11
$\begingroup$

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.

$\endgroup$
11
  • 2
    $\begingroup$ $z+\overline{z}$ has infinitely many zeros. $\endgroup$ Oct 25, 2014 at 22:05
  • 1
    $\begingroup$ Now the question is a duplicate of this one: mathoverflow.net/questions/77334/roots-of-bivariate-polynomials $\endgroup$ Oct 25, 2014 at 22:30
  • 4
    $\begingroup$ My understanding is the question I linked is about polynomials of two real variables $x,y$, which would be the same as polynomials in $z,\overline{z}$. $\endgroup$ Oct 25, 2014 at 22:35
  • 2
    $\begingroup$ @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. $\endgroup$ Oct 26, 2014 at 2:36
  • 3
    $\begingroup$ @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. $\endgroup$ Oct 26, 2014 at 3:49

3 Answers 3

21
$\begingroup$

$2n$ 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.

$\endgroup$
12
$\begingroup$

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.

vanishing loci of real and imaginary parts of a cubic polynomial

$\endgroup$
2
  • 3
    $\begingroup$ 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. $\endgroup$ Oct 27, 2014 at 7:57
  • 1
    $\begingroup$ @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. $\endgroup$
    – S. Carnahan
    Nov 5, 2014 at 1:05
4
$\begingroup$

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

$\endgroup$
1
  • 1
    $\begingroup$ 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? $\endgroup$ Oct 25, 2014 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.