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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.

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It seems unlikely that we could get a good characterization; for example, if $P=1$ and $Q=-1$ then $Z(P)=Z(Q)=\varnothing$ but $Z(P+Q)=\mathbb{C}$. Similarly if $Q=1-P$ then $Z(P)$ can be any finite subset of $\mathbb{C}$ but $Z(P+Q)=Z(1)=\varnothing$. – Zev Chonoles Jan 23 '12 at 16:00
Thank you, that is a fair point. What about if assume that P,Q, P+Q are non-constant and their zero sets are nonempty. Is there anything interesting to be said? – Henrik Jan 23 '12 at 16:38

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..

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