Let $z_0 \in \mathbb{D}$ and denote by $\theta_j = e^{2\pi i (j-1)/N},$ $j=1,...,N$. Consider $P_N = span \{1,z,z^2,...,z^{N-1} \}$ and the linear map $B : P_N \rightarrow \mathbb{C}^N$ defined by $$ B(p) = \left( \frac{p(\theta_j z_0)}{\sqrt{N}} \right)_{j=1}^N. $$ It's pretty easy to see that is an isomorphism, as is an injective map because the fundamental theorem of algebra and both vector spaces have the same dimension.
I'm trying to get the inverse but I'm really stuck with it.
Can anybody help me? Thank you very much.
Let $p(z) = \sum_{k=0}^{N-1}\,a_kz^k$ and $v = (v_1,\dots,v_N)$. You need to solve
$$\sum_{k=0}^{N-1}\,a_k\,{\left(\theta_j\,z_0\right)}^k = v_j\,\sqrt N \tag{$\forall j=1,2,\dots, N$}$$
for $(a_0,a_1,\dots,a_{N-1})$. This is a linear system of equations:
$$ \underbrace{\pmatrix{ 1 &\theta_1z_0 &{\left(\theta_1\,z_0\right)}^2 &\dots &{\left(\theta_1\,z_0\right)}^{N-1}\\ 1 &\theta_2z_0 &{\left(\theta_2\,z_0\right)}^2 &\dots &{\left(\theta_2\,z_0\right)}^{N-1}\\ \vdots &\vdots &\vdots &\ddots &\vdots \\ 1 &\theta_Nz_0 &{\left(\theta_N\,z_0\right)}^2 &\dots &{\left(\theta_N\,z_0\right)}^{N-1} }}_M \cdot \pmatrix{ a_0\\a_1\\\vdots\\a_{N-1}} =\sqrt{N}\pmatrix{ v_1\\v_2\\\vdots\\v_{N}} $$
and can be solved by traditional linear algebra methods $($e.g. computing $M^{-1})$. Notice that this is a Vandermonde matrix, which has an explicit inverse formula.