# contour integral representation of lagrange interpolating formula

how can i show that if the interpolation nodes are complex numbers $a_1,a_2,...,a_n$ and lie in some domain $G$, bounded by a piecewise-smooth contour $\gamma$, and if $f$ is a single-valued analytic function defined on the closure of $G$, then the Lagrange interpolation formula has the form $$L_n(z)=\frac{1}{2\pi i}\int_\gamma \frac{\omega (\zeta) - \omega(z)}{\zeta - z} \frac{f(\zeta)}{\omega(\zeta)} d\zeta,$$ where $$\omega(\zeta)=\prod_{k-1}^n (\zeta - a_i).$$

i know that i may use the Cauchy Integral Formula but i am stocked for a quite long time. thank you!

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Well the ordinary interpolation formula is true even in $\mathbb{C}$ where you have the following formula (using your notation) $$L_n (z) = \sum_{i = 1}^n \frac{\omega(z)f(a_i)}{(z-a_i)\omega'(a_i)}$$ Now you see that for analytic $F$ and $G$ with simple zero of $G$ at $z_0$ you have the residue as $Res( F/G, z_0) = F(z_0)/G'(z_0)$. Here observe that for a fixed $z\in \mathbb{C}$ the function of $\zeta$ given by $(\omega(\zeta)-\omega(z))/(\zeta -z)$ is analytic ( has removable singularity with $\omega'(z)$ at $\zeta = z$, hence replacing $$F(\zeta ) = \frac{(\omega(z)-\omega(\zeta))f(\zeta)}{z-\zeta},\ \ \ \ G(\zeta) = \omega(\zeta)$$ as $a_i$'s are simple zeros of $\omega$ we get it directly from Residue theorem that for $z \neq a_i \ \ \forall i$ $$\frac{1}{2\pi i} \int_\gamma \frac{(\omega(\zeta)-\omega(z))f(\zeta)}{(\zeta-z)\omega(\zeta)}d\zeta = \sum_{i=1}^n Res\Big(\frac{F}{G},a_i\Big) = \sum_{i = 1}^n \frac{\omega(z)f(a_i)}{(z-a_i)\omega'(a_i)} = L_n(z)$$