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Maybe it is not a standard term, a harmonic polynomial is $h_{n,m}(z,\bar{z})=f_n(z)+\overline{g_m(z)}$ where $f$ and $g$ are polynomials in $z\in\mathbb{C}$ of degrees $n,m$ respectively.We may assume that $n\geq m$.

Then how many roots of the equation $h_{n,m}(z,\bar{z})=0$ are there?

I heard a saying that it has at least $n$ roots using $\textit{Argument Principle}$.

Is it right?I do not know how to apply this theorem.I have a test on $z^2-\bar{z}=0$,$z^2-\bar{z}+1=0$.In these cases the conclusion holds.But I do not know how to get the general result.

Will someone be kind enough to give me some hints on this problem?And will someone be kind enough to show me how to solve such equations on $\textit{Mathematica}$?

Thank you very much!

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You are right if either $n>m$ or $n=m$ and the top degree coefficients of $f$ and $g$ have different absolute values (the roots have to be counted with their multiplicities, of course). However, $z^2+z+1+\bar z^2$, say, has no roots because every root $z$ must be real ($z^2+\bar z^2$ is always real) but $2x^2+x+1>0$ for all $x\in\mathbb R$.

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  • $\begingroup$ Thank you for your counterexample!But can you give me some more hints on your first statement?Thank you very much! $\endgroup$ – user14242 Jan 12 '12 at 13:19
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There is a conjecture posed by Wilmshurst in 1998, which says that when $1\leq m< n-1$, then $f$ has at most $m(m-1)+3n-2$ zeros. So far few cases were proved, for example $m=1$, then $f$ has at most $3n-2$ zeros proved by Khavinson and Swiatek. However when $m=n-3,n>=4$, there exist counterexamples to show zeros are more than the maximal valence by this conjecture. So this conjecture may not for every cases of $m,n$.

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