Originally Posted by

**Laurent** A piece of contribution, perhaps:

In dimension at least 2, you can easily find situations where $\displaystyle \det (f(z)+g(z))$ is identically zero. For instance, in dimension 2, choose $\displaystyle f(z)$ satisfying the assumptions, and define $\displaystyle g(z)$ by exchanging the columns of $\displaystyle f(z)$. Then the roots of their determinants are the same (hence negative), while $\displaystyle f(z)+g(z)$ has two identical columns hence is singular for any $\displaystyle z$. But maybe this is not allowed since all complex numbers are roots of $\displaystyle \det(f(z)+g(z))$...

You should specify what you mean by "monic" in your problem. There is no obvious definition, I think. In dimension 1, if you discard this hypothesis, the theorem becomes clearly false ($\displaystyle (2z+1)+(-z-2)=z-1$ has a positive root).