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?



Fixed points of a function $z\mapsto\overline{P(z)}$ of a complex variable
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](https://mathscinet.ams.org/mathscinet-getitem?mr=1933331)
Khavinson, Dmitry, Świa̧tek, Grzegorz ,
[On the number of zeros of certain harmonic polynomials](https://doi.org/10.1090/S0002-9939-02-06476-6),
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](https://www.ams.org/journals/notices/200806/tx080600666p.pdf), 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 $mn$?


