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?
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5$\begingroup$ $z=\overline{z}$ has more than 2 solutions... $\endgroup$– abxFeb 15, 2022 at 6:09
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5$\begingroup$ This paper seems to address the question, and the upper bound indeed appears to be $3n - 2$. I didn’t read it in detail though. I assume you meant to require $n > 1$. ams.org/journals/proc/2003-131-02/S0002-9939-02-06476-6/… $\endgroup$– Vik78Feb 15, 2022 at 6:16
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$\begingroup$ Thanks. Will look at it. Yes $\endgroup$– user159888Feb 15, 2022 at 6:38
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$\begingroup$ Great. Now it is more clear. Thanks $\endgroup$– user159888Feb 15, 2022 at 13:23
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1$\begingroup$ An obvious remark (certainly known to the OP but possibly useful for readers). These are the fixed points of $Q(z)=\overline{P(z)}$. Then $Q\circ Q=\bar{P}\circ P$, which is a polynomial of degree $n^2$. So if $n>1$, every fixed point is a zero of $(\bar{P}\circ P)(t)-t$, so the set of fixed points is finite and bounded above by $n^2$. $\endgroup$– YCorFeb 15, 2022 at 13:24
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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 $mn$?
answered Feb 15, 2022 at 13:53