# How many isolated roots can a polynomial in $z$ and $\overline{z}$ have?

This is probably well-known by I can't find it now online. My guess is that if the degree is $n$ then it's $2n$ but it's just a hunch.

EDIT: This is an edited version. Before I asked about roots without qualification.

• $z+\overline{z}$ has infinitely many zeros. – Christian Remling Oct 25 '14 at 22:05
• Now the question is a duplicate of this one: mathoverflow.net/questions/77334/roots-of-bivariate-polynomials – Christian Remling Oct 25 '14 at 22:30
• My understanding is the question I linked is about polynomials of two real variables $x,y$, which would be the same as polynomials in $z,\overline{z}$. – Christian Remling Oct 25 '14 at 22:35
• @Christian Remling: This question is not a duplicate. $P(z,\overline{z})=0$ when written in terms of $x,y$ is equivalent to TWO complex equations, not one. See details in my ans. – Alexandre Eremenko Oct 26 '14 at 2:36
• @Christian Remling: yes, but two polynomials of degree $d$ can have at most $d^2$ common zeros, and your representation does not help to see this. – Alexandre Eremenko Oct 26 '14 at 3:49

$2n$ is incorrect. The correct upper estimate is $n^2$ (if the number is finite). Indeed, let $P(z,\overline{z})$ be a polynomial of degree $n$. Writing $z=x+iy$ and $\overline{z}=x-iy$ we obtain one complex equation of the form $P^*(x,y)=0$, but one complex equation is equivalent to two real equations, each of degree $n$ so by Bézout theorem it has at most $n^2$ solutions, if finitely many.

Of course, such an equation can have infinitely many solutions, but if finitely many then at most $n^2$. This is best possible MR1443416.

On the other hand, there is a remarkable conjecture of Wilmshurst about polynomials of the form $P(z)-Q(\overline{z})$ where degrees $m,n$ of $P,Q$ are very unequal. If they are not equal, the number of solutions is finite. It is conjectured that when $m$ is bounded, the number of roots is at most linear in $n$. This is known only for $m=1$, in which case the number of solutions does not exceed $3n-2$. A conjecture of Wilmshurst says that in general at most $m(m-1)+3n-2$.

See MR2431564 for a survey of what is known.

EDIT: The conjecture of Wilmshurst is wrong, as stated, http://arxiv.org/abs/1308.6474, but the question remains wide open.

Alexandre Eremenko already described the failure of the $2n$ bound, but I thought I'd illustrate with an example. I did an unstructured manual search on cubic polynomials, and here's an example with 8 zeroes: $2z^3 + 4\bar{z}^3 - z^2 + z\bar{z} - \bar{z}^2 + z + 0.1 + 0.1i =0$. The graph shows vanishing loci of the real and imaginary parts, and the $Im(P(z,\bar{z}))=0$ locus is made of the components that are asymptotic to the $x$ and $y$ axes together with the lower right bubble.

• Some thing like $\prod_1^n((z+\bar{z}-k)$ is always real and is zero on $n$ vertical lines while $\prod_1^n((z-\bar{z}-ki)$ is always pure imaginary and zero on $n$ horizontal lines.Add to get something of degree $n$ with complex coefficients and $n^2$ isolated roots. A better challenge is to achieve this with integer coefficients. The references lead to one elegant construction doing this. – Aaron Meyerowitz Oct 27 '14 at 7:57
• @AaronMeyerowitz Your method yields integer coefficients quite easily. As long as you replace $k = 1,\ldots,n$ with an arrangement of Gaussian integers $ki$ that is symmetric across the real line, you get a polynomial with Gaussian integer coefficients that is Galois-invariant. – S. Carnahan Nov 5 '14 at 1:05

$z-\overline{z}=0\iff y=0\,,$ so it would seem infinitely many roots is possible.

• I was just going to comment about $z \bar z = 1$ but yours is even simpler.  If there's no positive-dimensional component then Bézout gives a quadratic upper bound. Can this bound be attained with only real solutions? – Noam D. Elkies Oct 25 '14 at 22:05