# Distribution of roots of complex polynomials

I generated random quadratic and cubic polynomials with coefficients in $\mathbb{C}$ uniformly distributed in the unit disk $|z| \le 1$. The distribution of the roots of 10000 of these polynomials are shown below (left: quadratic; right: cubic).

What explains these distributions? In particular: (1) Why the relative paucity of roots near the origin. (2) Why is the density concentrated in $\frac{1}{2} \le |z| \le 1$? (3) Why is the cubic distribution sharper than the quadratic?

• It's probably significant that the distribution of the product of the roots, and therefore that of the product of their absolute values, does not depend on the degree. – Niels J. Diepeveen Aug 17 '13 at 16:31
• Have you tried plotting the logarithm of the roots? This seems to show a density that is independent of the imaginary part and symmetric w.r.t. the real part. – Niels J. Diepeveen Aug 18 '13 at 2:39
• @NielsDiepeveen: Nice insight! But I still don't understand what's driving these distributions... – Joseph O'Rourke Aug 18 '13 at 14:01
• The distribution of roots is invariant under rotation. More precisely, under the transform $a_k \to e^{ik\theta} a_k$, a root $r$ of the polynomial $z^n + a_1 z^{n-1} + \cdots + a_n$ corresponds to a root $r e^{-i\theta}$ of the transformed polynomial. That's why the density is independent of the imaginary part of log of the root. – achille hui Aug 19 '13 at 12:19
• Relevant: math.ucr.edu/home/baez/week285.html – Alice Ryhl May 17 '14 at 8:30

The first two observations can be explained by the fact that if $r$ is the root of $a_nz^n+\cdots +a_0$ then $r^{-1}$ is the root of $a_0z^n+\cdots +a_n$. Since the joint probability density of the coefficients is symmetric under $a_k\mapsto a_{n-k}$ transformation the density of the roots should be symmetric under $r\mapsto r^{-1}$. It should be fairly straightforward to find asymptotic behavior of the density as $z\to0$ or $z\to\infty$. In the first case, $r\to a_0/a_1$ and the second case is symmetric under inversion.
The distribution around $|z|=1$ is less obvious, so I need to think more to explain your third observation.