Let $f(z)$ be a complex polynomial of degree at least $2$ and $R$ be a positive number such that $f(z) \neq 0$ for all $|z| \geq R$. Show that $\int_{|z|=R} \frac{dz}{f(z)}=0$ and deduce that $\sum_{k=1}^n \dfrac{1}{\prod_{j\neq k}(a_k-a_j)}=0$ where $a_i$ is $n$ distinct roots of $f$
I can indicate $\sum_{k=1}^n \dfrac{1}{\prod_{j\neq k}(a_k-a_j)}=0$ by Residue but I wonder why $\int_{|z|=R} \frac{dz}{f(z)}=0$?
Hint. Note that if $f(z)=\prod_{j=1}^n (z-a_j)$ where the complex numbers $a_j$ with $j=1,\dots,n$ are all distinct then $$\frac{1}{f(z)}=\sum_{k=1}^n\frac{A_k}{z-a_k}\quad \text{ where } A_k=\dfrac{1}{\prod_{j\neq k}(a_k-a_j)}.$$ Then for $R>\max_{j=1,\dots,n}{|a_j|}$, $$\int_{|z|=R} \frac{dz}{f(z)}=\sum_{k=1}^nA_k\int_{|z|=R} \frac{dz}{z-a_k}.$$ P.S. Do you know what the Residue at infinity is? Well in our case such residue is zero: $$\text{Res}\left(1/f;\infty\right)=-\text{Res}\left(\dfrac{1}{z^2f\left(\dfrac 1z\right)},0\right) =-\text{Res}\left(\dfrac{z^{n-2}}{\prod_{j=1}^n (1-a_jz)},0\right)=0$$ because $n\geq 2$.