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

Solution 1:

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.

Solution 2:

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.

Solution 3:

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.