An equality relation for complex numbers off the nonnegative real axis For every complex number $z$ off the nonnegative real axis there exist positive numbers $p_0,... ,p_n$ such that $\sum_{i=0}^n p_iz^i = 0$.
Finding difficulty in proceeding with the problem. Need some hints.
 A: If $z$ is in the left half plane, there is a quadratic polynomial with non-negative coefficients which has $z$ as a root. If $z$ is not in the left half plane, first find $n$ such that $z^n$ is in the left half plane, then choose a quadratic polynomial in $z^n$. 
A: It can be rephrased as, is every complex number other than a nonnegative real the root of a polynomial with only nonnegative coefficients?
The answer is yes. Consider $P_n:= (1+x)^n Q$ where $Q $ is a monic nonnegative real quadratic polynomial whose roots are not real (so that they are necessarily complex conjugates of each other); there exists $n$ st $P_n$ has only nonnegative (in fact, strictly positive) coefficients (this is a consequence of a result due to Poincaré before he became famous). If $z$ is not real, such a $Q$ exists with $Q(z) = 0$; if $r$ is a negative real, then obviously $x-r$ works. 
We even have that if $n$ is sufficiently large, then the distribution of coefficients of $P_n$ is log concave (strongly unimodal). And obviously, there is no uniform degree that will work for all nonreal $z$. 
To provide a sort of qualitative response to the query of Geoffrey Irving, as the argument of $z$ tends to zero, the minimal $n$ goes to infinity (true for the three possible versions of the problem, nonnegative coefficients, no gaps in the coefficients and nonnegative, and nonnegative and log concave). More can be said, but not in this space. 
