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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$ $$ \frac{1}{2\pi i} \int_\gamma \frac{(\omega(\zeta)-\omega(z))f(\zeta)}{(\zeta-z)\omega(\zeta)}d\zeta = \sum_{i=1}^n Res\Big(\frac{F}{G},a_i\Big) = \sum_{i = 1}^n \frac{\omega(z)f(a_i)}{(z-a_i)\omega'(a_i)} = L_n(z) $$

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