I've been looking for references/answers to this problem for several days and I couldn't find anything.


If we consider the closed unit ball $B$ in $\mathbb C^2$ then for any point $(z\_1,z\_2)\notin B$ we can find a polynomial $P$ of two variables with $$|P(z\_1,z\_2)|>\sup\_{(w\_1,w\_2)\in B}|P(w\_1,w\_2)|.$$ 


In fact $P$ can be chosen to be just a line, but this statement written with polynomials is related to a more general theory of polynomially convex sets. See for instance <https://www.encyclopediaofmath.org/index.php/Polynomial_convexity>.


My question is, can we find $P$ being symmetric i.e. $P(w\_1,w\_2)=P(w\_2,w\_1)$ for all $w\_1,w\_2\in\mathbb C$?


I found a non-costructive proof of the result but it uses several strong results. However, the question can be easily solved if we consider the polydisc of center zero and radius one instead of $B$. We can consider the polynomial $P\_n(w\_1,w\_2)=w\_1^n+w\_2^n$ with $n$ big enough such that the argument of $z\_1$ and $z\_2$ is almost the same. However, this argument does not work for $B$ since we can have that $z\_1$ and $z\_2$ both have modulus smaller than one, and as a consequence when $p$ is big the values of the polynomials $P\_n$ go to zero at $(z\_1,z\_2)$. 


Any reference would be appreciated. Thanks in advance.



Symmetric polynomial separating points
I think combining the two tricks in your post answers the question. Let $(x\_1, x\_2)$ be a point not in $B$. Set
$$\ell(z\_1, z\_2) = \frac{\overline{x\_1}}{\sqrt{|x\_1|^2+|x\_2|^2}} z\_1 + \frac{\overline{x\_2}}{\sqrt{|x\_1|^2+|x\_2|^2}} z\_2.$$
Then $\ell$ is a linear polynomial with $\ell(x\_1,x\_2)>1$ but $|\ell(y\_1, y\_2)| \leq 1$ on $B$.


Set $f\_n(z\_1, z\_2) = \ell(z\_1, z\_2)^n + \ell(z\_2, z\_1)^n$. Then $f$ is a symmetric polynomial and $|f| \leq 2$ on $B$, but $\lim \sup\_{n \to \infty} |f\_n(x\_1, x\_2)|$ is infinite.


