Magnitudes of roots of random polynomials I'm looking at the roots of random polynomials with integer coefficients, and constant term=leading term = 1.
Using the Mathematica code
randomCoeffs := 
 Table[RandomInteger[{-10, 10}], {i, RandomInteger[{1, 10}]}]
getRoots := 
 With[{list = 
    x /. NSolve[Fold[#1*x + #2 &, 1, randomCoeffs]*x + 1 == 0, 
      x]}, {Re[#], Im[#]} & /@ list // Point]
Graphics[{EdgeForm[Blue], White, Rectangle[{-11, -1}, {11, 1}], 
  Darker[Orange], Table[getRoots, {i, 10000}], Thick, Red, 
  Circle[{0, 0}, 4]}, ImageSize -> {400, 400}]

I get the picture:

I understand why a box of width $(10+1)*2$ works to bound the roots (see here).  But why are the non-real roots contained in a circle of only radius 4?
 A: The following considerations are too long for a comment.
They are not a full answer (in particular, they do not give a sharp bound); but they at least provide an explanation of why you find only complex roots of significantly lower magnitude.
You plot the roots of polynomials with real coefficients. That means that all complex roots come in conjugate pairs with identical magnitude.
Also, you consider polynomials $P$ of degree $n = 10$ and coefficients bounded by $10$, except for the leading and trailing coefficients. Hence, for the Mahler measure $M(P)$ of $P$ you have the bound $M(P) \le \lVert P\rVert_2 \le \sqrt{1 + 10^2 + \cdots + 10^2 + 1} = \sqrt{902}$; see Landau's inequality.
The Mahler measure is the product of the absolute values of all roots outside the unit circle. If there is such a complex root, also its complex conjugate is accounted for in the Mahler measure. Thus, you immediately get the bound $\lvert\alpha\rvert \le \sqrt[4]{902} < 5.49$ for each complex root $\alpha \in \mathbb C \setminus \mathbb R$ of $P$.
Intuitively speaking, there is only a certain "total of root magnitude" (and, BTW, root separation) that has to be distributed between the roots.
A real root can be a lonely individual, taking almost all of that reservoir (e.g., if the polynomial is $(x-\alpha)(x^n-1)$, where $\alpha$ can be anything, but the roots of unity are as tame as roots can be). But a complex root of a real polynomial has at least one equally "bad" sibling.

Note that, for what it's worth, $4$ is not the proper bound:
$$x^{10} + 7x^9 + 10x^8 - 10x^7 + 8x^6 - 9x^5 + 6x^4 - 3x^3 + 1$$
has roots at approximately $-3.9702 \pm 0.9473\operatorname{i}$, which have a magnitude of approximately $4.0817$. I have no deep insight why this polynomial is more "pathologic" than others; I simply got it by running enough random examples.
My first guess was that the trailing coefficient being $1$ is significant; at least, it implies the constraint $\prod_{P(\alpha)=0} \alpha = 1$. But several thousand runs with trailing coefficients in $\{-10, \dots, 10\}$ did not produce a "better" instance.
Next idea: integer coefficients in such a small range could be rather restrictive. But again, testing with uniformly random coefficients in the interval $[-10, 10]$ did not change anything.
So when it comes to a sharp(er) bound, I'm still clueless.
