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I’m interested in the zeroes of the complex function

$f(z,\bar{z}) = p(z) + \frac{1}{log(|z|)} q(z)$

where both $p$ and $q$ are polynomials of the complex variable $z$ (and are therefore holomorphic). I’m interested even in restricted cases, such as when both $p$ and $q$ are of low degree. I would also be very happy if anything could be said about the case where $p$ and $q$ are Kac-polynomials of high degree.

While I don’t know the literature well, I am aware that there are beautiful results relating to the clustering of zeroes on the unit circle in the context of random polynomials --- the simplest being Kac-polynomials. Presumably the factor of $\frac{1}{log(|z|)}$ spoils the usage of these results though.

I am also vaguely aware of Rouche’s theorem. However, since $f$ is not holomorphic, this result can not be used directly.

I would like to know how to count the zeroes of this function. Treating the coefficients as independent and identically distributed random variables would be fine too if that would be useful. To be honest I have no idea where to even start looking, so any and all suggestions are most welcome.

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  • $\begingroup$ What exactly do you mean by "count"? Estimate their number in terms of degrees of p, q? $\endgroup$ – Alexandre Eremenko Nov 17 '16 at 14:37
  • $\begingroup$ Yes, sorry, I am being rather vague because I would be happy with various statements. In the case where $p$ and $q$ are assumed to be random polynomials with iid coefficients, then I am interested in the distribution of zeroes in the complex plane. Presumably in this case one would need to assume that the degree $n$ of $p$ and $q$ is large and then take the large $n$ limit in order to make any kind of sharp statement. For low degree polynomials, I would be happy even with a simple condition for the existence of zeroes. $\endgroup$ – user41147 Nov 17 '16 at 15:43
  • $\begingroup$ I added a lower estimate. $\endgroup$ – Alexandre Eremenko Nov 18 '16 at 17:41
  • $\begingroup$ What are "Kac-polynbomials"? $\endgroup$ – Alexandre Eremenko Nov 19 '16 at 15:43
  • $\begingroup$ @AlexandreEremenko, thank you so much for your response. You have been enormously helpful. I have been out of action for the last few days so I have not been able to study your answer in any detail yet. However, I can at least quickly answer your question about Kac-polynomials. Terry Tao has a nice discussion on his blog. I also recently came across a question about them here on mathoverflow. $\endgroup$ – user41147 Nov 20 '16 at 20:52
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There is no general theory which applies here. However your problem can be restated as a problem about zeros of harmonic maps, or about fixed points of anti-holomorphic maps, if you rewrite your equation as $$\overline{z}=\exp(-2q(z)/p(z))/z.$$ Of course I am aware that this last equation has more solutions than the original one (in fact infinitely many in most cases) but still it is useful.

This type of problems (solutions of equations $\overline{z}=R(z)$, where $R$ is meromorphic) have been considered in

MR2431564 Khavinson, Dmitry; Neumann, Genevra, From the fundamental theorem of algebra to astrophysics: a "harmonious'' path. Notices Amer. Math. Soc. 55 (2008), no. 6, 666–675.

MR2676458 Bergweiler, Walter; Eremenko, Alexandre, On the number of solutions of a transcendental equation arising in the theory of gravitational lensing. (English summary) Comput. Methods Funct. Theory 10 (2010), no. 1, 303–324.

arXiv:1507.01704 Walter Bergweiler and Alexandre Eremenko, Green's function and anti-holomorphic dynamics on a torus.

The first paper treats the equations $$p(z)|z|+q(z)=0,\quad\mbox{and}\quad p(z)|z|^2+q(z)=0.$$ The methods developed in these papers also apply to the present problem.

EDIT. Using these methods, one can obtain that the number of solutions is at most $$3\max\{ m,n\}+2m,$$ where $m,n$ are degrees of $p,q$. This is sometimes exact: your equation with $p(z)=1,\; q(z)=1-z^n$ has $3n$ solutions.

Also when $m=n$, consider the equation $$n\log|z|=3\log 2\frac{z^n-1}{z^n+1}.$$ It is easy to see that for $n=1$ it has 5 real roots (1/2,1,2 plus two negative). Replacing $z$ by $z^n$ we obtain an equation with $5n$ roots. These examples are due to Walter Bergweiler. Using similar examples we could show that the estimate is exact for $m=0$, for $n\leq m$ and for $n=2m$.

EDIT2. Concerning the lower estimates, consider the function $$g(z)=\log|z|-q(z)/p(z).$$ It is easy to see that it defines a continuous (actually smooth) map of the Riemann sphere. This map has a topological degree, which is easily seen to be $\deg(q/p)=\max\{ m,n\}$. Therefore we have at least $\max\{ m,n\}$ solutions. This is exact for all $m,n$.

Here is a sketch of the proof. (Last updated on 11/24).

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  • $\begingroup$ Thanks a lot for your great answer! Your sketch of the proof has been very helpful indeed! $\endgroup$ – user41147 Nov 30 '16 at 13:44
  • $\begingroup$ I am interested to see where this can be used.. $\endgroup$ – Alexandre Eremenko Nov 30 '16 at 14:50

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