Find a set of polynomials $\{P_1, \dots, P_n\},$ all of whose coefficients are complex numbers, whose common zero set is $\{(3 + 2i, -i), (0, 1 - 4i)\}$ in $\mathbb{C}^2.$
When there's only one point, such as $\{(3 + 2i, -i)\},$ the answer is easy enough. We can use something like $P_1 = x - (3+2i)$ and $P_2 = y - (-i),$ and the zero sets of these polynomials will intersect in the given point. I tried something similar in the stated problem, but $P_1 = x - (3+2i), P_2 = y - (-i), P_3 = x - 0,$ and $P_4 = y - (1 - 4i)$ would have an empty common zero set.
I also tried $P_1 = (x - (3+2i))(x - 0)$ and $P_2 = (y - (-i))(y - (1 - 4i)),$ but this would have a common zero set that is too large: $$\{(3 + 2i,-i), (3 + 2i, 1 - 4i), (0, -i), (0, 1 - 4i)\}.$$
If we were working over $\mathbb{R}^2$ and looking for a common zero set that is a finite set of points, such as $\{(1, 2), (0, 5)\},$ then I would use $$P_1 = ((x-1)^2 + (y-2)^2)((x-0)^2 + (y-5)^2).$$ This works because $(x-1)^2$ and $(y-2)^2$ must be non-negative, so the only solution to $(x-1)^2 + (y-2)^2 = 0$ is $(1, 2).$ Something like this wouldn't work in $\mathbb{C}^2$ since we can't claim that $(x-1)^2$ must be non-negative.
So I've tried a few things and feel I am close, but I don't quite have the solution. While my question is for a specific set of points, I'm looking for a method that can be generalized to any finite set of points in $\mathbb{C}^2.$
We could take the $f_1 = x(x-(3+2i))$, $f_2= (x-(3+2i))(y-(1-4i))$, $f_3 = x(y-(-i))$, and $f_4=(y-(-i))(y-(1-4i))$.