One can see that algebraic numbers are dense in the complex plane by just looking at quadratic polynomials. I am interested in a "higher order" density of algebraic numbers. 


More specifically: is it known that if $D\_1,\ldots, D\_n \subset \mathbb{C}$ are disjoint disks, then there is an irreducible polynomial in $\mathbb{Z}[x]$ having at least one root in each $D\_i$?



Density of tuples of conjugate algebraic numbers
The idea by Yaakov Baruch works. Take any $q\_i\in D\_i\cap \mathbb Q[i]$ and take the polynomial 
$$
 g(x)=\prod\_{i=1}^n (x-q\_i)(x-\overline q\_i).
$$
It has rational coefficients.


If a polynomial $f(x)$ of degree $2n$ differs coefficientwise sufficiently small from $g(x)$, then $f(x)$ has a root in each $D\_i$ --- e.g., by Rouché's theorem. It remains to make $f(x)$ irreducible.


To perform this, one may use Eisenstein's criterion. Multiply $g(x)$ by a sufficiently large integer $N$ divisible by all the denominators of its coefficients. Then change its coefficients by 0 or 1 so that the leading one is odd, all others are even, and the last one is not divisible by 4. Finally, divide back by $N$.


