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I came across the following, well known, uniqueness result concerning polynomial interpolation, which is sated as follows:

Let $x_0,x_1,\dotsc,x_n$ be $n+1$ distinct points in $[a,b]$. There exists a unique polynomial $p$ of degree $n$ or less that interpolates $f(x)$ at the points $\{x_i\}$.

My question is twofold:

  1. Is there a complex version of the above result? Maybe something like: If $z_0, z_1,\dotsc,z_n$ are distinct points in a closed connected region, then there exists a complex polynomial $p$ of degree $n$ or less that interpolates $f(z)$, where $f(z)$ is holomorphic?
  2. If so, where might I be able to find the theorem?

I attempted searching online, but all of the results I came across concern themselves with $\mathbb{R}$.

Thank you!

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    $\begingroup$ This is true because of the following result or this one. Just assume there are 2 polynomials and they are different. $\endgroup$ Commented Oct 30, 2019 at 21:54

1 Answer 1

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Using this lemma

Lemma. A polynomial $P(z)$ of degree $n$ with $n+1$ zeros is $P(z)\equiv 0$.

we can assume that there exists $2$ polynomials $P_1(z) \ne P_2(z)$ such that $P_1(z_i)=P_2(z_i)=f(z_i), i=0..n$. Both polynomials are of degree $n$ or less. Then we construct the following polynomial $$P(z)=P_1(z) - P_2(z)$$ which is of degree $n$ or less and has $n+1$ zeros $$P(z_i)=P_1(z_i) - P_2(z_i)=f(z_i)-f(z_i)=0, i=0..n$$ But according to the lemma above, $P(z)\equiv 0$ or $P_1(z)\equiv P_2(z)$, which contradicts the assumption.

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