $a_1 = 0$.

Now take a sequence of complex numbers $c_n \to a_1$, $c_n \neq 0$, and use $c_n$ instead of $a_1$ and take limits.

Let
$$f(z)=\frac{1}{(z-a_1)(z-a_2)\cdots(z-a_n)}$$
and consider
$$\lim_{R\to\infty}\frac{1}{2\pi i}\int_{|z|=R}f(z)\,dz\ .$$
By residues (or the Cauchy Integral Formula), the limit is your expression. By estimating $|f(z)|$, the limit is zero. The conclusion follows.

Another kind of answer (note that $n\geq 2$).

Put $\displaystyle R(z)=\frac{1}{\prod_{i=1}^n (z-a_i)}$. Then we have
$\displaystyle R(z)=\sum_{j=1}^n \frac{c_j}{z-a_j}$.
We immediately get that $\displaystyle c_j=\frac{1}{\prod_{i\not =j}^n (a_j-a_i)}$. We have $\displaystyle zR(z)=\sum_{j=1}^n \frac{zc_j}{z-a_j}\to \sum_{j=1}^n c_j$ as $z\to \infty$, and as $n\geq 2$, $zR(z)\to 0$ as $z\to \infty$ and we are done.