This is probably well-known by I can't find it now online. My guess is that if the degree is $n$ then it's $2n$ but it's just a hunch.


EDIT:
This is an edited version. Before I asked about roots without qualification.



How many isolated roots can a polynomial in $z$ and $\overline{z}$ have?
$2n$ is incorrect. The correct upper estimate is $n^2$ (if the number is finite).
Indeed, let $P(z,\overline{z})$ be a polynomial of degree $n$. Writing $z=x+iy$ and $\overline{z}=x-iy$ we obtain one complex equation of the form $P^\*(x,y)=0$,
but one complex equation is equivalent to two real equations, each of degree $n$
so by Bézout theorem it has at most $n^2$ solutions, if finitely many.


Of course, such an equation can have infinitely many solutions, but if finitely many then
at most $n^2$. This is best possible MR1443416.


On the other hand, there is a remarkable conjecture of Wilmshurst
about polynomials of the form
$P(z)-Q(\overline{z})$ where degrees $m,n$ of $P,Q$ are very unequal. If they are
not equal, the number of solutions is finite. It is conjectured
that when $m$ is bounded, the number of roots is at most linear in $n$.
This is known only for $m=1$, in which case the number of solutions does not exceed
$3n-2$. A conjecture of Wilmshurst says that in general at most $m(m-1)+3n-2$.


See MR2431564 for a survey of what is known.


EDIT: The conjecture of Wilmshurst is wrong, as stated,
<http://arxiv.org/abs/1308.6474>, but the question remains wide open.


