Let me give an informal explanation using what little I know about complex analysis.

Suppose that $p(z)=a_{0}+...+a_{n}z^{n}$ is a polynomial with random complex coefficients and suppose that $p(z)=a_{n}(z-c_{1})\cdots(z-c_{n})$. Then take note that

$$\frac{p'(z)}{p(z)}=\frac{d}{dz}\log(p(z))=\frac{d}{dz}\log(z-c_{1})+...+\log(z-c_{n})=
\frac{1}{z-c_{1}}+...+\frac{1}{z-c_{n}}.$$

Now assume that $\gamma$ is a circle larger than the unit circle. Then

$$\oint_{\gamma}\frac{p'(z)}{p(z)}dz=\oint_{\gamma}\frac{na_{n}z^{n-1}+(n-1)a_{n-1}z^{n-2}+...+a_{1}}{a_{n}z^{n}+...+a_{0}}\approx\oint_{\gamma}\frac{n}{z}dz=2\pi in.$$

However, by the residue theorem,

$$\oint_{\gamma}\frac{p'(z)}{p(z)}dz=\oint_{\gamma}\frac{1}{z-c_{1}}+...+\frac{1}{z-c_{n}}dz=2\pi i|\{k\in\{1,\ldots,n\}|c_{k}\,\,\textrm{is within the contour}\,\,\gamma\}|.$$

Combining these two evaluations of the integral, we conclude that
$$2\pi i n\approx 2\pi i|\{k\in\{1,\ldots,n\}|c_{k}\,\,\textrm{is within the contour}\,\,\gamma\}|.$$ Therefore there are approximately $n$ zeros of $p(z)$ within $\gamma$, so most of the zeroes of $p(z)$ are within $\gamma$, so very few zeroes can have absolute value significantly greater than $1$. By a similar argument, very few zeroes can have absolute value significantly less than $1$. We conclude that most zeroes lie near the unit circle.

$\textbf{Added Oct 11,2014}$

A modified argument can help explain why the zeroes tend to be uniformly distributed around the circle as well. Suppose that $\theta\in[0,2\pi]$ and $\gamma_{\theta}$ is the pizza slice shaped contour defined by
$$\gamma_{\theta}:=\gamma_{1,\theta}+\gamma_{2,\theta}+\gamma_{3,\theta}$$ where

$$\gamma_{1,\theta}=([0,1+\epsilon]\times\{0\})$$

$$\gamma_{2,\theta}=\{re^{i\theta}|r\in[0,1+\epsilon]\}$$

$$\gamma_{3,\theta}=\cup\{e^{ix}(1+\epsilon)|x\in[0,\theta]\}.$$

Then $$\oint_{\gamma_{\theta}}\frac{p'(z)}{p(z)}dz=
\oint_{\gamma_{\theta,1}}\frac{p'(z)}{p(z)}dz+\oint_{\gamma_{\theta,2}}\frac{p'(z)}{p(z)}dz+\oint_{\gamma_{\theta,3}}\frac{p'(z)}{p(z)}dz$$

$$\approx O(1)+O(1)+\oint_{\gamma_{\theta,3}}\frac{p'(z)}{p(z)}dz$$

$$\approx O(1)+O(1)+\oint_{\gamma_{\theta,3}}\frac{na_{n}z^{n-1}+(n-1)a_{n-1}z^{n-2}+...+a_{1}}{a_{n}z^{n}+...+a_{0}}dz$$

$$\approx O(1)+O(1)+\oint_{\gamma_{\theta,3}}\frac{n}{z}dz\approx n i\theta$$.

Therefore, there should be approximately $\frac{i\theta}{2\pi}$ zeroes inside the pizza slice $\gamma_{\theta}$.

22

That's a great question. Have you taken a look at the pictures created by Dan Christensen (http://jdc.math.uwo.ca/roots/) and the stuff John Baez has written about it (http://math.ucr.edu/home/baez/roots/)? See also http://johncarlosbaez.wordpress.com/2011/12/11/the-beauty-of-roots/ and Jordan Ellenberg here: http://quomodocumque.wordpress.com/2010/01/09/what-do-roots-of-random-polynomials-look-like/

– Tom Leinster – 2014-10-02T21:24:53.807@TomLeinster: yeah I know about those. I am considering computing one more degree than Dan, so up to 25. It would take about 3 days using several computers. – Andrej Bauer – 2014-10-02T21:56:00.020

4For polynomials of the form $x^2+bx+c$, the absolute values of the non-real roots depend only on $c$, so that as $b$ varies, we get lots of roots all on a circle. Is this the same phenomenon writ small, and does it shed any light? – Steven Landsburg – 2014-10-02T22:05:40.373

@Steven, the

absolutevalues $\ \xi\ \eta\ $of the two roots (regardless of being real or not) travel on the positive-quarter branch of hyperbole $\ {(\xi\ \eta) : \xi\cdot\eta =|c|}\ $ when $\ b\ $ varies. (Your comment was strange to me). – Włodzimierz Holsztyński – 2014-10-03T01:59:39.527@WłodzimierzHolsztyński: Well, yes, but when the roots are non-real, they "travel" in the degenerate sense that they don't actually travel at all. – Steven Landsburg – 2014-10-03T02:02:53.053

1

Related: http://mathoverflow.net/questions/139804/distribution-of-roots-of-complex-polynomials . I thought I had seen other discussions of this phenomenon on here as well....

