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](https://mathoverflow.net/questions/426655/on-well-separated-circular-regions-in-the-riemann-sphere-and-complex-polynomials)



A statement on complex polynomials
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.


