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.



Volume of solution sets for polynomials in $\mathbb{C}[x]$
$\def\CC{\mathbb{C}}\def\RR{\mathbb{R}}$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](https://mathoverflow.net/questions/99527/) 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 $\prod\_{j=1}^n \tfrac{\pi}{j} = \tfrac{\pi^n}{n!}$.


