A statement on complex polynomials

Question

I have a feeling the following is true.

Assume that there are $n$ mutually disjoint closed disks $D_i$ in the complex plane and $n$ complex polynomials $p_i(z)$ of degree $n - 1$, with both types of objects indexed from $1$ to $n$, such that

condition: the polynomial $p_i(z)$ has exactly 1 complex root in each disk $D_j$, where $1 \leq j \leq n$ and $j \neq i$ and no other roots.

Are the $n$ polynomials $p_1(z), \ldots, p_n(z)$ necessarily linearly independent over $\mathbb{C}$? I think it is true.

So for example, if $n = 3$, then $p_1(z)$ has exactly $1$ root in $D_2$ and exactly $1$ root in $D_3$ and no other roots, and so on.

If my intuition is correct, this problem should be some kind of "dual" to the problem where you have $n$ polynomials, say $q_i(z)$, of degree $n - 1$, where all the roots of $q_i(z)$ are in the closed disk $D_i$ (we also assume that the closed disks $D_i$, for $i = 1, \dots, n$, are mutually disjoint). I know how to prove this possibly "dual" problem using the Grace-Walsh-Szego (accents?) coincidence theorem.

I did not run any numerical simulations yet. Any ideas or suggestions? I came across this problem by thinking about the Atiyah problem on configurations of points.

Edit: the conjecture in this post was proved to be false for $n \geq 3$ by Noam D. Elkies (see his answer below). However, for a follow-up question, please see On well separated circular regions in the Riemann sphere and complex polynomials

Answer 1

The conjecture is easily seen to be true for $n<3$. We give a counterexample for $n=3$.

Let $p_i = z^2 - \omega_i$ where the $\omega_i$ are the cube roots of unity. These are linearly dependent because they're in the $2$-dimensional subspace spanned by $1$ and $z^2$. But their zeros are the sixth roots of unity, forming a regular hexagon $H$ each of whose three long diagonals $d_i$ joins the roots of one of the $p_i$. Let $s_i$ be pairwise disjoint sides of $H$ such that $s_i$ is parallel to the corresponding $d_i$. It's easy to find discs $D_i$ such that each $D_i$ contains $s_i$ and is disjoint from $s_j$ for each $j \neq i$; for instance start from the discs with diameter $d_i$ and dilate each one by a factor $1.01$. Then each $p_i$ has no roots in $D_i$ and exactly one root in $D_j$ for each $j \neq i$, QEF.

A similar construction works for $n>3$ using $p_i = z^{n-1} - \omega_i$ where the $\omega_i$ are the $n$-th roots of unity. Note that these $p_i$ are not only linearly dependent but span a linear space of dimension only $2$, the minimum possible; moreover, their roots are all on the unit circle, so a fractional linear transformation such as $z = (z'+i) / (z'-i)$ makes all the roots real, which means that the $p_i$ can be taken to have real coefficients.