Show $P(z) $ is a polynomial of degree $n-1$ interpolating an Analytic function.

Let $C$ be a regular curve enclosing the distinct points $ω_1,ω_2,...ω_n$ and let $p(ω) = (ω −ω_1)(ω −ω_2) \cdots (ω −ω_n)$. Suppose that $f (ω)$ is analytic in a region that includes $C$. Show that

$$P(z) = \frac{1}{2 \pi i} \int_C \frac{f(ω)}{p(ω)} . \frac{p(ω) -p(z)}{ω -z} dω$$

is a polynomial of degree $n-1$ and $P(ω_i) = f(ω_i)$.

As $\omega - z$ divides $p(\omega) - p(z)$, this quotient is actually a polynomial: $$\frac{p(\omega) - p(z)}{\omega - z} = R(\omega,z) = \sum_{k=0}^{n-1} R_k(\omega)z^k.$$ Then, $$P(z) = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\frac{p(\omega) − p(z)}{\omega − z}d\omega = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\left(\sum_{k=0}^{n-1} R_k(\omega)z^k\right)d\omega = $$ $$ \sum_{k=0}^{n-1}\int_C\left(\frac1{2\pi i}\frac{f(\omega)}{p(\omega)} R_k(\omega)d\omega\right)z^k. $$ ($P$ is a polynomial)

On the other hand, $$P(\omega_k) = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\frac{p(\omega) − p(\omega_k)}{\omega − \omega_k}d\omega = \frac1{2\pi i}\int_C\frac{f(\omega)}{p(\omega)}\frac{p(\omega) − 0}{\omega − \omega_k}d\omega = $$ $$ = \frac1{2\pi i}\int_C\frac{f(\omega)}{\omega − \omega_k}d\omega = f(\omega_k) $$ by the Cauchy integral formula.

This directly follows from the residue theorem: $$ P(z)=\sum_i \frac{f(\omega_i)}{\prod_{j\ne i}(\omega_i-\omega_j)}\frac{-p(z)}{\omega_i-z} $$ which can be written as $$ P(z)=\sum_i \frac{f(\omega_i)}{\prod_{j\ne i}(\omega_i-\omega_j)}\prod_{j\ne i} (z-\omega_j) $$ The last product $\prod_{j\ne i} (z-\omega_j)$ is a polynomial of degree $n-1$ which cancels at $z=\omega_k$ for $k\ne i$. $P(z)$ is thus a polynomial of degree $n-1$ with $P(\omega_i)=f(\omega_i)$.