TITLE: The zero set of sums of polynomials
QUESTION [5 upvotes]: As I am new to this forum, please correct me if this post is not appropriate. In that case I apologize. 
Let $P(z)$ and $Q(z)$ be polynomials with coefficients in $\mathbb{C}$, furthermore let $Z(P)$ and $Z(Q)$ denote their zero sets. What can be said about $Z(P+Q)$? 
Without imposing any further restrictions on $P$ and $Q$. I see that  $Z(P)  \cap   Z(Q) \subset Z(P+Q)$. Or if we additionally assume that one of the polynomials dominates $P+Q$ in the sense that $|P(z)|\geq|P(z)+Q(z)|$  for all $z\in \mathbb{C}$, then clearly also $Z(P)\subset Z(P+Q)$ holds. 
Without imposing to harsh restrictions (Very vague, I know) on the involved polynomials, what can be said? 
Lastly, I really appreciate any help from you.

REPLY [3 votes]: I highly doubt that you can say anything in general, and here is a very heuristic argument.
Let $f(z)=a_9z^9+...+a_1z+a_0$ be any polynomial of degree at most 9. 
Let $P(z)=a_9z^9+...+a_5z^5=z^5(a_9z^4+...+a_6z+a_5)$ and let $Q(z)=a_4z^4+...+a_1z+a_0$
Then $Z(P)$ and $Z(Q)$ are well known, and they can be computed with radicals, while $Z(P+Q)$ cannot be in general computed with radicals....
P.S. While $P,Q$ have a pretty different asymptotic behavior, this can also be fixed:
$$f(z)=a_5z^2+...+a_1z+a_0= (\frac{a_5}{2}z^5+a_0)+(\frac{a_5}{2}z^5+a_4z^4+...+a_1z)$$
P.P.S. This implies that for those examples $Z(P+Q) \cap {\mathbb Q}(Z(P),Z(Q)) = \emptyset$. Anyhow, results of the type under certain conditions $Z(P+Q)$ is a subset of the convex hull of $Z(P) \cup Z(Q)$ might hold, Roche Theorem is the closest thing I can think of..