Efficients method for finding a zero of a multilinear complex polynomial in an specified region Let P be a given multilinear polynomial in $\mathbb{C}[z_1,\dots,z_n]$ and $D\subset \mathbb{C}$ be a given disc in the complex plane. Does there exist an efficient method for checking that $P$ has a zero in $D^n$ or not?
(A polynomial is called multilinear if its linear in each of its variables separately, for example $P(x,y)=xy-x+y+1$ is multilinear.)
 A: Assume that the disk $D \in \mathbb{C}$ is the ball $B(z_0,r)$ with center $z_0$ and radius $r>0$. By the transformation
$$\zeta_k~=~ \frac{z_k-z_0}{r},~~k\in[n]~=~\{1,\ldots,n\},$$
the zeros of the multilinear polynomial  $P(z_1,\ldots,z_n)$ on $D^n$ correspond to the zeros of the
multilinear polynomial   $Q(\zeta_1,\ldots,\zeta_n)$ on   $B(0,1)^n$, and the coefficients $q_{I}$ for $I\in2^{[n]}$ are easy to calculate in terms of the  coefficients $p_{I}$ of $P$.
It is now evident that a necessary condition for the existence of a  zero  of $Q$ in $B(0,1)^n$ is
$$  \sum \limits_{I\in 2^{[n]}\backslash \emptyset}  |q_I|  ~\geq~|q_\emptyset|.$$
Unfortunately this is far from being a sufficient condition for large $n$. Because $Q$ generically has $2^n$ coefficients, but is controlled by only $n$ variables the non-constant terms heuristically perform a random walk in the complex plane. The maximal stepsize is achieved if all $|\zeta_k|=1$ and for equal $|q_{I}|=s$ the distance traveled will be
$2^{n/2}\,s$. For this to be 1 requires  $s=2^{-n/2}$, or
$$  \sum \limits_{I\in 2^{[n]}\backslash \emptyset} \frac{|q_I|}{|q_\emptyset|} ~\geq~2^{n/2}.$$
Probably for a sufficiently large right hand side of the given order of magnitude continuity and control over $n$ variables suffices to guarantee a solution. For right hand sides   between 1 and  $2^{n/2}$ solutions are unlikely, and  probably difficult to find if they exist.