– usul – 2014-10-03T08:00:51.62711

"The zeros of random polynomials cluster uniformly near the unit circle" http://www-old.newton.ac.uk/preprints/NI04017.pdf

– Benjamin Dickman – 2014-10-03T08:42:43.5833

Related: http://math.stackexchange.com/q/206890/622

– Asaf Karagila – 2014-10-03T09:40:53.95018I suspect your constraints on the coefficients are similar to taking the $100$th root of a relatively small number: $\sqrt[100]{42}\approx 1.038$. Try multiplying the coefficients of $x^i$ by $2^{100-i}$ to see if it makes a difference. – Henry – 2014-10-04T13:53:53.793

4

George Lowther proves that the roots "becomes concentrated on the unit circle" in his answer to the MO question, "Distribution of roots of complex polynomials" (as linked by

– Joseph O'Rourke – 2014-10-04T14:10:56.090usul).2Since the roots and coefficients of a polynomial are Fourier transforms of each other, another way to rephrase the question that I find interesting is to ask why the Fourier transform of a random real vector is another random vector most of whose components have unit magnitude. I don't have an answer as to why but maybe this will help someone find another intuitive explanation. – Mehrdad – 2014-10-05T09:47:13.653

4If we fix the degree at $n$ and the coefficients are iidrvs then the distribution is invariant under $p(z) \mapsto z^np(1/z)$, which means that the distribution of zeros is invariant under $z \mapsto 1/z$. I wonder if there is an equally simple way to see that there has to be some rotational symmetry. – François G. Dorais – 2014-10-05T20:54:36.327

25I'm sorry, but with all due respect for the other contributions to this site of the OP and most answerers , this is definitely not a great question. It is not precise, does not admit any definitive answer, and is kind of trivial: Henry's comment already demystifies it completely, and Joseph Van Name shows that it's juts a very simple exercise in complex analysis. Yet another example of the stack exchange system gone crazy. – Joël – 2014-10-06T12:51:36.310

3I know very little about the math going on in the background but a colleague demystified this for me by saying that basically it is a special feature of the

basisyou have chosen, namely the monomials. If you choose a different basis or even just weight each monomial by a factor, the roots will tend to congregate on a different set. Basically although a) any polynomial can arise and b) you chose them randomly... They aren't as generic as as you think; they've in fact been chosen in a special way. – Spencer – 2014-10-06T14:46:01.9471@Joël: While I don't disagree with the craziness, I disagree with the reasoning. While seemingly plausible, Henry's comment is arguably incorrect. While Joseph's answer is enlightening, it does not explain the observed rotational symmetry. There is a lot more to this question than meets the eye... – François G. Dorais – 2014-10-07T01:11:25.293

14@Joël: I myself wouldn't rate this question as highly as it is, I think we're all victims of the scale-free networks. If I had to guess what makes it interesting: it is easy to understand (and thereby very likely not "research level") and it sparks ideas in people's minds because it looks like it is within reach of a good coffee break discussion. – Andrej Bauer – 2014-10-07T06:51:44.670

@Francois Dorais. I added to my answer some remarks that do explain why there should be rotational symmetry. Unfortunately it is not very rigorous at this point since the denominator could be near zero close to the contour. But I would agree that there is definitely more than meets the eye (just see Terry Tao's comments on my answer). – 35093731895230467514051 – 2014-10-12T00:22:09.010

1I have a theorem for this in my habilitation from 1993 (Interpolation Points and Zeros of Polynomials in Approximation Theory, R. Grothmann, Habil. at KU Eichstätt). It is not a central result there and I never published this corollary anywhere else, I must admit. Such things seem to get rediscovered often. – Rene – 2014-10-12T17:32:49.183

1a quick answer to the title's question is: if z has modulus 1, all its powers have modulus 1 too, and a random linear combination of them (with coefficients iid chosen in [a,b]) has more chances to vanish. For $n$ large compared to |a|+|b|, $|z|\neq 1$ puts a strong constraint on the possible vanishing linear combinations of the powers of z. – Pietro Majer – 2014-10-13T07:56:50.593

1I think it is not difficult to make the above argument into an elementary rigorous estimate of the set of coefficients $(a_0,\dots,a_n)\in[a,b]^{n+1}$ of polynomials that have a zero $1-\epsilon <|z|< 1+\epsilon,$ and with a bit more care, with $\alpha<\mathrm{arg}z<\beta$, to explain quantitatively the uniform concentration on the unit circle. – Pietro Majer – 2014-10-13T08:07:06.430

1@PietroMajer: please post an answer if you have a nice one. – Andrej Bauer – 2014-10-13T08:20:10.247