On the number of roots of $p(z,\bar z)$

complex numberscomplex-analysispolynomialsroots

Let $p$ be a polynomial with complex (or even real) coefficients in the variables $z$, $\bar z$, where $\bar z$ is the conjugate of $z$.

What can we say on the number of complex roots of $p$?

Clearly $p$ can have both zero roots (e.g. $p(z,\bar z) = z + \bar z – i$) or infinite roots (e.g. $p(z,\bar z) = z + \bar z$). I'd like to know if we can say anything when the number of roots is known to be finite. Clearly this relates in some way to the number of real solutions to the system
$$
\begin{cases}
P(x,y) = 0, \\
Q(x,y) = 0,
\end{cases}
$$
where $P,Q$ are polynomials with real coefficients.

I know some vague things about Bezout's Theorem, but since this is not my field I'm not sure if this result can be applied to this specific case.
Besides, maybe there are some better results for this kind of polynomials which I have failed to find.

Thanks in advance to everyone who will help!

Best Answer

Let $\,n\,$ be the total degree of $\,p\,$ in $\,z, \bar z\,$, and let $\,q(z, \bar z) = \overline{p(z, \bar z)}=\bar p (\bar z, z)\,$ where $\,\bar p\,$ is the polynomial $\,p\,$ with all coefficients replaced by their complex conjugates.

Then $\,p(z, \bar z) = 0 \iff q(z, \bar z) = 0\,$. Eliminating $\,\bar z\,$ between the two equations using polynomial resultants gives a polynomial equation in $\,z\,$ alone $\,r(z) = \text{res}_{\bar z}(p, q) = 0\,$ of degree $\,\le n^2\,$.

  • If $\,\text{deg}(r) \lt 1\,$ the equation $\,p(z, \bar z)=0\,$ has either no roots, or infinitely many roots.

  • If $\,\text{deg}(r) \ge 1\,$ the roots of $\,p(z, \bar z)\,$ are among the roots of $\,r(z)\,$, at most $\,n^2\,$ of them.

For example, a few simple equations solved this way can be found in my answers 1, 2, 3.

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