TITLE: Fixed points of a function $z\mapsto\overline{P(z)}$ of a complex variable
QUESTION [6 upvotes]: The equation $z^2=\overline{z}$ has four zeros and this example motivates us to generalize the problem to this form;
How many zeros does the equation $P(z)=\overline{z}$ have if $P(z)$ is a polynomial of degree $n>1?$ Can we find the bound for the number of zeros of this problem? The example motivate us to conjecture that it may be at most $2n,$ if not  at most $2n+n-2=3n-2.$ I am suggesting mere by intuition! May I request you to share your thoughts on this?

REPLY [9 votes]: Function $z\mapsto\overline{P(z)}$ has at most $3d-2$ fixed points, where $d\geq 2$
is the degree of $P$, and this is best possible. This remarkable result is due to Khavinson and Świa̧tek,
MR1933331
Khavinson, Dmitry, Świa̧tek, Grzegorz ,
On the number of zeros of certain harmonic polynomials,
Proc. Amer. Math. Soc. 131 (2003), no. 2, 409–414.
and it was later generalized to rational functions, and to some transcendental
functions. There is a survey of related results:
D. Khavinson and G. Neumann, From the fundamental theorem of algebra to astrophysics:
a “harmonious” path, Notices Amer. Math. Soc. 55 (2008), no. 6, 666–675.
Let me mention a major unsolved problem: let $p$, $q$ be polynomials
of degrees $m>n$. How many solutions can the equation
$$\overline{p(z)}=q(z)$$
have? Can one do better than the Bézout estimate $mn$?