Let $P\in\mathbb{C}[X]$ be a complex polynomial of degree $n\geq 2$ with complex roots $\alpha\_1, \alpha\_2,\ldots, \alpha\_n$. My question is about the existence of a formula for the variance of the roots of $P$ in terms of the coefficients of $P$. First, let me fix some notations.


I will denote by $m=m(P)$ the arithmetic mean of the roots
$$m=\frac{\alpha\_1+\alpha\_2+\cdots+\alpha\_n}{n},$$
and by $v=v(P)$, the variance
$$v=\frac{1}{n}\sum\_{i=1}^n\vert \alpha\_i-m\vert^2,$$
that is, the arithmetic mean of the squared distance to the mean, which is a quantity of great interest in probability theory.
Note that $v=0$ if and only if $P$ has one root of multiplicity $n$.


Le us write 
$$P(X)=a\_nX^n+a\_{n-1}X^{n-1}+\cdots+a\_0=a\_n\prod\_{i=1}^n(X-\alpha\_i).$$


Clearly, the arithmetic mean can be expressed in terms of the coefficients as
$$m=-\frac{a\_{n-1}}{na\_n}.$$
Consider now the problem of expressing the variance $v$ in terms of the coefficients by a formula involving only *arithmetic operations, radicals and conjugation*
(in what follows, a formula means such a formula).
By a standard computation, the variance can be written as
$$v=(n-1)\vert m\vert^2-\frac{2}{n}Re(\sum\_{i<j}\alpha\_i\bar{\alpha}\_j).$$
So my question becomes :


**Is it possible to find a formula for
$$Re(\sum\_{i<j}\alpha\_i\bar{\alpha}\_j)$$
in terms of the coefficients of $P$ for any given degree $n$, or is there some theoritical obstruction as for expressing the roots themselves ?**


I've worked out the cases $n=2$ and $n=3$ for which the formula looks like that for the roots themselves; so my basic intuition is that it will not be possible in general.


Here are those formulas :


Without loss of generality, we can focus on monic polynomial ($a\_n=1$).


The case $n=2$ is particularly simple : a direct calculation shows that
$$v=\frac{\vert \alpha\_1-\alpha\_2\vert^2}{4}=\frac{\vert (\alpha\_1+\alpha\_2)^2-4\alpha\_1\alpha\_2\vert}{4}=\frac{\vert a\_1^2-4a\_0\vert}{4}.$$
In other words, $v=\vert \Delta \vert/4$, where $\Delta$ stands for the discrimant of $P$.


Before considering the case $n=3$, let me make a few observations. For any complex number $t$, the roots of $Q(X)=P(X+t)$ are $\alpha\_i-t$, $i=1\ldots n$, and the variance is invariant by translation, hence $v(Q)=v(P)$. On the other hand, the coefficients of $Q$ can be expressed in terms of the coefficients of $P$ by means of arithmetic operations. So, w.l.o.g. one can concentrate on the problem of expressing $v(Q)$ in terms of the coefficients of $Q$ for some well-chosen $t$.
A natural choice is to set $t=m$ since, by doing this, we obtain a polynomial $Q$ with $m(Q)=0$. 


These remarks show that it is sufficient to consider the case where $P$ is monic, with mean zero. In this case, 
$$nv(P)=-2Re(\sum\_{i<j}\alpha\_i\bar{\alpha}\_j).$$


Starting directly from Cardan's formulas for the roots of $$P(X)=X^3-pX-q,$$ I found the following expression :
$$v=\sqrt[3]{\left\vert\frac{q+\sqrt{\Delta}}{2}\right\vert^2}+\sqrt[3]{\left\vert\frac{q-\sqrt{\Delta}}{2}\right\vert^2},$$
where $\Delta=q^2-4p^3/27$ is the discriminant of the associated quadratic equation.



Variance of the roots of a complex polynomial
You may notice that your formula is very similar to the formula for the roots of the polynomial. That is not an accident - they are basically equally difficult. So in particular there is no formula in radicals for $n\geq 5$, and you don't want to see the formula in radicals for $n=4$.


proof: Take any formula in terms of arithmetic operations, radicals and conjugation. Because the other operations commute with conjugation, we can assume that conjugation comes first - i.e. the function is an algebraic function, expressible using radicals, of $a\_0,\dots, a\_n, \overline{a\_0}, \dots, \overline{a\_n}$.


We can express $a\_0,\dots, a\_n, \overline{a\_0}, \dots, \overline{a\_n}$ as algebraic functions of $\alpha\_1\dots, \alpha\_n, \overline{\alpha\_1},\dots,\overline{\alpha\_n}$. The mean is also an algebraic function of these numbers. So we are seeking an identity of two algebraic functions. Here is the key point: The subset of $\mathbb C^{2n}$ with coordinates $\alpha\_1,\dots,\alpha\_n,\beta\_1,\dots,\beta\_n$ defined by the relations $\beta\_k=\overline{\alpha\_k}$ is Zariski dense, so any algebraic equation that holds there holds for all $\alpha\_i,\beta\_i$.


Now we're essentially done, because we have a question of pure Galois theory. We have three fields:


$K = \mathbb C( \alpha\_1,\dots, \alpha\_n,\beta\_1,\dots,\beta\_n)$


$L= \mathbb C( a\_1,\dots, a\_n, b\_1,\dots b\_n) $ (where $b\_i$ are the symmetric polynomials of the $\beta\_i$.)


$M = \mathbb C( a\_1,\dots, a\_n, b\_1,\dots b\_n, \sum\_{i=1}^n \alpha\_i \beta\_i) $


We want to know whether $M$ is a radical extension of $L$. This happens if and only if its normal closure is solvable. It is contained in $K$, which has Galois group $S\_n \times S\_n$ over $M$. The subgroup that fixes $\sum\_{i=1}^n a\_i b\_i$ is clearly the diagonal $S\_n$. For $n>2$, the diagonal subgroup contains no normal subgroup, and so the normal closure of $M$ over $L$ is $K$. For $n>4$, this is not solvable.


