Lecture Notes

For $n \geq 2$, let $z_1, \cdots, z_n$ be $n$ distinct points on $\C$, with $|z_i| < 1$. Let $P(z) = (z-z_1)\cdots (z - z_n)$. Prove that $\oint_{|z|=1} \frac{1}{P(z)} dz = 0.$ Hint: change variable $w = 1/z$, so that the integrand in the interior of disk $|w| < 1$ has no poles.

Remark: this also holds for more general numerators. For polynomial $Q(z)$ with $\deg Q \leq n-2$, we have $\oint_{|z|=1} \frac{Q(z)}{P(z)} dz = 0.$ You can prove this more general form (instead of the above one) if you want.