Volume of solution sets for polynomials in $\mathbb{C}[x]$ Denote $\pmb{a}=(a_1,\dots,a_d)\in\mathbb{R}^d$ and consider the set
$$\mathcal{E}_d=\{\pmb{a}\in\mathbb{R}^d: \text{each root $\xi$ of $x^d+a_dx^{d-1}+\cdots+a_2x+a_1=0$ lies in $\vert\xi\vert<1$}\}.$$
In the reference shown below, Fam proved that the $d$-dimensional Lebesgue measure satisfies
$$\lambda_d(\mathcal{E}_d)=2^d\prod_{k=1}^{\lfloor\frac{d}2\rfloor}\left(1+\frac1{2k}\right)^{2k-d}.$$
I'd like to propose a complex version here. Denote $\pmb{c}=(c_1,\dots,c_d)\in\mathbb{C}^d$ and consider the set
$$\mathcal{S}_d=\{\pmb{c}\in\mathbb{C}^d: \text{each root $\xi$ of $x^d+c_dx^{d-1}+\cdots+c_2x+c_1=0$ lies in $\vert\xi\vert<1$}\}.$$
Now, let's ask:

QUESTION. What is the $2d$-dimensional Lebesgue measure
  $$\lambda_{2d}(\mathcal{S}_d)?$$

For contrast, $\lambda_1(\mathcal{E}_1)=2$ while $\lambda_2(\mathcal{S}_1)=\pi$.
Reference.
A. T. Fam,The volume of the coefficient space stability domain of monic polynomials, Proc. IEEE Int. Symp.Circuits and Systems, 2 (1989), pp. 1780–1783.
 A: $\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$The answer is $\tfrac{\pi^n}{n!}$. It is certainly a surprise to have the answer come out so simple!
Let $\phi : \CC^n \to \CC^n$ be the map which takes $(z_1, z_2, \ldots, z_n)$ to the elementary symmetric functions $(e_1, e_2, \ldots, e_n)$ where $e_k = \sum_{1 \leq i_1 < i_2 < \cdots < i_k \leq n} z_{i_1} z_{i_2} \cdots z_{i_n}$. Let $D$ be the unit disc in $\CC$. So you want to compute the volume of $\phi(D^n)$, also known as $\int_{\phi(D^n)} \mathrm{Vol}$.
The map $D^n \to \phi(D^n)$ is $n!$ to $1$ and, since $\phi$ is complex analytic, $\phi$ is orientation preserving. So
$$\int_{\phi(D^n)} \mathrm{Vol} = \frac{1}{n!} \int_{D^n} \phi^{\ast}(\mathrm{Vol}) = \frac{1}{n!} \int_{D^n} \det J^{\RR}_{\phi}$$
where $J^{\RR}_{\phi}$ is the Jacobian of $\phi$ considered as a smooth map $\RR^{2n} \to \RR^{2n}$. I will write $J^{\CC}_{\phi}$ when I instead want the $n \times n$ matrix of complex numbers coming from thinking of $\phi$ as a complex analytic map $\CC^n \to \CC^n$.
The relation between these two notions is this: $\det J^{\RR}_{\phi} = |\det J^{\CC}_{\phi}|^2$. (This is just linear algebra -- if $L/K$ is a degree $d$ field extension, $f: L^n \to L^n$ is a linear map and $g: K^{dn} \to K^{dn}$ is the linear map gotten by identifying $L$ with $K^d$, then $\det g = N_{L/K}(\det f)$.) We have the well known identity 
$$\det J^{\CC}_{\phi}(z_1, \ldots, z_n) = \prod_{i<j} (z_i - z_j) = \sum_{w \in S_n} (-1)^w z_1^{w(1)-1} z_2^{w(2)-1} \cdots z_n^{w(n)-1}.$$
Here is the first reference I found.
So we need to compute
$$\frac{1}{n!} \int_{D^n} \sum_{u \in S_n} (-1)^u z_1^{u(1)-1} z_2^{u(2)-1} \cdots z_n^{u(n)-1} \overline{\sum_{v \in S_n} (-1)^v z_1^{v(1)-1} z_2^{v(2)-1} \cdots z_n^{v(n)-1}}.$$
We can distribute the product to get a sum of $(n!)^2$ terms of the form $\int_{D^n} \prod z_j^{a_j} \overline{z_j}^{b_j}$.  If $a_j \neq b_j$, the integral on $z_j$ is zero, so we reduce to 
$$\frac{1}{n!} \sum_{u \in S_n} \int_{D^n} (z_1 \overline{z_1})^{u(1)-1}  (z_2 \overline{z_2})^{u(2)-1} \cdots  (z_n \overline{z_n})^{u(n)-1}.$$
All $n!$ terms have the same integral, so we are reduced to the one integral
$$\int_{D^n} \prod (z_j \overline{z_j})^{j-1} = \prod_j \int_D (z \overline{z})^{j-1}.$$
Switching to polar coordinates,
$$\int_D (z \overline{z})^{j-1} = 2 \pi \int_{r=0}^1 r^{2j-1} dr = \frac{\pi}{ j}.$$
So the final answer is $\prod_{j=1}^n \tfrac{\pi}{j} = \tfrac{\pi^n}{n!}$.
